Integration

MATH314: Lent 2006

Lecture times: weeks 16–20,Monday 15:00, Bowland North Seminar Room 19;Thursday 17:00, Faraday Seminar Room 2;Friday 12:00, Furness B11.

Tutorials: Weeks 17–20,Tuesday 13.00, Fylde B49; Thursday 10.00, Fylde B67.

Lecturer: Dr John Orr; email jorr@math.unl.edu ; Tel (5)93945; Office B21, Fylde Coll.

Office hours: tba.

Assessment: 10% coursework, based on four assignments, and 90% from Final examina-tion. In the examination, section A consists of questions worth a total of 50 marks, butthe total section A mark of each student is capped at 40, and section B consists of twoquestions from a choice of three, so that each student can gain at most 60 marks fromsection B.

Exam papers: http://www.maths.lancs.ac.uk/department/study/exams/archive/level3The exam papers are in .pdf format and can be read using acroread.

Website: Notes, graphs, tutorial questions and solutions will be posted onhttp://www.math.unl.edu/∼jorr/math314

Aims: The aim of this course is to introduce the Lebesgue integral for functions on thereal line. The course features a classical approach in constructing Lebesgue measure onthe line and in defining the integral. The convergence theorems are proved, taking thebounded convergence theorem as fundamental. Applications to some classical convergenceproblems, including the Fourier inversion theorem, illustrate the power of the results.

Description: In this module we construct Lebesgue measure on the line, which extendsthe idea of the length of an interval. This is used to define an integral which is shown tohave good properties under pointwise convergence.

The power of the convergence theorems will be illustrated in applications to someclassical limit problems, including the analysis of Fourier integrals which are fundamentalto probability theory and differential equations.

Syllabus: Lebesgue’s definition of the integral. Integral of a step function. Subsets of thereal line; open sets and countable sets. Measure of an open set. Measurable sets and nullsets.

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Integrable functions. Lebesgue integral of a bounded function. Lebesgue bounded con-vergence theorem. Lebesgue integral of an unbounded function. Dominated convergencetheorem; monotone convergence theorem.

Applications of the convergence theorems. Wallis’s product for π. The Fourier cosineinversion formula.

MATH 314 Contents

§0. Introduction. The problem of area. Dirichlet’s comb. Lebesgue’s

definition of the integral. Measure. Convergence problem. Applications.

§1. Covering and measure. 1.1 Countable sets. 1.4 Rationals and irrationals.Archimedes’ axiom. 1.5 Cantor’s theorem. 1.6 Subsets of the real line. Open sets.1.7 Covering. Measure of an open set. 1.8 Coverings by open sets, Heine–BorelCovering Theorem. 1.9 Outer measure, null sets. 1.10 Subadditivity of measureon open sets. 1.11 Inner measure and measurable sets. 1.12 Overlap lemma. 1.13Subadditivity of outer measure. 1.14. Main theorem. 1.15∗ Lebesgue measurablesets form a σ-algebra.§2. The Lebesgue integral. 2.1 Measurable functions. 2.2 Continuous func-tions are measurable. 2.3 Definition of the integral of a bounded measurablefunction. 2.4 Fundamental theorem of calculus. 2.5 Integration of unboundedfunctions. 2.6 Properties of integral. 2.7∗ L1 and the Hilbert space L2.§3. Convergence theorems. 3.1 Lebesgue bounded convergence theorem.Bessel function. 3.2 Monotone Convergence Theorem. 3.3 Inequality of themeans. 3.4 Gaussian integral. 3.5 Wallis’s product for π. 3.6 Lindemann’s the-orem, 3.7 Dominated Convergence Theorem. 3.7 Series for Bessel function. 3.8∗

Completeness theorem.§4. Fourier integrals. 4.1 Fourier integrals. 4.2 Integrated Inversion theorem.4.3∗ L1 cosine inversion theorem. 4.4∗ Plancherel cosine formula.

∗ The topics that are marked with an asterisk will not be covered in lectures, nor examined;

their purpose is to provide links between the content of the preceding chapter and other

courses such as MATH313 and 317. In the notes there are also a few proofs that will be

covered in detail in lectures, but will not be examined as bookwork since they are tricky

to remember. There are also a few historical remarks. These extra sections start beneath

horizontal rules, and finish with square boxes.

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Recommended Text

M. Capinski and E. Kopp, Measure, Integration and Probability , Second edition, Springer,London, 2004.

Other books of interest

E.C. Titchmarsh, The Theory of Functions, Second Edition, OUP Oxford, 1939.R.L. Schilling, Measures, Integrals and Martingales, Cambridge University Press, Cam-bridge, 2005.H.S. Bear, A Primer of Lebesgue Integration, Academic Press London, 1995.F. Jones, Lebesgue integration on Euclidean space, Jones and Bartlett, Boston Mass, 1993.H. Lebesgue, Lecons sur L’Integration, Second Edition, Gauthier–Villars, Paris, 1950.

§0. Introduction

At the end of this chapter, you should be able to:

understand the basic concepts of Lebesgue’s definition of the integral;

appreciate Dirichlet’s comb function.

Integration is concerned with working out areas, for instance the area under the graphof a real function. Once one has discovered how to do this, one can proceed to calculateareas of fields, volumes of wine jars and so forth. So we begin with the fundamentalproblem.

0.1 PROBLEM. How to define the area under the graph of a bounded real function?

Riemann introduced an integral as in MATH210. Suppose that f : [a, b] → R is abounded function on a bounded interval, so that |f(x)| ≤ M for all x. One begins bypartitioning the domain of f by introducing

P = a = x0 < x1 < x2 < · · · < xn = b. (0.1)

Now consider the function f(x) when x is restricted to lie in [xj , xj+1] and introduceλj = inff(x) : x ∈ [xj , xj+1] and Λj = supf(x) : x ∈ [xj , xj+1] such that

λj ≤ f(x) ≤ Λj on [xj , xj+1].

One can draw a rectangle with height λj on [xj , xj+1], which lies below the graphof f, and draw a larger rectangle with height Mj on the same base, which lies above thegraph of f. Certainly we wish to define

(area of rectangle) = (base) × (height);

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and whatever precise interpretation is to be put upon the notion of area, we should have

(total area of inscribed rectangles) ≤ (area under graph of f ) ≤ (total area of outer rectangles).

Consequently, for any integral we should have

n−1∑j=0

λj(xj+1 − xj) ≤∫ b

a

f(x)dx ≤n−1∑j=0

Λj(xj+1 − xj).

CONTINUOUS. A real function f is continuous at x if for each ε > 0 there exists δ > 0

such that |f(y)− f(x)| < ε whenever |x− y| < δ. Equivalently, for any sequence (xn) such

that xn → x as n→∞, it follows that f(xn) → f(x) as n→∞.

0.2 PROPOSITION. Suppose that f : [a, b] → R is continuous at each x ∈ [a, b]. Then

f is bounded and when one refines the partition so that maxj(xj+1 − xj) → 0, the total

area of the outer rectangles and the total area of the inscribed rectangles converge to a

common limit.

This limit we define to be∫ b

af(x) dx, the Riemann integral of f. Indeed, the Riemann

integral may be defined for any function f such that the outer and inner rectangles have

areas converging to a common limit under refinement.

This integral is useful for some purposes, but has nasty properties under pointwise

convergence of sequences of functions.

0.3 EXAMPLE (Dirichlet’s Comb.) For each k = 1, 2, . . . we introduce the function

fk(x) = 1, for x = m/2k,m = 0, 1, 2, . . . , 2k;

0, else. (0.2)

The graph of fk consists of 2k + 1 spikes with no width, and it is easy to verify that∫01fk(x)dx = 0.

The function f realised as the pointwise limit of the fk has the form

f(x) = limk→∞

fk(x) = 1, for x = m/2k, 0 ≤ m ≤ 2k, k ≥ 0;

0, else.

This function f has a graph which resembles a comb with infinitely many teeth of zero

width. One can show that f is discontinuous at all points of (0, 1) and does not have a

Riemann integral.

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This example may appear unnatural; but pointwise convergence is an important meansof defining functions, and examples of this kind occur frequently in the theory of Fourierseries. So a new definition of the integral is required: an integral that takes greatest accountof the properties of the integrand on large sets, and that can neglect bad behaviour onsmall sets.

Such an integral was introduced by Lebesgue in his thesis of 1902. (Note that hisname is spelt with a ‘g’ and not a ‘q’.) The detailed properties of this integral will bedescribed in this course, yet they depend upon simple principles which will be stated now.

For a bounded function f : [a, b] → R, we partition the range by introducing

P = A = y0 < y1 < y2 < · · · < yn = yn+1 = B. (0.3)

We then consider the set Ek = x : yy < f(x) ≤ yk+1, where f takes values between yk

and yk+1. This set may have complicated properties, but we have good information aboutthe behaviour of f upon Ek.

A real function is said to be simple, if it takes only finitely many values. To approx-imate f, we introduce the lower simple function sL and the upper simple function sU ,

where sL(x) = yk and sU (x) = yk+1 on Ek. These have sL ≤ f ≤ sU , so we should have

(area under graph of sL) ≤ (area under graph of f ) ≤ (area under graph of sU . )

In order to calculate the areas under the graphs of simple functions, we need to measurethe size of sets such as Ek, which may be of complicated shape. We shall introduce thenotion of Lebesgue measure m(Ek), which generalizes the notion of length, and which isdefined for all sets Ek that arise from reasonable (measurable) functions. Once we havedone this, can use the principle

area = (measure of base) × (height)

to calculate the area under the graph of a simple function. We obtain on adding up therespective areas∫

sL(x)dx =n∑

k=0

ykm(Ek) ≤n∑

k=0

yk+1m(Ek) =∫sU (x)dx;

and the integral should lie between these sums. Finally we can show that the areas underthe graphs of the lower and upper simple function converge to a common limit as thepartition refines. This limit gives us the Lebesgue integral.

For a collection A of subsets E of a non-empty set Ω, we can consider the followingconditions.

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(σ1) Ω ∈ A.(σ2) If E ∈ A, then Ω \ E ∈ A.(σ3) If En ∈ A for n = 1, 2, . . . , then ∪∞n=1En also belongs to A.

In chapter one we shall see that the notion of length of an interval may be generalizedto measure m(E), defined for a wide class of subsets E of a bounded interval (a, b). Suchsets are called Lebesgue measurable, and include all the sets that occur in elementaryanalysis. When Ω = (a, b), the set A of all measurable subsets of (a, b) satisfies (σ1), (σ2)and (σ3). The fundamental property of Lebesgue measure is its countable additivity.

0.4 THEOREM (Lebesgue). If (En) (n = 1, 2, . . . ) is a sequence of disjoint measurable

sets in (a, b), then their union E = ∪∞n=1En is measurable with measure

(Σ) m(E) =∞∑

n=1

m(En). (0.4)

The proof makes use of the Heine–Borel Theorem, a fundamental result about openintervals and closed sets in the real line. Although the steps are relatively simple, thearchitecture of the proof is subtle. The structural diagram provided below should help thereader fit the parts of the proof together.

Equipped with this measure, we then attempt to integrate a correspondingly wideclass of real functions f , called the measurable functions. This class includes continuousfunctions and their limits under standard convergence operations involving sequences. Theintegral will be constructed in chapter two, using the method sketched above.

0.5 THEOREM (Lebesgue). Let f be a bounded measurable function on a bounded

interval [a, b]. Then f has a well–defined integral which satisfies the following.

(P ) Positivity:∫ b

af(x) dx ≥ 0 if f ≥ 0.

(Lin) Linearity:∫ b

a

(λf(x) + µg(x))dx = λ

∫ b

a

f(x) dx+ µ

∫ b

a

g(x) dx, (0.5)

for scalars λ and µ.

(C) Furthermore, if f : [a, b] → R is continuous, then it is bounded and measurable,

and its Lebesgue integral equals its Riemann integral.

A key feature of the Lebesgue integral is absolute convergence; when f is Lebesgueintegrable,

∫|f(x)|dx < ∞. This property make it easy to extend the definition of the

integral to a unbounded measurable functions on the line without ambiguity.

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For any probability space (Ω,P), the probability P(E) of an event E gives a notion of

the measure of E. The Lebesgue integral may be defined for bounded measurable functions

(random variables) on any probability space (Ω,P) ; thus one can define the expectation

EX of any bounded random variable X by∫ΩXdP. The proof of the main existence

theorem 2.3 mainly uses the additivity of Lebesgue measure.

In chapter 3 we consider the convergence problem:

0.6 PROBLEM. Let (fn) (n = 1, 2, . . . ) be a sequence of integrable functions with

fn(x) → f(x) as n→∞. Under what conditions does∫fn(x) dx→

∫f(x) dx ?

The Lebesgue integral is well suited to treating this question. The main sufficient

conditions for convergence of sequences of integrals are stated in Lebesgue’s bounded,

monotone, and dominated convergence theorems, stated as Theorems 3.1, 3.2 and 3.7

below. An example of where we shall apply them is in evaluating the Gaussian integral as

the limit ∫ √n

−√

n

(1− x2

n

)n

dx→∫ ∞

−∞e−x2dx (n→∞). (0.6)

The left-hand side may be computed using Wallis’s product for π. Chapter 3 contains

examples of this kind, some of which are of historical significance.

The convergence theorems also hold for the Lebesgue integral∫ΩXdP on a probability

space (Ω,P). The proofs extend in the obvious way.

Fourier integrals

An important tool in probability theory and differential equations is the Fourier trans-

form

f(t) =∫ ∞

−∞e−ixtf(x) dx (t ∈ R). (0.7)

The characteristic function of a probability density function is defined by a similar (but

slightly different) formula. In chapter 4 we consider the application of the Lebesgue conver-

gence theorems to the theory of the Fourier integral. By introducing Lebesgue integrable

functions, it is possible to obtain simpler and more complete statements of the main in-

version theorems for the Fourier transform.

Notation. Throughout, R denotes the real numbers, and x, t are real variables. We

reserve m for Lebesgue measure.

