Proof of Sections of Product-Measurable Sets and Maps Are Measurable, and Insertion Maps into Products Are Measurable

Two elementary facts are used throughout. (P) Let $\Phi:A\to A'$ be a map between sets and let $\mathcal{S}$ be a $\sigma$-algebra on $A$. The family of subsets $S'$ of $A'$ with $\Phi^{-1}(S')\in\mathcal{S}$ is a $\sigma$-algebra on $A'$, because $\Phi^{-1}(A')=A$, $\Phi^{-1}(A'\setminus S')=A\setminus\Phi^{-1}(S')$ and $\Phi^{-1}(\bigcup_nS'_n)=\bigcup_n\Phi^{-1}(S'_n)$. Hence, if a $\sigma$-algebra $\mathcal{S}'$ on $A'$ is the $\sigma$-algebra generated by a family $\mathcal{E}$ of subsets of $A'$ and $\Phi^{-1}(E)\in\mathcal{S}$ for every $E\in\mathcal{E}$, then $\mathcal{S}'$ is contained in the $\sigma$-algebra of sets with measurable preimage, that is, $\Phi$ is measurable with respect to $\mathcal{S}$ and $\mathcal{S}'$. In particular, a map into a product $X\times Y$ is measurable for $\mathcal{F}\otimes\mathcal{G}$ as soon as the preimages of all measurable rectangles $A\times B$ ($A\in\mathcal{F}$, $B\in\mathcal{G}$) are measurable, the rectangles generating the product $\sigma$-algebra. (Q) A composition of measurable maps is measurable, the preimage of a set under the composition being the preimage under the inner map of its preimage under the outer map.

Claim 1. Fix $x\in X$ and let $\mathcal{S}$ be the family of subsets $S$ of $X\times Y$ with $S_x\in\mathcal{G}$. It contains every measurable rectangle $A\times B$, whose section $(A\times B)_x$ is $B$ if $x\in A$ and empty otherwise, both members of $\mathcal{G}$. It is a $\sigma$-algebra on $X\times Y$: $(X\times Y)_x=Y\in\mathcal{G}$; $((X\times Y)\setminus S)_x=Y\setminus S_x$; and $(\bigcup_nS_n)_x=\bigcup_n(S_n)_x$, since $y\in(\bigcup_nS_n)_x$ if and only if $(x,y)\in S_n$ for some $n$. Hence $\mathcal{S}$ contains the $\sigma$-algebra generated by the rectangles, which is $\mathcal{F}\otimes\mathcal{G}$, so $S_x\in\mathcal{G}$ for every $S\in\mathcal{F}\otimes\mathcal{G}$. The statement for $S^{y}$ follows in the same way, the section $(A\times B)^{y}$ being $A$ if $y\in B$ and empty otherwise.

Claim 2. For $S\subseteq X\times Y$ one has $\iota_x^{-1}(S)=\{y:(x,y)\in S\}=S_x$, which lies in $\mathcal{G}$ for every $S\in\mathcal{F}\otimes\mathcal{G}$ by claim 1; so $\iota_x$ is measurable. Likewise $(\iota^{y})^{-1}(S)=S^{y}\in\mathcal{F}$, so $\iota^{y}$ is measurable.

Claim 3. The map $y\mapsto\Psi(x,y)$ is the composition $\Psi\circ\iota_x$, measurable by claim 2 and (Q); likewise $x\mapsto\Psi(x,y)$ is $\Psi\circ\iota^{y}$. For a $[0,\infty]$-valued $\Theta$ with $\{\Theta>a\}\in\mathcal{F}\otimes\mathcal{G}$ for every real $a$, the set $\{y:\Theta(x,y)>a\}$ is the section at $x$ of $\{\Theta>a\}$, hence lies in $\mathcal{G}$ by claim 1; so the section satisfies the criterion, and symmetrically for the sections at $y$.

Claim 4. By (P) it suffices to check the preimages of the measurable rectangles $A\times C$ with $A\in\mathcal{F}\otimes\mathcal{G}$ and $C\in\mathcal{H}$, which generate $(\mathcal{F}\otimes\mathcal{G})\otimes\mathcal{H}$. Under $(y,z)\mapsto((x,y),z)$ the preimage of $A\times C$ is $A_x\times C$, a measurable rectangle of $\mathcal{G}\otimes\mathcal{H}$ because $A_x\in\mathcal{G}$ by claim 1. Under $(x,z)\mapsto((x,y),z)$ it is $A^{y}\times C$, a measurable rectangle of $\mathcal{F}\otimes\mathcal{H}$ because $A^{y}\in\mathcal{F}$ by claim 1. Under $(x,y)\mapsto((x,y),z)$ it is $A$ if $z\in C$ and empty otherwise, a member of $\mathcal{F}\otimes\mathcal{G}$. The consequence for $\Psi$ follows by (Q), each partial map being the composition of $\Psi$ with the corresponding insertion map.

Claim 5. Under $(x,y)\mapsto x$ the preimage of $A\in\mathcal{F}$ is the rectangle $A\times Y$, and under $(x,y)\mapsto y$ the preimage of $B\in\mathcal{G}$ is $X\times B$; so the projections are measurable. Under $(x,y)\mapsto(y,x)$ the preimage of a rectangle $B\times A$ ($B\in\mathcal{G}$, $A\in\mathcal{F}$) is the rectangle $A\times B$, so the map is measurable by (P). Under $((x,y),z)\mapsto(z,x)$ the preimage of a rectangle $C\times A$ ($C\in\mathcal{H}$, $A\in\mathcal{F}$) is $(A\times Y)\times C$, a measurable rectangle of $(\mathcal{F}\otimes\mathcal{G})\otimes\mathcal{H}$ since $A\times Y\in\mathcal{F}\otimes\mathcal{G}$; so the map is measurable by (P).