Chapter 6

Integration on Product Spaces

So far we have been concerned with measurable functions of one variable and their integration. In this chapter we consider measurable functions of more than one variable, that is, functions on measurable spaces of the form X: = X₁ × X₂ × · · · × X_k with appropriate σ-algebras and measures on them. We shall restrict our study for the case of k = 2. Thus, the idea is to construct a new σ-algebra and a measure on X₁ × X₂ using the measure spaces X₁ and X₂, and see how integration on the product space is related to the integration on the component spaces. For this purpose we shall use most of the concepts and basic theorems introduced in the previous chapters.

6.1  Motivation

Let (X1,A1,μ1) and (X2,A2,μ2) be measure spaces. We would like to have a σ-algebra A1⊗A2 on X₁ × X₂ and a measure μ, called the product measure on A1⊗A2 with the following properties:

(1)A1⊗A2⊇{A×B:A∈A1,B∈A2}.

(2)μ(A×B)=μ1(A)μ2(B)∀A∈A1,B∈A2.

Let f be a non-negative extended real valued function on X₁ × X₂. For each x ∈ X₁ and y ∈ X₂, let f_x: X₂ → [0, ∞] and f^y: X₁ → [0, ∞] be defined by

fx(u)=f(x,u),u∈X2,fy(v)=f(v,y),v∈X1,

respectively. We would like to show that, if f is measurable, then

(3)f_x and f^y are measurable with respect to A2 and A1, respectively, for each x ∈ X₁ and y ∈ X₂,

(4)the functions g: X₁ → [0, ∞] and h: X₂ → [0, ∞] defined by

g(x)=∫X2fx(y)dμ2(y),x∈X1,

h(y)=∫X1fy(x)dμ1(x),y∈X2,

respectively, are measurable, and

(5)∫X1×X2fdμ=∫X1gdμ1=∫X2hdμ2.

The equalities in (5) are written, sometimes, as

∫X1×X2f(x,y)dμ(x,y)=∫X1(∫X2f(x,y)dμ2(y))dμ1(x)=∫X2(∫X1f(x,y)dμ1(x))dμ2(y).

We would also like to have results in (3), (4), (5) above for a complex valued measurable function f, whenever f is integrable with respect to the product measure.

We shall show that the existence of a product measure with the required properties will be guaranteed whenever μ₁ and μ₂ are σ-finite measures.

6.2  Product σ-algebra and Product Measure

Let (X1,A1,μ1) and (X2,A2,μ2) be measure spaces.

Definition 6.2.1 Sets of the form A × B with A∈A1 and B∈A2 are called measurable rectangles, and the σ-algebra generated by the family of all measurable rectangles is called the product σ-algebra. We denote the product σ-algebra by A1⊗A2.♢

Definition 6.2.2 For any set E⊆ X₁ × X₂ and (x, y) ∈ X₁ × X₂, let

Ex:={v∈X2:(x,v)∈E},Ey:={u∈X1:(u,y)∈E}.

The sets E_x and E^y are called x-section and y-section, respectively, of the set E.♢

It can be easily seen that

E⊆F⇒Ex⊆FxandEy⊆Fy.

In the following lemma, we state some easily verifiable facts.

Lemma 6.2.3 Let E be a measurable rectangle, say E = A × B with A∈A1 and B∈A2. Then,

Ex={B,x∈A,∅,x∉A,Ey={A,y∈B,∅,y∉B.

Further, the following are true.

(i)Ex∈A2 and Ey∈A1 for all (x, y) ∈ X₁ × X₂.

(ii)The functions x↦μ2(Ex) and y↦μ1(Ey) are measurable.

(iii)∫X1μ2(Ex)dμ1=μ1(A)μ2(B)=∫X2μ1(Ey)dμ2.

The results in the above lemma prompt us to ask whether the following statements are true for every E∈A1⊗A2:

(a)Ex∈A2, Ey∈A1 for every (x, y) ∈ X₁ × X₂,

(b)x↦μ2(Ex) and y↦μ1(Ey) are measurable,

(c)∫X1μ2(Ex)dμ1(x)=∫X2μ1(Ey)dμ2(y), and

(d)E↦∫X1μ2(Ex)dμ1(x) is a measure on A1⊗A2.

