(X, S, Μ) and (Y,T,Ν) Be Two Measure Spaces. in Order to Define the Prod

Chapter 4. Product Measures

§1. Product Measures

Let (X, S, µ) and (Y, T, ν) be two measure spaces. In order to define the product of µ and ν, we first define an outer measure on X × Y in terms of µ and ν. For each subset E of X × Y , we define ∞ X ∞ τ(E) := inf µ(Aj)ν(Bj): Aj ∈ S, Bj ∈ T ∀ j ∈ IN and E ⊆ ∪j=1Aj × Bj . j=1 Theorem 1.1. The set function τ is an outer measure on X × Y .

Proof. Clearly, τ(∅) = 0. Moreover, if E ⊆ F ⊆ X × Y , then τ(E) ≤ τ(F ). It remains ∞ to prove that τ is countably subadditive. Suppose E ⊆ ∪k=1Ek ⊆ X × Y . We wish to P∞ P∞ P∞ show τ(E) ≤ k=1 τ(Ek). This is true if k=1 τ(Ek) = ∞. Suppose k=1 τ(Ek) < ∞. k k Fix ε > 0. For each k ∈ IN, there exist sets Aj ∈ S and Bj ∈ T for j ∈ IN such that ∞ k k Ek ⊆ ∪j=1Aj × Bj and ∞ X ε µ(Ak)ν(Bk) < τ(E ) + . j j k 2k j=1 ∞ ∞ k k Thus, E ⊆ ∪k=1 ∪j=1 Aj × Bj and

∞ ∞ ∞ ∞ X X k k X k X τ(E) ≤ µ(Aj )ν(Bj ) ≤ (τ(Ek) + ε/2 ) = ε + τ(Ek). k=1 j=1 k=1 k=1 P∞ Since ε > 0 could be arbitrary, we conclude that τ(E) ≤ k=1 τ(Ek).

Let A denote the σ-algebra of τ-measurable subsets of X × Y and let

µ × ν = τ|A.

We call µ × ν the product measure of µ and ν. A subset E of X × Y is said to be µ × ν-measurable if and only if it is τ-measurable.

Theorem 1.2. If A ∈ S and B ∈ T , then A × B is τ-measurable and

(µ × ν)(A × B) = µ(A)ν(B).

Proof. In order to show that A × B is τ-measurable, it suﬃces to prove that for any G ⊆ X × Y with τ(G) < ∞ the following inequality holds:

τ(G) ≥ τ(G ∩ (A × B)) + τ(G ∩ (A × B)c).

1 ∞ Fix ε > 0. There exist sets Aj ∈ S and Bj ∈ T for j ∈ IN such that G ⊆ ∪j=1Aj × Bj P∞ and j=1 µ(Aj)ν(Bj) < τ(G) + ε. It follows that

∞ G ∩ (A × B) ⊆ ∪j=1(Aj ∩ A) × (Bj ∩ B)

and c ∞ c c G ∩ (A × B) ⊆ ∪j=1 ((Aj ∩ A) × (Bj ∩ B )) ∪ ((Aj ∩ A ) × Bj) .

Consequently,

τ(G ∩ (A × B)) + τ(G ∩ (A × B)c) ∞ ∞ ∞ X X c X c ≤ µ(Aj ∩ A)ν(Bj ∩ B) + µ(Aj ∩ A)ν(Bj ∩ B ) + µ(Aj ∩ A )ν(Bj) j=1 j=1 j=1 ∞ ∞ ∞ X X c X = µ(Aj ∩ A)ν(Bj) + µ(Aj ∩ A )ν(Bj) = µ(Aj)ν(Bj) < τ(G) + ε. j=1 j=1 j=1

Thus, τ(G ∩ (A × B)) + τ(G ∩ (A × B)c) < τ(G) + ε for all ε > 0. Therefore,

τ(G ∩ (A × B)) + τ(G ∩ (A × B)c) ≤ τ(G) ∀ G ⊆ X × Y.

