> [!definition]- > > Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be [[Measure Space|measure spaces]]. A **measurable rectangle** is a [[Set|set]] of the form $A \times B$ where $A \in \cm, B \in \cn$. Then the collection $\ce$ of measurable rectangles is an [[Elementary Family|elementary family]], and the set $\alg$ of finite disjoint unions of measurable rectangles is an [[Algebra|algebra]]. > > Let $\pi: \ce \to [0, \infty]$ where $\pi(A \times B) = \mu(A) \cdot \nu(B)$ and extend $\pi$ to $\alg$ by > $ > \pi(A \times B) = \sum_{i = 1}^{n}\pi(A_i \times B_i) \quad > A \times B = \bigcup_{i = 1}^{n}A_i \times B_i \text{ disjoint} > $ > then $\pi$ is a [[Premeasure|premeasure]] on $\alg$. > > The smallest [[Sigma Algebra|sigma algebra]] $\cm(\alg)$ generated by $\alg$ is the [[Product Sigma Algebra|product sigma algebra]] $\cm \otimes \cn$. There exists a [[Measure Space|measure]] $\mu \times \nu$ on $\cm \otimes \cn$ such that $\mu \times \nu(A \times B) = \mu(A) \cdot \nu(B)$ for all $A \in \cm, B \in \cn$, known as the **product** of $\mu$ and $\nu$. If $\mu$ and $\nu$ are $\sigma$-finite, then this measure is unique. > > *Proof*. > > ### Elementary Family > > Let $A \times B, C \times D \in \ce$ be two measurable rectangles, then > $ > \begin{align*} > (A \times B) \cap (C \times D) &= \bracs{(a, b): a \in A, b \in B} \cap \\ > & \quad \bracs{(c, d): c \in C, d \in D} \\ > &= \bracs{(x, y): (x \in A, y \in B) \text{ and }(x \in C, y \in D)} \\ > &= \bracs{(x, y): x \in A \cap C, y \in B \cap D} \\ > &= (A \cap D) \times B \cap D > \end{align*} > $ > the intersection of two rectangles is another rectangle, and > $ > (A \times B)^c = (A \times B^c) \cup (A^c \times B) \cup \bracs{A^c \cap B^c} > $ > the complement of one is a finite disjoint union of other rectangles, as seen > > | $A^C \times B^C$ | $A \times B^C$ | $A^C \times B^C$ | > | ---------------- | -------------- | ---------------- | > | $B \times A^C$ | $A \times B$ | $B \times A^C$ | > | $A^C \times B^C$ | $A \times B^C$ | $A^C \times B^C$ | > > Therefore $\ce$ is an [[Elementary Family|elementary family]], and the collection $\alg$ of FDUs of $\ce$ is an [[Algebra|algebra]]. > > Since $\ce \subset \alg$, $\cm(\ce) \subseteq \cm(\alg)$. Let $\bigcup_{i = 1}^{n}A_i \times B_i \in \alg$, then since a $\sigma$-algebra is closed under finite unions, $\bigcup_{i = 1}^{n}A_i \times B_i \in \cm(\ce)$. So $\cm(\ce) = \cm(\alg)$. > > Let $A \in \cm$ or $B \in \cn$, then $A \times Y$ and $B \times X$ are both in $\ce$. So $\cm(\ce) \supseteq \cm \otimes \cn$. Since each rectangle > $ > A \times B = \pi_X^{-1}(A) \cap \pi_Y^{-1}(B) \in \cm \otimes \cn > $ > is in the product algebra, $\cm(\alg) = \cm(\ce) = \cm \otimes \cn$. Therefore the $\sigma$-algebra generated by $\alg$ coincides with the product $\sigma$-algebra. > > ### Premeasure > > Define $\pi: \ce \to [0, \infty]$ by $\pi(A \times B) = \mu(A) \cdot \nu(B)$, and extend $\pi$ to $\alg$ by > $ > \pi\paren{\bigcup_{i = 1}^{n}A_i \times B_i} = \sum_{i = 1}^{n}\pi(A_i \times B_i) > $ > To see that $\pi$ is well-defined and is a premeasure, let $\seq{A_n \times B_n}$ be a [[Sequence|sequence]] of measurable rectangles such that $\bigcup_{n \in \nat}A_n \times B_n = A \times B$ is a disjoint union, then > $ > \begin{align*} > \chi_{A \times B} &= \sum_{n \in \nat}\chi_{A_n \times B_n} \\ > R(x, y) = \chi_{A}(x) \cdot \chi_{B}(y) &= \sum_{n \in \nat}\chi_{A_n}(x) \cdot \chi_{B_n}(y) > \end{align*} > $ > their [[Indicator Function|indicator functions]] add up. Here $\chi_{A}, \chi_{B}$ is an abuse of notation where $(a, b) \in A$ if $a \in A$ and $(a, b) \in B$ if $b \in B$. Now, fix an $x \in A$, then by the [[Monotone Convergence Theorem]], the sum may be interchanged with the integral: > $ > \begin{align*} > R(x, y) &= \sum_{n \in \nat}\chi_{A_n}(x) \cdot \chi_{B_n}(y) \\ > \int_B R(x, y)d \nu(y) &= \int_B\braks{\sum_{n \in \nat}\chi_{A_n}(x) \cdot \chi_{B_n}(y)}d\nu(y) \\ > \int_B\chi_{A}(x) \cdot \chi_{B}(y)d\nu(y)&= \sum_{n \in \nat}\braks{\int_B \chi_{A_n}(x) \cdot \chi_{B_n}(y)d\nu(y)} \\ > \chi_{A}(x) \cdot \nu(B)&= \sum_{n \in \nat}\braks{\chi_{A_n}(x) \cdot \int_{B}\chi_{B_n}(y)d\nu(y)} \\ > \chi_A(x) \cdot \nu(B)&= \sum_{n \in \nat}\chi_{A_n}(x) \cdot \nu(B_n) > \end{align*} > $ > Applying the same idea across $x$ gives > $ > \begin{align*} > \int_A \chi_{A}(x) \cdot \nu(B) d\mu(x) &= \int_A\braks{\sum_{n \in \nat}\chi_{A_n}(x) \cdot \nu(B_n)}d\mu(x) \\ > &= \sum_{n \in \nat}\braks{\int_A\chi_{A_n}(x) \cdot \nu(B_n)d\mu(x)} \\ > &= \sum_{n \in \nat}\braks{\nu(B_n) \cdot \int_A\chi_{A_n}(x)d\mu(x)} \\ > \mu(A) \cdot \nu(B) &= \sum_{n \in \nat}\mu(A_n) \cdot \nu(B_n) \\ > \end{align*} > $ > Together, these results yield that > $ > \mu(A) \cdot \nu(B) = \int_A\int_B R(x, y) d\nu(y) d\mu(x) = \sum_{n \in \nat}\mu(A_n) \cdot \nu(B_n) > $ > Since the above result holds for finite sums as well, $\pi$ is both well-defined and a premeasure on $\alg$. > > ### Measure > > By [[Carathéodory's Extension Theorem]], there is an extension of $\pi$ into a [[Complete Measure|complete measure]] on $\cm \otimes \cn$ satisfying $\pi(A \times B) = \mu(A) \cdot \nu(B)$. > > If $\mu$ and $\nu$ are $\sigma$-finite, then there exists $\seq{A_i}$ and $\seq{B_i}$ such that $X = \bigcup_{i \in \nat}A_i$, $Y = \bigcup_{i \in \nat}B_i$ and $\mu(A_i) < \infty$, $\mu(B_i) < \infty$ for all $i \in \nat$. This means that > $ > X \times Y = \bigcup_{i \in \nat}A_i \times Y = \bigcup_{i \in \nat}\bigcup_{j \in \nat}A_i \times B_j > $ > where $\pi(A_i \times B_j) = \mu(A_i) \cdot \nu(B_j) < \infty$ and $\pi$ is $\sigma$-finite, in which case $\pi$ is the unique measure on $\cm \otimes \cn$ satisfying $\pi(A \times B) = \mu(A) \cdot \nu(B)$. > [!definition] > > Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be measure spaces. Let $E \subset X \times Y$, then the $x$-section and $y$-section of $E$ are > $ > E_x = \bracs{x: (x, y) \in E} \quad E^y = \bracs{y: (x, y) \in E} > $ > Let $f$ be a [[Function|function]] on $X \times Y$, then the $x$-section and $y$-section of $f$ are > $ > f_x(y) = f^y(x) = f(x, y) > $ > Then $(\chi_{E})_x = \chi_{E_x}$ and $(\chi_E)^y = \chi_{E^y}$. > [!theorem] > > Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be measure spaces, then > - If $E \in \cm \otimes \cn$, then $E_x \in \cn$ and $E^y \in \cm$ for all $x \in X, y \in Y$. > - If $f: X \times Y \to \complex$ is $\cm \otimes \cn$-[[Measurable Function|measurable]], then $f_x$ is $\cn$-measurable and $f^y$ is $\cm$-measurable. > > *Proof*. Let > $ > \mathcal{R} = \bracs{E \in \cm \otimes \cn: E_x \in \cn, E^y \in \cm \ \forall x \in X, y \in Y} > $ > be the collection of sets in $\cm \otimes \cn$ such that all of their $x$-sections and $y$-sections are $\mu$ and $\nu$-measurable. > > Let $E \in \mathcal{R}$, then for any $x \in X, y \in Y$, $(E^c)_x = (E_x)^c$ and $(E^c)^y = (E^y)^c$. Let $\seq{E_n} \subset \mathcal{R}$, then for any $x \in X, y \in Y$, > $ > \paren{\bigcup_{n \in \nat}E_n}_x = \bigcup_{n \in \nat}(E_n)_x \quad > \paren{\bigcup_{n \in \nat}E_n}^y = \bigcup_{n \in \nat}(E_n)^y > $ > So $\mathcal{R}$ is a $\sigma$-algebra. Since each measurable rectangle is in $\mathcal{R}$, $\mathcal{R} \supseteq \cm(\ce) = \cm \otimes \cn$. > > Let $f: X \times Y \to \complex$ be a $\cm \otimes \cn$-measurable function. Then for any [[Borel Sigma Algebra|Borel set]] $E \in \cb_\complex$, $x \in X$ and $y \in Y$, > $ > \begin{align*} > (f_x)^{-1}(E) &= \bracs{y \in Y: f_x(y) \in E} \\ > &= \bracs{y \in Y: f(x, y) \in E} \\ > &= \bracs{y \in Y: (x, y) \in f^{-1}(E)} \\ > &= \braks{f^{-1}(E)}_x \\ > (f^y)^{-1}(E) &= \braks{f^{-1}(E)}^y > \end{align*} > $ > Since $f^{-1}(E) \in \cm \otimes \cn$, $\braks{f^{-1}(E)}_x \in \cn$ and $\braks{f^{-1}(E)}^y \in \cm$ are both measurable sets. Therefore $f_x$ is $\cn$-measurable and $f_y$ is $\cm$-measurable.