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

In the setting of Measure Spaces and the Lebesgue Integral: Standing Notation, let $(Z,d)$ be a complete and separable metric space and $(Y,d_{Y})$ a separable metric space, with Borel $\sigma$-algebras $\mathcal{B}(Z)$ and $\mathcal{B}(Y)$; let $q:Z\to Y$ be measurable with respect to $\mathcal{B}(Z)$ and $\mathcal{B}(Y)$; let $\pi$ be a Borel measure on $(Z,d)$ with $\pi(Z)=1$, with image measure $\nu=q_{\#}\pi$; and let conditional kernels of $\pi$ given $q$ be those of The Conditional Kernel of a Probability Measure Given a Measurable Map §conditional-kernel, with $\kappa_{y}=\kappa(y,\cdot)$. For $y\in Y$ the singleton $\{y\}$ is closed in $(Y,d_{Y})$, since for $w\ne y$ the open ball about $w$ of radius $d_{Y}(w,y)$ misses $y$, hence belongs to $\mathcal{B}(Y)$ by claim 1 of Borel Measurability and Bounded Integration on a Metric Space, so the fibre $q^{-1}(\{y\})$ belongs to $\mathcal{B}(Z)$. For a measure $\lambda$ and a function $h$ integrable with respect to it we also write $\int h(t)\,\lambda(dt)$ for $\int h\,d\lambda$.

1. (Existence)¶ There is a conditional kernel of $\pi$ given $q$.

2. (Uniqueness)¶ If $\kappa$ and $\kappa'$ are conditional kernels of $\pi$ given $q$, the set of the $y\in Y$ with $\kappa_{y}\ne\kappa'_{y}$ belongs to $\mathcal{B}(Y)$ and has $\nu$-measure $0$.

3. (Fibres)¶ If $\kappa$ is a conditional kernel of $\pi$ given $q$, the set of the $y\in Y$ with $\kappa_{y}(q^{-1}(\{y\}))=1$ belongs to $\mathcal{B}(Y)$ and has $\nu$-measure $1$.

4. (Bounded functions)¶ If $\kappa$ is a conditional kernel of $\pi$ given $q$ and $f:Z\to\mathbb{R}$ is measurable with respect to $\mathcal{B}(Z)$ and bounded, then $f$ is integrable with respect to $\pi$ and to every $\kappa_{y}$, the function $y\mapsto\int_{Z}f\,d\kappa_{y}$ is measurable with respect to $\mathcal{B}(Y)$, bounded and integrable with respect to $\nu$, and

$$\int_{Z}f\,d\pi=\int_{Y}\Bigl(\int_{Z}f\,d\kappa_{y}\Bigr)\nu(dy).$$

5. (Nonnegative functions)¶ Let $\kappa$ be a conditional kernel of $\pi$ given $q$ and let $f:Z\to\mathbb{R}$ be measurable with respect to $\mathcal{B}(Z)$, with $f\ge0$. Then the set $N$ of the $y\in Y$ for which $f$ is not integrable with respect to $\kappa_{y}$ belongs to $\mathcal{B}(Y)$, the function $g:Y\to\mathbb{R}$ equal to $\int_{Z}f\,d\kappa_{y}$ for $y\notin N$ and to $0$ for $y\in N$ is measurable with respect to $\mathcal{B}(Y)$, and $f$ is integrable with respect to $\pi$ if and only if $\nu(N)=0$ and $g$ is integrable with respect to $\nu$; in that case $\int_{Z}f\,d\pi=\int_{Y}g\,d\nu$.