Proof of Disintegration of a Probability Measure on a Product with a Polish Factor: Existence and Almost-Everywhere Uniqueness of the Conditional Kernel

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

Throughout, $\mathcal{T}_{d}$ is the collection of the subsets of $Z$ that are open in $(Z,d)$, a topology on $Z$ by Metric Open Sets Form a Topology, and "open", "closed" and "compact" for subsets of $Z$ refer to $(Z,\mathcal{T}_{d})$. We write $\mathbb{Q}_{>0}$ for the set of positive rational numbers, $B_{d}(q,s)$ for the open ball (open ball) and $\bar{B}_{d}(q,s)=\{z\in Z:d(q,z)\le s\}$ for the closed ball. For $A\in\mathcal{Y}$ and $B\in\mathcal{B}(Z)$ the rectangle $A\times B$ belongs to $\mathcal{Y}\otimes\mathcal{B}(Z)$ by Product Sigma-Algebra. Since $\pi(Y\times Z)=1$, every value of $\pi$ lies in $[0,1]$ by Basic Properties of a Measure §monotone; the same holds for every probability measure used below. "Measurable" for a real-valued function on $Y$ means measurable with respect to $\mathcal{Y}$ and $\mathcal{B}(\mathbb{R})$, and "a.e." and "null" refer to $\mu$.

Part 1: existence.

Step 0 (the second marginal). The projection $(y,z)\mapsto z$ 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, so by claim 1 of Image Measures, Measures with Densities, and Change of Variables the formula $\rho(B)=\pi(Y\times B)$, for $B\in\mathcal{B}(Z)$, defines a measure $\rho$ on $(Z,\mathcal{B}(Z))$ with $\rho(Z)=1$; it is a Borel measure on $(Z,d)$. Moreover $Z\neq\varnothing$, since otherwise $Y\times Z=\varnothing$ and $\pi(Y\times Z)=0$.

Step 1 (an increasing sequence of compact sets). For each $m\in\mathbb{N}$, Ulam's Theorem: a Finite Borel Measure on a Complete Separable Metric Space is Tight §tight, applied to the finite Borel measure $\rho$ on the complete separable metric space $(Z,d)$ with $\varepsilon=1/m$, shows that the set $\mathcal{A}_{m}$ of the compact sets $L\subseteq Z$ with $\rho(Z\setminus L)\le 1/m$ is nonempty. By Axiom of Countable Choice, applied to the family $(\mathcal{A}_{m})_{m\in\mathbb{N}}$ of nonempty subsets of the set of all subsets of $Z$, there is a sequence $(L_{m})_{m\in\mathbb{N}}$ with $L_{m}\in\mathcal{A}_{m}$ for every $m$; that is, each $L_{m}$ is compact and $\rho(Z\setminus L_{m})\le 1/m$. Put $K_{m}=L_{1}\cup\dots\cup L_{m}$. Each $K_{m}$ is compact: if $(O_{i})_{i\in I}$ is a family of open sets with $K_{m}\subseteq\bigcup_{i\in I}O_{i}$, then for each $j\le m$ the criterion Compact Subset Criterion via Open Covers in the Ambient Space gives a finite $J_{j}\subseteq I$ with $L_{j}\subseteq\bigcup_{i\in J_{j}}O_{i}$; the set $J=J_{1}\cup\dots\cup J_{m}$ is finite by claim 3 of Peeling an Element off a Finite Set, and Unions of Finite Sets and induction, and $K_{m}\subseteq\bigcup_{i\in J}O_{i}$, so $K_{m}$ is compact by the same criterion. By Compact Subsets of a Metric Space are Closed and Borel §closed and Compact Subsets of a Metric Space are Closed and Borel §borel each $K_{m}$ is closed and belongs to $\mathcal{B}(Z)$. Clearly $K_{m}\subseteq K_{m+1}$, and $\rho(Z\setminus K_{m})\le\rho(Z\setminus L_{m})\le 1/m$ by Basic Properties of a Measure §monotone. Let $K_{\infty}=\bigcup_{m}K_{m}\in\mathcal{B}(Z)$. For every $m$ we have $\rho(Z\setminus K_{\infty})\le\rho(Z\setminus K_{m})\le 1/m$, so $\rho(Z\setminus K_{\infty})=0$, since a positive real number exceeds some $1/m$ by claim 3 of The Archimedean Property of the Real Numbers. Consequently, by Basic Properties of a Measure §monotone,

$$\pi\bigl(A\times(Z\setminus K_{\infty})\bigr)\le\pi\bigl(Y\times(Z\setminus K_{\infty})\bigr)=\rho(Z\setminus K_{\infty})=0\qquad(A\in\mathcal{Y}).\tag{1}$$