LIMINF and LIMSUP. Let (aj) (j = 1, 2, . . . ) be a bounded sequence of real numbers.

Then Aj = infak : k ≥ j gives an increasing sequence which is bounded above, and we

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can definelim infj→∞

aj = α = limj→∞

Aj ,

Similarly, Bj = supak : k ≥ j gives a decreasing sequence which is bounded below, andwe can define

lim supj→∞

aj = β = limj→∞

Bj .

The sequence (aj) converges if and only if α = β ; in this case, its limit is the commonvalue of these quantities.

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§1. Covering and measure

At the end of this chapter, you should be able to:

prove basic properties of countable sets;

understand Archimedes’ axiom and Cantor’s uncountability theorem;

prove the structure theorem for open sets;

prove covering lemmas for open sets;

understand the statement (not proof) of the Heine–Borel Covering Theo-

rem;

understand outer measure and use it in proofs;

understand the notion of inner measure;

prove Lebesgue’s theorem on countable additivity of measure.

1.1 Counting

SURJECTIVE. Let ϕ : T → S be a function. Then ϕ is surjective (or onto) if for

each s ∈ S there exists t ∈ T such that ϕ(t) = s. We shall often write ψ : T →→ S to

emphasize this.

INJECTIVE. Let ϕ : T → S be a function. Then ϕ is injective if ϕ(t1) = ϕ(t2) implies

t1 = t2.

COUNTABLE SETS. A set S is countable, if there exists a counting function ψ : N → S

such that for each s ∈ S there exists n ∈ N with ψ(n) = s. We do not require the counting

function to be injective (one–to–one), so each point may be counted many times over.

Bearing this in mind, we see that any finite set is countable. One can easily show that the

integers Z form a countable set.

1.1 LEMMA. The set N× N of ordered pairs of natural numbers is countable.

Proof. The required counting function is obtained by going along the cross–diagonals

of the following matrix

(1, 1) (1, 2) (1, 3) (1, 4) . . .

(2, 1) (2, 2) (2, 3) (2, 4) . . .

(3, 1) (3, 2) (3, 3) (3, 4) . . .

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(4, 1) (4, 2) (4, 3) (4, 4) . . . (1.1)

The elements in this array may be listed according to the numbers of steps required toreach them, as in

k : 1 2 3 4 5 6 7 . . .

ψ(k) : (1, 1) (1, 2) (2, 1) (3, 1) (2, 2) (1, 3) (1, 4) . . . . (1.2)

One can verify that every point in the array is eventually covered, so ψ is surjective.

1.2 COROLLARY. The set Q of rational numbers is countable.

Proof. A rational number q may be expressed as a fraction q = k/n where k is aninteger, and n a positive integer. Hence there is a surjective map Z × N →→ Q, definedby (k, n) 7→ k/n. Further, there is a surjective map N →→ Z, since Z is countable.Composing these, we obtain a surjective map N × N →→ Z × N. By Lemma 1.1, N × Nis countable, so we can form a composition of surjective maps

N →→ N× N →→ Z× N →→ Q; (1.3)

one checks that this is surjective.

1.3 PROPOSITION. (i) If T is countable, and S is a subset of T, then S is also

countable.

(ii) If S and T are countable, then their Cartesian product S × T is countable.

(iii) If (Tj)j∈N is a sequence of sets with Tj countable for each j, then their union

∪∞j=1Tj is also countable.

Proof. (i) Since T is countable, there exists a counting function φ : N →→ T. Tocount out the smaller set S, we take s ∈ S and define ψ : T →→ S by

ψ(u) = u, if u ∈ S;

s, if u ∈ T − S. (1.4)

The composition ψ φ : N →→ T →→ S gives the required counting function.(ii) The work for this is essentially contained in the proof of Lemma 1.1. Let φ :

N →→ S and ψ : N →→ T be counting functions, and form the map N × N →→ S × T

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by (n,m) 7→ (φ(n), ψ(m)). We can now use the map of Lemma 1.1 to form a compositionof surjections

N →→ N× N →→ S × T. (1.5)

(iii) One can consider the set T as being ordered like the words in a dictionary, withTj representing the set of words containing j. To count all the elements of T we introduceφj : N →→ Tj and combine these to form a map N×N →→ T by (j, n) 7→ φj(n) ∈ Tj ⊆ T.

Thus we can form a surjection N →→ N× N →→ T.

1.3 EXAMPLE. The set of English words is countable.

For the purpose of this exercise, an English word as a finite and ordered string of lettersfrom the Roman alphabet, such as misunderestimate. We let Sj be the set of words withj letters, which has 26j members. The set of words is S =

⋃∞j=1 Sj , a countable union of

finite sets. By Proposition 1.3 (iii), S is countable.

1.4 On rationals and irrationals

RATIONALS. A real number x is said to be rational if it can be written in the formx = p/q, with p and q integers; it is irrational otherwise.

Examples of rationals include: 1.4, 3,−5.6, 1/3 etc.Examples of irrationals include:√

2 , as shown by Pythagoras;√3 , as shown by Theodorus c. 400BC;

π and the base e of natural logarithms, as shown by Lambert in 1761.Open problem: Let γ = limn→∞(1+1/2+1/3+ · · ·+1/n− log n) be Euler’s constant.

Is γ rational?

1.4 ARCHIMEDES’ AXIOM. Between any two distinct real numbers lies some rational

number; or, equivalently, for any δ > 0 there exists some integer n with nδ > 1.

This axiom is incorporated into all constructions of the real numbers from the ratio-nals. It states that the rational numbers are dense in the real numbers.

1.5 The set of real numbers is uncountable.

1.5 CANTOR’S THEOREM. The set of real numbers is uncountably infinite.

Cantor’s diagonal argument shows that the real numbers are not countable. The de-tails of this are contained in exercise 1.5. In these notes an alternative, indirect, proof isgiven. The very existence of Lebesgue measure proves that the real numbers are uncount-

able. See Proposition 1.9 and Lemma 1.10.

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1.5 Exercise. (i) Show that each real number x with 0 ≤ x < 1 may be expresseduniquely as a decimal x = 0.a1a2a3 . . . , where aj = 0, 1, 2, . . . , 9 ; excluding recurringdecimals such as 0.0124999999 . . . , which may be expressed as finite decimals such as0.0125.

(ii) Let (xj), (j = 1, 2, . . . ) be any list of real numbers, and let the decimal expansionof xj give the entries of the jth row of the matrix [Xjk]j,k=1,2,.... Show that

ak = Xkk + 1, for Xkk = 0, 1, 2, . . . , 7;

0, for Xkk = 8, 9

gives a number x = 0.a1a2a3 . . . in [0, 1) that is not on the list.(iii) Deduce that the real numbers are uncountable.

ALGEBRAIC NUMBERS are the real numbers x that may be expressed as the root ofsome non–trivial polynomial equation with integral coefficients; otherwise, x is said to betranscendental.

Examples of algebraic numbers include all rationals (for p/q is a root of qX − p = 0 );surds such as

√2 (a root of X2− 2 = 0 ); and roots of polynomials of degree higher than

four, which cannot generally be expressed as surds.Given a unit length in the plane, it is of interest to determine which real numbers one

can construct as lengths, using only straight edge and compass, with standard operations.(For instance, one can construct

√2 as the hypotenuse of a right-angled isosceles triangle

with short sides of unit length.) Hilbert showed that all the numbers that may be soconstructed are algebraic. See theorem 3.6 below.

To exhibit specific examples of transcendental numbers is difficult. Examples include:e (Hermite 1872); π (Lindemann); Liouville numbers such as

∑∞n=1 10−n!.

1.6 Subsets of the line

BOUNDED SETS. A subset F of the real line R is said to be bounded, if there existsM < ∞ such that |x| ≤ M for all x ∈ F ; or equivalently, if there exists M < ∞ suchthat F ⊆ [−M,M ].

OPEN SETS. A subset E of the real line R is said to be open if, for each x ∈ E, thereexists ε > 0 such that (x− ε, x+ ε) ⊆ E. Any interval of the form (a, b) is open, and weshall show that any open set is a countable union of such open intervals.

CLOSED SETS. A subset S ⊆ R is said to be closed if its complement Sc = R \ S isopen.

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Note that a set may be neither open nor closed; further, the sets R and ∅ are both

open and closed.

1.6 LEMMA (i) The set (a, b) is open for each a < b.

(ii) If Ej is open for each j in some index set J, then their union ∪j∈JEj is open.

(iii) Let E and F be open sets. Then their intersection E ∩ F is also open.

Proof. (ii) If x ∈⋃

j Ej , then there is some j such that x ∈ Ej . Since Ej is open,

there exists ε > 0 such that (x−ε, x+ε) ⊆ Ej . But Ej ⊆ ∪kEk, so (x−ε, x+ε) ⊆ ∪kEk.

(iii) Tutorial exercise.

Regarding (iii), we observe that the intersection of infinitely many open sets may not

be open.

The structure of open sets.

1.6 PROPOSITION. Let E be a non–empty open subset of R. Then there are mutually

disjoint and non–empty open intervals (aj , bj) with E = ∪j(aj , bj). This representation

is unique (up to reordering), and the family of open intervals in the union is finite or

countably infinite.

Proof. We wish to split up E into sets Ej which are connected line segments. To do

this, we introduce an equivalence relation, called path–connectedness, on E. For x and y

in E, we define

x ∼ y, if [x, y] ⊆ E or [y, x] ⊆ E. (1.6)

We need to have two conditions here to allow for the possibilities x ≤ y and y ≤ x,

respectively. The relation ‘∼ ’ has the following properties:

Symmetric, for if x ∼ y then clearly y ∼ x ;

Reflexive, x ∼ x since [x, x] = x ⊆ E, for any x ∈ E ; and

Transitive, when x ∼ y and y ∼ z it follows that x ∼ z. In checking this we can

suppose without much loss that x < y < z. Then we have [x, y] ⊆ E and [y, z] ⊆ E, so

their union has [x, z] ⊆ E.

By the general theory of relations, E may be expressed as the union of disjoint

equivalence classes Ej , that are uniquely determined by the relation. We shall prove

that the Ej are open intervals.

To see that they are open, we take x ∈ Ej , for some j. Then x ∈ E, and since E

is open, there exists ε >> 0 such that (x − ε, x + ε) ⊆ E. We can check in each case

that this neighbourhood is actually contained in Ej . For instance, if x < y < x+ ε, then

we have [x, y] ⊆ (x − ε, x + ε) ⊆ E and hence x ∼ y, which implies that x and y lie in

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the same equivalence class Ej . The union of the open intervals (x − ε, x + δ) that have

the element x and that are contained in Ej plainly gives an open interval; this union

must exhaust Ej . To check this, consider (x1, x2), where x2 = supy : [x, y] ⊂ E and

x1 = infz : [z, x] ⊂ E.We now prove that there are only countably many Ej . The simplest way to count

the intervals Ej is to select some rational number qj ∈ Ej ; here we invoke Archimedes’

Axiom. Each qj , so chosen, determines the equivalence class and, by Corollary 1.2, and

Proposition 1.3 (i) there are only countably many such rationals.

REMARK. The natural way to draw a given collection of intervals is by ordering them

by their left-hand endpoints. But there are uncountably many possible left-hand endpoints,

so this does not give a satisfactory counting scheme for the equivalence classes. Hence in

the preceding proof we count intervals by using rational elements.

1.6 LEBESGUE MEASURE of an open set is the sum of the lengths of its constituent

intervals; if E = ∪∞j=1Ej , as in Proposition 1.6, with Ej = (aj , bj), then

m(E) =∞∑

j=1

(bj − aj). (1.7)

On account of Dirichlet’s Theorem, the sum does not depend upon the order of the intervals

and hence the definition is unambiguous.

1.6 Exercise. If E is an open subset of (a, b), where a and b are finite, then m(E) ≤b− a.

1.7 Covering

Let K be a subset of R, and let Ij : j ∈ J be a collection of subsets of R. We say

that Ij : j ∈ J is a cover of K if K ⊆⋃

j∈J Ij . We abbreviate this by saying that the

Ij cover K. We shall sometimes speak of the length of the cover, meaning the number of

covering sets Ij ; this should be distinguished from the measure of the covering sets.

1.7 COVERING LEMMA. Suppose that (a, b) ⊆⋃n

k=1(ak, bk), where the (ak, bk) are

not necessarily disjoint. Then the total measure of the sets in the cover is as large as the

measure of the set covered, so

(b− a) ≤n∑

k=1

(bk − ak). (1.8)

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Proof. We prove this statement by induction on n, the length of the cover. The resultis clear for covers of length n = 1. Suppose that the result is valid for all covers of length< n, and consider (a, b) ⊆ ∪n

j=1(aj , bj). First suppose that there is some index j such thatwe still have a cover on discarding (aj , bj) ; then we say that (aj , bj) is redundant. Sincethe sum in (1.8) increases as we take more intervals, we can discard redundant intervalsfrom the cover and then invoke the induction hypotheses to obtain the stated result.

Otherwise, there are no redundant intervals in the cover. We order them so thata1 ≤ a < a2 < b1 < b2, and we have (a2, b) ⊆

⋃nk=2(ak, bk). This is a cover of length

n− 1, so by the induction hypothesis b− a2 ≤∑n

k=2(bk − ak). We deduce that

b− a = b− a2 + a2 − a ≤ b1 − a1 +n∑

k=2

(bk − ak), (1.9)

as required.

