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UNIVERSITY OF CALICUT

SCHOOL OF DISTANCE EDUCATION

STUDY MATERIAL

M.Sc. MATHEMATICS

PAPER III

REAL ANALYSIS

Lessons prepared by

Bijumon R. ,

Lecturer (Senior Scale),

P.G. Dept. of Mathematics,

Mahatma Gandhi College, Iri t ty.

Copyright reserved

PAPER III : REAL ANALYSIS

CONTENTS

Unit I

Chapter 1 The Real and Complex Number System 5

2 Basic Topology: Metric Spaces 32

3 Continuity 41

4 Differentiation 60

Unit II

5 The Riemann-Stieltjes Integral 78

6 Sequences and Series of Functions 101

7 Lebesgue Measure 131

Unit III

8 The Lebesgue Integral 164

9 Differentiation and Integration 192

10 Measure and Integration 209

University Question Paper 2008 231

University Question Paper 2008 233

Chapter 1

THE REAL AND COMPLEX NUMBER SYSTEMS

5

We know that , the set of all integers, is integral domain and , the set of rational

numbers, is the smallest field containing (Ref. pp38: Chapter 8 SDE Study Material

Linear Algebra). But the rational number system is inadequate for many purposes, both as a

field and as an ordered set. For instance, next theorem says that, there is no rational p such

that 2 2.p This leads to the introduction of so-called “irrational numbers” which are often

written as infinite decimal expressions and are considered to be “approximated” by the

corresponding finite decimals. Thus the sequence

1,1.4,1.414,1.4142,

“tends to 2.” But unless the irrational number 2 has been clearly defined, the question

must arise: Just what is it that this sequence “tends to”?

This sort of question can be answered as soon as the so-called “real number system” is

constructed.

In the coming two theorems we show that the rational number system has certain gaps,

in spite of the fact that between any two rational there is another: If r s , then .2

r sr s

The real number system fills these gaps. This is the principal reason for the fundamental role

which it plays in analysis.

(1.1) Theorem The equation 2 2p has no solutions in rational numbers. Thus 2 is not a

rational number.

Proof

Suppose that p = m/n for ,m nZ is a rational number such that 2 2p . We assume that

we have cancelled any factors common to m and n, so that the integers m and n are are not

both even. Then 2 2p implies

2 22 ,m n . . . (1)

and shows that that 2m is an even integer, which implies that m is also even (For, otherwise

m is odd implies 2 1m k for some integer k implies

22 2 22 1 4 4 1 4( ) 1,m k k k k k which is also odd, contradicting the fact that 2m

UNIT I CHAPTER 1

6

is an even integer). Hence 2m is divisible by 4. It follows from (1) that 22n is also

divisible by 4 or 2n is divisible by 2 or 2n is even. Hence by the argument above, n is even.

Hence we arrives at the conclusion that both m and n are even, contrary to our assumption

that m and n are not both even. Hence 2 2p is not satisfied by a rational number p.

(1.2) Theorem

( i) 2: 0, 2A p p p has no largest number in .

( ii) 2: 0, 2B p p p has no smallest number in .

Proof

To prove the above, we associate with each rational 0p the number

2 1 2 2

.2 2

p pq p

p p

. . . (2)

Then

2

2

2

2 22 .

2

pq

p

. . . (3)

If ,p A then 2 2 0,p hence first equality in (2) shows that q p , and (3) shows

that 2 2.q Thus .q A

If ,p B then 2 2 0,p hence first and second equalities in (2) shows that 0 q p ,

and (3) shows that 2 2.q Thus .q B

2 Ordered Sets

(2.1) Definition Let S be a set. An order on S is a relation, denoted by <, with the

following two properties:

( i) If x S and y S the one and only one of the statements

, ,x y x y y x

is true.

( ii) If , ,x y z S , if ,x y y z , then .x z

(2.2) Definition An ordered set is a set S in which an order is defined.


7

Example: Show that is an ordered set if r s is defined to mean that s r is a positive

rational number.

Solution

(i) If r and s then either s r is a positive rational number or s r = 0 or

r s is positive number. i.e., one and only one of the statements

, ,x y x y y x hold.

(ii) If , ,r s t , and if ,r s s t , then s r is a positive rational number and r t

is a positive rational number implies s t s r r s is a positive rational

number implies .r t

(2.3) Definition Suppose S is an ordered set, and .E S If there exists a S such that

, ,x x E we say that E is bounded above, and call an upper bound of E.

(2.4) Definition Suppose S is an ordered set, ,E S and E is bounded above. Suppose

there exists an S with the following properties:

( i) is an upper bound of E.

( ii) If then is not an upper bound of E.

Then is called the least upper bound (or supremum) of E, and we write

= sup E.

(2.5) Definition Suppose S is an ordered set, and .E S If there exists a S such that

, ,x x E we say that E is bounded below, and call a lower bound of E.

(2.6) Definition Suppose S is an ordered set, ,E S and E is bounded below. Suppose

there exists an S with the following properties:

( iii) is a lower bound of E.

( iv) If then is not a lower bound of E.

Then is called the greatest lower bound (or infimum) of E, and we write

= inf E.

(2.7) Examples Example 1 Consider the sets A and B [in Theorem (1.2)] as subsets of the ordered set .

The set A is bounded above. In fact, the upper bounds of A are exactly the members of B.

Since B contains no smallest element, A has no least upper bound in .

UNIT I CHAPTER 1

8

Similarly, B is bounded below: The set of all lower bounds of B consists of A and of all

r with 0.r Since A has no largest element, B has no greatest lower bound in .

Example 2 If = sup E exists, then may or may not be a member of E. For example,

let 1 : 0E r r and 2 : 0 .E r r Then

1 2sup sup 0,E E

and 1 20 ,0 .E E

Example 3 Let 1

: .E nn

Then sup 1E E and inf 0 .E E

(2.8) Definition An ordered set S is said to have the least upper bound property if the

following is true:

If ,E S E and E is bounded above, then sup E exists in S.

(2.9) Example Describe the least upper bound property of ordered sets. Verify whether the

ordered field of rationals have this property. (UQ 2008)

Solution

Consider the set 2: 0, 2A p p p which is discussed in Theorem (1.2) and Example

1 in (2.7). The set A is bounded above. In particular, A is bounded above by each element

of 2: 0, 2B p p p discussed in Theorem (1.2). But there is no least upper bound

for the set A as A has no largest element and B has no smallest number in , by Theorem

(1.2), and p with 2 2p is not rational by Theorem (1.1). Hence:

does not have the least upper bound property.

(2.10) Theorem Suppose S is an ordered set with the least-upper bound property,

, ,B S B and B is bounded below. Let L be the set of all lower bounds of B. Then

sup L

exists in S, and inf .B

In particular, inf B exists in S.

Proof Since B is bounded below, L is not empty. Since L consists of exactly those y S

which satisfy the inequality ,y x x B we see that every x B is an upper bound of L.


9

Thus L is bounded above. Since, by the hypothesis, S is an ordered set with least upper

bound property, we obtain that L has supremum in S; call it .

If then [by Definition (2.4)] is not an upper bound of L, hence .B It

follows that .x x B Thus .L

If then ,L since is an upper bound of L.

We have shown that L but L if . In other words, is a lower bound of

B, but is not if . This means that inf .B

3 FIELDS

(3.1) Definition A field is a set F with two operations, called addition and multiplication,

which satisfy the following so-called “field axioms” (A), (M), and (D):

(A) Axioms for addition

(A1) If xF and yF, then their sum x y F .

(A2) Addition is commutative: x y y x xF and .yF

(A3) Addition is associative: ( ) ( )x y z x y z , ,x y zF .

(A4) F contains an element 0 such that 0 x x x F .

(A5) To every xF corresponds an element x F such that ( ) 0.x x

(B) Axioms for multiplication

(A1) If xF and yF, then their product xyF .

(A2) Multiplication is commutative: xy yx xF and .yF

(A3) Multiplication is associative: ( ) ( )xy z x yz , ,x y zF .

(A4) F contains an element 1 0 such that 1x x x F .

(A5) If xF and 0x then there exists an element 1

xF such that

11.x

x

(D) The distributive law

( )x y z xy xz , ,x y zF .

UNIT I CHAPTER 1

10

(3.2) Remarks

One usually write (in any field)

2 3, , , , , , 2 , 3 ,x

x y x y z xyz x x x xy

in place of

1( ), , ( ) , ( ) , , , , ,x y x x y z xy z xx xxx x x x x x

y

(3.3) Examples

The field axioms clearly hold in if addition and multiplication have their

customary meaning. Thus Q is a field. Similarly, R and C are fields. Z is not a field.

(3.4) Proposition The axioms for addition imply the following statements:

(a) (Cancellation law) If x y x z then .y z

(b) (Uniqueness of zero element) If x y x then 0.y

(c) (Uniqueness of inverse element) If 0x y then .y x

(d) ( ) .x x

Proof

(a) Suppose x y x z . Now

0 ,y y by (A4)

( ) ,x x y using (A2) and (A5)

( ),x x y using (A3)

( ),x x z using the assumption

( ) ,x x z using (A3)

0 ,z using (A2) and (A5)

,z using (A4)

(b) Take z = 0 in (a) above to obtain (b).

(c) Take z = x in (a) above to obtain (c).

(d) Since 0,x x by (c) above, taking x in place of x and x in place of y, we obtain

( ) .x x

(3.5) Proposition The axioms for multiplication imply the following statements.

(a) (Cancellation law) If 0x and xy xz then .y z


11

(b) (Uniqueness of zero element) If 0x and xy x then 1.y

(c) (Uniqueness of inverse element) If 0x and 1x y then 1

.yx

(d) If 0x then 1

.1

x

x

Proof

(a) Suppose 0x and xy xz . Now

1 ,y y by (B4)

1,x y

x

using (B2) and (B5)

1( ),xy

x using (B3)

1( ),xz

x using the assumption

1,x z

x

using (B3)

1 ,z using (B2) and (B5)

,z using (B4)

(b) Take z = 1 in (a) above to obtain (b).

(c) Take 1

zx

in (a) above to obtain (c).

(d) Since 1

1xx , by (c) above, taking

1

x in place of x and x in place of y, we obtain

1.

1x

x

(3.6) Proposition The field axioms imply the following statements, for any , ,x y zF :

(a) 0x = 0.

(b) If 0x and 0y then 0.xy

(c) (x)y = (xy) = x(y),

(d) (x)(y) = xy.

UNIT I CHAPTER 1

12

Proof

(a) 0 0 (0 0) 0 .x x x x Hence (3.4)(b) implies that 0 0,x and (a) holds.

(b) Next, assume 0x , 0y , but 0xy . Then (a) gives

1 1 1 11 0 0,xy

y x y x

a contradiction. Thus (b) holds.

(c) The first equality in (c) comes from

( ) 0 0,x y xy x x y y

combined with (3.4)(c); the other half of (c) is proved in the same way.

(d) Finally,

( )( ) [ ( )] [ ( )]x y x y xy xy

by (c) and (3.4)(d).

(3.7) Definition An ordered field is a field F which is also an ordered set, such that

( i) x y x z if , ,x y zF and ,y z

( ii) 0xy if , , 0,x y x F and 0.y

If 0,x we call x positive; if 0,x x is negative.

(3.8) Example Q is an ordered field.

(3.9) Proposition The following statements are true in every ordered field.

(a) If 0x then 0x and vice versa.

(b) If 0x and y z then .xy xz

(c) If 0x and y z then .xy xz

(d) If 0x then 2 0.x In particular, 1 0.

(e) If 0 x y then 1 1

0 .y x


13

Proof

(a) If 0x then 0 0,x x x so that 0.x If 0x then 0 0,x x x

so that 0.x This proves (a).

(b) Since ,z y we have 0,z y y y hence ( ) 0,x z y and therefore

( ) 0 .xz x z y xy xy xy

This proves (b).

(c) By (a) and (b), and Proposition (3.6) (c),

[ ( )] ( )( ) 0,x z y x z y

so that ( )x z y < 0, hence .xz xy

(d) If 0x , Part (ii) of Definition (3.7) gives 2 0.x If 0x , then 0x , hence, again

by Part (ii) of Definition (3.7), 2( ) 0.x But 2 2( ) 0,x x by Proposition (3.6)(d).

Hence 2 0.x Since 21 1 , 1 0.

(e) Note that if 0y and 0,v then 0.yv But 1

1 0.yy

Hence 1

0.y (For, other

wise, 1

0y forces us to say that

10,y

y

by the note, which is not the case here).

Likewise, 1

0.x If we multiply both sides of inequality x y by the positive quantity,

1 1,

x y

we obtain

1 1.

y x

4 THE REAL FIELD

(4.1) Theorem There exists an ordered field which has the least upper bound property.

Moreover, contains as a subfield.

Proof The second statement means that is a subset of and that the operations of

addition and multiplication in , when applied to members of , coincide with the usual

operations on rational numbers; also, the positive rational numbers are positive elements of

UNIT I CHAPTER 1

14

. The members of are called real numbers. Now the rest of the proof is done by

construction from . We shall divide the construction into several steps.

Step 1 The members of will be a certain subsets of , called cuts. A cut is, by definition,

any set with the following three properties:

(I) is not empty and .

(II) If , ,p q and ,q p then .q

(III) If ,p then p r for some .r

Notation The letters , , ,p q r will always denote rational numbers, and

, , , will denote cuts.

Note that (III) simply says that has no largest member; (II) implies two facts which will

be used freely:

If p and q then .p q

If r and r s then .s

Step 2 Define to mean: is a proper subset of .

Claim: < is an order [Ref. Definition (2.1)]

(i) If and it is clear that . (As a proper subset of a proper subset is a

proper subset).

(ii) It is also clear that at most one of the three relations

, ,

can hold for any pair , . To show that at least one holds, assume that the first two fail.

Then is not a subset of . Hence there is a p with .p If ,q it follows that

q p (since ),p hence ,q by (II). Thus . Since , we conclude:

Thus is now an ordered set.

Step 3 The ordered set has the least upper bound property.


15

To prove this, let A be a nonempty subset of , and assume that is an upper

bound of A. Define to be the union of all .A In other words, p if and only if

p for some .A

Claim: and sup .A

Since A is not empty, there exists an 0 .A This 0 is not empty. Since

0 , . Next (since , A ), and therefore . Thus satisfies

property (I). To prove (II) and (III), pick .p Then 1p for some 1 .A If

,q p then 1,q hence q ; this proves (II). If 1 is so chosen that ,r p we see

that r (since 1 ), and therefore satisfies (III).

Thus .

It is clear that .A

Suppose . Then there is an s and that .s Since s , s for

some .A Hence , and is not an upper bound of A.

This gives the desired result: sup .A

Step 4 If and we define : , .r s r s

We define *0 to be the set of all negative rational numbers. It is clear that *0 is a cut.

Verification of the fact that the axioms for addition [see Definition (3.1)] hold in ,

with *0 playing the role of 0:

(A1) We have to show that is a cut. It is clear that is a nonempty subset of .

Take , .r s Then r s r s for all choices of , .r s Thus .r s It

follows that has property (I).

Pick .p Then ,p r s with , .r s If ,q p then ,q s r so

,q s and ( ) .q q s s Thus (II) holds. Choose t so that .t r Then

p t s and t s . Thus (III) holds.

(A2) : , .r s r s By the same definition, : , .s r r s Since

, ,r s s r r s we have .

(A3) As above, this follows from the associative law in .

UNIT I CHAPTER 1

16

(A4) If r and *0 ,s then ,r s r hence r s . Thus *0 . To obtain the

opposite inclusion, pick ,p and pick , .r r p Then *0 ,p r and

*( ) 0 .p r p r Thus *0 . We conclude that *0 .

(A5) Fix . Let be the set of all p with the following property:

There exists 0r such that .p r

In other words, some rational number smaller than p fails to be in .

Claim: and *0 .

If s and 1,p s then 1 ,p hence .p So is not empty. If

,q then .q So . Hence satisfies (I).

Pick ,p and pick 0r , so that .p r If ,q p then ,q r p r

hence .q r Thus ,q and (II) holds. Put .2

rt p Then ,t p and

,2

rt p r so that .t Hence satisfies (III). That is, we have proved that

.

If and ,s then ,s hence , 0.r s r s Thus *0 .

To prove the opposite inclusion, pick *0 ,v put .2

vw Then w >0, and there is an

integer n such that nw but ( 1) .n w (Note that this depends on the fact that has

the Archimedean property). Put ( 2) .p n w Then ,p since ,p w and

.v nw p

Thus *0 .

We conclude that *0 .

This will of course be denoted by .

Step 5 Having proved that the addition defined in Step 4 satisfies Axioms (A) of Definition

(3.1), it follows that Proposition (3.4) is valid in , and we can prove one of the

requirements of Definition (3.7):

If , , and , then .


17

Indeed, it is obvious from the definition of + in that ; if we had

, the cancellation law [Proposition (3.4)] would imply .

It also follows that *0 if and only if *0 .

Step 6 Multiplication is a little more bothersome than addition in the present context, since

products of negative rationals are positive. For this reason we confine ourselves first to

*: 0 .

If and , we define

: for some choice of , , 0, 0p p rs r s r s

We define *1 : 1 .q q

Then the axioms (M) and (D) of Definition (3.1) hold, with in place of F and

with *1 in place of 1:

The proofs are so similar to the ones given in detail in Step 4 that we omit them.

Note, in particular, that the second requirement of Definition (3.7) holds: If *0

and *0 then *0 .

Step 7 We complete the definition of multiplication by setting * * *0 0 0 , and by

setting

* *

* *

* *

( )( ) if 0 , 0

[( ) ] if 0 , 0

[ ( )] if 0 , 0 .

The products on the right were defined in Step 6.

Having proved (in Step 6) that the axioms (M) hold in , it is now perfectly simple

to prove them in , by repeated application of the identity ( ) which is a part of

Proposition (3.4) (See Step 5).

The proof of the distributive law

( )

UNIT I CHAPTER 1

18

breaks into cases. For instance, suppose * * *0 , 0 , 0 . Then ( ) ( ),

and (since we already know that the distributive law holds in )

( ) ( ).

But ( ) . Thus

( ).

The other cases are handled in the same way.

We have now completed the proof that is an ordered field with the least upper

bound property.

Step 8 We associate with each ,r the set * : .r p p r Clearly *r is a cut;

that is, * .r These cuts satisfy the following relations:

(a) ** * ,r s r s

(b) ** * ,r s rs

(c) * *r s if and only if .r s

To prove (a), choose * *.p r s Then ,p u v where , .u r v s Hence

,p r s which says that *.p r s

Conversely, suppose *.p r s Then .p r s Choose t so that

2 ,t r s p put

, .r r t s s t

Then * *, ,r r s s and ,p r s so that * *.p r s This proves (a). The proof of

(b) is similar.

If ,r s then *,r s but *;r r hence * *.r s

If * *r s , then there is a *p s such that *.p r Hence ,r p s so that

.r s This proves (c).

Step 9 Claim: is a subfield of .


19

We saw in Step 8 that the replacement of the rational numbers r by the corresponding

“rational cuts” *r preserves sums, products, and order. This fact may be expressed by

saying that the ordered field is isomorphic to the ordered field * whose elements are

rational cuts. Of course, *r by no means the same as r, but the properties we are concerened

with (arithmetic and order) are the same in the two fields.

It is the above identification of with * which allows us to regard as a

subfield of .

(4.2) Theorem

(a) (Archimedean property of . ) If , ,x y and 0x , then there is a positive

integer n such that

.nx y

(b) (Density of in . ) If , ,x y and ,x y then there exists a p

such that .x p y i.e., between any two real numbers there is a rational one.

Proof

(a) Let : .A nx n If (a) were false, then y would be an upper bound of A. But

then A has a least upper bound in . Put sup .A Since 0, ,x x and x is

not an upper bound of A. Hence x mx for some .m But then ( 1) ,m x A

which is impossible, since is an upper bound of A.

(b) Since ,x y we have 0,y x and (a) furnishes a positive integer n such that

( ) 1.n y x

Apply (a) again, to obtain positive integers 1m and 2m such that 1 ,m nx 2m nx .

Then

2 1.m nx m

Hence there is an integer m (with 2 1)m m m such that

1 .m nx m

If we combine these inequalities, we obtain

1 .nx m nx ny

UNIT I CHAPTER 1

20

Since 0,n it follows that

.m

x yn

This proves (b) with m

pn

.

(4.3) Theorem (Existence of nth roots of positive reals) For every real x > 0 and every

integer n > 0 there is one and only one real y such that ny x and 0.y (UQ 2006)

Remarks

The number y above is written n x or 1/ .nx

In particular, 2 is real number (Taking x = 2 and n = 2)

Proof of the theorem That there is at most one such y is clear, since 1 20 y y

implies 1 2 .n ny y

Let : 0, .nE t t t x

If 1

xt

x

then 0 1.t Hence .nt t x Thus ,t E and E is not empty.

If 1t x then ,nt t x so that .t E Thus 1 x is an upper bound of E.

Hence Theorem (4.1) implies the existence of

sup .y E

To prove that ny x we will show that each of the inequalities ny x and ny x

leads to a contradiction.

The identity 1 2 1( )( )n n n n nb a b a b b a a yields the

inequality

1( )n n nb a b a nb

when 0 .a b

Assume .ny x Choose h so that 0 1h and

1.

( 1)

n

n

x yh

n y


21

Put , .a y b y h Then

1 1( ) ( ) ( 1) .n n n n ny h y hn y h hn y x y

Thus ( ) ,ny h x and .y h E Since ,y h y this contradicts the fact that y is an upper

bound of E.

Assume .ny x Put

1.

n

n

y xk

ny

Then 0 .k y If ,t y k we conclude that

1( ) .n n n n n ny t y y k kny y x

Thus ,nt x and .t E It follows that y k is an upper bound of E.

But ,y k y which contradicts the fact hat y is the least upper bound of E.

Hence ,ny x and the proof is complete.

(4.4) Corollary If a and b are positive real numbers and n is a positive integer, then

1/ 1/ 1/( ) .n n nab a b

Proof Put 1/ 1/, .n na b Then

( ) ,n n nab

since multiplication is commutative [Axiom (M2) in Definition (3.1)]. The uniqueness

assertion of Theorem (4.3) shows therefore that

1/ 1/ 1/( ) .n n nab a b

UNIT I CHAPTER 1

22

(4.5) The Relation between Real numbers and Decimals

Let 0x be real. Let 0n be the largest integer such that 0 .n x (Note that the

existence of 0n depends on the Archimedean property of . ) Having chosen 0n ,

1 1, , ,kn n let kn be the largest integer such that

10 .

10 10

k

k

nnn x

Let E be the set of these numbers

10 ( 0,1,2, )

10 10

k

k

nnn k (1)

Then sup .x E The decimal expansion of x is

0 1 2 3 .kn n n n n (2)

Conversely, for any infinite decimal (2) the set E of numbers (1) is bounded above, and (2)

is the decimal expansion of sup .E

5 THE EXTENDED REAL NUMBER SYSTEM

(5.1) Definition The extended real number system consists of the real field and two

symbols, and . We preserve the original order in , and define

.x x

It is then clear that is an upper bound of every subset of the extended real

number system, and that every nonempty subset has a least upper bound. If,

for example, E is a nonempty set of real numbers which is not bounded

above in , then sup E in the extended real number system.

Similarly, is a lower bound of every subset of the extended real number

system, and that every nonempty subset has a greatest lower bound. If, for

example, E is a nonempty set of real numbers which is not bounded below in

, then inf E in the extended real number system.


23

The extended real number system does not form a field, but it is customary to

make the following conventions:

(a) If x then

, , 0.x x

x x

(b) If 0x then ( ) , ( ) .x x

(c) If 0x then ( ) , ( ) .x x

When it is desired to make the distinction between real numbers on the one hand

and the symbols and on the other quite explicit, the former are called

finite.

6 THE COMPLEX FIELD

(6.1) Definition A complex number is an ordered pair ( , )a b of real numbers. We denote

the set of complex numbers by .

Ordered means that ( , )a b and ( , )b a are regarded as distinct if .a b

Let ( , ), ( , )x a b y c d be two complex numbers. We write x y if and only if

a c and .b d

(6.2) Theorem The addition and multiplication in the set of complex numbers defined by

( , ).x y a c b d

( , )xy ac bd ad bc

makes it into a field, with (0,0) and (1,0) in the role of 0 and 1.

Proof We simply verify the field axioms, as listed in Definition (3.1).

Let ( , ), ( , ), ( , ).x a b y c d z e f

(A1) is clear.

UNIT I CHAPTER 1

24

(A2) ( , )x y a c b d , by the definition of addition in

( , ),c a d b by the commutativity of addition in .

.y x

(A3) ( ) ( , ) ( , )x y z a c b d e f

( , ),a c e b d f by the associativity of addition in .

( , ) ( , )a b c e d f

( ).x y z

(A4) 0 ( , ) (0,0) ( , ) .x a b a b x

(A5) Put ( , ).x a b Then

( ) ( , ) ( , ) (0,0) 0.x x a b a b

(M1) is clear.

(M2) ( , )xy ac bd ad bc , by the definition of multiplication in

( , )ca db da cb , by the commutativity of multiplication in .

.yx

(M3) ( ) ( )( , )xy z ac bd ad bc e f

( , )ace bde adf bcf acf bdf ade bce

( , )( , )a b ce df cf de

( ).x yz

(M4) 1 (1,0)( , ) ( , ) .x a b a b x

(M5) If 0x then ( , ) (0,0),a b which means that at least one of the real numbers a, b is

different from 0. Hence 2 2 0,a b by Proposition (3.9)(d), and we can define

2 2 2 2

1( , ).

a b

x a b a b


25

Then 2 2 2 2

1( , )( , ) (1, 0) 1.

a bx a b

x a b a b

(D) ( ) ( , )( , )x y z a b c e d f

( , )ac ae bd bf ad af bc be

( , ) ( , )ac bd ad bc ae bf af be

.xy xz

THE REAL FIELD AS A SUBFIELD OF THE COMPLEX FIELD.

(6.3) Theorem For any real numbers a and b, we have

( , 0) ( ,0) ( , 0), ( , 0)( ,0) ( , 0).a b a b a b ab

The proof of the above Theorem is trivial. Theorem (6.3) shows that the complex numbers of

the form ( , 0)a have the same arithmetic properties as the corresponding real numbers a.

We can therefore identify ( , 0)a with a. This identification gives us the real field as a

subfield of the complex field.

(6.4) Definition (0, 1).i

(6.5) Theorem 2 1.i

Proof 2 (0, 1)(0, 1) ( 1, 0) (1, 0) 1.i

(6.6) Theorem If , ,a b then ( , ) .a b a bi

Proof ( , 0) ( , 0)(0, 1)a bi a b

( , 0) (0, )a b

( , ).a b

(6.7) Definition If , ,a b and ,z a bi then the complex number z a bi is called

the conjugate of z. The numbers a and b are the real part and the imaginary part of z,

respectively.

We shall occasionally write Re( ), Im( ).a z b z

UNIT I CHAPTER 1

26

(6.8) Theorem If z and w are complex, then

(a) .z w z w

(b) .zw z w

(c) 2Re( ), 2 Im( ).z z z z z i z

(d) zz is real and positive (except when z = 0).

Proof (a), (b) and (c) are quite trivial. To prove (d), write ,z a bi and note that

2 2.zz a b

Figure: Vector representation of complex numbers (addition and subtraction)

(7.1) Definition If z is a complex number, its absoloute value z is the nonnegative

square root of ;zz i.e., 1/ 2

.z zz

(7.2) Remarks

The existence (and uniqueness) of z follows from Theorem (4.3) and part (d) of

Theorem (6.8).

When x is real, then ,x x hence 2 .x x Thus , if 0

, if 0

x xx

x x

(7.3) Theorem Let z and w be complex numbers, then

(a) 0z unless z = 0, 0 0.

(b) .z z unless z = 0, 0 0.


27

(c) .zw z w

(d) Re .z z

(e) .z w z w

Proof (a) and (b) are trivial.

(c) Put ,z a bi ,w c di with , , , .a b c d Then

2 2 22 2 2 2 2 2( ) ( ) ( )( ) .zw ac bd ad bc a b c d z w

From the above, we have 22

.zw z w Now (c) follows from the uniqueness assertion of

Theorem (4.3).

(d) Note that 2 2 2 ,a a b hence

2 2 2 .a a a b

i.e., Re .z z

(e) Note that zw is the conjugate of ,zw so that 2Re( ).zw zw zw Hence

2

( )( )z w z w z w zz zw zw ww

2 2

2Re( )z zw w

2 2

2 ,z zw w using (d) above

2 2

2 ,z z w w since w w

2

.z w

Now (e) follows by taking square roots.

(7.4) Notation If 1, , nx x are complex numbers, we write

UNIT I CHAPTER 1

28

11

.n

n jj

x x x

(7.5) Theorem (Schwarz inequality) If 1, , na a and 1, , nb b are complex numbers,

then

22 2

1 1 1

.n n n

j j j jj j j

a b a b

(UQ 2006)

Proof

Put 2 2

1 1 1

, , .n n n

j j j jj j j

A a B b C a b

Case (i) If 0B , then 22

1 0,nb b implies 1 0,nb b and hence

1

0,n

j jj

C a b

and we get the result (In this case equality occurs).

Case (ii) Assume 0.B Then

2

1 1

,n n

j j j j j jj j

Ba Cb Ba Cb Ba Cb

by the Definition (7.1)

2 222

1 1 1 1

n n n n

j j j j j jj j j j

B a BC a b BC a b C b

22B A BCC BCC C B

22B A B C

2.B AB C

Since each term in the L.H.S. is nonnegative, we see that 20.B AB C

Since 0,B it follows that 2

0.AB C

i.e.,

22 2

1 1 1

0,n n n

j j j jj j j

a b a b

the desired inequality.

*(7.6) Spherical representation of complex number

Let the complex plane P be tangent to the unit sphere S such that the complex number z = 0

is the south pole (denoted by O) on S and let G be the north pole (See figure). Then OG, the

diameter of S, is perpendicular to P. Along OG draw a line and consider it as axis. Then G


29

is on the positive axis at (0, 0, 1). Corresponding to any point P(x,y) on P the line GP

intersecting S at one and only one point P(,,), where

.1

,1

,1 22

22

2222

yx

yx

yx

y

yx

x

Thus we can represent any

complex number in P by a point on the

sphere S. The point G itself

corresponds to the point at infinity of

the plane. The set of all points of the

complex plane including the point at

infinity is called entire complex

plane or the extended complex plane.

The above method for mapping the

plane into the sphere is called stereographic projection. The sphere is called Riemann

sphere.

7 EUCLIDEAN SPACES

Let k be a positive integer. k is the set of all ordered k-tuples x = (x1, x2, … , xk)

where xi . Then k

is a vector space over under the following operations of

addition and scalar multiplication: For 1 1, , , , , kk kx x y y x = y = and

1 1, , k kx y x y x + y =

1, , kx x x =

(For details refer the study material Linear Algebra, Page 101) The zero element of k is

0, , 0 ,0 = the k tuple with all its coordinates 0.

(7.7) Definitions Inner product (or scalar product) of x and y is

1

k

i ii

x y

x y =

and the norm of x is 1/ 2

1/ 2 2

1

.k

ii

x

x = x x =

UNIT I CHAPTER 1

30

(7.8) Theorem Suppose , , kx y z and . Then

(a) 0;x

(b) 0x if and only if ,x = 0 the zero vector;

(c) ; x x

(d) ; x y x y

(e) ; x y x y

(f) . x z x y y z

Proof

(a) By definition, 1/ 2

1/ 2 2

1

,k

ii

x

x = x x = and hence 0.x

(b) 0x if and only if

1/ 2

2

1

0k

ii

x

if and only 2

1

0k

ii

x

if and only if 2 0ix for

1, ,i k if and only if 0ix for 1, ,i k if and only if

1, , 0, , 0kx x x = = 0.

(c) 1/ 2 1/ 2

2 2

1 1

.k k

i ii i

x x

x = x

(d) 1

,k

i ii

x y

x y by the definition of the norm

1/ 2 1/ 2

2 2

1 1

,k k

i ii i

x y

by Schwarz inequality

. x y

(e) 2

, x y x y x y by the definition of the norm

2 x x x y + y y

2 2

2 x x y + y , using (d) above

2

, x + y so that (e) is proved.


31

(f)

a b

x z x y + y z

, a b by (e) above

. x y y z

(7.9) Corollary . x x

Proof 1 , x x by taking 1 in (c) of Theorem 7.2

1 x

. x

(7.10) Example Suppose , .k k a b Find kc and 0r such that

2 x a x b if and only if .r x c (UQ 2006)

___________________

Chapter 2

BASIC TOPOLOGY - METRIC SPACES

32

(1.1) Definition A set X, whose elements we shall call points, is said to be a metric space

if with any two points p and q of X there is associated a real number ( , ),d p q called the

distance from p to q, such that

(a) ( , ) 0d p q if p q ; ( , ) 0;d p p

(b) ( , ) ( , );d p q d q p

(c) ( , ) ( , ) ( , ),d p q d p r d r q for any .r X

Any function :d X X with the above properties is called a distance function or

a metric.

(1.2) Remark Every subset Y of a metric space X is a metric space in its own right, with

the same distance function. For it is clear that if conditions (a) to (c) of Definition (1.1) hold

for , , ,p q r X they also hold if we restrict p, q, r to lie in Y.

(1.3) Example

In the Euclidean space k , d defined by

( , )d x y x y , kx y

is a distance function.

Verification: Here kX .

(a) ( , ) 0d x y x y if x y , by (b) of Theorem 7.2 of Chapter 1.

(b) ( , )d x y x y

y x

y x , by Corollary 7.3 in Chapter 1

( , ).d y x

(b) ( , )d x y x y

x y y z , by (f) of Theorem 7.2 Chapter 1

BASIC TOPOLOGY-METRIC SPACES

33

( , ) ( , )d d x z z y

In particular, 1 (the real line) and 2 (the plane or the complex plane) are metric spaces.

Also note that in 1 and 2 the norm is just the absolute value of the corresponding real or

complex number.

(1.4) Remark Every subset of a Euclidean space is a metric space, by Remark (1.2).

(1.5) Definitions

By the segment ( , )a b we mean the set of all real numbers x such that .a x b

i.e., ( , ) : .a b x a x b

By the interval [ , ]a b we mean the set of all real numbers x such that .a x b

i.e., [ , ] : .a b x a x b

The half open interval [ , )a b is [ , ) : .a b x a x b

The half open interval ( , ]a b is ( , ] : .a b x a x b

If i ia b for 1, ,i k , the set of all points 1, , kkx x x = whose

coordinates satisfy the inequalities 1i i ia x b i k is called a k-cell. In

particular, a 1-cell is an interval, a 2-cell is a rectangle, etc.

If kx and 0r , the open ball B with center at x and radius r is defined to the

set of all ky such that .r y x

If kx and 0r , the closed ball B with center at x and radius r is defined to the

set of all ky such that .r y x

A set kE is convex if

1 ) E x +( y

whenever Ex , ,Ey and 0 1.

(1.6) Proposition Balls are convex.

UNIT I CHAPTER 2

34

Proof Let B be an open ball with center at x and radius r . Let By , ,Bz and

0 1. We have to show that 1 ) B y +( z ; for this we have to show that

1 ) .r y +( z x

Now By , Bz implies r y x and r z x . Also

1 ) ( ) 1 )( ) y +( z x y x +( z x

1 ) y x +( z x

1 )r r +(

,r

as desired.

The same proof applies to closed balls with < changed to .

(1.7) Proposition k- cells are convex.

(1.8) Definitions Let X be a metric space. All points and sets mentioned below are

understood to be elements and subsets of X:

A neighborhood of a point p with radius r is the set

( ) : ( , ) .rN p q X d p q r

A point p is a limit point of the set E if every neighborhood of p contains a

point q p such that ;q E in other words, p is a limit point of the set E if every

neighborhood of p intersects E at a point other than the point p.

If p E and p is not a limit point of E, then p is called an isolated point of E.

E is closed if every limit point of E is a point of E.

A point p is an interior point of E if there is a neighborhood N of p such that

.N E

E is open if every point of E is an interior point of E.