This shows that A × B is τ-measurable. We have (µ × ν)(A × B) ≤ µ(A)ν(B). To prove µ(A)ν(B) ≤ (µ × ν)(A × B), for ∞ j ∈ IN we let Aj and Bj be sets in S and T respectively such that A × B ⊆ ∪j=1Aj × Bj. It follows that ∞ X χA(x)χB(y) ≤ χAj (x)χBj (y), x ∈ X, y ∈ Y. j=1 Integrating both sides of the above inequality ﬁrst with respect to ν and then with respect to µ, we obtain Z Z µ(A)ν(B) = χA(x)χB(y) dν(y) dµ(x) X Y ∞ ∞ X Z Z X ≤ χAj (x)χBj (y) dν(y) dµ(x) = µ(Aj)ν(Bj). j=1 X Y j=1

Therefore, µ(A)ν(B) ≤ (µ × ν)(A × B).

∞ A subset F of X×Y is said to be a σ-set, if F can be represented as ∪j=1Aj ×Bj, where Aj ∈ S and Bj ∈ T for j ∈ IN. By Theorem 1.2, a σ-set is (µ × ν)-measurable. Suppose

2 that E is a (µ × ν)-measurable subset of X × Y . By the deﬁnition of the product measure, for any ε > 0 there exists a σ-set F such that E ⊆ F and (µ × ν)(F ) < (µ × ν)(E) + ε. Let E be a subset of X × Y . For x ∈ X, the x-section of E is the subset of Y given by

Ex := {y ∈ Y :(x, y) ∈ E}.

For y ∈ Y , the y-section of E is the subset of X given by

Ey := {x ∈ X :(x, y) ∈ E}.

Regarding sections of sets, the following identities hold: y y (a) (∪i∈I Ai)x = ∪i∈I (Ai)x and (∪i∈I Ai) = ∪i∈I (Ai) ; y y (b) (∩i∈I Ai)x = ∩i∈I (Ai)x and (∩i∈I Ai) = ∩i∈I (Ai) ; y y y (c) (A \ B)x = Ax \ Bx and (A \ B) = A \ B . In what follows, (X, S, µ) and (Y, T, ν) are assumed to be complete measure spaces.

Theorem 1.3. Let E be a µ × ν-measurable subset of X × Y with (µ × ν)(E) < ∞.

Then for µ-almost every x ∈ X the set Ex is ν-measurable and the function x 7→ ν(Ex) is integrable over X, and Z (µ × ν)(E) = ν(Ex) dµ(x). X Similarly, for ν-almost every y ∈ Y the set Ey is µ-measurable and the function y 7→ µ(Ey) is integrable over Y , and Z (µ × ν)(E) = µ(Ey) dν(y). Y Proof. It suﬃces to establish the ﬁrst formula. The proof goes by steps.

Step 1. Assume E = A × B, where A ∈ S and B ∈ T . Then Ex = B if x ∈ A and

Ex = ∅ if x∈ / A. Hence, ν(Ex) = ν(B)χA(x). It follows that

Z Z ν(Ex) dµ(x) = ν(B)χA(x) dµ(x) = µ(A)ν(B) = (µ × ν)(E). X X

∞ Step 2. Suppose that E is a σ-set, that is, E = ∪n=1(An × Bn), where An ∈ S and

Bn ∈ T for n ∈ IN. Let En := An × Bn. We may assume that En (n ∈ IN) are mutually ∞ P∞ disjoint. Then Ex = ∪n=1(En)x is ν-measurable and ν(Ex) = n=1 ν((En)x). Therefore,

∞ ∞ Z X Z X ν(Ex) dµ(x) = ν((En)x) dµ(x) = (µ × ν)(En) = (µ × ν)(E). X n=1 X n=1

3 ∞ Step 3. Suppose that E = ∩n=1En, where each En is a σ-set, En+1 ⊆ En for n ∈ IN and (µ × ν)(E1) < ∞. By Step 2 we have

