Proof of Pairings into a Product, the Graph of a Measurable Map, and Couplings Concentrated on a Graph

Each result cited is universally quantified over the data in its own statement, and is applied below with the data named at each use.

Fix the measurable space $(Y,\mathcal{Y})$ and the metric space $(Z,d_{Z})$ of the statement. The Borel $\sigma$-algebra $\mathcal{B}(Z)$ is a $\sigma$-algebra on $Z$ by claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra, being generated by the open subsets (Borel Sigma-Algebra of a Metric Space).

Step 1 (claim 1, pairings). Let $(\Omega,\mathcal{O})$ be a measurable space and let $F:\Omega\to Y$ and $G:\Omega\to Z$ be measurable as in claim 1. Let $\mathcal{K}$ be the family of those $E\subseteq Y\times Z$ with $(F,G)^{-1}(E)\in\mathcal{O}$. We check the three properties of Sigma-Algebra and Measurable Space for $\mathcal{K}$ on $Y\times Z$, using the same properties of $\mathcal{O}$ on $\Omega$. First, $(F,G)^{-1}(Y\times Z)=\Omega\in\mathcal{O}$. Second, if $E\in\mathcal{K}$ then $(F,G)^{-1}\bigl((Y\times Z)\setminus E\bigr)=\Omega\setminus(F,G)^{-1}(E)\in\mathcal{O}$. Third, for a sequence $(E_{m})_{m\in\mathbb{N}}$ in $\mathcal{K}$, $(F,G)^{-1}\bigl(\bigcup_{m}E_{m}\bigr)=\bigcup_{m}(F,G)^{-1}(E_{m})\in\mathcal{O}$. So $\mathcal{K}$ is a $\sigma$-algebra on $Y\times Z$. For a measurable rectangle $A\times B$ with $A\in\mathcal{Y}$ and $B\in\mathcal{B}(Z)$,

$$(F,G)^{-1}(A\times B)=\{\omega\in\Omega:F(\omega)\in A,\ G(\omega)\in B\}=F^{-1}(A)\cap G^{-1}(B),$$

and $F^{-1}(A)\in\mathcal{O}$, $G^{-1}(B)\in\mathcal{O}$ by the measurability of $F$ and $G$ (Measurable Function and Real-Valued Measurable Function); as $\mathcal{O}$ is closed under finite intersections (Sigma-Algebra and Measurable Space), $A\times B\in\mathcal{K}$. Thus $\mathcal{K}$ contains the family of measurable rectangles, whose generated $\sigma$-algebra is $\mathcal{Y}\otimes\mathcal{B}(Z)$ by Product Sigma-Algebra; by the minimality in claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra, applied on the set $Y\times Z$ to the family of measurable rectangles and the $\sigma$-algebra $\mathcal{K}$, $\mathcal{Y}\otimes\mathcal{B}(Z)\subseteq\mathcal{K}$. That is, $(F,G)^{-1}(E)\in\mathcal{O}$ for every $E\in\mathcal{Y}\otimes\mathcal{B}(Z)$, so $(F,G)$ is measurable with respect to $\mathcal{O}$ and $\mathcal{Y}\otimes\mathcal{B}(Z)$.

For the special case, let $S:Y\to Z$ be measurable with respect to $\mathcal{Y}$ and $\mathcal{B}(Z)$. The identity map of $Y$ is measurable with respect to $\mathcal{Y}$ and $\mathcal{Y}$, since the preimage of every $A\in\mathcal{Y}$ is $A$. Applying what was just proved with $(\Omega,\mathcal{O})=(Y,\mathcal{Y})$, $F=\mathrm{id}$ and $G=S$ shows that $(\mathrm{id},S)$ is measurable with respect to $\mathcal{Y}$ and $\mathcal{Y}\otimes\mathcal{B}(Z)$. The argument of this step used nothing about $(Y,\mathcal{Y})$ beyond its being a measurable space; it therefore also applies with any other measurable space in place of $(Y,\mathcal{Y})$, which is how it is used in Step 3.

Step 2 (the diagonal). Assume now that $(Z,d_{Z})$ is separable. We show that the diagonal $\Delta=\{(z,z')\in Z\times Z:z=z'\}$ belongs to $\mathcal{B}(Z)\otimes\mathcal{B}(Z)$, the product $\sigma$-algebra of $(Z,\mathcal{B}(Z))$ with itself. If $Z=\varnothing$ then $\Delta=\varnothing$, which belongs to every $\sigma$-algebra on $Z\times Z$ (Sigma-Algebra and Measurable Space). Otherwise, by Separable Metric Space choose a countable subset $D\subseteq Z$ that is dense in $Z$ (Dense Subset of a Topological Space), so that its closure is $Z$. The set $D$ is nonempty: for a point $z\in Z$, which lies in the closure of $D$, Characterization of the Closure in a Metric Space by Open Balls (implication from 1 to 3, with $A=D$, $x=z$ and $\varepsilon=1$) provides a point of $D$. Hence, by Countable Set, there is a sequence $(q_{n})_{n\in\mathbb{N}}$ in $Z$ whose set of terms is $D$. For each natural number $m\ge1$ let

$$U_{m}=\bigcup_{n\in\mathbb{N}}B_{d_{Z}}(q_{n},1/m)\times B_{d_{Z}}(q_{n},1/m),$$

with the open balls of Open Ball in a Metric Space. We claim that $\Delta=\bigcap_{m\ge1}U_{m}$.