Step 2 (a countable algebra generating $\mathcal{B}(Z)$). Since $(Z,d)$ is separable, fix a countable set $D\subseteq Z$ that is dense in $(Z,\mathcal{T}_{d})$. It is nonempty: pick $z\in Z$ (Step 0); then $z$ lies in the closure of $D$, which is $Z$, and $Z$ is an open set containing $z$, so $Z\cap D\ne\varnothing$ by Closure of a Subset of a Topological Space. Hence $\bigl((Z,d),D\bigr)$ is a separable metric datum in the sense of Borel Sets and Measurable Maps in a Separable Metric Space; let $\mathcal{E}$ be the family of the open balls $B_{d}(q,s)$ with $q\in D$ and $s\in\mathbb{Q}_{>0}$, which is countable by claim 1 of that lemma. Let $\mathcal{G}_{\mathrm{cl}}$ be the family of the closed balls $\bar{B}_{d}(q,s)$ with $q\in D$ and $s\in\mathbb{Q}_{>0}$. The set $\mathbb{Q}_{>0}$ is countable by claim 2 of The Integers and the Rational Numbers are Countable and claim 3 of Basic Properties of Countable Sets, so $D\times\mathbb{Q}_{>0}$ is countable by claim 1 of Products and Powers of Countable Sets, and $\mathcal{G}_{\mathrm{cl}}$, its image under $(q,s)\mapsto\bar{B}_{d}(q,s)$, is countable by claim 4 of Basic Properties of Countable Sets. The family $\mathcal{G}_{\mathrm{cp}}=\{K_{m}:m\in\mathbb{N}\}$, the image of $\mathbb{N}$ under $m\mapsto K_{m}$, is countable by claims 1 and 4 of Basic Properties of Countable Sets. Let $\mathcal{G}=\mathcal{E}\cup\mathcal{G}_{\mathrm{cl}}\cup\mathcal{G}_{\mathrm{cp}}$. Putting $\mathcal{X}_{1}=\mathcal{E}$, $\mathcal{X}_{2}=\mathcal{G}_{\mathrm{cl}}$ and $\mathcal{X}_{i}=\mathcal{G}_{\mathrm{cp}}$ for $i\ge3$, we obtain a family $(\mathcal{X}_{i})_{i\in\mathbb{N}}$ of countable subsets of the set of all subsets of $Z$ whose union is $\mathcal{G}$, so $\mathcal{G}$ is countable by A Countable Union of Countable Sets is Countable. It is nonempty, as $K_{1}\in\mathcal{G}$, so by Countable Set there is a sequence $(G_{i})_{i\in\mathbb{N}}$ whose set of terms is $\mathcal{G}$.

For a set $G\subseteq Z$ write $G^{1}=G$ and $G^{0}=Z\setminus G$. For $n\in\mathbb{N}$ and $\epsilon=(\epsilon_{1},\dots,\epsilon_{n})\in\{0,1\}^{n}$ let $P_{\epsilon}=G_{1}^{\epsilon_{1}}\cap\dots\cap G_{n}^{\epsilon_{n}}$. For $z\in Z$ let $\epsilon^{n}(z)\in\{0,1\}^{n}$ have $i$-th entry $1$ exactly when $z\in G_{i}$; then $z\in P_{\epsilon}$ if and only if $\epsilon=\epsilon^{n}(z)$, so the sets $P_{\epsilon}$, $\epsilon\in\{0,1\}^{n}$, are pairwise disjoint and cover $Z$. Let $\mathcal{U}_{n}$ be the family of the sets $\bigcup_{\epsilon\in S}P_{\epsilon}$ with $S\subseteq\{0,1\}^{n}$ (the empty union being $\varnothing$). It is finite, being the image of the finite set of subsets of $\{0,1\}^{n}$. It contains $Z$ (take $S=\{0,1\}^{n}$); the complement of $\bigcup_{\epsilon\in S}P_{\epsilon}$ is $\bigcup_{\epsilon\notin S}P_{\epsilon}$, and the union of the members given by $S$ and $S'$ is the member given by $S\cup S'$. Since $P_{\epsilon}=P_{(\epsilon,1)}\cup P_{(\epsilon,0)}$ with $(\epsilon,1),(\epsilon,0)\in\{0,1\}^{n+1}$, we have $\mathcal{U}_{n}\subseteq\mathcal{U}_{n+1}$. Let $\mathcal{U}=\bigcup_{n}\mathcal{U}_{n}$; it is countable by claim 2 of Basic Properties of Countable Sets and A Countable Union of Countable Sets is Countable. It contains $Z$, so it is nonempty, and by Countable Set we fix a sequence $(U_{j})_{j\in\mathbb{N}}$ whose set of terms is $\mathcal{U}$; for $B\in\mathcal{U}$ let $j_{B}$ be the least $j\in\mathbb{N}$ with $U_{j}=B$. The family $\mathcal{U}$ is closed under complements, and if $B\in\mathcal{U}_{n}$ and $B'\in\mathcal{U}_{n'}$ then both lie in $\mathcal{U}_{\max(n,n')}$, so $B\cup B'\in\mathcal{U}$. Hence $\mathcal{U}$ is also closed under $B\cap B'=Z\setminus\bigl((Z\setminus B)\cup(Z\setminus B')\bigr)$ and under $B'\setminus B=B'\cap(Z\setminus B)$; in particular $\mathcal{U}$ is a $\pi$-system in the sense of Dynkin's Pi-Lambda Theorem, and $\varnothing\in\mathcal{U}$. Each $G_{i}$ is the union of the $P_{\epsilon}$ with $\epsilon\in\{0,1\}^{i}$ and $\epsilon_{i}=1$, so $\mathcal{G}\subseteq\mathcal{U}$.

We show that the $\sigma$-algebra generated by $\mathcal{U}$, denoted $\sigma(\mathcal{U})$, is $\mathcal{B}(Z)$. Open balls are open by Open Ball in a Metric Space is Open, closed balls are closed by claim 3 of Elementary Properties of the Closed Ball in a Metric Space, and open and closed sets belong to $\mathcal{B}(Z)$ by claim 1 of Borel Measurability and Bounded Integration on a Metric Space; with Step 1, $\mathcal{G}\subseteq\mathcal{B}(Z)$. As $\mathcal{B}(Z)$ is a $\sigma$-algebra, closed under complements and finite unions and intersections (Sigma-Algebra and Measurable Space), every $P_{\epsilon}$ and every member of $\mathcal{U}$ belongs to $\mathcal{B}(Z)$, so $\sigma(\mathcal{U})\subseteq\mathcal{B}(Z)$ by claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra. Conversely $\mathcal{E}\subseteq\mathcal{U}\subseteq\sigma(\mathcal{U})$, so by the same claim and claim 2 of Borel Sets and Measurable Maps in a Separable Metric Space, $\mathcal{B}(Z)=\sigma(\mathcal{E})\subseteq\sigma(\mathcal{U})$.

