Proof of Conditional Kernels of a Borel Probability Measure on a Polish Space Given a Borel Map: Existence, Uniqueness and Concentration on the Fibres

Each result cited is universally quantified over the data in its own statement.

Throughout, $\mathcal{B}(Y)\otimes\mathcal{B}(Z)$ is the product $\sigma$-algebra on $Y\times Z$, generated by the rectangles $A\times B$ with $A\in\mathcal{B}(Y)$ and $B\in\mathcal{B}(Z)$, and $(Y,\mathcal{B}(Y))$ is a measurable space by Borel Sigma-Algebra of a Metric Space and claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra.

Step 1 (the graph measure). Define $T:Z\to Y\times Z$ by $T(z)=(q(z),z)$. Let $\mathcal{S}$ be the family of the sets $E\subseteq Y\times Z$ with $T^{-1}(E)\in\mathcal{B}(Z)$. Since $T^{-1}(Y\times Z)=Z$, $T^{-1}\bigl((Y\times Z)\setminus E\bigr)=Z\setminus T^{-1}(E)$ and $T^{-1}\bigl(\bigcup_{m}E_{m}\bigr)=\bigcup_{m}T^{-1}(E_{m})$, and $\mathcal{B}(Z)$ is a $\sigma$-algebra, $\mathcal{S}$ is a $\sigma$-algebra on $Y\times Z$ (Sigma-Algebra and Measurable Space). For $A\in\mathcal{B}(Y)$ and $B\in\mathcal{B}(Z)$ we have $T^{-1}(A\times B)=B\cap q^{-1}(A)\in\mathcal{B}(Z)$, by the measurability of $q$. So $\mathcal{S}$ contains every rectangle, and by claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra it contains $\mathcal{B}(Y)\otimes\mathcal{B}(Z)$; that is, $T$ is measurable with respect to $\mathcal{B}(Z)$ and $\mathcal{B}(Y)\otimes\mathcal{B}(Z)$. By claim 1 of Image Measures, Measures with Densities, and Change of Variables, $\hat{\pi}(E)=\pi(T^{-1}(E))$ defines a probability measure $\hat{\pi}$ on $(Y\times Z,\mathcal{B}(Y)\otimes\mathcal{B}(Z))$, and

$$\hat{\pi}(A\times B)=\pi\bigl(B\cap q^{-1}(A)\bigr)\qquad(A\in\mathcal{B}(Y),\ B\in\mathcal{B}(Z)).\tag{1}$$

In particular $\hat{\pi}(Y\times Z)=1$, and the first marginal of $\hat{\pi}$ is $A\mapsto\hat{\pi}(A\times Z)=\pi(q^{-1}(A))=\nu(A)$. Hence Disintegration of a Probability Measure on a Product with a Polish Factor: Existence and Almost-Everywhere Uniqueness of the Conditional Kernel applies to the measurable space $(Y,\mathcal{B}(Y))$, the complete separable metric space $(Z,d)$ and the measure $\hat{\pi}$, and the first marginal called $\mu$ there is $\nu$.

Step 2 (conditional kernels given $q$ are the kernels of $\hat{\pi}$). Let $\kappa$ be a probability kernel from $(Y,\mathcal{B}(Y))$ to $(Z,\mathcal{B}(Z))$. We claim that $\kappa$ is a conditional kernel of $\pi$ given $q$ in the sense of The Conditional Kernel of a Probability Measure Given a Measurable Map §conditional-kernel if and only if $\hat{\pi}=\nu\otimes\kappa$. If $\kappa$ is a conditional kernel of $\pi$ given $q$, then by (1)

$$\hat{\pi}(A\times B)=\int_{Y}\mathbf{1}_{A}(y)\,\kappa(y,B)\,\nu(dy)\qquad(A\in\mathcal{B}(Y),\ B\in\mathcal{B}(Z)),$$

so $\hat{\pi}=\nu\otimes\kappa$ by Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §rectangles, applied to the probability measure $\nu$ on $(Y,\mathcal{B}(Y))$. Conversely, if $\hat{\pi}=\nu\otimes\kappa$, then by (1) and Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §composite, $\pi\bigl(B\cap q^{-1}(A)\bigr)=\hat{\pi}(A\times B)=(\nu\otimes\kappa)(A\times B)=\int_{Y}\mathbf{1}_{A}(y)\,\kappa(y,B)\,\nu(dy)$ for all $A\in\mathcal{B}(Y)$ and $B\in\mathcal{B}(Z)$, which is the defining identity of a conditional kernel of $\pi$ given $q$.