1.8 PROPOSITION. Lebesgue measure is countably additive on open sets. Let (En)(n = 1, 2, . . . ) be a sequence of disjoint open sets and let E =

⋃∞n=1En. Then

(Σ) m(E) =∞∑

n=1

m(En). (1.10)

Proof. By Proposition 1.6, each En may be written as a disjoint union of openintervals En = ∪∞k=1(a

(n)k , b

(n)k ). Consequently we can write

E =∞⋃

n=1

En =∞⋃

n=1

∞⋃k=1

(a(n)k , b

(n)k ); (1.11)

this is a disjoint union of open intervals, since the En are themselves disjoint. By the verydefinition of Lebesgue measure we can write

m(E) =∞∑

n,k=1

(b(n)k − a

(n)k ), (1.12)

and we can reorder the sum by Dirichlet’s Theorem to get

m(E) =∞∑

n=1

∞∑k=1

(b(n)k − a

(n)k ) =

∞∑n=1

m(En). (1.13)

1.8 Covering by open sets

15

There are various statements which one can make about the measure of an open set Ewhich are evident when E is a finite union of open intervals, but which are less clear whenE is an infinite union of open intervals. The Heine–Borel covering theorem enables usto replace infinite unions of intervals by finite unions. The Heine–Borel covering theoremhas two important consequences in the construction of Lebesgue measure, namely Lemmas1.10 and 1.12.

1.8 HEINE–BOREL COVERING THEOREM. Suppose that K is a closed and bounded

subset of real line R, and let (Ij)j∈J be a collection of open sets that covers K. Then

there exists a finite subcollection of the open sets Ij that also covers K.

Proof by bisection.∗ Suppose K ⊆ [a, b], and let c be the midpoint of [a, b]. If wecan cover the left half K∩ [a, c] by finitely many Ij , and the right half K∩ [c, b] by finitelymany Ij , then we can combine these covers to cover all of K by finitely many Ij , and weare done. Otherwise, we select the half that we cannot cover with finitely many Ij , takingthe right half if we cannot cover either half finitely.

This gives a set K ∩ [a1, b1] which is covered by all the Ij ; we let c1 be the midpointof [a1, b1] and attempt to cover K ∩ [a1, c1] and K ∩ [c1, b1] finitely. By the choice of[a1, b1], we cannot cover both halves with finitely many Ij , so we pick the (right) half thatwe cannot finitely cover.

Proceeding in this way, we obtain ak and bk with:

(i) (ak) is an increasing sequence, bounded above by b ;

(ii) 0 ≤ bk − ak ≤ 2−k(b− a) ;

(iii) [ak, bk] ∩K cannot be covered by finitely many Ij ; but

(iv) K ⊆ ∪jIj .

By condition (i), ak → α as k →∞, for some α ∈ [a, b]. If α is not in K, then sinceK is a closed set with open complement, [ak, bk] does not intersect K for k sufficientlylarge, by (ii). But this contradicts (iii). So α ∈ K, and by (iv), there exists j with α ∈ Ij .Further, Ij is open, so there exists ε >> 0 with (α − ε, α + ε) ⊆ Ij . But then by (ii) wehave

[ak, bk] ⊆ (α− ε, α+ ε) ⊆ Ij (1.14)

for all sufficiently large k, contradicting (iii).

COMPACTNESS∗. The property of a set of real numbers, that every open cover has afinite subcover, is called compactness. This property is characterized by the Heine–BorelCovering Theorem and its converse, Theorem 1.8’.∗ Not covered in lectures.

16

1.8 CONVERSE TO THE HEINE–BOREL COVERING THEOREM. Suppose that K is a

subset of the real line which is either not closed or not bounded. Then K has an open

cover that has no finite subcover.

Proof. This is essentially contained in tutorial exercise 8.

1.9 Outer measure

We have seen how to define the Lebesgue measure of an open set, and shown that

Lebesgue measure is countably additive on the collection of open sets. We shall extend

the definition of Lebesgue measure to a much more general class of sets, the measurable

sets, and then prove that the measure is countably additive.

OUTER or EXTERIOR MEASURE. Let E be an open set as in Proposition 1.5. The

Lebesgue measure of E is∑

j(bj − aj). For any subset F of R, the exterior measure of

F is the infimum of the measures of all open sets containing F ; that is

me(F ) = infm(E) : F ⊆ E, E open. (1.15)

This is also called the outer measure of E, and gives the size of a set, as viewed from the

outside. The outer measure of a bounded set is clearly finite; whereas the outer measure

of an unbounded set may be infinite, and R has infinite outer measure.

NULL SETS of R are the sets that have zero outer measure. These are the sets F

such that for each ε > 0 there is an open set E with F ⊂ E and m(E) < ε.

If a property P (x) of real numbers holds, except when x belongs to a set E with

me(E) = 0, then we say that P holds almost everywhere (with respect to Lebesgue

measure).

1.9 PROPOSITION. Let F be a countable subset of R. Then me(F ) = 0.

Proof. Given ε > 0, we aim to produce an open cover E for F of measure less

than ε. We shall give a slightly laborious proof, so that we can easily calculate m(E). We

consider the principal case in which F = xn : n = 1, 2, . . . , where the xn are distinct

real numbers. To cover x1, we let E1 = (x1 − ε/4, x1 + ε/4). If x2 ∈ E1, then we take

E2 = ∅ ; otherwise we take E2 to be an open interval containing x2, of measure less than

ε/4, and disjoint from E1. Now if x3 ∈ E1 ∪ E2 , then we take E3 = ∅ ; otherwise we

take E3 to be an open interval containing x3, of measure less than ε/8, and disjoint from

E1 ∪ E2.

17

Proceeding in this way, we obtain disjoint open sets En with m(En) ≤ ε/2n, andsuch that xn ∈ ∪n

k=1Ek for n = 1, 2, . . . . Hence F ⊆ ∪∞n=1En, and E = ∪∞k=1Ek gives usan open set covering F. Since the En are disjoint, we can use the elementary Proposition1.8 and the geometric series formula to show that

m(E) =∞∑

n=1

m(En) ≤∞∑

n=1

ε/2n = ε. (1.16)

1.9 COROLLARY. The rational numbers are a null set, so me(Q) = 0.

Proof. In view of 1.2, we can apply the preceding Proposition. This result is mildlysurprising, when one considers Archimedes’ Axiom.

1.9 EXAMPLE. The Cantor set is an uncountable null set.

Solution.∗ We construct the Cantor set C by successive deletions of subintervals of[0, 1]. Details are left to the reader as an exercise.

Let E1 = (1/3, 2/3) be the middle third of [0, 1]. Now let E2 = (1/9, 2/9)∪ (7/9, 8, 9)be the middle thirds of the two intervals which form [0, 1] \ E1. Now let

E3 = (1/27, 2/27) ∪ (7/27, 8/27) ∪ (19/27, 20/27) ∪ (25/27, 26/27)

be the set made up of the middle thirds of [0, 1] \ (E1 ∪ E2). At the kth stage in theconstruction, we let Ek be the union of the open intervals which form the middle thirdsof the set [0, 1] \

⋃k−1j=1 Ej .

(i) Show that the Ej are disjoint open sets, and that the measure of Ej is 2j−13−j .

(ii) Deduce that the total measure of the open set E = ∪∞j=1Ej is one, so that C hasmeasure zero.

The Cantor set is uncountable.

This fact may be proved using the diagonal argument of 1.5. By the ternary expansionof a real number we mean an expression of the form

x = r0 +∞∑

j=1

3−jrj ,

where r0 is an integer, and rj = 0, 1 or 2 for j = 1, 2, . . . .Exercise.

(i) Show that there are only countably many x such that the ternary expansion isfinite; that is rk = 0 for all k ≥ K for some K <∞.

18

(ii) Show that there are only countably many x such that the tenary expansion hasrk = 2 for all k ≥ K, for some K <∞.

(iii) Let S be the set of x ∈ [0, 1] such that the tenary expansion is not of the forms(i) or (ii), and such that rk = 0 or 2, for all k. (Thus 1 does not appear in the tenaryexpansion.) Show that S is uncountable.

(iv) Show that S is contained in C, and hence that C is uncountable.

Subadditivity of Lebesgue measure on open sets

The next result asserts that Lebesgue measure of open sets is subadditive, so that thetotal measure of open sets covering a given set is greater than or equal to the measure ofthe set covered. This is not true for coverings by arbitrary subsets of the line, as Example1.13 shows.

1.10 MAIN LEMMA. If E and En (n = 1, 2, . . . ) are open sets, not necessarily disjoint,

with E ⊆⋃∞

n=1En, then

m(E) ≤∞∑

n=1

m(En). (1.17)

Proof. We write E = ∪∞k=1(ak, bk), where the (ak, bk) are disjoint; and similarlyEn = ∪t(a

(n)t , b

(n)t ). The measure of E is m(E) =

∑k(bk−ak), and the case that presents

greatest difficulty is when this is a convergent infinite series. Given ε > 0, we take L <∞sufficiently large that

∑∞k=L+1(bk − ak) < ε. We shrink the set ∪L

k=1(ak, bk) by pushingthe endpoints of its constituent intervals inwards to form

K =L⋃

k=1

[ak + ε/L, bk − ε/L]. (1.18)

Now K is closed and bounded, and possesses most of the measure of E. Further, K iscovered by the En, so K ⊆

⋃n

⋃t(a

(n)t , b

(n)t ). By the Heine–Borel Covering Theorem, we

need only finitely many intervals to cover K ; so there exist N <∞ and T <∞ such that

K ⊆ ∪Nn=1 ∪T

t=1 (a(n)t , b

(n)t ). (1.19)

By an obvious extension of the Covering Lemma 1.7, the total measure of the coveringintervals is greater than the measure of the set covered, so

m(K) =L∑

k=1

(bk − ak − 2ε/L) ≤N∑

n=1

T∑t=1

(b(n)t − a

(n)t ) ≤

N∑n=1

m(En), (1.20)

19

which leads toL∑

k=1

(bk − ak) ≤ 2ε+∞∑

n=1

m(En). (1.21)

By the choice of L, we deduce that

m(E) =L∑

k=1

(bk − ak) +∞∑

k=L+1

(bk − ak) ≤ 3ε+∞∑

n=1

m(En); (1.22)

and since ε > 0 was arbitrary, this inequality gives the stated result.

1.10 COROLLARY. If (Fn) (n = 1, 2, . . . ) is a sequence of null sets, then their union

F =⋃∞

n=1 Fn is also null.

1.10 Exercise. Deduce Proposition 1.9 from the Main Lemma 1.10.

Inner measure and Lebesgue measure

1.11 INNER MEASURE of a subset F of the bounded interval (a, b) is

mi(F ) = (b− a)−me(F c), (1.23)

where F c = (a, b)\F denotes the complement of F in (a, b). To make this definition seemreasonable, one can consider what remains of (a, b) after one removes a cover of F c.

1.11 PROPOSITION. Let F be any subset of the bounded interval (a, b). Then its

inner measure satisfies

0 ≤ mi(F ) ≤ me(F ) ≤ (b− a). (1.24)

This lemma says that the inner measure of a set is always nonnegative, and less thanor equal to the outer measure.

Proof. To begin with, F ⊆ (a, b), so me(F ) ≤ (b − a). Similarly, me(F c) ≤ b − a,

and we deduce that the inner measure of F has mi(F ) = b− a−me(F c) ≥ 0.The most substantial result is the middle inequality, which we prefer to transform into

a statement involving only outer measure. The inequality mi(F ) ≤ me(F ) is equivalentto

b− a ≤ me(F ) +me(F c). (1.25)

This is plausible in view of the fact that (a, b) = F ∪ F c. To prove it, we take, for ε > 0,open sets G and U with F ⊆ G and F c ⊆ U ; these are chosen to be good covers in thesense that

m(G) ≤ me(F ) + ε and m(U) ≤ me(F c) + ε. (1.26)

20

Now (a, b) ⊆ F ∪ F c ⊆ G ∪ U, so by the Main Lemma 1.10, we have

b− a = m(a, b) ≤ m(G) +m(U) ≤ me(F ) +me(F c) + 2ε, (1.27)

whence the result.

LEBESGUE MEASURE. If a subset F of (a, b) has mi(F ) = me(F ), then we say that

F is (Lebesgue) measurable and write m(F ) = me(F ) for the common value of inner

and outer measure. This defines the Lebesgue measure of F. By definition, F ⊆ (a, b) is

measurable if and only if me(F ) +me(F c) = b − a ; so F is measurable if and only if F c

is measurable, where F c is the complement with respect to (a, b) .

Why outer measure? Outer measure estimates the size of a set, as viewed from the

outside. It is a key feature of Lebesgue’s theory, and two features of its definition require

comment:

(i) The outer measure of any set F of real numbers is defined. It is technically

advantageous to have some notion of the size of an arbitrary set, for it will turn out that

not all sets are Lebesgue measurable.

(ii)∗ We allow coverings by countable collections of disjoint open sets. On account of

Example 1.13, it is futile to allow coverings by arbitrary collections of sets; and the Heine–

Borel Covering Theorem suggests that open sets give the right coverings to consider. On

the other hand, the Jordan content, defined using finite coverings by open intervals,

mJ(F ) = inf n∑

j=1

(bj − aj) : F ⊆ ∪nj=1(aj , bj)

(1.28)

has different properties. Note that the set F = Q ∩ (0, 1) has outer measure me(F ) = 0,

but Jordan content mJ(F ) = 1. Countable sets can thus appear to have large size. Using

Jordan content does not lead to a satisfactory theory of integration; one ends up with

something resembling the Riemann integral.

Why inner measure? One is led to the notion of inner measure by the complementation

axiom (σ2). Complements with respect to a bounded interval are considered to avoid

nonsensical calculations involving ∞−∞. It is quite reasonable to compare the notions of

size of a set as viewed from the inside and the outside; it is like comparing the legroom to the

length of a car. Proofs of the properties of inner measure will be carried out by transforming

them into corresponding statements involving outer measure of the complement. It is

simpler to think in terms of coverings, and these relate directly to outer measure.

21

It can be shown that non–measurable sets exist, but their construction depends uponthe higher axioms of set theory. Open sets and closed sets, and the sets that arise from themthrough complementation, unions and intersections involving sequences, are all Lebesguemeasurable. Once one has established these basic facts, one can regard non–measurablesets as exotic, and not worry about them unduly.

Covering lemmas for measure

1.12 LEMMA ON OVERLAPS. Let E and F be bounded open sets with (a, b) ⊆ E∪F,where a and b are finite. Then the measure of their intersection has

m(E ∩ F ) ≤ m(E) +m(F )− (b− a). (1.29)

The content of this result is that if small open sets cover (a, b), then the size of theiroverlap cannot be too big.

