Proof of Conditioning in Two Steps: Composing a Conditional Kernel Given a Map with a Conditional Kernel Given a Function of It

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 metric spaces, the maps $q$ and $s$, the measure $\pi$, the image measures $\nu_{1}=q_{\#}\pi$ and $\nu_{2}=(s\circ q)_{\#}\pi=s_{\#}\nu_{1}$, and the kernels $\kappa$ and $\lambda$ of the statement. Two facts recalled in the statement are used throughout: for every $w\in Y_{2}$, $\lambda_{w}$ is a Borel measure on $(Y_{1},d_{1})$ with $\lambda_{w}(Y_{1})=1$; and for every $B\in\mathcal{B}(Z)$, the function $\kappa(\cdot,B):Y_{1}\to\mathbb{R}$ is measurable with respect to $\mathcal{B}(Y_{1})$ and the Borel $\sigma$-algebra of the real line, with values in $[0,1]$ (definition of a probability kernel, applied to $\kappa$).

Step 1 (values in $[0,1]$). Fix $w\in Y_{2}$ and then $B\in\mathcal{B}(Z)$. By claim 6(a) of Borel Measurability and Bounded Integration on a Metric Space, applied to the metric space $(Y_{1},d_{1})$ and the Borel measure $\lambda_{w}$, the constant functions $0$ and $1$ on $Y_{1}$ are integrable with respect to $\lambda_{w}$, with integrals $0\cdot\lambda_{w}(Y_{1})=0$ and $1\cdot\lambda_{w}(Y_{1})=1$. Since $0\le\kappa(y,B)\le1$ for every $y\in Y_{1}$, the monotonicity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, applied on the measure space $(Y_{1},\mathcal{B}(Y_{1}),\lambda_{w})$, gives $0\le(\kappa\circ\lambda)(w,B)\le1$. Moreover, by claim 6(c) of Borel Measurability and Bounded Integration on a Metric Space (same data, $f=\kappa(\cdot,B)$, $M=1$), $(\kappa\circ\lambda)(w,B)$ equals the integral of the nonnegative measurable function $\kappa(\cdot,B)$ in the sense of Lebesgue Integral of a Nonnegative Measurable Function; call this fact (F).

Step 2 (each $(\kappa\circ\lambda)(w,\cdot)$ is a probability measure). Fix $w\in Y_{2}$ and write $\rho(B)=(\kappa\circ\lambda)(w,B)$ for $B\in\mathcal{B}(Z)$. By Step 1, $\rho$ maps $\mathcal{B}(Z)$ into $[0,1]\subseteq[0,\infty]$. Since each $\kappa_{y}$ is a probability measure on $(Z,\mathcal{B}(Z))$, $\kappa(y,\varnothing)=0$ and $\kappa(y,Z)=1$ for every $y\in Y_{1}$, so $\rho(\varnothing)$ and $\rho(Z)$ are the integrals of the constant functions $0$ and $1$ with respect to $\lambda_{w}$, that is, $\rho(\varnothing)=0$ and $\rho(Z)=1$ by Step 1.

For countable additivity, let $(B_{j})_{j\in\mathbb{N}}$ be a sequence of pairwise disjoint members of $\mathcal{B}(Z)$ with union $B$. For each $y\in Y_{1}$, countable additivity of the measure $\kappa_{y}$ (Measure, Measure Space, and Probability Measure) gives $\kappa(y,B)=\sum_{j}\kappa(y,B_{j})$; as this sum is the real number $\kappa(y,B)\le1$, the definition of the sum of a sequence in $[0,\infty]$ in Measure, Measure Space, and Probability Measure shows that the partial sums are bounded above and $\kappa(y,B)$ is their least upper bound. For $n\in\mathbb{N}$ let $g_{n}:Y_{1}\to\mathbb{R}$ be the partial sum $g_{n}(y)=\sum_{j\le n}\kappa(y,B_{j})$. Each $g_{n}$ is measurable with respect to $\mathcal{B}(Y_{1})$ by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, applied on $(Y_{1},\mathcal{B}(Y_{1}))$ to the measurable functions $\kappa(\cdot,B_{j})$; it is nonnegative, and $g_{n}(y)\le g_{n+1}(y)$ since $\kappa(y,B_{n+1})\ge0$; and $\sup_{n}g_{n}(y)=\kappa(y,B)$ for every $y$. By the standing convention on measurability, these real-valued measurable functions are measurable also as $[0,\infty]$-valued functions. Hence Monotone Convergence Theorem, applied on $(Y_{1},\mathcal{B}(Y_{1}),\lambda_{w})$ to the sequence $(g_{n})$, gives, for the integrals of nonnegative functions,