Step 3 (inner approximation by closed and by compact members of $\mathcal{U}$). Call a set $B\subseteq Z$ closed-approximable if there is a sequence $(C_{k})_{k\in\mathbb{N}}$ of closed members of $\mathcal{U}$ with $C_{k}\subseteq C_{k+1}$ for every $k$ and $\bigcup_{k}C_{k}=B$. We show that every member of $\mathcal{U}$ is closed-approximable.

(a) Let $\mathcal{G}^{\pm}$ consist of the members of $\mathcal{G}$ and their complements; all lie in $\mathcal{U}$. Each is closed or open: the sets $\bar{B}_{d}(q,s)$ and $K_{m}$ are closed, $B_{d}(q,s)$ is open, $Z\setminus B_{d}(q,s)$ is closed and $Z\setminus\bar{B}_{d}(q,s)$, $Z\setminus K_{m}$ are open, by Closed Subset of a Topological Space. A closed member $B$ is closed-approximable by the constant sequence $C_{k}=B$. Let $V\in\mathcal{U}$ be open and let $\mathcal{K}_{V}$ be the family of the members of $\mathcal{G}_{\mathrm{cl}}$ contained in $V$. We claim $\bigcup\mathcal{K}_{V}=V$. One inclusion is clear. Let $z\in V$. By claim 1 of Borel Sets and Measurable Maps in a Separable Metric Space there are $p\in D$ and $t\in\mathbb{Q}_{>0}$ with $z\in B_{d}(p,t)\subseteq V$. As $t-d(p,z)>0$, claim 2 of The Archimedean Property of the Real Numbers gives $n\in\mathbb{N}$ with $t<n\bigl(t-d(p,z)\bigr)$; then $n\ge2$ (for $n=1$ this would say $d(p,z)<0$), and $d(p,z)<t(n-1)/n=:t'$. Now $t'\in\mathbb{Q}_{>0}$ and $t'<t$, so $\bar{B}_{d}(p,t')\subseteq B_{d}(p,t)\subseteq V$; thus $\bar{B}_{d}(p,t')\in\mathcal{K}_{V}$ and $z\in\bar{B}_{d}(p,t')$. If $V=\varnothing$, take $C_{k}=\varnothing$. Otherwise $\mathcal{K}_{V}$ is nonempty and countable (claim 3 of Basic Properties of Countable Sets), so there is a sequence $(F_{j})_{j\in\mathbb{N}}$ whose set of terms is $\mathcal{K}_{V}$, and $C_{k}=F_{1}\cup\dots\cup F_{k}$ is closed by claim 2 of Complements, Unions and Intersections of Closed Sets in a Topological Space, belongs to $\mathcal{U}$, increases with $k$ and has union $V$. So every member of $\mathcal{G}^{\pm}$ is closed-approximable.

(b) If $B,B'$ are closed-approximable through $(C_{k})$ and $(C'_{k})$, then $B\cup B'$ and $B\cap B'$ are closed-approximable through $(C_{k}\cup C'_{k})$ and $(C_{k}\cap C'_{k})$: these are closed members of $\mathcal{U}$ by claims 2 and 3 of Complements, Unions and Intersections of Closed Sets in a Topological Space, they increase with $k$, every point of $B\cup B'$ lies in some $C_{k}$ or some $C'_{k}$, and a point of $B\cap B'$ lying in $C_{k}$ and in $C'_{k'}$ lies in $C_{k''}\cap C'_{k''}$ for $k''=\max(k,k')$; the reverse inclusions are clear.

(c) The empty set is closed-approximable by the constant sequence $\varnothing$, which is closed by claim 1 of Complements, Unions and Intersections of Closed Sets in a Topological Space. By (a), (b) and induction, every $P_{\epsilon}$, being a finite intersection of members of $\mathcal{G}^{\pm}$, and every finite union of such sets is closed-approximable; that is, every member of $\mathcal{U}$ is.

For $j\in\mathbb{N}$ let $\mathcal{C}_{j}$ be the set of the sequences $(C_{k})_{k\in\mathbb{N}}$ of closed members of $\mathcal{U}$ with $C_{k}\subseteq C_{k+1}$ for every $k$ and $\bigcup_{k}C_{k}=U_{j}$; it is nonempty, as $U_{j}$ is closed-approximable. By Axiom of Countable Choice, applied to the family $(\mathcal{C}_{j})_{j\in\mathbb{N}}$ of nonempty subsets of the set of all sequences in $\mathcal{U}$, there is a sequence $(c_{j})_{j\in\mathbb{N}}$ with $c_{j}\in\mathcal{C}_{j}$ for every $j$. For $B\in\mathcal{U}$ let $(C_{B,k})_{k\in\mathbb{N}}$ be the sequence $c_{j_{B}}$, which belongs to $\mathcal{C}_{j_{B}}$ with $U_{j_{B}}=B$, and put $E_{B,k}=C_{B,k}\cap K_{k}$. Then $E_{B,k}\in\mathcal{U}$, $E_{B,k}$ is compact by claim 1 of Compactness of Intersections with Closed Sets and of Level Sets of Semicontinuous Functions, $E_{B,k}\subseteq E_{B,k+1}$, $E_{B,k}\subseteq B$, and $\bigcup_{k}E_{B,k}=B\cap K_{\infty}$: a point of $B\cap K_{\infty}$ lies in some $C_{B,k}$ and some $K_{k'}$, hence in $E_{B,k''}$ for $k''=\max(k,k')$.