If $(z,z)\in\Delta$ and $m\ge1$, then $z$ lies in the closure of $D$, so Characterization of the Closure in a Metric Space by Open Balls (implication from 1 to 3, with $A=D$, $x=z$ and $\varepsilon=1/m$) gives $a\in D$ with $d_{Z}(z,a)<1/m$; write $a=q_{n}$. By symmetry of the metric (Metric Space), $d_{Z}(q_{n},z)<1/m$, so $z\in B_{d_{Z}}(q_{n},1/m)$ and $(z,z)\in U_{m}$. Conversely, let $(z,z')\in U_{m}$ for every $m\ge1$. For each $m$ there is $n$ with $d_{Z}(q_{n},z)<1/m$ and $d_{Z}(q_{n},z')<1/m$, so by symmetry and the triangle inequality (Metric Space) $d_{Z}(z,z')\le d_{Z}(z,q_{n})+d_{Z}(q_{n},z')<2/m$. Thus the real number $d_{Z}(z,z')\ge0$ is less than $2/m$ for every $m\ge1$; if it were positive, a natural number $m>2/d_{Z}(z,z')$ would give a contradiction, so $d_{Z}(z,z')=0$, and $z=z'$ by property 2 of Metric Space. This proves the claim.

Each open ball $B_{d_{Z}}(q_{n},1/m)$ is open by Open Ball in a Metric Space is Open, hence belongs to $\mathcal{B}(Z)$ by claim 1 of Borel Measurability and Bounded Integration on a Metric Space; so each $B_{d_{Z}}(q_{n},1/m)\times B_{d_{Z}}(q_{n},1/m)$ is a measurable rectangle and belongs to $\mathcal{B}(Z)\otimes\mathcal{B}(Z)$ by Product Sigma-Algebra and claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra. Since a $\sigma$-algebra is closed under countable unions and countable intersections (Sigma-Algebra and Measurable Space), each $U_{m}$ and then $\Delta$ belong to $\mathcal{B}(Z)\otimes\mathcal{B}(Z)$.

Step 3 (claim 2, the graph is measurable). Let $(Z,d_{Z})$ be separable and $S:Y\to Z$ measurable with respect to $\mathcal{Y}$ and $\mathcal{B}(Z)$. Define $\Psi:Y\times Z\to Z\times Z$ by $\Psi(y,z)=(S(y),z)$, that is, $\Psi=(S\circ\mathrm{pr}_{Y},\mathrm{pr}_{Z})$. For $B\in\mathcal{B}(Z)$, $(S\circ\mathrm{pr}_{Y})^{-1}(B)=\mathrm{pr}_{Y}^{-1}(S^{-1}(B))$, where $S^{-1}(B)\in\mathcal{Y}$ by measurability of $S$ and then $\mathrm{pr}_{Y}^{-1}(S^{-1}(B))\in\mathcal{Y}\otimes\mathcal{B}(Z)$ by measurability of $\mathrm{pr}_{Y}$ (claim 5 of Sections of Product-Measurable Sets and Maps Are Measurable, and Insertion Maps into Products Are Measurable); so $S\circ\mathrm{pr}_{Y}$ is measurable with respect to $\mathcal{Y}\otimes\mathcal{B}(Z)$ and $\mathcal{B}(Z)$ (Measurable Function and Real-Valued Measurable Function). Also $\mathrm{pr}_{Z}$ is measurable with respect to $\mathcal{Y}\otimes\mathcal{B}(Z)$ and $\mathcal{B}(Z)$ by claim 5 of Sections of Product-Measurable Sets and Maps Are Measurable, and Insertion Maps into Products Are Measurable. The argument of Step 1, applied with the measurable space $(Z,\mathcal{B}(Z))$ in place of $(Y,\mathcal{Y})$, with $(\Omega,\mathcal{O})=(Y\times Z,\mathcal{Y}\otimes\mathcal{B}(Z))$, $F=S\circ\mathrm{pr}_{Y}$ and $G=\mathrm{pr}_{Z}$, shows that $\Psi$ is measurable with respect to $\mathcal{Y}\otimes\mathcal{B}(Z)$ and $\mathcal{B}(Z)\otimes\mathcal{B}(Z)$. For $(y,z)\in Y\times Z$ one has $(y,z)\in\Gamma_{S}$ iff $z=S(y)$ iff $\Psi(y,z)\in\Delta$, so $\Gamma_{S}=\Psi^{-1}(\Delta)$, which belongs to $\mathcal{Y}\otimes\mathcal{B}(Z)$ by Step 2 and the measurability of $\Psi$.