Our first attempt in this chapter is to prove that the above results are true provided μ₁ and μ₂ are σ-finite. For this, we shall make use of another easily verifiable proposition.

Proposition 6.2.4 For every E⊆ X₁ × X₂,

(Ec)x=Exc,(Ec)y=(Ey)c

and for E_n⊆ X₁ × X₂, n∈N,

(∪En)x=∪(En)x,(∪En)y=∪(En)y.

Further, if {En:n∈N} is a disjoint family of subsets of X₁ × X₂, then {(En)x:n∈N} and {Eny:n∈N} are disjoint families.

Notation: For n∈N, if E_n ∈ X₁ × X₂, then we shall use the notations E_n,x and Eny for (E_n)_x and (E_n)^y, respectively.

Theorem 6.2.5 Let E∈A1⊗A2. Then for every (x, y) ∈ X₁ × X₂, we have Ex∈A2 and Ey∈A1.

Proof. Let S be the family of all E∈A1⊗A2 such that Ex∈A2 for all x ∈ X₁. We have to show that S=A1⊗A2. For this, it is enough to show that S is a σ-algebra containing all measurable rectangles. We have already observed that if E is a measurable rectangle, then Ex∈A1 and Ey∈A2 for every (x, y) ∈ X₁ × X₂ so that S contains all measurable rectangles. The fact that S is a σ-algebra follows from Proposition 6.2.4.

Similarly we see that the family of all E∈A1⊗A2 such that Ey∈A1 for all y ∈ X₂ is A1⊗A2.▮

Theorem 6.2.6 Suppose (X1,A1,μ1) and (X2,A2,μ2) are σ-finite measure spaces. Then for every E∈A1⊗A2, the functions

x↦μ2(Ex),y↦μ1(Ey)

are measurable with respect to A1 and A2, respectively, and

∫X1μ2(Ex)dμ1(x)=∫X2μ1(Ey)dμ2(y).

For the proof of the above theorem we shall also make use of a lemma (Lemma 6.2.9) whose statement requires the following two definitions.

Definition 6.2.7 A subset of X₁ × X₂ is called an elementary set if it is a disjoint union of a finite number of measurable rectangles.♢

Definition 6.2.8 Let X be a set. A family S of subsets of a set X is called a monotone class if it has the following two properties:♢

(1)If Ai∈S and A_i⊆ A_i+1 for all i∈N, then ∪Ai∈S,

(2)If Ai∈S and Ai⊇Ai+1 for all i∈N, then ∩Ai∈S.♢

We observe the following:

(a)Given any family F of subsets of X, there exists a smallest monotone class containing F, called the monotone class generated by F.

(b)A1⊗A2 is a monotone class containing all elementary sets.

Notation: Given any family F of subsets of X, the monotone class generated by F is denoted by MF.

Now, we state the required lemma; its proof is given at the end of this subsection.

Lemma 6.2.9 The σ-algebra A1⊗A2 is the smallest monotone class containing all elementary sets.

Proof of Theorem 6.2.6. Let S be the class of all E∈A1⊗A2 such that the conclusions in the theorem hold. We prove that S=A1⊗A2 so that the proof will be complete.

The following facts can be verified easily:

(a)S contains all measurable rectangles.

(b)S contains all elementary sets.

(c)S contains finite disjoint unions of its members.

We claim that S has also the following properties:

(i)If En∈S with E_n⊆ E_n+1 for all n∈N, then ∪En∈S.

(ii)If {E_n} is a disjoint family in S, then ∪En∈S.

(ii)If En∈S such that En⊇En+1 for all n∈N and there exists a measurable rectangle A × B satisfying E₁⊆ A × B, μ₁(A) < ∞ and μ₂(B) < ∞, then ∩En∈S.

Note that, if μ₁ and μ₂ are finite measures, then (i) and (iii) imply that S is a monotone class (containing all measurable rectangles), so that by Lemma 6.2.9, S=A1⊗A2, and hence the proof is complete in this case.

Proof of (i): Let En∈S with E_n⊆ E_n+1 for all n∈N. By the definition of S, for each n∈N, the functions

x↦μ2(En,x),y↦μ1(Eny)

are measurable functions and

∫X1μ2(En,x)dμ1(x)=∫X2μ1(Eny)dμ2(x).