Step 4 (densities with values in $[0,1]$). For $B\in\mathcal{B}(Z)$ define $\lambda_{B}(A)=\pi(A\times B)$ for $A\in\mathcal{Y}$. Then $\lambda_{B}(\varnothing)=0$, and for pairwise disjoint $A_{m}\in\mathcal{Y}$ the rectangles $A_{m}\times B$ are pairwise disjoint with union $\bigl(\bigcup_{m}A_{m}\bigr)\times B$, so $\lambda_{B}$ is a measure on $(Y,\mathcal{Y})$. As $A\times B\subseteq A\times Z$, Basic Properties of a Measure §monotone gives $\lambda_{B}(A)\le\mu(A)$; so $\lambda_{B}$ is finite and $\lambda_{B}(A)=0$ whenever $\mu(A)=0$. The measure $\mu$ is $\sigma$-finite, as $\mu(Y)=1$ (Measure, Measure Space, and Probability Measure).

Let $B\in\mathcal{U}$. By The Radon-Nikodym Theorem for a Finite Measure and a Sigma-Finite Measure, and Uniqueness of Densities §existence there is a measurable $\tilde{h}_{B}:Y\to\mathbb{R}$ with $\tilde{h}_{B}\ge0$ and $\lambda_{B}(A)=\int_{Y}\mathbf{1}_{A}\tilde{h}_{B}\,d\mu$ for every $A\in\mathcal{Y}$. For $n\in\mathbb{N}$ let $t_{n}=1+1/n$ and $A_{n}=\{y\in Y:t_{n}\le\tilde{h}_{B}(y)\}$, which belongs to $\mathcal{Y}$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §markov. The function $\mathbf{1}_{A_{n}}\tilde{h}_{B}$ is measurable by claims 1 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, it is nonnegative, and the set where it is at least $t_{n}$ is $A_{n}$, since $t_{n}>0$. By The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §markov and Basic Properties of a Measure §monotone,

$$t_{n}\,\mu(A_{n})\le\int_{Y}\mathbf{1}_{A_{n}}\tilde{h}_{B}\,d\mu=\lambda_{B}(A_{n})\le\mu(A_{n}),$$

and as $\mu(A_{n})\le1$ is real this gives $\mu(A_{n})/n\le0$, so $\mu(A_{n})=0$. If $\tilde{h}_{B}(y)>1$, claim 3 of The Archimedean Property of the Real Numbers gives $n$ with $1/n<\tilde{h}_{B}(y)-1$, so $y\in A_{n}$; hence $\{\tilde{h}_{B}>1\}$ is contained in the null set $\bigcup_{n}A_{n}$ (The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-union). Let $h_{B}=\min(\tilde{h}_{B},1)$, measurable by claims 1 and 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, with values in $[0,1]$ and equal to $\tilde{h}_{B}$ off $\bigcup_{n}A_{n}$. For $A\in\mathcal{Y}$ the nonnegative measurable functions $\mathbf{1}_{A}h_{B}$ and $\mathbf{1}_{A}\tilde{h}_{B}$ agree a.e., so by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison

$$\pi(A\times B)=\lambda_{B}(A)=\int_{Y}\mathbf{1}_{A}h_{B}\,d\mu\qquad(A\in\mathcal{Y},\ B\in\mathcal{U}),\tag{2}$$

that is, $h_{B}$ is a density of $\lambda_{B}$ with respect to $\mu$. Consequently, for every $j\in\mathbb{N}$ the set $\mathcal{H}_{j}$ of the measurable functions $h:Y\to[0,1]$ with $\pi(A\times U_{j})=\int_{Y}\mathbf{1}_{A}h\,d\mu$ for every $A\in\mathcal{Y}$ is nonempty. By Axiom of Countable Choice, applied to the family $(\mathcal{H}_{j})_{j\in\mathbb{N}}$ of nonempty subsets of the set of all functions from $Y$ to $\mathbb{R}$, there is a sequence $(f_{j})_{j\in\mathbb{N}}$ with $f_{j}\in\mathcal{H}_{j}$ for every $j$. From now on, for $B\in\mathcal{U}$, $h_{B}$ denotes $f_{j_{B}}$; since $U_{j_{B}}=B$, it is a measurable function $Y\to[0,1]$ satisfying (2).

Step 5 (countably many identities almost everywhere).

(5.1) The constant function $1$ is measurable by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and $\int_{Y}\mathbf{1}_{A}\cdot1\,d\mu=\mu(A)=\pi(A\times Z)=\lambda_{Z}(A)$ by The Integral of an Indicator Function is the Measure of the Set. So $1$ and $h_{Z}$ are densities of $\lambda_{Z}$, and by The Radon-Nikodym Theorem for a Finite Measure and a Sigma-Finite Measure, and Uniqueness of Densities §uniqueness the set $N_{Z}=\{y:h_{Z}(y)\ne1\}$ belongs to $\mathcal{Y}$ and $\mu(N_{Z})=0$.

