Proof of Head Maps of a Noise-Optimal Coupling: the Head of the Target is a Function of the Source

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

Conventions and order of choices. For $z\in X\times X$ we write $x=\pi_{1}(z)$ and $y=\pi_{2}(z)$ as in Probability Measures on a Hilbert Space Transported in the Noise Norm: Standing Notation §background. The noise weights $a_{k}$ of Probability Measures on a Hilbert Space Transported in the Noise Norm: Standing Notation §weights are positive by Weight Sequences and the Weighted Fisher Information Relative to a Diagonal Gaussian Measure on a Hilbert Space §weights; for $v\in\mathbb{R}^{n}$ let $A_{n}v=(a_{1}v_{1},\dots,a_{n}v_{n})\in\mathbb{R}^{n}$. A noise-optimal coupling lies in $\Pi^{a}(\mu,\nu)$ by Noise-Optimal Couplings §optimal, which is contained in $\Pi(\mu,\nu)$ by Couplings of Finite Noise Cost and Their Noise Cost §couplings; so $(\pi_{1})_{\#}\pi=\mu$ and $(\pi_{2})_{\#}\pi=\nu$ by Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §coupling. We fix a density $f$ of $\mu$ with respect to $\gamma_{c}$; by The Radon-Nikodym Theorem for a Finite Measure and a Sigma-Finite Measure, and Uniqueness of Densities it is a function $f:X\to\mathbb{R}$ with $f\ge0$, measurable with respect to $\mathcal{B}(X)$, that is, Borel. Let $t_{n}=Q_{n}\circ\pi_{1}$ and $\mu_{n}=(t_{n})_{\#}\pi$; for $A\in\mathcal{B}(X)$, $\mu_{n}(A)=\pi(\pi_{1}^{-1}(Q_{n}^{-1}(A)))=\mu(Q_{n}^{-1}(A))$, so $\mu_{n}=(Q_{n})_{\#}\mu$. Let $\psi=(\pi_{1},p_{n}\circ\pi_{2}):X\times X\to X\times\mathbb{R}^{n}$; its components are Borel by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §product-sigma, Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity and claim 4 of Borel Measurability and Bounded Integration on a Metric Space, so $\psi$ is measurable with respect to $\mathcal{B}(X\times X)$ and $\mathcal{B}(X)\otimes\mathcal{B}(\mathbb{R}^{n})$ by Pairings into a Product, the Graph of a Measurable Map, and Couplings Concentrated on a Graph §pairing (with $(X\times X,\mathcal{B}(X\times X))$, $(X,\mathcal{B}(X))$ and $\mathbb{R}^{n}$ in place of $(\Omega,\mathcal{O})$, $(Y,\mathcal{Y})$ and $Z$). The choices are made in this order: conditional kernels $\chi$ and $\lambda$ (Step 1); the modification $\chi'$ of $\chi$ (Step 2); the composite kernel $\kappa$ (Step 3); the map $T_{n}$, defined from $\chi'$ alone (Step 4); and then, only for the verification, a map $S_{w}$ for each $w$ in a set $W$ (Step 5).

Step 1 (Two conditional kernels). The space $X\times X$ is a real Hilbert space by Properties of the Product of Two Real Inner Product Spaces §hilbert, so $(X\times X,d)$ is complete by Real Hilbert Space §hilbert, and it is separable by Properties of the Product of Two Real Inner Product Spaces §separable, $(X,d)$ being separable by Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §space, which also gives that $(X,d)$ is complete. The map $\pi_{1}$ is Borel by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §product-sigma, and $Q_{n}$ by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity. By Conditional Kernels of a Borel Probability Measure on a Polish Space Given a Borel Map: Existence, Uniqueness and Concentration on the Fibres §existence, applied with $(X\times X,d)$, $(X,d)$, $\pi_{1}$ and $\pi$ in place of $(Z,d)$, $(Y,d_{Y})$, $q$ and $\pi$ (image measure $(\pi_{1})_{\#}\pi=\mu$), there is a conditional kernel $\chi$ of $\pi$ given $\pi_{1}$; by the same claim, applied with $(X,d)$, $(X,d)$, $Q_{n}$ and $\mu$ in place of $(Z,d)$, $(Y,d_{Y})$, $q$ and $\pi$ (image measure $(Q_{n})_{\#}\mu=\mu_{n}$), there is a conditional kernel $\lambda$ of $\mu$ given $Q_{n}$. Write $\chi_{x}=\chi(x,\cdot)$ and $\lambda_{w}=\lambda(w,\cdot)$.