Proof.† We reduce to the case in which the covering sets E and F are finite unionsof open intervals. The result in that case is geometrically evident, and may be provedrigorously by induction on the number of intervals in the cover.

Suppose that E = ∪∞j=1(aj , bj) and F = ∪∞k=1(ck, dk) are, respectively, disjoint unionsof open intervals. Take ε >> 0 and shrink (a, b) to the closed bounded setK = [a+ε/2, b−ε/2]. Then K ⊆ E ∪ F, so by the Heine–Borel Covering Theorem, we can cover K withfinitely many of the intervals (aj , bj) and (ck, dk). More precisely, we can choose intervalsso that:

(i) K ⊆ Q ∪R, where Q is a union of finitely many (aj , bj), andR is the union of finitely many (ck, dk) ;

(ii) the open set Q′ = E \Q ; and(iii) the open set R′ = F \R .

The finite version of the Lemma shows that

m(Q ∩R) ≤ m(Q) +m(R)−m(K). (1.30)

Since E = Q ∪Q′ and F = R ∪R′, it is easily seen that E ∩ F ⊆ (Q ∩R) ∪Q′ ∪R′ ;and all the sets here are open. From Lemma 1.10 we deduce that

m(E ∩ F ) ≤ m(Q ∩R) +m(Q′) +m(R′). (1.31)

By the preceding inequality this is

m(E ∩ F ) ≤ m(Q) +m(R)−m(K) +m(Q′) +m(R′) (1.32)

≤ m(E) +m(F )− (b− a) + ε. (1.33)

22

† This proof will not be examined as bookwork.

1.13 LEMMA (Subadditivity of outer measure). Let F and Fn (n = 1, 2, . . . ) be

subsets of R, and suppose that F ⊆⋃∞

n=1 Fn. Then the total outer measure of the covering

sets is greater than or equal to the measure of the set they cover, so

me(F ) ≤∞∑

n=1

me(Fn). (1.34)

1.13 EXAMPLE. It is important that we take the (Fn)∞n=1 to be a sequence of setsin Lemma 1.13. The result is false for arbitrary families. For example, [0, 1] has outermeasure me[0, 1] = 1 ; whereas we have [0, 1] = ∪x∈[0,1]x, where in this uncountableunion the covering sets have outer measure me(x) = 0.

Proof of Lemma 1.13. We aim to reduce to considering open sets. Given ε > 0, wecover each Fn by an open set En, which is ever–so–slightly larger, so that Fn ⊆ En, withm(En) ≤ me(Fn) + ε/2n. The open set E =

⋃∞n=1En covers F, since

F ⊆∞⋃

n=1

Fn ⊆∞⋃

n=1

En = E, (1.35)

so F has outer measure me(F ) ≤ m(E). But we can estimate the measure of E byapplying Lemma 1.10 to the union ∪∞n=1En, to get

m(E) ≤∞∑

n=1

m(En) ≤∞∑

n=1

(me(Fn) + ε/2n). (1.36)

An exercise in geometric series reduces this to the inequality

me(F ) ≤ m(E) ≤∞∑

n=1

me(Fn) + ε, (1.37)

which yields the stated result.

1.14 MAIN THEOREM. Lebesgue measure is countably additive. Let (a, b) be a

bounded interval in R, let (En)∞n=1 be a sequence of disjoint measurable subsets of (a, b)and let E = ∪∞n=1En be their union. Then E is measurable and

(Σ) m(E) =∞∑

n=1

m(En). (1.38)

23

Proof.† There are two sources of difficulty in proving this result, namely: the sets

can have complicated structure, and the union involved is countably infinite. The former

presents the more serious challenge, and will be treated first. We shall suppose that only

a finite union is involved; there is no real loss in taking E = E1 ∪ E2, where E1 and E2

are disjoint measurable sets. We require to prove that

mi(E) = me(E) = m(E1) +m(E2).

Outer measures are considered first. We have E ⊆ E1∪E2 so by Lemma 4.7, me(E) ≤me(E1) +me(E2). Recalling that E1 and E2 are measurable, we deduce that

me(E) ≤ m(E1) +m(E2). (1.39)

Inner measures are considered next. We aim to prove

mi(E) ≥ m(E1) +m(E2), (1.40)

which is equivalent to

me(Ec) ≤ m(Ec1) +m(Ec

2)− (b− a). (1.41)

Since E1 and E2 are disjoint, we have E1 ∩ E2 = ∅, so (a, b) = Ec1 ∪ Ec

2. To reduce

to the context of the lemma 1.12 on overlaps, we cover Ec1 by an open set G, and Ec

2 by

an open set U with

m(G) ≤ m(Ec1) + ε, m(U) ≤ m(Ec

2) + ε (1.42)

and (a, b) ⊆ G ∪ U. From Lemma 1.7 it follows that

m(G ∩ U) ≤ m(G) +m(U)− (b− a) (1.43)

≤ m(Ec1) +m(Ec

2)− (b− a) + 2ε. (1.44)

Now we observe that Ec = Ec1 ∩ Ec

2 ⊆ G ∩ U, where G ∩ U is an open set; so me(Ec) ≤m(G ∩ U). When this is combined with the previous inequality, we obtain (1.41), up to

an arbitrary ε > 0.

By Proposition 1.11, we have mi(E) ≤ me(E), so (1.39) and (1.40) together show

that mi(E) and me(E) are equal, and take the common value m(E1) +m(E2). Hence E

is measurable.

24

Infinite unions. Suppose that E = ∪∞k=1Ek, where the Ek are disjoint measurablesets. An induction argument shows that

Sn = ∪nk=1Ek, (1.45)

being the union of finitely many measurable sets, is measurable with

m(Sn) =n∑

k=1

m(Ek).

Outer measure of E may be estimated using Lemma 1.12. We have E ⊆ ∪∞k=1Ek ,and hence

me(E) ≤∞∑

k=1

me(Ek) =∞∑

k=1

m(Ek). (1.46)

Inner measure of E is estimated using the Sn. We have Sn ⊆ E, and hence mi(E) ≥m(Sn) ; indeed, Ec ⊆ Sc

n, so me(Ec) ≤ me(Scn). Using the finite case of the Theorem, we

see that

mi(E) ≥ me(Sn) =n∑

k=1

m(Ek), (1.47)

and letting n→∞, we deduce that

mi(E) ≥∞∑

k=1

m(Ek). (1.48)

Combined with the estimate (1.46) on outer measure, we have

mi(E) = me(E) =∞∑

k=1

m(Ek), (1.49)

so E is measurable with the stated measure.† The parts of this proof involving inner measure will not be examined as bookwork.

1.15 REMARK. In applications of measure theory to classical analysis it is rarelynecessary to check that sets are measurable, since most constructions involve countableunions or intersections of open sets.

1.15 The Lebesgue measurable sets form a σ-algebra.∗

The remaining results of this chapter extend, to sequences of intersecting sets, themeasurability results that we have achieved for disjoint measurable sets.∗ This section will not be covered in lectures, nor examined.

25

1.15 LEMMA. If E and F are measurable sets, then their intersection E ∩ F is also

measurable.

Proof. We reduce to simpler cases. If E is measurable, and F = (a, b), then it is

clear that E ∩ (a, b) is measurable.

If F is open, then we can invoke 1.10 and write F =⋃∞

k=1(ak, bk), a union of disjoint

open intervals. By the previous observation, E ∩ (ak, bk) is measurable for each k, and so

by Theorem 1.14,

E ∩ F =∞⋃

k=1

(F ∩ (ak, bk)

)(1.50)

is measurable, as a union of countably many disjoint measurable sets.

Now for F any measurable subset of (a, b), we can introduce open sets J and G with

F ⊆ J and F c ⊆ G with

m(J) +m(G) ≤ (b− a) + 2ε. (1.51)

Then E ∩ F ⊆ E ∩ J, where this latest set is measurable, and

(E ∩ F )c = Ec ∪ F c = (F ∩ Ec) ∪ F c ⊆ (J ∩ Ec) ∪G. (1.52)

Taking the measures of these sets, we obtain

me(E ∩ F ) +me((E ∩ F )c) ≤ m(E ∩ J) +m(J ∩ Ec) +m(G) (1.53)

= m(J) +m(G) ≤ (b− a) + 2ε. (1.54)

1.15 PROPOSITION. If E and F are measurable, not necessarily disjoint, then their

union E ∪ F is also measurable.

Proof. Our argument is based upon the identity E ∪ F = (E ∩ F c) ∪ F, which is

a union of disjoint sets. The set F c is measurable, so by the Lemma 1.15, E ∩ F c is

measurable. The disjoint union of two measurable sets is measurable, by Theorem 1.14,

whence the result.

We extend the theory of measure to unbounded subsets as follows. A subset F of R is

said to be Lebesgue measurable if for each n ∈ N the set F∩(−n, n) is a measurable subset

of (−n, n). We define the Lebesgue measure of such a set to be limn→∞m(F ∩ (−n, n))

with the convention that this may be infinite. The statements of the preceding results hold

for this extended definition of the Lebesgue measure, where they are intelligible.

26

1.15 THEOREM. If (En) (n = 1, 2, . . . ) is any sequence of measurable subsets, then

their union E = ∪∞n=1En is a measurable set. The family of Lebesgue measurable sets

forms a σ-algebra of subsets of R containing the collection of open sets.

Proof. It suffices to check the axiom (σ3). Let (En) be a sequence of measurablesets, not necessarily disjoint. We can extend the previous Proposition 1.15 to show thatfinite unions of measurable sets are measurable. Hence

F1 =E1, (1.56)

F2 =(E1 ∪ E2) \ E1,

F3 =(E1 ∪ E2 ∪ E3) \ (E1 ∪ E2),

. . . . . .

gives a sequence of disjoint measurable subsets with ∪∞n=1En = ∪∞n=1Fn. By Theorem 1.14,this countable union of disjoint measurable sets is measurable.

27

§2. The Lebesgue integral

At the end of this chapter, you should be able to:

understand the notion of a measurable function, and prove that continuous

functions are measurable;

define the integral of a bounded measurable function;

extend the definition of the Lebesgue integral to unbounded functions and

understand its properties.

We begin this chapter by introducing a very general class of real functions, called

measurable functions, which are used in integration. We then show that any bounded

measurable function on a bounded real interval has a Lebesgue integral. This will be

constructed using the approximation procedure sketched in the introduction, where func-

tions are approximated by simple functions introduced by partitioning the range. We then

extend the definition of the integral to suitable unbounded measurable functions, and to

functions defined on typical measurable subsets of the line.

To define the integral we shall need mainly the conclusions of Theorem 1.14 and 1.15

which establish that Lebesgue measure is countably additive. This point is important, as it

allows one to generalize results to integrals (expectations) defined over probability spaces.

2.1 MEASURABILITY. A function f : [a, b] → R is said to be measurable if for each

c ∈ R the set

x ∈ [a, b] : f(x) > c (2.1)

is measurable. This set is often denoted by [f > c].

2.1 PROPOSITION. Any continuous real function is measurable.

Proof. Let λ be real and consider the set x ∈ (a, b) : f(x) > λ. We shall show that

this is open, and hence measurable. When x has f(x) > λ, we can take δ = (f(x)−λ)/2 >

0. Then by definition of continuity, there exists ε > 0 such that

|f(x)− f(y)| < δ whenever |x− y| < ε. (2.2)

Hence, if |x− y| < ε, then we have

f(y) = f(x)− (f(x)− f(y)) > f(x)− δ = (f(x) + λ)/2 > λ. (2.3)

Consequently, we have (x− ε, x+ ε) ⊆ [f > λ], and our set is open.

28

2.2∗ PROPOSITION. (i) Let f and g be measurable functions and λ ∈ R. Then the

functions λf, |f |, f+g, f.g and max(f, g) defined by pointwise operations are measurable

(ii) Let (fn(x)) be a sequence of measurable functions on [a, b]. Then

F (x) = lim supn→∞ fn(x) defines a measurable function of x ∈ [a, b], in the sense that

E = x : F (x) = ∞ is a measurable set, and the restriction F |(R\E) is a measurable

function.

(iii) Let g be a measurable function and f a continuous real function. Then the

composition f g is a measurable function.

Proof. (i) To show that the sum of measurable functions is measurable, we note thatif f(x) + g(x) > λ or equivalently g(x) > λ − f(x), then by Archimedes’ axiom we canfind a rational q with g(x) > q and f(x) > λ − q. Conversely, we have f(x) + g(x) > λ,

whenever such a q exists. Hence

[f + g > λ] =⋃q∈Q

([f > λ− q] ∩ [g > q]

)(2.4)

is measurable by Theorem 1.15, being a countable union of measurable sets.To show that the product of measurable functions is measurable, we note that g2

is measurable, since [g2 > λ] = [g >√λ] ∪ [g < −

√λ] for λ >> 0. Then we write the

product asfg = (f + g)2/4− (f − g)2/4, (2.5)

the difference of two measurable functions.(ii) See proof of the bounded convergence theorem 3.1.(iii) See proof of Proposition 2.1.

Due to the above facts, the usual operations of elementary analysis always produce mea-surable functions. In practice one seldom takes the trouble to check that a function ismeasurable if it arises from continuous functions under standard limiting operations.

INTEGRALS. The integral of a simple function

f(x) =n∑

j=1

cjIEj (x), (2.6)

where IEjis the indicator function of measurable set Ej , is defined to be∫

f(x) dx =n∑

j=1

cjm(Ej). (2.7)

29

One again, the integral is the area under the graph of f , and the area of a rectangle is themeasure of the base times the height.

We can define the Lebesgue integral of any bounded measurable function on [a, b] byapproximating it by simple functions.

THEOREM 2.3. Let f be a bounded measurable function on the interval [a, b]. Then

f has a well-defined integral over [a, b] ; so that, when f is a simple function, this coincides

with the previous definition.