(5.2) Let $\mathcal{P}$ be the set of pairs $(B,B')\in\mathcal{U}\times\mathcal{U}$ with $B\cap B'=\varnothing$; it is countable by claim 1 of Products and Powers of Countable Sets and claim 3 of Basic Properties of Countable Sets. For $(B,B')\in\mathcal{P}$ we have $B\cup B'\in\mathcal{U}$, the function $h_{B}+h_{B'}$ is measurable (claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions) and nonnegative, and for $A\in\mathcal{Y}$, by Linearity and Monotonicity of the Lebesgue Integral §nonnegative, (2) and Basic Properties of a Measure §additivity applied to the disjoint sets $A\times B$ and $A\times B'$,

$$\int_{Y}\mathbf{1}_{A}(h_{B}+h_{B'})\,d\mu=\int_{Y}\mathbf{1}_{A}h_{B}\,d\mu+\int_{Y}\mathbf{1}_{A}h_{B'}\,d\mu=\pi(A\times B)+\pi(A\times B')=\pi\bigl(A\times(B\cup B')\bigr).$$

So $h_{B}+h_{B'}$ and $h_{B\cup B'}$ are densities of $\lambda_{B\cup B'}$, and by The Radon-Nikodym Theorem for a Finite Measure and a Sigma-Finite Measure, and Uniqueness of Densities §uniqueness the set $N_{B,B'}=\{y:h_{B\cup B'}(y)\ne h_{B}(y)+h_{B'}(y)\}$ belongs to $\mathcal{Y}$ and has $\mu(N_{B,B'})=0$. The set $\mathcal{P}$ is nonempty, as $(\varnothing,\varnothing)\in\mathcal{P}$, so by Countable Set there is a sequence $(p_{j})_{j\in\mathbb{N}}$ whose set of terms is $\mathcal{P}$. Write $N_{p}=N_{B,B'}$ for $p=(B,B')\in\mathcal{P}$, and let $N_{\mathrm{add}}=\bigcup_{j\in\mathbb{N}}N_{p_{j}}$, which is the union of the $N_{B,B'}$ over $(B,B')\in\mathcal{P}$. Each $N_{p_{j}}$ belongs to $\mathcal{Y}$, so $N_{\mathrm{add}}\in\mathcal{Y}$ because $\mathcal{Y}$ is a $\sigma$-algebra (Sigma-Algebra and Measurable Space), and $\mu(N_{\mathrm{add}})\le\sum_{j\in\mathbb{N}}\mu(N_{p_{j}})=0$ by Basic Properties of a Measure §subadditivity; so $\mu(N_{\mathrm{add}})=0$, and in particular $N_{\mathrm{add}}$ is null. If $y\notin N_{\mathrm{add}}$ and $B\subseteq B'$ are members of $\mathcal{U}$, then $(B,B'\setminus B)\in\mathcal{P}$ with union $B'$, so

$$h_{B'}(y)=h_{B}(y)+h_{B'\setminus B}(y)\ge h_{B}(y).\tag{3}$$

(5.3) Let $B\in\mathcal{U}$. Let $g_{B}(y)=\sup_{k}h_{E_{B,k}}(y)$; it takes values in $[0,1]$ and is measurable by claim 1 of Measurability of Countable Suprema, Bounded Pointwise Limits, Monotone Functions, and Continuous Functions, applied with the bound $1$. For $k\in\mathbb{N}$ let $s_{B,k}=\max(h_{E_{B,1}},\dots,h_{E_{B,k}})$, measurable by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions and induction on $k$. At every $y$ the sequence $(s_{B,k}(y))_{k}$ is nondecreasing with supremum $g_{B}(y)$, and for $y\notin N_{\mathrm{add}}$, (3) applied to $E_{B,i}\subseteq E_{B,k}$ for $i\le k$ gives $s_{B,k}(y)=h_{E_{B,k}}(y)$. Let $A\in\mathcal{Y}$. The measurable functions $\mathbf{1}_{A}s_{B,k}$ (claims 1 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions) are nondecreasing in $k$ at every point, with supremum $\mathbf{1}_{A}g_{B}$, so by Monotone Convergence Theorem

$$\int_{Y}\mathbf{1}_{A}g_{B}\,d\mu=\sup_{k}\int_{Y}\mathbf{1}_{A}s_{B,k}\,d\mu=\sup_{k}\int_{Y}\mathbf{1}_{A}h_{E_{B,k}}\,d\mu=\sup_{k}\pi(A\times E_{B,k}),$$

where the second equality is The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison, the two integrands agreeing off the null set $N_{\mathrm{add}}$, and the third is (2). The sets $A\times E_{B,k}$ increase with $k$ and have union $A\times(B\cap K_{\infty})$ by Step 3, so by Basic Properties of a Measure §continuity-below the last supremum is $\pi\bigl(A\times(B\cap K_{\infty})\bigr)$. By Basic Properties of a Measure §additivity, Basic Properties of a Measure §monotone and (1), $\pi(A\times B)=\pi\bigl(A\times(B\cap K_{\infty})\bigr)+\pi\bigl(A\times(B\setminus K_{\infty})\bigr)=\pi\bigl(A\times(B\cap K_{\infty})\bigr)$. Hence $\int_{Y}\mathbf{1}_{A}g_{B}\,d\mu=\lambda_{B}(A)$ for every $A\in\mathcal{Y}$, so $g_{B}$ and $h_{B}$ are densities of $\lambda_{B}$, and by The Radon-Nikodym Theorem for a Finite Measure and a Sigma-Finite Measure, and Uniqueness of Densities §uniqueness the set $N^{B}=\{y:g_{B}(y)\ne h_{B}(y)\}$ belongs to $\mathcal{Y}$ and has $\mu(N^{B})=0$.