Step 2 (A version of $\chi$ concentrated on every fibre). By Conditional Kernels of a Borel Probability Measure on a Polish Space Given a Borel Map: Existence, Uniqueness and Concentration on the Fibres §fibres, with the data used for $\chi$ in Step 1, the set $G=\{x\in X:\chi_{x}(\pi_{1}^{-1}(\{x\}))=1\}$ belongs to $\mathcal{B}(X)$ and $\mu(G)=1$, the fibres $\pi_{1}^{-1}(\{x\})$ being Borel by the preamble of that corollary; let $N_{1}=X\setminus G$, so $\mu(N_{1})=1-1=0$ by Basic Properties of a Measure §differences. For $z_{0}\in X\times X$ let $\delta_{z_{0}}(B)=1$ if $z_{0}\in B$ and $\delta_{z_{0}}(B)=0$ if $z_{0}\notin B$, for $B\in\mathcal{B}(X\times X)$. This is a probability measure: $\delta_{z_{0}}(\varnothing)=0$, $\delta_{z_{0}}(X\times X)=1$, and for a sequence of pairwise disjoint Borel sets $B_{m}$ the point $z_{0}$ lies in at most one $B_{m}$, so $\delta_{z_{0}}(\bigcup_{m}B_{m})$ and $\sum_{m}\delta_{z_{0}}(B_{m})$ are both $1$ if $z_{0}$ lies in some $B_{m}$ and both $0$ otherwise. Let $D:X\to X\times X$, $D(x)=(x,x)$, which is measurable with respect to $\mathcal{B}(X)$ and $\mathcal{B}(X\times X)$ by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §pairing, applied with $(X,\mathcal{B}(X))$ and $S=T$ the identity of $X$ (measurable, since the preimage of each Borel set is that set); then $\delta_{(x,x)}(B)=\mathbf{1}_{D^{-1}(B)}(x)$. Define

$$\chi'(x,B)=\chi(x,B)\quad(x\in G),\qquad\chi'(x,B)=\delta_{(x,x)}(B)\quad(x\in N_{1}),\qquad B\in\mathcal{B}(X\times X).$$

Each $\chi'_{x}=\chi'(x,\cdot)$ is a probability measure on $(X\times X,\mathcal{B}(X\times X))$, and for fixed $B$ one has $\chi'(\cdot,B)=\mathbf{1}_{G}\,\chi(\cdot,B)+\mathbf{1}_{N_{1}}\,\mathbf{1}_{D^{-1}(B)}$, which is measurable with respect to $\mathcal{B}(X)$ by claims 1, 2 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. So $\chi'$ is a probability kernel from $(X,\mathcal{B}(X))$ to $(X\times X,\mathcal{B}(X\times X))$ (Probability Kernels Between Measurable Spaces §kernel). For $A\in\mathcal{B}(X)$ and $B\in\mathcal{B}(X\times X)$ the functions $x\mapsto\mathbf{1}_{A}(x)\chi(x,B)$ and $x\mapsto\mathbf{1}_{A}(x)\chi'(x,B)$ are measurable (same claims) and agree off the $\mu$-null set $N_{1}$; the first is integrable with integral $\pi(B\cap\pi_{1}^{-1}(A))$ because $\chi$ is a conditional kernel, so by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison the second is integrable with the same integral. Hence $\chi'$ is a conditional kernel of $\pi$ given $\pi_{1}$ (The Conditional Kernel of a Probability Measure Given a Measurable Map §conditional-kernel). Moreover:

