Proof of The Product σ-Algebra: Rectangles, Generators, Sections, Projections, Pairings and the Swap

Throughout, functions given by an expression, such as $y\mapsto(x,y)$ or $(x,y)\mapsto(y,x)$, exist as in Sets and Maps: Ordinary Notation (v), sets of rectangles such as $\{A\times B:A\in\mathcal{C},\ B\in\mathcal{D}\}$ are sets of subsets of $X\times Y$ as images (Sets and Maps: Ordinary Notation (vii)), and preimages are computed elementwise with Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii). Write $\mathrm{Rect}=\mathrm{Rect}(\mathcal{F},\mathcal{G})$. By definition, $\mathcal{F}\otimes\mathcal{G}=\sigma(\mathrm{Rect})$, a $\sigma$-algebra on $X\times Y$ containing $\mathrm{Rect}$ (Measurable Rectangles and the Product σ-Algebra (ii), Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iii)). Every measurability claim into $\mathcal{F}\otimes\mathcal{G}$ below is checked on generators, by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (i).

Rectangles form a $\pi$-system. Every element of $\mathrm{Rect}$ is a subset of $X\times Y$ by Measurable Rectangles and the Product σ-Algebra (i). For $A,A'\in\mathcal{F}$ and $B,B'\in\mathcal{G}$ (Rectangles, Sections and the Swap: Set Identities (i)),

$$(A\times B)\cap(A'\times B')=(A\cap A')\times(B\cap B').$$

Since $A\cap A'\in\mathcal{F}$ and $B\cap B'\in\mathcal{G}$ (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix)), this is again a measurable rectangle, and $\mathrm{Rect}$ is a $\pi$-system on $X\times Y$ (π-System (i)). Moreover $X\in\mathcal{F}$ and $Y\in\mathcal{G}$ (σ-Algebra and Measurable Space (i)), so $X\times Y\in\mathrm{Rect}$. This proves (i).

Projections. Let $\mathrm{pr}_{1}:X\times Y\to X$ and $\mathrm{pr}_{2}:X\times Y\to Y$ be the projections $\mathrm{pr}_{1}(x,y)=x$ and $\mathrm{pr}_{2}(x,y)=y$ (Projection, Evaluation and Currying Maps (i)). For $A\in\mathcal{F}$ and $B\in\mathcal{G}$,

$$\mathrm{pr}_{1}^{-1}(A)=A\times Y,\qquad \mathrm{pr}_{2}^{-1}(B)=X\times B,$$

and both are measurable rectangles, as $X\in\mathcal{F}$ and $Y\in\mathcal{G}$; so they lie in $\mathcal{F}\otimes\mathcal{G}$. By Measurable Map, and the Set of Measurable Maps (i), $\mathrm{pr}_{1}$ is $(\mathcal{F}\otimes\mathcal{G})/\mathcal{F}$-measurable and $\mathrm{pr}_{2}$ is $(\mathcal{F}\otimes\mathcal{G})/\mathcal{G}$-measurable. This proves (vi); the argument applies to any two measurable spaces.

Pairings. Let $\varphi:W\to X\times Y$, $\varphi(w)=(g(w),h(w))$. Then $\mathrm{pr}_{1}\circ\varphi=g$ and $\mathrm{pr}_{2}\circ\varphi=h$. If $\varphi$ is $\mathcal{K}/(\mathcal{F}\otimes\mathcal{G})$-measurable, then $g$ and $h$ are measurable as composites of $\varphi$ with the measurable projections (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (ii)). Conversely, let $g$ be $\mathcal{K}/\mathcal{F}$-measurable and $h$ be $\mathcal{K}/\mathcal{G}$-measurable. For $A\in\mathcal{F}$ and $B\in\mathcal{G}$,

$$\varphi^{-1}(A\times B)=\{w\in W:g(w)\in A,\ h(w)\in B\}=g^{-1}(A)\cap h^{-1}(B),$$

which lies in $\mathcal{K}$, since $g^{-1}(A),h^{-1}(B)\in\mathcal{K}$ (Measurable Map, and the Set of Measurable Maps (i), Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix)). As $\mathcal{F}\otimes\mathcal{G}=\sigma(\mathrm{Rect})$, $\varphi$ is $\mathcal{K}/(\mathcal{F}\otimes\mathcal{G})$-measurable by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (i). This proves (vii), again for any two measurable spaces.

Rectangles of generators. Let $\mathrm{Rect}_{0}=\{A\times B:A\in\mathcal{C},\ B\in\mathcal{D}\}$. Since $\mathcal{C}\subseteq\mathcal{F}$ and $\mathcal{D}\subseteq\mathcal{G}$, we have $\mathrm{Rect}_{0}\subseteq\mathrm{Rect}$, hence $\sigma(\mathrm{Rect}_{0})\subseteq\sigma(\mathrm{Rect})=\mathcal{F}\otimes\mathcal{G}$ (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (v)). For the reverse inclusion, $\sigma(\mathrm{Rect}_{0})$ is a $\sigma$-algebra on $X\times Y$ containing $\mathrm{Rect}_{0}$ (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iii)); regard $\mathrm{pr}_{1}$ as a function from the measurable space $(X\times Y,\sigma(\mathrm{Rect}_{0}))$ to $(X,\mathcal{F})$. For $A\in\mathcal{C}$ we have $\mathrm{pr}_{1}^{-1}(A)=A\times Y\in\mathrm{Rect}_{0}$, because $Y\in\mathcal{D}$. Since $\mathcal{F}=\sigma(\mathcal{C})$, $\mathrm{pr}_{1}$ is $\sigma(\mathrm{Rect}_{0})/\mathcal{F}$-measurable by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (i), that is,

$$A\times Y=\mathrm{pr}_{1}^{-1}(A)\in\sigma(\mathrm{Rect}_{0})\qquad\text{for every }A\in\mathcal{F}.$$