Step 4 (claim 3, a coupling on the graph). Let $(Z,d_{Z})$ be separable, and let $\mu$, $S$ and $\pi$ be as in claim 3; write $\Phi=(\mathrm{id},S):Y\to Y\times Z$, which is measurable with respect to $\mathcal{Y}$ and $\mathcal{Y}\otimes\mathcal{B}(Z)$ by Step 1, so that the image measure $\Phi_{\#}\mu$, $E\mapsto\mu(\Phi^{-1}(E))$, is a probability measure on $(Y\times Z,\mathcal{Y}\otimes\mathcal{B}(Z))$ by claim 1 of Image Measures, Measures with Densities, and Change of Variables (with $(Y,\mathcal{Y},\mu)$ and $T=\Phi$). By Step 3, $\Gamma_{S}\in\mathcal{Y}\otimes\mathcal{B}(Z)$.

First, $\pi$ does not see the complement of the graph. The measure $\pi$ is finite, so Basic Properties of a Measure §differences (on $(Y\times Z,\mathcal{Y}\otimes\mathcal{B}(Z),\pi)$, with the set $\Gamma_{S}$) gives $\pi\bigl((Y\times Z)\setminus\Gamma_{S}\bigr)=1-1=0$. Let $H\in\mathcal{Y}\otimes\mathcal{B}(Z)$. The sets $H\cap\Gamma_{S}$ and $H\setminus\Gamma_{S}$ belong to $\mathcal{Y}\otimes\mathcal{B}(Z)$, are disjoint and have union $H$, so $\pi(H)=\pi(H\cap\Gamma_{S})+\pi(H\setminus\Gamma_{S})$ by Basic Properties of a Measure §additivity; and $0\le\pi(H\setminus\Gamma_{S})\le\pi\bigl((Y\times Z)\setminus\Gamma_{S}\bigr)=0$ by Basic Properties of a Measure §monotone. Hence

$$\pi(H)=\pi(H\cap\Gamma_{S})\qquad\text{for every }H\in\mathcal{Y}\otimes\mathcal{B}(Z).$$

Now fix $E\in\mathcal{Y}\otimes\mathcal{B}(Z)$. The set $\Phi^{-1}(E)$ belongs to $\mathcal{Y}$, and $E'=\mathrm{pr}_{Y}^{-1}(\Phi^{-1}(E))=\{(y,z)\in Y\times Z:(y,S(y))\in E\}$ belongs to $\mathcal{Y}\otimes\mathcal{B}(Z)$ by claim 5 of Sections of Product-Measurable Sets and Maps Are Measurable, and Insertion Maps into Products Are Measurable. We have $E\cap\Gamma_{S}=E'\cap\Gamma_{S}$: if $(y,z)\in\Gamma_{S}$ then $z=S(y)$, so $(y,z)\in E$ exactly when $(y,S(y))\in E$, that is, exactly when $(y,z)\in E'$. Using the last display twice (with $H=E$ and $H=E'$) and the hypothesis $(\mathrm{pr}_{Y})_{\#}\pi=\mu$,

$$\pi(E)=\pi(E\cap\Gamma_{S})=\pi(E'\cap\Gamma_{S})=\pi(E')=\pi\bigl(\mathrm{pr}_{Y}^{-1}(\Phi^{-1}(E))\bigr)=\mu\bigl(\Phi^{-1}(E)\bigr)=(\Phi_{\#}\mu)(E).$$

As $E$ was arbitrary, $\pi=\Phi_{\#}\mu=(\mathrm{id},S)_{\#}\mu$.

Finally, $\mathrm{pr}_{Z}$ and $S$ are measurable (claim 5 of Sections of Product-Measurable Sets and Maps Are Measurable, and Insertion Maps into Products Are Measurable and the hypothesis), so $(\mathrm{pr}_{Z})_{\#}\pi$ and $S_{\#}\mu$ are image measures on $(Z,\mathcal{B}(Z))$ by claim 1 of Image Measures, Measures with Densities, and Change of Variables. Since $\mathrm{pr}_{Z}(\Phi(y))=\mathrm{pr}_{Z}(y,S(y))=S(y)$ for every $y\in Y$, one has $\Phi^{-1}(\mathrm{pr}_{Z}^{-1}(B))=S^{-1}(B)$ for every $B\in\mathcal{B}(Z)$, and therefore, by what was just proved,

$$\bigl((\mathrm{pr}_{Z})_{\#}\pi\bigr)(B)=\pi\bigl(\mathrm{pr}_{Z}^{-1}(B)\bigr)=\mu\bigl(\Phi^{-1}(\mathrm{pr}_{Z}^{-1}(B))\bigr)=\mu\bigl(S^{-1}(B)\bigr)=(S_{\#}\mu)(B).$$

So $(\mathrm{pr}_{Z})_{\#}\pi=S_{\#}\mu$. $\square$