Claim 1. By Disintegration of a Probability Measure on a Product with a Polish Factor: Existence and Almost-Everywhere Uniqueness of the Conditional Kernel §existence there is a probability kernel $\kappa$ from $(Y,\mathcal{B}(Y))$ to $(Z,\mathcal{B}(Z))$ with $\hat{\pi}=\nu\otimes\kappa$; by Step 2 it is a conditional kernel of $\pi$ given $q$.

Claim 2. If $\kappa$ and $\kappa'$ are conditional kernels of $\pi$ given $q$, then $\hat{\pi}=\nu\otimes\kappa=\nu\otimes\kappa'$ by Step 2, and Disintegration of a Probability Measure on a Product with a Polish Factor: Existence and Almost-Everywhere Uniqueness of the Conditional Kernel §uniqueness shows that the set of the $y\in Y$ with $\kappa_{y}\ne\kappa'_{y}$ belongs to $\mathcal{B}(Y)$ and has $\nu$-measure $0$.

Claim 3. Let $\kappa$ be a conditional kernel of $\pi$ given $q$, so that $\hat{\pi}=\nu\otimes\kappa$ by Step 2. Let $Y\times Y$ carry the product metric $d_{Y\times Y}$, a metric by claim 1 of The Product Metric is a Metric, and let $\Delta=\{(y,y):y\in Y\}$. The set $\Delta$ is closed in $Y\times Y$ for the topology of the subsets that are open in $(Y\times Y,d_{Y\times Y})$: given $(y,y')\notin\Delta$, put $r=d_{Y}(y,y')/2>0$; if $d_{Y\times Y}\bigl((y,y'),(w,w')\bigr)<r$, then $d_{Y}(y,w)<r$ and $d_{Y}(y',w')<r$ by claim 3 of The Product Metric is a Metric, so by the triangle inequality $d_{Y}(w,w')\ge d_{Y}(y,y')-d_{Y}(y,w)-d_{Y}(y',w')>0$ and $(w,w')\notin\Delta$. Thus $(Y\times Y)\setminus\Delta$ is open in $(Y\times Y,d_{Y\times Y})$ by Open Subset of a Metric Space, so $\Delta$ is closed (Closed Subset of a Topological Space) and belongs to the Borel $\sigma$-algebra $\mathcal{B}(Y\times Y)$ of $(Y\times Y,d_{Y\times Y})$ by claim 1 of Borel Measurability and Bounded Integration on a Metric Space. As $(Y,d_{Y})$ is separable, The Borel Sigma-Algebra of a Product of Two Separable Metric Spaces is the Product Sigma-Algebra §product, applied with both factors equal to $(Y,d_{Y})$, gives $\mathcal{B}(Y\times Y)=\mathcal{B}(Y)\otimes\mathcal{B}(Y)$, so $\Delta\in\mathcal{B}(Y)\otimes\mathcal{B}(Y)$.

Define $\Phi:Y\times Z\to Y\times Y$ by $\Phi(y,z)=(y,q(z))$. Exactly as for $T$ in Step 1, the family of the sets $E\subseteq Y\times Y$ with $\Phi^{-1}(E)\in\mathcal{B}(Y)\otimes\mathcal{B}(Z)$ is a $\sigma$-algebra containing every rectangle $A\times A'$ with $A,A'\in\mathcal{B}(Y)$, because $\Phi^{-1}(A\times A')=A\times q^{-1}(A')$ is a rectangle of $\mathcal{B}(Y)\otimes\mathcal{B}(Z)$; so by claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra that family contains $\mathcal{B}(Y)\otimes\mathcal{B}(Y)$. Hence

$$G=\Phi^{-1}(\Delta)=\{(y,z)\in Y\times Z:q(z)=y\}\in\mathcal{B}(Y)\otimes\mathcal{B}(Z).$$

For every $z\in Z$ we have $T(z)=(q(z),z)\in G$, so $T^{-1}(G)=Z$ and $\hat{\pi}(G)=\pi(Z)=1$. The section of $G$ at $y\in Y$ is $G_{y}=\{z:q(z)=y\}=q^{-1}(\{y\})$. Let $\varphi(y)=\kappa(y,G_{y})=\kappa_{y}\bigl(q^{-1}(\{y\})\bigr)$. By Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §sections, $\varphi$ takes values in $[0,1]$, is measurable with respect to $\mathcal{B}(Y)$ and is integrable with respect to $\nu$, and by Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §composite