Proof. Our tactics are to approximate the function f by simple functions such thatthe integrals will approximate the area under the graph of f. Let A and B be constantswith A ≤ f(x) ≤ B for all a < x < b. We partition the range [A,B] by introducing thepartition

P = A = y0 < y1 < y2 < y3 < · · · < yk < · · · < yn = yn+1 = B, (2.8)

and introduce a corresponding partition of the domain (a, b) by

Ek =x : yk ≤ f(x) < yk+1, (0 ≤ k ≤ n− 1); (2.9)

En =x : f(x) = yn.

The sets Ek are clearly disjoint; and they are measurable, since f is measurable. Weintroduce simple functions that are constant on the Ek and satisfy fL(x) ≤ f(x) ≤ fU (x).Evidently the lower simple function

fL(x) =n∑

k=0

ykIEk(x) (2.10)

and the upper simple function

fU (x) =n∑

k=0

yk+1IEk(x) (2.11)

fit the bill. These functions may be integrated to give a lower estimate

s =∫ b

a

fL(x) dx =n∑

k=0

ykm(Ek) (2.12)

and an upper estimate

S =∫ b

a

fU (x) dx =n∑

k=0

yk+1m(Ek) (2.13)

30

for the area under the graph of f, respectively. We can ensure that these estimates areclose together by making the partition sufficiently fine. Indeed, given ε > 0, we can takeenough terms in the partition to ensure that yk+1 − yk ≤ ε for all 0 ≤ k ≤ n. In thiscircumstance

0 ≤ S − s =n∑

k=0

(yk+1 − yk)m(Ek) ≤ εn∑

k=0

m(Ek). (2.14)

To estimate this latest expression we need to use properties of Lebesgue measure. TheEk are disjoint and measurable subsets with ∪n

k=0Ek = (a, b) ; so, by Theorem 1.14, theprevious inequality becomes

0 ≤ S − s ≤ εm(a, b) = ε(b− a). (2.15)

This gives an upper (S = S(P )) and a lower (s = s(P )) approximation to the integralwhich depends upon P . We can refine the partition to P ⊆ P ′ ⊆ P ′′ ⊆ . . . by addingpoints. We wish to define the Lebesgue integral of f to be the common value of the limits∫ b

a

f(x) dx = limPS = lim

Ps (2.16)

as the partition refines, so that maxk(yk+1−yk) → 0. To check that this is a good definition,we must ensure that the value of this limit does not depend upon the choice of partition.To do this we show

( any lower sum) ≤ (any upper sum), (2.17)

even when they originate from different partitions.Let P = y0 < y1 < y2 < · · · < yn = yn+1 be a partition with lower sum s and

upper sum S, as above; let P ′ = y′0 < y′1 < · · · < y′` = y′`+1 be another partitionwith lower sum s′ and upper sum S′. We combine these partitions to form a commonrefinement P ′′ = y0, y′0, y1, y′1, . . . , yn+1, y

′`+1, with lower sum s′′ and upper sum S′′.

We shall prove thats ≤ s′′ ≤ S′′ ≤ S, s′ ≤ s′′ ≤ S′′ ≤ S′. (2.18)

The desired inequalities s ≤ S′ and s′ ≤ S follow directly.One can show that refining a partition increases lower sums and diminishes upper

sums. This may be established by induction of the number of terms in the refined partition;the induction step is to show that adding one point to a partition increases the lower sumand diminishes the upper sum. So let us add to P a point y with yk < y < yk+1. The setEk = x : yk ≤ f(x) < yk+1 is split into the union of the disjoint measurable sets

E′k = x : yk ≤ f(x) < y, E′′k = x : y ≤ f(x) < yk+1. (2.19)

31

To form the new lower sum, we replace the summand ykm(Ek) by the larger quantity

ykm(E′k) + ym(E′′k ) ≥ yk(m(E′k) +m(E′′k )) = ykm(Ek). (2.20)

Similarly, in the upper sum, yk+1 is replaced by a smaller quantity

ym(E′k) + yk+1m(E′′k ) ≤ yk+1(m(E′k) +m(E′′k )) = yk+1m(Ek). (2.21)

This concludes the proof of the existence theorem.

In a few cases, one can use the defining construction to obtain a value for the Lebesgueintegral.

2.3 EXAMPLE.∫ 1

0x1/3 dx = 3/4.

Solution. For each integer n ≥ 1 we introduce a partition of the range by yk = k/n

for k = 0, 1, . . . , n. Then the corresponding partition of the domain is given by the sets

Ek = x : k/n ≤ x1/3 < (k + 1)/n = x : k3/n3 ≤ x < (k + 1)3/n3 (2.22)

for k = 0, 1, . . . , n− 1. A lower sum for the integral is

s =n−1∑k=0

k

nm(Ek) =

n−1∑k=0

k

n

(k + 1)3− k3n3

. (2.23)

This reduces by algebra to

s =1n4

n−1∑k=1

(3k3 + 3k2 + k) =1n4

(34(n− 1)2n2 +

36(n− 1)n(2n− 1) +

12(n− 1)n

). (2.24)

For large n, only the terms of degree 4 make a significant contribution. As n → ∞,

s → 3/4 ; and the corresponding upper sum has the same limit. Hence the integral takesthe value 3/4.

Remark. This method for calculating integrals was standard in the sixteenth century.It was from such considerations that the standard formulæ of integral calculus were firstdiscovered. The algebraic formulæ for sums of powers may be found by the method ofdifferences, or may be verified by induction.

2.3 COROLLARY. Let f : [a, b] → R be a continuous function on a closed and bounded

interval. Then f has a Lebesgue integral.

32

Proof. By results of MATH210 Real Analysis, such an f is bounded; moreover, such

an f is measurable by Proposition 2.1. Hence Theorem 2.3 applies.

2.3 REMARKS. (i) We have shown: Let f be a bounded measurable function on [a, b]

and ε > 0. Then there exists a simple function s with |s(x)− f(x)| < ε for all x ∈ [a, b].

The converse is also true.

(ii) This characterization of measurable functions provides a means of proving Propo-

sition 2.1 which is less painful than a direct argument.

2.4 FUNDAMENTAL THEOREM OF CALCULUS. Let f : [a, b] → R be a continuous

function on a closed and bounded interval. Then f has a Lebesgue integral. Further, the

indefinite integral

F (x) =∫ x

a

f(t) dt (2.25)

defines a differentiable function of x on (a, b), with F ′(x) = f(x)

Proof. A standard result of real analysis shows that such a function is bounded, and

we know from Proposition 2.1 that f is measurable. Later we shall show that the Lebesgue

integral has the same value as the Riemann integral defined for such a function.

The second statement is a weak version of the Fundamental Theorem of Calculus.

The right-hand derivative at x may be found by considering the difference quotient

F (y)− F (x)y − x

=1

y − x

∫ y

x

f(t) dt. (2.26)

Given ε > 0 we can take δ > 0 so that

f(x)− ε < f(t) < f(x) + ε

whenever x < t < y < x+ δ; hence by Q15

(f(x)− ε)(y − x) ≤∫ y

x

f(t) dt ≤ (f(x) + ε)(y − x),

so

f(x)− ε ≤ F (y)− F (x)y − x

≤ f(x) + ε.

Letting y → x+, we deduce that F ′(x) = f(x).

Complement. Let G be any function that is continuous on [a, b] and differentiable

on (a, b) , and such that G′(x) = f(x) for all a < x < b. Then there exists a constant C

33

such that G(x) = F (x) +C. This follows directly from the mean value theorem applied to

F (x)−G(x). One can use this fact to evaluate integrals by the usual rules of calculus.

COROLLARY. Let f be a continuous function on [a, b]. Then the Riemann and

Lebesgue integrals of f over [a, b] are equal.

Proof. Let F (x) =∫ x

af(t) dt be the Lebesgue indefinite integral and G(x) =∫ x

af(t) dt be the Riemann indefinite integral. Then by the fundamental theorem of calculus

F ′(x)−G′(x) = f(x)− f(x) = 0, and hence by the mean value theorem F (x)−G(x) = C,

a constant. But F (a) − G(a) = 0; hence we have F (x) − G(x) = 0 for all x and hence

F (a) = F (b), as required.

Truncation. A bounded function on [a, b] has its graph contained in a bounded rect-

angle. One can truncate a possibly unbounded real function by modifying it so as to ensure

that its graph is contained in suitable bounded rectangle.

2.5 Integrals of unbounded functions

We define the Lebesgue integral of an unbounded measurable function to be the limit

of the integrals of its truncations, if this exists. At each stage in the construction, one

ensures that integrals are absolutely convergent; when f is integrable, its modulus |f | is

also integrable. This is of some importance for the Lebesgue convergence theorems below.

Integration of unbounded functions on (a, b).

Suppose that f is measurable on a bounded interval (a, b), and that f is unbounded.

Then f may or may not be integrable. We suppose that f ≥ 0 and investigate further by

introducing the truncation (f)N (x) = minf(x), N, which is clearly measurable and, by

design, is bounded. By Theorem 2.3, (f)N is integrable. Further the sequence of integrals∫ b

a(f)N (x) dx is increasing, so there are two possibilities, either:

(i) the sequence of integrals is unbounded, and∫ b

a(f)N dx→∞ as N →∞ ; or

(ii) the sequence is bounded above, and converges to its supremum.

In case (ii), we say that f is integrable and we define∫ b

a

f(x) dx = limN→∞

∫ b

a

(f)N (x) dx. (2.27)

In general, if f : (a, b) → ∞ is measurable, we write f = f+ − f−, where we have

introduced the positive measurable functions

f+(x) = maxf(x), 0, f−(x) = max−f(x), 0. (2.28)

34

If both f+ and f− are integrable, then we say that f is integrable and define∫ b

a

f(x) dx =∫ b

a

f+(x) dx−∫ b

a

f−(x) dx. (2.29)

Integrals over subsets of the real line. It is a straightforward matter to extend thedefinition of the Lebesgue integral from integrals over bounded intervals to R. A measur-able function f is integrable over R if

∫ N

−N|f(x)|dx is a bounded sequence. The Lebesgue

integral∫∞−∞ f(x) dx is defined to be limN

∫ N

−Nf(x) dx. Further, if E is any measurable

subset of R, then we define the integral of f over E to be∫E

f(x) dx =∫ ∞

−∞IE(x)f(x) dx, (2.30)

for all functions f such that f(x)IE(x) is integrable over R. If E is a null set, then∫Ef(x) dx = 0. For the purposes of integration theory, one can afford to ignore null sets.

Complex functions. The Lebesgue integral of a complex-valued function is defined inthe obvious way by adding real and imaginary parts∫

f(x) dx =∫<f(x) dx+ i

∫=f(x) dx (2.31)

when the real functions on the right-hand side are both integrable.

2.6 Properties of the integral

2.6 PROPOSITION. The Lebesgue integral has the following properties.

(i) Absolute convergence: for f any integrable function,∫|f(x)|dx <∞.

(ii) Positivity: if f ≥ 0, then∫f(x) dx ≥ 0.

(iii) Linearity: for any integrable functions f and g and any scalars λ and µ,∫(λf(x) + µg(x)) dx = λ

∫f(x) dx+ µ

∫g(x) dx. (2.32)

(iv) Triangle inequality:

(4)∣∣∣∫ f(x) dx

∣∣∣≤ ∫|f(x)|dx; (2.33)

(v) Additivity: for disjoint measurable sets E and F,∫E∪F

f(x) dx =∫

E

f(x) dx+∫

F

f(x) dx. (2.34)

35

Proof. The basic properties of the integral for simple random variables were estab-lished in MATH313. The Lebesgue integral is the limit of integrals of simple functions,and thus the Lebesgue integral inherits similar properties.

2.6 PROPOSITION∗. Let f be an integrable function. Then its indefinite integral

F (x) =∫ x

−∞ f(t)dt is continuous.

Solution. Let x ∈ R and ε > 0. Then

F (x+ h)− F (x) =∫ x+h

x

f(t) dt, (2.35)

and we can find N <∞ so large that the truncation (f)N (x) = minN,max−N, f(x)has

∫|f − (f)N |dt < ε. Hence

|F (x+ h)− F (x)| ≤∣∣∣∫ x+h

x

(f)N (t) dt∣∣∣ +

∣∣∣∫ x+h

x

(f(t)− (f)N (t)) dt∣∣∣ (2.36)

≤ 2N |h|+ ε. (2.37)

We can make this < 2ε by taking |h| < ε/(2N).

2.6 REMARK. All functions f that are defined by reasonable formulæ are measur-able, and integrability reduces to consideration of |f |, so we have a comparison test:

(i) to show that f is integrable, find g integrable with |f | ≤ g ;

(ii) to show that f is not integrable, find h not integrable with 0 ≤ h ≤ |f |.In carrying out estimates, one can use the heuristic principle comparing functions of

large x :exponentials beat large powers beat small powers beat logarithms

For instance, one has ex ≥ x2 ≥√x ≥ log x for all sufficiently large x > 0.

2.6 EXAMPLE. For which real s is f(x) = x−s integrable over (0, 1) or (0, 1)?

(i)∫ ∞

1

dxxs

= 1/(s− 1) s > 1;

diverges for s ≤ 1.

(ii)∫

0

1dxxs

= 1/(1− s), s < 1;

diverges for s ≥ 1.

36

2.6(iii) EXAMPLE. The function f(x) = sinx/x is not integrable over (0,∞), although

limT→∞

∫ T

0

sinxx

dx =π

2. (2.38)

Solution. The function f is continuous on (0,∞) and is actually bounded in modu-

lus by one. Hence we can integrate f over any bounded subinterval of [0,∞) ; the problem

is that∫∞0|f(x)|dx diverges. To see this, we let n ≥ 1 be an integer and consider

nπ ≤ x ≤ (n+ 1)π. Then 1/x ≥ 1/(n+ 1)π and so by periodicity of the sine function we

have ∫ (n+1)π

nπ

| sinx|x

dx ≥ 1(n+ 1)π

∫ (n+1)π

nπ

| sinx|dx =2

π(n+ 1). (2.39)

Summing this inequality over n we have

∫ (N+1)π

π

| sinx|x

dx ≥N∑

n=1

2π(n+ 1)

. (2.40)

The series here diverges to infinity as N →∞, by comparison with the divergent harmonic

series∑

1/n.