35

The complement of E (denoted by cE or ~ E ) is the set : .cE p X p E

E is perfect if E is closed and if every point of E is a limit point of E.

E is bounded if there is a real number M and a point q X such that ( , )d p q M

for all .p E

E is dense in X if every point of X is a limit point of E, or a point of E (or both).

(1.9) Remarks

In 1 neighborhoods are segments.

In 2 neighborhoods are interiors of circles.

(1.10) Theorem Every neighborhood is an open set.

Proof Consider a neighborhood ( ),rE N p and let .q E Then there is a positive real

number h such that

( , ) .d p q r h

For all points s such that ( , ) ,d q s h we have then

( , ) ( , ) ( , ) ,d p s d p q d q s r h h r

so that .s E Thus q is an interior point of E.

(1.11) Theorem If p is a limit point of a set E, then every neighborhood of p contains

infinitely many points of E.

Proof Suppose there is a neighborhood N of p which contains only a finite number of

points of E. Let 1, , nq q be those points of ,N E which are distinct from p, and put

1min ( , ) min ( , ) :1 .m m

m nr d p q d p q m n

The minimum of a finite set of positive numbers is clearly positive, so that 0.r

The neighborhood ( )rN p contains no point q of E such that ,q p so that p

is not a limit point of E. This contradiction establishes the theorem.

UNIT I CHAPTER 2

36

(1.12) Corollary A finite point set has no limit points.

Proof Let E be a finite point set. Suppose E has a limit point, say p. The by Theorem

(1.11) every neighborhood of p must contain infinitely many points of E, which is not

possible because E is a finite set. Hence E has no limit point.

(1.13) Examples Let us consider the following subsets of 2 :

(a) The set of all complex z such that 1.z

(b) The set of all complex z such that 1.z

(c) A finite set.

(d) The set of all integers.

(e) Consider the set 1

: 1,2,3, .E nn

Let us note that this set E has a limit

point (namely, z = 0) but that no point of E is a limit point of E; here note the

difference between having a limit point and containing one.

(f) The set of all complex numbers (that is, 2 ).

(g) The segment ( , )a b .

Let us note that (d), (e), (g) can be regarded also as subsets of 1.

Some properties of the above sets are tabulated below:

Closed Open Perfect Bounded

(a) No Yes No Yes

(b) Yes No Yes Yes

(c) Yes No No Yes

(d) Yes No No No

(e) No No No Yes

(f) Yes Yes Yes No

(g) No No Yes

In (g), we left the second entry blank. The reason is that the segment ( , )a b is not open if we

regard it as a subset of 2 , but it is not a subset of 1.

(1.14) Theorem Let E be a (finite or infinite) collection of sets .E Then

.C

CE E


37

Proof If ,

C

x E

then ,x E

hence ,x E hence ,Cx E

so that

.Cx E

Thus .C

CE E

Conversely, if ,Cx E

then ,Cx E

hence ,x E

hence ,x E

so

that .

C

x E

Thus .

C

CE E

Hence .C

CE E

(1.15) Theorem A set E is open if and only if its complement is closed.

Proof First, suppose CE is closed. Choose .x E Then ,Cx E and x is not a limit point

of CE . Hence there exists a neighborhood N of x such that ,CE N i.e., .N E

Thus x is an interior point of E, and E is open.

Next, suppose E is open. Let x be a limit point of CE . Then every neighborhood of

x contains a point of CE , so that x is not an interior point of E. Since E is open, this

means that .Cx E It follows that CE is closed.

(1.16) Corollary A set F is closed if and only if its complement is open.

(1.17) Theorem

(a) For any collection G of open sets, G

is open.

(b) For any collection F of open sets, F

is closed.

(c) For any finite collection 1, , nG G of open sets, 1

n

ii

G

is open.

(d) For any finite collection 1, , nF F of closed sets, 1

n

ii

F

is closed.

UNIT I CHAPTER 2

38

Proof

(a) To show that G

is open, it is enough to show that each point of G

is an interior

point. Let ,x G

then x G for some . Since G is open , x is an interior point

of G , hence x is an interior point of G

, and so G

is open.

(b) By Theorem (1.14),

,

C

CF F

(1)

and CF

is open, by Theorem (1.15). Hence (a) implies that R.H.S. of (1) is open. Hence its

complement F

is closed.

(c) Let 1

.n

ii

x G

Then ix G for 1, ,i n . Since each iG is open, x is an interior point

of each iG and hence there exist neighborhoods iN of x, with radii ,ir such that

i ix N G , 1, ,i n . Put

1min , , ,nr r r

and let N be the neighborhood of x of radius r. Then ix N G for 1, ,i n , so that

1

,n

ii

x N G

and hence 1

n

ii

G

is open.

(d) By Theorem (1.14),

1 1

,i

Cn n

Ci

i i

F F

(2)

and i

CF is open, by Theorem (1.15). Hence (c) implies that R.H.S. of (2) is open. Hence its

complement 1

n

ii

F

is closed.

(1.18) Example (Example to show that intersection of infinite collection of open sets need

not be open) In parts (c) and (d) of the preceding theorem, the finiteness of the collections is

essential. For let 1 1

, 1,2,3, .nG nn n

Then nG , being a segment in 1 is a


39

neighborhood in 1 and by Theorem (1.10), is an open subset of 1. But,

1 1

, 0nn n

Gn n

is a finite set and hence is not an open subset of 1.

(1.19) Example (Example to show that union of infinite collection of closed sets need not be

closed) The set of rational numbers is the union of a countable collection of closed sets each

of which contains exactly one rational number. i.e.,

.a

a

Though each a is closed, is not closed (For, we can find a sequence in which

converges to 2 . )

(1.20) Definition If X is a metric space, if ,E X and if E denotes the set of all limit

points of E in X, then the closure of E is the set .E E E

(1.21) Theorem If X is a metric space and ,E X then

(a) E is closed,

(b) E E if and only if E is closed,

(c) E F for every closed set F X such that .E F

By (a) and (c), E is the smallest closed subset of X that contains E.

Proof of the theorem

(a) To show that E is closed, we show that its complement C

E is open.

If p X and C

p E , then p E implies (by means of Definition (1.20)) that p is

neither a point of E nor a limit point of E. Hence p has a neighborhood which does not

intersect E. This neighborhood of p will be wholly contained in C

E , showing that p is

an interior point of C

E . Since p is an arbitrary point of C

E , this shows that C

E is

open. Hence its complement E is closed.

(b) By (a), E is closed, hence E E implies that E is closed. Conversely, if E is closed,

then E E (by Definition of closed set in (1.8) and Definition (1.20)), hence .E E

(c) If F is closed and ,F E then F F , hence .F E Thus .F E

UNIT I CHAPTER 2

40

(1.22) Theorem Let E be a nonempty set of real numbers which is bounded above. Let

sup .y E Then .y E Hence y E if E is closed.

Proof

Case 1) If y E then .y E

Case 2) Assume .y E Since sup ,y E for every 0h there exists then a point x E

such that ,y h x y for otherwise y h would be an upper bound of E. Thus y is a

limit point of E. Hence .y E

(1.23) Definitions

Suppose ,E Y X where X is a metric space:

To say that E is an open subset of X means that to each point p E there is

associated a positive number r such that the conditions ( , ) ,d p q r q X imply

that q E .

E is an open relative to Y if to each point p E there is associated a positive

number r such that the conditions ( , ) ,d p q r q Y imply that q E .

(1.24) Theorem Suppose .Y X A subset E of Y is open relative to Y if and only if

E Y G for some open subset G of X.

Proof Suppose E is open relative to Y. To each p E there is a positive number pr such

that the conditions ( , ) ,pd p q r q Y imply that q E . Let

| ( , )p pV q X d p q r

and define

.pp E

G V

Then G is an open subset of X, by Theorems (1.10) and (1.15).

Since ,pp V p E it is clear that .E G Y

By our choice of pV , we have pV Y E p E , so that .G Y E Thus

,E G Y and one half of the theorem is proved.

Conversely, if G is open in X and ,E G Y every p E has a neighborhood

.pV G Then ,pV Y E so that E is open relative to Y.

___________________

Chapter 3

CONTINUITY

41

1 Limits of Functions

(1.1) Definition Let X and Y be metric spaces; suppose ,E X f maps E into Y, and p

is a limit point of E. We write ( )f x q as ,x p or

lim ( )x p

f x q

if there is a point q Y with the following property: For every 0 there exists a 0 such

that

( ( ), )Yd f x q

for all points x E for which

0 ( , ) .Xd x p

The symbols Xd and Yd refer to the distances in X and Y, respectively.

If X and/or Y are replaced by the real line , the complex plane , or by some

Euclidean spaces ,k the distances Xd and Yd are of course replaced by absolute values, or

by appropriate norms.

It should be noted that ,p X but that p need not be a point of E in the above definition.

Moreover, even if ,p E we may very well have ( ) lim ( ).x p

f p f x

Figure: Limit at c of the given function f from to is L

UNIT I CHAPTER 3

42

(1.2) Theorem Let X, Y, E, f, and p be as in Definition (1.1). Then

lim ( )x p

f x q

(1)

if and only if

lim ( )nn

f p q

(2)

for every sequence np in E such that

, lim .n nn

p p p p

(3)

Proof

Suppose (1) holds. Choose np in E satisfying (3). Let 0 be given. Then there exists

0 such that

( ( ), )Yd f x q

if x E and 0 ( , ) .Xd x p Also, there exists N such that n N implies

0 ( , ) .X nd p p

Thus for n N , we have

( ( ), ) ,Y nd f p q

which shows that (2) holds.

Conversely, suppose (1) is false. Then there exist some 0 such that for every 0 there

exists a point x E (depending on ), for which

( ( ), )Yd f x q

but 0 ( , ) .Xd x p Taking 1

1,2, ,n nn

we thus find a sequence in E satisfying

(3) for which (2) is false.

(1.3) Corollary If f has a limit at p, this limit is unique.

Proof

Suppose f has a limit at p. Also, suppose that lim ( )x p

f x q

and lim ( ) .x p

f x r

Claim .q r

By Theorem (1.2), lim ( )x p

f x q

implies that

lim ( )nn

f p q

(4)

CONTINUITY

43


, lim .n nn

p p p p

(5)

Similarly,

lim ( )x p

f x r

implies that

lim ( )nn

f p r

(6)


, lim .n nn

p p p p

(7)

(5) and (6) says that the sequence ( )nf p converges to q and r, and by the uniqueness of

limit of sequence, we have .q r

(1.4) Definition Let ,f g be complex functions defined on E. Then f g , fg and f

g

are defined pointwise as follows:

( ) ( ) ( )f g x f x g x

( ) ( ) ( )fg x f x g x

( )

( ) ,( )

f f xx

g g x

provided ( ) 0.g x

Similarly, if f, g map E into ,k then f + g , fg and f

g are defined pointwise as follows:

( ) ( ) ( )x x x f + g f g

UNIT I CHAPTER 3

44

( ) ( ) ( )x x x f g f g

( ) ( ),x x f f where is a real number.

(1.5) Theorem Suppose ,E X a metric space, p is a limit point of E, f and g are

complex functions on E, and

lim ( )x p

f x A

, lim ( ) .x p

g x B

Then

(a) lim( )( ) ;x p

f g x A B

(b) lim( )( ) ;x p

fg x AB

(c) lim ( ) ,x p

f Ax

g B

if 0.B

Proof

(a) By Theorem (1.2), lim ( )x p

f x A

and lim ( )x p

g x B

implies that

lim ( )nn

f p A


, lim .n nn

p p p p

and lim ( )nn

g p B


, lim .n nn

p p p p

Now lim ( ) lim ( )n nn n

A B f p g p

lim ( ) ( ) ,n nn

f p g p

by the property of limit of sequences

lim ( ),nn

f g p

by the definition of f g

showing again by Theorem (1.2) that lim( )( ) .x p

f g x A B

(b) and (c) can be similarly proved, using the properties of limit of sequences.

(1.6) Remark If f, g map E into ,k then (a) remains true, and (b) becomes

CONTINUITY

45

lim( )( )x p

b x

f g A B.

2 Continuous Functions

(2.1) Definition Suppose X and Y are metric spaces, ,E X ,p E f maps E into Y.

Then f is said to be continuous at the point p if for every 0 there exists a 0 such

that

( ( ), ( ))Yd f x f p

for all points x E for which

( , ) .Xd x p

(2.2) Remark f is said to be continuous at the point p if

( i) ( )f p is defined i.e., f is defined at the point p;

( ii) lim ( )x p

f x

exists; and

( iii) lim ( ) ( ).x p

f x f p

(2.3) Definition If f is continuous at every point of E, then f is said be continuous on

E.

If p is an isolated point of E, then every function f which has E as its

domain of definition is continuous at p. For, no matter which 0 we

choose, we can pick 0 so that the only point x E for which

( , ) .Xd x p is ;x p then

( ( ), ( )) ( ( ), ( )) 0 .Y Yd f x f p d f p f p

(2.4) Theorem In the situation given in Definition (2.1), assume also that p is a limit point

of E. Then f is continuous at p if and only if lim ( ) ( ).x p

f x f p

Proof Compare Definitions (1.1) and (2.1).

(2.5) Theorem Suppose X, Y, Z are metric spaces, ,E X f maps E into Y, g maps

the range of f, ,f E into Z, and h is the mapping of E into Z defined by

( ) ( ( )) .h x g f x x E

If f is continuous at a point p E and if g is continuous at the point ( ),f p then h is

continuous at p.

UNIT I CHAPTER 3

46

i.e., continuous function of a continuous function is continuous.

The function h is called the composition or the composite of f

and g. The notation h g f is frequently used in this context.

Proof

Let 0 be given. Since g is continuous at ( ),f p there exists 0 such that

( ), ( )Zd g y g f p (8)

if , ( )Yd y f p and ( ).y f E

Since f is continuous at ,p corresponding to 0 there exists a 0 such that

( ( ), ( ))Yd f x f p

if ( , )Xd x p and x E .

Since ( ) ( ( )) ,h x g f x x E from (8), we have

( ), ( ) ( ( )), ( )Z Zd h x h p d g f x g f p (9)

if ( , )Xd x p and x E .

Thus h is continuous at p.

Figure Fig (a) and (b) show two functions that are discontinuous at x = c but continuous at

all other x points. Fig (c) gives a continuous function.

(2.6) Theorem (Characterization of continuity) A mapping f of a metric space X into a

metric space Y is continuous on X if and only if 1f V is open in X for every open set V

in Y.

CONTINUITY

47

Proof

Note that 1 | ( ) .f V x X f x V

Suppose f is continuous on X and V is an open set in Y. To show that 1f V is open, it

is enough to show that every point of 1f V is an interior point of 1f V . For this,

suppose p X and ( )f p V . Since V is open, there exists 0 such that y V if

( ( ), ) ;Yd f p y (10)

and since f is continuous at p, there exists a 0 such that

( ( ), ( ))Yd f x f p (11)

for all points x E for which ( , ) .Xd x p

From (10) and (11), we have ( )f x V as soon as ( , ) .Xd x p i.e., 1x f V as soon as

( , ) .Xd x p This shows p has a neighborhood 1| ( , ) ;XN x X d x p f V

hence p is an interior point of 1f V . Since p is an arbitrary point of 1f V , this shows

that each point of 1f V is an interior point and hence 1f V is an open set.

Conversely, suppose 1f V is open in X for every open set V in Y. We have to show

that f is continuous on X. i.e., to show that f is continuous at every point of X. Fix p X

and 0 , let

| ( , ( ))YV y Y d y f p .

Then V , being a neighborhood, is open; hence by assumption 1f V is open in X ; hence

each point of 1f V is an interior point, so in particular p is an interior point, so that there

exists 0 such that 1x f V as soon as ( , ) .Xd x p But if 1x f V , then

( ) ,f x V so that ( ( ), ( ))Yd f x f p . This shows that f is continuous at p.

This completes the proof.

(2.7) Corollary A mapping f of a metric space X into a metric space Y is continuous if

and only if 1f C is closed in X for every closed set C in Y.

Proof Corollary follows from the theorem, since a set is closed if and only if its complement

is open, and since 1 1C

Cf E f E

for every .E Y

UNIT I CHAPTER 3

48

(2.8) Theorem Let f and g be complex continuous functions on a metric space X. Then

, ,f g fg and f

g are continuous on X. (In the last case, we must of course assume that

( ) 0, .g x x X )

Proof

Case 1) At isolated points of X there is nothing to prove.

Case 2) At limit points, the statement follows from Theorems (1.5)and (2.4). We prove only

one and leaving the other to exercises.

By Theorem (2.4), f and g continuous functions implies lim ( ) ( )x p

f x f p

and

lim ( ) ( ).x p

g x g p

To show that f g is continuous, again by Theorem (2.4), it is enough to

show that lim ( ) ( ).x p

f g x f g p

Now lim ( ) lim ( ) ( ) ,x p x p

f g x f x g x

by the definition of f g

lim ( ) lim ( ),x p x p

f x g x

by Theorem (1.5)

( ) ( )f p g p

( ),f g p by the definition of f g

(2.9) Theorem

(a) Let 1, , kf f be real functions on a metric space X, and let f be the mapping of X into

k defined by

1( ) ( ), , ( ) ;kx f x f x x X f

then f is continuous if and only if each of the functions 1, , kf f is continuous.

(b) If f and g are continuous mappings of X into ,k then f + g and f g are continuous

on X.

The functions 1, , kf f are called the components of f. Note that f + g is a mapping into

,k whereas f g is real function on X.

Proof Part (a) follows from the inequalities

1/ 22

1

( ) ( ) ( ) ( ) ( ) ( ) ,k

j j i ii

f x f y x y f x f y

f f

for 1, , .j k Part (b) follows from (a) and Theorem (2.8).

CONTINUITY

49

(2.10) Examples If 1, , kx x are the coordinates of the point ,kx the functions i

defined by

( ) ki ix x x

are continuous on k , since the inequality

( ) ( )i i x y x y

shows that we may take in Definition (2.1). The functions i are some times called the

coordinate functions.

Repeated application of Theorem (2.8) then shows that every monomial

1 21 2

knn nkx x x (12)

where 1, , kn n are nonnegative integers, is continuous on k . The same is true of constant

multiples of (12), since constants are evidently continuous. It follows that every polynomial

P, given by

1 2

1, , 1 2 ,k

k

nn n kn n kP c x x x x x (13)

is continuous on k . Here the coefficients 1, , kn nc are complex numbers, 1, , kn n are

nonnegative integers, and the sum in (13) has finitely many terms.

Furthermore, every rational function in 1, , kx x , that is, every quotient of two

polynomials of the form (13), is continuous on k wherever the denominator is different

from zero.

From the triangle inequality one sees easily that

.k x y x y x, y (14)

Hence the mapping x x is a continuous real function on k .

If now f is a continuous mapping from a metric space X into k , and if is defined

on X by setting ( ) ( ) ,p p f it follows, by Theorem (2.5), that is a continuous real

function on X.

3 Continuity and Compactness

(3.1) Definition A mapping f of a set E into k is said to be bounded if there is a real

number M such that ( ) .x M x E f

UNIT I CHAPTER 3

50

(3.2) Theorem Suppose f is a continuous mapping of a compact metric space X into a

metric space Y. Then ( )f X is compact.

Proof

Let V be an open cover of ( )f X . Since f is continuous, Theorem (2.6) shows that each

of the sets 1( )f V is open. Since X is compact, the open cover 1( )f V

has a finite

subcover consisting of 1

1 1( ), , ( )n

f V f V , such that

1

1 1( ) ( ).n

X f V f V (15)

Since 1( )f f E E for every E Y , (15) implies that

1

,n

f X V V (16)

showing that f X can be covered by a finite subcover of the given cover V . Hence

f X is compact.

(3.3) Theorem If f is a continuous mapping of a compact metric space X into k , then

Xf is closed and bounded. Thus, f is bounded.

Proof By the previous Theorem, Xf is compact. We know that a compact set in k is

closed and bounded. Hence Xf is closed and bounded. In particular, f is bounded.

(3.4) Theorem Suppose f is a continuous real function on a compact metric space X, and

sup ( ) | , inf ( ) | .M f p p X m f p p X (16)

Then there exists points ,a b X such that ( )f a M and ( ) .f b m

(Note that in the above M is the least upper bound of the set ( ) |f p p X

and m is the greatest lower bound of the same set.)

The conclusion of the theorem may also be stated as follows:

“ There exist points a and b in X such that ( ) ( ) ( ) ;f a f x f b x X that

is, f attains its maximum at a and minimum at b. ”

Proof

By Theorem (3.3), ( )f X is a closed and bounded set of real numbers; hence by supremum

and infimum properties of , supremum and infimum exists for the bounded set

( ) ( ) |f X f p p X and since ( )f X is closed the supremum M and infimum m are

CONTINUITY

51

elements of ( ) ( ) |f X f p p X . Hence there exist elements ,a b X such that

( )f a M and ( ) .f b m

(3.5) Theorem Suppose f is a continuous 1-1 mapping of a compact metric space X into a

metric space Y. Then the inverse mapping 1f defined on Y by

1 ( )f f x x x X

is a continuous mapping of Y onto X.

Proof

Applying Theorem (2.6) to 1f in place of f, we see that it suffices to prove that

f V is an open set in Y for every open set V in X. Fix such a set V.

The complement CV of the open set V is closed in X, hence compact (As “closed

subsets of compact sets are compact”); hence, by Theorem (3.2), Cf V is a compact subset

of Y and so is closed in Y (As “compact subsets of metric spaces are closed”). Since f is

one-to-one and onto, f V is the complement of Cf V . Hence f V is open, proving

that 1f is a continuous mapping of Y onto X.

(3.6) Definition Let f be a mapping of a metric space X into a metric space Y. We say

that f is uniformly continuous on X if for every 0 there exists a 0 such that

( ( ), ( ))Yd f p f q (16)

for all points ,p q X for which ( , ) .Xd p q

Differences between continuity and uniform continuity

Uniform continuity is a property of a function on set, whereas continuity can be

defined at single point. To ask whether a given function is uniformly continuous at a

certain point is meaningless.

If f is continuous on X, then it is continuous at each point of X, so it is possible to

find, for each 0 and for each point p X , a number 0 having the property

specified in Definition (2.1). This depends on and on p . If f is however,

unifiormly continuous on X, then it is possible, for each 0 , to find one number

0 which will do for all points p of X. This depends on only.

(3.7) Theorem Let f be a continuous mapping of a compact metric space X into a metric

space Y. Then f is uniformly continuous on X. (UQ 2006)

UNIT I CHAPTER 3

52

Proof Let 0 be given. Since f is continuous, we can associate to each point p X a

positive number ( )p such that

, ( , ) ( )Xq X d p q p implies ( ( ), ( )) .2

Yd f p f q

(16)

Let ( )J p be the set of all q X for which

1

( , ) ( ).2

Xd p q p (17)

Since ( ),p J p the collection of all sets ( )J p is an open cover of X; and since X is

compact, there is a finite set of points 1, , np p in X, such that

1( ) ( ).nX J p J p (18)

We put

1

1min ( ), , ( ) .

2np p (19)

Then 0. (This is one point where the finiteness of the covering, inherent in the definition

of compactness, is essential. The minimum of a finite set of positive numbers is positive,

whereas the infimum of an infinite set of positive numbers may very well be 0; for example

1inf : 0.n

n

)

Now let q and p be points of X, such that ( , ) .Xd p q By (18), there is an

integer m, 1 ,m n such that ( );mp J p hence

1

( , ) ( ),2

X m md p p p (20)

and we also have

1

( , ) ( , ) ( , ) ( ) ( ).2

X m X X m m md q p d p q d p p p p

Finally, (16) shows that therefore

( ( ), ( )) ( ( ), ( )) ( ( ), ( )) .Y Y m Y md f p f q d f p f p d f q f p


(3.8) Theorem Let E be a noncompact set in 1. Then

(a) there exists a continuous function on E which is not bounded;

(b) there exists a continuous and bounded function on E which has no maximum.

If, in addition, E is bounded, then

CONTINUITY

53

(c) there exists a continuous function on E which is not uniformly continuous.

Proof

Suppose first that E is bounded.

(a) Since E is not compact, E bounded implies that E is not closed, so that there exists a

limit point 0x of E which is not a point of E. Consider

0

1( ) .f x x E

x x

(21)

This is continuous on E, since the constant function 1 and 0x x are continuous and their

quotient is also continuous by Theorem (2.8). But f is unbounded.

(b) To see that (21) is not uniformly continuous, let 0 and 0 be arbitrary, and choose

a point x E such that 0 .x x Taking t close enough to 0 ,x we can then make the

difference ( ) ( )f t f x greater than , although .t x Since this is true for every 0 ,

f is uniformly continuous on E.

(c) The function g defined by

2

0

1( )

1g x x E

x x

(22)

is continuous on E, and is bounded, since 0 ( ) 1.g x It is clear that

sup ( ) | 1,g x x E

whereas ( ) 1g x for all x E . Thus g has no maximum on E.

Now suppose that E is unbounded.

(a) Then ( )f x x establishes (a).

(b) The function h defined by

2

2( )

1

xh x x E

x

(23)

establishes (b), since

sup ( ) | 1h x x E

and ( ) 1h x for all x E .

Assertion (c) would be false if boundedness were omitted from the hypotheses. For,

let E be the set of all integers. Then every function defined on E is uniformly continuous

on E. To see this, we need merely take 1 in Definition (3.6).

UNIT I CHAPTER 3

54

(3.9) Example Let X be the half-open interval [0,2 ) on the real line, and let f be the

mapping of X onto the circle Y consisting of all points whose distance from the origin is 1,

given by

( ) cos , sin 0 2 .t t t t f (24)

The trigonometric functions cosine and sine are continuous. Also, they are periodic with

period 2 . These results show that f is a continuous 1-1 mapping of X onto Y.

However, the inverse mapping (which exists, since f is one-to-one and onto) fails to be

continuous at the point (1, 0) = f(0). Of course, X is not compact in this example.

Another argument to the fact that f 1

is not continuous, for if : f 1

is continuous, then

1 Yf is compact implies [0, 2 )X is compact which is false.

4Continuity and Connectedness

(4.1) Theorem If f is a continuous mapping of a metric space X into a metric pace Y, and

if E is a connected subset of X, then ( )f E is connected.

Proof

Assume, on the contrary, that ( )f E is not connected. Then ( ) ,f E A B where A

and B are nonempty separated subsets of Y. Put 1 1( ), ( ).G E f A H E f B

Then ,E G H and neither G nor H is empty.

Since A A (the closure of A), we have 1( );G f A the latter set is closed, since f

is continuous; hence 1( ).G f A It follows that ( ) .f G A Since ( )f H B and

,A B we conclude that .G H

The same argument shows that .G H Thus G and H are separated and

implies that E is not connected, contrary to the fact that E is connected.

We need the following theorem A for proving Theorem (4.2).

Theorem A: A subset E of the real line 1 is connected if and only if it has the following

property: If , ,x E y E and ,x z y then .z E

(4.2) Theorem (Intermediate value theorem) Let f be a continuous real function on the

interval [ , ].a b If ( ) ( )f a f b and if c is a number such that ( ) ( ),f a c f b then there

exists a point ( , )x a b such that ( ) .f x c

CONTINUITY

55

A similar result holds, of course, if ( ) ( ).f a f b Roughly speaking, the theorem says

that a continuous real function assumes all intermediate values on an interval.

Proof By Theorem A above, [ , ]a b is connected; hence by Theorem (4.1) shows that

[ , ]f a b is connected subset of 1, and the assertion follows if we appeal once more to

Theorem A.

(4.3) Theorem At first glance, it might seem that Theorem (4.2) has a converse. That is,

one might think that if for any two points 1 2x x and for any number c between 1( )f x and

2( )f x there is a point 1 2( , )x x x such that ( ) ,f x c then f must be continuous (Refer the

example in the coming section.

5 Discontinuity

(5.1) Definition Let f be defined on ( , ).a b Consider any point x such that .a x b We

define ( )f x , the right hand limit at x, by

( )f x q

if ( )nf t q as ,n for all sequences nt in ,x b such that .nt x

(5.2) Definition Let f be defined on ( , ).a b Consider any point x such that .a x b We

define ( )f x , the left hand limit at x, by

( )f x r

if ( )nf t r as ,n for all sequences nt in ,a x such that .nt x

(5.3) Remark Let f be defined on ( , ).a b Consider any point x such that .a x b It can

be seen that lim ( )t x

f t

, the limit of f at x, exists if and only if

( ) ( ) lim ( ).t x

f x f x f t

(5.4) Definition Let f be defined on ( , ).a b If f is discontinuous at a point x, and if

( )f x and ( )f x exist, then f is said to have a discontinuity of the first kind, or a

simple discontinuity, at x. Otherwise the discontinuity is said to be of the second kind.

There are two ways in which a function can have a simple discontinuity: either

( )f x ( )f x (in which case the value of ( )f x is not considered), or

( )f x ( ) ( ).f x f x

UNIT I CHAPTER 3

56

(5.5) Examples

Example 1 Define f by

1 when rational

( )0 when irrational

xf x

x

Then at any point x, neither ( )f x nor ( )f x exist, hence f has a discontinuity of the

second kind at every point x.


2 when 3 2

( ) 2 when 2 0

2 when 0 1

x x

f x x x

x x

At the point x = 0, (0 ) 2f while (0 ) 2f exist, hence (0 )f and (0 )f exist, but

they are not equal. Hence f has a simple discontinuity at x = 0. But f is continuous at

every other point of ( 3,1).


1sin when 0

( )

0 when 0

xf x x

x

At the point x = 0, neither (0 )f nor (0 )f exist, hence f has a discontinuity of the

second kind at x = 0. But f is continuous at every other than x = 0, since sin y is a

continuous function and 1

when 0xx

is also continuous function, hence their composition

1sin

x is continuous at 0.x

6 Monotonic Functions

(6.1) Definition Let f be real on ( , ).a b Then f is said to be monotonically increasing on

( , )a b if a x y b implies ( ) ( ).f x f y

(6.2) Definition Let f be real on ( , ).a b Then f is said to be monotonically decreasing on

( , )a b if a x y b implies ( ) ( ).f x f y

(6.3) Theorem Let f be monotonically increasing on ( , ).a b Then ( )f x and ( )f x exist

at every point ( , ).x a b More precisely,

CONTINUITY

57

sup ( ) : ( , ) ( ) ( ) ( ) inf ( ) : ( , ) .f t t a x f x f x f x f t t x b (1)

Furthermore, if a x y b , then

( ) ( ).f x f y (2)

(UQ 2006)

Proof

Since f is monotonically increasing, the set ( ) :f t a t x is bounded above by ( ).f x

Therefore, by the supremum property of the real line, ( ) :f t a t x has a least upper

bound which we shall denote by A. i.e., sup ( ) : .A f t a t x Clearly,

( ).A f x

Claim 1: ( ).A f x (i.e., the left hand limit of f at x is A).

Let 0 be given. It follows from the definition of A as a least upper bound that there

exists 0 such that a x x and

( ) .A f x A (3)

Since f is monotonic increasing, we have

( ) ( )f x f t A .x t x (4)

Combining (3) and (4), we see that

( ) .f t A x t x

Hence ( ) ,f x A and the claim is proved.

Since f is monotonically increasing, the set ( ) :f t x t b is bounded below by

( ).f x Therefore, by the infimum property of the real line, ( ) :f t x t b has a greatest

lower bound which we shall denote by B. i.e., inf ( ) : .B f t x t b Clearly,

( ) .f x B

Claim 2: ( ).B f x (i.e., the right hand limit of f at x is B).

Let 0 be given. It follows from the definition of B as a greatest lower bound that there

exists 0 such that x x b and

( ) .B f x B (5)

Since f is monotonic increasing, we have

( ) ( ).B f t f x .x t x (6)

Combining (5) and (6), we see that

UNIT I CHAPTER 3

58

( ) .f t B x t x

Hence ( ) ,f x B and the claim is proved.

Next, if a x y b , we see from (1) that

( ) inf ( ) : ( , ) inf ( ) : ( , ) .f x f t t x b f t t x y (7)

In the above, the last equality is obtained by applying (1) to ( , )a y in place of ( , ).a b

Similarly,

( ) sup ( ) : ( , ) sup ( ) : ( , ) .f y f t t a y f t t x y (8)

Comparison of (7) and (8) gives (2).

(6.4) Theorem Let f be monotonically decreasing on ( , ).a b Then ( )f x and ( )f x exist

at every point ( , ).x a b More precisely,

inf ( ) : ( , ) ( ) ( ) ( ) sup ( ) : ( , ) .f t t a x f x f x f x f t t x b

Furthermore, if a x y b , then

( ) ( ).f x f y

Proof is left to the exercise.

(6.5) Corollary Monotonic functions have no discontinuities of the second kind.

(6.6) Remark The above Corollary implies that every monotonic function is discontinuous

at a countable number of points at most. A simpler proof is given in the following Theorem.

(6.7) Theorem Let f be monotonic on ( , ).a b Then the set of points of ( , )a b at which f is

discontinuous is at most countable. (UQ 2006)

Proof Suppose, for the sake of definiteness, that f is increasing, and let

( , ) : is discontinuous at .E x a b f x

Claim: E is countable.

Note that x is point of E implies that x is a point at which f is discontinuous. By Corollary

(6.5) f has no discontinuities of the second kind. Hence the discontinuity at x is the

discontinuity of the first kind. Hence two cases arise:

Case 1) ( )f x ( )f x

or Case 2) ( )f x ( ) ( ).f x f x

CONTINUITY

59

The second case is not possible, since f is monotonic increasing (Think yourself!). Hence at

the discontinuous point x, we have ( )f x ( )f x .

Hence with every point we associate a rational number ( )r x such that

( ) ( ) ( ).f x r x f x

Since 1 2x x implies 1 2( ) ( ),f x f x we see that 1 2( ) ( )r x r x if 1 2.x x

We have thus established a 1-1 correspondence between the set E and a subset of the

set of rational numbers. Since the set of rational numbers is countable, this shows that E is

countable. i.e., the set of points of ( , )a b at which f is discontinuous is at most countable.

7 Infinite Limits and Limits at Infinity

(7.1) Definition For any real c, the set of real numbers x such that x c is called a

neighborhood of and is written ( , ).c Similarly, ( , )c is a neighborhood of .

(7.2) Definition Let f be a real function defined on E. We say that

( )f t A as ,t x

where A and x are extended real number system, if for every neighbourhood U of A there

is a neighborhood V of x such that ,V E and such that ( ) , .f t U t V E t x

(7.3) Theorem Let f and g be defined on E. Suppose

( )f t A , ( )g t B as .t x

Then

(a) ( )f t A implies .A A

(b) ( ) .f g t A B

(c) ( ) .fg t AB

(d) ( ) ,f A

tg B

provided the right members of (b), (c), and (d) are defined.

Note that , 0 , ,0

A

are not defined.

___________________

Chapter 4

DIFFERENTIATION

60

1 The Derivative of a Real Function

(1.1) Definition Let f be defined (and real-valued) on [ , ].a b For any [ , ]x a b form the

quotient

( ) ( )

( ) , ,f t f x

t a t b t xt x

(1)

and define

( ) lim ( ),t x

f x t

(2)

provided this limit exists in accordance with Definition 1.1 in Chapter 3 Continuity.

We thus associate with the function f a function f whose domain is the set of

points x at which the limit (2) exists; f is called the derivative of f . In particular ( )f x

is the derivative of f at x.

If f is defined at a point x, we say that f is differentiable at x. If f is defined

at every point of a set [ , ],E a b we say that f is differentiable on E.

(1.2) Definition/Remark

In (2), if we consider ( ),x the right hand limit of at x, in place of lim ( )t x

t

, then

( ) lim ( ),t x

f x t

(3)

and ( )f x is the right hand derivative of f at the point x.

In (2), if we consider ( ),x the left hand limit of at x, in place of lim ( )t x

t

, then

( ) lim ( ),t x

f x t

(4)

and ( )f x is the left hand derivative of f at the point x.

At the end points a and b, the derivative, if it exists, is a right-hand or left-hand

derivative, respectively.

If f is defined on a segment ( , )a b and if ,a x b then ( )f x is defined by (1) and

(2), as above. But ( )f a and ( )f b are not defined in this case.

DIFFERENTIATION

61

(1.3) Example Consider the

function ( )f x x defined over .