Z ν((E1)x) dµ(x) = (µ × ν)(E1) < ∞. X

Hence, ν((E1)x) < ∞ for µ-almost every x ∈ X. Moreover, (En+1)x ⊆ (En)x for n ∈ IN. ∞ Thus, Ex = ∩n=1(En)x is ν-measurable and

Z Z ν(Ex) dµ(x) = lim ν((En)x) dµ(x) = lim (µ × ν)(En) = (µ × ν)(E). X n→∞ X n→∞

Step 4. Suppose (µ × ν)(E) = 0. There exists a sequence (En)n=1,2,... of σ-sets such ∞ that E ⊆ ∩n=1En, En+1 ⊆ En for n ∈ IN, (µ × ν)(E1) < ∞, and limn→∞(µ × ν)(En) = 0. ∞ ∞ Let F := ∩n=1En. Then Fx = ∩n=1(En)x is ν-measurable for each x ∈ X and

Z Z ν(Fx) dµ(x) = lim ν((En)x) dµ(x) = lim (µ × ν)(En) = 0. X n→∞ X n→∞

Hence, ν(Fx) = 0 for µ-almost every x ∈ X. But Ex ⊆ Fx and (X, S, µ) is complete.

Therefore, for µ-almost every x ∈ X, Ex is ν-measurable and ν(Ex) = 0. Consequently,

Z ν(Ex) dµ(x) = 0 = (µ × ν)(E). X

Step 5. The general case. Since (µ×ν)(E) < ∞, There exists a sequence (En)n=1,2,... ∞ of σ-sets such that E ⊆ F := ∩n=1En,(µ × ν)(F \ E) = 0, En+1 ⊆ En for n ∈ IN, and

(µ × ν)(E1) < ∞. By Step 3 and Step 4, for µ-almost every x ∈ X, Fx and (F \ E)x are

ν-measurable, so Ex = Fx \ (F \ E)x is ν-measurable. Moreover,

Z Z (µ × ν)(F ) = ν(Fx) dµ(x) and (µ × ν)(F \ E) = ν(Fx \ Ex) dµ(x). X X

Subtracting the second equality from the ﬁrst one, we obtain

Z (µ × ν)(E) = ν(Ex) dµ(x). X

The proof is complete.

4 §2. The Fubini-Tonelli Theorem

Throughout this section, (X, S, µ) and (Y, T, ν) are assumed to be two complete measure spaces. Let f be a function from X × Y to IR. For each ﬁxed x ∈ X, we deﬁne

fx(y) := f(x, y), y ∈ Y,

and for each ﬁxed y ∈ Y deﬁne

f y(x) := f(x, y), x ∈ X.

Theorem 2.1. Let f : X × Y → IR be a (µ × ν)-integrable function. Then fx is ν- integrable for µ-almost every x ∈ X, f y is µ-integrable for ν-almost every y ∈ Y , the R R y function g : x 7→ Y fx dν is µ-integrable, the function h : y 7→ X f dµ is ν-integrable, and Z Z Z Z Z f d(µ × ν) = f(x, y) dν(y) dµ(x) = f(x, y) dµ(x) dν(y). X×Y X Y Y X Proof. The proof goes by steps.

Step 1. If f = χE, where E is (µ × ν)-measurable and (µ × ν)(E) < ∞, then our assertions are true, by Theorem 1.3. Pm Step 2. Our assertions are valid if f = j=1 cjχEj is a simple function with cj ∈ IR and (µ × ν)(Ej) < ∞ for each j = 1, . . . , m.

