Proof of Replacing the Tail of a Probability Measure on a Hilbert Space by an Independent Diagonal Gaussian Tail: Same Head and Convergence in the Quadratic Wasserstein Distance

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

Fix $n\in\mathbb{N}$ until Step 5. For maps $T:S_{1}\to S_{2}$ and $R:S_{2}\to S_{3}$ between measurable spaces, each measurable, and a measure $\lambda$ on $S_{1}$, one has $(R\circ T)^{-1}(C)=T^{-1}(R^{-1}(C))$ for every measurable $C\subseteq S_{3}$, so that the image measures of claim 1 of Image Measures, Measures with Densities, and Change of Variables satisfy

$$(R\circ T)_{\#}\lambda=R_{\#}(T_{\#}\lambda).\tag{$\dagger$}$$

Let $\rho$ be the product metric on $\mathbb{R}^{n}\times X$, so that $\rho((u,w),(u',w'))=\max\{\lVert u-u'\rVert,|w-w'|\}$, using $d_{E}(u,u')=\lVert u-u'\rVert$ from Euclidean Space and Lebesgue Measure: Standing Notation §space; its Borel $\sigma$-algebra is $\mathcal{B}(\mathbb{R}^{n})\otimes\mathcal{B}(X)$, as recalled in the statement. On $X\times X$ we use the norm, distance and coordinate maps $\pi_{1},\pi_{2}$ of Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §pairs; by Properties of the Product of Two Real Inner Product Spaces §coordinates, $\pi_{1}$ and $\pi_{2}$ are linear with $|\pi_{i}z|\le|z|$, and by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §product-sigma, $\mathcal{B}(X\times X)=\mathcal{B}(X)\otimes\mathcal{B}(X)$ and $\pi_{1},\pi_{2}$ are Borel. The function $q:X\to\mathbb{R}$, $q(y)=|y|^{2}$, is Borel, as recorded in The Second Moment of a Borel Probability Measure on a Hilbert Space and the Probability Measures with Finite Second Moment.

Step 1 (a coupling). The measures $\mu,\gamma_{c}\in\mathcal{P}(X)$ have finite mass, so by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §product-measure the product measure $\lambda=\mu\otimes\gamma_{c}$ of Existence and Uniqueness of the Product Measure is a Borel measure on $X\times X$ with $(\pi_{1})_{\#}\lambda=\gamma_{c}(X)\,\mu=\mu$ and $(\pi_{2})_{\#}\lambda=\mu(X)\,\gamma_{c}=\gamma_{c}$; it is a probability measure, since $\lambda(X\times X)=\mu(X)\gamma_{c}(X)=1$.

Define $j_{n}:X\times X\to\mathbb{R}^{n}\times X$ by $j_{n}(z)=(p_{n}(\pi_{1}z),Q_{n}\pi_{2}z)$. For $z,z'\in X\times X$, Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity (the maps $p_{n}$ and $Q_{n}$ are Lipschitz with constant $1$) and the linearity of $\pi_{i}$ give

$$\rho\bigl(j_{n}(z),j_{n}(z')\bigr)\le\max\{|\pi_{1}(z-z')|,|\pi_{2}(z-z')|\}\le|z-z'|,$$

so $j_{n}$ is continuous (take $\delta=\varepsilon$), hence Borel with respect to $\mathcal{B}(X\times X)$ and $\mathcal{B}(\mathbb{R}^{n})\otimes\mathcal{B}(X)$ by claim 3 of Borel Measurability and Bounded Integration on a Metric Space. Let $T_{n}=\Phi_{n}\circ j_{n}:X\times X\to X$, so that

$$T_{n}(x,g)=p_{n}^{*}(p_{n}(x))+Q_{n}g=P_{n}x+Q_{n}g\qquad(x,g\in X),$$

using $P_{n}=p_{n}^{*}\circ p_{n}$ from Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §coordinates. Since $\Phi_{n}$ is Borel by Head and Tail of a Diagonal Gaussian Measure on a Hilbert Space are Independent §synthesis, $T_{n}$ is Borel by claim 4 of Borel Measurability and Bounded Integration on a Metric Space. Let $\Psi_{n}=(\pi_{1},T_{n}):X\times X\to X\times X$, that is, $\Psi_{n}(x,g)=(x,P_{n}x+Q_{n}g)$; it is Borel by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §pairing. Let $\pi^{(n)}=(\Psi_{n})_{\#}\lambda$, a Borel probability measure on $X\times X$ by claim 1 of Image Measures, Measures with Densities, and Change of Variables, so $\pi^{(n)}\in\mathcal{P}(X\times X)$.