$$\int_{Y_{1}}\kappa(y,B)\,\lambda_{w}(dy)=\sup_{n}\int_{Y_{1}}g_{n}\,d\lambda_{w}.$$

By additivity of the nonnegative integral in claim 1 of Linearity and Monotonicity of the Lebesgue Integral (on the same measure space, by induction on $n$), $\int_{Y_{1}}g_{n}\,d\lambda_{w}=\sum_{j\le n}\int_{Y_{1}}\kappa(y,B_{j})\,\lambda_{w}(dy)$. By fact (F) of Step 1, applied with $B$ and with each $B_{j}$, the nonnegative integrals of $\kappa(\cdot,B)$ and $\kappa(\cdot,B_{j})$ are the real numbers $\rho(B)$ and $\rho(B_{j})$. So $\rho(B)$ is the least upper bound of the partial sums $\sum_{j\le n}\rho(B_{j})$, which are real and bounded above by $\rho(B)$; by the definition of the sum in Measure, Measure Space, and Probability Measure, $\sum_{j}\rho(B_{j})=\rho(B)$. Thus $\rho$ is a measure on $(Z,\mathcal{B}(Z))$ with $\rho(Z)=1$, a probability measure.

Step 3 (measurability in $w$). Fix $B\in\mathcal{B}(Z)$ and define $f:Y_{2}\times Y_{1}\to\mathbb{R}$ by $f(w,y)=\kappa(y,B)$, where $Y_{2}\times Y_{1}$ carries the product $\sigma$-algebra $\mathcal{B}(Y_{2})\otimes\mathcal{B}(Y_{1})$ of Product Sigma-Algebra. Then $f=\kappa(\cdot,B)\circ p_{1}$ with $p_{1}(w,y)=y$; $p_{1}$ is measurable with respect to $\mathcal{B}(Y_{2})\otimes\mathcal{B}(Y_{1})$ and $\mathcal{B}(Y_{1})$ by claim 5 of Sections of Product-Measurable Sets and Maps Are Measurable, and Insertion Maps into Products Are Measurable, and $\kappa(\cdot,B)$ is measurable with respect to $\mathcal{B}(Y_{1})$ and the Borel $\sigma$-algebra of the real line, so $f$ is measurable with respect to $\mathcal{B}(Y_{2})\otimes\mathcal{B}(Y_{1})$ by claim 4 of Borel Measurability and Bounded Integration on a Metric Space (with the metric space $(Y_{1},d_{1})$, $\Omega=Y_{2}\times Y_{1}$, the map $p_{1}$ in the role of $Y$, and $g=\kappa(\cdot,B)$). Also $|f|\le1$. Apply Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §bounded with $(Y_{2},\mathcal{B}(Y_{2}))$ and $(Y_{1},\mathcal{B}(Y_{1}))$ in place of $(Y,\mathcal{Y})$ and $(Z,\mathcal{Z})$, the probability kernel $\lambda$ in place of $\kappa$, the probability measure $\nu_{2}$ in place of $\mu$, and this $f$: the function $w\mapsto\int_{Y_{1}}f(w,y)\,\lambda_{w}(dy)=(\kappa\circ\lambda)(w,B)$ is measurable with respect to $\mathcal{B}(Y_{2})$. Together with Step 2, the definition of a probability kernel shows that $\kappa\circ\lambda$ is a probability kernel from $(Y_{2},\mathcal{B}(Y_{2}))$ to $(Z,\mathcal{B}(Z))$. This is the first assertion.