Nevertheless, we can use integration by parts to write∫ T

π/2

sinxx

dx =[− cosx

x

]T

π/2−

∫ T

π/2

cosxx2

dx,

where the function cosx/x2 is Lebesgue integrable on [π/2,∞). Hence the limit

limT→∞

∫ T

0

sinxx

dx =∫ π/2

0

sinxx

dx−∫ ∞

π/2

cosxx2

dx (2.41)

exists.

To find the stated value of the improper integral above, one can use complex analysis.

The idea is to integrate eiz/z round a suitable semicircular contour.

L1 and Hilbert space L2

We now introduce some spaces of functions commonly used in analysis. Let (a, b) be

an interval in the real line, possibly unbounded. The space L1(a, b) of complex-valued

integrable functions is the set of Lebesgue measurable functions f on (a, b) such that∫ b

a|f(x)|dx is finite.

37

Similarly, we define L2(a, b) to be the Lebesgue measurable functions f with |f(x)|2

integrable on (a, b). The spaces L1(a, b) and L2(a, b) are vector spaces over C underpointwise addition of functions. For p = 1, 2, one introduces the norm

‖f‖p =(∫ b

a

|f(x)|p dx)1/p

(f ∈ Lp(a, b)). (2.42)

One shows that ‖f + g‖p ≤ ‖f‖p + ‖g‖p for functions f, g ∈ Lp(a, b).

Hilbert space. For L2(a, b), the standard inner product is

〈f, g〉 =∫ b

a

f(x)g(x) dx, (2.43)

which satisfies, for f, g, h ∈ L2(a, b) and λ ∈ C:(i) 〈f, g〉 = 〈g, f〉;(ii) 〈λf, g〉 = λ〈f, g〉;(iii) 〈f + h, g〉 = 〈f, g〉+ 〈h, g〉;(iv) 〈f, f〉 > 0, unless f = 0 almost everywhere.The space L2(a, b) with this inner product is an important example of a Hilbert space.

Whereas it is easy to verify the properties (i)-(iv), completeness requires a complicatedproof which appears as Theorem 3.8.

The Cauchy–Schwarz inequality asserts that∣∣∣∫ f(x)g(x) dx∣∣∣ ≤ (∫

|f(x)|p dx)1/2(∫

|g(x)|2 dx)1/2

. (2.44)

38

§3. Convergence Theorems

At the end of this chapter, you should be able to:

prove the bounded convergence theorem, and deduce the monotone con-

vergence theorem;

understand how the dominated convergence theorem follows from the

bounded convergence theorem;

apply the convergence theorems in standard examples such as the series

for the zeroth order Bessel function or the zeta function;

derive Wallis’s product for π and deduce the value of the Gaussian integral.

The Lebesgue convergence theorems fully justify the development of the theory of theLebesgue integral.

Their input consists of:(a) sequences of integrable functions with fn(x) → f(x); and(b) various bounds on the size of the functions.

Their output consists of the limit formula∫fn(x) dx→

∫f(x) dx.

The conclusions are most useful in applications, where functions are frequently definedby summation of series or other limiting operations.

ALMOST EVERYWHERE. Let (fn) be a sequence of measurable functions on [a, b].We say that fn(x) converges to f(x) almost everywhere if the set E, of values of x suchthat fn(x) does not converge to f(x), is null.

In effect, the convergence theorems clarify the relationship between almost everywhereconvergence and convergence of the integrals. The method of proof is to split up the regionof integration into a set of large measure where we have good control of the behaviourof the functions, and a complementary set of small measure on which the functions arecontrolled by weaker conditions. Thereby one can control the average behaviour of thefunctions. The convergence theorems have simple statements and can be applied easily. Inour treatment, the bounded convergence theorem is regarded as fundamental, and otherconvergence theorems are deduced from it.

Before considering the proofs in detail, the reader may find it helpful to review theproof of Chebyshev’s inequality of MATH313; for the convergence theorems follow fromsimilar estimates.

The geometrical interpretation of the hypotheses in the bounded convergence theoremis that the fn have graphs in a common bounded rectangle [a, b]× [−M,M ].

39

3.1 BOUNDED CONVERGENCE THEOREM. Let (fn(x)) (n = 1, 2, . . . ) be a sequence

of measurable functions on the bounded interval [a, b] such that:

(i) |fn(x)| ≤M, for all x ∈ [a, b] and all n;

(ii) fn(x) → f(x) as n→∞, for almost all x ∈ [a, b].

Then f is measurable and∫ b

a

fn(x) dx→∫ b

a

f(x) dx as n→∞. (3.1)

Proof. We first settle the measurability question. For each real λ and integer k ≥ 1

we introduce

Fk =x : fn(x) ≤ λ+

1k

for infinitely many n

=∞⋂

N=1

∞⋃n=N

x : fn(x) ≤ λ+

1k

, (3.2)

which is measurable by Theorem 1.15 since the functions fn are measurable. Further, we

have

x : f(x) ≤ λ =∞⋂

k=1

x : f(x) ≤ λ+

1k

=

∞⋂k=1

Fk

since fn(x) → f(x) as n→∞.

Further, |f(x)| is bounded by M, as

|f(x)| = limn|fn(x)| ≤M.

To establish convergence of the integrals we use linearity to show that

∣∣∣∫ b

a

f(x) dx−∫ b

a

fn(x) dx∣∣∣ =

∣∣∣∫ b

a

(f(x)− fn(x)) dx∣∣∣ (3.3)

≤∫ b

a

∣∣f(x)− fn(x)|dx, (3.4)

where the last step follows from the triangle inequality. We introduce the function gn(x) =

|f(x)− fn(x)| ; this is measurable, positive and bounded by 2M. Our aim is to show that∫ b

agn(x)dx → 0. By the assumptions, we know that gn(x) → 0 except on a set E with

m(E) = 0. We can afford to ignore the null set E, since E does not contribute to the

integrals.

40

We aim to split up the range of integration into a good set, on which gn(x) is small,and a bad set, which is small in measure. To do this we introduce, for ε > 0 the sets

E1 =x : ε ≥ g1(x), g2(x), g3(x), . . . ; (3.5)

E2 =x : g1(x) > ε ≥ g2(x), g3(x), g4(x), . . . ;

E3 =x : g2(x) > ε ≥ g3(x), g4(x), . . . ;

. . . . . .

Now the sets Ek are measurable, by Theorem 1.15, and they are disjoint; one notesthat gn−1(x) > ε for x ∈ En, whereas gn−1(x) ≤ ε for x ∈ Ek with k < n. Further, theEk cover (a, b). Indeed, since gn(x) → 0, for each x there exists a smallest index k suchthat gn(x) ≤ ε for all n ≥ k ; then x ∈ Ek. (If something happens, then it must happen fora first time.) By Theorem 1.14,

∑∞k=1m(Ek) = m(∪∞k=1Ek) = b− a. Consequently there

exists an index K such that Fn =⋃∞

k=n+1Ek has measure m(Fn) ≤ ε for all n ≥ K.

Now Fn is our bad set; fortunately Fn has small measure, and on Fn we have therough bound |gn(x)| ≤ 2M. On the good set Gn = ∪n

k=1Ek we have the much betterestimate gn(x) ≤ ε. We can split our integral as∫ b

a

gn(x) dx =∫

Gn

gn(x) dx+∫

Fn

gn(x) dx (3.6)

and use the bounds on the good and bad sets to get∫ b

a

gn(x) dx ≤ εm(Gn) + 2Mm(Fn) (3.7)

≤ ε(b− a) + 2Mε. (3.8)

Hence∫ b

agn(x) dx→ 0 as n→∞ ; and we deduce that

∫ b

afn(x) dx→

∫ b

af(x) dx.

3.1 REMARK When applying the convergence theorems one should always work with

sequences and check that the hypotheses are fulfilled.

In the Tutorial Exercises, there are examples of sequences of functions fn that convergepointwise to some f, but such that the integrals

∫fn do not converge to

∫f. To ensure

that limiting sets and functions are measurable, we need to work with sequences, and notwith nets or limits over continuous parameters.

3.1 EXAMPLE. The Bessel function of order zero

J0(t) =∫ 2π

0

cos(t cos θ)dθ2π

(t ∈ R) (3.9)

41

is a continuous function of t .

Solution. It suffices to take t ∈ R and show that J0(tn) → J0(t) for any sequence oftn with tn → t as n→∞. Our neutral notation helps us to avoid confusion of variables.Here fn(θ) := cos(tn cos θ) is a continuous, and hence measurable, function of θ, and it isbounded with

|fn(θ)| ≤ 1 (n = 1, 2, . . . , θ ∈ [0, 2π]), (3.10)

since −1 ≤ cosx ≤ 1. By continuity of the cosine function, we have

fn(θ) = cos(tn cos θ) → cos(t cos θ) = f(θ)

as tn → t, for each θ ∈ [0, 2π].By the Bounded Convergence Theorem we can take limits

J0(tn) =∫ 2π

0

fn(θ)dθ2π

→∫ 2π

0

f(θ)dθ2π

= J0(t) (tn → t), (3.11)

and deduce that J0(t) is continuous at t.

In the following result, we remove the hypothesis of boundedness, and substitutemonotonicity. We recall that an increasing sequence of real numbers is either boundedabove, in which case the sequence converges, or the sequence is unbounded and hencediverges to infinity.

3.2 MONOTONE CONVERGENCE THEOREM. Let (fn(x)) be a sequence of positive

integrable functions on a measurable set E such that:

(i) the sequence is increasing, so f1(x) ≤ f2(x) ≤ f3(x) ≤ . . . ;

(ii) the sequence of integrals is bounded above, so∫

Efn(x) dx ≤ K for all n,

where K <∞.

Then there is a function f in L1(E) with fn(x) → f(x) as n → ∞, for almost all

x ∈ E, and ∫E

fn(x) dx→∫

E

f(x) dx as n→∞. (3.12)

Proof. For each x, the sequence (fn(x)) is increasing, and hence has a limit f(x) ≥ 0 ;this may be infinite, but we shall show that it is actually finite almost everywhere.

There is no loss in generality in proving this result in the special case of integrals overthe real line; for otherwise one can cut down to the smaller set E by multiplying by the in-dicator function of E. The proof of this particular case also involves a truncation argument.

42

For each M <∞ we cut down the functions fn to (fn)M (x) = I[−M,M ](x) minfn(x),M,which have graphs contained in the rectangle [−M,M ]× [0,M ].

For each fixed M <∞, it is evident that the truncations converge, so that (f)M (x) =

limn→∞(fn)M (x). Further, the truncations (fn)M are uniformly bounded for each M,

with 0 ≤ (fn)M (x) ≤ M. By the bounded convergence theorem, we deduce that (f)M is

measurable and ∫ M

−M

(f)M (x) dx = limn→∞

∫ M

−M

(fn)M (x) dx. (3.13)

We can now remove the boundedness constraint introduced by truncation. First, the

function f is measurable, since its truncations are measurable, and it is integrable. This

follows from the previous formula since by hypothesis (ii), there exists K <∞ such that∫ M

−M

(fn)M (x) dx ≤∫ ∞

−∞fn(x) dx ≤ K (3.14)

for all n and M. To determine the value of the integral of f, we exploit the fact that∫ M

−M(fn)M (x)dx is increasing in n and in M. Hence we can change the order of limits and

write ∫ ∞

−∞f(x) dx = lim

M→∞

∫ M

−M

(f)M (x) dx (3.15)

= limM→∞

limn→∞

∫ M

−M

(fn)M (x) dx

= supM,n

∫ M

−M

(fn)M (x) dx

= limn→∞

limM→∞

∫ M

−M

(fn)M (x) dx (3.16)

= limn→∞

∫ ∞

−∞fn(x) dx. (3.17)

3.2 REMARK. In most applications one knows f in advance; the force of the conclusion

is that the limit function is integrable and one can obtain the value of its integral. If (i)

holds, but (ii) does not, then∫fn diverges to infinity and f is not integrable. Identities

of the form ∞ = ∞ are meaningful, but seldom illuminating.

3.3 INEQUALITY OF THE MEANS. Let a1, . . . , an be positive numbers, and let G =

(a1 . . . an)1/n be their geometric mean and A = (a1 + · · ·+an)/n be their arithmetic mean.

Then the inequality G ≤ A holds, with equality if and only if all the aj are equal.

43

3.4 EXAMPLE. (Euler) The Gaussian integral is given by∫ ∞

0

e−x2 dx =√π

2.

Solution. This integral occurs in various branches of Mathematics; a notable in-

stance is in the probability density function e−x2/2/√

2π for a standard Gaussian random

variable. On account of this, there are several methods of evaluating the integral. Our

approach uses Wallis’s product for π in Theorem 3.5 below. We start by introducing the

functions

fn(x) = I(0,√

n)(x)(1− x2/n)n. (3.18)

We claim that the fn(x) increase monotonically to f(x) = e−x2 as n→∞. The value of

the limit is clear from elementary properties of the exponential function; to prove that our

sequence is increasing we recall the inequality of the means for positive numbers:

(geometric mean) ≤ (arithmetic mean). (3.19)

We apply this to the list of n+ 1 functions(1− x2

n

), . . . ,

(1− x2

n

), 1,

and we deduce the inequality

((1− x2

n

)n

× 1)1/(n+1)

≤ 1n+ 1

(n(1− x2

n

)+ 1

)(0 ≤ x ≤

√n), (3.20)

from whence (1− x2

n

)n/(n+1)

≤(1− x2

n+ 1

)(0 ≤ x ≤

√n),

so (1− x2

n

)n

≤(1− x2

n+ 1

)n+1

(0 ≤ x ≤√n). (3.21)

For fixed x > 0, we can take n so large that the preceding terms are all positive, and we

deduce that the fn(x) are ultimately increasing.