Also we have

En,x⊆En+1,x,Eny⊆En+1y∀n∈N

so that

μ2(En,x)→μ2(∪En,x)=μ2((∪En)x),

μ1(Eny)→μ1(∪Eny)=μ1((∪En)y)

as n → ∞. Hence, the functions

x↦μ2((∪En)x),y↦μ1((∪En)y)

are measurable, and by MCT (Theorem 4.2.15), we have

limn→∞∫X1μ2(En,x)dμ1(x)=∫X1μ2((∪En)x)dμ1(x),

limn→∞∫X2μ1(Eny)dμ2(y)=∫X2μ1((∪En)y)dμ2(y).

Therefore, ∪En∈S.

Proof of (ii): Let {E_n} be a disjoint family in S. Then ∪En=∪Fn where Fn=∪i=1nEi, n∈N. Since each F_n is a finite disjoint union of members of S, we have Fn∈S for every n∈N. Also, F_n⊆ F_n+1 for every n∈N. Hence, by (i), ∪En=∪Fn∈S.

Proof of (iii): As in (i), the functions

x↦μ2(En,x),y↦μ1(Eny)

are measurable functions and

∫X1μ2(En,x)dμ1(x)=∫X2μ1(Eny)dμ2(y).

Since A×B⊇E1, we have

E1,x⊆(A×B)x,E1y⊆(A×B)y.

Hence, by Lemma 6.2.3,

μ2((A×B)x)=μ2(B)χA(x)<∞,

μ1((A×B)y)=μ1(A)χB(y)<∞.

Also, the condition En⊇En+1 for all n∈N, consequently, the facts

En,x⊇En+1,x,Eny⊇En+1y∀n∈N,

imply the convergence

μ2(En,x)→μ2(∩En,x)=μ2((∩En)x),

μ1(Eny)→μ1(∩Eny)=μ1((∩En)y)

as n → ∞. Hence, the functions

x↦μ2((∩En)x),y↦μ1((∩En)y)

are measurable. Again since

μ2(En,x)≤μ2((A×B)x)=μ2(B)χA(x),

μ1(Eny)≤μ1((A×B)y)=μ1(A)χB(y)

with

∫X1μ2(B)χA(x)dμ1(x)=μ2(B)μ1(A)=∫X2μ1(A)χB(y)dμ2(y)

for all n∈N, by DCT (Theorem 5.1.13),

limn→∞∫X1μ2(En,x)dμ1(x)=∫X1μ2((∩En)x)dμ1(x),

limn→∞∫X2μ1(Eny)dμ2(y)=∫X2μ1((∩En)y)dμ2(y).

Therefore, ∩En∈S.

Now, since μ₁ and μ₂ are σ-finite measures, there exist disjoint families {X1(n):n∈N} and {X2(n):n∈N} of measurable subsets of X₁ and X₂, respectively, such that

X1=⋃n=1∞X1(n),X2=⋃m=1∞X2(m)

with μ1(X1(n))<∞ and μ2(X2(m))<∞ for all n,m∈N. For E∈A1⊗A2, let

En,m:=E∩(X1(n)×X2(m)),n,m∈N.

Clearly, E is a disjoint union of {En,m:n,m∈N}. Let

A:={E∈A1⊗A2:En,m∈S∀n,m∈N}.

By (i), (ii), and (iii), it can be seen (verify) that A is a monotone class containing all elementary sets. Hence, by Lemma 6.2.9, A=A1⊗A2. Thus, E∈A1⊗A2 implies En,m∈S for all n,m∈N. Again by (ii) above, E=∪En,m∈S. That is, for every E∈A1⊗A2, the conclusions of the theorem hold. Thus, we have proved that S=A1⊗A2, which completes the proof.▮

The following corollary is immediate from Theorem 6.2.6.

Corollary 6.2.10 Let (X1,A1,μ1) and (X2,A2,μ2) be σ-finite measure spaces and E∈A1⊗A2. If μ₂(E_x) = 0 for a.a. x ∈ X₁, then μ₁(E_y) = 0 for a.a. y ∈ X₂.