Let $N_{0}=N_{Z}\cup N_{\mathrm{add}}\cup\bigcup_{j\in\mathbb{N}}N^{U_{j}}$; since the terms of $(U_{j})$ are exactly the members of $\mathcal{U}$, $N^{B}\subseteq N_{0}$ for every $B\in\mathcal{U}$. The set $N_{0}$ is the union of the sequence $N_{Z},N_{\mathrm{add}},N^{U_{1}},N^{U_{2}},\dots$ of members of $\mathcal{Y}$ of $\mu$-measure $0$, so $N_{0}\in\mathcal{Y}$ because $\mathcal{Y}$ is a $\sigma$-algebra (Sigma-Algebra and Measurable Space), and $\mu(N_{0})=0$ by Basic Properties of a Measure §subadditivity.

Step 6 (a premeasure on $\mathcal{U}$ for each $y\notin N_{0}$). Fix $y\in Y\setminus N_{0}$ and let $\tau_{y}(B)=h_{B}(y)\in[0,1]$ for $B\in\mathcal{U}$.

(6.1) $\tau_{y}(Z)=1$, as $y\notin N_{Z}$.

(6.2) $\tau_{y}(B\cup B')=\tau_{y}(B)+\tau_{y}(B')$ for disjoint $B,B'\in\mathcal{U}$, as $y\notin N_{\mathrm{add}}$. Taking $B=B'=\varnothing$ gives $\tau_{y}(\varnothing)=2\tau_{y}(\varnothing)$, so $\tau_{y}(\varnothing)=0$.

(6.3) If $B\subseteq B'$ in $\mathcal{U}$, then $\tau_{y}(B'\setminus B)=\tau_{y}(B')-\tau_{y}(B)$ and $\tau_{y}(B)\le \tau_{y}(B')$, by (6.2).

(6.4) For $B_{1},\dots,B_{r}\in\mathcal{U}$, $\tau_{y}(B_{1}\cup\dots\cup B_{r})\le \tau_{y}(B_{1})+\dots+\tau_{y}(B_{r})$: for $r=2$, $B_{1}\cup B_{2}$ is the disjoint union of $B_{1}$ and $B_{2}\setminus B_{1}$, so (6.2) and (6.3) give $\tau_{y}(B_{1}\cup B_{2})=\tau_{y}(B_{1})+\tau_{y}(B_{2}\setminus B_{1})\le \tau_{y}(B_{1})+\tau_{y}(B_{2})$; the general case follows by induction on $r$.

(6.5) For $B\in\mathcal{U}$, $\tau_{y}(B)=h_{B}(y)=g_{B}(y)=\sup_{k}\tau_{y}(E_{B,k})$, as $y\notin N^{B}$. Hence for every real $\eta>0$ there is $k$ with $\tau_{y}(B)-\tau_{y}(E_{B,k})\le\eta$, since otherwise $\tau_{y}(B)-\eta$ would be an upper bound of the $\tau_{y}(E_{B,k})$ smaller than their least upper bound.

(6.6) Continuity at $\varnothing$. Let $(R_{n})_{n\in\mathbb{N}}$ be a sequence in $\mathcal{U}$ with $R_{n+1}\subseteq R_{n}$ for every $n$ and $\bigcap_{n}R_{n}=\varnothing$. We show that $\tau_{y}(R_{n})$ converges to $0$. By (6.3) the sequence $(\tau_{y}(R_{n}))$ is nonincreasing and bounded below by $0$; let $\delta\ge0$ be its greatest lower bound, and suppose, for a contradiction, that $\delta>0$. Then $\tau_{y}(R_{n})\ge\delta$ for every $n$. With $\delta$ fixed, for each $n$ let $k_{n}$ be the least $k\in\mathbb{N}$ with $\tau_{y}(R_{n})-\tau_{y}(E_{R_{n},k})\le\delta2^{-n-1}$, which exists by (6.5). Put $W_{n}=E_{R_{1},k_{1}}\cap\dots\cap E_{R_{n},k_{n}}\in\mathcal{U}$. Then $W_{n+1}\subseteq W_{n}$ and $W_{n}\subseteq E_{R_{n},k_{n}}\subseteq R_{n}$. Since $R_{n}\subseteq R_{i}$ for $i\le n$,

$$R_{n}\setminus W_{n}=\bigcup_{i=1}^{n}\bigl(R_{n}\setminus E_{R_{i},k_{i}}\bigr)\subseteq\bigcup_{i=1}^{n}\bigl(R_{i}\setminus E_{R_{i},k_{i}}\bigr),$$

so by (6.3) and (6.4)

$$\tau_{y}(R_{n}\setminus W_{n})\le\sum_{i=1}^{n}\bigl(\tau_{y}(R_{i})-\tau_{y}(E_{R_{i},k_{i}})\bigr)\le\sum_{i=1}^{n}\delta2^{-i-1}<\frac{\delta}{2},$$

and therefore, by (6.3), $\tau_{y}(W_{n})=\tau_{y}(R_{n})-\tau_{y}(R_{n}\setminus W_{n})>\delta-\delta/2>0$. By (6.2), $W_{n}\ne\varnothing$ for every $n$. Each $E_{R_{i},k_{i}}$ is compact, hence closed by Compact Subsets of a Metric Space are Closed and Borel §closed, so every $W_{n}$ is closed by claim 3 of Complements, Unions and Intersections of Closed Sets in a Topological Space, and $W_{1}=E_{R_{1},k_{1}}$ is compact. The sets $Z\setminus W_{n}$ are open, and since $\bigcap_{n}W_{n}\subseteq\bigcap_{n}R_{n}=\varnothing$ they cover $W_{1}$. By Compact Subset Criterion via Open Covers in the Ambient Space there is a finite $J\subseteq\mathbb{N}$ with $W_{1}\subseteq\bigcup_{n\in J}(Z\setminus W_{n})$; $J$ is nonempty because $W_{1}\ne\varnothing$. Let $n^{*}$ be the largest element of $J$. For $n\in J$ we have $W_{n^{*}}\subseteq W_{n}$, so $Z\setminus W_{n}\subseteq Z\setminus W_{n^{*}}$, and hence $W_{1}\subseteq Z\setminus W_{n^{*}}$. But $\varnothing\ne W_{n^{*}}\subseteq W_{1}$, a contradiction. Thus $\delta=0$. Given a real $\eta>0$ there is then $n_{0}$ with $\tau_{y}(R_{n_{0}})<\eta$, and $0\le \tau_{y}(R_{n})\le \tau_{y}(R_{n_{0}})<\eta$ for $n\ge n_{0}$; so $\tau_{y}(R_{n})$ converges to $0$.