The left hand limit at x = 0 is 1,

while the right hand limit is 1. As

they are different, f is not

differentiable at x = 0. For the

points 0,x both left and right

hand limits coincide and the value

is 1. Hence the derivative of f at

0x is 1. Similarly, it can be

seen that f is differentiable for the points 0,x and the derivative at each of those points is 1.

i.e., f differentiable at all points except x = 0.

(1.4) Theorem Let f be defined on [ , ].a b If f is differentiable at a point [ , ]x a b ,

then f is continuous at x.

Proof

Note that ( ) ( )

( ) ( ) .f t f x

f t f x t xt x

( ) ( )

lim ( ) ( ) limt x t x

f t f xf t f x t x

t x

( ) ( )

lim lim ,t x t x

f t f xt x

t x

by Theorem (1.5) of Chapter 3

( ) 0f x

0.

Hence lim ( ) ( ),t x

f t f x

showing that f is continuous at x.

(1.5) Theorem Suppose f and g defined on [ , ]a b and are differentiable at a point

[ , ]x a b . Then , ,f g fg and f

g are differentiable at x, and

(a) ( ) ( ) ( ) ( );f g x f x g x

(b) ( ) ( ) ( ) ( ) ( ) ( );fg x f x g x f x g x

UNIT I CHAPTER 4

62

(c) 2

( ) ( ) ( ) ( )( ) ,

( )

f g x f x g x f xx

g g x

provided ( ) 0.g x

Proof

(a) ( ) ( )

( ) ( ) limt x

f g t f g xf g x

t x

( ) ( ) ( ) ( )

lim limt x t x

f t f x g t g x

t x t x

,

by Theorem (1.5) of Chapter 3

( ) ( ).f x g x

(b) Let .h fg Then

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .h t h x f t g t g x g x f t f x

Dividing the above by ,t x we obtain

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) .h t h x g t g x f t f x

f t g xt x t x t x

Hence

( ) ( ) ( ) ( ) ( ) ( )

( ) lim lim ( ) lim lim ( ) limt x t x t x t x t x

h t h x g t g x f t f xh x f t g x

t x t x t x

( ) ( ) ( ) ( ).f x g x g x f x

i.e., ( ) ( ) ( ) ( ) ( ) ( )fg x f x g x f x g x is proved.

(c) Let .f

hg

Then

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( )

f t f x f t g x f x g th t h x

g t g x g t g x

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

g x f t f x f x g t g x

g t g x

Hence

( ) ( ) 1 ( ) ( ) ( ) ( )

( ) ( ) .( ) ( )

h t h x f t f x g t g xg x f x

t x g t g x t x t x

Hence

DIFFERENTIATION

63

( ) ( )

( ) limt x

h t h xh x

t x

1 ( ) ( ) ( ) ( )

lim lim ( ) lim lim ( ) lim( ) ( )t x t x t x t x t x

f t f x g t g xg x f x

g t g x t x t x

2

( ) ( ) ( ) ( )

( )

g x f x g x f x

g x

(1.6) Examples

Example 1 The derivative of any constant function is clearly 0.

Example 2 If f is defined by

( ) ,f x x

then ( ) 1.f x

Example 3 Let n be a non-negative integer. If f is defined by ( ) ,nf x x then repeated

application of (b) in Theorem above shows that f is differentiable and 1( ) .nf x nx

Example 4 Let n be a negative integer. If f is defined by ( ) , 0nf x x x then repeated

application of (c) above shows that f is differentiable and 1( ) .nf x nx

Remark In view Examples 3 and 4 above, every polynomial is

differentiable, and so is every rational function, except at the points

where the denominator is zero.

(1.7) Theorem (Chain rule for differentiation) Suppose f is continuous on [ , ]a b , ( )f x

exists at some point [ , ]x a b , g is defined on an interval I which contains the range of f,

and g is differentiable at the point ( ).f x If

( ) ( ( )) ,h t g f t a t b

then h is differentiable at x, and

( ) ( ( )) ( ).h x g f x f x (5)

Proof

Let ( ).y f x By the definition of the derivative,

( ) ( )

( ) lim .t x

f t f xf x

t x

Hence

( ) ( ) ( ) ( ) ,f t f x t x f x u t (6)

UNIT I CHAPTER 4

64

where [ , ],t a b and ( ) 0u t as t x

Similarly,

( ) ( ) ( ) ( ) ,g s g y s y g y v s (7)

where s I , ( ) 0v s as .s y

Let ( ).s f t

Then

( ) ( ) ( ( )) ( ( ))h t h x g f t g f x

( ) ( ) ( ) ( ) ,f t f x g y v s using (7)

( ) ( ) ( ) ( ) ,t x f x u t g y v s using (6)

Hence if ,t x

( ) ( )

( ) ( ) ( ) ( ) .h t h x

g y v s f x u tt x

(8)

Letting t x , we see that ,s y by the continuity of f, so that the right side of (8) tends to

( ) ( ),g y f x which gives (5).

(1.8) Examples

Example 1 Let f be defined by

1sin 0

( )

0 0

x xf x x

x

Noting that the derivative of sin x is cos .x Also note, by the chain rule for differentiation

(Theorem (1.7)), that the derivative for 0x of 1

sinx

is 2

1 1cos .

x x

Applying Theorem

(1.5), we obtain

1 1 1

( ) sin cos 0 .f x xx x x

(9)

At 0x , these Theorems do not apply any longer, since 1

sinx

is not defined there, and we

appeal directly to the definition: for 0,t

( ) (0) 1

sin .0

f t f

t t

As 0,t this does not tend to any limit, so that (0)f does not exist.

Example 2 Let f be defined by

DIFFERENTIATION

65

2 1sin 0

( )

0 0

x xf x x

x

As above, applying Theorems (1.7) and (1.5), we obtain

1 1

( ) 2 sin cos 0 .f x x xx x

(10)

At 0x , we appeal directly to the definition: for 0,t

( ) (0) 1

sin .0

f t ft t

t t

As 0,t we see that

(0) 0.f

Thus f is differentiable at all points x, but f is not a continuous function, since 1

cosx

in

(10) does not tend to a limit as 0,x so that 0

lim ( )x

f x

doesn’t exist.

2 Mean Value Theorems

(2.1) Definition Let f be a real function defined on a metric space X. We say that f has a

local maximum at a point p X if there exists 0 such that ( ) ( )f q f p q X with

( , )d p q .

(2.2) Definition Let f be a real function defined on a metric space X. We say that f has a

local minimum at a point p X if there exists 0 such that ( ) ( )f q f p q X with

( , )d p q .

(2.3) Theorem Let f be defined on [ , ]a b ; if f has a local maximum at a point ( , ),x a b

and if ( )f x exists, then ( )f x = 0.

The analogous statement for local minima is “Let f be defined on [ , ]a b ; if f has

a local minimum at a point ( , ),x a b and if ( )f x exists, then ( )f x = 0” and is true.

Proof of the Theorem

Choose 0 in accordance with Definition (2.1), so that

.a x x x b

If ,x t x then

UNIT I CHAPTER 4

66

( ) ( )

0.f t f x

t x

Letting ,t x we see that

( ) 0.f x (11)

If ,x t x then

( ) ( )

0.f t f x

t x

Letting ,t x we see that

( ) 0.f x (12)

Combining (11) and (12), we obtain ( )f x = 0.

(2.4) Theorem (Generalized Mean Value Theorem) If f and g are continuous real

functions on [ , ]a b which are differentiable in ( , )a b , then there is a point ( , )x a b at which

[ ( ) ( )] ( ) [ ( ) ( )] ( ).f b f a g x g b g a f x

Note that differentiability is not required at the endpoints.

Proof

Put

( ) [ ( ) ( )] ( ) [ ( ) ( )] ( ) .h t f b f a g t g b g a f t a t b

Then h is continuous on [ , ]a b , h is differentiable in ( , )a b , and

( ) ( ) ( ) ( ) ( ) ( ).h a f b g a f a g b h b (13)

To prove the theorem, we have to show that ( ) 0h x for some ( , )x a b .

Case 1) If h is constant, this holds for every ( , )x a b .

Case 2) If ( ) ( )h t h a for some ( , ),t a b let x be a point on [ , ]a b at which h attains its

maximum (Ref. Theorem (3.4) of Chapter 3). By (13), ( , )x a b , and Theorem (2.3) shows

that ( ) 0.h x

Case 3) If ( ) ( )h t h a for some ( , ),t a b let x be a point on [ , ]a b at which h attains its

minimum (Ref. Theorem (3.4) of Chapter 3). By (13), ( , )x a b , and Theorem (2.3) shows

that ( ) 0.h x

(2.5) Theorem (“The” Mean Value Theorem) If f is a real continuous function on [ , ]a b

which is differentiable in ( , )a b , then there is a point ( , )x a b at which

( ) ( ) ( ).f b f a b a f x

DIFFERENTIATION

67

Proof

Take ( )g x x in the previous theorem. Then

[ ( ) ( )] ( ) [ ( ) ( )] ( )f b f a g x g b g a f x

becomes

( ) ( ) ( ).f b f a b a f x

(2.6) Theorem Suppose f is differentiable in ( , )a b .

(a) If ( ) 0f x for all ( , ),x a b then f is monotonically increasing.

(b) If ( ) 0f x for all ( , ),x a b then f is constant.

(c) If ( ) 0f x for all ( , ),x a b then f is monotonically decreasing.

Proof

Consider the equation

2 1 2 1( ) ( ) ( ),f x f x x x f x (14)

which is valid, for each pair of numbers 1 2,x x in ( , )a b , for some x between 1x and 2.x

(a) If ( ) 0f x for all ( , ),x a b then 2 1x x implies 2 1 0x x implies 2 1 ( ) 0x x f x

implies [by (14)], 2 1( ) ( ) 0f x f x implies 2 1( ) ( )f x f x implies f is monotonically

increasing.

(b) If ( ) 0f x then by (14), 2 1( ) ( ) 0f x f x for each pair of numbers 1 2,x x in ( , )a b ,

implies 2 1( ) ( )f x f x implies f is constant.

(c) If ( ) 0f x for all ( , ),x a b then 2 1x x implies 2 1 0x x implies 2 1 ( ) 0x x f x

implies [by (14)], 2 1( ) ( ) 0f x f x implies 2 1( ) ( )f x f x implies f is monotonically

decreasing.

3 The Continuity of Derivatives

(3.1) Theorem Suppose f is a real differentiable function on [ , ]a b and suppose

( ) ( ).f a f b Then there is a point ( , )x a b such that ( ) .f x

Proof

Put ( ) ( ) .g t f t t

Then ( ) 0,g a so that 1( ) ( )g t g a for some 1 ( , ),t a b and ( ) 0,g b so that

2( ) ( )g t g b for some 2 ( , ).t a b Hence g attains its minimum on [ , ]a b (Ref. Theorem

UNIT I CHAPTER 4

68

(3.4) of Chapter 3) at some point x such that .a x b By Theorem (2.3), ( ) 0.g x

Hence ( ) .f x

(3.2) Theorem Suppose f is a real differentiable function on [ , ]a b and suppose

( ) ( ).f a f b Then there is a point ( , )x a b such that ( ) .f x

Proof Similar to the proof of Theorem (3.1).

(3.3) Corollary If f is differentiable function on [ , ]a b , then f cannot have any simply

discontinuities on [ , ]a b .

But f may very well have discontinuities of the second kind.

4 L’Hospital’s Rule

The limit of the quotient ( )

( )

f x

g x as ax can be evaluated using the quotient rule

( )

( ),

( ) ( )

limlim

lim

x a

x ax a

f xf x

g x g x

provided ( ) 0.limx a

g x

However, the limit problems:

2

20 3

( ) ( )sin 9lim , lim , lim

6x x x a

x ax x f f

x x ax x

. . . (15)

cannot be solved using the quotient rule as the rule requires that the limit of the denominator

be different from 0.

It doesn’t mean that the limits in (15) do not exist, only that the quotient rule will not

determine them.

From the basic analysis course, the reader is familiar with the way of verifying

0

sinlim 1x

x

x .

Also, the algebraic technique of factoring yields

2

23 3 3

3 39 3 6lim lim lim .

3 2 2 56x x x

x xx x

x x xx x

Would it not be nice to have a standard procedure for handling all problems of this type?

That is too much to hope for. However, there is a simple rule that works well on a wide

DIFFERENTIATION

69

variety of such problems. It is known as l’Hôpital’s Rule (pronounced Lö’pëtäl); also called

l’Hospital rule or Lhospital rule. The rule was introduced by the French analyst and geometer

Guillaume François Antoine de l’Hospital (1661-1704).

Example 1 Use l’Hôpital’s rule to show that 0

sinlim 1x

x

x .

Solution

Here limits of both the numerator and denominator is 0. Therefore sin

0lim x

xx

is in the 0/0

form.

Now

1

coslim

sinlim

00

x

x

x

xx , using l’Hôpital’s Rule and noting that derivative of sin x is

cos x and that of x is 1.

1lim

coslim

0

0

x

xx

, using quotient rule for limits

.11

1

Example 2 Find xx e

x

lim

Solution

Both x and xe tend to as .x Hence limit is in / form.

xxxx ee

x 1limlim

, applying l’Hôpital’s Rule

= 0.

(4.1) Theorem: L’Hospital’s Rule Suppose f and g are real and differentiable in

( , )a b , and ( ) 0g x for all ( , )x a b , where .a b Suppose

( )

( )

f xA

g x

as .x a (16)

If

( ) 0f x and ( ) 0g x as ,x a (17)

or if

( )g x as ,x a (18)

then

UNIT I CHAPTER 4

70

( )

( )

f xA

g x as .x a (19)

The analogous statement is of course also true if ,x b or if ( )g x in (18).

Let us note that we now use the limit concept in the extended sense of Definition in Section 7

of the previous chapter.

Proof

We first consider the case in which .A Choose a real number q such

that ,A q and then choose r such that .A r q By (16) there is a point ( , )c a b such

that a x c implies

( )

.( )

f xr

g x

(20)

If ,a x y c then Theorem (1.7) shows that there is a point ( , )t x y such that

( ) ( ) ( )

.( ) ( ) ( )

f x f y f tr

g x g y g t

(21)

Suppose (17) holds. Letting x a in (21), we see that

0 ( )

0 ( )

f yr

g y

i.e., ( )

.( )

f yr q a y c

g y (22)

Now suppose that (18) holds. Keeping y fixed in (21), we can choose a point 1 ( , )c a y

such that ( ) ( )g x g y and ( )g x 0 if 1.a x c ( ) ( ) 0 and ( ) 0g x g y g x implies

( ) ( )0.

( )

g x g y

g x

Multiplying (21) by

( ) ( ),

( )

g x g y

g x

we obtain

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

g x g y f x f y g x g yr

g x g x g y g x

( ) ( ) ( )

( ) ( )

f x f y g yr r

g x g x

1

( ) ( ) ( ).

( ) ( ) ( )

f x g y f yr r a x c

g x g x g x (23)

If we let x a in (23), we get

( ) 1 1

lim ( ) lim ( ) lim( ) ( ) ( )x x x

f xr rg y f y

g x g x g x

DIFFERENTIATION

71

i.e., ( )

lim( )x

f xr

g x , as

1lim 0,

( )x g x using (18).

Hence there is a point 2 1( , )c a c such that

2

( ).

( )

f xq a x c

g x (24)

Summing up, (22) and (24) shows that for any q, subject only to the condition ,A q there is

a point 2c such that

2

( )if .

( )

f xq a x c

g x (25)

In the same manner, if ,A and p is chosen so that ,p A we can find a

point 3c such that

3

( ),

( )

f xp a x c

g x (26)

and (19) follows from these two statements (25) and (26).

5 Derivatives of Higher Order

(5.1) Definition If f has a derivative f on an interval, and if f is itself differentiable,

we denote the derivative of f by f and call f the second derivative of f. Continuing in

this manner, we obtain functions

(3) ( ), , , , , ,nf f f f f

each of which is the derivative of the preceding one. ( )nf is called the nth derivative, or the

derivative of the order n, of f.

In order for ( ) ( )nf x to exist at a point x , ( 1) ( )nf t must exist in a neighborhood of

x (or in a one-sided neighborhood, if x is an endpoint of the interval on which f is defined),

and ( 1)nf must be differentiable at x. Since ( 1)nf must exist in a neighborhood of x,

( 2)nf must be differentiable at that neighborhood.

(5.2) Taylor’s Theorem Suppose f is a real function on [ , ],a b n is a positive integer,

( 1)nf is continuous on [ , ],a b ( 1)( )nf t exists for any ( , ).t a b Let , be distinct points

of [ , ],a b and define

UNIT I CHAPTER 4

72

( )1

0

( )( ) ( ) .

!

knk

k

fP t t

k

(27)

Then there exists a point x between and such that

( ) ( )

( ) ( ) ( ) .!

nnf x

f Pn

(28)

(UQ 2006)

Remarks

For n = 1, Taylor’s theorem is just the mean value theorem: For

n = 1, (28) becomes

(1)

1( )( ) ( ) ( )

1!

f xf P (28A)

and from (28) with n = 1, we have

( ) (0)1 1

0

0

( ) ( ) ( )( ) ( ) ( ) 1 ( )

! 0! 1

kk

k

f f fP f

k

Substituting this in (28A), we get the mean value theorem

( ) ( ) ( )( )f f f x

i.e., ( ) ( ) ( )( ).f f f x

In general, the theorem shows that f can be approximated by a

polynomial of degree 1,n and that (28) allows us to estimate the

error, if we know bounds on ( ) ( ) .nf x

Proof of Taylors Theorem

Let M be the number defined by

( ) ( ) ( )nf P M (29)

and put

( ) ( ) ( ) ( ) .ng t f t P t M t a t b (30)

We have to show that ( )! ( )nn M f x for some x between and .

Differentiating (30) successively, we get

1( ) ( ) ( ) ( ) ( ) ( ) ( )n ndg t f t P t M t f t P t nM t

dt

(31)

2

2

2( ) ( ) ( ) ( ) ( ) ( ) ( 1) ( )n nd

g t f t P t M t f t P t n n M tdt

(32)

DIFFERENTIATION

73

1

( 1) ( 1) ( 1)

1( ) ( ) ( ) ( )

nn n n n

n

dg t f t P t M t

dt

( 1) ( 1)( ) ( ) ( 1) [ ( 2)] ( )n nf t P t n n n n M t (33)

Also, we have

( ) ( ) ( )( ) ( ) ( ) ( )n

n n n n

n

dg t f t P t M t

dt

( ) ( ) 0 !,nf t Mn since by (27) ( ) ( )nP t is a

polynomial of degree 1,n and hence

( ) ( ) 0,nP t and ( ) !.n

n

n

dt n

dt

i.e., ( ) ( )( ) ( ) ! ( ).n ng t f t n M a t b (34)

Hence the proof will be completed if we can show that ( ) ( ) 0ng x for some x between

and .

Note that for 0, , 1,k n

( )

( ) ( )( ) ! termsinvolving ( )

!

kk f

P t k tk

implies ( ) ( )( ) ( )k kP f for 0, , 1.k n

From (30), we have

( ) ( ) ( ) ( )ng f P M

( ) ( ) 0.f f

From (31) , we have

1( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0.ng f P nM f P f f

From (32) , we have

2( ) ( ) ( ) ( 1) ( ) ( ) ( ) ( ) ( ) 0.ng f P n n M f P f f

From (33) , we have

( 1) ( 1) ( 1)( ) ( ) ( ) ( 1) [ ( 2)] ( )n n ng f P n n n n M

( 1) ( 1) ( 1) ( 1)( ) ( ) ( ) ( ) 0.n n n nf P f f

UNIT I CHAPTER 4

74

i.e., ( 1)( ) ( ) ( ) 0.ng g g

Also, from (30), we have

( ) ( ) ( ) ( )ng f P M

= 0, using (29).

i.e., we have ( ) ( ) 0g g with g continuous and differentiable, hence by mean value

theorem, there is some 1 ,x such that

1( ) ( ) ( )( )g g g x

i.e., 10 ( )( )g x

implies 1( ) 0.g x

Now, we have ( )g 1( ) 0g x , g continuous and differentiable, hence by applying mean

value theorem for the function g , there is some 2 1,x x such that 2( ) 0.g x After n

steps we arrive at the conclusion that ( ) ( ) 0nng x for some 1,n nx x , i.e., for some nx

between and .

(5.3) Example Suppose f is defined in a neighborhood of x, and suppose ( )f x exists.

Show that 20

( ) ( ) 2 ( )lim ( ).h

f x h f x h f xf x

h

Show by an example that the limit may

exist even if ( )f x does not. (UQ 2006)

6 Differentiation of Vector Valued Functions

(6.1) Remarks Definition (1.1) applies without any change to complex functions f defined

on [ , ],a b and Theorems (1.4) and (1.5), as well as their proofs, remain valid. If 1f and 2f

are the real and imaginary parts of f, that is, if

1 2( ) ( ) ( )f t f t i f t

for ,a t b where 1( )f t and 2( )f t are real, then we clearly have

1 2( ) ( ) ( );f x f x i f x (35)

also, f is differentiable at x if and only if both 1f and 2f are differentiable at x.

Passing to vector-valued functions in general, i.e., to functions f which map [ , ]a b

into some k , we may still apply Definition (1.1) to define ( ).xf The map ( )t in (1) is

DIFFERENTIATION

75

now, for each t, a point in ,k and the limit in (2) is taken with respect to the norm of .k

In other words, ( )xf is that point of k (if there is one) for which

( ) ( )

lim ( ) 0,t x

t xx

t x

f ff (36)

and f is again a function with values in .k

If 1, , kf f are the components of f, as defined in Theorem (2.9) of the previous

chapter, then

1, , ,kf f f (37)

and f is differentiable at a point x if and only if each of the functions 1, , kf f is

differentiable at x.

Theorem (1.5) is true in this context as well, and so is Theorem(1.5)(a) and (b), if

fg is replaced by the inner product .f g

(6.2) Examples

Example 1 (Example to show that mean value theorem fails to be true for complex-valued

functions).

Define, for real x,

( ) cos sin .ixf x e x i x (38)

Then

(2 ) (0) cos2 sin 2 cos0 sin 0 1 1 0,f f i i (39)

but

( ) ,ixf x ie (40)

so that ( ) 1f x for all real x. Thus mean value theorem fails in this case.

Example 2 (Example to show that L’Hospitals rule fails to be true for complex-valued

functions).

On the segment (0, 1), define ( )f x x and

22( ) .

i

xg x x x e (41)

0

( )lim

( )x

f x

g x is in the 0/0 form. Since 1ite for all real t, we see that

UNIT I CHAPTER 4

76

20 0

2

( )lim lim

( ) ix xx

f x x

g xx x e

20

1lim

1

ixxxe

(Dividing throughout by x)

= 1. (42)

Now to evaluate 0

( )lim

( )x

f x

g x

:

22

( ) 1 2 0 1 ,

i

xi

g x x e xx

(43)

so that

2 2

( ) 2 1 1.i

g x xx x

(44)

Hence

( ) 1

( ) ( ) 2

f x x

g x g x x

(45)

and so

0

( )lim 0.

( )x

f x

g x

(46)

By (42) and (46), L’Hospital’s rule fails in this case. Note also that ( ) 0g x on (0, 1), by

(44).

However, there is a consequence of the mean value theorem which, for purposes of

applications, is almost as useful as Mean Value Theorem (2.5), and which remains true for

vector-valued functions: From Theorem (2.5) it follows that

( ) ( ) ( ) sup ( ) .a x b

f b f a b a f x

(47)

(6.3) Theorem Suppose f is a continuous mapping of [ , ]a b into k and f is differentiable

in ( , ).a b Then there exists ( , )x a b such that

( ) ( ) ( ) ( ) .b a b a x f f f (UQ 2006)

Proof

Put ( ) ( ),b az = f f and define

( ) ( ) .t t a t b = z f

DIFFERENTIATION

77

Then is a real-valued continuous function on [ , ]a b which is differentiable in ( , ).a b The

mean value theorem shows therefore that

( ) ( ) ( ) ( ) ( ) ( )b a b a x b a x = = z f

for some x ( , ).a b On the other hand,

2

( ) ( ) ( ) ( ) .b a b a = z f z f = z z = z

The Schwarz inequality now gives

2

( ) ( ) ( ) ( ) .b a x b a x z z f z f

Hence ( ) ( ) ,b a x z f

i.e., ( ) ( ) ( ) ( ) .b a b a x f f f

___________________

Chapter 5

THE RIEMANN-STIELTJES INTEGRAL

78

1 Definition and Existence of the Integral

(1.1) Definition Let [ , ]a b be a given interval. By a partition P of [ , ]a b we mean a

finite set of points 0 1, , , ,nx x x where

0 1 1 .n na x x x x b

We write

1 1, , .i i ix x x i n

Now suppose f is a bounded real function defined on [ , ]a b . Corresponding to each partition

P of [ , ]a b we put

1sup ( ) : , ,i i iM f x x x x

1inf ( ) : , ,i i im f x x x x

1

( , ) ,n

i ii

U P f M x

1

( , ) ,n

i ii

L P f m x

and

_

sup ( , ) : is a partition of [ , ] ,b

a

fdx L P f P a b (1)

inf ( , ) : is a partition of [ , ] .b

a

fdx U P f P a b (2)

_

b

a

fdx is called lower Riemann integral of f over [ , ]a b and b

a

fdx is called upper

Riemann integral of f over [ , ]a b .


79

(1.2) Definition A bounded function f is said to be Riemann integrable on [ , ]a b , if

_

.bb

a a

fdx fdx This common value (called the Riemann integral of f over [ , ]a b ) is

denoted by

b

a

fdx

or by ( ) ,b

a

f x dx

or by ( ) ,b

a

f x dx

(1.3) Notation The set of Riemann integrable functions is denoted by .

(1.4) Remark In the above, since f is bounded, there exist two numbers, m and M , such

that

( ) .m f x M a x b

Hence, for every partition P,

( ) ( , ) ( , ) ( ),m b a L P f U P f M b a (UQ 2008)

so that the numbers ( , )L P f and ( , )U P f form a bounded set.

Verification:

( ) .m f x M a x b

implies 1,2, , .i im m M M i n

implies 1,2, , .i i i i i im x m x M x M x i n

implies 1 1 1 1

n n n n

i i i i i ii i i i

m x m x M x M x

implies ( ) ( , ) ( , ) ( ).m b a L P f U P f M b a

This shows that the upper and lower integrals are defined for every bounded

function f. The question of their equality, and hence the question of the integrability of f, is a

more delicate one. Instead of investigating it separately for the Riemann integral, we shall

immediately consider a more general situation.

UNIT I CHAPTER 5

80

(1.5) Definition Let be a monotonically increasing function on [ , ]a b (since ( )a and

( )b are finite, it follows that is bounded on [ , ]a b (with lower bound ( )a and upper

bound ( )b ). Corresponding to each partition P of [ , ]a b , we write

1( ) ( ).i i ix x

Since is monotonically increasing, 0.i For any real function f which is bounded on

[ , ]a b we put



1

( , , ) ,n

i ii

U P f M

1

( , , ) ,n

i ii

L P f m

and

_

sup ( , , ) : is a partition of [ , ] ,b

a

f d L P f P a b (3)

inf ( , , ) : is a partition of [ , ] .b

a

f d U P f P a b (4)

(1.6) Definition In the above, if _

,bb

a a

f d f d we denote their common value by

b

a

f d (5)

or sometimes by

( ) ( )b

a

f x d x (6)

and this is the Riemann-Stieltjes integral (or simply the Stieltjes integral) of f with respect

to , over [ , ]a b .

If (5) exists, we say that f is integrable with respect to , in the Riemann sense,

and write ( ).f

Special Case: By taking ( ) ,x x the Riemann integral is seen to be a special case of the

Riemann-Stieltjes integral.


81

(1.7) Definition We say that the partition *P is a refinement of P if *P P (i.e., if every

point of P is a point of *P ). Given two partitions, 1P and 2 ,P we say that *P is their

common refinement if *1 2.P P P

(1.8) Theorem If *P is a refinement of P, then

*( , , ) ( , , )L P f L P f (7)

and

*( , , ) ( , , ).U P f U P f (8)

Proof

To prove (7), suppose first that *P contains just one point more than P. Let this extra point

be *,x and suppose *1 ,i ix x x where 1ix and ix are two consecutive points of P. Put

*1 1inf ( ) : ,iw f x x x x

*2 inf ( ) : , iw f x x x x

Also, as before 1inf ( ) : , .i i im f x x x x

Clearly 1 iw m and 2 ,iw m and hence

*( , , ) ( , , )L P f L P f

* *1 1 2 1( ) ( ) ( ) ( ) ( ) ( )i i i i iw x x w x x m x x

* *1 1 2( ) ( ) ( ) ( ) 0.i i i iw m x x w m x x

If *P contains k points more than P, we repeat this reasoning k times, and arrive at

(7). The proof of (8) is analogous.

(1.9) Theorem _

.bb

a a

f d f d

Proof

Let *P be the common refinement of two partitions 1P and 2.P By Theorem (1.8),

* *1 2( , , ) ( , , ) ( , , ) ( , , ).L P f L P f U P f U P f

Hence

1 2( , , ) ( , , ).L P f U P f (9)

If 2P is fixed and the supremum is taken over all 1,P then (9) gives

UNIT I CHAPTER 5

82

1 1 2sup ( , , ) : is a partition of [ , ] ( , , ).L P f P a b U P f

i.e., 2_

( , , ).b

a

f d U P f (10)

Now take the infimum over all 2 ,P then (10) gives

2 2_

inf ( , , ) : is a partition of [ , ]b

a

f d U P f P a b

i.e., _

.bb

a a

f d f d

(1.10) Theorem ( )f on [ , ]a b if and only if for every 0 there exists a partition P

such that

( , , ) ( , , ) .U P f L P f (11)

(UQ 2006, 2008)

Proof

Assume that for every 0 there exists a partition P such that ( , , ) ( , , ) .U P f L P f

For every partition P we have

_

( , , ) ( , , ).bb

a a

L P f f d f d U P f

Hence

_

0 ( , , ) ( , , ).b b

a a

f d f d U P f L P f (12)

By assumption, ( , , ) ( , , ) .U P f L P f Hence (12) gives

_

0 .b b

a a

f d f d (13)

Since 0 is arbitrary and can be made very small, (13) gives

_

0b b

a a

f d f d

or _

.b b

a a

f d f d

i.e., ( )f .


83

Conversely, suppose ( )f , and let 0 be given. ( )f implies

_

.bb b

a a a

f d f d f d Since

inf ( , , ) : is a partition of [ , ] ,b

a

f d U P f P a b

we have 2 2

b b

a a

f d f d

is not a lower bound of the set

( , , ) : is a partition of [ , ]U P f P a b and hence there exist a partition 2P such that

2( , , ) .2

b

a

U P f f d

Also, since _

sup ( , , ) : is a partition of [ , ] ,b

a

f d L P f P a b we have 2

b

a

f d

_ 2

b

a

f d

is not an upper bound of the set ( , , ) : is a partition of [ , ]L P f P a b and

hence there exist a partition 1P such that

1( , , ).2

b

a

f d L P f

i.e., corresponding to 0 there exist partitions 1P and 2P such that

2( , , ) ,2

b

a

U P f f d

(14)

1( , , ) .2

b

a

f d L P f

(15)

We choose P to be the common refinement of 1P and 2P . i.e., 1 2.P P P Then

Theorem (1.8), together with (14) and (15), shows that

2 1( , , ) ( , , ) ( , , ) ( , , ) ,2

b

a

U P f U P f f d L P f L P f

so that for this partition P, we have

( , , ) ( , , ) .U P f L P f

(1.11) Theorem

(a) If

( , , ) ( , , )U P f L P f (16)

UNIT I CHAPTER 5

84

for some partition P and some , then (16) holds (with the same ) for every

refinement of P.

(b) If (16) holds for 0 , , nP x x and if ,i is t are arbitrary points in 1[ , ],i ix x then

1

( ) ( ) .n

i i ii

f s f t

(c) If ( )f and the hypotheses of (b) hold, then

1

( ) .bn

i ii a

f t f d

Proof

(a) By Theorem (1.8), if *P is a refinement of P, then

*( , , ) ( , , )L P f L P f

and *( , , ) ( , , ).U P f U P f

Hence * *( , , ) ( , , ) ( , , ) ( , , )U P f L P f U P f L P f and (a) is

proved.

(b) Suppose (16) holds for 0 , , nP x x and ,i is t are arbitrary points in 1[ , ].i ix x Then,

since 1sup ( ) : , ,i i iM f x x x x 1inf ( ) : , ,i i im f x x x x we have ( )i im f s and

( )i if t M i.e., both ( )if s and ( )if t lie in , ,i im M so that ( ) ( ) .i i i if s f t M m

Thus

1

( ) ( ) ( , , ) ( , , ) .n

i i ii

f s f t U P f L P f

Hence (b) is proved.

(c) ( )f implies that for every 0 there exists a partition P such that

( , , ) ( , , ) .U P f L P f

From the discussion of (b), we have ( )i i im f s M and hence

1

( , , ) ( ) ( , , )n

i ii

L P f f s U P f

(17)

Also, we have

( , , ) ( , , )L P f f d U P f (18)

From (17) and (18), we obtain


85

1

( ) ( , , ) ( , , ) .bn

i ii a

f t f d U P f L P f

(1.12) Theorem If f is continuous on [ , ]a b then ( )f on [ , ]a b .

Proof

[ , ]a b , being a closed and bounded set, is compact. f is continuous on the compact set

[ , ]a b and hence f is uniformly continuous on [ , ]a b . Let 0 be given. Choose 0 so

that

( ) ( ) .b a

Since f is uniformly continuous on [ , ]a b , corresponding to the 0 there exists a 0

such that

( ) ( )f x f t (19)

if [ , ], [ , ],x a b t a b and .x t

If P is any partition on [ , ]a b such that ix for all i, then (19) implies that

1, ,i iM m i n (20)

and therefore

1 1

( , , ) ( , , ) ( ) ( ) ( ) .n n

i i i ii i

U P f L P f M m b a

Hence by Theorem (1.10), ( )f .

Recall from Chapter 4: Theorem (Intermediate value theorem) Let f be a

continuous real function on the interval [ , ].a b If ( ) ( )f a f b and if c is a

number such that ( ) ( ),f a c f b then there exists a point ( , )x a b such

that ( ) .f x c

(1.13) Theorem If f is monotonic on [ , ]a b , and if is continuous on [ , ]a b , then

( )f . (We still assume, of course, that is monotonic).

Proof

Let 0 be given. For any positive integer n, choose a partition such that

( ) ( )

1, , .i

b ai n

n

This is possible, by intermediate value theorem, since is continuous.

We suppose that f is monotonically increasing (the proof is analogous in the other

case). Then

UNIT I CHAPTER 5

86

1( ), ( ) 1, , ,i i i iM f x m f x i n

so that

11

( ) ( )( , , ) ( , , ) ( ) ( )

n

i ii

b aU P f L P f f x f x

n

( ) ( )

( ) ( )b a

f b f an

if n is taken large enough. Hence, by Theorem (1.10), ( )f .

(1.14) Theorem Suppose f is bounded on [ , ]a b , f has only finitely many points of

discontinuities on [ , ]a b , and is continuous at every point at which f is discontinuous.

Then ( )f .

Proof

Let 0 be given. Put

sup ( ) : [ , ] .M f x x a b

Let E be the set of points at which f is discontinuous. Since f has only finitely many points

of discontinuities, E is finite. Also, is continuous at every point of E. Since E is finite

we can cover E by finitely many disjoint intervals [ , ] [ , ]j ju v a b such that the sum of the

corresponding differences ( ) ( )j jv u is less than . Furthermore, we can place these

intervals in such a way that every point of ( , )E a b lies in the interior of some [ , ]j ju v .

Remove the segments ( ,j ju v ) from [ , ]a b . The remaining set K (being closed and

bounded) is compact. Hence f is uniformly continuous on K, and hence corresponding to

the given 0 there exists a 0 such that

( ) ( )f s f t if , , .s K t K s t

Now form a partition 0 1, , , nP x x x of [ , ]a b as follows:

Each ju occurs in P. Each jv occurs in P. No point of any segment ( ,j ju v ) occurs in P. If

1ix is not one of the ju , then .ix

Note that 2i iM m M for every i, and i iM m unless 1ix is one of the ju .