Step 4 (the composite measure $\nu_{2}\otimes\lambda$ is an image of $\nu_{1}$). Let $\Phi:Y_{1}\to Y_{2}\times Y_{1}$, $\Phi(y)=(s(y),y)$. The identity map of $Y_{1}$ is measurable with respect to $\mathcal{B}(Y_{1})$ and $\mathcal{B}(Y_{1})$, as the preimage of each set is the set itself (Measurable Function and Real-Valued Measurable Function), and $s$ is measurable with respect to $\mathcal{B}(Y_{1})$ and $\mathcal{B}(Y_{2})$. So Pairings into a Product, the Graph of a Measurable Map, and Couplings Concentrated on a Graph §pairing, applied with the measurable space $(Y_{2},\mathcal{B}(Y_{2}))$ in place of $(Y,\mathcal{Y})$, the metric space $(Y_{1},d_{1})$ in place of $(Z,d_{Z})$, $(\Omega,\mathcal{O})=(Y_{1},\mathcal{B}(Y_{1}))$, $F=s$ and $G$ the identity of $Y_{1}$, shows that $\Phi$ is measurable with respect to $\mathcal{B}(Y_{1})$ and $\mathcal{B}(Y_{2})\otimes\mathcal{B}(Y_{1})$. By claim 1 of Image Measures, Measures with Densities, and Change of Variables, applied to $(Y_{1},\mathcal{B}(Y_{1}),\nu_{1})$ and $T=\Phi$, the image measure $\Phi_{\#}\nu_{1}$, $E\mapsto\nu_{1}(\Phi^{-1}(E))$, is a probability measure on $(Y_{2}\times Y_{1},\mathcal{B}(Y_{2})\otimes\mathcal{B}(Y_{1}))$.

Let $\nu_{2}\otimes\lambda$ be the composite measure of Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §composite, with the same data as in Step 3. Let $A'\in\mathcal{B}(Y_{2})$ and $C\in\mathcal{B}(Y_{1})$. Then $\Phi^{-1}(A'\times C)=\{y\in Y_{1}:s(y)\in A',\ y\in C\}=C\cap s^{-1}(A')$, so

$$(\Phi_{\#}\nu_{1})(A'\times C)=\nu_{1}\bigl(C\cap s^{-1}(A')\bigr)=\int_{Y_{2}}\mathbf{1}_{A'}(w)\,\lambda(w,C)\,\nu_{2}(dw),$$

the second equality being the defining identity of a conditional kernel for $\lambda$, read with $(Y_{1},d_{1})$, $(Y_{2},d_{2})$, $s$, $\nu_{1}$ and $s_{\#}\nu_{1}=\nu_{2}$ in place of $(Z,d_{Z})$, $(Y,d_{Y})$, $q$, $\pi$ and $\nu$, and with the sets $A'$ and $C$ in place of $A$ and $B$. Therefore Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §rectangles, with the data of Step 3 and the measure $\Phi_{\#}\nu_{1}$ in the role of $\pi$, gives

$$\Phi_{\#}\nu_{1}=\nu_{2}\otimes\lambda .$$

Step 5 (the conditional-kernel identity). By the definition of a conditional kernel, read with $Y_{2}$, $s\circ q$ (measurable, as recalled in the statement), $\pi$ and $(s\circ q)_{\#}\pi=\nu_{2}$ in place of $Y$, $q$, $\pi$ and $\nu$, and given Step 3, it remains to show that for all $A\in\mathcal{B}(Y_{2})$ and $B\in\mathcal{B}(Z)$

$$\pi\bigl(B\cap(s\circ q)^{-1}(A)\bigr)=\int_{Y_{2}}\mathbf{1}_{A}(w)\,(\kappa\circ\lambda)(w,B)\,\nu_{2}(dw).$$