(1) For every $x\in X$ and $C\in\mathcal{B}(X)$, $\chi'_{x}(\pi_{1}^{-1}(C))=\mathbf{1}_{C}(x)$. For $x\in N_{1}$ this holds because $(x,x)\in\pi_{1}^{-1}(C)$ exactly when $x\in C$. Let $x\in G$. If $x\in C$, then $\pi_{1}^{-1}(\{x\})\subseteq\pi_{1}^{-1}(C)$, so $1=\chi_{x}(\pi_{1}^{-1}(\{x\}))\le\chi_{x}(\pi_{1}^{-1}(C))\le1$ by Basic Properties of a Measure §monotone. If $x\notin C$, then $\pi_{1}^{-1}(C)\subseteq(X\times X)\setminus\pi_{1}^{-1}(\{x\})$, a set of $\chi_{x}$-measure $1-1=0$ by Basic Properties of a Measure §differences, so $\chi_{x}(\pi_{1}^{-1}(C))=0$ by Basic Properties of a Measure §monotone.

Step 3 (A conditional kernel given $t_{n}$). Apply Conditioning in Two Steps: Composing a Conditional Kernel Given a Map with a Conditional Kernel Given a Function of It §composition with $(X\times X,d)$, $(X,d)$, $(X,d)$, $\pi_{1}$, $Q_{n}$ and $\pi$ in place of $(Z,d_{Z})$, $(Y_{1},d_{1})$, $(Y_{2},d_{2})$, $q$, $s$ and $\pi$; there $\nu_{1}=(\pi_{1})_{\#}\pi=\mu$ and $\nu_{2}=(Q_{n}\circ\pi_{1})_{\#}\pi=\mu_{n}$, and we take the conditional kernel $\chi'$ of $\pi$ given $\pi_{1}$ (Step 2) and the conditional kernel $\lambda$ of $\mu$ given $Q_{n}$ (Step 1) in place of $\kappa$ and $\lambda$. So

$$\kappa(w,B)=\int_{X}\chi'(x,B)\,\lambda_{w}(dx)\qquad(w\in X,\ B\in\mathcal{B}(X\times X))$$

defines a probability kernel $\kappa$ from $(X,\mathcal{B}(X))$ to $(X\times X,\mathcal{B}(X\times X))$ that is a conditional kernel of $\pi$ given $Q_{n}\circ\pi_{1}=t_{n}$. Write $\kappa_{w}=\kappa(w,\cdot)$. The integrand $x\mapsto\chi'(x,B)$ is Borel with values in $[0,1]$, so the integral above is also its integral as a nonnegative function, by claim 6(c) of Borel Measurability and Bounded Integration on a Metric Space ($\lambda_{w}$ being a Borel measure of mass $1$ on $(X,d)$); we use this identification below without further comment.

(2) For every $w\in X$, $(\pi_{1})_{\#}\kappa_{w}=\lambda_{w}$. For $C\in\mathcal{B}(X)$, by (1) and The Integral of an Indicator Function is the Measure of the Set, $\kappa_{w}(\pi_{1}^{-1}(C))=\int_{X}\mathbf{1}_{C}\,d\lambda_{w}=\lambda_{w}(C)$.

Step 4 (The map $T_{n}$). For $k\in\{1,\dots,n\}$ let $\eta_{k}:X\times X\to\mathbb{R}$, $\eta_{k}(z)=y_{k}$; it is Borel by claim 4 of Borel Measurability and Bounded Integration on a Metric Space, being the composite of $\pi_{2}$ (Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §product-sigma) with the coordinate $x\mapsto x_{k}$ (Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity). The functions $\eta_{k}^{+}=\max(\eta_{k},0)$ and $\eta_{k}^{-}=\max(-\eta_{k},0)$ are nonnegative and Borel by claims 1, 2 and 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. By Conditional Kernels of a Borel Probability Measure on a Polish Space Given a Borel Map: Existence, Uniqueness and Concentration on the Fibres §nonnegative, applied with $(X\times X,d)$, $(X,d)$, $\pi_{1}$, $\pi$, the conditional kernel $\chi'$ and $\eta_{k}^{+}$ (respectively $\eta_{k}^{-}$) in place of $(Z,d)$, $(Y,d_{Y})$, $q$, $\pi$, $\kappa$ and $f$, the set $N_{k}^{+}$ (respectively $N_{k}^{-}$) of the $x\in X$ for which $\eta_{k}^{+}$ (respectively $\eta_{k}^{-}$) is not integrable with respect to $\chi'_{x}$ belongs to $\mathcal{B}(X)$, and the function $g_{k}^{+}$ (respectively $g_{k}^{-}$) equal to $\int\eta_{k}^{+}\,d\chi'_{x}$ (respectively $\int\eta_{k}^{-}\,d\chi'_{x}$) off that set and to $0$ on it is measurable with respect to $\mathcal{B}(X)$. Put