Hence as in the proof of Theorem (1.12),

( , , ) ( , , ) ( ) ( ) 2 .U P f L P f b a M

Since is arbitrary, by Theorem (1.10), ( )f .


87

(1.15) Remark In the above Theorem, if f and have a common point of discontinuity,

then f need not be in ( ) .

(1.16) Theorem Suppose ( )f on [ , ]a b , ,m f M is continuous on [ , ],m M

and ( ) ( ( ))h x f x on [ , ]a b . Then ( )h on [ , ]a b .

Proof

Choose 0 . Since is continuous on the compact set [ , ],m M it is uniformly continuous on

[ , ],m M hence there exists 0 such that and ( ) ( )s t if s t and

, [ , ].s t m M

Since ( )f , there is a partition 0 1, , , nP x x x of [ , ]a b such that

2( , , ) ( , , ) .U P f L P f (21)

Let



*1sup ( ) : , ,

i i iM h x x x x

*1inf ( ) : , .

i i im h x x x x

Divide the numbers 1, , n into two classes: i A if i iM m and i B if

i iM m .

For i A , our choice of shows that * * .i i

M m

For i B , * * 2 ,i i

M m K

where sup ( ) : .K t m t M

On i B i iM m implies

i i i ii B i B

M m

(22)

and by (21), (22) becomes

2i i i i

i B i B

M m

(23)

so that .ii B

It follows that

UNIT I CHAPTER 5

88

* * * *( , , ) ( , , )i i i ii i

i A i B

U P h L P h M m M m

( ) ( ) 2 ( ) ( ) 2 .b a K b a K

Since was arbitrary, by Theorem (1.10), ( )f .

2 Properties of the Integral

(2.1) Theorem

(a) If 1 ( )f and 2 ( )f on [ , ]a b , then

1 2 ( ),f f

( )cf for every constant c, and

1 2 1 2 ,b b b

a a a

f f d f d f d

.b b

a a

cfd c f d

(b) If 1 2( ) ( )f x f x on [ , ]a b , then

1 2 .b b

a a

f d f d

(c) If ( )f on [ , ]a b and if ,a c b then ( )f on [ , ]a c and on [ , ],c b and

.c b b

a c a

f d f d f d

(d) If ( )f on [ , ]a b and if ( )f x M on [ , ]a b , then

( ) ( ) .b

a

f d M b a

(e) If 1( )f and 2( )f , then 1 2( )f and

1 2 1 2;b b b

a a a

f d f d f d

if ( )f and c is a positive constant, then ( )f c and

.b b

a c

f d c c f d

Proof


89

(a) If 1 2f f f and P is any partition of [ , ]a b , we have

1 2 1 2( , , ) ( , , ) ( , , ) ( , , ) ( , , ) ( , , )L P f L P f L P f U P f U P f U P f (24)

If 1 ( )f and 2 ( )f , let 0 be given. There are partitions 1P and 2P such that

1 1 1 1( , , ) ( , , )U P f L P f

and 2 2 2 2( , , ) ( , , ) .U P f L P f

These inequalities persist if 1P and 2P are replaced by their common refinement P. Then

(24) implies

1 1 2 2 1 1 2 2( , , ) ( , , ) ( , , ) ( , , ) ( , , ) ( , , ) 2 ,L P f U P f U P f U P f L P f L P f

hence by Theorem (1.10), ( )f . i.e., 1 2 ( ).f f

With the same P we have

1 1( , , )U P f f d

and 2 2( , , )U P f f d ;

hence (24) implies

1 2 1 2( , , ) ( , , ) ( , , ) 2 .f d U P f U P f U P f f d f d

Since was arbitrary, we conclude that

1 2 .f d f d f d (25)

If we replace 1f and 2f in (25) by 1f and 2 ,f the inequality is reversed, and the

equality is proved.

The proofs of the other assertions of Theorem (2.1) are so similar that we omit the

details. In part (c) the point is that we may restrict ourselves to partitions which contain the

point c, in approximating .f d

(2.2) Theorem If ( )f and ( )g on [ , ]a b , then

(a) ( );fg

(b) ( )f and .b b

a a

f d f d

Proof

(a) Take 2( ) .t t Being a polynomial, is continuous. Also, given ( )f . By

Theorem (1.16) if ( ) ( ( ))h t f t , then ( )h . Note that 2( ) ( ( )) ( )h t f t f t and

hence 2 ( ).f

UNIT I CHAPTER 5

90

Also, note the identity

2 24 ( ) ( )fg f g f g .

( )f and ( )g implies by Theorem (2.1)(a) that ( ),f g and

( ).f g Also by the discussion just above, 2

( )f g and 2

( ).f g

Again, by Theorem (2.1)(a) , 2 2

( )f g f g and so 4 ( )fg and hence

( ).fg

(b) Take ( ) .t t Then is continuous. Also, given ( )f . By Theorem (1.16) if

( ) ( ( ))h t f t , then ( )h . Note that ( ) ( ( )) ( )h t f t f t and hence ( ).f

Choose 1,c so that

0.c f d

Then

,f d c f d cf d f d

since .cf f

(2.3) Theorem The unit step function I is defined by

0 0 ,( )

1 0 .

xI x

x

(2.4) Theorem If ,a s b f is bounded on [ , ]a b , f is continuous at s, and

( ) ( ),x I x s then

( ).b

a

f d f s

Proof

Consider partitions 0 1 2 3, , , ,P x x x x where 0 ,x a and 1 2 3 .x s x x b Then

As usual, we take 1sup ( ) : , ,i i iM f x x x x


1 1 0 2 2 1 3 3 21

( , , ) ( ) ( ) ( ) ( ) ( ) ( )n

i ii

U P f m M x x M x x M x x

1 2 3 20 0 1 0 1 1 ;M M M M


91

and 1 1 0 2 2 1 3 3 21

( , , ) ( ) ( ) ( ) ( ) ( ) ( )n

i ii

U P f m m x x m x x m x x

1 2 3 20 0 1 0 1 1 .m m m m

Since f is continuous at s, we have lim ( ) ( )x s

f x f s

and hence we see that 2M and 2m

converge to ( )f s as 2 .x s

Hence _

sup ( , , ) : is a partition of [ , ] ( )b

a

f d L P f P a b f s

inf ( , , ) : is a partition of [ , ] ( ).b

a

f d U P f P a b f s

Hence _

b

a

f d ( )b

a

f s f d and the common value is the Riemann integral

( ).b

a

f d f s

(2.5) Theorem Suppose 0nc for 1,2,3, , nc converges, ns is a sequence of

distinct points in ( , ),a b and

1

( ) ( ).n nn

x c I x s

(26)

Let f be continuous on [ , ].a b Then

1

( ).b

n nna

f d c f s

(27)

Proof

The comparison test shows that the series (26) converges for every x. Its sum ( )x is

evidently monotonic, and ( ) 0, ( ) .na b c

Let 0 be given, and choose N so that

1

.nN

c

Put

11

( ) ( ),N

n nn

x c I x s

21

( ) ( ).n nN

x c I x s

11

( )b b N

n nna a

f d f d c I x s

UNIT I CHAPTER 5

92

1

( )bN

n nn a

c f d I x s

, by Theorem (2.1)(e)

1

( )N

n nn

c f s

, by Theorem (2.4).

i.e., 11

( ).b N

n nna

f d c f s

(28)

Since 2 2( ) ( ) ,b a

2 ,b

a

f d M (29)

where we take sup ( ) : [ , ] .M f x x a b

Since 1 2, it follows from Theorem (2.1)(e) that 1 2.b b b

a a a

f d f d f d Hence it

follows from (29) that

1 ,b b

a a

f d f d M

and it follows from (28) that

1

( ) .b N

n nna

f d c f s M

(30)

If we let ,N we obtain 1

( ).b

n nna

f d c f s

(2.6) Theorem Assume increases monotonically and (i.e., is Riemann

integrable, not Riemann-Steiltjes integrable). Let f be a bounded real function on [ , ].a b

Then ( )f if and only if .f In that case

( ) ( ) .b b

a a

f d f x x dx (31)

(UQ 2006)

Proof

Let 0 be given and apply Theorem (1.10) to . Then there is a partition 0, , nP x x

of [ , ]a b such that

( , ) ( , ) .U P L P (32)

The mean value theorem furnishes points 1[ , ]i i it x x such that


93

( )i i it x

for 1, , .i n If 1[ , ],i i is x x then

1

( ) ( ) ,n

i i ii

s t x

(33)

by (32) and Theorem (1.11)(c). Put sup ( ) : [ , ] .M f x x a b Since

1 1

( ) ( ) ( )n n

i i i i ii i

f s f s t x

it follows from (33) that

1 1 1 1

( ) ( ) ( ) ( ) ( ) ( ) ( )n n n n

i i i i i i i i i i ii i i i

f s f s s x f s t x f s s x

1 1 1

( ) ( ) ( ) ( )n n n

i i i i i i ii i i

M t x M s x M s t x M

i.e., 1 1

( ) ( ) ( ) .n n

i i i i ii i

f s f s s x M

(34)

In particular,

1

( ) , ,n

i ii

f s U P f M

for all choices of 1[ , ],i i is x x so that

, , , .U P f U P f M

The same argument leads from (34) to

, , , .U P f U P f M

Thus

, , , .U P f U P f M (35)

Now note that (32) remains true if P is replaced by any refinement. Hence (35) also remains

true. We conclude that

( ) ( ) .b b

a a

f d f x x dx M

But is arbitrary. Hence

( ) ( ) ,b b

a a

f d f x x dx (36)

UNIT I CHAPTER 5

94

for any bounded f. The equality of the lower integrals follows from (34) in exactly the same

way. This completes the proof of the theorem.

(2.7) Example Evaluate 2

0

cos (sin ).x d x

Solution Take sin x . Then increases monotonically in 0 to 2

. Also, cos x is

bounded on 0, .2

Hence by Theorem (2.6),

2 2 2

2

0 0 0

cos (sin ) ( ) ( ) cos cos cosb b

a a

x d x f d f x x dx x x dx x dx

1

,2 2 4

by the reduction formula

/ 2

0

1 3 5 2. . . .1, when is odd.

2 4 3

cos

1 3 5 1. . . , when is even.

2 4 2 2

n

n n nn

n n n

x dx

n n nn

n n n

(2.8) Example Evaluate 1

0

x d x x where x denotes the largest integer not greater

than x. (UQ 2003)

Solution Take x x . Then decreases monotonically in 0 to 1 and x x

increases monotonically in 0 to 1. Also, x is bounded on 0, 1 . Hence by Theorem (2.6),

1 1 1

0 0 0

( ) ( ) 1 ,b b

a a

x d x x x d x x f d f x x dx x dx as ( ) 0 1.d

x xdx

12

0

1.

2 2

x

(2.9) Theorem (change of variable) Suppose is a strictly increasing continuous

function that maps an interval [ , ]A B onto [ , ].a b Suppose is monotonically increasing on

[ , ]a b and ( )f on [ , ].a b Define and g on [ , ]A B by

( ) ( ( )), ( ) ( ( )).y y g y f y (37)


95

Then ( )g and

.B b

A a

g d f d (38)

Proof

To each partition 0 , , nP x x of [ , ]a b corresponds a partition 0, , nQ y y

of [ , ]A B , so that ( ).i ix y All partitions of [ , ]A B are obtained in this way. Since the

values taken by f on 1[ , ]i ix x are exactly the same as those taken by g on 1[ , ],i iy y we see

that

, , , , ,U Q g U P f , , , , .L Q g L P f (39)

Since ( )f , P can be chosen so that both , ,U P f and , ,L P f are close to

b

a

f d . Hence (39), shows that , ,U Q g and , ,L Q g are close to b

a

f d and this

combined with Theorem (1.10), shows that ( )g and .B b

A a

g d f d This completes

the proof.

Special Case: Take ( ) .x x Then . Assume on [ , ]A B . If Theorem (2.6)

is applied to the left side of (38), we obtain

( ) ( ) ( ) .b B

a A

f x dx f y y dy (40)

3 Integration and Differentiation

(3.1) Theorem Let f on [ , ]a b . For ,a x b put

( ) ( ) .x

a

F x f t dt

Then F is continuous on [ , ]a b ; furthermore, if f is continuous at a point 0x of [ , ]a b , then

F is differentiable at 0x , and

0 0( ) ( ).F x f x

Proof

Since f , f is bounded. Suppose ( )f t M for .a t b If ,a x y b then

UNIT I CHAPTER 5

96

( ) ( ) ( ) ( ) ( ) ( ),y yx

a a x

F y F x f t dt f t dt f t dt M y x

by Theorem (2.1)(c) and (d). Given 0 , we see that

( ) ( ) ,F y F x

provided that .y xM

This proves continuity (and, in fact, uniform continuity) of F.

Now suppose f is continuous at 0.x Given 0 , choose 0 such that

0( ) ( )f t f x

if 0 ,t x and .a t b Hence, if

0 0 0x s x t x and ,a s t b

we have, by Theorem (2.1)(d),

0 0

( ) ( ) 1( ) [ ( ) ( )] .

t

s

F t F sf x f u f x du

t s t s

It follows that 0 0( ) ( ).F x f x

(3.2) Fundamental Theorem of Calculus If f on [ , ]a b and if there is a

differentiable function F on [ , ]a b such that ,F f then

( ) ( ) ( ).b

a

f x dx F b F a

Proof

Let 0 be given. Choose a partition 0 , , nP x x of [ , ]a b so that

, , .U P f L P f The mean value theorem furnishes points 1[ , ]i i it x x such that

1( ) ( ) ( )i i i iF x F x f t x

for 1, , .i n Thus

1

( ) ( ) ( ).n

i ii

f t x F b F a

It now follows from Theorem (1.11)(c) that

( ) ( ) ( ) .b

a

F b F a f x dx

Since this holds for every 0 , the proof is complete.


97

(3.3) Theorem (Integration by Parts) Suppose F and G are differentiable functions

on [ , ]a b , F f , and G g . Then

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .b b

a a

F x g x dx F b G b F a G a f x G x dx

Proof

Put ( ) ( ) ( )H x F x G x and apply Fundamental Theorem of Calculus [Theorem (3.2)] to H

and its derivative. Substituting in Theorem (3.2), we have

( ) ( ) ( ).b

a

H x dx H b H a

Then, noting that ( ) ( ) ( ) ( ) ( ),H x F x g x f x G x we have

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )b

a

F x g x f x G x dx F b G b F a G a

i.e., ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .b b

a a

F x g x dx F b G b F a G a f x G x dx

4 Integration of Vector-Valued Functions

(4.1) Definition Let 1, , kf f be real functions on [ , ]a b , and let 1, , kf ff = be the

corresponding mapping of [ , ]a b into .k If increases monotonically on [ , ]a b , to say

that f ( ) means that ( )jf for 1, , .j k If this is the case, we define

1 , , .b b b

ka a a

d f d f d

f =

In other words, b

a

d f is the point in k whose jth coordinate is .b

ja

f d

It is clear that parts (a), (c) and (e) of Theorem (2.1) are valid for these vector-valued

integrals; we simply apply the earlier results to each coordinate. The same is true of

Theorems (2.6), (3.1) and (3.2).

(4.2) Theorem (Analogue of Theorem (3.2)) If f and F map [ , ]a b into ,k if f

on [ , ]a b , and if F = f , then

UNIT I CHAPTER 5

98

( ) ( ) ( ).b

a

t dt b a f F F

We need the following Theorem (3.5) from Chapter 3 in the proof of the Theorem

(4.3) Theorem Suppose f is a continuous 1-1 mapping of a compact metric space X into a

metric space Y. Then the inverse mapping 1f defined on Y by

1 ( )f f x x x X

is a continuous mapping of Y onto X.

(4.4) Theorem (Analogue of Theorem (2.2)(b)) If f maps [ , ]a b into k and if

f ( ) for some monotonically increasing function on [ , ]a b , then f ( ) , and

.b b

a a

d d f f (41)

Proof

If 1, , kf f are the components of f , then

1

1/ 22 2 .

kf f f (42)

By Theorem (1.16), each of the functions 2

if ( ) ; hence so does their sum [For details

Refer the proof of the Theorem (2.2)]. Since 2x (being a polynomial) is a continuous

function of x, Theorem (4.3) shows that the square-root function (which is the inverse

function of 2x ) is continuous on [0, ],M for every real M. Note that the function f in (42)

is the composition of the square root function and the function 1

2 2

kf f ( ) . Hence,

if we apply Theorem (1.16) once more, (42) shows that f ( ) .

To prove (41), put 1, , ,ky yy where .j jy f d Then we have ,d y f

and

2 2 .i j j j jy y f d y f d y

By the Schwarz inequality,

( ) ( ) ;j jy f t t a t b y f (43)

hence Theorem (2.1)(b) implies


99

2

.d y y f (44)

If y = 0, (41) is trivial. If 0,y division of (44) by y gives (41).

5 Rectifiable Curves

(5.1) Definition A continuous mapping of an interval [ , ]a b into k is called a curve

in k . To emphasize the parameter interval [ , ]a b , we may also say that is a curve on

[ , ]a b .

If is one-to-one, is called an arc.

If ( ) ( ),a b is said to be a closed curve.

(5.2) Remark It should be noted that we define a curve to be a mapping, not a point set. Of

course, with each curve in k there is associated a subset of k , namely the range of ,

but different curves may have the same range.

(5.3) Definition We associate to each partition 0 , , nP x x of [ , ]a b and to each curve

on [ , ]a b the number

11

, ( ) ( ) .n

i ii

P x x

The ith term in this sum is the distance (in k ) between the points 1( )ix and ( )ix . Hence

,P is the length of a polygonal path with vertices at 0 1( ), ( ), , ( ),nx x x in this order.

As our partition becomes finer and finer, this polygon approaches the range of more and

more closely. This makes it seen reasonable to define the length of as

sup , : is a partition of [ , ]P P a b .

If , we say that is rectifiable.

In certain cases, is given by a Riemann integral.

(5.4) Theorem If is continuous on [ , ]a b , then is rectifiable, and

( ) .b

a

t dt

Proof

UNIT I CHAPTER 5

100

If 1 ,i ia x x b then

1 1

1( ) ( ) ( ) ( ) .i i

i i

x x

i ix x

x x t dt t dt

Hence

, ( ) .b

a

P t dt (45)

To prove the opposite inequality, let 0 be given. Since is continuous on [ , ]a b , is

uniformly continuous on [ , ]a b , and hence there exists 0 such that

( ) ( )s t if s t .

Let 0 , , nP x x be a partition of [ , ]a b , with .ix i If 1 ,i ix t x it follows that

( ) ( ) .it x

Hence

1

( ) ( )i

i

x

i i ix

t dt x x x

1

( ) ( ) ( )i

i

x

i ix

t x t dt x

1 1

( ) ( ) ( )i i

i i

x x

i ix x

t dt x t dt x

1( ) ( ) 2 .i i ix x x

If we add these inequalities, we obtain

( ) , 2 ( )b

a

t dt P b a

2 ( ),b a by taking supremum over all

partitions P.

Since was arbitrary,

( ) .b

a

t dt


Chapter 6

SEQUENCES AND SERIES OF FUNCTIONS

101

1 Discussion of Main Problem

(1.1) Definition Suppose , 1,2,3,nf n is a sequence of functions defined on a set E,

and suppose that the sequence of numbers ( )nf x converges for every .x E We can then

define a function f by

( ) lim ( ) .nn

f x f x x E

(1)

Under these circumstances we say that nf converges on E and that f is the limit,

or the limit function, of nf . We shall say that “ nf converges of f pointwise on E’ if

(1) holds.

Similarly, if ( )nf x converges for every x E , and if we define

1

( ) ( )nn

f x f x x E

(2)

the function f is called the sum of the series .nf

The main problem which arises is to determine whether important properties of

functions are preserved under the limit operations (1) and (2). For instance, if the functions

nf are continuous, or differentiable, or integrable, is the same true of the limit function?

What are the relations between nf and ,f say, or between the integrals of nf and that of

f?

Recall that f is continuous at x means

lim ( ) ( ).t x

f t f x

(3)

i.e., lim ( ) (lim ).t x t x

f t f t

Hence, to ask whether the limit of a sequence of continuous functions (i.e., lim nn

f

) is

continuous is the same as to ask whether

lim lim ( ) lim lim ( ),n nt x n n t x

f t f t

(4)

UNIT I CHAPTER 6

102

i.e., whether the order in which limit processes are carried out is immaterial. On the left side

of (3), we first let ,n then ;t x on the right side, t x first, then .n

T h e f o l l o w i n g e x a m p l e s s h o w t h a t l i m i t

p r o c e s s e s c a n n o t i n g e n e r a l b e i n t e r c h a n g e d

w i t h o u t a f f e c t i n g t h e r e s u l t .

(1.2) Example For 1,2, 3, , 1,2,3, ,m n let

, .m n

ms

m n

Then, for every fixed n,

,

1 1lim lim 1,

11

m nm m

sn

m

so that

,lim lim 1.m nn m

s

(5)

On the other hand, for every fixed m,

,lim lim 0,m nn n

ms

m n

so that

,lim lim 0.m nm n

s

(6)

(1.3) Example (Convergent series of continuous functions may have a discontinuous

sum) Let

2

2( ) real; 1,2, ,

1n n

xf x x n

x

Then each nf , being the quotient of polynomials, is continuous.

Consider


103

2

20 0

( ) ( ) .

1n n

n n

xf x f x

x

(7)

Since (0) 0,nf we have (0) 0.f

For 0,x

22

2 20 0

1

1 1n n

n n

xx

x x

, a geometric series with

common ration 2

1.

1 x

2

2 2 2 2

2 2

2 2

1 1 11 .

1 1 111 1

xx x x x

x x

x x

Hence from (7), we have

2

0 0 ,( )

1 0 ,

xf x

x x

(8)

Clearly f is not continuous at x = 0. This example shows that a convergent series of

continuous functions may have a discontinuous sum.

(1.4) Example (Everywhere discontinuous limit function, which is not Riemann-

integrable)

For 1,2, 3, ,m put

2

( ) lim cos ! .n

mn

f x m x

When !m x is an integer, cos !m x is 1 implies 2

cos ! 1n

m x and hence ( ) 1.mf x For

all other values of x, 1 cos ! 1m x implies 2

lim cos ! 0n

nm x

, using the Theorem :

“If 1,x then lim 0.n

nx

”

i.e., 1 when ! is an integer,

( )0 when ! is not an integer,

m

m xf x

m x

UNIT I CHAPTER 6

104

For irrational x, !m x is not an integer, and hence from the above ( ) 0 ;mf x m

hence ( ) 0.f x

For rational x, say ,p

xq

where p and q are integers, we see that !m x is an

integer if ,m q so that ( ) lim ( ) lim 1 1.mm m

f x f x

!m x is not an integer, and hence from the above ( ) 0 ;mf x m hence

( ) lim ( ) lim 0 0.mm m

f x f x

i.e., 0 when is irrational,

( )1 when is rational,

xf x

x

i.e., 2 0 when is irrational,

lim lim cos !1 when is rational.

n

m n

xm x

x

We have thus obtained an every where discontinuous limit function, which is not Riemann-

integrable.

(1.5) Example Let

sin

( ) real; 1,2, ,n

nxf x x n

n (9)

Then

sin

( ) lim ( ) lim 0.nn n

nxf x f x

n

implies ( ) 0.f x

From (9), cos

( ) cos ,n

n nxf x n nx

n

so that nf does not converge to f . For instance,

(0) cos 0 .nf n n n

Hence lim (0) lim ,nn n

f n

whereas (0) 0.f


105

i.e., lim (0) lim (0).n nn n

f f

Hence limit and differentiation cannot in general be interchanged.

(1.6) Example (Limit of the integral need not be equal to the integral of the limit, even if

both are finite)

Let

2 2( ) (1 ) 0 1, 1,2, .nnf x n x x x n (10)

For 0 1,x we have

lim ( ) 0,nn

f x

using the Theorem : “ If 0p and , then lim 0.(1 )nn

n

p

”

Since (0) 0,nf we see that lim (0) 0.nn

f

Hence, we have

lim ( ) 0 0 1 .nn

f x x

(11)

A simple calculation shows that

1

2

0

1(1 ) .

2 2

nx x dxn

(Hint: Put 2x t , then

2xdx dt and limits of integration are t = 0 and t = 1.)

Thus, inspite of (11),

21

0

( )2 2

n

nf x dx

n

as .n

If, in (10), we replace 2n by ,n (11) still holds, but we now have

1

0

1 1lim ( ) lim lim ,

22 2 22

nn n n

nf x dx

n

n

whereas 1 1

0 0

lim ( ) 0 0.nn

f x dx dx

UNIT I CHAPTER 6

106

Thus the limit of the integral need not be equal to the integral of the limit, even if both are

finite.

I n t h e c o m i n g s e c t i o n , w e d e f i n e a n e w m o d e o f

c o n v e r g e n c e w h i c h e n a b l e s u s t o i n t e r c h a n g e t h e

l i m i t p r o c e s s e s .

2 Uniform Convergence

(2.1) Definition We say that a sequence of functions , 1,2,3,nf n converges

uniformly on E to a function f if for every 0 there is an integer N such that n N

implies

( ) ( )nf x f x (12)

for all .x E

(2.2) Remarks

Every uniformly convergent sequence of functions is pointwise convergent.

The difference between pointwise convergence and uniform convergence:

If nf converges pointwise on E, then there exists a function f such that , for every

0 , and for every ,x E there is an integer N , depending on and on x, such that

(12) holds if n N ; if nf converges uniformly on E, then there exists a function

f such that , for every 0 , there is an integer N , depending only on , such that

(12) holds if n N .

(2.3) Definition We say that the series ( )nf x converges uniformly on E if the

sequence ns of nth partial sums defined by

1

( ) ( )n

i ni

f x s x

converges uniformly on E.

(2.4) Theorem The sequence of functions ,nf defined on E, converges uniformly on E

if and only if for every 0 there exists an integer N such that , ,m N n N x E implies


107

( ) ( )n mf x f x . (13)

Proof

Suppose nf converges uniformly on E, and let f be the limit function. Then there is an

integer N such that ,n N x E implies

( ) ( ) ,2

nf x f x

so that

( ) ( ) ( ) ( ) ( ) ( )n m n mf x f x f x f x f x f x ( ) ( ) ( ) ( )n mf x f x f x f x

if , ,m N n N x E .

Conversely, suppose the Cauchy condition holds. By Theorem: “In ,k every

Cauchy sequence converges”, for every x the sequence ( )nf x converges, say to ( ).f x

Thus the sequence nf converges on E, to f. We have to prove that the convergence is

uniform.

Let 0 be given, and choose N such that (13) holds. Fix n, and let m in

(13). Since ( ) ( )mf x f x as m , this gives

( ) ( )nf x f x

for every , ,n N x E which shows that nf converges on E, to f uniformly.

(2.5) Theorem Suppose

lim ( ) ( ) .nn

f x f x x E

Put

sup ( ) ( ) .n nx E

M f x f x

.

Then nf f uniformly on E if and only if 0nM as .n

Proof

nf f uniformly on E if and only if for every 0 there is an integer N such that

n N implies

UNIT I CHAPTER 6

108

( ) ( )nf x f x (12)

for all .x E

This is if and only if

sup ( ) ( )n nx E

M f x f x

for n N

if and only if 0nM for n N

if and only if (from the definition of limit of sequences) 0nM as .n

(2.6) Theorem (Weierstrass M (Majorant) Test) Suppose nf is a sequence of

functions defined on E, and suppose

( ) , 1, 2, 3, .n nf x M x E n

Then nf converges uniformly on E if nM converges.

Proof

If nM converges, then, its sequence of nth partial sums 1

n

ii

M

converges and hence the

sequence 1

n

ii

M

is Cauchy. Hence for 0 , with m n with m and n very large,

1 1

m n

i ii i

M M

implies

1

m

ii n

M

implies

1 1

( ) .m m

i ii n i n

f x M x E

i.e., 1 1

( ) ( )m n

i ii i

f x f x x E


109

when m and n very large. Hence by Theorem (2.4) the sequence of functions 1

n

ii

f

(i.e.,

the sequence of nth partial sums) is uniformly convergent. Hence, the series nf converges

uniformly on E . This completes the proof.

3 Uniform Convergence and Continuity

(3.1) Theorem Suppose nf f uniformly on a set E in a metric space. Let x be a limit

point of E, and suppose that

lim ( ) 1,2,3, .n nt x

f t A n

(13)

Then nA converges, and

lim ( ) lim .nt x n

f t A

(14)

i.e., lim lim ( ) lim lim ( ).n nt x n n t x

f t f t

(15)

(UQ 2006)

Proof

Let 0 be given. By the uniform convergence of nf , by Theorem (2.4), there exists

there exists an integer N such that , ,m N n N t E implies

( ) ( )n mf t f t . (16)

Letting t x in (16), we obtain

n mA A

for , ,m N n N so that nA is a Cauchy sequence and therefore converges, say to A.

Next,

( ) ( ) ( ) ( ) .n n n nf t A f t f t f t A A A (17)

We first choose n such that

( ) ( )3

nf t f t

(18)

UNIT I CHAPTER 6

110

for all t E (the same n for all t is possible by the uniform convergence), and such that

.3

nA A

(19)

Then, for this n, we choose a neighborhood V of x such that

( )3

n nf t A

(20)

if , .t V E t x [(20) is possible because lim ( ) ].n nt x

f t A

Substituting the inequalities (18) to (20) into (17), we see that

( ) ,f t A (21)

provided , .t V E t x This is equivalent to saying that lim ( ) .t x

f t A

i.e.,

lim ( ) lim ,nt x n

f t A

as desired.

(3.2) Theorem (Corollary to Theorem (3.1)) Suppose nf is a sequence of continuous

functions on E, and if nf f uniformly on E, then f is continuous on E.

Proof

nf is a sequence of continuous functions on E implies that each nf is continuous at each

point of E. Let x be a point on E. We have to show that f is continuous at x, then since x is

arbitrary point, it follows that f is continuous on E.

Since each nf is continuous at x, we have

lim ( ) ( ) 1,2,3, .n nt x

f t f x n

Then, since nf f uniformly on E, using the previous theorem we have

lim ( ) lim ( ).nt x n

f t f x

i.e., lim ( ) ( ),t x

f t f x

showing that f is continuous at x.


111

The converse of Theorem (3.2) is not true; i.e., a sequence of continuous

functions may converge to a continuous function, although the convergence is

not uniform.

(3.3) Theorem Suppose K is compact, and

(a) nf is a sequence of continuous functions on K,

(b) nf converges pointwise to a continuous function f on K,

(c) 1( ) ( )n nf x f x for all , 1,2,3,x K n

Then nf f uniformly on K.

Proof

Put .n ng f f Then ng is continuous, 0ng pointwise, and 1.n ng g We have to

prove that 0ng uniformly on K.

Let 0 be given. Let

| ( ) .n nK x K g x

Since ng is continuous, 1 [ , )n nK g is closed, and hence is compact. Since

1n nf f we have 1n ng g and hence 1.n nK K Fix .x K Since ( ) 0,ng x we see

that nx K if n is sufficiently large. Thus .nx K In other words, nK is empty.

Hence NK is empty for some N. It follows that 0 ( )ng x x K and .n N This

proves the theorem.

(3.4) Remark In the above theorem compactness is really needed. For instance, if

1

( ) 0 1; 1,2,3,1

nf x x nnx

then ( ) 0nf x monotonically in (0, 1), but the convergence is not uniform.

(3.5) Definition If X is a metric space, ( )X will denote the set of all complex valued,

continuous, bounded functions with domain X.

UNIT I CHAPTER 6

112

We associate with each ( )f X its supremum norm

sup ( ) : .f f x x X

Since f is assumed to be bounded, by the supremum property of , .f

( i) 0f if and only if sup ( ) : 0f x x X if and only if ( ) 0f x for all x X

if and only if ( ) 0f x for all x X if and only if 0,f the zero function.

( ii) Note that, for all x X

( ) ( ) ( ) ( ) ( )f g x f x g x f x g x f g .

Hence sup ( ) :f g x x X f g

i.e., .f g f g

For ( )f X and ( )g X define ,d f g f g . Then d is a metric on ( ) :X For,

( i) , 0 0 0 .d f g f g f g f g

( ii) , , .d f g f g f g g f d g f

( iii) , , , .d f g f g f h h g f h h g d f h d h g

Thus ( )X is a metric space.

Now Theorem (2.5) can be rephrased as follows:

(3.6) Theorem (Theorem (2.5)) A sequence nf converges to f with respect to the

metric of ( )X if and only if nf f uniformly on X.

Closed subsets of ( )X are called uniformly closed, the closure of a

set ( )XA is called its uniform closure, and so on.

(3.7) Theorem The above metric makes ( )X into a complete metric space.


113

Proof

We have to show that every Cauchy sequence in ( )X is convergent. Let nf be a Cauchy

sequence in ( )X . This means that to each 0 there corresponds an N such that

n mf f if n N and m N .

i.e., sup ( ) :n mf f x x X if n N and m N .

i.e., ( ) ( )n mf x f x if x X n N and m N .

Hence it follows by Theorem (2.4) that there is a function f with domain X to which

nf converges uniformly. By Theorem (3.2), since each nf is continuous, f is continuous.

Since there is an n such that ( ) ( ) 1 ,nf x f x x X and hence

( ) 1 ( ) ,nf x f x x X so that nf is bounded. Hence, f is bounded.

Thus ( ),f X and since nf f uniformly on X, we have 0nf f

as .n

4 Uniform Convergence and Integration

(4.1) Theorem Let be monotonically increasing on [ , ].a b Suppose ( )nf on

[ , ],a b for 1,2,3,n and suppose nf f uniformly on [ , ].a b Then ( )f on

[ , ],a b and

lim .b b

nn

a a

f d f d

(22)

(The existence of the limit is part of the conclusion.)

Proof

It suffices to prove this for real nf . Put

sup ( ) ( ) ,n nf x f x (23)

the supremum being taken over .a x b Then

,n n n nf f f

so that the upper and lower integrals of f satisfy

UNIT I CHAPTER 6

114

_

.b b

n n n na a

f d f d f d f d (24)

Hence

_

0 2 ( ) ( ) .nf d f d b a (25)

Since nf f uniformly on [ , ],a b using Theorem (2.5) , we have 0n as n and

hence from (25), the upper and lower integrals of f are equal.

Thus ( )f . Another application of (24) now yields

( ) ( ) .b b

n na a

f d f d b a (26)

Since nf f uniformly on [ , ],a b using Theorem (2.5) , we have 0n as n and

hence from (26), we have

lim .b b

nn

a a

f d f d

(4.2) Corollary If ( )nf on [ , ],a b and if

1

( ) ( ) ,nn

f x f x a x b

the series converging uniformly on [ , ],a b then

1

.b b

nna a

f d f d

(27)

In other words, the series may be integrated term by term.

5 Uniform Convergence and Differentiation

(5.1) Theorem Suppose nf is a sequence of functions, differentiable on [ , ]a b and such

that 0( )nf x converges from some point 0 [ , ].x a b If nf converges uniformly on

[ , ]a b , then nf converges uniformly on [ , ]a b , to a function f, and


115

( ) lim ( ) .nn

f x f x a x b

(28)

(UQ 2006)

Proof

Let 0 be given. Choose N such that , ,n N m N implies

0 0( ) ( )2

n mf x f x

(29)

and

( ) ( ) .2( )

n mf t f t a t bb a

(30)

If we apply the mean value theorem to the function n mf f , (30) shows that

( ) ( ) ( ) ( )2( ) 2

n m n m

x tf x f x f t f t

b a

(31)

for any x and t on [ , ]a b , if , .n N m N The inequality

0 0 0 0( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )n m n m n m n mf x f x f x f x f x f x f x f x

implies, by (29) and (31), that

( ) ( ) , , ,n mf x f x a x b n N m N

so that nf converges uniformly on [ , ]a b . Let

( ) lim ( ) .nn

f x f x a x b

Let us now fix a point [ , ]x a b and define

( ) ( ) ( ) ( )

( ) , ( )n nn

f t f x f t f xt t

t x t x

(32)

for , .a t b t x Then

lim ( ) ( ) 1,2,3,n nt x

t f x n

(33)

The first inequality in (31) shows that

( ) ( ) , ,2( )

n t t n N m Nb a

UNIT I CHAPTER 6

116

so that n converges uniformly, for .t x Since nf converges uniformly to f, we

conclude from (32) that

lim ( ) ( )nn

t t

(34)

uniformly for , .a t b t x

If we now apply Theorem (3.1) to n , (33) and (34) show that

lim ( ) lim ( );nt x n

t f x

and this is (28), by the definition of ( ).t

(5.2) Theorem There exists a real continuous function on the real line which is nowhere

differentiable.