Fix $A$ and then $B$. Define $h:Y_{2}\times Y_{1}\to\mathbb{R}$ by $h(w,y)=\mathbf{1}_{A}(w)\,\kappa(y,B)$. The indicator $\mathbf{1}_{A}$ is measurable on $(Y_{2},\mathcal{B}(Y_{2}))$ by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and the projection $p_{2}(w,y)=w$ is measurable with respect to $\mathcal{B}(Y_{2})\otimes\mathcal{B}(Y_{1})$ and $\mathcal{B}(Y_{2})$ by claim 5 of Sections of Product-Measurable Sets and Maps Are Measurable, and Insertion Maps into Products Are Measurable, so $\mathbf{1}_{A}\circ p_{2}$ is measurable by claim 4 of Borel Measurability and Bounded Integration on a Metric Space (with the metric space $(Y_{2},d_{2})$). The function $(w,y)\mapsto\kappa(y,B)$ is the measurable $f$ of Step 3. Hence $h$, their pointwise product, is measurable with respect to $\mathcal{B}(Y_{2})\otimes\mathcal{B}(Y_{1})$ by claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and $|h|\le1$.

For each $w\in Y_{2}$, $h(w,\cdot)=\mathbf{1}_{A}(w)\,\kappa(\cdot,B)$ is a real multiple of a function integrable with respect to $\lambda_{w}$, so by claim 2 of Linearity and Monotonicity of the Lebesgue Integral (with $a=\mathbf{1}_{A}(w)$ and $b=0$) $\int_{Y_{1}}h(w,y)\,\lambda_{w}(dy)=\mathbf{1}_{A}(w)\,(\kappa\circ\lambda)(w,B)$. Thus Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §bounded, with the data of Step 3 and $h$ in place of $f$, shows that $h$ is integrable with respect to $\nu_{2}\otimes\lambda$ and

$$\int_{Y_{2}\times Y_{1}}h\,d(\nu_{2}\otimes\lambda)=\int_{Y_{2}}\mathbf{1}_{A}(w)\,(\kappa\circ\lambda)(w,B)\,\nu_{2}(dw).$$

By Step 4 the left-hand side is $\int h\,d(\Phi_{\#}\nu_{1})$. By claim 2 of Image Measures, Measures with Densities, and Change of Variables, applied to $(Y_{1},\mathcal{B}(Y_{1}),\nu_{1})$, $T=\Phi$ and the measurable real function $g=h$, which is integrable with respect to $\Phi_{\#}\nu_{1}$, the function $h\circ\Phi$ is integrable with respect to $\nu_{1}$ and

$$\int_{Y_{2}\times Y_{1}}h\,d(\Phi_{\#}\nu_{1})=\int_{Y_{1}}h(\Phi(y))\,\nu_{1}(dy).$$

For $y\in Y_{1}$, $h(\Phi(y))=\mathbf{1}_{A}(s(y))\,\kappa(y,B)=\mathbf{1}_{s^{-1}(A)}(y)\,\kappa(y,B)$, since $s(y)\in A$ exactly when $y\in s^{-1}(A)$. The set $s^{-1}(A)$ belongs to $\mathcal{B}(Y_{1})$ by measurability of $s$, so the defining identity of a conditional kernel for $\kappa$ (given $q$, with image measure $q_{\#}\pi=\nu_{1}$), applied with the sets $s^{-1}(A)\in\mathcal{B}(Y_{1})$ and $B$, gives

$$\int_{Y_{1}}\mathbf{1}_{s^{-1}(A)}(y)\,\kappa(y,B)\,\nu_{1}(dy)=\pi\bigl(B\cap q^{-1}(s^{-1}(A))\bigr)=\pi\bigl(B\cap(s\circ q)^{-1}(A)\bigr),$$

the last equality because $q^{-1}(s^{-1}(A))=(s\circ q)^{-1}(A)$. Chaining the last three displays with Step 4 proves the required identity, so $\kappa\circ\lambda$ is a conditional kernel of $\pi$ given $s\circ q$. $\square$