Its first marginal: $\pi_{1}\circ\Psi_{n}=\pi_{1}$, so by ($\dagger$), $(\pi_{1})_{\#}\pi^{(n)}=(\pi_{1})_{\#}\lambda=\mu$.

Its second marginal: $\pi_{2}\circ\Psi_{n}=T_{n}=\Phi_{n}\circ j_{n}$, so by ($\dagger$), $(\pi_{2})_{\#}\pi^{(n)}=(\Phi_{n})_{\#}((j_{n})_{\#}\lambda)$. For $A\in\mathcal{B}(\mathbb{R}^{n})$ and $B\in\mathcal{B}(X)$ we have $j_{n}^{-1}(A\times B)=p_{n}^{-1}(A)\times Q_{n}^{-1}(B)$, a measurable rectangle ($p_{n}$, $Q_{n}$ being Borel), so

$$(j_{n})_{\#}\lambda(A\times B)=\mu\bigl(p_{n}^{-1}(A)\bigr)\,\gamma_{c}\bigl(Q_{n}^{-1}(B)\bigr)=\bigl((p_{n})_{\#}\mu\bigr)(A)\,\bigl((Q_{n})_{\#}\gamma_{c}\bigr)(B).$$

The measure $(j_{n})_{\#}\lambda$ on $\mathcal{B}(\mathbb{R}^{n})\otimes\mathcal{B}(X)$ therefore has the defining rectangle values of the product of the probability (hence $\sigma$-finite) measures $(p_{n})_{\#}\mu$ and $(Q_{n})_{\#}\gamma_{c}$, and by the uniqueness in Existence and Uniqueness of the Product Measure, $(j_{n})_{\#}\lambda=(p_{n})_{\#}\mu\otimes(Q_{n})_{\#}\gamma_{c}$. Hence $(\pi_{2})_{\#}\pi^{(n)}=\mu^{(n)}$. As $\mu^{(n)}$ is a Borel probability measure on $X$, $\mu^{(n)}\in\mathcal{P}(X)$, and $\pi^{(n)}$ is a coupling of $\mu$ and $\mu^{(n)}$: $\pi^{(n)}\in\Pi(\mu,\mu^{(n)})$.

Step 2 (the cost of the coupling). Let $a_{n}=q\circ Q_{n}:X\to\mathbb{R}$, $a_{n}(y)=|Q_{n}y|^{2}$, Borel by claim 4 of Borel Measurability and Bounded Integration on a Metric Space, with $0\le a_{n}(y)\le|y|^{2}$ because $|Q_{n}y|\le|y|$ by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity. Let $\kappa(z)=|\pi_{1}z-\pi_{2}z|^{2}$ be the Borel integrand of the quadratic cost. For $(x,g)\in X\times X$,

$$\kappa(\Psi_{n}(x,g))=|x-P_{n}x-Q_{n}g|^{2}=|Q_{n}x-Q_{n}g|^{2}\le\bigl(|Q_{n}x|+|Q_{n}g|\bigr)^{2}\le2a_{n}(x)+2a_{n}(g),$$

by the triangle inequality and $(s+t)^{2}\le2s^{2}+2t^{2}$ for real $s,t$. By claim 2 of Image Measures, Measures with Densities, and Change of Variables (change of variables for nonnegative functions), the monotonicity and additivity of the integral of nonnegative measurable functions (Linearity and Monotonicity of the Lebesgue Integral, in force by Measure Spaces and the Lebesgue Integral: Standing Notation §background), and claim 2 of Image Measures, Measures with Densities, and Change of Variables once more together with the marginals of Step 1,

$$I(\pi^{(n)})=\int_{X\times X}\kappa\circ\Psi_{n}\,d\lambda\le2\int_{X\times X}a_{n}\circ\pi_{1}\,d\lambda+2\int_{X\times X}a_{n}\circ\pi_{2}\,d\lambda=2\int_{X}a_{n}\,d\mu+2\int_{X}a_{n}\,d\gamma_{c}.$$

