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

  1. 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 $\bracs{z \in X|(z, y) \in E}\in \cn$.

  2. 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).$\square$