The University of Sydney

School of Mathematics and Statistics

Solutions to Tutorial 4 (Week 5)

MATH3969: Measure Theory and Fourier Analysis (Advanced) Semester 2, 2014

Web Page: http://www.maths.usyd.edu.au/u/UG/SM/MATH3969/

Lecturer: Daniel Daners

Material covered

(1) integrals of non-negative functions

(2) interchanging limits and integrals

(3) applications of the monotone convergence theorem

(4) simple substitution formulae

Outcomes

After completing this tutorial you should

(1) be able to prove elementary properties of integrals

(2) know how to apply the monotone convergence theorem

(3) have an appreciation of conditions allowing to interchange limits and integrals

Summary of essential material

A simple function ϕ : X → Y is a function with finite range {a0, a1, . . . , an}. Let Ak :=ϕ−1[{ak}. If Y = R

N or CN (or some other vector space), then

ϕ =n

∑

k=0

ak1Ak(1)

is a linear combination of indicator functions for the disjoint sets Ak. If (X,A, µ) is a measurespace and Y a metric space (subset of RN most of the time), then ϕ is measurable if and onlyif Ak ∈ A for all k = 0, 1, . . . , n.

If ϕ : X → [0,∞) is a simple measurable function of the form (1) with Ak, k = 0, . . . , n,disjoint, we define the integral

∫

X

ϕdµ :=n

∑

k=0

akµ(Ak). (2)

If ϕ, ψ : X → [0,∞) are measurable simple functions, then

•

∫

X

ϕ+ ψ dµ =

∫

X

ϕdµ+

∫

X

ψ dµ;

•

∫

X

αϕdµ = α

∫

X

ϕdµ for all α ≥ 0;

• If 0 ≤ ϕ ≤ ψ, then 0 ≤

∫

X

ϕdµ ≤

∫

X

ψ dµ.

Copyright c© 2014 The University of Sydney 1

If f : X → [0,∞] is measurable we define∫

X

f dµ := sup{

∫

X

ϕdµ : ϕ simple and measurable and 0 ≤ ϕ ≤ f}

. (3)

The two definitions coincide for simple functions (see Question 1 below).

One of the key properties of the abstract Lebesgue integral is the monotone conver-

gence theorem: Let fn : X → [0,∞] be measurable functions such that 0 ≤ fn ≤ fn+1

for all n ∈ N. Then

limn→∞

∫

X

fn dµ =

∫

X

limn→∞

fn dµ.

Note that due to the monotonicity limn→∞ fn(x) exists in [0,∞] for all x ∈ X.’A corollary is Fatou’s Lemma: Let fn : X → [0,∞] be arbitrary measurable functions.Then

∫

X

lim infn→∞

fn dµ ≤ lim infn→∞

∫

X

fn dµ.

Questions to complete during the tutorial

1. Show that the definitions (2) and (3) coincide for simple functions f . Use the definitionof a supremum for that.

Solution: Suppose f is a simple measurable function. Denote the integral defined in

(2) by

∫ old

X

and the one in (2) by

∫ new

X

. If we set ϕ0 := f , then 0 ≤ ϕ0 ≤ f and by

definition of an supremum∫ new

X

f dµ ≥

∫ old

X

ϕ0 dµ =

∫ old

X

f dµ.

If 0 ≤ ϕ ≤ f and ϕ is simple and measurable, then from the properties of integrals of asimple function we have

∫ old

X

ϕdµ ≤

∫ old

X

f dµ.

Hence

∫ old

X

f dµ is an upper bound for all 0 ≤ ϕ ≤ f and hence by definition of a

supremum∫ new

X

f dµ = sup{

∫

X

ϕdµ : ϕ simple and measurable and 0 ≤ ϕ ≤ f}

≤

∫ old

X

f dµ.

Combining the two inequalities we get equality between the two different definitions.

2. Let fn(x) := 2nxe−nx2

for all x ≥ 0 and n ∈ N. Then fn(x) → 0 for all x ≥ 0 as n→ ∞.

Show that

∫ ∞

0

fn(x) dx 6→ 0.