The following theorem leads to the definition of the product measure.

Theorem 6.2.11 Suppose that (X1,A1,μ1) and (X2,A2,μ2) are σ-finite measure spaces. For E∈A1⊗A2, let

μ(E):=∫X1μ2(Ex)dμ1(x)=∫X2μ1(Ey)dμ2(y).

Then μ is a measure on A1⊗A2.

Proof. Clearly, μ(∅)=0. Let {En:n∈N} be a disjoint family in A1⊗A2. Then, by Proposition 6.2.4, we have

μ(∪En)=∫X1μ2((∪En)x)dμ1(x)=∫X1μ2(∪En,x)dμ1(x).

Now, using the fact that {En,x:n∈N} is a disjoint family in A2 and the monotone convergence theorem, we have

μ(∪En)=∫X1∑n=1∞μ2(En,x)dμ1(x)=∑n=1∞∫X1μ2(En,x)dμ1(x)=∑n=1∞μ(En).

This completes the proof.▮

Definition 6.2.12 The measure μ in Theorem 6.2.11 is called the product measure on A1⊗A2, and it is denoted by μ₁ × μ₂.♢

Proof of Lemma 6.2.9. The proof involves two main steps:

Step (i): Let F be an algebra on a set X, MF be the monotone class generated by F and AF be the σ-algebra generated by F. Then MF is a σ-algebra and MF=AF.

Step (ii): The family E of all elementary sets in A1⊗A2 is an algebra.

Since AE, the σ-algebra generated by E, is A1⊗A2, results in Step 1 and Step 2 will imply the required result ME=A1⊗A2.

Proof of Step (i): Since MF is a monotone class, for showing that MF is a σ-algebra, it is enough to show that it is an algebra, because, in that case, for any (A_n) in MF,

⋃n=1∞An=⋃n=1∞(⋃k=1nAk)∈MF,

as

⋃k=1nAk∈MFand⋃k=1nAk⊆⋃k=1n+1Ak∀n∈N.

Let M~:={A:Ac∈MF}. Then it can be seen that M~ is a monotone class containing F. Hence, MF⊆M~. Thus,

A∈MF⇒Ac∈MF.

Next, we show that MF is closed under finite unions, that is, to show that A,B∈MF implies A∪B∈MF; equivalently, to show that for every A∈MF, A^=MF, where

A^:={B∈MF:A∪B∈MF}.

So, let A∈MF. Note that A^ is a monotone class. Therefore, for showing that A^=MF, it is enough to show that F⊆A^. So, let C∈F. Note that F⊆C^, because, D∈F implies D∪C∈F⊆MF. Thus, C^ is a monotone class containing F. Since MF is the smallest monotone class containing F, we obtain C^=MF. Thus, A∈MF=C^ so that C∈A^, proving that F⊆A^. Thus, the proof of Step (i) is completed.

Proof of Step (ii): It can be easily seen that E is closed under finite disjoint unions and finite intersections. Also, for any measurable rectangle A₁ × A₂,

(A1×A2)c=(A1c×A2)∪(A1×A2c)

so that (A₁ × A₂)^c is a disjoint union of two measurable rectangles. Hence, (A1×A2)c∈E. Since each member A of E is a finite disjoint union of measurable rectangles, say A=∪i=1nRi, where {R_i: i = 1, …, n} is a disjoint family of rectangles, we have Ac=⋂i=1nRic∈E. Now, if A,B∈E, then

A∪B=(A∖B)∪B=(A∩Bc)∪B,

which is a finite disjoint union of members of E. Thus, E is closed under finite unions as well.▮

6.3  Fubini’s Theorem

Let (X1,A1,μ1) and (X2,A2,μ2) be measure spaces.

Definition 6.3.1 Let f be a function defined on X₁ × X₂ taking values in another set Y. For each x ∈ X₁, the function f_x: X₂ → Y defined by

fx(y)=f(x,y),y∈X2,

is called the x-section of f, and for each y ∈ X₂, the function f^y: X₁ → Y defined by

fy(x)=f(x,y),x∈X1,

is called the y-section of f.♢

Proposition 6.3.2 Let f be a measurable function on X₁ × X₂ taking values in a topological space Y. Then for each (x, y) ∈ X₁ × X₂, f_x and f^y are measurable with respect to A2 and A1, respectively.