Step 3. Suppose f is integrable and f ≥ 0. Then there exists a sequence (φn)n=1,2,... of measurable simple functions such that 0 ≤ φn ≤ φn+1 ≤ f for all n ∈ IN and limn→∞ φn(x, y) = f(x, y) for all (x, y) ∈ X × Y . Since f is integrable, each φn is

integrable. By Step 2, (φn)x is ν-integrable for µ-almost every x ∈ X, the function R gn : x 7→ Y φn(x, y) dν(y) is µ-integrable over X, and Z Z gn(x) dµ(x) = φn d(µ × ν). X X×Y By the monotone convergence theorem, for µ-almost every x ∈ X, Z Z lim gn(x) = lim φn(x, y) dν(y) = f(x, y) dν(y) = g(x). n→∞ n→∞ Y Y

Note that 0 ≤ gn ≤ gn+1 for all n ∈ IN. By using the monotone convergence theorem again we obtain Z Z Z Z g(x) dµ(x) = lim gn(x) dµ(x) = lim φn d(µ × ν) = fd(µ × ν). X n→∞ X n→∞ X×Y X×Y

5 The assertion on h can be proved analogously. Step 4. In general, suppose that f is a (µ × ν)-integrable function on X × Y . We have f = f + − f −, where f + := max{f, 0} and f − := max{−f, 0}. Then f + and f − are integrable. By Step 3, our assertions are valid for f + and f −, and thereby valid for f.

Theorem 2.2. Let (X, S, µ) and (Y, T, ν) be two σ-ﬁnite complete measure spaces, and let f : X × Y → IR be a (µ × ν)-measurable function. If one of the iterated integrals R R R R X Y |f| dν dµ or Y X |f| dµ dν is ﬁnite, then the function f is integrable and Z Z Z Z Z f d(µ × ν) = f(x, y) dν(y) dµ(x) = f(x, y) dµ(x) dν(y). X×Y X Y Y X

Proof. It suffices to consider the case f ≥ 0. Since (X, S, µ) and (Y, T, ν) are σ-finite measure spaces, the product measure µ × ν is also σ-finite. There exists a sequence

(φn)n=1,2,... of (µ × ν)-integrable functions such that 0 ≤ φn ≤ φn+1 ≤ f for all n ∈ IN and limn→∞ φn(x, y) = f(x, y) for all (x, y) ∈ X × Y . By Theorem 2.1, for each n ∈ IN we have Z Z Z φn d(µ × ν) = φn(x, y) dν(y) dµ(x). X×Y X Y By the monotone convergence theorem we obtain Z Z Z Z Z Z f d(µ × ν) = lim φn d(µ × ν) = lim φn dν dµ = f dν dµ. X×Y n→∞ X×Y n→∞ X Y X Y R R Thus, if the integral X Y f dν dµ is ﬁnite, then f is a (µ × ν)-integrable function. Hence, the desired result follows from Theorem 2.1.

Let us consider the following example. Suppose that f is the function deﬁned on [0, 1] × [0, 1] given by

( 2 2 x −y if (x, y) 6= (0, 0), f(x, y) = (x2+y2)2 0 if (x, y) = (0, 0).

Then we have

Z 1 Z 1 Z 1h −x ix=1 Z 1 dy π f(x, y) dx dy = 2 2 dy = − 2 = − 0 0 0 x + y x=0 0 1 + y 4 and Z 1 Z 1 Z 1h y iy=1 Z 1 dx π f(x, y) dy dx = 2 2 dy = 2 = . 0 0 0 x + y y=0 0 1 + x 4

6 §3. The Lebesgue Measure on IRn

n Let λ be the Lebesgue measure on IR. For n ∈ IN, the Lebesgue measure λn on IR is deﬁned by λ1 := λ and λn := λ × λn−1 for n > 1.

In what follows, n is ﬁxed and λn is abbreviated as λ. Let Λ denote the collection of all Lebesgue measurable subsets of IRn.