Proof

Define

( ) 1 1x x x (35)

and extend the definition of ( )x to all real x by requiring that

( 2) ( ).x x (36)

Then, for all s and t,

( ) ( ) .s t s t (37)

In particular, is continuous on . Define

0

3( ) (4 ).

4

nn

n

f x x

(38)

Since 0 1, Theorem (2.6) shows that the series (38) converges uniformly on . By

Theorem (3.2), f is continuous on .

Now fix a real number x and a positive integer m. Put

1

42

mm

(39)

where the sign is so chosen that no integer lies between 4m x and 4mmx . This can be

done, since 1

42

mm Define

(4 ( )) (4 )

.n n

mn

m

x x

(40)


117

When ,n m then 4nm is an even integer, so that 0.n When 0 ,n m (37) implies

that 4 .nn

Since 4 ,mm we conclude that

0

( ) ( ) 3

4

nmm

nnm

f x f x

1

0

13 3 3 1 .

2

mm n m

n

As m , 0.m It follows that f is not differentiable at x.

6 Equicontinuous Families of Functions

(6.1) Definition Let nf be a sequence of functions defined on a set E.

We say that nf is point wise bounded on E if the sequence ( )nf x is bounded for

every ,x E i.e., if there exists a finite-valued function defined on E such that

( ) ( ) , 1,2,3,nf x x x E n .

We say that nf is uniformly bounded on E if there exists a number M such that

( ) , 1,2,3, .nf x M x E n

(6.2) Remark If nf is pointwise bounded on E and 1E is a countable subset of E, it is

always possible to find a subsequence knf such that ( )

knf x converges for every 1.x E

This can be done by the diagonal process which is used in the proof of (coming) Theorem

(6.6).

(6.3) Example

Let

( ) sin 0 2 , 1,2,3, .nf x nx x n

Suppose there exists a sequence kn such that sin kn x converges, for every [0, 2 ].x In

that case we must have

1lim sin sin 0 0 2 ;k kk

n x n x x

UNIT I CHAPTER 6

118

hence

2

1lim sin sin 0 0 2 .k kk

n x n x x

(41)

By Lebesgue’s theorem concerning integration of boundedly convergent sequences, (41)

implies

2

2

10

lim sin sin 0.k kk

n x n x dx

(42)

But a simple calculation shows that

2

2

10

sin sin 2 ,k kn x n x dx

which contradicts (42).

(6.4) Example

Let

2

2 2( ) 0 1, 1,2,3, .

(1 )n

xf x x n

x nx

Then ( ) 1,nf x so that nf is uniformly bounded on [0, 1]. Also

lim ( ) 0 0 1 ,nn

f x x

but

1

1 1,2,3, ,nf nn

so that no subsequence can converge uniformly on [0, 1].

(6.5) Definition

A family F of complex functions f defined on a set E in a metric space X is said be

equicontinuous on E if for every 0 there exists a 0 such that

( ) ( )f x f y

whenever ( , ) , , ,d x y x E y E and f F . Hence d denotes the metric of X.

Every member of an equicontinuous family is uniformly continuous.

The sequence in Example (6.4) is not equicontinuous.


119

(6.6) Theorem If nf is a pointwise bounded sequence of complex functions on a

countable set E , then nf has a subsequence knf such that ( )

knf x converges for

every .x E

Proof

Let , 1,2,3,ix i be the points of E, arranged in a sequence. Since 1( )nf x is bounded,

there exists a subsequence, which we shall denote by 1, ,kf such that 1, 1( )kf x converges

as .k

Let us now consider sequences 1 2 3, , , ,S S S which we represent by the array

1 1,1 1,2 1,3 1,4

2 2,1 2,2 2,3 2,4

3 3,1 3,2 3,3 3,4

:

:

:

S f f f f

S f f f f

S f f f f

and which have the following properties:

(a) nS is a subsequence of 1nS for 2,3,4,n

(b) , ( )n k nf x converges, as k (the boundedness of ( )n nf x makes it

possible to choose nS in this way);

(c) The order in which the functions appear is the same in each sequence; i.e., if one

function precedes another in 1,S they are in the same relation in every ,nS until

one or the other is deleted. Hence, when going from one row in the above array

to the next below, functions may move to the left but never to the right.

We now go down the diagonal of the array; i.e., we consider the sequence

1,1 2,2 3,3 4,4:S f f f f

By (c), the sequence S (except possibly its first n 1 terms) is a subsequence of nS for

1,2,3,n Hence (b) implies that , ( )n n if x converges, as , .in x E

(6.7) Theorem If K is a compact metric space, if ( )nf K for 1,2,3,n , and if

nf converges uniformly on K, then nf is equicontinuous on K.

Proof

Let 0 be given. Since nf converges uniformly, there is an integer N such that

UNIT I CHAPTER 6

120

n N implies

( ) ( )nf x f x

for all .x E Since sup ( ) ( ) | ,n N n Nf f f x f x x X we have

.n Nf f n N (43)

Since continuous functions are uniformly continuous on compact sets, there is a 0 such

that

( ) ( )i if x f y (44)

if 1 i N and ( , ) .d x y

If n N and ( , ) ,d x y it follows that

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 .n n n N N N N nf x f y f x f x f x f y f y f y (45)

By (44) and (45), we have, for all n

( ) ( ) 3n nf x f y

whenever ( , ) ,d x y ,x E y E and ( )nf K . Hence [by the Definition (6.5)] nf is

equicontinuous on K.

(6.8) Theorem If K is compact, if ( )nf K for 1,2,3,n , and if nf is pointwise

bounded and equicontinuous on K, then

(a) nf is uniformly bounded on K,

(b) nf contains a uniformly convergent subsequence. (UQ 2006)

Proof

(a) Let 0 be given and choose 0 , in accordance with Definition (6.5), so that

( ) ( )n nf x f y (46)

for all n, provided ( , ) .d x y

Since K is compact, there are finitely many points 1, , rp p in K such that to every

x K corresponds at least one ip with ( , ) .id x p Since nf is pointwise bounded, there

exist iM such that ( )n i if p M for all n. If 1max , , ,rM M M then

( )f x M for every .x K This proves (a).


121

(b) Let E be a countable dense subset of K. Theorem (6.6) shows that nf has a

subsequence inf such that ( )

inf x converges for every .x E

Put ,in if g to simplify the notation. We shall prove that ig converges uniformly

on K.

Let 0 , and pick 0 as in the beginning of the proof. Let

( , ) | ( , ) .V x y K d x y

Since E is dense in K, and K is compact, there are finitely many points 1, , mx x in E such

that

1( , ) ( , ).mK V x V x (47)

Since ( )ig x converges for every ,x E there is an integer N such that

( ) ( )i s j sg x g x (48)

whenever , , 1 .i N j N s m

If ,x K (47) shows that ( , )sx V x for some s , so that

( ) ( )i i sg x g x

for every i. If i N and j N , it follows from (48) that

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )i j i i s i s j s j s jg x g x g x g x g x g x g x g x

3 .


7 The Stone-Weierstrass Theorem

(7.1) Theorem (Weierstrass theorem) If f is a continuous complex function on [ , ]a b ,

there exists a sequence of polynomials nP such that

lim ( ) ( )nn

P x f x

uniformly on [ , ]a b . If f is real, the nP may be taken real.

This is the form in which the theorem was originally discovered by Weierstrass.

Proof

UNIT I CHAPTER 6

122

We may assume, without loss of generality, that [ , ]a b = [0, 1]. We may also assume that

(0) (1) 0.f f For if the theorem is proved for this case, consider

( ) ( ) (0) [ (1) (0)] 0 1 .g x f x f x f f x

Here (0) (1) 0,g g and if g can be obtained as the limit of a uniformly convergent

sequence of polynomials, it is clear that the same is true for f, since f g is a polynomial.

Furthermore, we define ( )f x to be zero for x outside [0, 1]. Then f is uniformly

continuous on the whole line.

We put

2( ) (1 ) 1,2,3, ,nn nQ x c x n (49)

where nc is chosen so that

1

1

( ) 1 1,2,3, ,nQ x dx n

(50)

We need some information about the order of magnitude of nc . Since

1

1 12 2 2

1 0 0

(1 ) 2 (1 ) 2 (1 )n

n n nx dx x dx x dx

1

2

0

2 (1 )n

nx dx †

4

3 n

1

,n

it follows from (50) that

.nc n (51)

† The inequality 2 2(1 ) 1nx nx which we used above is easily shown to be true by

considering the function

2 2(1 ) 1nx nx

which is zero at 0x and whose derivative is positive on (0, 1).

For any 0, (51) implies


123

2( ) (1 ) 1 ,nnQ x n x (52)

so that 0nQ uniformly in 1x .

Now set

1

1

( ) ( ) ( ) 0 1 .n nP x f x t Q t dt x

(53)

Our assumptions about f show, by a simple change of variable, that

1

( ) ( ) ( )x

n nx

P x f x t Q t dt

1

0

( ) ( )nf t Q t x dt , by putting x t u

and the last integral is clearly a polynomial in x. Thus nP is a sequence of polynomials,

which are real if f is real.

Given 0 , we choose 0 such that y x implies

( ) ( ) .2

f y f x

Let sup ( ) : [ , ] .M f x x a b

Using (50), (52), and the fact that ( ) 0,nQ x we see that for 0 1,x

1

1

( ) ( ) ( ) ( ) ( )n nP x f x f x t f x Q t dt

1

1

( ) ( ) ( )nf x t f x Q t dt

1

1

2 ( ) ( ) 2 ( )2

n n nM Q t dt Q t dt M Q t dt

24 (1 )2

nM n

for all large enough n, which proves the theorem.

(7.2) Corollary For every interval [ , ]a a there is a sequence of real polynomials nP such

that (0) 0nP and such that

UNIT I CHAPTER 6

124

lim ( )nn

P x x

uniformly on [ , ].a a

Proof

By Theorem (7.1), there exists a sequence *

nP of real polynomials which converges to x

uniformly on [ , ].a a In particular, *(0) 0n

P as .n The polynomials

* *( ) ( ) (0) 1,2,3,n nnP x P x P n

have desired properties: (0) 0nP and such that

lim ( )nn

P x x

uniformly on [ , ].a a

(7.3) Definition A family A of complex functions defined on a set E is said to be an

algebra (complex algebra) if

( i) A is closed under addition:

i.e., ,f g f g A A A

( ii) A is closed under multiplication:

i.e., ,fg f g A A A

( iii) A is closed under scalar multiplication:

i.e., cf f A A and complex constant c.

Example: The set of all polynomials is an algebra.

(7.4) Definition A family A of complex functions defined on a set E is said to be a real

algebra if

( i) A is closed under addition:

i.e., ,f g f g A A A

( ii) A is closed under multiplicaton:

i.e., ,fg f g A A A

( iii) A is closed under scalar multiplication:

i.e., cf f A A and real constant c.


125

(7.5) Definition If A has the property that f A whenever 1,2,3,nf n A and

nf f uniformly on E, then A is said to be uniformly closed.

(7.6) Definition Let B be the set of all functions which are limits of uniformly convergent

sequences of members of A . Then B is called the uniform closure of A .

Remark: Since the set of all polynomials is an algebra, the Weierstrass

theorem may be stated by saying that the set of continuous functions on

[ , ]a b is the uniform closure of the set of polynomials on [ , ]a b .

(7.7) Theorem Let B be the uniform closure of an algebra A of bounded functions. Then

B is a uniformly closed algebra.

Proof

If f B and gB , (by the Definition (7.6) of B ) there exist uniformly convergent

sequences ,n nf g such that ,n nf f g g and ,n nf g A A. Since we are dealing

with bounded functions, it is easy to show that

, , ,n n n n nf g f g f g fg cf cf

where c is any constant, the convergence being uniform in each case.

Hence (again by the Definition (7.6) of B ) f g fg B, B, and cf B , so that B is an

algebra.

Also B , being the closure of a set, is (uniformly) closed.

(7.8) Definition Let A be a family of functions on a set E. Then A is said to separate

points on E if to every pair of distinct points 1 2,x x E there corresponds a function f A

such that 1 2( ) ( ).f x f x

(7.9) Definition Let A be a family of functions on a set E. If to each x E there

corresponds a function gA such that ( ) 0,g x we say that A vanishes at no point of E.

(7.10) Examples

The algebra of all polynomials in one variable is clearly has the properties in

Definitions (7.8) and (7.9).

UNIT I CHAPTER 6

126

An example of an algebra which does not separate points is the set of all even

polynomials, say on [ 1, 1], since ( ) ( )f x f x for every even function f, so that for

two distinct points x and x there is no polynomial such that ( ) ( ).f x f x

(7.11) Theorem Suppose A is an algebra of functions on a set E, A separates points on

E, and A vanishes at no point of E. Suppose 1 2,x x are distinct points of E, and 1 2,c c are

constants (real if A is a real algebra). Then A contains a function f such that

1 1 2 2( ) , ( ) .f x c f x c

Proof

o A separates points on E, implies that there exist a function gA such that

1 2( ) ( )g x g x .

o A vanishes at no point of E, hence corresponding to 1x there exist a function hA

such that 1( ) 0h x .

o A vanishes at no point of E, hence corresponding to 2x there exist a function kA

such that 2( ) 0k x .

Define the functions u and v by

1 2( ) , ( ) .u gk g x k v gh g x h

Then u v A, A,

1 1 1 1 1( ) ( ) ( ) ( ) ( ) 0u x g x k x g x k x ,

2 2 2 1 2 2 1 2( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0,u x g x k x g x k x g x g x k x

1 2( ) 0, ( ) 0.v x v x

Hence the function

1 2

1 2( ) ( )

c v c uf

v x u x

has the properties:

1 1 2 11 1

1 2

( ) ( )( )

( ) ( )

c v x c u xf x c

v x u x and 1 2 2 2

2 21 2

( ) ( )( ) .

( ) ( )

c v x c u xf x c

v x u x

(7.12) Theorem (Stone’s generalization of the Weierstrass theorem) Let A be an algebra

of real continuous functions on a compact set K. If A separates points on K and if A


127

vanishes at no point of K, then the uniform closure B of A consists of all real continuous

functions on K.

Proof

The proof is divided into four steps.

Step 1 If f B, then f B.

Proof

Let

sup ( ) :a f x x K (54)

and let 0 be given. By Corollary (7.2) there exist real numbers 1, , nc c such that

1

.n

ii

i

c y y a y a

(55)

Since B is an algebra, the function

1

ni

ii

g c f

is a member of B . By (54) and (55), we have

( ) ( ) .g x f x x K

Since B is uniformly closed, this shows that f B.

Step 2 If f B, and gB, then max( , )f g B and min( , )f g B.

Proof

By max( , )f g we mean the function h defined by

( ) if ( ) ( )

( )( ) if ( ) ( )

f x f x g xh x

g x f x g x

and is given by

,2 2

f gf gh

and by min( , )f g we mean the function l defined by

( ) if ( ) ( )

( )( ) if ( ) ( )

f x f x g xl x

g x f x g x

UNIT I CHAPTER 6

128

and is given by

.2 2

f gf gl

Step 2 follows from Step 1 and the identities

max( , ) ,2 2

f gf gf g

and min( , ) .2 2

f gf gf g

By iteration, the result can of course be extended to any finite set of functions: If

1, , nf f B , then 1max( , , )nf f B and 1min( , , )nf f B .

Step 3 Given a real function f, continuous on K, a point ,x K and 0, there exists a

function xg B such that ( ) ( )xg x f x and

( ) ( ) .xg t f t t K (56)

Proof

Since A B and A satisfies the hypotheses of Theorem (7.11) so does B . Hence, for every

y K , we can find a function yh B such that

( ) ( ), ( ) ( ).y yh x f x h y f y (57)

By the continuity of h, there exist an open set yJ containing y, such that

( ) ( ) .y yh t f t t J (58)

Then the collection :yJ y K form an open cover for K. Since K is compact, this cover

has a finite sub cover, and hence there is a finite set of points 1, , ny y such that

1

.ny yK J J (59)

Put

1

max , , .nx y yg h h

By Step 2, xg B , and the relations (57) to (59) show that xg has the other required

properties.


129

Step 4 Given a real function f, continuous on K, and 0, there exists a function hB

such that

( ) ( ) .h x f x x K (60)

Since B is uniformly closed, this statement is equivalent to the conclusion of the theorem.

Proof

Let us consider the functions ,xg for each ,x K constructed in step 3. By the continuity of

,xg there exist open sets xV containing x, such that

( ) ( ) .x xg t f t t V (61)

Since K is compact, there exists a finite set of points 1, , mx x such that

1

.mx xK V V (62)

Put

1

min , , .mx xh g g

By step 2, hB , and (56) implies

( ) ( ) ,h t f t t K (63)

whereas (61) and (62) imply

( ) ( ) .h t f t t K (64)

From (63) and (64), we have

( ) ( )h t f t t K

i.e., ( ) ( ) .h t f t t K

Hence (60) is obtained.

(7.13) Definition An algebra A is said to be self adjoint if for every f A its complex

conjugate f A , where f is defined by ( ) ( ).f x f x

(7.14) Theorem Suppose A is a self adjoint algebra of complex continuous functions on a

compact set K, A separates points on K, and A vanishes at no point of K. Then the

uniform closure B of A consists of all complex continuous functions on K. In other words,

A is dense in ( ).K

Proof

UNIT I CHAPTER 6

130

Let A be the set of all real functions on K which belong to A .

If f A and ,f u iv with u, v real, then 2 ,u f f and since A is self adjoint,

we see that .uA If 1 2 ,x x there exists f A such that 1 2( ) 1, ( ) 0;f x f x hence

2 10 ( ) ( ) 1,u x u x which shows that A separates points on K. If ,x K then ( ) 0g x

for some g A , and there is a complex number such that ( ) 0;g x if

,f g ,f u iv it follows that ( ) 0;u x hence A vanishes at no point of K.

Thus A satisfies the hypothesis of Theorem (7.12). It follows that every real

continuous function on K lies in the uniform closure of A , hence lies in B . If f is a

complex continuous function on K, ,f u iv then uB , vB , hence f B. This

completes the proof.

___________________

Chapter 7

LEBESGUE MEASURE

131

(1.1) Intervals

The simplest sets of real numbers are the intervals. We define open interval ( , )a b to be the

set ( , ) : .a b x a x b We always take ,a b but we also consider the infinite

intervals ( , ) :a x a x and ( , ) :a x x b . Some times we write ( , )

for , the set of all real numbers. We define closed interval [ , ]a b to be the set

[ , ] : .a b x a x b For closed intervals we take a and b finite but always assume that

.a b The half open interval ( , ]a b to be the set ( , ] : .a b x a x b The half

open interval [ , )a b to be the set [ , ) : .a b x a x b

(1.2) Length of an interval

Let I be an interval of any type above. Then the length ( )l I of the interval I is

defined, as usual, to be the difference of the endpoints of the interval. For example, length of

the interval (0, 1) is 1 and that of [0, 1] is also 1.

Length is an example of a set function, that is, a function that associates an extended

real number to each set in some collection of sets.

In the case of length the domain is the collection of all intervals.

We should like to extend the notion of length to more complicated sets than intervals.

For instance, we could define the length of an open set to be the sum of the lengths the open

intervals of which it is composed.

(1.3) Definition The system of real numbers can be extended by including two

elements and . This enlarged set is called the extended real number system.

(1.4) Definition A set function m that assigns to each set E in some collection M of

sets of real numbers a nonnegative extended real number m(E), or simply mE, is called the

measure of E if it satisfies the following properties:

i. mE is defined for each set E of real numbers; that is, M = P ( ), the power

set of .

ii. For an interval I, ( ).mI l I

iii. If nE is a sequence of disjoint sets (for which m is defined),

.n nm E mE

UNIT II CHAPTER 7

132

iv. m is translation invariant; that is, if E is a set for which m is defined and if

| ,E y x y x E the set obtained by replacing each point x in E by the

point ,x y then ( ) .m E y mE

Unfortunately, it is impossible to construct a set function having all four of these

properties. So first property must be weakened so that mE need not be defined for all sets E

of real numbers. We shall want mE to be defined for as many sets as possible and will find it

convenient to require the family M of sets for which m is defined to be a algebra.

(1.5) Definition We say that m is a countably additive measure if

( i) it is a nonnegative extended real valued function whose domain of definition is a

algebra M of sets (of real numbers) and

( ii) e n nm E mE for each sequence nE of disjoint sets in M .

Our goal in the coming sections will be the construction of a countably additive

measure which is translation invariant and has the property that ( )mI l I for each interval I.

(1.6) Definition Let A and B two sets. Then \A B is

the set of all elements in A that are not in B. i.e.,

\ : , .A B x x A x B

In Figure 1, shaded region indicates the set \A B .

(1.7) Notations

For a set E, we denote the complement of E by E or some times by ~ .E .

\A B is also denoted by .A B

(1.8) Remark \ .A B A B [Ref. Figure 1].

(1.9) Definition (Algebra) A collection of subsets of X is called an algebra of sets (or a

Boolean algebra) if

( i) A, B .A B

( ii) A .A

(1.10) Remark If is an algebra, then

( iii) A, B .A B

Proof A, B , ,A B by (ii) above

,A B by (i)

LEBESGUE MEASURE

133

,A B by (ii)

[Here A B denotes the complement of A B

,A B since A B A B by De Morgan’s law

,A B since ,A A and .B B

(1.11) Remark If a collection of subsets of X satisfies (ii) and (iii) above, then by De-

Morgan’s law (proceeding as in the proof above, replacing by and vice-versa) it also

satisfies (i) and hence is an algebra.

(1.12) Remark By taking unions two at a time, we see that

( iv) 1, , nA A , then 1 nA A

Also,

( v) 1, , nA A , then 1 nA A .

(1.13) Definition ( -Algebra) An algebra of sets is called a -algebra (or Borel

field), if every union of countable collection of sets in is again in .

i.e., if iA is a sequence of sets, then [in addition to (i) to (ii) of Definition (1.9)]

( iv) 1

.ii

A

By De-Morgan’s law, from (i) proceeding as in the proof of (iii), we have

( v) 1

.ii

A

(i.e., intersection of countable collection of sets in is again in ).

(1.14) Theorem Given any collection C of subsets of X, there is a smallest -algebra that

contains C ; i.e., there is a -algebra containing C such that if B is any -algebra

containing C , then B.

The smallest -algebra containing C is called the -algebra generated by C .

The proof of the theorem is not part of the syllabus and is omitted.

(1.15) Proposition Let be an algebra of subsets and iA be a sequence of sets in .

Then there is a sequence iB of sets in such that n mB B for n m and

1 1

.i ii i

B A

UNIT II CHAPTER 7

134

Moreover, 1 1,B A and for 1 2 11, \ .n n nn B A A A A Hence n nB A for each n.

The proof of the theorem is not part of the syllabus and is omitted.

(1.16) Problems

Let m be a countably additive measure defined for all sets in a algebra M .

1. If A and B are two sets in M with

,A B then .mA mB (This property is

known as monotonicity)

Proof

Since ,A B we can write

\ .B A B A [Ref. Figure 2]

Also note that \ .A B A By

Definition (1.5), since A and \B A are disjoint,

\ \m A B A mA m B A

i.e., \mB mA m B A

implies ,mB mA since \ 0m B A as m is nonnegative function.

2. Let iA be any sequence of sets in M. Then .i im A mA

Proof By Proposition (1.15), there is a sequence iB of sets in M such that

n mB B for n m and 1 1

.i ii i

B A

Hence

1 1

i ii i

m A m B

1

ii

mB

, as m is countably additive measure and iB

are mutually disjoint.

LEBESGUE MEASURE

135

1

,ii

mA

as i iB A by Proposition (1.15), and by 1

above .i i i iB A mB mA

3. If there is a set A in M such that ,mA then 0.m

Proof A A . Hence

mA mA m , as m is countably additive and A and are mutually disjoint.

Since ,mA we subtract mA from each side and get 0.m

4. Let , if is infinite

number of elemets in , if is finite

EnE

E E

Then n is a countably additive set function that is translation invariant and defined for all

sets of real numbers. This measure is called the counting measure.

( i) Clearly n is a nonnegative extended real valued function whose domain of

definition is a algebra M of sets (of real numbers)

( ii) n nn E nE for each sequence nE of disjoint sets ( nE ’s may be finite

sets or infinite sets) in M .

( iii) n is translation invariant; that is, if E is a set for which n is defined and if

| ,E y x y x E the set obtained by replacing each point x in E by the

point ,x y then

Case 1: E is a finite set. Then E y is also a finite set and contains the

same number of element of E. Hence

number of elemets in .nE E n E y

Case 2: E is an infinite set, then E y is also infinite. Hence

.nE n E y

2 Outer Measure

UNIT II CHAPTER 7

136

For each set A of real numbers consider the countable collection nI of open

intervals that cover A, that is, collection for which ,nA I and for each such collection

consider the sum of the length of the intervals in the collection. Since the lengths are positive

numbers, this sum is uniquely defined independently of the order of the terms.

(2.1) Definition Let nI of open intervals that cover A. The outer measure *m A of A

to be infimum of all sums discussed above. That is,

*m A inf .nA In

l I

(2.2) Remarks

* 0.m

If ,A B then * * .m A m B

Each set consisting of a single point has outer measure 0.

Proof Let { }x be a singleton set. Then for 0 , ( , )x x is an open

cover of { }x so that *{ }m x inf 2nA In

l I

. Since is arbitrary, it

follows that *{ } 0.m x

(2.3) Proposition The outer measure of an interval is its length. (UQ)

Proof We begin with the case in which we have a closed finite interval, say [ , ].a b Since the

open interval ( , )a b contains [ , ]a b for each positive , we have

*[ , ] ( , ) 2 .m a b l a b b a

Since

*[ , ] 2m a b b a

for each positive (in particular for every positive near to 0), we must have

*[ , ] .m a b b a

Claim: *[ , ] .m a b b a

This is equivalent to show that if nI is any countable collection of open intervals covering

[ , ],a b then

.nl I b a (1)

By the Heine-Borel Theorem, any collection of open intervals covering [ , ]a b contains a finite

subcollection that also covers [ , ]a b , and since the sum of the lengths of the finite

LEBESGUE MEASURE

137

subcollection is no greater than the sum of the lengths of the original collection, it suffices to

prove the inequality (1) for finite collections nI that cover [ , ]a b . Since a is contained in

nI , there must be one of the nI ’s that contains a. Let this be the interval 1 1( , ).a b We

have 1 1.a a b If 1 ,b b then 1 [ , ],b a b and since 1 1 1( , ),b a b there must be an interval

2 2( , )a b in the collection nI such that 1 2 2( , );b a b that is 2 1 2.a b b Continuing in this

fashion, we obtain a sequence 1 1( , ), , ( , )k ka b a b from the collection nI such that

1 .i i ia b b

Since nI is a finite collection, our process must terminate with some interval

( , ).k ka b But it terminates only if ( , );k kb a b that is if .k ka b b Thus

n i il I l a b

1 1 1 1( ) ( ) ( )k k k kb a b a b a

1 1 2 2 1 1( ) ( ) ( )k k k k kb a b a b a b a

1,kb a

since 1.i ia b But kb b and 1 ,a a and so we have 1 ,kb a b a whence

( ).nl I b a This shows that *[ , ] .m a b b a

If I is any finite interval, then given 0, there is a closed interval J I such that

.l J l I Hence

* * * .l I l J m J m I m I l I l I

Thus for each 0,

* ,l I m I l I

and so * .m I l I

If I is an infinite interval, then given any real number , there is a closed interval

J I with .l J Hence * * .m I m J l J Since *m I for each ,

* .m I l I

(2.4) Proposition Let nA be a countable collection of sets of real numbers. Then

* * .n nm A m A (1)

(UQ 2008)

UNIT II CHAPTER 7

138

Property (1) is called subadditivity.

Proof

If one of the sets nA has infinite outer measure, say *nm A , the inequality holds

trivially.

If *nm A is finite for all n, then, given 0, there is a countable collection ,n i

iI of

open intervals such that ,n n i

i

A I and *, 2 .n

n i n

i

l I m A

Now the collection , ,,

n i n in i i

n

I I is countable, being the union of a countable

number of countable collections, and covers .nA Thus

* *, ,

,

2 nn n i n i n

n i n i n

m A l I l I m A

* .nm A

Since was an arbitrary positive number,

* * .n n

n

m A m A

(2.5) Corollary If A is countable, * 0.m A (UQ 2008)

Proof

If A is countable, then the set A is of the form 1 2 3, , , , , .nA x x x x Then A can be

expressed as the countable union of pariwise disjoint singleton sets

1 2 3, , , , ,nx x x x ..

i.e., 1 2 3 1 2 3, , , , ,n nA x x x x x x x x

Now by Proposition (2.4),

* * * 0 0,n n

n nn

m A m x m x

since * 0 fornm x 1,2,3,n

(2.6) Corollary The set [0, 1] is not countable.

LEBESGUE MEASURE

139

Proof

If [0, 1] is countable, then by Corollary (2.5), its measure must be 0 which is false. Hence

[0, 1] is not countable.

(2.7) Proposition Given any set A and any 0, there is an open set O such that

A O and * * .m O m A There is a G G such that A G and * * .m A m G

(2.8) Problems

1. Let A be the set of rational numbers between 0 and 1, and let nI be a finite

collection of open intervals covering A. Then 1.nl I

Here (0, 1) .A Let nI be a finite collection of open intervals

covering A. Since A is dense in (0, 1) , nI will be a cover for (0, 1) also. Now

*(0, 1)m inf ,nA Jn

l J

infimum taken over all open covering nJ of (0, 1) . Since nI is a cover for

(0, 1) , the above infimum will not be greater than nl I .

If 1,nl I the above forces us to have *(0, 1) 1,m which is a contradiction to

the fact that *(0, 1) 1,m by Proposition (2.3). Hence 1.nl I

2. Prove that *m is translation invariant.

Proof Note that if nx I is a cover of x + A, then nI is a cover of A and

.n nl x I l I Hence

*m x A inf infn nx A x I A In n

l x I l I

* .m A

3. Prove that if * 0m A , then * * .m A B m B

Proof Suppose * 0m A . Then, since B A B , we have

* * * * *m B m A B m A m B m B implies * * .m A B m B

3 Measurable Sets

(2.9) Definition A set E is said to be measurable if for each set A we have

* * * .m A m A E m A E

(2.10) Remarks

UNIT II CHAPTER 7

140

Since we always have * * * ,m A m A E m A E we see that E is measurable

if (and only if) for each A we have * * * .m A m A E m A E

Since the definition of measurability is symmetric in E and E , we have E

measurable whenever E is.

Clearly and are measurable.

(2.11) Lemma If * 0,m E then E is measurable.

Proof

Suppose * 0.m E Let A be any set. Then ,A E E and so * * 0.m A E m E Also

,A A E and so

* * * * ,m A m A E m A E m A E

and therefore, by the Remark above, E is measurable.

(2.12) Lemma If 1E and 2E are measurable, so is 1 2.E E

Proof

Let A be any set. Since 2E is measurable, we have

* * *1 1 2 1 2 ,m A E m A E E m A E E

and since 1 2 1 2 1[ ] [ ],A E E A E A E E [Ref. Figure] we have

* * *1 2 1 2 1( ) ( ).m A E E m A E m A E E

Thus

* * * * *1 2 1 2 1 2 1 1 2( ) ( ) ( ) ( )m A E E m A E E m A E m A E E m A E E

* * *1 1( ) ( ) ,m A E m A E m A

LEBESGUE MEASURE

141

by the measurability of 1.E Since 1 2 1 2~ ,E E E E this shows that 1 2E E is

measurable.

(2.13) Corollary to Lemma (2.12) The family Μ of measurable sets is an algebra of sets.

Proof

( i) 1E Μ and 2E Μ implies 1E and 2E are measurable and by Lemma (2.12)

1 2E E is measurable, i.e., 1 2 .E E Μ

( ii) By the second remark in (2.10), we have E is measurable implies E is measurable.

i.e., EΜ implies EΜ.

Hence, by Definition (1.9), Μ is an algebra.

(2.14) Corollary to (2.15) If 1E and 2E measurable, then 1 2~E E is measurable.

Proof 1E and 2E measurable implies 1E Μ and 2E Μ implies (since Μ an algebra of

sets) 1E Μ and 2E Μ implies 1 2E E Μ [Ref Remark (1.10)] implies 1 2~E E Μ ,

since 1 2 1 2~ .E E E E

(2.16) Lemma Let A be any set, and 1, , nE E a finite sequence of disjoint measurable

sets. Then

* *

11

( ).n n

i i

ii

m A E m A E

Proof We prove the Lemma by induction on n. It is clearly true for n = 1, and we assume it

is true if we have n – 1 sets iE . Since the iE are disjoint sets, we have

1

n

i n n

i

A E E A E

and 1

1 1

.n n

i n i

i i

A E E A E

Hence the measurability of nE implies that

1

* * *

1 1

n n

i n i

i i

m A E m A E m A E

1

* *

1

n

n i

i

m A E m A E

,

by our assumption of Lemma for n – 1 sets.

UNIT II CHAPTER 7

142

(2.17) Theorem The collection Μ of measurable sets is a -algebra; that is, the

complement of a measurable set is a measurable and the union (and intersection) of a

countable collection of measurable sets is measurable. Moreover, every set with outer

measure zero is measurable.

Proof We have already observed in Corollary (2.13) that Μ is an algebra of sets, and so we

have only to prove that if a set E is the union of a countable collection of measurable sets it

is measurable. By Proposition such an E must be union of a sequence nE of pairwise

disjoint measurable sets. Let A be any set, and let 1

.n

n i

i

F E

Then nF is measurable, and

.nF E Hence

* * * * * .n n nm A m A F m A F m A F m A E

By Lemma (2.16)

* *

1

( ).n

n i

i

m A F m A E

Thus

* * *

1

( ) ( ).n

i

i

m A m A E m A E

Since the left side of this inequality is independent of n, we have

* * *

1

( ) ( )i

i

m A m A E m A E

* *( ) ( )m A E m A E

by the countable subadditivitiy of *.m

(2.18) Lemma The interval ( , )a is measurable.

Proof

Take ( , )E a , then ( , ].E a For any set A, we have to show that

* * * .m A m A E m A E

i.e., to show that * * *( , ) ( , ] .m A m A a m A a

Let A be any set, 1 2( , ), ( , ].A A a A A a

Clearly, * * *1 2.m A m A m A

LEBESGUE MEASURE

143

It remains to show that

* * *1 2 .m A m A m A

If * ,m A then there is nothing to prove.

If * ,m A then, given 0, there is a countable collection nI of open intervals which

cover A and for which

*( ) .nl I m E

Let ( , ), ( , ].n n n nI I a I I a Then andn nI I are intervals (or empty) and

* *( ) ( ) ( ) .n n n n nl I l I l I m I m I

Since 1 ,nA I we have

* * *1 ,n nm A m I m I

and, since 2 ,nA I we have

* * *2 .n nm A m I m I

Thus

* * * *1 2 n nm A m A m I m I

*( ) .nl I m A

But was an arbitrary positive number, and so we must have * * *1 2 .m A m A m A


BOREL SET

The union of a any collection of open sets is open, the intersection of any finite

collection of open sets is open, but the intersection of a countable collection of open sets need

not be open. For example: For each n , the set 1 1

,a an n

is open, but

1

1 1, { }

n

a a an n

is not open.