$$\int_{Y}\varphi(y)\,\nu(dy)=(\nu\otimes\kappa)(G)=\hat{\pi}(G)=1 .$$

The constant function $1$ is integrable with respect to the Borel measure $\nu$ on $(Y,d_{Y})$, with $\int_{Y}1\,\nu(dy)=\nu(Y)=1$, by claim 6(a) of Borel Measurability and Bounded Integration on a Metric Space. By Linearity and Monotonicity of the Lebesgue Integral §integrable, $1-\varphi$ is integrable with $\int_{Y}\bigl(1-\varphi(y)\bigr)\,\nu(dy)=0$. The function $1-\varphi$ is measurable and takes values in $[0,1]$, so by claim 6(c) of Borel Measurability and Bounded Integration on a Metric Space its integral as a nonnegative function is also $0$, and by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §vanishing the set $E=\{y\in Y:\varphi(y)\ne1\}$ is $\nu$-null. The set $\mathbb{R}\setminus\{1\}$ is open in the real line with the absolute-value metric by Open Subset of a Metric Space, since for $t\ne1$ the open ball about $t$ of radius $|t-1|>0$ does not contain $1$; so it belongs to $\mathcal{B}(\mathbb{R})$ by claims 1 and 2 of Borel Measurability and Bounded Integration on a Metric Space; hence $E=\varphi^{-1}(\mathbb{R}\setminus\{1\})\in\mathcal{B}(Y)$, and being null it has $\nu(E)=0$ by Basic Properties of a Measure §monotone. The set of the $y$ with $\kappa_{y}(q^{-1}(\{y\}))=1$ is $Y\setminus E\in\mathcal{B}(Y)$, and $\nu(Y\setminus E)=\nu(Y)-\nu(E)=1$ by Basic Properties of a Measure §differences.

Claims 4 and 5. Let $\kappa$ be a conditional kernel of $\pi$ given $q$, so that $\hat{\pi}=\nu\otimes\kappa$ by Step 2, and let $f:Z\to\mathbb{R}$ be measurable with respect to $\mathcal{B}(Z)$. Define $F:Y\times Z\to\mathbb{R}$ by $F(y,z)=f(z)$. The projection $(y,z)\mapsto z$ is measurable with respect to $\mathcal{B}(Y)\otimes\mathcal{B}(Z)$ and $\mathcal{B}(Z)$ by claim 5 of Sections of Product-Measurable Sets and Maps Are Measurable, and Insertion Maps into Products Are Measurable, so $F$, its composite with $f$, is measurable with respect to $\mathcal{B}(Y)\otimes\mathcal{B}(Z)$ and $\mathcal{B}(\mathbb{R})$ by claim 4 of Borel Measurability and Bounded Integration on a Metric Space. For every $y\in Y$ the function $F(y,\cdot)$ is $f$, and $F\circ T=f$. By claim 2 of Image Measures, Measures with Densities, and Change of Variables, applied to $\hat{\pi}$, the image measure of $\pi$ under $T$, $F$ is integrable with respect to $\hat{\pi}$ if and only if $f$ is integrable with respect to $\pi$, and then

$$\int_{Y\times Z}F\,d\hat{\pi}=\int_{Z}f\,d\pi .\tag{2}$$

For claim 4, let $f$ be bounded; then $F$ is bounded. By Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §bounded, applied to $\nu$, $\kappa$ and $F$: for every $y$ the function $f$ is integrable with respect to $\kappa_{y}$; the function $y\mapsto\int_{Z}f\,d\kappa_{y}$ is measurable with respect to $\mathcal{B}(Y)$ and bounded; $F$ is integrable with respect to $\nu\otimes\kappa=\hat{\pi}$; that function is integrable with respect to $\nu$; and $\int_{Y\times Z}F\,d\hat{\pi}=\int_{Y}\bigl(\int_{Z}f\,d\kappa_{y}\bigr)\nu(dy)$. Since $F$ is integrable with respect to $\hat{\pi}$, $f$ is integrable with respect to $\pi$, and (2) gives the displayed identity of claim 4.

For claim 5, let $f\ge0$; then $F\ge0$, and since $F(y,\cdot)=f$, the set $N$ and the function $g$ of claim 5 are exactly the set and the function that Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §nonnegative attaches to $F$, $\nu$ and $\kappa$. By that clause, $N\in\mathcal{B}(Y)$, $g$ is measurable with respect to $\mathcal{B}(Y)$, and $F$ is integrable with respect to $\nu\otimes\kappa=\hat{\pi}$ if and only if $\nu(N)=0$ and $g$ is integrable with respect to $\nu$, in which case $\int_{Y\times Z}F\,d\hat{\pi}=\int_{Y}g(y)\,\nu(dy)$. Combining this with the equivalence and the identity (2) above, $f$ is integrable with respect to $\pi$ if and only if $\nu(N)=0$ and $g$ is integrable with respect to $\nu$, and in that case $\int_{Z}f\,d\pi=\int_{Y}g(y)\,\nu(dy)$. $\blacksquare$