Topics In Analysis

10.10.2 Product Measure

Definition 10.10.6 Let

and

be two measure spaces. A measurable rectangle is a set A×B ⊆ X×Y where A ∈S and B ∈ℱ. An elementary set will be any subset of X × Y which is a finite union of disjoint measurable rectangles. S×ℱ will denote the smallest σ algebra of sets in P(X ×Y ) containing all elementary sets.

Example 10.10.7 It follows from Lemma 12.1.2 or more easily from Corollary 12.1.3 that the elementary sets form an algebra.

Definition 10.10.8 Let E ⊆ X × Y,

Ex = {y \in Y : (x,y) \in E },

Ey = {x \in X : (x,y) \in E }.

These are called the x and y sections.

Theorem 10.10.9 If E ∈S×ℱ, then E_x ∈ℱ and E^y ∈S for all x ∈ X and y ∈ Y .

Proof: Let

ℳ = {E \subseteq S \times ℱ such that for all x \in X, Ex \in ℱ,

and for all y \in Y, Ey \in S.}

Then ℳ contains all measurable rectangles. If E_i ∈ℳ,

(\cup\inftyi=1Ei)x = \cup \inftyi=1(Ei)x \in ℱ.

Similarly,

(\cup \infty Ei)y = \cup \infty Ey \in S. i=1 i=1 i

It follows ℳ is closed under countable unions.

If E ∈ℳ,

(EC ) = (Ex )C \in ℱ. x

Similarly,

^y ∈S. Thus ℳ is closed under complementation. Therefore ℳ is a σ-algebra containing the elementary sets. Hence, ℳ⊇S×ℱbecause S×ℱ is the smallest σ algebra containing these elementary sets. But ℳ⊆S×ℱby definition and so ℳ = S×ℱ. This proves the theorem.

It follows from Lemma 12.1.2 that the elementary sets form an algebra because clearly the intersection of two measurable rectangles is a measurable rectangle and

(A \times B)∖ (A0 \times B0) = (A ∖ A0)\times B \cup (A \cap A0)\times (B ∖B0 ),

an elementary set.

Theorem 10.10.10 If (X,S,μ) and (Y,ℱ,λ) are both finite measure spaces (μ(X), λ(Y ) < ∞), then for every E ∈S×ℱ,

a.) x → λ(E_x) is μ measurable, y → μ(E^y) is λ measurable

b.) ∫ _Xλ(E_x)dμ = ∫ _Yμ(E^y)dλ.

Proof: Let

ℳ = {E \in S \times ℱ such that both a.) and b.) hold}.

Since μ and λ are both finite, the monotone convergence and dominated convergence theorems imply ℳ is a monotone class.

Next I will argue ℳ contains the elementary sets. Let

E = \cupni=1Ai \times Bi

where the measurable rectangles, A_i × B_i are disjoint. Then

\int \int \sumn λ(Ex) = Y XE (x,y)dλ = Y XAi\timesBi (x,y)dλ n \int i=1 n = \sum X (x,y)dλ = \sum X (x) λ(B ) i=1 Y Ai\timesBi i=1 Ai i

which is clearly μ measurable. Furthermore,

\int \int \sumn \sumn λ (Ex )dμ = XAi (x)λ(Bi)dμ = μ(Ai)λ(Bi). X X i=1 i=1

Similarly,

\int y \sumn Y μ(E )dλ = μ(Ai)λ (Bi ) i=1

and y → μ

is λ measurable and this shows ℳ contains the algebra of elementary sets. By the monotone class theorem, ℳ = S×ℱ. This proves the theorem.

One can easily extend this theorem to the case where the measure spaces are σ finite.

Theorem 10.10.11 If (X,S,μ) and (Y,ℱ,λ) are both σ finite measure spaces, then for every E ∈S×ℱ,

a.) x → λ(E_x) is μ measurable, y → μ(E^y) is λ measurable.

b.) ∫ _Xλ(E_x)dμ = ∫ _Yμ(E^y)dλ.

Proof: Let X = ∪_n=1^∞X_n,Y = ∪_n=1^∞Y _n where,

X \subseteq X ,Y \subseteq Y ,μ(X ) < \infty, λ(Y ) < \infty. n n+1 n n+1 n n

Let

Sn = {A\cap Xn : A \in S},ℱn = {B \cap Yn : B \in ℱ }.

Thus (X_n,S_n,μ) and (Y _n,ℱ_n,λ) are both finite measure spaces.

Claim: If E ∈S×ℱ, then E ∩ (X_n × Y _n) ∈S_n ×ℱ_n.

Proof: Let

ℳn = {E \in S \times ℱ : E \cap (Xn \timesYn ) \in Sn \times ℱn }.

Clearly ℳ_n contains the algebra of elementary sets. It is also clear that ℳ_n is a monotone class. Thus ℳ_n = S×ℱ.