By the monotone convergence theorem∫∞0fn(x)dx→

∫∞0f(x)dx ; that is,

∫ √n

0

(1− x2

n

)n

dx→∫ ∞

0

e−x2 dx (n→∞). (3.22)

44

To evaluate the limit on the left-hand side, we substitute x =√n sin θ. This gives us

the limit√n

∫ π/2

0

cos2n+1 θ dθ →∫ ∞

0

e−x2 dx (n→∞). (3.23)

In the notation of the following theorem, the left-hand side is√nI2n+1, and we shall

identify the limit of this expression as n→∞.

3.5 THEOREM (Wallis). Let In =∫ π/2

0cosn θ dθ, for integers n ≥ 0. The recurrence

relation

(i) In =(n− 1)n

In−2 (n ≥ 2)

holds, and consequently:

(ii) I2n =(2n− 1)(2n− 3) . . . 1

(2n)(2n− 2) . . . 2π

2, (n ≥ 0); (3.24)

(iii) I2n−1 =(2n− 2)(2n− 4) . . . 2(2n− 1)(2n− 3) . . . 3

, (n = 1, 2, . . . );

(iv)√nI2n+1 →

√π

2(n→∞).

Proof. We observe that by elementary calculus I0 = π/2 and I1 = 1. To establish

the recurrence relation, we write

In =∫ π/2

0

cosn−1 θ cos θ dθ (3.25)

and integrate by parts to get

In =[cosn−1 θ sin θ

]π/2

0+

∫ π/2

0

(n− 1) cosn−2 θ sin 2θ dθ. (3.26)

Using the identity sin 2θ = 1− cos 2θ, we can reduce this to

In = (n− 1)∫ π/2

0

cosn−2 θ dθ − (n− 1)∫ π/2

0

cosn θ dθ (n > 1), (3.27)

so on recalling the definition of the integral we write this as

In = (n− 1)In−2 − (n− 1)In.

On simplifying, we achieve that stated result.

45

Using the recurrence relation repeatedly, we obtain (ii) for the even integers and (iii)for the odd integers. Taking the ratio of these expressions, we have

I2n

I2n−1=

(2n− 1)2(2n− 3)2 . . . 32(2n− 2)2(2n− 4)2 . . . 22

π

212n

=π

4nI22n−1. (3.28)

To obtain the desired result (iv), it suffices to show that I2n/I2n−1 → 1 as n→∞.

It is clear from the defining formula (3.29) that I2n−2 ≥ I2n−1 ≥ I2n, so

2n2n− 1

=I2n−2

I2n≥ I2n−1

I2n≥ 1. (3.29)

Letting n→∞ we deduce that I2n−1/I2n → 1, by a sandwich argument. This concludesthe proof of (iv) and identifies the limit in Example 3.2.

Not squaring the circle

The integral∫01(1−x2)1/2dx = π/4 is the area of a quadrant of the unit circle. Wallis

considered the more general integrals

J(p, q) =∫

0

1(1− x1/p)qdx

which include the previous as the special case q = 1/2, p = 1/2. When q is a positiveinteger, one can expand the integrand using the binomial theorem for integer coefficients,as in Pascal’s triangle, and obtain an explicit formula for J(p, q) in terms of p and q. Atthe time of Wallis, the general binomial theorem was not known, but Newton subsequentlydiscovered the general formula

(1 + y)s = 1 + sy +s(s− 1)

2!y2 +

s(s− 1)(s− 2)3!

y3 + . . . (−1 < y < 1; s ∈ R).

Taking y = −x2 and s = 1/2 in this formula, one can obtain a series formula for J(1/2, 1/2),hence a formula for π. Wallis was presumably hoping to find something similar, but why?

As a means of computing π, formula (iv) is singularly useless since it involves toomany products and converges slowly. It is reasonable to suppose that Wallis was aimingto prove:

3.6 LINDEMANN’S THEOREM (1882). One cannot square the circle: given a circle of

unit radius in the plane, it is not possible to construct, with straight edge and compass by

using only standard operations, a square that bounds an equal area.

Lindemann showed that√π is not algebraic; that is, there does not exist a non-zero

polynomial with integral coefficients with√π as a root. However, Hilbert showed that any

46

number that may be constructed from a unit length using straight edge and compass isalgebraic.

The problem of squaring the circle was raised in Antiquity; many unsuccessful at-tempts to solve it led to valuable results. Questions of this kind are undergoing a revivalof interest.

Periods.∗ A period is a complex number such that the real and imaginary parts areequal to absolutely convergent integrals of rational functions with rational coefficients overa set that is determined by finitely many inequalities involving rational functions.

3.6. EXAMPLE. The following numbers are periods.

(i)√

2 =∫

[2x2≤1]

dx; (ii) log 2 =∫

1

2dxx

;

(iii) π =∫∫

[x2+y2≤1]

dxdy = 2∫−1

1√

1− x2 dx.

3.6 LEMMA. There are only countably many periods, and hence there exists a complex

number that is not a period.

3.6 CONJECTURE. (Kontsevich, Zagier). The base e of natural logarithms is not a

period.

3.6 PROBLEM. Exhibit a complex number that is not a period.

This problem is likely to be difficult.

3.7 The Dominated Convergence Theorem

3.7 DOMINATED CONVERGENCE THEOREM. Let (fn(x)) be a sequence of measurable

functions on a measurable set E such that:

(i) |fn(x)| ≤ F (x) for all x ∈ E, where F ∈ L1(E) ; and

(ii) fn(x) → f(x) as n→∞, for all x ∈ E.Then the limit function f belongs to L1(E) and∫

E

fn(x) dx→∫

E

f(x) dx (n→∞). (3.30)

Proof. As before, one can reduce to the case in which E = R by multiplying thefunctions by the indicator function IE . The limit function f is certainly measurable,and since |f(x)| = limn→∞ |fn(x)| ≤ F (x), it is integrable. To evaluate its integral, we

47

introduce further truncations which will reduce the problem to the context of the boundedconvergence theorem.

The key idea is that, since the dominating function F is integrable, everything takesplace in the region of the plane (x, y) : −F (x) ≤ y ≤ F (x) which is of finite area. Mostof this area lies in a square of side 2M for some M <∞. Specifically, given ε > 0 we let

(F )M (x) = I[−M,M ](x) minM,F (x), (3.31)

where M <∞ is chosen so large that

0 ≤∫ ∞

−∞

F (x)− (F )M (x)

dx < ε. (3.32)

By truncating F horizontally and vertically, we truncate all of the integrands in (3.30) atone fell swoop: the truncations

(fn)M (x) = I[−M,M ](x)max−M,minM,fn(x)

satisfy

|fn(x)− (fn)M (x)| ≤ F (x)− (F )M (x); (3.33)

from whence ∫ ∞

−∞

∣∣fn(x)− (fn)M (x)∣∣ dx < ε (n = 1, 2, . . . ), (3.34)

and similarly ∫ ∞

−∞

∣∣f(x)− (f)M (x)∣∣ dx < ε. (3.35)

Further, the truncations (fn)M are bounded, with (fn)M (x) → (f)M (x), where

(f)M (x) = I[−M,M ](x) max−M,minM,f(x)

. (3.36)

By the bounded convergence theorem, it holds that

limn→∞

∫ M

−M

(fn)M (x) dx =∫ M

−M

(f)M (x) dx. (3.37)

We can conclude the proof by using the triangle inequality to show that∣∣∣∫ f −∫fn

∣∣∣ (3.38)

≤∣∣∣∫ (f − (f)M )

∣∣∣ +∣∣∣∫ ((f)M − (fn)M )

∣∣∣ +∣∣∣∫ ((fn)M − fn)

∣∣∣. (3.39)

48

On the right-hand side, the first and third integrals are each less than ε, by (3.34) and(3.35) ; and by (3.37) the middle integral is also less than ε, for all sufficiently large n.

3.7 EXAMPLE. For each real α ,∫ ∞

0

sinαxex − 1

dx =∞∑

k=1

α

k2 + α2. (3.40)

This equation is used in number theory to establish identities for the Riemann zetafunction.

Solution. The rough idea is to expand the integrand f(x) as a geometric series

f(x) =sinαxex − 1

=sinαx

ex(1− e−x)=

∞∑k=1

sinαx e−kx, (3.41)

so that ∫ ∞

0

sinαxex − 1

dx =∞∑

k=1

∫ ∞

0

e−kx sinαxdx. (3.42)

To justify this, we introduce the functions

fn(x) =n∑

k=1

e−kx sinαx (n = 1, 2, . . . ;x > 0), (3.43)

that are the partial sums of the geometric series. Since the common ratio of our geometricseries is 0 < e−x < 1 for x > 0, we have convergence, and

fn(x) → f(x) =sinαxex − 1

(x > 0), (3.44)

as n→∞ as in (ii) of the dominated convergence theorem.Our choice of dominating function is

F (x) =| sinαx|ex − 1

, (3.45)

as suggested by (3.44). Using the triangle inequality in (3.43) one shows that |fn(x)| ≤F (x) for all x and n ; so it remains to check that F is integrable. To do this, we note thatF decays like e−x as x→∞, so is integrable over (1,∞). For small values of x > 0, onecan use Maclaurin series to show

0 ≤ F (x) =| sinαx|ex − 1

=|α|x

(1 + x+O(x2))− 1= |α|+O(x); (3.46)

49

hence F (x) is continuous on [0, 1], and consequently integrable on (0, 1). One can alter-

natively argue that ex ≥ 1 + x and | sinαx| ≤ |α|x for 0 ≤ x ≤ 1 to reach the same

conclusion.

By the Dominated Convergence Theorem, we have∫∞0fn(x) dx →

∫∞0f(x) dx as

n→∞ ; this gives us (3.42).

To conclude the calculation, we only need to know that∫ ∞

0

e−kx sinαxdx =α

α2 + k2(k > 0;α ∈ R). (3.47)

This elementary integral may be found by repeated integration by parts.

3.7 REMARKS. (i) The convergence theorems for integrals are deduced in this course

from the countable additivity property (Σ) of Lebesgue measure. One can reverse the

argument and deduce (Σ) from the conclusions of the convergence theorems.

(ii) The bounded convergence theorem applies to bounded functions on bounded in-

tervals.

(iii) The monotone convergence theorem applies to increasing sequences of functions

(which are quite different from sequences of increasing functions; for example, think of

fn(x) = xn ).

(iv) The dominated convergence theorem is particularly useful when dealing with

Fourier integrals. See chapter 4 below.

(v) The dominated convergence theorem implies the monotone and the bounded con-

vergence theorems; this justifies the alternative name ‘general convergence theorem.’

(vi) There are corresponding results for series which justify integration through the

summation sign.

3.7 EXAMPLE. The power series expansion of the Bessel function of order zero is

J0(t) =∫ 2π

0

cos(t cos θ

) dθ2π

=∞∑

n=0

(−1)n t2n

22n(n!)2(t ∈ R). (3.48)

Solution. In the defining integral, we expand the integrand as a power series in

t cos θ, namely

J0(t) =∫ 2π

0

cos(t cos θ)dθ2π

=∫ 2π

0

∞∑n=0

(−1)nt2n

(2n)!cos2n θ

dθ2π. (3.49)

(Recall the Maclaurin series cosx = 1− x2/2! + x4/4! + . . . .)

50

Anticipating the proof, we interchange the order of summation and integration and

write

J0(t) =∞∑

n=0

(−1)nt2n

(2n)!

∫ 2π

0

cos2n θdθ2π. (3.50)

Provided that we can find a dominating function for the partial sums of the series is

absolutely convergent, then the Dominated Convergence Theorem will justify the change

of order of summation and integration which was carried out above. We let

fN (θ) =N∑

n=0

(−1)nt2n

(2n)!cos2n θ,

which are bounded by

|fN (θ)| ≤∞∑

n=0

t2n

(2n)!= cosh t.

So, recalling (3.24), we substitute the value I2n = (2n)!/22n(n!)2 for the cosine integral in

the nth term and obtain

J0(t) =∞∑

n=0

(−1)nt2n

22n(n!)2. (3.51)

This series converges absolutely for all real t, as is clear from the ratio test.

3.8 Completeness∗

Cauchy’s convergence criterion asserts that a sequence (an) of complex numbers con-

verges, if and only if for each ε > 0 there exists n0(ε) such that |an − am| < ε whenever

n,m ≥ n0(ε). This condition is often stated as:

(an) converges ⇐⇒ |an − am| → 0 as n,m→∞.

On account of this property, one says that the complex numbers are complete.

The Lebesgue spaces Lp (p = 1, 2) also have the important property of completeness,

so that a sequence (fn) of Lp functions with ‖fn − fm‖LP → 0 as n,m→∞ converges in

Lp norm to some f ∈ Lp. We prove this in the case of L2.

3.8 COMPLETENESS THEOREM. Let (fn) (n = 1, 2, . . . ) be a sequence of functions

in L2 with ∫ ∞

−∞

∣∣fn(x)− fm(x)∣∣2 dx→∞ (n,m→∞.) (3.52)

Then there exists f ∈ L2 such that fn → f in L2.