Similarly, the intersection of any collection of closed sets is closed and the union of

any finite collection of closed sets is closed, but the union of a countable collection of closed

sets need not be closed. For example:

The set of rational numbers is the union of a countable collection of closed sets each of

which contains exactly one rational number. i.e.,

UNIT II CHAPTER 7

144

.a

a

Though each a is closed, is not closed (For, we can find a sequence in which

converges to 2 . )

Thus if we are interested in -algebras of sets that contain all of the closed

sets, we must consider more general types of sets than the open and closed sets. This leads us

to the following definition:

(2.19) Definition (Borel set) The collection B of Borel sets is the smallest -algebra

which contains all of the open sets.

Such a smallest -algebra exists by Proposition (1.14).

(2.20) Remark

The collection B of Borel sets is the smallest -algebra that contains all closed sets

and the smallest -algebra that contains the open intervals.

(2.21) Definition A set which is a countable union of closed sets is called an F ( F for

closed, for sum).

(2.22) Examples/Remarks

Every countable set is an F -set. In particular, every closed set is an F -set.

A countable union of sets in F is again in F .

Each open interval is in F , because any open interval ( , )a b can be written as

1

1 1( , ) ,

n

a b a bn n

.

Hence, also, each open set is in F .

(2.23) Definition A set is a G if it is the countable intersection of open sets ( G for open,

for durchschnitt).

(2.24) Examples/Remarks

The complement of an F is a G , and conversely.

The F and G are Borel sets.

LEBESGUE MEASURE

145

(2.25) Theorem Every Borel set is measurable. In particular,

each open set is measurable.

each closed set is measurable.

(UQ 2006)

Proof

( , )a is measurable by Lemma (2.18). Since Μ is a -algebra, the complement of ( , )a is

also measurable. i.e., ( , ] ~ ( , )a a is measurable. In particular, for each n the

intervals 1

, bn

are measurable and hence elements in Μ . Since Μ is a -algebra,

their countable union 1

1,

n

bn

is again in Μ . Since ( , )b can be written as

1

1( , ) , ,

n

b bn

we can say that ( , )b is in Μ . i.e., ( , )b measurable.

An open interval ( , )a b can be written as ( , ) ( , ) ( , )a b b a . By the

discussion above ( , )b and ( , )a are in Μ and since Μ is an algebra their

intersection is again in Μ . Hence ( , )a b is in Μ . i.e., each open interval ( , )a b is

measurable.

Each open set is the union of a countable number of open intervals. By the

discussion above each open interval is in Μ and since Μ is a -algebra, the

countable union of open intervals is in Μ . Hence each open set is measurable.

Each closed set is the complement of an open set. By the discussion above each open

set is in Μ and since Μ is a -algebra, the complement of open set must be in Μ .

Hence each closed set is measurable.

Thus Μ is a -algebra containing the open sets and must therefore contain the family

B of Borel sets, since B is the smallest -algebra containing the open sets [Ref. Definition

(2.19)]

Note: The theorem also follows immediately from the fact that Μ is a -

algebra containing each interval of the form ( , )a and the fact that B is the

smallest -algebra containing all such intervals.

UNIT II CHAPTER 7

146

4 Properties of Lebesgue Measure

(3.1) Definition If E is a measurable set, we define the Lebesgue measure mE to be the

outer measure of E. Thus m is the set function obtained by restricting the set function *m

to the family Μ of measurable sets.

(3.2) Remark

mE inf .nE In

l I

(3.3) Proposition Let iE be a sequence of measurable sets. Then

.i im E mE (1)

If the sets nE are pairwise disjoint, then

.i im E mE (2)

Property (1) is called the countable suadditivity.

Proof The inequality is simply a restatement of the subadditivity of *m given

By Proposition (2.4), we have

* *n nm A m A (3)

for a countable collection nA of sets of real numbers. Here iE be a sequence of

measurable sets and by Definition (3.1) m is the restriction of the set function *m to the

family Μ of measurable sets; hence (3) becomes

.i im E mE

It remains to show that if the sets nE are pairwise disjoint, then

.i im E mE

Case 1) If iE is a finite sequence of disjoint measurable sets, then Lemma (2.16) with

A implies that

,i im E mE

and so Μ is finitely additive.

Case 2) Let iE be an infinite sequence of pairwise disjoint measurable sets. Then

1 1

,n

i i

i i

E E

and so

LEBESGUE MEASURE

147

1 1

n

i i

i i

m E m E

1

,n

i

i

mE

by Case 1 above.

Since the left side of this inequality is independent of n, we have

11

.i i

ii

m E mE

i.e., 1 1

.i i

i i

mE m E

(4)

The reverse inequality follows from countable subadditivity (1), and hence from (1) and (4)

we have

11

.n

i i

ii

m E mE

(3.4) Proposition Let nE be an infinite decreasing sequence of measurable sets, that is,

a sequence with 1n nE E for each n. Let 1mE be finite. Then

1

lim .i nn

i

m E mE

Proof Let 1

,i

i

E E

and let 1\ .i i iF E E Then

1

1

~ ,i

i

E E F

and the sets iF are pairwise disjoint. Hence

1

1

~ i

i

m E E m F

1

,i

i

mF

as iF are pairwise disjoint and by (2) of Prop(3.3)

1

1

~ .i i

i

m E E

Since 1

,i

i

E E

1E E and hence we can write 1 1 ~E E E E where E and

1 ~E E are pairwise disjoint. Hence

UNIT II CHAPTER 7

148

1 1 ~ .mE mE m E E

Similarly, since 1 ,i iE E we can write 1 1~i i i iE E E E where 1iE and 1~i iE E are

pairwise disjoint and hence

1 1~ .i i i imE mE m E E

Since 1 ,imE mE we have

1 1~m E E mE mE

and 1 1~ .i i i im E E mE mE

Thus 1 1

1

i i

i

mE mE mE mE

1

1

limn

i in

i

mE mE

1lim nn

mE mE

1 lim .nn

mE mE

Since 1 ,mE we have

lim .nn

mE mE

(3.5) Definition Let ,A B two sets. Then the symmetric difference of A and B is the

set given by ~ ~ .A B A B B A

(3.6) Proposition (First Principle of Littlewood) Let E be a given set. Then the

following five statements are equivalent:

i. E is measurable.

ii. Given 0, there is an open set O E with * ~ .m O E

iii. Given 0, there is a closed set F E with * ~ .m E F

iv. There is a G in G with ,E G * ~ 0.m G E

v. There is a F in F with ,F E * ~ 0.m E F

If *m E is finite, the above statements are equivalent to:

vi. Given 0, there is a finite union U of open intervals such that * .m U E

LEBESGUE MEASURE

149

Proof

(i) implies (ii): Let E be a measurable set of finite measure. Since *m A inf ,nA In

l I

for 0 , *m E cannot be a lower bound of the set :n nl I A I and hence there

exists a countable collection nI of open intervals such that * .nl I m E

Let 1

.n

n

O I

Then * * .nm O l I m E

Since \O E O E a disjoint union, we have * * * \m O m E m O E . Thus

* * *\ .m O E m O m E

Now we assume that E is a measurable set of infinite measure. For each n , let

( , ).nE E n n Then each nE has finite measure and by what proved above there is an

open set, say nO , n nO E such that * \ .2

n n nm O E

Let

1

.n

n

O O

Then O is open,

O E and since 1

~ ~n n

n

O E O E

we get

* * *

11

~ ~ ~ .2

n n n n nnn

m O E m O E m O E

Hence (i) implies (ii) is proved.

(ii) implies (iii): E is measurable. Hence its complement E is measurable. Then by (ii),

there is an open set O such that O E with * ~ .m O E Take .F E Then F E

and ~ ~ .O E E F Hence * ~ .m E F i.e., (ii) implies (iii) is proved.

Similarly, taking complements, (iii) implies (ii) is proved.

(ii) implies (iv): By (ii) for each n we can find an open set nO E such that

* 1~ .nm O E

n Let

1

.n

n

G O

Then G is a G set, E G and

* * 1~ ~ .nm G E m O E

n

This is true for each n. So * ~ 0.m G E

(iii) implies (v) and (v) implies (iii) similarly.

UNIT II CHAPTER 7

150

(i) implies (vi): Suppose E is measurable. Then by (ii), given 0, there is an open set

O E with * ~ .2

m O E

The open set O can be considered as the disjoint union of

open intervals , 1,2,iI i i.e., 1

.ii

O I

* * *

1 11

.i i i

i ii

m O m I m I l I

Since

*m O is finite, given 0, there exists n such that 1

.2

i

i n

l I

We write 1

.ii

U I

Then ~ ~E U U E E U

1 1

~ ~i i

i i

I E E I

1

~ ~ .i

i

O E O I

Hence * * *

1

~ ~ i

i

m E U m O E m O I

1

* *

1

~n

i

i

m O E m I

*

1

~ .i

i

m O E l I

(3.7) Problem Show that if 1E and 2E are

measurable, then

1 2 1 2 1 2.m E E m E E mE mE

Solution

1 2 1 2 1 2 2 1~ ~E E E E E E E E is a

pairwise disjoint union.

1 1 2 1 2~E E E E E is a pairwise disjoint union.

2 1 2 2 1~E E E E E is a pairwise disjoint union.

Hence

1 2 1 2 1 2 2 1~ ~m E E m E E m E E m E E (1)

1 1 2 1 2~mE m E E m E E (2)

2 1 2 2 1~mE m E E m E E (3)

LEBESGUE MEASURE

151

Adding (2) and (3), we have

1 2 1 2 1 2 1 2 2 1~ ~mE mE m E E m E E m E E m E E

implies

1 2 1 2 1 2 ,mE mE m E E m E E using (1).

Next example says that there is an uncountable set having measure 0.

(3.8) Example The cantor ternary set has measure zero. (UQ 2006)

Proof

We first describe Cantor set.

Let 0 [0, 1].E Remove the segment 1 23 3

, , and let 1 21 3 3

[0, ] [ , 1].E Remove the

middle thirds of these intervals, and let

3 6 7 81 22 9 9 9 9 9 9

[0, ] [ , ] [ , ] [ , 1].E

Continuing in this way, we obtain a sequence of compact sets ,nE such that

(a) 1 2 3 ;E E E

(b) nE is the union of 2n intervals, each of length 3 .n

The set

1

nn

P E

is called the Cantor set. P is clearly compact, and by the Theorem “If K is a collection of

compact subsets of a metric space X such that the intersection of every finite subcollection of

K is nonempty, then K

is nonempty” P is nonempty.

Now we come to our problem:

Clearly * 1 22 1, 2, 3, ;

3 3

n nn

nm E n

Since 1

nn

P E

, we have

,nP E n

so that * 2lim 0.

3

n

nm P

i.e., the measure of cantor ternary set is zero.

UNIT II CHAPTER 7

152

5 Existence of Nonmeasurable set

In this section we are going to show the existence of a nonmeasurable set.

(4.1) Definition If x and y are real numbers in [0, 1), we define the sum modulo 1 of x

and y , denoted by x y , as follows:

, if 1

1, if 1

x y x yx y

x y x y

is a commutative and associative binary operation taking pairs of numbers in [0, 1)

into numbers in [0, 1).

(4.2) Lemma Let [0, 1)E be a measurable set. Then for each [0, 1)y the set E y is

measurable and .m E y mE

Proof

Take [0, 1)y . Let 1 [0, 1 )E E y and 2 [1 , 1).E E y

Then 1E and 2E are disjoint measurable sets whose union is E, and so

1 2.mE mE mE

Now 1 1E y E y , and so 1 1 1,m E y m E y mE

since m is translation invariant,

and so 1E y is measurable. Also 2 1 ( 1)E y E y and so

2 2 2( 1) ,m E y m E y mE

and so 2E y is also measurable.

But 1 2E y E y E y

and the sets 1E y and 2E y are disjoint measurable sets,

so by Lemma (2.12), E y is measurable. Also,

1 2m E y m E y m E y

, since 1E y and 2E y are disjoint

1 2mE mE

.mE

LEBESGUE MEASURE

153


(4.3) Construction of a nonmeasuarble set

Step 1: We define a relation on [0, 1) as follows:

x ~ y if x y is a rational number. It is easy to see that ~ is an

equivalence relation. Hence ~ partitions [0, 1) into equivalence classes, i.e., classes such that

any two elements of one class differ by a rational number, while any two elements of different

classes differ by an irrational number (Ref. Figure). By the axiom of choice1 there is a set P

which contains exactly one element from each equivalence class.

In the above figure, x ~ y and ( x and y are in the same classe) x y is a

rational number; ~x z and ( x and z are in different classes) x z is an

irrational number; z ~ w and z w is a rational number; ~z p and z p is

an irrational number.

Step 2: Let 0i i

r

be an enumeration of the rational numbers in [0, 1) with 0 0,r and

define .i iP P r Then 0 .P P

Claim: i jP P if .i j

Let .i jx P P Then i i j jx p r p r with ip and jp belonging to P. But

i j j ip p r r is a rational number, whence ~i jp p , i.e., ip and jp are in the same class.

Since P has only one element from each equivalence class, we must have .i j This

implies that i jP P if .i j

UNIT II CHAPTER 7

154

i.e., 0i i

P

is a pair wise disjoint sequences of sets. On the other hand, each real number x

in [0, 1) is in some equivalence class and so is equivalent to an element in P. But if x

differs from an element in P by the rational number ir , then .ix P Thus [0, 1).iP

Since each iP is a translation modulo 1 of P, each iP will be measurable if P is measuarble

and will have the same measure i.e., imP mP for all i. But if this were the case,

1 1

[0, 1) ,i

i i

m mP mP

(**)

and the right side is either zero or infinite, depending on whether mP is zero or positive. But

this is impossible since [0, 1)m 1, and consequently P cannot be measurable. i.e., we have

constructed a nonmeasurable set P. We arrive at the following result:

(4.4) Theorem If m is a countably additive, translation invariant measure defined on a

-algebra containing the set P seen above, then [0, 1)m is either zero or infinite.

Proof If m is a countably additive, translation invariant measure defined on a -algebra

containing the set P means measure of P is zero or positive. Now theorem follows from

(**) in the construction (4.3).

1Axiom of choice: Let be any collection of nonempty sets. Then there is a function F

defined on which assigns to each set A an element F (A) in A.

The above equivalent to say that we have a set, say P, in which contains exactly

one element from each set A .

6 Measurable Functions

(5.1) Proposition Let f be an extended real-valued function whose domain is

measurable. Then the following statements are equivalent:

i. For each real number the set : ( )x f x is measurable.

ii. For each real number the set : ( )x f x is measurable.

iii. For each real number the set : ( )x f x is measurable.

iv. For each real number the set : ( )x f x is measurable.

LEBESGUE MEASURE

155

These statements imply

v. For each extended real number the set : ( )x f x is measurable.

Proof

Let the domain of f be D. Given that D is measurable.

(i) implies (iv): Note that : ( ) : ( )x f x D x f x .

D and : ( )x f x are measurable implies the difference : ( )D x f x is also

measurable. That is, : ( )x f x is measurable. That is, (i) implies (iv).

(iv) implies (i): Note that : ( ) : ( )x f x D x f x .

D and : ( )x f x are measurable implies the difference : ( )D x f x is also

measurable. That is, : ( )x f x is measurable. That is, (iv) implies (i).

As above, (ii) implies (iii) and (iii) implies (ii) can be proved.

(i) implies (ii): Note that 1

1: ( ) : ( )

n

x f x x f xn

.

(i) implies that 1

: ( )x f xn

is measurable for any natural number n and, since

intersection of a sequence of measurable sets is measurable, we have 1

1: ( )

n

x f xn

is

measurable. This implies that : ( )x f x is measurable. That is, (i) implies (ii) is proved.

(ii) implies (i): Note that 1

1: ( ) : ( )

n

x f x x f xn

.

(ii) implies that 1

: ( )x f xn

is measurable for any natural number n and, since union

of a sequence of measurable sets is measurable, we have 1

1: ( )

n

x f xn

is measurable.

This implies that : ( )x f x is measurable. That is, (ii) implies (i) is proved.

That we have shown that the first four statements are equivalent.

Next we have to show that first four statements imply the fifth statement:

Case 1) If is a real number:

: ( ) : ( ) : ( ) ,x f x x f x x f x

UNIT II CHAPTER 7

156

and so (ii) and (iv) implies (v)

Case 2) If :

1

: ( ) : ( ) ,n

x f x x f x n

and so (ii) implies (v).

Case 3) If :

1

: ( ) : ( ) ,n

x f x x f x n

and so (iv) implies (v).

We have seen that, in any case (ii) and (iv) implies (v).

(5.2) Definition An extended real valued function f is said to be (Lebesgue) measurable

if its domain is measurable and if it satisfies one of the first four statements of Proposition

(5.1).

(5.3) Example Prove that 2

2, 0( )

, 0

xf x

x x

is measurable. (UQ 2006)

Solution

Case 1) 0:

: ( ) 0x f x is measurable.

Case 2) 0 2:

: ( ) [0, ]x f x is measurable.

Case 3) 2:

: ( ) ( , ]x f x is measurable.

2

2y x

0

Graph of f

LEBESGUE MEASURE

157

i.e., for any , : ( )x f x is measurable. Hence f is measurable.

The following results (6.4) to (6.6) which are needed in the coming propositions are

already known to the student and their proofs are omitted here.

(5.4) Proposition Every ordered field contains (sets isomorphic to) the natural numbers,

the integers, and the rational numbers.

(5.5) Axiom of Archimedes Given any real number x, there is an integer n such that x

.x n

(5.6) Corollary to Axiom of Archimedes Between any two real numbers is a rational;

that is, if ,x y then there is a rational r with .x r y

(5.7) Proposition Let c be a constant and f and g be two measurable real valued

functions defined on the same domain. Then the functions , , , ,f c cf f g g f and fg

are measurable.

Proof

(i) f and g are measurable implies their domain is measurable and it satisfies one of the first

four statements of Proposition (5.1). We use condition (iii) of Proposition (5.1). Note that

: ( ) : ( ) : ( )x f c x x f x c x f x c .

Since f is measurable, using condition (iii) of Proposition (5.1), we have : ( )x f x c is

measurable. Hence : ( )x f c x is measurable. Hence the function f c is

measurable.

(ii)

Case 1: For c = 0, : ( )x cf x for 0 and : ( )x cf x for 0.

In either case : ( )x cf x is measurable since we know that and are measurable

sets.

Case 2: For positive c

: ( ) : ( ) : ( )x cf x x cf x x f xc

.

UNIT II CHAPTER 7

158

Since f is measurable, using condition (iii) of Proposition (5.1), we have : ( )x f xc

is

measurable. Hence : ( )x cf x is measurable. Hence the function cf is measurable,

for positive c.

Case 3: for negative c

: ( ) : ( ) : ( )x cf x x cf x x f xc

.

Since f is measurable, using condition (i) of Proposition (5.1), we have : ( )x f xc

is

measurable. Hence : ( )x cf x is measurable. Hence the function cf is measurable,

for negative c.

(iii) If ( ) ( ) ,f x g x then ( ) ( )f x g x and by the Corollary (5.6) to the axiom of

Archimedes there is a rational number r such that

( ) ( ).f x r g x

Hence

: ( ) ( ) : ( ) : ( ) .x f x g x x f x r x g x r

Since the rational are countable, this set if measurable and hence f g is measurable. Since

1g g is measurable when g is, we have f g is measurable.

(iv) The function 2f is measurable, since

Case 1) For 0 ; 2: ( ) : ( ) : ( )x f x x f x x f x is

measurable (being the union of two measurable sets)

and Case 2) For 0, 2: ( )x f x D is measurable, since D the domain

of f is measurable.

Also note that

2 2 21

2fg f g f g

(1)

In (1), RHS is measurable, as f g is measurable by (i) above and hence by the discussion

just above 2 2 2, ,f g f g are measurable and so is

2 2 21.

2f g f g

Hence fg is

measurable.

LEBESGUE MEASURE

159

(5.8) Basic Definitions 1

Let nx is a sequence of real numbers.

The limit superior is defined by

lim inf sup inf sup : : .n k kn k n

x x x k n n

The symbols lim and lim sup are both used for the limit superior.

The limit inferior is defined by

lim supinf sup inf : : .n k kk nn

x x x k n n

(5.9) Remark A real number l is the limit superior of the sequence nx if and only if

( i) given 0 , there exists n such that ,kx l k n and

( ii) given 0 and given n, there exists k n such that .kl x

(5.10) Remarks

The extended real number is the limit superior of the sequence nx if and

only if given and n there exists k n such that .kx

The extended real number is the limit superior of the sequence nx if and

only if lim .nx

lim lim .n nx x

If nx is a convergent sequence, then lim lim lim .n n nx x x

(5.11) Basic Definitions 2

Let nf be a sequence of functions.

The function 1 2sup , , , nh f f f is defined by

1 2( ) sup ( ), ( ), , ( ) .nh x f x f x f x

The function 1 2inf , , , ng f f f is defined by

1 2( ) inf ( ), ( ), , ( ) .ng x f x f x f x

The function 1 2sup sup , , , ,n nn

f f f f is defined by

1 2sup ( ) sup ( ), ( ), , ( ),n nn

f x f x f x f x

UNIT II CHAPTER 7

160

The function 1 2inf inf , , , ,n nn

f f f f is defined by

1 2inf ( ) inf ( ), ( ), , ( ),n nn

f x f x f x f x

lim nf is the limit superior function 1lim inf sup , , ,n k kn k n

f f f

defined by

1 1lim ( ) inf sup ( ), ( ), , inf sup ( ), ( ), ,: :n k k k kn k n

f x f x f x f x f x k n n

lim is also denoted by lim sup.

lim nf is the limit inferior function 1lim supinf , , ,n k kk nn

f f f

defined by

1 1lim ( ) supinf ( ), ( ), , sup inf ( ), ( ), ,: :n k k k kk nn

f x f x f x f x f x k n n

lim is also denoted by lim inf.

(5.12) Theorem Let nf be a sequence of measurable functions (with the same domain

of definition). Then the functions 1sup , , nf f , 1inf , , nf f , sup ,nn

f , inf ,nn

f ,

lim nf , and lim nf are all measurable.

Proof

(i) If h is defined by 1( ) sup ( ), , ( ) ,nh x f x f x then

1

: ( ) : ( ) .n

i

i

x h x x f x

(2)

As 1, , nf f are measurable, : ( )ix f x is measurable for 1, , .i n Being the union

of measurable sets, 1

: ( )n

i

i

x f x

is measurable and hence : ( )x h x is measurable

and so h is measurable.

(ii) 1inf , , nf f is measurable can be proved similarly.

(iii) If g is defined by 1( ) sup ( ), , ( ), ,ng x f x f x then

1

: ( ) : ( ) .n

n

x g x x f x

(2)

LEBESGUE MEASURE

161

As 1, , ,nf f are measurable, : ( )nx f x is measurable for 1, 2, .n Being the

countable union of measurable sets, 1

: ( )n

n

x f x

1

: ( )n

i

i

x f x

is measurable and

hence : ( )x g x is measurable and so g is measurable.

(iv) 1 2inf inf , , , ,n nn

f f f f is measurable can be proved similarly.

(v) By the discussion above, inf function and sup function are measurable. Hence

1lim supinf , , ,n k kk nn

f f f

is measurable.

(vi) Similarly, 1lim inf sup , , ,n k kn k n

f f f

is measurable.

(5.13) Definition A property is said to hold almost everywhere (abbreviated a.e.) if the

set of points where it fails to hold is a set of measure zero.

We say that f g a.e. if f and g have the same domain and

: ( ) ( ) 0.m x f x g x (i.e., the measure of the set : ( ) ( )x f x g x is 0.

We say that nf converges to g a.e. if there is a set E of measure zero such that

( )nf x converges to ( )g x for each x not in E.

(5.14) Proposition If f is measurable function and f g a.e., then g is measurable.

Proof

Let : ( ) ( ) .E x f x g x Since f g a.e., we have 0.mE

Now

: ( ) : ( ) : ( ) \ : ( ) .x g x x f x x E g x x E g x

The first set of the right is measurable, since f is a measurable function. The last two sets

on the right are measurable, since they are subsets of E and 0.mE Thus : ( )x g x is

measurable for each , and so g is measurable.

(5.15) Proposition Let f be a measurable function defined on an interval [ , ],a b and

assume that f takes the values only on a set of measure zero. Then given 0, we can

find a step function g and a continuous function h such that

f g and f h

UNIT II CHAPTER 7

162

except on a set of measure less than ; i.e., : ( ) ( )m x f x g x and

: ( ) ( ) .m x f x h x If in addition ,m f M then we may choose the functions g

and h so that m g M and .m h M

The above proposition tells us that a measurable function is “almost” a

continuous function.

(5.16) Definition If A is any set, we define the characteristic function A of the set A

to be the function given by

1 if

( )0 if

A

x Ax

x A

The function A is measurable if and only if A is measurable.

Thus the existence of a nonmeasurable set implies the existence of a nonmeasuarble function.

(5.17) Definition A real-valued function is called simple if it is measurable and

assumes only a finite number of values.

If is simple and has the values 1, , n then

1

,n

i Aii

where : ( )i iA x x . The sum, product, and difference of two simple functions are

simple.

7 Littlewood’s Three Principles

1. Every (measurable) set is nearly a finite union of intervals. Various forms of the first

principle is given by Proposition (3.6)

2. Every (measurable) function is nearly continuous. One version of the second principle

is given by Proposition (5.15).

3. Every convergent sequence of (measurable) functions is nearly uniformly convergent.

One version of the third principle is given by the following Proposition.

LEBESGUE MEASURE

163

(6.1) Proposition Let E be a measurable set of finite measure, and nf a sequence of

measurable functions defined on E. Let f be a real-valued function such that for each x in

E we have ( ) ( ).nf x f x Then given 0 and 0, there is a measurable set A E with

mA and an integer N such that for all x A and all ,n N

( ) ( ) .nf x f x

Proof

Let

: ( ) ( ) ,n nG x E f x f x

and set

: ( ) ( ) for some .n n n

n N

E G x E f x f x n N

We have 1 ,N NE E and for each x E there must be some NE to which x does not

belong, since ( ) ( ).nf x f x Thus ,NE and so, by Proposition (3.4), lim 0.NmE

Hence given 0, there is N such that ;NmE that is,

: ( ) ( ) for some .nm x E f x f x n N

If we write A for this NE , then mA and

: ( ) ( ) for all .nA x E f x f x n N

(6.2) Remark If, as in the hypothesis of the proposition, we have ( ) ( )nf x f x for each

x, we say that the sequence nf converges pointwise to f on E. If there is a subset B of E

with mB = 0 such that nf f pointwise on \ ,E B we say that nf f a.e. on E. The

following is the trivial modification of the last proposition:

(6.3) Proposition [Trivial modification of Proposition (6.1)] Let E be a measurable set of

finite measure, and nf a sequence of measurable functions that converge to a real-valued

function f a.e. on E. Then, given 0 and 0, there is a set A E with ,mA and

an N such that for all x A and all ,n N

( ) ( ) .nf x f x

___________________

Chapter 8

LEBESGUE INTEGRAL

The Riemann Integral Equation Chapter (Next) Section 8

(8.1) The Riemann Integral

In this section we review Riemann integral using the notations that are used in this chapter.

Let f be a bounded real-valued function defined on the interval [ , ]a b and let

0 1 na b

be a subdivision (some times called partition) of [ , ]a b . Then for each subdivision we can

define the sums

1

1

( )n

i i i

i

S M

and 1

1

( ) ,n

i i i

i

s m

where 1

sup ( ),ixi i

M f x

1

inf ( ).ixi i

m f x

We define the upper Riemann integral of f by

( ) inf

b

a

f x dx S

with the infimum taken over all possible subdivisions (partitions) of [ , ]a b . Similarly, we

define the lower Riemann integral of f by

_

( ) sup .

b

a

f x dx s

The upper integral is always at least as large as the lower integral.

If upper Riemann integral and lower Riemann integral of f are equal, then we say that f is

Riemann integrable and the common value is the Riemann integral of f. The Riemann

integral of f is denoted by

( )

b

a

f x dx

to distinguish it from the Lebesgue integral, which we consider in this chapter.

LEBESGUE INTEGRAL

165

The upper and lower Riemann Integrals in terms of step functions

(8.2) Definition By a step function we mean a function which has the form

1( ) ,i i ix c x

for some subdivision of [ , ]a b and some set of constant .ic Then the integral of the step

function is given by

1

1

( ) .

b n

i i i

ia

x dx c

(8.3) With the above in mind we see that the upper Riemann integral of f in terms of the

step functions is

( ) inf ( )

b b

a a

f x dx x dx (8.3)

for all step functions ( ) ( ).x f x

(8.4) The lower Riemann integral of f in terms of the step functions is

_

( ) sup ( )

b b

a a

f x dx x dx (8.4)

for all step functions ( ) ( ).x f x

(8.5) Problem Show that if

0 irrational

( )0 rational ,

xf x

x

then ( )

b

a

f x dx b a and

_

( ) 0.

b

a

f x dx

[Hint: For the step functions 1,

( )0, elsewhere

a x bx

and ( ) 0x x , we have

( ) ( ) ( ).x f x x ]

UNIT III CHAPTER 8

166

The Lebesgue Integral of a Bounded Function

over a Set of Finite Measure

(8.6) Definition The function E defined by

1( )

0E

x Ex

x E

is called the characteristic function of E.

(8.7) Definition A linear combination

1

( ) ( )n

i Eii

x a x

(1)

is called a simple function if the sets iE are measurable. This representation for is not

unique. However, we note that a function is simple if and only if it is measurable and

assumes only a finite number of values.

(8.8) Definition If is a simple function and 1, , na a the set of nonzero values of ,

then

1

,n

i Aii

a

where : ( ) .i iA x x a

This representation for is called the canonical representation, and it is characterized by

the fact that the iA are disjoint and the ia distinct and nonzero.

(8.9) Definition (Integral of a simple function) If is a simple function that vanishes

outside a set of finite measure, we define the integral of by

1

( )n

i i

i

x dx a mA

when has the canonical representation 1

.n

i Aii

a

(8.10) Remark Read and understand the underlined words because now the integral of the

simple function is given by means of its canonical representation (We will see in Remark

(8.13) that this is not necessary). Now, by the Definition (8.9), the integral of the simple

function cannot be determined immediately using (1) of Definition (8.7); to evaluate the

LEBESGUE INTEGRAL

167

integral we have first to write in some canonical representation as in Definition (8.8). This

will be clear to the student from the proof of the coming Lemma (8.11).

We sometimes abbreviate the expression for this integral to . If E is any

measurable set, we define

.E

E

(8.11) Lemma Let 1

,n

i Eii

a

with i jE E for i j . Suppose each set iE is a

measurable set of finite measure. Then

1

.n

i i

i

a mE

(UQ 2006)

Proof

To determine using Definition (8.9) we have first to find the canonical representation of

[Read Remark (8.10)]

To determine the canonical representation, we consider the sets

: ( ) .a i

a ai

A x x a E

[Ref. the following figure]

UNIT III CHAPTER 8

168

The Venn Diagram is drawn as for 3x E , 3

1

( ) ( ) ,n

i Eii

x a x a

[since

1 3( )

0 3Ei

ix

i

] and assuming that 3a a . Similarly, assuming 7a a and 9a a ,

we have ( )x a for 7 9,x E E ; Similarly, ( )x b for 1 2 12, ,x E E E ; ( )x c for

4 6, .x E E

Then the canonical representation of is

A aa .

Hence by Definition (8.9),

( ) ax dx amA (2)

Since iE is pairwise disjoint, by the additivity of m, we have

a i i

a ai

amA a mE

, (3)

Hence from (2), we have

( ) .i ix dx a mE

(8.12) Proposition Let and be simple functions which vanish outside a set of finite

measure. Then

(a) ,a b a b

and

(b) if a.e., then

.

Proof

Let 1

,n

i Aii

a

where : ( )i iA x x a and 1

,m

i Bii

b

where : ( )i iB x x b be the

canonical representations of and , respectively. Then iA and iB be the sets

occurring in canonical representations of and .

Let 0A and 0B be the sets where and are zero.

LEBESGUE INTEGRAL

169

We consider the intersections i jA B . These intersections form a finite disjoint

collection of measurable sets and we denote the collection by 1

n

k kE

.

Then

1

N

k Ekk

a

, with ,i jE E for .i j (4)

1

,N

k Ekk

b

with ,i jE E for .i j (5)

(4) and (5) need not be the canonical representations of and . But we are now in a

position to use Lemma (8.11), since ,i jE E for .i j

i.e., by Lemma (8.11), we have

1

N

k k

k

a mE

and 1

N

k k

k

b mE

.

Hence

1 1 1

N N N

k k k k k k k k

k k k

a b a a mE b b mE aa mE bb mE

(6)

Also, from (4) and (5)

1

N

k k Ekk

a b aa bb

,

UNIT III CHAPTER 8

170

whence, again by Lemma (8.11) , it follows that

1

.N

k k k

k

a b aa bb mE

(7)

R.H.S. of (6) and (7) are the same. Hence,

.a b a b

Proof of the second statement:

a.e. implies 0 a.e. implies is a non-

negative simple function. Then if the canonical representation of is 1

,n

i Cii

c

where : ( ) ,i iC x x c then is a non-negative implies 0.ic

Also, by Definition (8.9),

1

n

i i

i

c mC

. (8)

Hence by noting that 0ic and m is a nonnegative function the above implies

0. (9)

Taking 1, 1,a b part (a) implies

.

Hence by (9), we have

0.

i.e., .

(8.13) Remark Putting 1, 0a b in (7) of the proof above, it follows that, if

1

,n

i Eii

a

then 1

,n

i i

i

a mE

and so the restriction of Lemma (8.11) that the sets iE be

disjoint is unnecessary.

Let f be a bounded real valued function and E a measurable set of finite measure.

By analogy with the Riemann integral we consider for simple functions and the

numbers

inf :

E

f

(usually denoted by inf

fE

)

LEBESGUE INTEGRAL

171

and sup : ,

E

f

(usually denoted by sup

f E

)

and ask when these two numbers are equal. The answer is given by the following proposition:

(8.14) Proposition Let f be defined and bounded on a measurable set E with mE

finite. In order that

inf ( )f

E

x dx sup ( )

f E

x dx

for all simple functions and , it is necessary and sufficient that f be measurable.

Proof Let f be bounded by M and suppose that f is measurable.

f be bounded by M implies ( )f x M implies ( )M f x M x

or ( ) .M f x M x

Then we consider the sets

( 1)

: ( ) ,k

kM k ME x f x n k n

n n

that are measurable, disjoint, and have union E.

Some particular sets kE :

For k n , ( 1)

: ( )n

E x M f x Mnn

;For k n , ( 1)

: ( )n

E x M f x Mnn

Since kE are disjoint and have union E, we have

.n

k

k n

mE mE

The simple functions defined by

( )( )n

n E xkk n

Mx k

n

and ( )( ) ( 1)n

n E xkk n

Mx k

n

satisfy ( ) ( ) ( ).n nx f x x

Thus

* *

inf ( ) ( )n

n kf

k nE E

Mx dx x dx kmE

n

UNIT III CHAPTER 8

172

The reason for the inequality “*

”: inf ( ) inf ( ) : , is simplef

E E

x dx x dx f

and

( ) ( ) : , is simple .n

E E

x dx x dx f

The equality “ *

” follows from Remark (8.13)

.

and

sup ( ) ( ) ( 1) ,n

n kf k nE E

Mx dx x dx k mE

n

whence

0 inf ( ) sup ( ) .n

kf f k nE E

M Mx dx x dx mE mE

n n

Since n is arbitrary, we have

0 inf ( ) sup ( ) 0.f fE E

x dx x dx

Hence

inf ( ) sup ( ) 0f fE E

x dx x dx

implies

inf ( ) sup ( )f fE E

x dx x dx

implies the condition is sufficient.

