15.8 Product Measures

Definition 15.8.1 (Product Measure). Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be measure spaces, then there exists a measure $\mu \otimes \nu: \cm \times \cn \to [0, \infty]$ such that:

For each $E \in \cm$ and $F \in \cn$, $\mu \otimes \nu(E \times F) = \mu(E)\nu(F)$.
For any measure $\lambda: \cm \otimes \cn \to [0, \infty]$, $\lambda \le \mu$. For any $A \in \cm \otimes \cn$ with $\mu(A) < \infty$, $\lambda(A) = \mu(A)$. In particular, if $\mu$ is $\sigma$-finite, then $\lambda = \mu$.

The measure $\mu \otimes \nu$ is the product of $\mu$ and $\nu$.

Proof. Let

\[\mathcal{R}= \bracs{E \times F|E \in \cm, F \in \cn}\]

then $\mathcal{R}$ is an elementary family by Proposition 14.4.3. Let

\[\alg = \bracs{\bigsqcup_{i = 1}^n E_j \times F_j \bigg | \seqf{E_j \times F_j} \subset \mathcal{R} \text{ pairwise disjoint}}\]

then $\alg$ is a ring over $X \times Y$. For each pairwise disjoint collection $\seqf{E_j \times F_j}\subset \mathcal{R}$, let

\[\mu \otimes \nu\paren{\bigsqcup_{j = 1}^n E_j \times F_j}= \sum_{j = 1}^{n} \mu(E_{j}) \mu(F_{j})\]

then $\mu \otimes \nu$ is well-defined and finitely additive on $\alg$.

Let $A \times B \in \mathcal{A}$ and $\seq{A_n \times B_n}\subset \mathcal{R}$ such that $A \times B = \bigsqcup_{n \in \natp}A_{n} \times B_{n}$, then for any $x \in X$ and $y \in Y$,

\[\one_{A \times B}(x, y) = \sum_{n \in \natp}\one_{A_n \times B_n}(x, y) = \sum_{n \in \natp}\one_{A_n}(x)\one_{B_n}(y)\]

By the Monotone Convergence Theorem, for any $y \in Y$,

\begin{align*}\mu(A)\one_{B}(y)&= \int_{X} \one_{A}(x)\one_{B}(y)\mu(dx) = \sum_{n \in \natp}\int_{X} \one_{A_n}(x)\one_{B_n}(y) \mu(dx) \\&= \sum_{n = 1}^{\infty} \mu(A_{n})\one_{B_n}(y)\end{align*}

By the Monotone Convergence Theorem again,

\begin{align*}\mu(A)\nu(B)&= \int_{Y} \mu(A)\one_{B}(y) \nu(dy) = \sum_{n = 1}^{\infty} \int_{Y} \mu(A_{n})\one_{B_n}\nu(dy) \\&= \sum_{n = 1}^{\infty} \mu(A_{n})\nu(B_{n})\end{align*}

Therefore $\mu \otimes \nu$ is a premeasure on $\alg$. By Carathéodory’s Extension Theorem, there exists a measure $\mu \otimes \nu$ satisfying (1) and (U).$\square$

Lemma 15.8.2. Let $(X, \cm)$ and $(Y, \cn)$ be measurable spaces, then:

For any $E \in \cm \otimes \cn$, $x \in X$, and $y \in Y$, $\bracs{z \in Y|(x, z) \in E}\in \cm$ and $\bracs{z \in X|(z, y) \in E}\in \cn$.
For any measure space $(Z, \cf)$, $(\cm \otimes \cn, \cf)$-measurable function $f: X \times Y \to Z$, $x \in X$, and $y \in Y$, $f(x, \cdot)$ is $(\cn, \cf)$-measurable and $f(\cdot, y)$ is $(\cm, \cf)$-measurable.

Proof. (1): Let $\alg \subset \cm \otimes \cn$ be the collection of all sets that satisfy (1), then

\[\alg \supset \bracs{E \times F| E \in \cm, F \in \cn}\]

For any $E \in \alg$, $\bracs{z \in Y|(y, z) \in E}^{c} = \bracs{z \in Y|(y, z) \in E^c}$, so $E^{c} \in \alg$ as well. For any $\seq{E_n}\subset \alg$,

\[\bracs{z \in Y \bigg| (y, z) \in \bigcup_{n \in \natp}E_n}= \bigcup_{n \in \natp}\bracs{z \in Y | (y, z) \in E_n}\]

so $\bigcup_{n \in \natp}E_{n} \in \alg$. Therefore $\alg$ is a $\sigma$-algebra, and $\alg = \cm \otimes \cn$.

