Proof of The Head Given the Tail is Absolutely Continuous for a Measure with a Density Relative to a Diagonal Gaussian Measure

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

Throughout, $K$ and $k_{n}$ are the objects of The Conditional Law of the Head Given the Tail for a Measure with a Density Relative to a Diagonal Gaussian Measure on a Hilbert Space, formed for the present $c$, $n$, $\mu$ and $f$; the measure $\nu_{n}$ of that lemma is the present $\nu_{n}=(Q_{n})_{\#}\mu$. By the kernel claim of that lemma $K$ is a probability kernel from $(X,\mathcal{B}(X))$ to $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$, so $K_{w}=K(w,\cdot)$ is a probability measure on $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$ for every $w\in X$ by Probability Kernels Between Measurable Spaces §kernel. To avoid a clash with the kernel $\lambda$, we write $\mathcal{L}_{n}$ for the Lebesgue measure on $\mathcal{B}(\mathbb{R}^{n})$ that Absolutely Continuous Probability Measure on Euclidean Space §ac calls $\lambda_{n}$. The order of choices is: $\lambda$ is given; from $K$ we build a second conditional kernel $\lambda'$ of $\mu$ given $Q_{n}$ (Steps 1 to 3), compare it with $\lambda$ (Step 4), and then compute the heads (Steps 5 and 6).

Step 1 (Translation maps). Define $\Phi:X\times\mathbb{R}^{n}\to X$ by $\Phi(w,u)=p_{n}^{*}(u)+w$, and for $w\in X$ define $\iota_{w}:\mathbb{R}^{n}\to X$ by $\iota_{w}(u)=\Phi(w,u)$. The map $p_{n}^{*}$ is linear, since $p_{n}^{*}(u)-p_{n}^{*}(u')=\sum_{k=1}^{n}(u_{k}-u'_{k})e_{k}=p_{n}^{*}(u-u')$ by vector space arithmetic. Hence for $(w,u),(w',u')\in X\times\mathbb{R}^{n}$ we have $\Phi(w,u)-\Phi(w',u')=(w-w')+p_{n}^{*}(u-u')$, and by the triangle inequality together with $|p_{n}^{*}(v)|=\lVert v\rVert$ for $v\in\mathbb{R}^{n}$ (Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity)

$$|\Phi(w,u)-\Phi(w',u')|\le|w-w'|+\lVert u-u'\rVert\le 2\,d_{X\times\mathbb{R}^{n}}\bigl((w,u),(w',u')\bigr),$$

where $d_{X\times\mathbb{R}^{n}}$ is the product metric and the last inequality is claim 2 of The Product Metric is a Metric. So $\Phi$ is continuous from $(X\times\mathbb{R}^{n},d_{X\times\mathbb{R}^{n}})$ to $(X,d)$ (given $\varepsilon>0$, the choice $\delta=\varepsilon/2$ works), hence measurable with respect to the Borel $\sigma$-algebra of $X\times\mathbb{R}^{n}$ and $\mathcal{B}(X)$ by claim 3 of Borel Measurability and Bounded Integration on a Metric Space; that Borel $\sigma$-algebra is $\mathcal{B}(X)\otimes\mathcal{B}(\mathbb{R}^{n})$ by The Borel Sigma-Algebra of a Product of Two Separable Metric Spaces is the Product Sigma-Algebra §product, $(X,d)$ being separable by Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §space and $(\mathbb{R}^{n},d_{E})$ by Euclidean Space is a Separable Metric Space §separable. Likewise $|\iota_{w}(u)-\iota_{w}(u')|=|p_{n}^{*}(u-u')|=\lVert u-u'\rVert$, so each $\iota_{w}$ is continuous, hence Borel by claim 3 of Borel Measurability and Bounded Integration on a Metric Space.

For $B\in\mathcal{B}(X)$ let $E_{B}=\Phi^{-1}(B)$, which belongs to $\mathcal{B}(X)\otimes\mathcal{B}(\mathbb{R}^{n})$ by the measurability of $\Phi$. Its section at $w\in X$ is $(E_{B})_{w}=\{u\in\mathbb{R}^{n}:p_{n}^{*}(u)+w\in B\}=\iota_{w}^{-1}(B)$.