Suppose now that n and

inf ( ) sup ( ) .f fE E

x dx x dx l

Then, given n, 1

ln

cannot be a lower bound implies there is a simple function n such that

( ) ( )n x f x and 1

( ) .2

n

E

x dx ln

Similarly, 1

ln

cannot be an upper bound and this

implies there is a simple function n such that ( ) ( )nf x x and 1

( ) .2

n

E

x dx ln

Now

1( )

2n

E

l x dxn

and 1

( )2

n

E

l x dxn

implies 1

( ) ( ) .n n

E E

x dx x dxn

i.e., for given n, there are simple functions n and n such that

LEBESGUE INTEGRAL

173

( ) ( ) ( )n nx f x x

and

1

( ) ( ) .n n

E E

x dx x dxn

Note that being the simple functions, n ‟s and n 's are measurable. Hence by Theorem

(5.12) of the previous chapter, we have the functions

* inf nn

and * sup nn

are measurable, and

* *( ) ( ) ( ).x f x x (10)

Now the set

* *: ( ) ( )x x x

is the union of the sets

* * 1: ( ) ( ) .v x x x

v

But each v is contained in the set 1

: ( ) ( ) ,n nx x xv

and this latter set has measure

less than .v

n Since n is arbitrary, 0,vm and so 0.m Thus * * except on a set of

measure zero, and hence by (10) * *f except on a set of measure zero. Thus f is

measurable by Proposition (5.14) of the previous chapter, and the condition is also necessary.

(8.15) Definition (Integral of a Bounded Measurable Function)If f be a bounded

measurable function defined on a measurable set E with mE finite, we define the

(Lebesgue) integral of f over E by

( ) inf ( )f

E E

f x dx x dx

for all simple functions f .

i.e., ( ) inf ( ) : issimplefunction and .

E E

f x dx x dx f

UNIT III CHAPTER 8

174

(8.16) Notation We sometimes write the integral as .

E

f If [ , ],E a b we write

b

a

f

instead of

[ , ]

.

a b

f If f is a bounded measurable function that vanishes outside a set E of

finite measure, we write f for .

E

f Note that .E

E E

f f

(8.17) Proposition (Corollary of Proposition (8.14))

Let f be a bounded function defined on [ , ].a b If f is a Riemann integral on [ , ],a b then it

is measurable and

( ) ( ) .

b b

a a

f x dx f x dx

i.e., Lebesgue integral is in fact a generalization of a the Riemann integral.

Proof

By (8.4),

_

( ) sup ( )

b b

a a

f x dx x dx

for all step functions ( ) ( ).x f x Since every step function is also a simple function, we

have for all simple functions 1 f

1

1

sup ( ) sup ( )

b b

f fa a

x dx x dx

.

Hence for all simple functions 1 f

1

1_

( ) sup ( )

b b

fa a

f x dx x dx

(8.5)

Similarly, considering (8.3), for all simple functions 1 f

11

inf ( ) ( ) .

bb

fa a

x dx f x dx

(8.6)

But, 1 111

sup ( ) inf ( )

b b

ff a a

x dx x dx

(8.7)

Combining (8.5) to (8.7), we obtain

LEBESGUE INTEGRAL

175

1 111_

( ) sup ( ) inf ( ) ( ) .

bb b b

ffa a a a

f x dx x dx x dx f x dx

(8.8)

If f is a Riemann integral on [ , ],a b then

_

( ) ( ) ,

bb

a a

f x dx f x dx so from (8.8), we

obtain 1 111

sup ( ) inf ( )

b b

ff a a

x dx x dx

and hence by Proposition (8.14) f is measurable. Also, using the Definition (8.15), we

have

( ) ( ) .

b b

a a

f x dx f x dx

(8.18) Proposition If f and g are bounded measurable functions defined on a set E of

finite measure, then:

i. .

E E E

af bg a f b g

ii. If f g a.e., then

.

E E

f g

iii. (a) If f g a.e., then

.

E E

f g

(b) Hence .f f

iv. If ( ) ,A f x B then

.

E

AmE f BmE

v. If A and B are disjoint measurable sets of finite measure, then

.

A B A B

f f f

UNIT III CHAPTER 8

176

Proof

(i) Note that if is a simple function so is ,a and conversely if 0,a then a is simple.

For 0a there is nothing to prove.

For 0,a

inf ( )a a f

E E

af a x dx

, by Definition (8.15)

inf ( ) inf ( ) .f f

E E E

a x dx a x dx a f

If 0,a

inf ( )a a f

E E

af a x dx

, using Definition (8.15)

inf ( )f

E

a x dx

, since 0a

sup ( )f E

a x dx

, since 0a

inf ( )f

E

a x dx

, since is measurable and using Proposition (8.14)

E

a f , by Definition (8.15) (1)

[In the above and are used to indicate the change in the inequality. and are

used to indicate the change of inf and sup].

If 1 is a simple function 1f , and 2 a simple function, 2 ,g then 1 2 is

a simple function 1 2.f g Hence

1 2 ,

E E

f g (2)

since by definition inff g

E E

f g

.

Also, by (a) of Proposition (8.12),

1 2 1 2

E E E

.

Hence (2) becomes

LEBESGUE INTEGRAL

177

1 2.

E E E

f g (3)

Since the infimum on the right side is ,f g we have

.

E E E

f g f g (4)

On the other hand, 1 f and 2 g imply 1 2 is a simple function not greater than

.f g Hence

1 2 ,

E E

f g (5)

since sup .f gE E

f g

Again by Proposition (8.12),

1 2 1 2.

E E E

Hence (5) becomes

1 2.

E E E

f g (6)

Since now the supremum of the right side is ,f g we have

,

E E E

f g f g (7)

hence, (4) and (7) gives

.

E E E

f g f g (8)

Using (1) and (8) it follows that

.

E E E E E

af bg af bg a f b g

(ii) To prove (ii), it now suffices to show that

UNIT III CHAPTER 8

178

0.

E

f g

Since 0f g a.e., it follows that if ,f g 0 a.e. From this it follows [referring to

(9) of Proposition (8.12)] that

0,

E

whence

inf 0.f g

E E

f g

Similarly, if ,f g 0 a.e. and 0,

E

and hence

sup 0,f gE E

f g

whence (ii) is proved.

(iii) (a) The above proof also serves to establish (iii). [Details: f g a.e. implies 0f g

a.e. If ,f g 0 a.e. and 0,

E

and hence

sup 0,f gE E

f g

(9)

and by (i)

E E E

f g f g , so (9) gives

0

E E

f g

i.e., .

E E

f g

(iii) (b) Now, f f a.e. implies [by (iii)(a)] that

.

E E

f f (10)

f f a.e. implies [by (iii)(a)] that .

E E

f f

LEBESGUE INTEGRAL

179

Also, by (i) .

E E

f f Hence, we have

.

E E

f f (11)

From (10) and (11), we have .f f

(iv)

( )A f x implies, by (iii)(a),

E E

A f implies, by (i), 1

E E

A f implies ,

E

AmE f

since 1 .

E

mE Similarly, ( )f x B implies, by (iii)(a),

E E

f B implies, by (i),

1

E E

f B implies .

E

f BmE This proves (iv).

(v) A B

A B A B

f f

,A B

A B

f

since A B A B when A and

B are disjoint.

A B

A B

f f

A B

A B A B

f f

, using (i)

A B

A B

f f

.

A B

f f

(8.19) Proposition (Bounded Convergence Theorem) Let nf be a sequence of

measurable functions defined on a set E of finite measure and suppose that there is a real

number M such that ( )nf x M for all n and all x. If ( ) lim ( )nf x f x for each x in E,

then lim .n

E E

f f

UNIT III CHAPTER 8

180

Proof

The proof of this propostion furnishes a nice illustration of the use of Littlewood‟s “three

principles”. The construction of the propostion would be trivial if nf converged to f

uniformly, Littlewood‟s three principles states that if nf converges to f pointwise, then

nf is “nearly” uniformly convergent to f. A precise version of this principle is given by

Propsotion in the previous chapter, which states that, given 0, there is an N and a

measurable set A E with 4

mAM

such that for n N and \x E A we have

( ) ( ) .2

nf x f xmE

Thus

n n

E E E

f f f f , by (i) of Proposition (8.18)

n

E

f f , by (iii) of Proposition (8.18)

\

n n

E A A

f f f f

\

1 2 12

E A A

MmE

2 ,2 4

mE MmE M

since \

1 \

E A

m E A mE

.2 2

Since 0 is arbitrary, we have

lim .n

E E

f f

(8.20) Proposition A bounded function f on [ , ]a b is Riemann integrable, if and only if

the set of points at which f is discontinuous has measure zero.

Hint of the proof

LEBESGUE INTEGRAL

181

Let f be a bounded function on [a, b], and let h be the upper envelope of f. Then

show that ( ) .

b b

a a

f x dx h [ If f is a step function, then h except at a finite

number of points, and so ( ) .

bb

a a

h f x dx But there is a sequence n of step functions

such that n .h By Proposition (8.19), we have lim

b b

n

a a

h ( ) .

b

a

f x dx ] Use this to

prove Proposition (8.20).

2 The Integral of a Nonnegative Function

(8.21) Definition (Integral of a Nonnegative Measurable Function) If f is a nonnegative

measurable function defined on a measurable set E, we define

sup ,h fE E

f h

where h is a bounded measurable function such that : ( ) 0m x h x is finite.

(8.22) Proposition If f and g are nonnegative measurable functions, then:

i. , 0.

E E

cf c f c

ii. .

E E E

f g f g

iii. If f g a.e., then

.

E E

f g

Proof

(i)

For 0,c

sup ,ch cfE E

cf ch

where ch is a bounded measurable function such

that : ( ) 0m x ch x

UNIT III CHAPTER 8

182

suph f E

ch

suph f E

c h

, by (i) of Proposition (8.18)

suph f E

c h

.

E

c f

For 0,c

supch cfE E

cf ch

suph f E

ch

suph f E

c h

, by Proposition (8.18)

infh f

E

c h

, since 0,c

sup ,g f E

c h

since inf suph f g fE E

h g

for bounded measurable

functions g and h.

.

E

c f

(ii)

If ( ) ( )h x f x and ( ) ( ),k x g x we have ( ) ( ) ( ) ( ),h x k x f x g x and so

.

E E E

h k f g

Taking suprema, we have

.

E E E

f g f g

On the other hand, let l be a bounded measurable function which vanishes outside a set of

finite measure and which is not greater than .f g Then we define the functions h and k

by setting

( ) min ( ), ( )h x f x l x

LEBESGUE INTEGRAL

183

and ( ) ( ) ( ).k x l x h x

We have

( ) ( )h x f x

and ( ) ( ),k x g x

while h and k are bounded by the bound for l and vanish where l vanishes. Hence

,

E E E E E

l h h f g and so

.

E E E

f g f g

Hence (ii) is proved.

Part (iii) follow directly from Proposition (8.18).

(8.23) Theorem (Fatou’s Lemma) If nf is a sequence of nonnegative measurable

functions and ( ) ( )nf x f x almost everywhere on a set E, then

lim .n

E E

f f

Proof

Without loss of generality, we may assume that the convergence is everywhere, since integrals

over sets of measure zero are zero. Let h be a bounded measurable function which is not

greater than f and which vanishes outside a set E of finite measure. Define a function nh

by setting

( ) min ( ), ( )n nh x h x f x

Then nh is bounded by the bound (say M) for h and vanishes outside E . Then ( )nh x M

Also ( ) ( )nh x h x for each x in E . Thus we have

lim n

E E E

h h h

by Bounded Convergence Theorem

(Proposition (8.19)),

lim .n

E

f

i.e., lim .n

E E

h f

UNIT III CHAPTER 8

184

Taking the supremum over h, we get (using Definition (8.21))

lim .n

E E

f f


(8.24) Monotone Convergence Theorem Let nf be an increasing sequence of

nonnegative measurable functions, and let lim nf f a.e. Then

lim .nf f

Proof

By Fatou‟s Lemma , we have

lim .nf f (1)

But for each n we have ,nf f and so .nf f But this implies

lim .nf f (2)

Since lim limn nf f , (1) and (2) gives

lim lim .n nf f f f (3)

(3) tells us that all the inequalities are equal and hence

lim lim ,n nf f f

and hence lim nf exists and

lim .nf f

(8.25) Corollary Let nu be a sequence of nonnegative measurable functions, and let

1

.n

n

f u

Then

LEBESGUE INTEGRAL

185

1

.n

n

f u

i.e., 1 1

.n n

n n

u u

Proof Take 1

.n

n k

k

f u

Since each ku is nonnegative, nf is an increasing sequence of

nonnegative measurable functions, and lim nf f . Then by Monotone Convergence

Theorem (8.24),

lim nf f

1

limn

k

k

u

1

limn

k

k

u

, using Proposition (8.22)

1

n

n

u

.

(8.26) Proposition Let f be a nonnegative function and iE a disjoint sequence of

measurable sets. Let .iE E Then

.iE Ei

f f

Proof

Let .i Eiu f

Then E Eif f f

Eii

f , since iE are disjoint

Eii

f ,

,iu

and so

UNIT III CHAPTER 8

186

i

E E

f u

i

i E

u , using Corollary (8.25)

.Eii E

f

.i Ei

f

(8.27) Definition A nonnegative measurable function f is called integarble over the

measurable set E if

.

E

f

(8.28) Proposition Let f and g be two nonnegative measurable functions. If f is

integarble over E and ( ) ( )g x f x on E, then g is also integarble on E, and

.

E E E

f g f g

Proof

By Proposition (8.22) in this section,

.

E E E

f f g g (1)

Since the left side is finite, the terms on the right must also be finite and so g is integarble.

Also from (1) above, we have

.

E E E

f g f g

(8.29) Proposition Let f be a nonnegative function which is integarble over a set E. Then

given 0 there is a 0 such that for every set A E with mA we have

.

A

f

Proof

The proposition would be trivial if f were bounded.

LEBESGUE INTEGRAL

187

Set

( ), if ( )

( ), otherwise

n

f x f x nf x

n

Then each nf is bounded (with a bound n). For each x as n we have ( ) ( )nf x f x .

i.e., nf converges to f at each point. By the Monotone Convergence Theorem there is an N

such that ,2

N

E E

f f

and .2

N

E E

f f

i.e., .2

N

E

f f

Choose .2N

If

,mA we have

N N

A A A

f f f f

N

E

f f NmA

.2 2

3 The General Lebesgue Integral

(8.30) Definition The positive part f of a function f is the function

0;f f

i.e., f is given by

( ) max ( ), 0 .f x f x

The negative part f of a function f is the function

0;f f

i.e., f is given by

( ) max ( ), 0 .f x f x

(8.31) Remark If f is measurable, so are f and f . We have

f f f

and .f f f

UNIT III CHAPTER 8

188

(8.32) Definition A measurable function f is said to be integrable over E, if f and f

are both integrable over E. In this case we define

.

E E E

f f f

(8.33) Proposition Let f and g be integrable over E. Then

i. The function cf is integrable over E, and .

E E

cf c f

ii. The function f g is integrable over E, and

.

E E E

f g f g

iii. If f g a.e., then .

E E

f g

iv. If A and B are disjoint measurable sets contained in E, then

.

A B A B

f f f

Proof

Part (i) follows directly from the definition of the integral and Proposition (8.22). To

prove part (ii), we first note that if 1f and 2f are nonnegative integrable functions with

1 2 ,f f f then 2 1.f f f f For Proposition (8.22) tells us that

2 1,f f f f

and so

1 2.f f f f f

But, if f and g are integarble, so are f g and ,f g and

.f g f g f g Hence

f g f g f g

f g f g

LEBESGUE INTEGRAL

189

.f g

Part (iii) follows from part (ii) and the fact that the integral of a nonnegative integrable

function is nonnegative. For (iv) we have

A B

A B

f f

A Bf f , as A and B are disjoint implies

A B A B

A B

f f .


A and B are disjoint

(8.34) Lebesgue Convergence Theorem Let g be integrable over E and let nf be

a sequence of measurable functions such that nf g on E and for almost all x in E we

have ( ) lim ( ).nf x f x Then

lim .n

E E

f f (UQ 2006)

Proof

The function ng f is nonnegative, and so by Fatou‟s Lemma

lim .n

E E

g f g f

Since ,f g f is integrable, and we have

lim ,n

E E E E

g f g f

whence

lim .n

E E

f f

Similarly, considering ng f , we get

UNIT III CHAPTER 8

190

lim ,n

E E

f f

and the theorem follows.

(8.35) Theorem (Generalization of Lebesgue Convergence Theorem) Let ng be a

sequence of integrable functions which converges a.e. to an integrable function g. Let nf

be a sequence of measurable functions such that n nf g and nf converges to f a.e. If

lim ,ng g

then

lim .nf f

Convergence in Measure

(8.36) Definition A sequence nf of measurable functions is said to converge to f in

measure if, given 0, there is an N such that for all n N we have

: ( ) ( ) .nm x f x f x

It follows directly from this definition and Proposition (6.1) of Chapter 7 that if

nf is a sequence of measurable functions defined on a measurable set E of

finite measure and nf f a.e., then nf converges to f in measure.

An example of a sequence nf that converges to zero in measure in [0, 1] but

such that ( )nf x does not converge for any x in [0, 1] can be constructed as

follows: Let 2 , 0 2 ,v vn k k and set

1, if [ 2 , ( 1)2 ]

( )0, otherwise

v v

n

x k kf x

Then 2

: ( )nm x f xn

,

and so 0nf in measure, although for any x in [0, 1], the sequence

( )nf x has the value 1 for arbitrarily large values of n and so does not

converge.

LEBESGUE INTEGRAL

191

(8.37) Proposition Let nf be a sequence of measurable functions that converges in

measure to f. Then there is a subsequence nkf that converges to almost everywhere.

(UQ 2006)

Proof

Given v, there is an integer n, such that for all vn n we have

: ( ) ( ) 2 2 .v vnm x f x f x

Let : ( ) ( ) 2 .vv nE x f x f x Then, if ,v

v k

x E

we must have ( ) ( ) 2 vnv

f x f x for

,v k and so ( ) ( ).nvf x f x Hence ( ) ( )nv

f x f x for any 1

.v

k v k

x A E

But

12 .kv v

v kv k

mA m E mE

Hence mA = 0.

(8.38) Corollary Let nf be a sequence of measurable functions defined on a measurable

set E of finite measure. Then nf converges to f in measure if and only if every

subsequence of nf has in turn a subsequence that converges almost everywhere to f.

(8.39) Proposition Fatou‟s Lemma and the Monotone and Lebesgue Convergence

Theorems remain valid if „convergence a.e.‟ is replaced by „convergence in measure‟.

Chapter 9

DIFFERENTIATION AND INTEGRATION

1 Differentiation of Monotone Functions

(1.1) Definition Let be a collection of intervals. We say that covers a set E in the

sense of Vitali, if for each 0 and any x in E, there is an interval I such that x I

and ( ) .l I The intervals may be open, closed or half-open, but we do not allow degenerate

intervals consisting of only one point.

(1.2) Lemma (Vitali) Let E be a set of finite outer measure and a collection of

intervals that cover E in the sense of VItali. Then, given 0 , there is a finite disjoint

collection 1, , NI I of intervals in such that

*

1

~ .N

n

n

m E I

Proof

It suffices to prove the lemma in the case that each interval in is closed, for

otherwise we replace each interval by its closure and observe that the set of endpoints of

1, , NI I has measure zero.

Let O be an open set of finite measure containing E. Since is a Vitali covering of

E, we may assume without loss of generality that each I is contained in O. We choose a

sequence nI of disjoint intervals of by induction as follows: Let 1I be any interval in ,

and suppose that 1, , nI I has already been chosen. Let nk be the supremum of the lengths

of the intervals of that do not meet any of the intervals 1, , nI I . Since each I is

contained in O, we have .nk mO Unless 1

,n

i

i

E I

we can find 1nI with

1

1

2n nl I k and 1nI disjoint from 1, , nI I .

Thus we have a sequence nI of disjoint intervals of , and since ,iI O we

have .nl I mO Hence we can find an integer N such that

1

.5

n

N

l I


193

Let

1

~ .N

n

n

R E I

The lemma will be established if we can show that * .m R Let x be an arbitrary point of

R. Since 1

N

n

n

I

is a closed set not containing x, we can find an interval I which

contains x and whose length is so small that I does not meet any of the intervals 1, , NI I .

If now iI I for ,i n we must have 12 .n nl I k l I Since lim 0,nl I the

interval I must meet at least one of the intervals .nI Let n be the smallest integer such that

I meets .nI We have ,n N and 1 2 .n nl I k l I Since ,x I and I has a point in

common with ,nI it follows that the distance from x to the midpoint of nI is atmost

512 2

.n nl I l I l I Thus x belongs to the interval nJ having the same midpoint as nI

and five times the length. Thus we have shown that

1

.n

N

R J

Hence

*

1 1

5 .n n

N N

m R l J l J


The four quantities - derivates

0

( ) ( )( ) lim ,

h

f x h f xD f x

h

0

( ) ( )( ) lim ,

h

f x f x hD f x

h

0

( ) ( )( ) lim ,

h

f x h f xD f x

h

0

( ) ( )( ) lim .

h

f x f x hD f x

h

UNIT III CHAPTER 9

194

Clearly, we have ( ) ( )D f x D f x and ( ) ( ).D f x D f x

If

( ) ( ) ( ) ( ) ,D f x D f x D f x D f x we say that f is differentiable at x and

define ( )f x to be the common value of the derivates at x.

If ( ) ( ),D f x D f x we say that f has a right-hand derivative at x and define

( )f x to be their common value.

If ( ) ( ),D f x D f x we say that f has a left-hand derivative at x and define ( )f x

to be their common value.

(1.3) Proposition If f is continuous on [ , ]a b and one of its derivates (say )D is every

where nonnegative on [ , ]a b , then f is nondecreasing on [ , ]a b ; i.e., ( ) ( )f x f y for .x y

(1.4) Theorem Let f be an increasing real-valued function on the interval [ , ]a b .Then f is

differentiable almost everywhere. The derivative f is measurable, and

( ) ( ) ( ).

b

a

f x dx f b f a (UQ 2006)

Proof

Let us show that the sets where any two derivates are unequal have measure zero. We

consider only the set E where ( ) ( ),D f x D f x the sets arising from other combinations of

derivates being similarly handled. Now the set E is the union of the sets

, : ( ) ( )u vE x D f x u v D f x

for all rational u and v. Hence it suffices to prove that *, 0.u vm E

Let *,u vs m E and. choosing 0, enclose ,u vE in an open set O with .mO s For

each point , ,u vx E there is an arbitrarily small interval [ , ]x h x contained in O such that

( ) ( ) .f x f x h vh

By Lemma (1.2) we can choose a finite collection 1, , NI I of them whose interiors cover

a subset A of ,u vE of outer measure greater than .s Then, summing over these

intervals, we have

1 1

( ) ( )N N

n n n n

n n

f x f x h v h

vmO


195

( ).v s

Now to each point y A is the left endpoint of an arbitrarily small interval ( , )y y k that is

contained in some nI and for which ( ) ( ) .f y k f y uk Using Lemma (1.2) again, we

can pick out a finite collection 1, , MJ J of such intervals such that their union contains a

subset of A of outer measure greater than 2 .s Then summing over these intervals yields

1

( ) ( )M

i i i i

n

f y k f y u k

( 2 ).u s

Each interval iJ is contained in some interval nI , and if we sum over those i for which

iJ ,nI we have

( ) ( ) ( ) ( ),i i i n n nf y k f y f x f x h

since f is increasing. Thus

1 1

( ) ( ) ( ) ( ) ,N M

n n n i i i

n i

f x f x h f y k f y

and so

( ) ( 2 ).v s u s

Since this is true for each positive , we have .vs us But ,u v and so s must be zero.

This shows that

0

( ) ( )( ) lim

h

f x h f xg x

h

is defined almost everywhere and that f is differentiable whenever g is finite. Let

1( ) ( ) ( ) ,n ng x n f x f x

where we set ( ) ( )f x f b for .x b Then ( ) ( )ng x g x for almost all x, and so g is

measurable. Since f is increasing, we have 0.ng Hence by Fatou’s Lemma

1lim lim ( )

b b b

n n

a a a

g g n f x f x dx

1 1

lim

b an n

b a

n f n f

UNIT III CHAPTER 9

196

1

lim ( )

an

a

f b n f

( ) ( ).f b f a

This shows that g is integarble and hence finite almost everywhere. Thus f is differentiable

a.e. and g f a.e.

2 Functions of Bounded Variation

Let us denote

, if 0

0, if 0

r rr

r

.r r r

Then r and r are non-negative and .r r r

For example, if r = 5, then 5r and 5 5 0;r r r if r = 5, then 0r and

5 0 5.r r r

Let f be a real-valued function defined on the interval [ , ],a b and let

0 1 ka x x x b be any subdivision of [ , ].a b

Define

1

1

( ) ( )k

i i

i

p f x f x

1

1

( ) ( )k

i i

i

n f x f x

1

1

( ) ( ) .k

i i

i

t n p f x f x

We have ( ) ( ) ,f b f a p n

since ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ).1 1 1 01 1 1

k k kp n f x f x f x f x f x f x f x f xi i ii i i k

i i i

Set

sup ,P p

sup ,N n


197

sup ,T t

where we take the suprema over all possible subdivisions of [ , ].a b We clearly have

.P T P N We call P, N, T the positive, negative, and total variations of f over [ , ].a b

We some times write , ( ),b ba aT T f etc., to denote the dependence on the interval [ , ]a b or on

the function f. If T , we say that f is of bounded variation over [ , ].a b This notion

is abbreviated by writing .f BV

(2.1) Notation baP denotes s u pp where we take the suprema over all possible

subdivisions of [ , ].a b xaP denotes sup p where we take the suprema over all possible

subdivisions of [ , ].a x Similarly, baN , x

aN etc denote suprema over all possible subdivisions

of [ , ]a b or [ , ].a x

(2.2) Lemma If f is of bounded variation on [ , ]a b , then

b b ba a aT P N

and

( ) ( ) .b ba af b f a P N

Proof

For any subdivision of [ , ]a b

( ) ( ),p n f b f a

and taking suprema over all possible subdivisions, we obtain

( ) ( ).P N f b f a

Also

( ) ( ) .t p n p p f b f a

Taking surpema, we obtain

2 ( ) ( ) .T P f b f a P N

(2.3) Theorem A function f is of bounded variation on [ , ]a b if and only if f is the

difference of two monotone real-valued functions on [ , ]a b .

Proof

Let f be of bounded variation, and set ( ) xag x P and ( ) .x

ah x N Then g and h are

monotone increasing functions which are real valued, since 0 x x ba a aP T T and

UNIT III CHAPTER 9

198

0 .x x ba a aN T T But ( ) ( ) ( ) ( )f x g x h x f a by Lemma (2.2). Since ( )h f a is a

monotone function, we have f expressed as the difference of two monotone functions.

On the other hand, if f g h on [ , ]a b with g and h increasing, then for any

subdivision we have

1 1 1( ) ( ) ( ) ( ) ( ) ( )i i i i i if x f x g x g x h x h x

( ) ( ) ( ) ( ).g b g a h b h a

Hence

( ) ( ) ( ) ( ) ( ).baT f g b h b g a h a

(2.4) Corollary If f is of bounded variation on [ , ]a b , then ( )f x exists for almost all

[ , ].x a b

3 Differentiation of an Integral

(3.1) Definition If f is an integrable function on [ , ]a b , we define its indefinite integral

to be the function F defined on [ , ]a b by

( ) ( ) .

x

a

F x f t dt

(3.2) Lemma If f is an integrable function on [ , ]a b , then the function F defined by

( ) ( )

x

a

F x f t dt

is a continuous function of bounded variation on [ , ]a b . (UQ 2006)

Proof

Clearly F is continuous (Details are left to the exercise). To show that F is of bounded

variation, let 0 1 ka x x x b be any subdivision of [ , ]a b . Then

1

1 1 11 1

( ) ( ) ( ) ( )

x xi ik k k

i i

i i ix xi i

F x F x f t dt f t dt

( ) .

b

a

f t dt

Thus ( ) ( ) .

bb

a

a

T F f t dt


199

(3.3) Lemma If f is integrable on [ , ]a b , and

( ) 0

x

a

f t dt

for all [ , ],x a b then ( ) 0f t a.e. in [ , ]a b .

Proof

Suppose ( ) 0f x on a set E of positive measure. Then, by Proposition (4.4) in Chapter 7

Lebesgue Meausre, there is a closed set F E with 0.mF Let ( , ) \ .O a b F Then either

0,

b

a

f or else

0 ,

b

a F O

f f f

and 0.

O F

f f

But O is the disjoint union of a countable collection ( , )n na b of open intervals, and so by

Proposition: “Let f be a nonnegative function and iE a disjoint sequence of measurable

sets. Let .iE E Then iE Ei

f f ” , we have

.

bn

O an

f f

Thus for some n we have

0,

bn

an

f

and so either

0,

an

a

f

or

0.

bn

a

f

UNIT III CHAPTER 9

200

In any case we see that if f is positive on a set of positive measure, then for some [ , ]x a b

we have

0.

x

a

f

Similarly, if f negative on a set of positive measure, i.e., ( ) 0f x on a set E of positive

measure, then for some [ , ]x a b we have

0.

x

a

f

i.e., in any case ( ) 0f x on a set of positive measure implies 0.

x

a

f

Hence by contraposition † , it follows that ( ) 0

x

a

f t dt for all [ , ],x a b then ( ) 0f t a.e. in

[ , ]a b .

† Contraposition: If p and q are two statements, then p q and ~ ~q p are logically

equivalent. Hence to show that p q it is enough to show that ~ ~q p . In this Lemma

the atement p : ( ) 0

x

a

f t dt for all [ , ],x a b

statement q : ( ) 0f t a.e. in [ , ]a b .

statement ~ q : ( ) 0f t on a set of positive measure

statement ~ p : ( ) 0

x

a

f t dt for some [ , ].x a b

(3.4) Lemma If f is bounded and measurable on [ , ]a b and

( ) ( ) ( ),

x

a

F x f t dt F a

then ( ) ( )F x f x for almost all x in [ , ]a b . (UQ 2006)

Proof

By Lemma (3.2), F is bounded variation over [ , ]a b , and so ( )F x exists for almost all x in

[ , ]a b . Let .f K Then setting


201

( ) ( )

( ) ,n

F x h F xf x

h

with 1

,hn

we have

1

( ) ( ) ,

x h

n

x

f x f t dth

and so .nf K

Since ( ) ( )nf x F x a.e., the bounded convergence theorem implies that

0

1( ) lim ( ) lim ( ) ( )

c c c

nh

a a a

F x dx f x dx F x h F x dxh

0

1 1lim ( ) ( )

c h a h

ha a

F x dx F x dxh h

( ) ( ) ( ) ,

c

a

F c F a f x dx

since F is continuous. Hence

( ) ( ) 0

c

a

F x f x dx

for all [ , ],c a b and so

( ) ( )F x f x a.e. by Lemma (3.3).


(3.5) Theorem Let f be an integrable function on [ , ]a b , and suppose that

( ) ( ) ( ) .

x

a

F x F a f t dt

Then ( ) ( )F x f x for almost all x in [ , ]a b .

Proof

Without loss of generality, we may assume 0.f Let nf be defined by ( ) ( ),nf x f x if

( ) ,f x n and ( )nf x n if ( ) .f x n Then 0,nf f and so

( )

x

n n

a

G x f f

UNIT III CHAPTER 9

202

is an increasing function of x, which must have a derivative almost everywhere, and this

derivative will be nonnegative. Now by Lemma (3.4)

( )

x

n n

a

df f x

dx a.e.,

and so

( )

x

n n

a

dF x G f

dx

( )nf x a.e.

Since n is arbitrary,

( ) ( )F x f x a.e.

Consequently,

( ) ( ) ( ) ( ).

b b

a a

F x dx f x dx F b F a

Thus by Theorem (1.4) we have

( ) ( ) ( ) ( )

b b

a a

F x dx F b F a f x dx

and

( ) ( ) 0.

b

a

F x f x dx

Since ( ) ( ) 0,F x f x this implies that ( ) ( ) 0F x f x a.e., and so ( ) ( )F x f x a.e.

4 Absolute Continuity

(4.1) Definition A real-valued function f defined on [ , ]a b is said to be absolutely

continuous on [ , ]a b if, given 0 , there is a 0 such that

1

( ) ( )n

i i

i

f x f x

for every finite collection ( , )i ix x of nonoverlapping intervals with

1

.n

i i

i

x x


203

(4.2) Remarks

An absolutely continuous function is continuous.

Every indefinite integral is absolutely continuous, by the Proposition : “Let f be a

nonnegative function which is integrable over a set E. Then given 0 there is a

0 such that for every set A E with mA we have .

A

f ”

(4.3) Lemma If f is absolutely continuous on [ , ]a b , then it is of bounded variation on

[ , ]a b .

Proof

Let be the in the definition of absolute continuity that corresponds to 1. Then any

subdivision of [ , ]a b can be split (by inserting fresh division points, if necessary) into K sets

of intervals, each of total length less than , where K is the largest integer less than

( )1 .

b a

Thus for any subdivision we have ,t K and so .T K

(4.4) Corollary If f is absolutely continuous, then f has a derivative almost every where.

Proof

By Lemma (4.3), f is absolutely continuous implies f is of bounded variation and then by

Corollary (2.4), f has a derivative almost every where.

(4.5) Lemma If f is absolutely continuous on [ , ]a b and ( ) 0f x a.e., then f is

constant.

Proof

We wish to show that ( ) ( )f a f c for any [ , ]c a b . Let ( , )E a c be set of measure c a

in which ( ) 0f x , and let and be arbitrary positive numbers. To each x E there is an

arbitrarily small interval [ , ]x x h contained in [ , ]a c such that ( ) ( ) .f x h f x h By

Lemma (1.2) we can find a finite collection [ , ]k kx y of nonoverlapping intervals of this sort

which cover all of E except for a set of measure less than , where is the positive

number corresponding to in the definition of the absolute continuity of f. If we label the

kx so that 1k kx x , we have

0 1 1 2 1n ny a x y x y c x

and 1

0

.n

k k

k

x y

UNIT III CHAPTER 9

204

Now

0

( ) ( ) ( )n

k k k k

k

f y f x y x

( )c a

by the way the intervals [ , ]k kx y were constructed, and

1

0

( ) ( )n

k k

k

f x f y

by the absolute continuity of f. Thus

1

0 1

( ) ( ) ( ) ( ) ( ) ( )n n

k k k k

k k

f c f a f x f y f y f x

( ).c a

Since and are arbitrary positive numbers, ( ) ( )f c f a = 0.

(4.6) Theorem A function F is an indefinite integral if and only if it is absolutely

continuous.

Proof

If F is an indefinite integral, then F is absolutely continuous by the Proposition : “Let f be

a nonnegative function which is integrable over a set E. Then given 0 there is a 0

such that for every set A E with mA we have .

A

f ”

Suppose on the other hand that F is absolutely continuous on [ , ]a b . Then F is of

bounded variation, and we may write

1 2( ) ( ) ( ),F x F x F x

where the functions iF are monotone increasing. Hence ( )F x exists almost everywhere and

1 2( ) ( ) ( ).F x F x F x

Thus

1 2 1 2( ) ( ) ( ) ( ) ( )F x dx F b F b F a F a

by Theorem (1.4), and ( )F x is integrable. Let

( ) ( ) .

x

a

G x F t dt


205

Then G is absolutely continuous and so is the function .f F G It follows from Theorem

(3.5) that ( ) ( ) ( ) 0f x F x G x a.e., and so f is constant by Lemma (4.5). Thus

( ) ( ) ( ).

x

a

F x F t dt F a (1)

i.e., F is an indefinite integral.

(4.7) Corollary Every absolutely continuous function is the indefinite integral of its

derivative.

Proof

Proof is obvious from (1) of Theorem (4.6), for if F is absolutely continuous function then

( ) ( ) ( ),

x

a

F x F t dt F a which says that F is the indefinite integral of its derivative F .

5 Convex Functions

(5.1) Definition A function defined on an open interval ( , )a b is said to be convex if

for each , ( , )x y a b and each , 0 1 we have

(1 ) ( ) (1 ) ( ).x y x y

If we look at the graph of in 2 , this condition can be formulated geometrically by

saying that each point on the chord between , ( )x x and , ( )y y is above the graph of .

(5.2) Lemma If is convex on ( , )a b and if , , ,x y x y are points of ( , )a b with

x x y and ,x y y then the chord over ( , )x y has larger slope than the chord over

( , );x y i.e.,

( ) ( ) ( ) ( )

.y x y x

y x y x

(5.3) Definition If the upper and lower left-hand derivatives D f and D f of a function

f are equal and finite at a point x, we say that f is differentiable on the left at x and call

this common value the left-hand derivative at x. Similarly, we say that f is differentiable on

the right at x if D f and D f are equal there.