$$b_{k}=\mathbf{1}_{X\setminus(N_{k}^{+}\cup N_{k}^{-})}\,\bigl(g_{k}^{+}-g_{k}^{-}\bigr),$$

measurable by claims 1, 2 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and $T_{n}(x)=(b_{1}(x),\dots,b_{n}(x))$. Then $T_{n}:X\to\mathbb{R}^{n}$ is Borel by claims 2 and 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets. Let $\Gamma_{T_{n}}=\{(x,v)\in X\times\mathbb{R}^{n}:v=T_{n}(x)\}$, which belongs to $\mathcal{B}(X)\otimes\mathcal{B}(\mathbb{R}^{n})$ by Pairings into a Product, the Graph of a Measurable Map, and Couplings Concentrated on a Graph §graph-measurable (with $(X,\mathcal{B}(X))$, $(\mathbb{R}^{n},d_{E})$ and $T_{n}$ in place of $(Y,\mathcal{Y})$, $(Z,d_{Z})$ and $S$; $(\mathbb{R}^{n},d_{E})$ is separable by Euclidean Space is a Separable Metric Space §separable), and let $G_{n}=\{z\in X\times X:p_{n}(y)=T_{n}(x)\}$. A point $z$ lies in $G_{n}$ exactly when $\psi(z)=(x,p_{n}(y))\in\Gamma_{T_{n}}$, so $G_{n}=\psi^{-1}(\Gamma_{T_{n}})\in\mathcal{B}(X\times X)$.

(3) Let $x\in X$ and $c\in\mathbb{R}^{n}$, and suppose that some $F\in\mathcal{B}(X\times X)$ with $\chi'_{x}(F)=1$ satisfies $\pi_{1}(z)=x$ and $p_{n}(\pi_{2}(z))=c$ for every $z\in F$. Then $T_{n}(x)=c$ and $\chi'_{x}((X\times X)\setminus G_{n})=0$. Indeed $\chi'_{x}((X\times X)\setminus F)=1-1=0$ by Basic Properties of a Measure §differences, and for $k\le n$ we have $\eta_{k}=c_{k}$ on $F$. So $\eta_{k}^{+}$ agrees $\chi'_{x}$-almost everywhere with the constant function $\max(c_{k},0)$, which is integrable with respect to $\chi'_{x}$, a Borel measure of mass $1$ on $(X\times X,d)$, with integral $\max(c_{k},0)$ by claim 6(a) of Borel Measurability and Bounded Integration on a Metric Space; by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison, $\eta_{k}^{+}$ is integrable with respect to $\chi'_{x}$ with integral $\max(c_{k},0)$. Likewise $\eta_{k}^{-}$ is integrable with integral $\max(-c_{k},0)$. Hence $x\notin N_{k}^{+}\cup N_{k}^{-}$ and $b_{k}(x)=\max(c_{k},0)-\max(-c_{k},0)=c_{k}$. Thus $T_{n}(x)=c$, every $z\in F$ satisfies $p_{n}(y)=c=T_{n}(x)$, so $F\subseteq G_{n}$, and $\chi'_{x}((X\times X)\setminus G_{n})\le\chi'_{x}((X\times X)\setminus F)=0$ by Basic Properties of a Measure §monotone.