51

Proof. It is not generally true to say that the fn(x) are pointwise convergent, soto produce our candidate for the limit we need to extract a subsequence which convergesalmost surely. We select an increasing sequence of integers n(k) (k = 1, 2, . . . ) such that∫ ∞

−∞

∣∣fn(x)− fm(x)∣∣2 dx ≤ 2−2k (n,m ≥ n(k)) (3.53)

for each integer k ≥ 1. For K <∞ we have( K∑k=1

∣∣fn(k)(x)− fn(k+1)(x)∣∣)2 =

K∑k=1

K∑`=1

∣∣fn(k)(x)− fn(k+1)(x)∣∣∣∣fn(`)(x)− fn(`+1)(x)

∣∣,(3.54)

where the expression increases as K increases. Hence∫ ∞

−∞

( K∑k=1

∣∣fn(k)(x)− fn(k+1)(x)∣∣)2 dx

=K∑

k=1

K∑`=1

∫ ∞

−∞

∣∣fn(k)(x)− fn(k+1)(x)∣∣∣∣fn(`)(x)− fn(`+1)(x)

∣∣ dx. (3.55)

We estimate this by using the Cauchy-Schwarz inequality for integrals

≤K∑

k=1

K∑`=1

(∫ ∞

−∞

∣∣fn(k)(x)−fn(k+1)(x)∣∣2 dx

)1/2(∫ ∞

−∞

∣∣fn(`)(x)−fn(`+1)(x)∣∣2 dx

)1/2

(3.56)

Then estimating each term, we get

≤K∑

k=1

K∑`=1

2−k2−` ≤( K∑

k=1

2−k)2, (3.57)

which is uniformly bounded by one. By the monotone convergence theorem 3.2, the inte-grands

(∑Kk=1 |fn(k)(x)− fn(k+1)(x)|

)2 converge almost everywhere as K →∞, and their

limit defines an integrable function. Hence∑

k(fn(k)(x)−fn(k+1)(x)) is absolutely conver-gent almost everywhere, and the limit limk→∞ fn(k)(x) is our f. We can define f(x) = 0on the null set on which the integrands diverge.

To check that fn converges to f in L2, we take n(k) > m to obtain∫ ∞

−∞

∣∣f(x)− fm(x)∣∣2 dx ≤

∫ ∞

−∞

(∣∣fn(k)(x)− fm(x)∣∣ +

∞∑`=k

∣∣fn(`+1)(x)− fn(`)

∣∣)2 dx. (3.58)

This expression is finite, so f ∈ L2 ; further, the right-hand side converges to zero asm→∞, so fm → f in L2.

COROLLARY 3.8. With the standard inner product (2.43), L2(R) forms a Hilbert space.

52

§4. Fourier transforms

At the end of this chapter, you should be able to:

establish that Fourier transform integrals are continuous;

calculate the Fourier integral of the Gaussian probability density function;

understand the statement of the Plancherel formula.

Lebesgue measure has the property of translation invariance, namely the measure of

a set does not change when it is moved along the real line. In terms of integrals this may

be expressed as

(T )∫ ∞

−∞f(t+ x) dx =

∫ ∞

−∞f(x) dx (t ∈ R, f ∈ L1(R)). (4.1)

For each t ∈ R the trigonometric functions cosxt and sinxt are 2π|t| -periodic. Fourier

showed that many functions can be analysed in terms of these waveforms.

The Fourier transform of an integrable function f is defined to be

f(t) =∫ ∞

−∞exp(−itx)f(x) dx (t ∈ R). (4.2)

It is conventional to have −itx in the exponent, so the formula for the Fourier trans-

form is is not quite the same as the definition of the characteristic function of a probability

density function as in MATH313. To avoid calculations with complex numbers, one often

uses the Fourier cosine transform instead.

4.1 PROPOSITION. Let f ∈ L1(0,∞) and let

ϕ(t) =∫ ∞

0

cos tx f(x) dx (t ∈ R) (4.3)

be its Fourier cosine transform. Then ϕ(t) is a bounded and continuous function of t ∈ Rwith |ϕ(t)| ≤

∫∞0|f(x)|dx.

Proof. First, | cos tx| ≤ 1. By the triangle inequality we have

|ϕ(t)| ≤∫ ∞

0

| cos tx||f(x)| dx

≤∫ ∞

0

|f(x)| dx,

so we have a bound that is independent of t.

53

Now let (tn) be a real sequence such that tn → t as n→∞. We require to prove thatϕ(tn) → ϕ(t). Since cos is continuous, we have

f(x) cos tnx→ f(x) cosxt (tn → t)

and we have a domination

|f(x) cosxtn| ≤ |f(x)| (n = 1, 2, . . . )

where |f(x)| is integrable. Hence by the Dominated Convergence Theorem∫ ∞

0

f(x) cos tnx dx→∫ ∞

0

f(x) cosxt dx (tn → t);

that is, ϕ(tn) → ϕ(t).

EXAMPLE 4.1. The Fourier cosine transform of e−x2/2 is√

(π/2)e−t2/2.

Solution. This may be evaluated by various methods, including the complex analysistechniques indicated in MATH313. Here we start with the identity∫ ∞

0

cosxt e−x2/2 dx =∫ ∞

0

∞∑n=0

(−1)n t2nx2n

(2n)!e−x2/2 dx (4.4)

involving the Maclaurin series for cosxt. The series for cosxt may be compared to theMaclaurin series for coshxt, giving a term-by-term bound

∣∣∣ ∞∑n=0

(−1)n t2nx2n

(2n)!

∣∣∣ ≤ ∞∑n=0

t2nx2n

(2n)!= coshxt, (4.5)

where the dominating function F (x) = coshxt e−x2/2 is integrable. By the DominatedConvergence Theorem, we can change the order of summation and integration to write∫ ∞

0

cosxt e−x2/2 dx =∞∑

n=0

(−1)n t2n

(2n)!

∫ ∞

0

x2ne−x2/2 dx. (4.6)

To reduce to a Gamma function, we set u = x2/2, and obtain

ϕ(t) =∞∑

n=0

(−1)n t2n

(2n)!2n− 1

2

∫ ∞

0

un− 12 e−u du (4.7)

=∞∑

n=0

(−1)n t2n2n− 1

2

(2n)!Γ(n+

12

). (4.8)

54

By Legendre’s duplication formula 2.10(f) of MATH313, with p = n+ 12 , this is

ϕ(t) =∞∑

n=0

(−1)n t2n2n− 1

2

(2n)!Γ(2n+ 1)Γ( 1

2 )22nΓ(n+ 1)

=√π

2

∞∑n=0

(−1)n

n!t2n

2n; (4.9)

which we recognise as the convergent series for√

(π/2)e−t2/2. Alternatively, one can useQ27 to obtain the integrals in (4.6).

4.2 INTEGRATED COSINE INVERSION FORMULA.

Let f belong to L1(0,∞) and let ϕ(t) be its Fourier cosine transform, defined by

ϕ(t) =∫ ∞

0

cos tx f(x) dx. (4.10)

Then f is determined by ϕ by the formula∫ x

0

f(u) du = limT→∞

2π

∫ T

0

sinxtt

ϕ(t) dt. (4.11)

Proof. This was partially proved in the MATH313 notes. The missing detail therecan now be filled in using the Dominated Convergence Theorem. We change the order ofintegration on the right-hand side to get∫ T

0

sin txt

∫ ∞

0

cos tuf(u)du dt =∫ ∞

0

∫ T

0

sin tx cos tut

dtf(u) du. (4.12)

This is easily justified as the iterated integrals are absolutely convergent since f ∈ L1. Weuse the trigonometric addition formula

2 sin tx cos tu = sin(x+ u)t+ sin(x− u)t (4.13)

to simplify the inner integral.

LEMMA 4.2. (i) ∣∣∣∫ T

0

sin tx cos tut

dt∣∣∣ ≤ π

2+

2π. (4.14)

(ii) As T →∞, ∫ T

0

sin tx cos tut

dt→ π

2I(−x,x)(u). (4.15)

Proof. We split the integral∫ T

0

sin tt

dt =∫ π/2

0

sin tt

dt+∫ T

π/2

sin tt

dt (4.16)

55

and use the simple bound 0 < (sin t)/t < 1 for 0 < t < 1 to obtain

0 ≤∫ π/2

0

sin tt

dt ≤∫ π/2

0

dt =π

2. (4.17)

In the final integral in (4.16), we integrate by parts to obtain∫ T

π/2

sin ttdt =

[− cos tt

]T

π/2−

∫ T

π/2

cos tt2

dt

=− cosTT

−∫ T

π/2

cos tt2

dt (4.18)

so that ∣∣∣∫ T

π/2

sin ttdt

∣∣∣ ≤ | cosT |T

+∫ T

π/2

| cos t|t2

dt

≤ | cosT |T

+∫ T

π/2

1t2dt

≤ | cosT |T

+[−1t

]T

π/2

=| cosT |T

+2π− 1T

≤ 2π. (4.19)

Hence by adding the inequalities (4.17) and (4.19) we obtain

∣∣∣∫ T

0

sin tt

dt∣∣∣ ≤ ∣∣∣∫ π/2

0

sin tt

dt∣∣∣ +

∣∣∣∫ T

π/2

sin tt

dt∣∣∣ ≤ π

2+

2π. (4.20)

Since sinxt cos tu = 2−1(sin(x+ u)t+ sin(x− u)t), we can easily deduce that∫ T

0

sin tx cos tut

dt =12

∫ T

0

sin(x+ u)tt

dt+12

∫ T

0

sin(x− u)tt

dt (4.21)

is bounded in modulus by π/2 + 2/π.(ii) We have ∫ T

0

sin(x+ u)tt

dt→π

2sgn(x+ u),∫ T

0

sin(x− u)tt

dt→π

2sgn(x− u) (T →∞). (4.22)

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Conclusion of the proof of Theorem 4.2. By the Lemma, we can apply thedominated convergence theorem to deduce that the inner integral converges as T →∞ to

π/2, if − x < u < x;∫ T

0

sin tx cos tut

dt→ π/4, if u = ±x; and

0, else. (4.23)

Of course, only the values 0 < u < x will make a contribution to our integral, which is∫ ∞

0

∫ T

0

sinxt cos tut

dtf(u) du→ π

2

∫ ∞

0

I(x,x)(u)f(u) du =π

2

∫ x

0

f(u) du (4.24)

in the limit as T →∞ , by the dominated convergence theorem. This is the left-hand sideof (4.11), as required.

4.3 LEMMA. For distinct real x and y we have

−1 ≤ sinx− sin yx− y

≤ 1. (4.25)

Proof. This may be deduced from the mean value theorem. Alternatively, one canuse the formula

sinx− sin y =∫ x

y

cos tdt, (4.26)

which, for y < x, leads to the inequality

| sinx− sin y| ≤∫ x

y

| cos t|dt ≤∫ x

y

1 dt = x− y (4.27)

since | cos t| ≤ 1.

4.3 COSINE INVERSION THEOREM∗. Suppose that f(x) belongs to L1(0,∞), and

suppose further that its Fourier cosine transform ϕ(t) also belongs to L1(0,∞). Then f

is almost everywhere equal to a continuous function, and the inversion formula

f(x) =2π

∫ ∞

0

cos txϕ(t) dt, (4.28)

holds at all points x of continuity of f.

Proof. By the integrated form of the inversion theorem, we deduce that∫ y

0

f(u) du = limT→∞

2π

∫ T

0

sin tyt

ϕ(t) dt =2π

∫ ∞

0

sin tyt

ϕ(t) dt, (4.29)

57

by using absolute convergence at the last. We wish to obtain a formula for f(x) at somepoint of continuity x. Suppose that (xn) is a sequence with xn → x as n→∞. Then

1xn − x

∫ xn

x

f(u) du =2π

∫ ∞

0

sin txn − sin txtxn − tx

ϕ(t) dt. (4.30)

As xn → x, the integrands have limit

sin txn − sin txtxn − tx

ϕ(t) → cos txϕ(t); (4.31)

and by Lemma 4.3 they are bounded by∣∣∣ sin txn − sin txtxn − tx

∣∣∣|ϕ(t)| ≤ |ϕ(t)| (t > 0;n ≥ 1), (4.32)

where the right-hand side is by hypothesis an integrable function of t > 0. Hence by thedominated convergence theorem

2π

∫ ∞

0

sin txn − sin txtxn − tx

ϕ(t) dt→ 2π

∫ ∞

0

cosxtϕ(t) dt (n→∞). (4.33)

The left-hand side of (4.30) must also converge to the same limit. Since f is continuousat x we have

1xn − x

∫ xn

x

f(u) du→ f(x) (xn → x). (4.34)

The expression g(x) = 2π

∫∞0

cos txϕ(t) dt defines a continuous function. If f(x) 6=g(x) on a set of positive measure, then one could find a < b with

0 6= 1b− a

∫ b

a

(g(x)− f(x)) dx;

but we have

1b− a

∫ b

a

(g(x)− f(x)) dx = limT→∞

∫ T

0

sin bt− sin atbt− at

(ϕ(t)− ϕ(t)) dt = 0. (4.35)

Hence f(x) = g(x) at almost all points x ∈ (0,∞)

REMARKS. (i) The dominated convergence theorem is particularly important in ap-plications to the Fourier cosine transform, and we often take the dominating function tobe F (x) = M |f(x)| where M is some constant.

(ii)It is simplest to define the Fourier cosine transform for functions f ∈ L1(0,∞),since the integral (4.3) is (absolutely) convergent and gives a continuous bounded function;

58

the Fourier cosine transform thus gives us a map L1(0,∞) → Cb[0,∞). Unfortunately,the transform ϕ(t) need not be integrable over (0,∞), which leads to the asymmetricalinversion formula (4.11), or the artificial hypotheses for Theorem 4.3.

L1-transform. When f is a probability density function, and ϕ is the real part of thecharacteristic function, this asymmetry is entirely natural and advantageous, whereby thelaws obeyed by random variables are transformed into different laws for the characteristicfunctions.

L2-transform. The formula (4.28) suggests that a function and its Fourier cosinetransform should have the same status, and we can ensure symmetry by changing thespace of functions to L2 and changing the interpretation of the integral that defines theFourier cosine transform. The following result shows that the Fourier transform can beinterpreted as an invertible map L2(0,∞) → L2(0,∞).

4.5 PLANCHEREL COSINE FORMULA∗. Suppose that f is a real function with f ∈L1(0,∞) and f ∈ L2(0,∞). Then its Fourier cosine transform ϕ also belongs to L2(0,∞).

If one merely supposes that f ∈ L2(0,∞), then f has a Plancherel cosine transform

ϕ(t) defined by the limit∫ T

0

f(x) cos txdx→ ϕ(t) (T →∞) (4.36)

in L2(0,∞). The Fourier-Plancherel transform satisfies∫ ∞

0

f(x)2 dx =2π

∫ ∞

0

ϕ(t)2 dt. (4.37)

59