Proof. Let G be an open set in the topological space in which f takes values. Then we see that, for each (x, y) ∈ X₁ × X₂,

{u∈X1:fy(u)∈G}={u∈X1:f(u,y)∈G}=[f−1(G)]y,

{v∈X2:fx(v)∈G}={u∈X1:f(x,v)∈G}=[f−1(G)]x.

Since f is measurable, by Proposition 6.2.4, both [f⁻¹(G)]^y and [f⁻¹(G)]_x are measurable sets. Hence the result.▮

By the above proposition (Proposition 6.3.2), if f is an extended real valued and non-negative measurable function on X₁ × X₂, then its sections f_x and f^y are measurable for each x ∈ X₁ and y ∈ X₂, respectively. Thus, the integrals ∫X2f(x,y)dμ2(y) and ∫X1f(x,y)dμ1(x) are well-defined.

Now, we prove the Fubini’s theorem for non-negative measurable functions, which is also known as Tonelli’s theorem.

Theorem 6.3.3 (Fubini’s theorem - I) Let (X1,A1,μ1) and (X2,A2,μ2) be σ-finite measure spaces and let f be an extended real valued and non-negative measurable function on X₁ × X₂. Then the functions

x↦∫X2fxdμ2y↦∫X1fydμ1

are measurable with respect to A1 and A2, respectively, and

∫X1×X2fd(μ1×μ2)=∫X1(∫X2f(x,y)dμ2(y))dμ1(x)=∫X2(∫X1f(x,y)dμ1(x))dμ2(y).

Proof. For x ∈ X₁ and y ∈ X₂, let

g(x)=∫X2fxdμ2,h(y)=∫X1fydμ1.

Let us consider first the case f=χE for some E∈A1⊗A2. Then we have

g(x)=∫X2χExdμ2=μ2(Ex),h(y)=∫X1χEydμ1=μ1(Ey).

Hence, for f=χE, the result is a consequence of Theorem 6.2.6. Next, let f be a non-negative simple measurable function. In this case, the result follows by using the linearity of integrals. Now, let f be any non-negative measurable function. Then, consider an increasing sequence (φ_n) of simple measurable functions which converges to f pointwise. If we take

gn(x)=∫X2φn,xdμ2,hn(y)=∫X1φnydμ1,

then, by MCT (Theorem 4.2.15), g_n → g and h_n → h pointwise. Again, applying MCT, we have the convergence

∫X1gndμ1→∫X1gdμ1,∫X2hndμ2→∫X2hdμ2.

Since

∫X1gndμ1=∫X2hndμ2=∫X1×X2fnd(μ1×μ2)∀n∈N,

by taking limit, we obtain

∫X1gdμ1=∫X2hdμ2=∫X1×X2fd(μ1×μ2).

This completes the proof.▮

Now, we state and prove the Fubini’s theorem for a complex measurable function f on X₁ × X₂.

Before stating the theorem, let us recall from Proposition 6.3.2 that, if f is a complex measurable function on X₁ × X₂, then its sections f_x and f^y are measurable for each x ∈ X₁ and y ∈ X₂, respectively. Thus, the integrals

∫X2|f(x,y)|dμ2(y),∫X1|f(x,y)|dμ1(x)

are well-defined. Also, we know from Theorem 6.3.3 that the functions

x↦∫X2|fx|dμ2y↦∫X1|fy|dμ1

are measurable with respect to A1 and A2, respectively.

Theorem 6.3.4 (Fubini’s theorem - II) Let (X1,A1,μ1) and (X2,A2,μ2) be σ-finite measure spaces and let f be a complex measurable function on the product measure space (X1×X2,A1⊗A2,μ1×μ2). Suppose that at least one of the integrals

∫X1(∫X2|fx|dμ2)dμ1,∫X2(∫X1|fy|dμ1)dμ2,∫X1×X2|f|d(μ1×μ2)

is finite. Then they are equal and the following results hold.