Suppose −∞ < aj < bj < ∞ for j = 1, . . . , n. Then the product (a1, b1)×· · ·×(an, bn) is called an open cell in IRn. Clearly, an open cell is Lebesgue measurable. An open subset of IRn can be represented as a countable union of open cells in IRn. Hence, an open set in IRn is Lebesgue measurable. Consequently, a closed set in IRn is Lebesgue measurable. Theorem 3.1. Let E be a Lebesgue measurable subset of IRn. Then for any ε > 0 there exists an open set U ⊇ E and a closed set F ⊆ E such that λ(U \E) < ε and λ(E \F ) < ε. Proof. (a) First, consider the case λ(E) < ∞. For given ε > 0, there exists a sequence n (Am)m=1,2,... of subsets of IR with the following properties: (1) each Am can be represented as Am1 × · · · × Amn, where Am1, ··· ,Amn are Lebesgue measurable subsets of IR; ∞ P∞ (2) E ⊆ A := ∪m=1Am; (3) m=1 λ(Am) < λ(E)+ε/2. For each Am, we can ﬁnd an open n m+1 ∞ subset Gm of IR such that Gm ⊇ Am and λ(Gm) < λ(Am) + ε/2 . Let U := ∪m=1Gm. Then U is an open set, U ⊇ E, and ∞ X λ(U \ E) ≤ λ(U \ A) + λ(A \ E) ≤ λ(Gm \ Am) + ε/2 < ε. m=1 ∞ Next, consider the case λ(E) = ∞. We express E as a disjoint union ∪j=1Ej, where each Ej is Lebesgue measurable and λ(Ej) < ∞. Let ε > 0 be given. For each j ∈ IN, j ∞ there exists an open set Uj ⊇ Ej such that λ(Uj \ Ej) < ε/2 . Let U := ∪j=1Uj. Then U is an open set, U ⊇ E, and ∞ X λ(U \ E) ≤ λ(Uj \ Ej) < ε. j=1 For any ε > 0, there is an open set U such that U ⊇ Ec and λ(U \ Ec) < ε. Let F := U c. Then F is a closed set and F ⊆ E. Moreover, E \ F = E ∩ F c = E ∩ U = U \ Ec. Hence, λ(E \ F ) = λ(U \ Ec) < ε.

Let B be the σ-algebra generated by open subsets of IRn. A member of B is called a Borel set. Since any open subset of IRn is Lebesgue measurable, B ⊆ Λ. In other n words, any Borel set in IR is Lebesgue measurable. Recall that a Gδ set is a countable

intersection of open sets and an Fσ set is a countable union of closed sets. Evidently, Gδ sets and Fσ sets are Borel sets.

7 Theorem 3.2. The following statements are equivalent for a subset E of IRn: (a) E is Lebesgue measurable;

(b) E = V \ N1, where V is a Gδ set and λ(N1) = 0;

Proof. Clearly, either (b) or (c) implies (a). By Theorem 3.1, for j = 1, 2,..., there exist

an open set Uj ⊇ E and a closed set Kj ⊆ E such that

λ(E \ Kj) < 1/j and λ(Uj \ E) < 1/j.

∞ ∞ Let H := ∪j=1Kj and V := ∩j=1Uj. Then V is a Gδ set and H is an Fσ set. Moreover, λ(V \ E) = 0 and λ(E \ H) = 0.

For E ⊆ IRn and t ∈ IRn, deﬁne E + t := {x + t : x ∈ E}.

Theorem 3.3. Let E ⊆ IRn and t ∈ IRn. Then λ∗(E + t) = λ∗(E). Consequently, if E is Lebesgue measurable, then E + t is also Lebegue measurable.

Proof. Suppose A is a product of measurable sets, that is, A = A1 × · · · × An, where n A1,...,An are Lebesgue measurable subsets of IR. Then for t = (t1, . . . , tn) ∈ IR we have

n n ∗ ∗ Y ∗ Y ∗ ∗ λ (A + t) = λ (A1 + t1) × · · · × (An + tn) = λ (Aj + tj) = λ (Aj) = λ (A). j=1 j=1

n Let E be a subset of IR . For given ε > 0, there exists a sequence (Am)m=1,2,... of ∞ P∞ ∗ products of measurable sets such that E ⊆ ∪m=1Am and m=1 λ(Am) ≤ λ (E) + ε. We ∞ have E + t ⊆ ∪m=1(Am + t). It follows that

∞ ∞ ∗ X X ∗ λ (E + t) ≤ λ(Am + t) = λ(Am) ≤ λ (E) + ε. m=1 m=1

Since ε > 0 could be arbitrary, we have λ∗(E + t) ≤ λ∗(E). A similar argument gives λ∗(E) = λ∗(E + t − t) ≤ λ∗(E + t). This shows λ∗(E + t) = λ∗(E).