(6.7) Countable additivity. Let $(B_{i})_{i\in\mathbb{N}}$ be pairwise disjoint members of $\mathcal{U}$ whose union $B$ belongs to $\mathcal{U}$. Put $R_{n}=B\setminus(B_{1}\cup\dots\cup B_{n})\in\mathcal{U}$. Then $R_{n+1}\subseteq R_{n}$, and $\bigcap_{n}R_{n}=\varnothing$ because each point of $B$ lies in some $B_{i}$ and hence not in $R_{n}$ for $n\ge i$. As $B$ is the disjoint union of $B_{1},\dots,B_{n},R_{n}$, (6.2) and induction give $\sum_{i=1}^{n}\tau_{y}(B_{i})=\tau_{y}(B)-\tau_{y}(R_{n})$, which converges to $\tau_{y}(B)$ by (6.6). So the series $\sum_{i}\tau_{y}(B_{i})$ converges and its sum is $\tau_{y}(B)$, in the sense of Series of Real Numbers §convergent.

By Step 2, (6.2) and (6.7), the family $\mathcal{U}$ of subsets of $Z$ and the function $\tau_{y}$ satisfy the hypotheses of Extension of a Finite Premeasure from an Algebra of Sets to the Generated Sigma-Algebra. By Extension of a Finite Premeasure from an Algebra of Sets to the Generated Sigma-Algebra §extension there is a measure on $\sigma(\mathcal{U})=\mathcal{B}(Z)$ that agrees with $\tau_{y}$ on $\mathcal{U}$, and by Extension of a Finite Premeasure from an Algebra of Sets to the Generated Sigma-Algebra §uniqueness there is only one; we call it $\kappa_{y}$. By (6.1), $\kappa_{y}(Z)=1$, so $\kappa_{y}$ is a probability measure on $(Z,\mathcal{B}(Z))$, and

$$\kappa_{y}(B)=h_{B}(y)\qquad(y\in Y\setminus N_{0},\ B\in\mathcal{U}).\tag{4}$$

For $y\in N_{0}$ put $\kappa_{y}=\rho$, a probability measure on $(Z,\mathcal{B}(Z))$ by Step 0. Define $\kappa(y,B)=\kappa_{y}(B)$ for $y\in Y$ and $B\in\mathcal{B}(Z)$; its values lie in $[0,1]$.

Step 7 (measurability and the rectangle identity). Let $\mathcal{L}$ be the family of those $B\in\mathcal{B}(Z)$ such that $y\mapsto\kappa(y,B)$ is measurable and

$$\pi(A\times B)=\int_{Y}\mathbf{1}_{A}(y)\,\kappa(y,B)\,\mu(dy)\qquad\text{for every }A\in\mathcal{Y}.\tag{5}$$

$\mathcal{U}\subseteq\mathcal{L}$: for $B\in\mathcal{U}$, (4) gives $\kappa(\cdot,B)=\mathbf{1}_{Y\setminus N_{0}}h_{B}+\rho(B)\mathbf{1}_{N_{0}}$, which is measurable by claims 1, 2 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, as $N_{0}\in\mathcal{Y}$. For $A\in\mathcal{Y}$ the nonnegative measurable functions $\mathbf{1}_{A}\kappa(\cdot,B)$ and $\mathbf{1}_{A}h_{B}$ agree off the null set $N_{0}$, so (5) follows from The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison and (2).

$\mathcal{L}$ is a $\lambda$-system in the sense of Dynkin's Pi-Lambda Theorem. (1) $Z\in\mathcal{U}\subseteq\mathcal{L}$. (2) Let $B\subseteq B'$ be members of $\mathcal{L}$. For each $y$, $\kappa_{y}(B'\setminus B)=\kappa_{y}(B')-\kappa_{y}(B)$ by Basic Properties of a Measure §differences, so $\kappa(\cdot,B'\setminus B)$ is measurable by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. For $A\in\mathcal{Y}$ we have $\mathbf{1}_{A}\kappa(\cdot,B')=\mathbf{1}_{A}\kappa(\cdot,B)+\mathbf{1}_{A}\kappa(\cdot,B'\setminus B)$ pointwise, by Basic Properties of a Measure §additivity for each $\kappa_{y}$, so by Linearity and Monotonicity of the Lebesgue Integral §nonnegative and (5) for $B$ and $B'$,

$$\int_{Y}\mathbf{1}_{A}\kappa(\cdot,B'\setminus B)\,d\mu=\pi(A\times B')-\pi(A\times B)=\pi\bigl(A\times(B'\setminus B)\bigr),$$