(i)f_x ∈ L¹(μ₂) for a.a. x ∈ X₁, f^y ∈ L¹(μ₁) for a.a. y ∈ X₂, and f ∈ L¹(μ₁ × μ₂),

(ii)the functions x↦∫X2fxdμ2 and y↦∫X1fydμ1 belong to L¹(μ₁) and L¹(μ₂), respectively.

(iii)The integrals

∫X1(∫X2fxdμ2)dμ1,∫X2(∫X1fydμ1)dμ2,∫X1×X2fd(μ1×μ2)

are equal.

Proof. Since |f_x| = |f|_x and |f^y| = |f|^y, by Theorem 6.3.3, the integrals

∫X1(∫X2|fx|dμ2)dμ1,∫X2(∫X1|fy|dμ1)dμ2,∫X1×X2|f|d(μ1×μ2)

are equal. Hence, if one of these integrals is finite, then all of them are finite. In particular, results in (i) hold.

To prove (ii) and (iii), first we assume that f ∈ L¹(μ₁ × μ₂) is real valued. Note that

∫X1×X2f+d(μ1×μ2)≤∫X1×X2|f|d(μ1×μ2)<∞.

Hence, the integrals

∫X1(∫X2(f+)xdμ2)dμ1,∫X2(∫X1(f+)ydμ1)dμ2,∫X1×X2f+d(μ1×μ2)

are equal and finite. In particular, the functions

x↦∫X2f+(x,y)dμ2(y)y↦∫X1f+(x,y)dμ1(x)

belong to L¹(μ₁) and L¹(μ₂), respectively. Hence, (ii) and (iii) hold with f⁺ in place of f. Similarly, we have the conclusions in (ii) and (iii) with f⁻ in place of f. Therefore, we have (ii) and (iii) for f as well.

The case for complex valued f follows by writing f as f = Re(f) + iIm(f) and applying the results for the real valued functions Re(f) and Im(f), and observing the facts that |Re(f)| ≤ |f|, |Im(f)| ≤ |f|, and the linearity of taking integrals.▮

6.4  Counter Examples

6.4.1  σ-finiteness condition cannot be dropped

Let X₁ = [0, 1] with Lebesgue measure μ₁ and X₂ = [0, 1] with counting measure μ₂. Let D: = {(x, y) ∈ X₁ × X₂: x = y}, the diagonal set. Since D=∩Dn, where

Dn:=⋃j=1n([j−1n,jn]×[j−1n,jn]),n∈N,

it follows that D is a measurable subset of X₁ × X₂. Note that for x ∈ [0, 1],

Dx={x},Dy={y},μ2(Dx)=1,μ1(Dy)=0,

so that

∫X1μ2(Dx)dμ1=μ1(X1)=1,∫X2μ1(Dy)dμ2=0.

Hence, the integrals involved in the definition of product measure are not equal for the measurable set D.

Also, taking f=χD, the characteristic function of D, we have

∫X1(∫X2fxdμ2)dμ1=∫X1(∫X2χDxdμ2)dμ1=∫X1μ2(Dx)dμ1=1,

∫X2(∫X1fydμ1)dμ2=∫X2(∫X1χDydμ1)dμ2=∫X2μ1(Dy)dμ2=0.

Thus, the iterated integrals in Fubini’s theorem are not equal for f=χD. Note that μ₂ is not σ-finite.

6.4.2  Product of complete measures need not be complete

Suppose (X1,A1,μ1) and (X2,A2,μ2) are complete σ-finite measure spaces such that there exists A∈A1 with μ₁(A) = 0 and there exists B⊆ X₂ such that B∉A2. Then we have A × B⊆ A × X₂ with

(μ1×μ2)(A×X2)=μ1(A)μ2(X2)=0.

But, A×B∉A1⊗A2, since for every x ∈ A, (A×B)x=B∉A2. Thus, μ₁ × μ₂ is not complete.

As an example, consider

(X1,A1,μ1)=(X2,A2,μ2)=(R,M,m).

We know that Q∈A1 with m(Q)=0 and there exists B⊆R such that B∉M. Thus, (R,M,m) is complete, whereas (R×R,M⊗M,m×m) is not complete. It can be shown that the completion of (R×R,M⊗M,m×m) is the Lebesgue measure space (R2,M2,m2).