Step 5 (Fibre maps). The ordered pair $(\mu,\nu)$ is noise-connected by The Noise Wasserstein Distance is a Metric on the Measures Noise-Connected to the Reference Measure: Existence of Noise-Optimal Couplings, Comparison with the Quadratic Wasserstein Distance and Lower Semicontinuity §connected. Apply Fibrewise Cyclical Monotonicity of a Noise-Optimal Coupling: Given the Tail, the Heads are Cyclically Monotone §monotone with $\mu$, $\nu$, $\pi$, $n$ and the conditional kernel $\kappa$ of $\pi$ given $t_{n}$ of Step 3; with $h_{n}$ and $\theta_{w}=(h_{n})_{\#}\kappa_{w}$ as in that lemma, it gives $W_{1}\in\mathcal{B}(X)$ with $\mu_{n}(W_{1})=1$ such that $\operatorname{supp}\theta_{w}$ is cyclically monotone for every $w\in W_{1}$. Apply The Head Given the Tail is Absolutely Continuous for a Measure with a Density Relative to a Diagonal Gaussian Measure §absolutely-continuous with $c$, $\gamma_{c}$, $\mu$, $f$, $n$ and the conditional kernel $\lambda$ of $\mu$ given $Q_{n}$ of Step 1; its $\nu_{n}=(Q_{n})_{\#}\mu$ is $\mu_{n}$, so it gives $W_{2}\in\mathcal{B}(X)$ with $\mu_{n}(W_{2})=1$ such that $(p_{n})_{\#}\lambda_{w}$ is absolutely continuous for every $w\in W_{2}$. Let $W=W_{1}\cap W_{2}\in\mathcal{B}(X)$; as $\mu_{n}(X\setminus W_{1})=\mu_{n}(X\setminus W_{2})=0$ by Basic Properties of a Measure §differences, Basic Properties of a Measure §subadditivity gives $\mu_{n}(X\setminus W)=0$.

By the definition of $h_{n}$ (its first $n$ components are $x_{1},\dots,x_{n}$ and its last $n$ are $a_{1}^{-1}y_{1},\dots,a_{n}^{-1}y_{n}$), with the concatenation map $\iota$ of Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pairs,

$$h_{n}(z)=\iota\bigl(p_{n}(x),(a_{1}^{-1}y_{1},\dots,a_{n}^{-1}y_{n})\bigr)\qquad(z\in X\times X).$$

Hence, by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections, the coordinate projections $\mathrm{pr}_{1},\mathrm{pr}_{2}$ of Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pairs satisfy $\mathrm{pr}_{1}(h_{n}(z))=p_{n}(x)$ and $\mathrm{pr}_{2}(h_{n}(z))=(a_{1}^{-1}y_{1},\dots,a_{n}^{-1}y_{n})$. Hence for $w\in X$ and $B\in\mathcal{B}(\mathbb{R}^{n})$, using (2),

$$\theta_{w}\bigl(\mathrm{pr}_{1}^{-1}(B)\bigr)=\kappa_{w}\bigl(h_{n}^{-1}(\mathrm{pr}_{1}^{-1}(B))\bigr)=\kappa_{w}\bigl(\pi_{1}^{-1}(p_{n}^{-1}(B))\bigr)=\lambda_{w}\bigl(p_{n}^{-1}(B)\bigr),$$

so $(\mathrm{pr}_{1})_{\#}\theta_{w}=(p_{n})_{\#}\lambda_{w}$; call it $m_{w}$. For $w\in W$, $m_{w}$ is absolutely continuous and $\operatorname{supp}\theta_{w}$ is cyclically monotone, so A Coupling on Euclidean Space with an Absolutely Continuous First Marginal and Cyclically Monotone Support is Induced by a Map §map, applied with $n$ (which satisfies $1\le n$) and $\theta_{w}$ in place of $d$ and $\theta$, gives a Borel map $S:\mathbb{R}^{n}\to\mathbb{R}^{n}$ with $\theta_{w}=(\mathrm{id},S)_{\#}m_{w}$; for each $w\in W$ we choose one such map and call it $S_{w}$.