(5.4) Proposition If is convex on ( , )a b , then is absolutely continuous on each

closed subinterval of ( , )a b . The right- and left-hand derivatives of exist at each point of

( , )a b and are equal to each other except on a countable set. The left- and right-hand

UNIT III CHAPTER 9

206

derivatives are monotone increasing functions, and at each point the left-hand derivative is

less than or equal to the right-hand derivative.

Proof

Let [ , ] ( , ).c d a b Then, by Lemma (5.2), we have

( ) ( ) ( ) ( ) ( ) ( )c a y x b d

c a y x b d

for , [ , ].x y c d Thus ( ) ( )y x M x y in [ , ]c d , and so is absolutely continuous

there.

If 0 ( , ),x a b then 0

0

( ) ( )x x

x x

is an increasing function of x by Lemma (5.2), and

so the limits as x approaches 0x from the right and from the left exist and are finite. Thus

is differentiable on the right and on the left at each point, and the left-hand derivative is

less than or equal to the right-hand derivative. If 0 0 0,x y x y and 0x y , then

0 0

0 0

( ) ( ) ( ) ( ),

x x y y

x x y y

and either derivative at 0x is less than or equal to either derivative at 0.y Consequently, each

derivative is monotone, and they are equal at a point if one of them is continuous there. Since

a monotone function can have only a countable number of discontinuities, they are equal

except on a countable set.

(5.5) Proposition If is a continuous function on ( , )a b and if one derivative (say D )

of is nondecreasing, then is convex.

Proof

Given x, y with ,a x y b define a function on [0, 1] by

( ) [ (1 ) ] ( ) (1 ) ( ).t ty t x t y t x

Our goal is to show that is nonpositive on [0, 1]. Now is continuous, and

(0) (1) 0. Moreover,

( ) ( ) ( ),D y x D y x

and so D is nondecreasing on [0, 1] .

Let be a point where assumes its maximum on [0, 1] . If 1, then

( ) (1) 0t on [0, 1] . Hence suppose that [0, 1).


207

Since has a local maximum at , we have ( ) 0.D But D was

nondecreasing, and so 0D on [0, ]. Consequently, is nonincreasing on [0, ], and

hence ( ) (0) 0. Thus the maximum of on [0, 1] is nonpositive, and so 0 on

[0, 1] .

(5.6) Proposition Let have a second derivative at each point of ( , )a b . Then is

convex on ( , )a b if and only if ( ) 0x for each ( , ).x a b

Let be a convex function on ( , )a b and 0 ( , ).x a b The line 0 0( ) ( )y m x x x

through 0 0, ( )x x is called a supporting line at 0x if it always lies below the graph of ,

that is, if

0 0( ) ( ) ( ).x m x x x

It follows from Lemma (5.2) that such a line is a supporting line if and only if its slope m lies

between the left- and right-hand derivatives at 0.x Thus, in particular, there is always at least

one supporting line at each point. This notion enables us to give a short proof for the

following proposition:

(5.7) Proposition (Jensen Inequality) Let be a convex function on ( , ) and f

an integrable function on [0, 1] . Then

( ( )) ( ) .f t dt f t dt

Proof

Let ( ) ,f t dt and let ( ) ( )y m x be the equation of a supporting line at .

Then

( ( )) ( ) ( ).f t m f t

Integrating both sides with respect to t gives the proposition.

(5.8) Corollary Let f be an integrable function on [0, 1] . Then

exp( ( )) exp ( ) .f t dt f t dt

(5.9) Definition A function defined on an open interval ( , )a b is said to be strictly

convex if for each , ( , )x y a b and each , 0 1 we have

(1 ) ( ) (1 ) ( ).x y x y

UNIT III CHAPTER 9

208

(5.10) Definition A function defined on an open interval ( , )a b is said to be concave if

is convex.

(5.11) Remark The only functions that are both convex and concave are the linear

functions.

(5.12) Definition If I is any interval, (open, closed, or half-open), we say that is

convex on I if is continuous on I and convex in the interior.

___________________

Chapter 10

MEASURE AND INTEGRATION

209

1 Measure Spaces

The purpose of the present chapter is to abstract the most important properties of

Lebesgue measure and Lebesgue integration. We shall do this by giving certain axioms

which Lebesgue measure satisfies and base our integration theory on these axioms. As a

consequence our theory will be valid for every system satisfying the given axioms.

(1.1) Definition A -algebraB is a family of subsets of a given set X which contains

and is closed with respect to complements and with respect to countable unions.

(1.2) Definition By a set function we mean a function which assigns an extended real

number to certain sets.

(1.3) Definition By a measurable space we mean a couple X ,B consisting of a set X

and B of subsets of X. A subset A of X is called measurable (or measurable with respect

to B ) if AB .

(1.4) Definition By a measure on a measurable space X ,B we mean a nonnegative set

function defined for all sets of B and satisfying ( ) 0 and

11

i iii

E E

(1)

for any sequence iE of disjoint measurable sets. By a measure space ,X ,B we mean a

measurable space X ,B together with a measure defined on B .

The property (1) of is often referred to by saying that is countably additive. We

also have that is finitely additive; i.e.,

11

,N N

i iii

E E

(2)

for disjoint sets iE belong to B , since we may set iE for .i N

UNIT III CHAPTER 10

210

(1.5) Examples

1. , m,M is a measure space, where is the set of real numbers, M the

Lebesgue measurable sets of real numbers, and m Lebesgue measure.

2. [0, 1] , m,M is a measure space, whereM is the measurable subsets of sets of

[0, 1] real numbers, and m Lebesgue measure.

3. , m,B is a measure space, where B is the class of Borel sets, and m

Lebesgue measure.

4. Let X be any uncountable set, B the family of those subsets which are either

countable or the complement of a countable set. Then B is a -algebra and we

can define a measure on it by setting 0A for each countable set and 1B for

each set whose complement is countable.

(1.6) Proposition If AB , BB and ,A B then

.A B

Proof

Since \ ,A B A

\B A B A

is a disjoint union, and hence we have

( \ ) ,B A B A A since ( \ ) 0.B A

(1.7) Proposition If iE B , 1E and 1,i iE E then

1

lim .i nni

E E

(3)

Proof

Set 1

.ii

E E

Then

1 11

\i ii

E E E E

and this is a disjoint union. Hence

1 11

\ .i ii

E E E E

Since

1 1\i i i iE E E E


211

is a disjoint union, we have

1 1\ .i i i iE E E E

Hence

1 11

i ii

E E E E

1

11

limn

i in i

E E E

1 lim ,nn

E E E

whence the proposition follows.

(1.8) Proposition If iE B , then

11

.i iii

E E

Proof

Let

1

1

\ .n

n n ii

G E E

Then n nG E and the sets nG are disjoint. Hence

,n nG E

while

1 11

.i n ni ni

E G E

(1.9) Definition A measure is called finite if ( ) .X It is called a -finite if there is

a sequence nX of measurable sets in B such that

1

nn

X X

and .nX

(1.10) Examples

1. Lebesgue measure on [0, 1] is of finite measure.

2. Lebesgue measure on , is a - finite measure.

3. The counting measure on an uncountable set is a measure that is not -finite.

UNIT III CHAPTER 10

212

(1.11) Definition A set E is said to be of finite measure if EB and .E A set E is

said to be of -finite measure if E is the union of a countable collection of measurable sets

of finite measure.

Any measurable set contained in a set of -finite measure is itself and the union of a

countable collection of sets of -finite measure is again of -finite measure. If is -

finite, then every measurable set is of -finite measure.

(1.12) Definition A measure is said to be semifinite if each measurable set of infinite

measure contains measurable sets of arbitrary large finite measure.

Every -finite measure is semifinite.

The measure that assigns 0 to countable subsets of an uncountable set X and

to the uncountable sets is not semifinite.

(1.13) Definition A measure space ,X ,B is said to be complete if B contains all

subsets of sets of measure zero, that is, if BB , 0,B and A B imply AB .

Lebesgue measure is complete.

Lebesgue measure restricted to the -algebra of Borel sets is not complete.

(1.14) Proposition If ,X ,B is a measure space, then we can find a complete measure

space 0 0,X ,B such that

( i) 0B B .

( ii) 0 .E E E B =

( iii) 0E E A B B = where BB and ,A C C B , 0.C

The measure space 0 0,X ,B is called the completion of ,X ,B .

(1.15) Definition If ,X ,B is a measure space, we say that a subset E of X is locally

measurable if E B B for each BB with .B The collection C of all locally

measurable sets is a -algebra containing B . The measure is called saturated if every

locally measurable set is measurable (i.e., is in B ).

Every -finite measure is saturated.

A measure can always be extended to a saturated measure, but unlike the

process of completion, the process of saturation is not uniquely

determined.


213

2 Measurable Functions

The concept of a measurable function on an abstract measurable space is almost

identical with that for functions of a real variable. Consequently, those propsoitons and

theorem whose proofs are essentially the same as those in Chapter 7 Lebesgue Measure are

stated without proof:

Throughout the section we assume that a fixed measurable space X ,B is given

(2.1) Proposition (Analogue of Proposition (5.1) in Chapter 7) Let f be an extended

real-valued function defined on X (where X ,B is a measurable space). Then the following

statements are equivalent:

i. For each real number the set : ( )x f x B .

ii. For each real number the set : ( )x f x B .

iii. For each real number the set : ( )x f x B .

iv. For each real number the set : ( )x f x B .

(2.2) Definition The extended real-valued function f defined on X is called measurable

(or measurable with respect to B ) if any one of the statements of Proposition (2.1) holds.

Recall that f g is defined by

( ) max ( ), ( )f g x f x g x .

(2.3) Theorem (Analogue of Proposition (6.7) and Theorem (6.12) in Chapter 7) If c

is a constant and the functions f and g are measurable, then so are the functions

, , , ,f c cf f g f g and .f g Moreover, if nf is a sequence of measurable

functions, then sup ,nn

f inf ,nn

f lim nf , and lim nf are all measurable.

Recall that a simple function is a finite linear combination

1

( ) ( )n

i Eii

x c x

of characteristic functions of measurable sets .iE

UNIT III CHAPTER 10

214

(2.4) Proposition Let f be a nonnegative measurable function. Then there is a sequence

n of simple functions with 1n n such that lim nf at each point of X. If f is

defined on a -finite measure space, then we may choose the functions n so that each

vanishes outside a set of finite measure.

(2.5) Proposition (Analogue of Proposition (6.14) in Chapter 7) If is a complete

measure and f is a measurable function, then f = g a.e. implies g is measurable.

The sets : ( )x f x are sometimes called ordinate sets for f. They increase with

. The following lemma states that, given a collection B of measurable sets that increase

with , we can find a measurable function f which nearly has these for ordinate sets in the

sense that

: ( ) : ( ) .x f x B x f x

(2.6) Lemma Suppose that to each in a dense set D of real numbers there is assigned a

set B B such that B B for . Then there is a unique measurable extended real-

valued function f on X such that f on B and f on \ .X B

Proof

For each x X define

( ) inf :f x D x B

where, as usual, inf . If x B , then ( ) .f x If x B , then x B for each

, and so ( ) .f x To show that f is measurable, we take and choose a

sequence n from D with n and lim .n Then

1

: ( ) .n

n

x f x B

For if ( ) ,f x then ( ) nf x for some n, and so n

x B for some n, then

( ) .nf x Thus the sets : ( )x f x are all measurable, and so f is measurable.

To prove the uniqueness of f, let g be any extended real-valued function with g

on B and g on B . Then x B implies ( ) ,g x and so

: : ( ) .D x B D g x

Since ( )g x implies that x B we have


215

: ( ) : .D g x D x B

Because of the density of D we have

( ) inf : ( )g x D g x

inf : ( )D g x

inf : ( ).D x B f x

(2.7) Proposition Suppose that for each in a dense set D of real numbers there is

assigned a set B B such that \ 0B B for . Then there is a measurable

function f such that f a.e. on B and f a.e. on \ .X B If g is any other

function with this property, then g = f a.e.

Proof

Let C be a countable dense subset of D, and set \N B B for and in C wth

. Then N is the countable union of sets of measure zero and so itself a set of measure

zero. Let ..B B N For and in C with we have

\ \ \ \ ,B B B N B N B B N as \N B B .

Thus .B B By Lemma (2.6) there is a measurable function f such that f on B

and f on B .

Let D , and choose a sequence n from C with n and lim .n Then

\ \ \ .n n n

B B B B N B B

Thus \n

n

P B B is a countable union of null sets and so a null set. Let .n

A B

Then inf nf on A, and \ .A B P Thus f almost everywhere on B . A

similar argument shows that f almost everywhere on B .

Let g be an extended real-valued function with g a.e. on B and g on B

for each .C Then g on B and g a.e. on B except for x in a null set .Q

Thus Q Q is a null set and we must have f g on \ .X Q

UNIT III CHAPTER 10

216

3 Integration

Many definitions and proofs of Chapter 8 Lebesgue Integral depend on only those properties

of Lebesgue measure which are also true for an arbitrary measure in an abstract measure

space and carry over to this case.

(3.1) Definition (Integral of a nonnegative simple function) If E is a measurable set and

a nonnegative simple function, we define

1

,n

i iiE

d c E E

where

1

( ) ( ).i

n

i Ei

x c x

It is easily seen that the value of this integral is independent of the representation of

which we use. If a and b are positive numbers and and nonnegative simple

functions, then

.a b a b

(3.2) Proposition Let f be a bounded measurable function which is

identically zero outside a measurable set E of finite measure. Then

inf ( ) : sup ( ) :

E E

x dx f x dx f

for all simple functions and if and only if f = g almost everywhere for some

measurable function g.

Proof Proceed as in Proposition (8.14) of Chapter 8 The Lebesgue Integral.

(3.3) Definition (Integral of a nonnegative simple function) Let f be a nonnegative

extended real-valued measurable function on the measure space ,X ,B . Then f d is

the supremum of all integrals d as ranges over all simple functions with 0 .f

i.e., 0inf

ff d d

It follows immediately from this definition that f g implies that f g and

.cf c f


217

We cannot immediately say that f g f g , we need the Proposition (3.6).

(3.4) Theorem (Fatou’s Lemma) If nf is a sequence of nonnegative measurable

functions that converge almost everywhere on a set E to a function f. Then

lim .n

E E

f f

Proof

Without loss of generality, we may assume that ( ) ( )nf x f x for each .x E From the

definition of f it suffices to show that, if is any nonnegative simple function with

,f then lim .n

E E

f

If ,

E

then there is a measurable set A E with A such that 0a on

A. Set

: ( ) .n kA x E f x a k n

Then nA is an increasing sequence of measurable sets whose union contains A, since

lim .nf Thus .nA Since ,n n

E

f a A we have lim .n

E E

f

If ,

E

then the set : ( ) 0A x E x is a measurable set of finite measure.

Let M be the maximum of , let be a given positive number, and set

: ( ) (1 ) ( ) .n kA x E f x x k n

Then nA is an increasing sequence of sets whose union contains A, and so \ nA A is a

decreasing sequence of sets whose intersection is empty. By Proposition (1.7)

lim \ 0,nA A and so we can find an n such that \ kA A for all .k n Thus for

k n

(1 )k k

E A Ak k

f f

\

(1 )

E A Ak

.

E E

M

UNIT III CHAPTER 10

218

Hence

lim .n

E E E

f M

Since is arbitrary,

lim ,n

E E

f

and this completes the proof.

(3.5) Monotone Convergence Theorem Let nf be an increasing sequence of

nonnegative measurable functions which converge almost everywhere to a function f and

suppose that .nf f n Then

lim .nf f

Proof

By Fatou’s Lemma , we have

lim .nf f (1)

But for each n we have ,nf f and so .nf f But this implies

lim .nf f (2)

Since lim limn nf f , (1) and (2) gives

lim lim .n nf f f f (3)

(3) tell us that lim lim ,n nf f f and hence lim nf exists and

lim .nf f

(3.6) Proposition If f and g are nonnegative measurable functions and a and b

nonnegative constants, then


219

.af bg a f b g

We have

0f

with equality only if 0f a.e.

UNIT III CHAPTER 10

220

Proof

To prove the first statement, let n and n be increasing sequences of simple functions

which converge to f and g. Then n na b is an increasing sequence of simple

functions which converge to .af bg By the Monotone Convergence Theorem, we have

lim n naf bg a b

lim n na b

.a f b g

Clearly 0.f If 0,f let 1

: ( ) .nA x f xn

Then we have 1

,nAf

n and so

0.nn AA Since the set where 0f is the union of the sets nA , it has measure zero.

(3.7) Corollary Let nf be a sequence of nonnegative measurable functions. Then

1 1

.n nn n

f f

(3.8) Definition A nonnegative function f is called integrable (over a measurable set E

with respect to ) if it is measurable and

.f d

(3.9) Definition An arbitrary function f is said to be integrable if both f and f are

integrable. In this case we define

.E E E

f f f

(3.10) Proposition If f and g are integrable functions and E is a measurable set, then

( i) 1 2 1 2 .E E E

c f c g c f c g

( ii) If h f and f is measurable then h is integrable.

( iii) If f g a.e., then .f g


221

(3.11) Lebesgue Convergence Theorem Let g be integrable over E, and suppose

that nf is a sequence of measurable functions such that on E

( ) ( )nf x g x

and such that almost everywhere on E

( ) ( ).nf x f x

Then

lim .nE E

f f

Proof

Apply Fatou’s Lemma to sequences ng f and .ng f

4 General Convergence Theorems

(4.1) Proposition Let X ,B be a measurable space, n a sequence of measures that

converge setwise to a measure , and nf a sequence of nonnegative measurable functions

that converge pointwise to the function f. Then

lim .n nf f d (UQ 2006)

Proof

Setwise convergence of n to implies that

lim .nd d

for any simple function . From the definition of f d it suffices to prove that

lim n nd f d for any simple function .f

Suppose .d Then vanishes outside a set E of finite measure. Let be a

positive number, and set

: ( ) (1 ) ( ) for all .n kE x f x x k n

Then nE is an increasing sequence of sets whose union contains E, and so \ nE E is a

decreasing sequence of measurable sets whose intersection is empty. Thus by Proposition

(1.7) there is an m such that \ .mE E Since \ lim \ ,m k mE E E E we may

UNIT III CHAPTER 10

222

choose n m so that \k mE E for .k n Since \ \ ,k mE E E E we have

\k kE E for .k n Thus

(1 )

k k

k k k k kE E

f d f d d

\

(1 )

k

k kE E E

d d

(1 ) ,kE

d M

where M is the maximum of . Thus

lim .k kE

f d d M d

Since was arbitrary,

lim k kE

f d d .

i.e., lim .k kE

d f d

The case when E

d is handled similarly.

(4.2) Proposition Let X ,B be a measurable space and n a sequence of measures on

B that converge setwise to a measure . Let nf and ng be two sequences of measurable

functions that converge pointwise to f and g. Suppose that n nf g and that

lim .n ng d g d

Then

lim .n nf d f d

Proof

Apply Proposition (4.1) to sequences n ng f and .n ng f

5 Signed Measures

(5.1) Definition By a signed measure on the measurable space X ,B we mean an

extended real-valued set function v defined for the sets of B and satisfying the following

conditions:


223

( i) v assumes at most one of the values + , .

( ii) ( ) 0.v

( iii) 11

i iii

v E vE

for any sequence iE of disjoint measurable sets, the equality

taken to mean that the series on the right converges absolutely if 1

ii

v E

is finite

and that is properly diverges otherwise.

(5.2) Definition We say that a set A is a positive set with respect to a signed measure v if

A is measurable and for every measurable subset E of A we have 0.vE

Every measurable subset of a positive set is again positive, and if we take the

restriction of v to a positive set we obtain a measure.

(5.3) Definition We say that a set A is a negative set with respect to a signed measure v if

A is measurable and for every measurable subset E of A we have 0.vE

(5.4) Definition A set that is both positive and negative with respect to v is called a null

set.

A measurable set is a null set if and only if every measurable subset of it has v

measure zero.

(5.5) Lemma Every measurable subset of a positive set is itself positive. The union of a

countable collection of positive sets is positive. (UQ 2008)

Proof

The first statement is trivially true by the definition of a positive set. To prove the second

statement, let A be the union of a sequence nA of positive sets. If E is any measurable

subset of A, set

1 1.n n nE E A A A

Then nE is a measurable subset of nA and so 0.nvE Since the nE are disjoint and

,nE E we have

1

0.nn

vE vE

Thus A is a positive set.

1 1.n n nE E A A A

UNIT III CHAPTER 10

224

(5.6) Lemma Let E be a measurable set such that 0 .vE Then there is a positive set

A contained in E with 0.vA

Proof

Either E itself is a positive set or it contains sets of negative measure. In the latter case let 1n

be the smallest positive integer such that there is a measurable set 1E E with

11

1.vE

n

Proceeding inductively, if 1

1

\k

jj

E E

is not already a positive set, let kn be the smallest

positive integer for which there is a measurable set kE such that

1

1

\k

k jj

E E E

and

1

.kk

vEn

If we set

1

\ ,kk

A E E

then

1

.kk

E A E

Since this is a disjoint union, we have

1

kk

vE vA vE

with the series on the right absolutely convergent, since vE is finite. Thus 1

kn converges,

and we have .kn Since 0kvE and 0,vE we must have 0.vA

To show that A is a positive set, let 0 be given. Since ,kn we may choose

k so large that 1

1 .kn

Since

1

\ ,jj

A E E


225

A can contain no measurable sets with measure less than 1

1 ,kn

which is greater than

. Thus A contains no measurable sets of measure less than . Since is an arbitrary

positive number, it follows that A can contain no sets of negative measure and so must be a

positive set.

(5.7) Proposition (Han Decomposition Theorem) Let v be a signed measure on the

measurable space X ,B . Then there is a positive set A and a negative set B such that

X A B and .A B

Proof

Without loss of generality we may assume that is the infinite value omitted by v. Let

be the supremum of vA over all sets A that are positive with respect to v. Since the empty

set is positive, 0. Let iA be a sequence of positive sets such that

lim ,ii

vA

and set

1

.ii

A A

By Lemma (5.5) the set A is itself a positive set, and so .vA But \ iA A A and so

\ 0.iv A A Thus

\ .i i ivA vA v A A vA

Hence ,vA and so ,vA and .

Let ,B A the complement of A, and suppose that E is a positive subset of B. Then

E and A are disjoint and E A is a positive set. Hence

,v E A vE vA vE

whence 0,vE since 0 . Thus B contains no positive subsets of positive measure

and hence no susbsets of positive measure by Lemma (5.6). Consequently, B is a negative

set.

(5.8) Definition A decomposition of X into two disjoint sets A and B such that A is

positive for v and B negative is called a Hahn decomposition for v. Proposition (5.7)

states the existence of a Hahn decomposition for each signed measure. Hahn decomposition

is not unique.

UNIT III CHAPTER 10

226

(5.9) Definition Two measures 1v and 2v on X ,B are said to be mutually singular (in

symbols 1 2v v ) if there are disjoint measurable sets A and B with X A B such that

1 2 0.v A v B

(5.10) Proposition Let v be a signed measure on the measurable space X ,B . Then

there are two mutually singular measures v and v on X ,B such that .v v v

Moreover, there is only one such pair of mutually singular measures.

Proof

By Proposition (5.7), the signed measure v has a Hahn decomposition, say ,A B with

X A B and .A B

We define two measures v and v on X ,B with v v v by setting

( ) ( )v E v E A

and

( ) ( ).v E v E B

The measures v and v on X ,B are mutually singular because

0v A v A A v and 0v A v A A v on the measurable sets

A and B that are disjoint and with X A B .

The uniqueness part is left to the exercise.

(5.11) Definition The decomposition of v given by the Proposition (5.10) is called the

Jordan decomposition of v. The measures v and v are called the positive and negative

parts (or variations) of v. Since v assumes at most one of the values and , either v

and v must be finite. If they are both finite, we call v a finite signed measure.

The measure v defined by

v E v E v E

is called the absolute value or total variation of v. A set E is positive for v if 0.v E It

is a null set if 0.v E


227

6 The Radon-Nikodym Theorem

(6.1) Definition Let X ,B be a fixed measurable space. If and are two measures

defined on X ,B , we say that and are mutually singular (and write ) if there are

disjoint sets A and B in B such that X A B and 0.A B

Despite the fact that the notion of singularity is symmetric in and , we some times

say that is singularity with respect to .

(6.2) Definition A measure is said to be absolutely continuous with respect to the

measure if 0A for each set A for which 0.A We use the symbolism for

absolutely continuous with respect to .

In the case of signed measures and , we say if and

if .

Whenever we are dealing with more than one measure, we must specify almost

everywhere with respect to the given measure, say and it is abbreviated as

a.e. .

If and a property holds a.e. , then it holds for a.e. .

Let be a measure and f a nonnegative measurable function on X. For E in B , set

.E

vE f d

Then v is a set function defined on B .

It follows from Corollary (3.7) that v is countably additive and hence a measure.

The measure v will be finite if and only if f is integrable.

Since the integral over a set of -measure zero is zero, we have v absolutely

continuous with respect to .

(6.3) Radon-Nidkodym Theorem Let ,X ,B be a -finite measure space, and let v

be a measure defined on B which is absolutely continuous with respect to . Then there is a

nonnegative measurable function f such that for each set E in B we have

.E

vE f d

UNIT III CHAPTER 10

228

The function f is unique in the sense that if g is any measurable function with this property

then g = f a.e. .

Proof

The extension from the finite to the -finite case is not difficult and is left to the student.

Thus we shall assume that is finite. Then is a signed measure for each rational

number . Let ,A B be a Hahn decomposition for , and take 0 0, .A X B

Now \ .B B B A Thus \ 0,B B and hence

\ 0.B B If , these imply \ 0,B B and so by Proposition (2.7)

there is a measurable function f such that for each rational we have f a.e. on A and

f a.e. on .B Since 0 ,B we may take f to be nonnegative.

Let E be an arbitrary set in B , and set

1 \ ,k kN N

kE E B B \ .kN

E E B

Then 0

,kk

E E E

and this union is disjoint modulo null sets. Thus

0

.kk

vE vE vE

Since 1 ,k kN N

kE B A we have 1k kN N

f on ,kE and so

1 .

k

k kk kN N

E

E f d E (1)

Since 1 ,k kk k kN N

E E E

we have

1 1 11 1

k

k k kk k k k k k k k kN N N N N

E

E E E E E f d E E E EN N

i.e., 1 1

.

k

k k k kE

E E f d E EN N


229

On E we have f a.e. If 0,E we must have ,E since E is

positive for each . If 0,E we have 0,E since . In either case

.E

E f d

Adding this equality and our previous inequalities gives

1 1

.E

E E f d E EN N

Since E is finite and N arbitrary, we must have

.E

E f d

(6.4) Definition The function f given by Radon-Nidkodym Theorem is called Radon-

Nidkodym derivative of v with respect to . It is denoted by .d

d

(6.5) Proposition (Lebesgue Decomposition) Let ,X ,B be a -finite measure

space [Ref. Definition (1.11)] and v a -finite measure defined on B . Then we can find a

measure 0 , singular with respect to , a measure 1 , absolutely continuous with respect to

, such that 0 1. The measures 0 and 1 are unique.

Proof

We prove the existence of 0 and 1 , leaving the uniqueness part to the exercise.

Since and v are -finite measures, so is the measure . Since both and

v are absolutely continuous with respect to , the Radon-Nidkodym Theorem asserts the

existence of nonnegative masurable functions f and g such that for each EB

E

E f d , E

E g d .

Let : ( ) 0A x f x and : ( ) 0 .B x f x Then X is the disjoint union of A and B,

0.B If we define 0 by

0 ,E E B

UNIT III CHAPTER 10

230

we have 0( ) 0A and so 0 . Let

1 .E A

E E A g d

Then 0 1, and we have only to show that 1 . Let E be a set of measure zero.

Then

0 ,E

E f d

and f = 0 a.e. on E. Since 0f on ,A E we must have 0.A E Hence

0,A E and so 1 0.E A E

___________________

UNIVERSITY QUESTION PAPER

231

M.Sc. (PREVIOUS) DEGREE EXAMINATION, MAY 2008

Mathematics

Paper III – REAL ANALYSIS

(2000 Admission Onwards)

(Regular – For SDE candidates)

Time: Three Hours Maximum: 120 Marks

Answer all questions from Part A.

Each question carries 4 marks.

Answer any four questions from Part B without omitting any Unit.


Part A

1. (i) Describe the least upper bound property of ordered sets. Verify whether the ordered

field of rationals have this property.

(ii) Let Y be the unit circle and let :[0, 2 )f Y be defined by ( ) (cos , sin ).f t t t

Verify that f is one-to-one and onto.

(iii) Let f be defined on [ , ]a b with ( )m f x M for all x. With the usual notations

prove that ( ) ( , ) ( , ) ( ),m b a L P f U P f M b a

(iv) Prove that Lebesgue measure of a singleton set is zero.

(v) If 1 2 1 21 2 1 2E E F Fa a b b then is it necessary that 1 1a b and 2 2 ?a b

Justify you answer.

(vi) Give an example of a measure which is different from the Lebesgue measure m

and satisfying .m

(6 4 = 24 marks)

Part B

UNIT I

2. (a) Define neighborhood of a point in a metric space. Show that every neighborhood is

an open set.

(b) Show that if p is a limit point of a set E then every neighborhood of p contains

infinitely many points of E.


232

3. (a) Let X, Y, Z be metric spaces, :f X Y and : .g Y Z Show that if f is

continuous at p and g is continuous at ( )f p then ( ) ( ( ))h x f g x is continuous at

p.

(b) Show that :f X Y is continuous at every point at x if and only if 1( )f V is

open in X whenever V is open in Y.

4. (a) Let f be defined on [ , ]a b and suppose that f has a local maximum at ( , ).x a b

Show that if ( )f x exists then ( ) 0.f x

(b) Let f be continuous on [ , ]a b and differentiable on ( , ).a b Show that there exists

( , )x a b such that ( ) ( ) ( ) ( ).f b f a b a f x

UNIT II

5. (a) Let be increasing on [ , ]a b . Show that ( )f on [ , ]a b if and only if for

every 0 there exists a partition P such that ( , , ) ( , , ) .U P f L P f

(b) Show that if f is continuous on [ , ]a b and is increasing on [ , ]a b then ( )f

on [ , ]a b .

6. (a) Define equicontinuous family of functions.

(b) Let K be a compact metric space and ( )nf K for 1,2,3,n show that:

(i) if nf is uniformly convergent in K then nf is equicontinuous.

(ii) if nf is equicontinous and point wise bounded then nf contains a

uniformly convergent subsequence.

7. (a) Let nA be a sequence of sets of real numbers and *m be the Lebesgue outer

measure. Show that * * .n nm A m A

(b) Prove that if A is countable, then * 0.m A

UNIT III

8. (a) State and prove Fatou’s Lemma.

(b) State and prove Lebesgue convergence theorem.


233

9. Let f be increasing real valued function on [ , ]a b . Show that

(a) f is differentiable a.e. on [ , ]a b .

(b) the derivative f is measurable.

10. (a) Define positive set with respect to a signed measure . Show that if E is

measurable and 0 ,E then E contains a positive set A with 0.E

(b) Prove that the union of a countable collection of positive sets is positive.

(4 24 = 96 marks)

_______________________

M.Sc. (PREVIOUS) DEGREE EXAMINATION, MARCH/APRIL 2006

Mathematics

Paper III – REAL ANALYSIS

(2000 Admission Onwards)

(Regular – For SDE candidates)

Time: Three Hours Maximum: 120 Marks

Answer all questions from Part A.


Answer any four questions from Part B without omitting any Unit.


Part A

1. (i) Prove that there is no rational number whose square is 12.

(ii) Assume that f is a continuous real function defined in ( , )a b such that

2 2

f x f yx yf

for all , ( , ).x y a b Prove that f is convex.

(iii) Evaluate 5

0

x d x x where x denotes the largest integer not greater than x.

(iv) Show that canotor ternary set has measure zero.


234

(v) Let 1

,n

i Eii

a

with i jE E for i j . Suppose each set iE is a measurable

set of finite measure. Show that 1

.n

i i

i

a mE

(vi) Prove that 2

2, 0( )

, 0

xf x

x x

is measurable.

(6 4 = 24 marks)

Part B

UNIT I

2. (a) For every real x > 0 and every integer n > 0 prove that there is one and only one

real y such that ny x . (8 marks)

(b) If 1, , na a and 1, , nb b are complex numbers, then show that

22 2

1 1 1

.n n n

j j j jj j j

a b a b

(8 marks)

(c) Suppose , .k k a b Find kc and 0r such that 2 x a x b if

and only if .r x c (8 marks)

3. (a) Let f be a continuous mapping of a compact metric space X into a metric space Y.

Show that f is uniformly continuous on X. (8 marks)

(b) Let f be monotonically increasing on ( , ).a b Show that ( )f x and ( )f x exist

at every point ( , ).x a b Further show that

sup ( ) ( ) ( ) ( ) inf ( )a t xa t x

f t f x f x f x f t

and if a x y b , then ( ) ( ).f x f y (8 + 4 =12 marks)

(c) Let f be monotonic on ( , ).a b Then the set of points of ( , )a b at which f is

discontinuous is at most countable. (4 marks)

4. (a) State and prove Taylor’s theorem in 1. (8 marks)

(b) Suppose f is a continuous mapping of [ , ]a b into k and f is differentiable in

( , ).a b Then there exists ( , )x a b such that ( ) ( ) ( ) ( ) .b a b a x f f f


235

(8 marks)

(c) Suppose f is defined in a neighborhood of x, and suppose ( )f x exists. Show

that 20

( ) ( ) 2 ( )lim ( ).h

f x h f x h f xf x

h

Show by an example that the limit

may exist even if ( )f x does not. (8 marks)

UNIT II

5. (a) Show that ( )f on [ , ]a b if and only if for every 0 there exists a

partition P such that ( , , ) ( , , ) .U P f L P f (12 marks)

(b) Assume increases monotonically and . Let f be a bounded real function

on [ , ].a b Then show that ( )f if and only if .f In that case

( ) ( ) .b b

a a

f d f x x dx (12 marks)

6. (a) ]Suppose nf f uniformly on a set E in a metric space. Let x be a limit

point of E, and suppose that lim ( ) 1,2,3, .n nt x

f t A n

Then show that nA

converges, and lim ( ) lim .nt x n

f t A

(12 marks)

(b) Suppose nf is a sequence of functions, differentiable on [ , ]a b and such that

0( )nf x converges from some point 0 [ , ].x a b If nf converges uniformly on

[ , ]a b , then nf converges uniformly on [ , ]a b , to a function f, and

( ) lim ( ) .nn

f x f x a x b

(12 marks)

7. (a) If K is compact, if ( )nf K for 1,2,3,n , and if nf is pointwise

bounded and equicontinuous on K, then show that

( i) nf is uniformly bounded on K.

( ii) nf contains a uniformly convergent subsequence. (12 marks)


236

(b) Prove that every Borel set is measurable. (6 marks)

(c) Construct a non-measurable set. (6 marks)

UNIT III

8. (a) State and prove Lebesgue (dominated) convergence theorem. Is the condition

“dominated” necessary. Justify your answer. (12 marks)

(b) Let nf be a sequence of measurable functions that converges in measure to f.

Show that there is a subsequence knf that converges to f almost everywhere.

(12 marks)

9. (a) Let f be an increasing real-valued function on the interval [ , ]a b .Prove that f

is differentiable almost everywhere and the derivative f is measurable, and

( ) ( ) ( ).

b

a

f x dx f b f a (12 marks)

(b) If f is an integrable function on [ , ]a b , show that F defined on [ , ]a b by

( ) ( )

x

a

F x f t dt is a continuous function of bounded variation on [ , ]a b .

(12 marks)

10. (a) If f is bounded and measurable on [ , ]a b and ( ) ( ) ( ),

x

a

F x f t dt F a then

show that ( ) ( )F x f x for almost all x in [ , ]a b . (12 marks)

(b) Let X ,B be a measurable space, n a sequence of measures that converge

setwise to a measure , and nf a sequence of nonnegative measurable functions

that converge pointwise to the function f. Prove that lim .n nf f d

(12 marks)

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