Step 2 (The kernel $\lambda'$). For $w\in X$ and $B\in\mathcal{B}(X)$ put

$$\lambda'(w,B)=K\bigl(w,(E_{B})_{w}\bigr)=K_{w}\bigl(\iota_{w}^{-1}(B)\bigr).$$

For fixed $w$, the function $\lambda'_{w}=\lambda'(w,\cdot)$ is the image measure of $K_{w}$ under the Borel map $\iota_{w}$, a probability measure on $(X,\mathcal{B}(X))$ by claim 1 of Image Measures, Measures with Densities, and Change of Variables. For fixed $B$, the function $w\mapsto\lambda'(w,B)=K(w,(E_{B})_{w})$ is measurable with respect to $\mathcal{B}(X)$ by Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §sections, applied with $(X,\mathcal{B}(X))$, $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$, the kernel $K$, the probability measure $\nu_{n}$ and the set $E_{B}$ in place of $(Y,\mathcal{Y})$, $(Z,\mathcal{Z})$, $\kappa$, $\mu$ and $E$. Hence $\lambda'$ is a probability kernel from $(X,\mathcal{B}(X))$ to $(X,\mathcal{B}(X))$ by Probability Kernels Between Measurable Spaces §kernel.

Step 3 ($\lambda'$ is a conditional kernel of $\mu$ given $Q_{n}$). Let $A,B\in\mathcal{B}(X)$ and let $E=(A\times\mathbb{R}^{n})\cap E_{B}$, which belongs to $\mathcal{B}(X)\otimes\mathcal{B}(\mathbb{R}^{n})$ because $A\times\mathbb{R}^{n}$ is a measurable rectangle (Product Sigma-Algebra) and a $\sigma$-algebra is closed under finite intersections. By Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §coordinates, $P_{n}=p_{n}^{*}\circ p_{n}$ and $Q_{n}x=x-P_{n}x$, so every $x\in X$ satisfies

$$p_{n}^{*}(p_{n}(x))+Q_{n}x=P_{n}x+(x-P_{n}x)=x .$$

Hence $k_{n}(x)=(Q_{n}x,p_{n}(x))$ lies in $E$ if and only if $Q_{n}x\in A$ and $p_{n}^{*}(p_{n}(x))+Q_{n}x\in B$, that is, if and only if $Q_{n}x\in A$ and $x\in B$; so $k_{n}^{-1}(E)=B\cap Q_{n}^{-1}(A)$. The section $E_{w}$ is $(E_{B})_{w}$ for $w\in A$ and $\varnothing$ for $w\notin A$, so $K(w,E_{w})=\mathbf{1}_{A}(w)\,\lambda'(w,B)$ for every $w\in X$, because $K(w,\varnothing)=0$. The map $k_{n}$ is Borel by The Conditional Law of the Head Given the Tail for a Measure with a Density Relative to a Diagonal Gaussian Measure on a Hilbert Space §good-tails, and by The Conditional Law of the Head Given the Tail for a Measure with a Density Relative to a Diagonal Gaussian Measure on a Hilbert Space §disintegration and the defining formula of Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §composite (with $(X,\mathcal{B}(X))$, $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$, $K$ and $\nu_{n}$ in place of $(Y,\mathcal{Y})$, $(Z,\mathcal{Z})$, $\kappa$ and $\mu$),

$$\mu\bigl(B\cap Q_{n}^{-1}(A)\bigr)=\mu\bigl(k_{n}^{-1}(E)\bigr)=\bigl((k_{n})_{\#}\mu\bigr)(E)=(\nu_{n}\otimes K)(E)=\int_{X}K(w,E_{w})\,\nu_{n}(dw)=\int_{X}\mathbf{1}_{A}(w)\,\lambda'(w,B)\,\nu_{n}(dw).$$

This is the identity of The Conditional Kernel of a Probability Measure Given a Measurable Map §conditional-kernel with $(X,d)$, $(X,d)$, $Q_{n}$, $\mu$ and $\nu_{n}=(Q_{n})_{\#}\mu$ in place of $(Z,d_{Z})$, $(Y,d_{Y})$, $q$, $\pi$ and $\nu$, the map $Q_{n}$ being Borel by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity. So $\lambda'$ is a conditional kernel of $\mu$ given $Q_{n}$.