In the same way, using $X\in\mathcal{C}$ and $\mathcal{G}=\sigma(\mathcal{D})$, $X\times B\in\sigma(\mathrm{Rect}_{0})$ for every $B\in\mathcal{G}$. Hence every measurable rectangle

$$A\times B=(A\times Y)\cap(X\times B)\qquad(A\in\mathcal{F},\ B\in\mathcal{G})$$

lies in $\sigma(\mathrm{Rect}_{0})$ (the identity by Rectangles, Sections and the Swap: Set Identities (i), membership by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix)). So $\mathrm{Rect}\subseteq\sigma(\mathrm{Rect}_{0})$, and $\mathcal{F}\otimes\mathcal{G}=\sigma(\mathrm{Rect})\subseteq\sigma(\mathrm{Rect}_{0})$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iv). This proves (ii).

Insertions. Fix $x\in X$ and let $i_{x}:Y\to X\times Y$, $i_{x}(y)=(x,y)$. For $A\in\mathcal{F}$ and $B\in\mathcal{G}$ (Rectangles, Sections and the Swap: Set Identities (iii), Rectangles, Sections and the Swap: Set Identities (i)),

$$i_{x}^{-1}(A\times B)=(A\times B)_{x}=\begin{cases}B,&x\in A,\\ \emptyset,&x\notin A,\end{cases}$$

and both $B$ and $\emptyset$ lie in $\mathcal{G}$ (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vi)). By Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (i), $i_{x}$ is $\mathcal{G}/(\mathcal{F}\otimes\mathcal{G})$-measurable. Likewise, for $y\in Y$ the function $i^{y}:X\to X\times Y$, $i^{y}(x)=(x,y)$, satisfies $(i^{y})^{-1}(A\times B)=(A\times B)^{y}=A$ if $y\in B$ and $=\emptyset$ otherwise, so it is $\mathcal{F}/(\mathcal{F}\otimes\mathcal{G})$-measurable. This proves (iii).

Sections of measurable sets. Let $E\in\mathcal{F}\otimes\mathcal{G}$, so $E\subseteq X\times Y$, and let $x\in X$ and $y\in Y$. By Rectangles, Sections and the Swap: Set Identities (iii),

$$E_{x}=i_{x}^{-1}(E),\qquad E^{y}=(i^{y})^{-1}(E),$$

and these lie in $\mathcal{G}$ and $\mathcal{F}$ respectively, by (iii) (and Measurable Map, and the Set of Measurable Maps (i)). This proves (iv).

Sections of measurable functions. For $x\in X$ and $y\in Y$, $f_{x}=f\circ i_{x}$ and $f^{y}=f\circ i^{y}$ (Precomposition Commutes with Pointwise Operations, and Sections of Functions Are Precompositions (iv)). By (iii) and Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (ii), $f_{x}$ is $\mathcal{G}/\mathcal{H}$-measurable and $f^{y}$ is $\mathcal{F}/\mathcal{H}$-measurable. This proves (v).

The swap. Since $s(z)=(\mathrm{pr}_{2}(z),\mathrm{pr}_{1}(z))$ for $z\in X\times Y$, and $\mathrm{pr}_{2}$ and $\mathrm{pr}_{1}$ are $(\mathcal{F}\otimes\mathcal{G})/\mathcal{G}$- and $(\mathcal{F}\otimes\mathcal{G})/\mathcal{F}$-measurable, the pairing step, applied with $(W,\mathcal{K})=(X\times Y,\mathcal{F}\otimes\mathcal{G})$ and the target $(Y\times X,\mathcal{G}\otimes\mathcal{F})$, shows that $s$ is $(\mathcal{F}\otimes\mathcal{G})/(\mathcal{G}\otimes\mathcal{F})$-measurable. With the roles of the two spaces exchanged, the swap $t:Y\times X\to X\times Y$, $t(y,x)=(x,y)$, is $(\mathcal{G}\otimes\mathcal{F})/(\mathcal{F}\otimes\mathcal{G})$-measurable, and

$$t\circ s=\mathrm{id}_{X\times Y},\qquad s\circ t=\mathrm{id}_{Y\times X},$$

so $s$ is a bijection with inverse $t$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (vii)). Hence, for $E\subseteq X\times Y$,

$$s[E]=t^{-1}(E),\qquad E=s^{-1}(s[E])$$

by Images and Preimages under a Bijection and Its Inverse (i) and Images and Preimages under a Bijection and Its Inverse (ii). If $E\in\mathcal{F}\otimes\mathcal{G}$, then $s[E]=t^{-1}(E)\in\mathcal{G}\otimes\mathcal{F}$ by measurability of $t$; if $s[E]\in\mathcal{G}\otimes\mathcal{F}$, then $E=s^{-1}(s[E])\in\mathcal{F}\otimes\mathcal{G}$ by measurability of $s$. This proves (viii).