Now let E ∈S×ℱ. By Theorem 10.10.10,

\int \int λ((E \cap (X \times Y )) )dμ = μ((E \cap (X \times Y ))y)dλ (10.10.48) Xn n n x Yn n n

(10.10.48)

where the integrands are measurable. Also

(E \cap (Xn \times Yn))x = \emptyset

if x

X_n and a similar observation holds for the second integrand in 10.10.48 if y

Y _n. Therefore,

\int \int λ((E \cap (X \times Y )))dμ = λ((E \cap(X \times Y )) )dμ X n n x Xn n n x \int y = Y μ((E \cap (Xn \times Yn)))dλ \int n = μ((E \cap(Xn \times Yn))y)dλ. Y

Then letting n →∞, the monotone convergence theorem implies b.) and the measurability assertions of a.) are valid because

λ(Ex) = ln\toim\infty λ((E \cap (Xn \times Yn))x) μ(Ey) = lim μ((E \cap (Xn \times Yn))y). n\to \infty

This proves the theorem.

This theorem makes it possible to define product measure.

Definition 10.10.12 For E ∈S×ℱ and (X,S,μ),(Y,ℱ,λ)σ finite, (μ×λ)(E) ≡ ∫ _Xλ(E_x)dμ = ∫ _Yμ(E^y)dλ.

This definition is well defined because of Theorem 10.10.11.

Theorem 10.10.13 If A ∈ S, B ∈ ℱ, then (μ × λ)(A × B) = μ(A)λ(B), and μ × λ is a measure on S×ℱ called product measure.

Proof: The first assertion about the measure of a measurable rectangle was established above. Now suppose

_i=1^∞ is a disjoint collection of sets of S×ℱ. Then using the monotone convergence theorem along with the observation that

_x ∩

_x = ∅,

\int (μ\times λ) (\cup \infty E ) = λ((\cup\infty E ) )dμ i=1 i X i=1 ix \int \int \sum\infty = λ(\cup\inftyi=1(Ei)x)dμ = λ ((Ei)x)dμ X Xi=1 \sum\infty \int = X λ((Ei)x)dμ i=\infty1 = \sum (μ \timesλ )(E ) i=1 i

This proves the theorem.

The next theorem is one of several theorems due to Fubini and Tonelli. These theorems all have to do with interchanging the order of integration in a multiple integral.

Theorem 10.10.14 Let f : X × Y →

be measurable with respect to S×ℱ and suppose μ and λ are σ finite. Then

\int \int \int \int \int X\timesY fd(μ\times λ) = X Y f(x,y)dλdμ = Y X f(x,y)dμd λ (10.10.49)

(10.10.49)

and all integrals make sense.

Proof: For E ∈S×ℱ,

\int \int XE (x,y)dλ = λ(Ex), XE (x,y)dμ = μ(Ey). Y X

Thus from Definition 10.10.12, 10.10.49 holds if f = X_E. It follows that 10.10.49 holds for every nonnegative simple function. By Theorem 9.3.9 on Page 644, there exists an increasing sequence,

, of simple functions converging pointwise to f. Then

\int \int f(x,y)dλ = nl\toim\infty fn(x,y)dλ, Y Y

\int \int f (x,y)dμ = lim fn(x,y)dμ. X n\to\infty X

This follows from the monotone convergence theorem. Since

\int x \to fn(x,y)dλ Y

is measurable with respect to S, it follows that x →∫ _Yf(x,y)dλ is also measurable with respect to S. A similar conclusion can be drawn about y →∫ _Xf(x,y)dμ. Thus the two iterated integrals make sense. Since 10.10.49 holds for f_n, another application of the Monotone Convergence theorem shows 10.10.49 holds for f. This proves the theorem.

Corollary 10.10.15 Let f : X × Y → ℂ be S×ℱ measurable. Suppose either ∫ _X ∫ _Y

dλdμ or ∫ _Y ∫ _X

dμdλ < ∞. Then f ∈ L¹(X × Y,μ × λ) and

\int \int \int \int \int fd(μ \times λ) = f dλdμ = fdμdλ (10.10.50) X\timesY X Y Y X

(10.10.50)

with all integrals making sense.

Proof: Suppose first that f is real valued. Apply Theorem 10.10.14 to f⁺and f⁻.10.10.50 follows from observing that f = f⁺ − f⁻; and that all integrals are finite. If f is complex valued, consider real and imaginary parts. This proves the corollary.

Suppose f is product measurable. From the above discussion, and breaking f down into a sum of positive and negative parts of real and imaginary parts and then using Theorem 9.3.9 on Page 644 on approximation by simple functions, it follows that whenever f is S×ℱ measurable, x → f

is μ measurable, y → f

is λ measurable.