Now suppose E is Lebesgue measurable. By Theorem 3.2, there exists a Gδ set ∗ V ⊇ E such that λ (V \ E) = 0. Clearly, V + t is a Gδ set and V + t ⊇ E + t. Moreover, λ∗((V + t) \ (E + t)) = λ∗(V \ E) = 0. Hence, E + t is Lebesgue measurable.

8 §4. The Lebesgue Integral on IRn

Theorem 4.1. If f is a Lebesgue integrable function on IRn, then for given ε > 0 there exist finitely many open cells I1,...,Im and real numbers c1, . . . , cm such that Z m X f − cjχIj dλ < ε. n IR j=1 Proof. Since f is Lebesgue integrable, for ε > 0 we can find finitely many measurable sets

E1,...,Es with ﬁnite measure and real numbers c1, . . . , cs such that the simple function Ps φ := arχE satisﬁes r=1 r Z ε f − φ dλ < . IRn 2 Ps Let M := r=1 |ar|. In light of the proof of Theorem 3.1, for each Er there exists a ∞ countable family of open cells Iri (i ∈ IN) such that Er ⊆ ∪i=1Iri and ∞ X ε λ(E ) ≤ λ(I ) < λ(E ) + . r ri r 2M i=1

Consequently, for each r ∈ {1, . . . , s}, there exists some kr ∈ IN such that

kr Z X ε χ − χ dλ < . Er Iri n 2M IR i=1

Ps Pkr R Let g := r=1 i=1 arχIri . Then we have IRn |f − g| dλ < ε.

Theorem 4.2. If f : IRn → IR is a Lebesgue integrable function, then for given ε > 0 there exists a continuous function g : IRn → IR of compact support such that Z

f − g dλ < ε. IRn

Proof. By Theorem 4.1, it suﬃces to establish the theorem for the case f = χI , where

I = (a1, b1) × · · · × (an, bn) is an open cell. Suppose −∞ < a < b < ∞. For m = 1, 2,..., (a,b) (a,b) let gm be the function on IR given by gm (t) = 1 for t ∈ [a + 1/m, b − 1/m], gm(t) = 0 for t ∈ IR \ (a, b),

(a,b) (a,b) gm (t) = m(t − a) for a ≤ t ≤ a + 1/m and gm (t) = m(b − t) for b − 1/m ≤ t ≤ b.

(a,b) Then gm is a continuous function supported on [a, b]. For m = 1, 2,..., deﬁne

n Y (ak,bk) n gm(x) := gm (xk), x = (x1, . . . , xn) ∈ IR . k=1

9 n Each gm is a continuous function on IR with compact support. It is easily seen that Z lim |f − gm| dλ = 0. m→∞ IRn R Hence, for given ε > 0, there exists some N ∈ IN such that IRn |f − gN | dλ < ε.

Theorem 4.3. If f : IRn → IR is a Lebesgue integrable function, then for t ∈ IRn, Z Z f(· − t) dλ = f dλ. IRn IRn

Proof. By Theorem 3.3, our assertion is valid if f is an integrable simple function. Con- sider the case f ≥ 0. In this case, there exists a sequence (φn)n=1,2,... of measurable simple functions such that 0 ≤ φn ≤ φn+1 ≤ f for all n ∈ IN and limn→∞ φn(x) = f(x) for all n n x ∈ IR . It follows that limn→∞ φn(x − t) = f(x − t) for all x ∈ IR . By the monotone convergence theorem we have Z Z Z Z f(· − t) dλ = lim φn(· − t) dλ = lim φn dλ = f dλ. IRn n→∞ IRn n→∞ IRn IRn

The general case is established by considering f = f + − f −.