Solution: A simple substitution shows that∫ ∞

0

2nxe−nx2

dx = −e−nx2

∣

∣

∣

∞

0

= 1

for all n ∈ N. Hence, even though the integrand converges to zero pointwise, the limit ofthe integral is not zero.

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3. Suppose that µ is a measure defined on the σ-algebra A of subsets of X. Let f : X →[0,∞) be a nonnegative measurable function with

∫

Xf dµ = C <∞. Show that for each

α > 0 we have

µ(

{x ∈ X : f(x) ≥ α})

≤C

α.

Solution: Let A := {x ∈ X : f(x) ≥ α}. Then

C =

∫

X

f dµ ≥

∫

X

1Af dµ ≥ α

∫

X

1A dµ = αµ(A).

Hence the claim follows by dividing by α > 0.

4. Let δa be the Dirac measure on a set X concentrated a ∈ X. It is defined on the powerset P(X), so every set and every function is measurable.

(a) Let ϕ =∑

n

k=0ak1Ak

be a non-negative simple function. Find

∫

X

ϕdδa

Solution: Since⋃

n

k=0Ak = X is a disjoint union there exists precisely one

m ∈ {0, . . . , n} such that a ∈ Am. From the definition of the Dirac measure wehave δa(Ak) = 0 if k 6= m and δa(Am) = 1. Hence by definition of the integral of asimple function

∫

X

ϕdδa =n

∑

k=0

akδa(Ak) = am = ϕ(a).

(b) Let f : X → [0,∞] be a function. From the definition of the integral of a non-

negative function, determine

∫

X

f dδa.

Solution: Let ϕ be a simple function such that 0 ≤ ϕ ≤ f . Then by definitionof the integral of f as the supremum over all such ϕ and part (a)

∫

X

ϕdδa = ϕ(a) ≤ f(a),

so∫

X

f dδa ≤ f(a).

To prove the reverse inequality we use the simple function ϕ defined to be f(a) atx = a and zero otherwise. Then clearly 0 ≤ ϕ ≤ f and

∫

X

f dδa ≥

∫

X

ϕdδa = f(a).

Both inequalities together imply the required identi

5. Let µ be the counting measure on N, that is,

µ(A) :=

{

card(A) if A is a finite set,

∞ if A is infinite.

3

(a) Let f : N → [0,∞), that is f(k) := ak ≥ 0 for all k ∈ N. Use the monotoneconvergence theorem to show that

∫

N

f dµ =∞∑

k=0

ak.

Solution: Let fn(k) := f(k) if 0 ≤ k ≤ n and fn(k) := 0 for k > n. Then fn is asimple function attaining the finitely many values a0, . . . , an and 0. In particular,

fn =n

∑

k=0

f(k)1{k} =n

∑

k=0

ak1{k}.

By definition of the integral of a simple function we have

∫

N

fn dµ =n

∑

k=0

akµ({k}) =n

∑

k=0

ak

since µ({k}) = 1. From the definition of fn we have fn ≤ fn+1 and limn→∞ fn(k) =f(k) = ak for all k ∈ N. By the monotone convergence theorem

∫

N

f dµ = limn→∞

∫

N

fn dµ = limn→∞

n∑

k=0

ak =∞∑

k=0

ak.

(b) Show that f : N → C is integrable with respect to the counting measure if andonly if

∑∞k=0

ak is absolutely convergent. Here ak := f(k).

Solution: Note that |f(k)| = |ak| for all k ∈ N. Using part (a) integrable isequivalent to

∫

N

|f | dµ =∞∑

k=0

|ak| <∞.

The latter is the definition of absolute convergence.

(c) Suppose that f : N → R is integrable with respect to the counting measure. Usethe definition of the general Lebesgue integral to show that

∫

N

f dµ =∞∑

k=0

ak,

where ak := f(k).