the subtraction being legitimate since $\pi(A\times B)\le1$ is real, and the last equality being Basic Properties of a Measure §differences for $A\times B\subseteq A\times B'$, whose difference is $A\times(B'\setminus B)$. So $B'\setminus B\in\mathcal{L}$. (3) Let $(B_{n})_{n\in\mathbb{N}}$ be a nondecreasing sequence in $\mathcal{L}$ with union $B$. For each $y$, $\kappa(y,B_{n})$ is nondecreasing in $n$ by Basic Properties of a Measure §monotone, and, its values lying in $[0,1]$, it converges to $\kappa(y,B)$, which is also its least upper bound, by Basic Properties of a Measure §continuity-below; hence $\kappa(\cdot,B)$ is measurable by claim 5 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. For $A\in\mathcal{Y}$ the measurable functions $\mathbf{1}_{A}\kappa(\cdot,B_{n})$ are nondecreasing in $n$ with supremum $\mathbf{1}_{A}\kappa(\cdot,B)$, so by Monotone Convergence Theorem, (5) for each $B_{n}$, and Basic Properties of a Measure §continuity-below for the increasing sets $A\times B_{n}$ with union $A\times B$,

$$\int_{Y}\mathbf{1}_{A}\kappa(\cdot,B)\,d\mu=\sup_{n}\int_{Y}\mathbf{1}_{A}\kappa(\cdot,B_{n})\,d\mu=\sup_{n}\pi(A\times B_{n})=\pi(A\times B).$$

So $B\in\mathcal{L}$.

By Dynkin's Pi-Lambda Theorem, applied to the $\pi$-system $\mathcal{U}$ and the $\lambda$-system $\mathcal{L}$, we get $\sigma(\mathcal{U})\subseteq\mathcal{L}$; with Step 2 and $\mathcal{L}\subseteq\mathcal{B}(Z)$ this gives $\mathcal{L}=\mathcal{B}(Z)$.

Step 8 (conclusion of existence). By Step 6 each $\kappa_{y}$ is a probability measure on $(Z,\mathcal{B}(Z))$, and by Step 7 each $\kappa(\cdot,B)$, $B\in\mathcal{B}(Z)$, is measurable with respect to $\mathcal{Y}$; so $\kappa$ is a probability kernel from $(Y,\mathcal{Y})$ to $(Z,\mathcal{B}(Z))$ by Probability Kernels Between Measurable Spaces §kernel. By Step 7, (5) holds for all $A\in\mathcal{Y}$ and $B\in\mathcal{B}(Z)$, which is the displayed identity of claim 1, and Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §rectangles, applied to the probability measure $\mu$, the kernel $\kappa$ and the measure $\pi$, gives $\pi=\mu\otimes\kappa$.

Part 2: uniqueness. Let $\kappa$ and $\kappa'$ be probability kernels from $(Y,\mathcal{Y})$ to $(Z,\mathcal{B}(Z))$ with $\pi=\mu\otimes\kappa=\mu\otimes\kappa'$, and let $\mathcal{U}$ be the family fixed in Step 2, with its enumeration $(U_{j})_{j\in\mathbb{N}}$; it is a countable $\pi$-system that contains $Z$ and satisfies $\sigma(\mathcal{U})=\mathcal{B}(Z)$. Let $B\in\mathcal{U}$. By Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §composite, for every $A\in\mathcal{Y}$,

$$\lambda_{B}(A)=\pi(A\times B)=(\mu\otimes\kappa)(A\times B)=\int_{Y}\mathbf{1}_{A}(y)\,\kappa(y,B)\,\mu(dy),$$

and likewise with $\kappa'$. By Probability Kernels Between Measurable Spaces §kernel, $\kappa(\cdot,B)$ and $\kappa'(\cdot,B)$ are measurable with values in $[0,1]$, so both are densities of the finite measure $\lambda_{B}$ (Step 4) with respect to $\mu$, and by The Radon-Nikodym Theorem for a Finite Measure and a Sigma-Finite Measure, and Uniqueness of Densities §uniqueness the set $M_{B}=\{y:\kappa(y,B)\ne\kappa'(y,B)\}$ belongs to $\mathcal{Y}$ and has $\mu(M_{B})=0$. Let $M=\bigcup_{j\in\mathbb{N}}M_{U_{j}}$, which is the union of the $M_{B}$ over $B\in\mathcal{U}$, since the terms of $(U_{j})$ are exactly the members of $\mathcal{U}$. Each $M_{U_{j}}$ belongs to $\mathcal{Y}$, so $M\in\mathcal{Y}$ because $\mathcal{Y}$ is a $\sigma$-algebra (Sigma-Algebra and Measurable Space), and $\mu(M)\le\sum_{j\in\mathbb{N}}\mu(M_{U_{j}})=0$ by Basic Properties of a Measure §subadditivity, so $\mu(M)=0$.

We claim that $M$ is exactly the set of $y$ with $\kappa_{y}\ne\kappa'_{y}$. If $y\in M$, then $\kappa_{y}(B)\ne\kappa'_{y}(B)$ for some $B$, so $\kappa_{y}\ne\kappa'_{y}$. If $y\notin M$, then $\kappa_{y}$ and $\kappa'_{y}$ are measures on $(Z,\mathcal{B}(Z))$ with $\kappa_{y}(Z)=\kappa'_{y}(Z)=1<\infty$ that agree on the $\pi$-system $\mathcal{U}$ generating $\mathcal{B}(Z)$, so $\kappa_{y}=\kappa'_{y}$ by claim 1 of Uniqueness of Finite Measures on a Generating Pi-System and the Density of the Exponential Law. Hence the set of $y$ with $\kappa_{y}\ne\kappa'_{y}$ is $M$, which belongs to $\mathcal{Y}$ and has $\mu$-measure $0$. $\blacksquare$