Step 4 (Comparison with $\lambda$). The metric space $(X,d)$ is complete and separable by Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §space. By Conditional Kernels of a Borel Probability Measure on a Polish Space Given a Borel Map: Existence, Uniqueness and Concentration on the Fibres §uniqueness, applied with $(X,d)$ in place of both $(Z,d)$ and $(Y,d_{Y})$, with $Q_{n}$, $\mu$ and $\nu_{n}$ in place of $q$, $\pi$ and $\nu$, and with the conditional kernels $\lambda$ and $\lambda'$, the set $N=\{w\in X:\lambda_{w}\ne\lambda'_{w}\}$ belongs to $\mathcal{B}(X)$ and $\nu_{n}(N)=0$.

Step 5 (Tails have zero head). Let $W_{p}=p_{n}^{-1}(\{0\})$, where $0$ is the origin of $\mathbb{R}^{n}$; it belongs to $\mathcal{B}(X)$, since $\{0\}\in\mathcal{B}(\mathbb{R}^{n})$ by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §finite-sets and $p_{n}$ is Borel by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity. Each coordinate $x\mapsto x_{k}=\langle x,e_{k}\rangle$ is linear, the inner product being linear in its first argument, so $p_{n}$ is linear; with $p_{n}(p_{n}^{*}(v))=v$ for $v\in\mathbb{R}^{n}$ (Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity) this gives, for every $x\in X$,

$$p_{n}(Q_{n}x)=p_{n}\bigl(x-p_{n}^{*}(p_{n}(x))\bigr)=p_{n}(x)-p_{n}\bigl(p_{n}^{*}(p_{n}(x))\bigr)=p_{n}(x)-p_{n}(x)=0 .$$

So $Q_{n}^{-1}(W_{p})=X$ and $\nu_{n}(W_{p})=\mu(X)=1$. Let $W_{1}=W_{p}\setminus N$, which belongs to $\mathcal{B}(X)$. Since $\nu_{n}$ is a probability measure, $\nu_{n}(X\setminus W_{p})=1-1=0$ by Basic Properties of a Measure §differences; as $X\setminus W_{1}=(X\setminus W_{p})\cup N$, Basic Properties of a Measure §subadditivity (applied to the sequence $X\setminus W_{p},N,\varnothing,\varnothing,\dots$) gives $\nu_{n}(X\setminus W_{1})\le0+0=0$, and then $\nu_{n}(W_{1})=1-0=1$ by Basic Properties of a Measure §differences.

Step 6 (The head of $\lambda_{w}$). Let $w\in W_{1}$ and $C\in\mathcal{B}(\mathbb{R}^{n})$. For $u\in\mathbb{R}^{n}$, linearity of $p_{n}$, $p_{n}\circ p_{n}^{*}=\mathrm{id}$ and $p_{n}(w)=0$ (as $w\in W_{p}$) give

$$p_{n}(\iota_{w}(u))=p_{n}(p_{n}^{*}(u))+p_{n}(w)=u+0=u ,$$

so $\iota_{w}^{-1}(p_{n}^{-1}(C))=C$. Since $w\notin N$, $\lambda_{w}=\lambda'_{w}$, and therefore

$$\bigl((p_{n})_{\#}\lambda_{w}\bigr)(C)=\lambda'_{w}\bigl(p_{n}^{-1}(C)\bigr)=K_{w}\bigl(\iota_{w}^{-1}(p_{n}^{-1}(C))\bigr)=K_{w}(C).$$

Thus $(p_{n})_{\#}\lambda_{w}=K_{w}$. If $\mathcal{L}_{n}(C)=0$, then $K(w,C)=0$ by The Conditional Law of the Head Given the Tail for a Measure with a Density Relative to a Diagonal Gaussian Measure on a Hilbert Space §absolutely-continuous, which holds for every $w\in X$. Hence $(p_{n})_{\#}\lambda_{w}$, a probability measure on $\mathbb{R}^{n}$ by claim 1 of Image Measures, Measures with Densities, and Change of Variables, is absolutely continuous in the sense of Absolutely Continuous Probability Measure on Euclidean Space §ac for every $w\in W_{1}$, where $W_{1}\in\mathcal{B}(X)$ and $\nu_{n}(W_{1})=1$. This proves claim 1.