Solution: We know that f+(k) = max{f(k), 0} = max{ak, 0} and f−(k) =max{−f(k), 0} = max{−ak, 0}. Hence f(k) = f+(k) − f−(k) for all n ∈ N. Bydefinition

∫

N

f dµ =

∫

N

f+ dµ−

∫

N

f− dµ =∞∑

k=0

f+(k)−∞∑

k=0

f−(k)

=∞∑

k=0

(

f+(k)− f−(k))

=∞∑

k=0

ak.

A similar argument applies if we replace R by C.

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Extra questions for further practice

6. (a) Show that if U ⊆ RN is open and has Lebesgue measure zero, then U = ∅.

Solution: Suppose that U is non-empty and let x ∈ U . Since U is open thereexists an open rectangle R with R ⊆ U . We know that 0 < m(R) ≤ m(U) 6= 0.By contrapositive the claim follows.

(b) Use (a) to show that if f : RN → [0,∞) is continuous and if∫

RN f(x) dx = 0, thenf(x) = 0 for all x ∈ R

N .

Solution: Suppose that f(x) = 0 for all x ∈ RN . Then obviously

∫

RN f(x) dx =0. We prove the converse by contrapositive and assume that f is continuous andnon-zero. Then the sets

Un := f−1[(1/n,∞)] = {x ∈ RN : f(x) > 1/n}

are open for every n ∈ N. Since

⋃

n∈N

Un = {x ∈ RN : f(x) 6= 0} 6= ∅

there exists n ∈ N such that Un 6= ∅ and so by (a) µ(Un) 6= 0. Hence

∫

RN

f(x) dx ≥

∫

RN

1Unf(x) dx ≥

1

n

∫

RN

1Undx =

1

nµ(Un) > 0.

Hence the integral is non-zero, proving the contrapositive.

(c) Show that for an arbitrary measurable function f : RN → [0,∞) we can have∫

RN f(x) dx = 0 without f being the zero function. What condition needs to besatisfied for the integral to be zero?

Solution: An example of a non-zero function with zero integral is f = 1A, whereA is a non-empty set of measure zero, for example a singleton A = {a}. Moregenerally, the integral is zero if {x ∈ R

N : f(x) 6= 0} has zero measure.

We show that the above condition is also necessary for∫

RN f(x) dx = 0. We give aproof by contrapositive and assume that {x ∈ R

N : f(x) 6= 0} has positive measure.Define Un as above and note that

limn→∞

m(Un) = m(

⋃

n∈N

Un

)

= m(

{x ∈ RN : f(x) 6= 0}

)

> 0.

Hence there exists n ∈ N such that m(Un) > 0 and so

∫

RN

f(x) dx ≥

∫

RN

1Unf(x) dx ≥

1

n

∫

RN

1Undx =

1

nµ(Un) > 0.

Hence the integral is non-zero, proving the contrapositive.

7. Let f : R → C be measurable.

(a) Show that the functions x → f(x − t) and x → f(−x) are measurable for everyt ∈ R.

Solution: We first note that as a consequence of the definition, the Lebesguemeasure is translation invariant, that is, m(t + A) = m(A) for every measurableset A ⊆ R and t ∈ R, where t+A := {t+ x : x ∈ A}. Similarly m(−A) = m(A) if

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we set −A := {−x : x ∈ A}. It is also clear that t + A and −A are measurable ifand only if A is measurable.

Fix t ∈ R and let A := {x ∈ R : f(x) > a}. Therefore f(x − t) > a if and onlyif x − t ∈ A, that is, x ∈ t + A. Since A is measurable also t + A is measurable.Hence x 7→ f(x − t) is measurable. Similarly f(−x) > a if and only if −x ∈ A,that is, x ∈ −A. Hence also x→ f(−x) is measurable.

(b) Prove that

∫ ∞

−∞

f(x− t) dx =

∫ ∞

−∞

f(x) dx and

∫ ∞

−∞

f(−x) dx =

∫ ∞

−∞

f(x) dx.

Hint: First assume f is a nonnegative simple function on R and that t ∈ R. Thendo the general case.

Solution: If ϕ =∑

n

k=0ak1Ak

is a non-negative simple function, then by thetranslation invariance of the Lebesgue measure

∫ ∞

−∞

ϕ(x− t) dx =m∑

k=0

akm(Ak − t) =m∑

k=0

akm(Ak) =

∫ ∞

−∞

ϕ(x) dx.