6.5  Problems

1. Prove Lemma 6.2.3.
2. Prove Proposition 6.2.4.

3. Let (X1,A1,μ1) and (X2,A2,μ2) be σ-finite measure spaces and let S be the family of all E∈A1⊗A2 such that the functions x↦μ2(Ex) and y↦μ1(Ey) are measurable with respect to A1 and A2, respectively, and ∫X1μ2(Ex)dμ1=∫X2μ1(Ey)dμ2. Show that

   (a)S contains all measurable rectangles.

   (b)S contains all elementary sets.

   (c)S contains finite disjoint unions of its members.

4. Supply details of the proof of Corollary 6.2.6.

5. Prove that if A1=Bm and A2=Bn, then Bm⊗Bn=Bm+n.

   [Hint: Observe: Every open set in Rm+n is a countable union of sets of the form A × B where A and B are rectangles in Rm and Rn, respectively, and prove: Bmm+n contains sets of the form A×Rn and Rm×B where A∈Bm and B∈Bn.]

6. Let (X1,A1,μ1) and (X2,A2,μ2) be σ-finite measure spaces. Given measurable functions f1:X1→R and f2:X2→R, let

   f(x,y)=f1(x)f2(y),(x,y)∈X1×X2.

   Prove the following:

   (a)f is measurable on the product space (X1×X2,A1×A2).

   (b)If f₁ ∈ L¹(μ₁) and f₂ ∈ L¹(μ₂), then f ∈ L¹(μ₁ × μ₂) and

   ∫X1×X2fd(μ1×μ2)=(∫X1f1dμ1)(∫X2f2dμ2).
7. Let I = [ − 1, 1] and f(x,y)=xy(x2+y2)2 for (x,y)∈I×I∖{(0,0)} and f(0, 0) = 0. Show that the integrals
   ∫I(∫If(x,y)dm(x))dm(y),∫I(∫If(x,y)dm(y))dm(x)

   exist and are equal, but ∫I×If(x,y)d(m×m)(x,y) does not exist.

8. Let I = [0, 1], S=I×I∖{(0,0)} and f:S→R be defined by
   f(x,y)={x2−y2(x2+y2)2,(x,y)∈I×I∖{(0,0)},0,(x,y)=(00).

   Show that the integrals

   ∫I(∫If(x,y)dm(x))dm(y),∫I(∫If(x,y)dm(y))dm(x)

   exist and are unequal.

9. Let f be a non-negative measurable function on a σ-finite measure space (X,A,μ) and S={(x,y)∈X×R:0≤y≤f(x)}. Show that S∈A×M and
   ∫Xf(x)dμ(x)=∫X×RχS(x,y)d(μ×m).

10. Using Fubini’s theorem prove that limτ→∞∫0τsin⁡xxdx=π2.

    [Hint: Use the relation 1x=∫0∞e−xydy.]

11. Let X₁ = [0, 1] = X₂ with Lebesgue measure and f:X×X→R be defined by

    f(x,y)={xyx2+y2if (x,y)≠(0,0),0if (x,y)=(0,0).

    Show that

    (a)f is not integrable,

    (b)∫X(∫Xf(x,y)dm(x))dm(y) and ∫X(∫Xf(x,y)dm(y))d(x) exist and are equal.

12. Let (X,A,μ) be a complete measure space and f: X → [0, ∞) be an integrable function. Let g(t): = μ({x ∈ X: f(x) ≥ t}),  t ≥ 0. Show that
    ∫Xfdμ=∫0∞g(t)dt.
    [Hint: Use the function h(t,x)=χEt(x), where E_t: = {x ∈ X: f(x) ≥ t} for (t, x) ∈ [0, ∞) × X.]

13. For f, g ∈ L¹((0, 2π]), extending them as 2π-periodic functions on R and using the notation introduced in the beginning of Section 5.3.1, define the convolution of f and g by

    (f∗g)(x):=12π∫02πf(x−y)g(y)dm(y),x∈R.

    Using Fubini’s theorem, prove that (f∗g)^(n)=f^(n)g^(n) for all n∈Z, where, for f ∈ L¹((0, 2π]), f^(n) denotes the n-th Fourier coefficient defined as in Problem 19 in Section 5.5.