Step 6 ($\pi(G_{n})=1$). Fix $w\in W$. By Pairings into a Product, the Graph of a Measurable Map, and Couplings Concentrated on a Graph §graph-measurable, with $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$, $(\mathbb{R}^{n},d_{E})$ and $S_{w}$ in place of $(Y,\mathcal{Y})$, $(Z,d_{Z})$ and $S$, the graph $\Gamma_{S_{w}}=\{(u,v):v=S_{w}(u)\}$ belongs to $\mathcal{B}(\mathbb{R}^{n})\otimes\mathcal{B}(\mathbb{R}^{n})$. The map $r:\mathbb{R}^{n+n}\to\mathbb{R}^{n}\times\mathbb{R}^{n}$, $r(v)=(\mathrm{pr}_{1}(v),\mathrm{pr}_{2}(v))$, is measurable with respect to $\mathcal{B}(\mathbb{R}^{n+n})$ and $\mathcal{B}(\mathbb{R}^{n})\otimes\mathcal{B}(\mathbb{R}^{n})$ by Pairings into a Product, the Graph of a Measurable Map, and Couplings Concentrated on a Graph §pairing, its components being Borel by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections. So $\Gamma'_{w}=r^{-1}(\Gamma_{S_{w}})=\{v\in\mathbb{R}^{n+n}:\mathrm{pr}_{2}(v)=S_{w}(\mathrm{pr}_{1}(v))\}$ belongs to $\mathcal{B}(\mathbb{R}^{n+n})$. For $u\in\mathbb{R}^{n}$ the point $(\mathrm{id},S_{w})(u)=\iota(u,S_{w}(u))$ has $\mathrm{pr}_{1}$-image $u$ and $\mathrm{pr}_{2}$-image $S_{w}(u)$ by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections, so it lies in $\Gamma'_{w}$; hence $\theta_{w}(\Gamma'_{w})=m_{w}(\mathbb{R}^{n})=1$. Let $H_{w}=h_{n}^{-1}(\Gamma'_{w})$, which belongs to $\mathcal{B}(X\times X)$ since $h_{n}$ is Borel (preamble of Fibrewise Cyclical Monotonicity of a Noise-Optimal Coupling: Given the Tail, the Heads are Cyclically Monotone). By Step 5, $z\in H_{w}$ exactly when $a_{k}^{-1}y_{k}=S_{w}(p_{n}(x))_{k}$ for $k\le n$, that is (multiplying by $a_{k}>0$) when $p_{n}(y)=A_{n}S_{w}(p_{n}(x))$. Now $\kappa_{w}(H_{w})=\theta_{w}(\Gamma'_{w})=1$, so by Basic Properties of a Measure §differences and the formula for $\kappa$,

$$0=\kappa_{w}\bigl((X\times X)\setminus H_{w}\bigr)=\int_{X}\chi'_{x}\bigl((X\times X)\setminus H_{w}\bigr)\,\lambda_{w}(dx).$$

The integrand is Borel and nonnegative, so by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §vanishing the set $E_{w}=\{x\in X:\chi'_{x}((X\times X)\setminus H_{w})=0\}$, which belongs to $\mathcal{B}(X)$ as the preimage of $\{0\}$ under that Borel function, satisfies $\lambda_{w}(X\setminus E_{w})=0$.

Let $x\in E_{w}$ and $F=H_{w}\cap\pi_{1}^{-1}(\{x\})\in\mathcal{B}(X\times X)$, since $\{x\}\in\mathcal{B}(X)$ (as in Step 2) and $\pi_{1}$ is Borel. By (1) with $C=\{x\}$ and Basic Properties of a Measure §differences, $\chi'_{x}((X\times X)\setminus\pi_{1}^{-1}(\{x\}))=0$; since $(X\times X)\setminus F$ is the union of $(X\times X)\setminus H_{w}$ and $(X\times X)\setminus\pi_{1}^{-1}(\{x\})$, Basic Properties of a Measure §subadditivity gives $\chi'_{x}((X\times X)\setminus F)=0$, so $\chi'_{x}(F)=1$. Every $z\in F$ satisfies $\pi_{1}(z)=x$ and $p_{n}(\pi_{2}(z))=A_{n}S_{w}(p_{n}(x))$. By (3) with $c=A_{n}S_{w}(p_{n}(x))$, $\chi'_{x}((X\times X)\setminus G_{n})=0$. Thus the Borel function $x\mapsto\chi'_{x}((X\times X)\setminus G_{n})$, with values in $[0,1]$, vanishes on $E_{w}$, and $X\setminus E_{w}$ is $\lambda_{w}$-null; by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-integral,

$$\kappa_{w}\bigl((X\times X)\setminus G_{n}\bigr)=\int_{X}\chi'_{x}\bigl((X\times X)\setminus G_{n}\bigr)\,\lambda_{w}(dx)=0\qquad(w\in W).$$