Similarly

∫ ∞

−∞

ϕ(−x) dx =m∑

k=0

akm(−Ak) =m∑

k=0

akm(Ak) =

∫ ∞

−∞

ϕ(x) dx.

Taking the supremum over all simple functions 0 ≤ ϕ ≤ f , the assertions followfor non-negative measurable functions. For general f use the definition and applythe formula to positive and negative parts and then to real and imaginary parts.

(c) Prove that

∫

b

a

f(x− t) dx =

∫

b−t

a−t

f(x) dx and

∫

b

a

f(−x) dx =

∫ −a

−b

f(x) dx.

Solution: If f is a measurable function on (a, b) we simply set g(x) := f(x) ifx ∈ (a, b) and g(x) = 0 otherwise. Then g is a measurable function and

∫ ∞

−∞

g(x) dx =

∫

b

a

f(x) dx.

Hence by the previous part

∫

b−t

a−t

f(x− t) dx =

∫ ∞

−∞

g(x− t) dx =

∫ ∞

−∞

g(x) dx =

∫

b

a

f(x) dx

and similarly

∫

b

a

f(−x) dx =

∫ ∞

−∞

g(−x) dx =

∫ ∞

−∞

g(x) dx =

∫

b

a

f(x) dx

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Challenge questions (optional)

8. Generalise the Dominated Convergence Theorem as follows. Assume that fk : X → K

(K = R or C) is measurable for every k ∈ N and that fk → f pointwise. Instead ofassuming that there is a single integrable function g : X → [0,∞] such that |fk(x)| ≤ g(x)for all k and x, we assume that for each k there is an integrable function gk(x) such that

(i) |fk(x)| ≤ gk(x) for all k and all x,

(ii) g(x) = limk→∞ gk(x) exists for each x ∈ X,

(iii)∫

Xgk dµ→

∫

Xg dµ <∞ as k → ∞.

Conclude that∫

Xfk dµ→

∫

Xf dµ.

Hint: Use that if (ak) and (bk) are two sequences of real numbers, and if ak → ℓ ∈ R,then lim infk→∞(ak + bk) = ℓ+ lim infk→∞ bk.

Solution: As the pointwise limit of measurable functions, the function f is measurable.Passing to the limit in (i) we get |f(x)| ≤ g(x) for all x ∈ X, so f is integrable. Moreover,

|fk(x)− f(x)| ≤ |fk(x)|+ |f(x)| ≤ gk(x) + g(x)

for all x ∈ X. Next note that by (ii)

2g(x) = limk→∞

(

gk(x) + gk(x)− |fk(x)− f(x)|)

= lim infk→∞

(

g(x) + gk(x)− |fk(x)− f(x)|)

for all x ∈ X. By the above gk(x) + g(x)− |fk(x)− f(x)| ≥ 0, and so by Fatou’s Lemmaand (iii)

2

∫

X

g dµ =

∫

X

lim infk→∞

(

g(x) + gk(x)− |fk(x)− f(x)|)

dµ

= lim infk→∞

∫

X

g(x) + gk(x)− |fk(x)− f(x)| dµ

= lim infk→∞

(

∫

X

g(x) + gk(x) dµ−

∫

X

|fk(x)− f(x)| dµ)

= limk→∞

(

∫

X

g(x) + gk(x) dµ)

+ lim infk→∞

(

−

∫

X

|fk(x)− f(x)| dµ)

= 2

∫

X

g dµ− lim supk→∞

∫

X

|fk(x)− f(x)| dµ.

If we rearrange the above we get

lim supk→∞

∫

X

|fk(x)− f(x)| dµ ≤ 0 ≤ lim infk→∞

∫

X

|fk(x)− f(x)| dµ

and hence

limk→∞

∣

∣

∣

∫

X

fk dµ−

∫

X

f(x) dµ∣

∣

∣≤ lim

k→∞

∫

X

|fk(x)− f(x)| dµ = 0.

This concludes the proof.

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