(2): For any $F \in \cf$,

\[\bracs{y \in Y|f(x, \cdot) \in F}= \bracs{y \in Y| (x, y) \in f^{-1}(F)}\in \cn\]

by (1).$\square$

Theorem 15.8.3 (Fubini-Tonelli Theorem). Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be $\sigma$-finite measure spaces, then

For any $f \in L^{+}(X \times Y)$, $[x \mapsto \int f(x, y)\nu(dy)] \in L^{+}(X)$, $[y \mapsto \int f(x, y)\mu(dx)] \in L^{+}(Y)$, and
\begin{align*}\int_{X \times Y}f(z)\mu \otimes \nu(dz)&= \int_{X}\int_{Y} f(x, y)\nu(dy)\mu(dx) \\&= \int_{Y}\int_{X} f(x, y)\mu(dx)\nu(dy)\end{align*}
For any normed space $E$ and $f \in L^{1}(X \times Y; E)$,
1. For almost every $x \in X$, $f(x, \cdot) \in L^{1}(Y; E)$. For almost every $y \in Y$, $f(\cdot, y) \in L^{1}(X; E)$.
2. $[x \mapsto \int f(x, y)\nu(dy)] \in L^{1}(X; E)$ and $[y \mapsto \int f(x, y)\mu(dx)] \in L^{1}(Y; E)$.
3. \begin{align*}\int_{X \times Y}f(z)\mu \otimes \nu(dz)&= \int_{X}\int_{Y} f(x, y)\nu(dy)\mu(dx) \\&= \int_{Y}\int_{X} f(x, y)\mu(dx)\nu(dy)\end{align*}

Proof. (1): First suppose that $\mu$ and $\nu$ are both finite. Let $\alg \subset \cm \otimes \cn$ be the collection of sets whose indicator functions satisfy (1), then $\alg$ contains all rectangles with finite measure. By linearity of the integral, for any $E, F\in \alg$, $E \setminus F \in \alg$. For any $\seq{E_n}\subset \alg$ and $E \in \alg$ such that $E_{n} \upto E$, by the Monotone Convergence Theorem,

\begin{align*}\int_{X \times Y}\one_{E}\mu \otimes \nu(dz)&= \limv{n}\int_{X \times Y}\one_{E_n}\mu \otimes \nu(dz) \\&= \limv{n}\int_{X}\int_{Y} \one_{E_n}(x, y)\nu(dy)\mu(dx) \\&= \int_{X}\int_{Y} \one_{E_n}(x, y)\nu(dy)\mu(dx)\end{align*}

so $\alg$ is a $\lambda$-system. By Dynkin’s $\pi$-$\lambda$ Theorem, $\alg = \cm \otimes \cn$.

Now suppose that $\mu$ and $\nu$ are $\sigma$-finite, then $\alg$ contains all sets in $\cm \otimes \cn$ with finite measure. Since $\mu$ and $\nu$ are $\sigma$-finite, there exists rectangles $\seq{A_n \times B_n}\subset \alg$ such that $\mu \otimes \nu(A_{n} \times B_{n})< \infty$ for all $n \in \natp$ and $A_{n} \times B_{n} \upto X \times Y$. Let $A \in \cm \otimes \cn$, then by the Monotone Convergence Theorem,

\begin{align*}\mu \otimes \nu(A)&= \limv{n}\mu \otimes \nu(A \cap (A_{n} \times B_{n})) \\&= \limv{n}\int_{X}\int_{Y} \one_{A \cap (A_n \times B_n)}(x, y)\nu(dy)\mu(dx) \\&= \int_{X}\int_{Y} \one_{A}(x, y)\nu(dy)\mu(dx)\end{align*}

Therefore $\alg = \cm \otimes \cn$.

Now let $F \subset L^{+}(X \times Y)$ be the set of functions satisfying (1), then $F \supset \Sigma^{+}(X, \cm)$ by linearity. Let $f \in L^{+}(X \times Y)$, then by Proposition 18.5.5, there exists $\seq{\phi_n}\subset \Sigma^{+}(X \times Y)$ such that $\phi_{n} \upto f$. In which case, by the Monotone Convergence Theorem, (1) also holds for $f$.$\square$

Jerry's Digital Garden

15.8 Product Measures

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Jerry's Digital Garden

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