Finally, since $\kappa$ is a conditional kernel of $\pi$ given $t_{n}$, the identity of The Conditional Kernel of a Probability Measure Given a Measurable Map §conditional-kernel with $A=X$ (so $t_{n}^{-1}(A)=X\times X$ and $\mathbf{1}_{A}=1$) and $B=(X\times X)\setminus G_{n}$ gives

$$\pi\bigl((X\times X)\setminus G_{n}\bigr)=\int_{X}\kappa\bigl(w,(X\times X)\setminus G_{n}\bigr)\,\mu_{n}(dw)=0,$$

by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-integral, the integrand being Borel with values in $[0,1]$ (so its integral is its integral as a nonnegative function, claim 6(c) of Borel Measurability and Bounded Integration on a Metric Space) and vanishing off the $\mu_{n}$-null set $X\setminus W$. Hence $\pi(G_{n})=1-0=1$ by Basic Properties of a Measure §differences.

Step 7 (Claim 1). Let $\sigma=\psi_{\#}\pi$, a probability measure on $(X\times\mathbb{R}^{n},\mathcal{B}(X)\otimes\mathcal{B}(\mathbb{R}^{n}))$ by claim 1 of Image Measures, Measures with Densities, and Change of Variables, and let $\mathrm{pr}_{X}:X\times\mathbb{R}^{n}\to X$ be the coordinate projection (called $\mathrm{pr}_{Y}$ in Pairings into a Product, the Graph of a Measurable Map, and Couplings Concentrated on a Graph). For $A\in\mathcal{B}(X)$, $\psi^{-1}(\mathrm{pr}_{X}^{-1}(A))=\pi_{1}^{-1}(A)$, so $\sigma(\mathrm{pr}_{X}^{-1}(A))=\pi(\pi_{1}^{-1}(A))=\mu(A)$; that is, $(\mathrm{pr}_{X})_{\#}\sigma=\mu$. Moreover $\sigma(\Gamma_{T_{n}})=\pi(\psi^{-1}(\Gamma_{T_{n}}))=\pi(G_{n})=1$ by Step 6. By Pairings into a Product, the Graph of a Measurable Map, and Couplings Concentrated on a Graph §graph, applied with $(X,\mathcal{B}(X))$, $(\mathbb{R}^{n},d_{E})$ (separable), $\mu$, $T_{n}$ and $\sigma$ in place of $(Y,\mathcal{Y})$, $(Z,d_{Z})$, $\mu$, $S$ and $\pi$, we get $\sigma=(\mathrm{id},T_{n})_{\#}\mu$, that is, $(\pi_{1},p_{n}\circ\pi_{2})_{\#}\pi=(\mathrm{id},T_{n})_{\#}\mu$ with the Borel map $T_{n}$ of Step 4. This proves claim 1.

Step 8 (Claim 2). Let $T_{n}$ now be any Borel map $X\to\mathbb{R}^{n}$ as in claim 1, and let $G_{n}=\{z\in X\times X:p_{n}(y)=T_{n}(x)\}$ and $\Gamma_{T_{n}}$ its graph, which belongs to $\mathcal{B}(X)\otimes\mathcal{B}(\mathbb{R}^{n})$ by Pairings into a Product, the Graph of a Measurable Map, and Couplings Concentrated on a Graph §graph-measurable, applied as in Step 4 to this $T_{n}$. As in Step 4, $z\in G_{n}$ exactly when $\psi(z)\in\Gamma_{T_{n}}$, so $G_{n}=\psi^{-1}(\Gamma_{T_{n}})$, and $(\mathrm{id},T_{n})(x)=(x,T_{n}(x))\in\Gamma_{T_{n}}$ for every $x\in X$, so $(\mathrm{id},T_{n})^{-1}(\Gamma_{T_{n}})=X$. Therefore

$$\pi(G_{n})=\bigl(\psi_{\#}\pi\bigr)(\Gamma_{T_{n}})=\bigl((\mathrm{id},T_{n})_{\#}\mu\bigr)(\Gamma_{T_{n}})=\mu(X)=1 .$$

This proves claim 2.