Write $\varepsilon_{n}$ for the right-hand side. Since $a_{n}\le q$, monotonicity gives $\varepsilon_{n}\le2M_{2}(\mu)+2M_{2}(\gamma_{c})$, with the second moments of The Second Moment of a Borel Probability Measure on a Hilbert Space and the Probability Measures with Finite Second Moment §moment; here $M_{2}(\mu)<\infty$ because $\mu\in\mathcal{P}_{2}(X)$, and $M_{2}(\gamma_{c})<\infty$ because $\gamma_{c}\in\mathcal{P}_{2}(X)$ by Moments of a Diagonal Gaussian Measure on a Hilbert Space: Coordinate Covariances, Finite Second Moment, and Exponential Moments of the Squared Norm §moment. Thus $I(\pi^{(n)})\le\varepsilon_{n}<\infty$.

Step 3 (claim 1). Since $\mu\in\mathcal{P}_{2}(X)$, $\pi^{(n)}\in\Pi(\mu,\mu^{(n)})$ and $I(\pi^{(n)})<\infty$, the converse part of Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §cost-finite gives $\mu^{(n)}\in\mathcal{P}_{2}(X)$.

For the head, note that $p_{n}$ is linear (each coordinate $x\mapsto\langle x,e_{k}\rangle$ is linear) and that $p_{n}(p_{n}^{*}(y))=y$ for $y\in\mathbb{R}^{n}$ by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity; hence $p_{n}(P_{n}y)=p_{n}(y)$ and $p_{n}(Q_{n}y)=p_{n}(y)-p_{n}(P_{n}y)=0$ for $y\in X$. Consequently $p_{n}(T_{n}(x,g))=p_{n}(x)+0=p_{n}(x)$, that is, $p_{n}\circ\pi_{2}\circ\Psi_{n}=p_{n}\circ\pi_{1}$ on $X\times X$. By ($\dagger$), applied twice to each three-fold composite, and Step 1,

$$(p_{n})_{\#}\mu^{(n)}=(p_{n}\circ\pi_{2}\circ\Psi_{n})_{\#}\lambda=(p_{n}\circ\pi_{1})_{\#}\lambda=(p_{n})_{\#}\bigl((\pi_{1})_{\#}\lambda\bigr)=(p_{n})_{\#}\mu .$$

This proves claim 1 for the arbitrary $n$ fixed above.

Step 4 (a bound on $W_{2}$). Both $\mu^{(n)}$ and $\mu$ belong to $\mathcal{P}_{2}(X)$, so $W_{2}(\mu^{(n)},\mu)$ is defined. With $\sigma(x,y)=(y,x)$ the swap of $X\times X$, Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §swap gives $\sigma_{\#}\pi^{(n)}\in\Pi(\mu^{(n)},\mu)$ and $I(\sigma_{\#}\pi^{(n)})=I(\pi^{(n)})$. By The Quadratic Wasserstein Distance on a Hilbert Space §distance and Step 2,

$$W_{2}(\mu^{(n)},\mu)^{2}\le I(\sigma_{\#}\pi^{(n)})=I(\pi^{(n)})\le\varepsilon_{n}.$$

Step 5 (claim 2). Let $\eta\in\{\mu,\gamma_{c}\}$. The function $q$ is integrable with respect to $\eta$, being nonnegative with $\int_{X}q\,d\eta=M_{2}(\eta)<\infty$ (Measure Spaces and the Lebesgue Integral: Standing Notation §integral). For every $y\in X$, by Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §coordinates the sequence $(X_{m})_{m\in\mathbb{N}}$ is exhausting, $P_{m}$ is the orthogonal projection onto $X_{m}$ and $Q_{m}y=y-P_{m}y$, so Exhausting Sequences of Finite-Dimensional Subspaces in a Separable Real Hilbert Space, and Their Projections §tail gives that $(Q_{m}y)_{m\in\mathbb{N}}$ converges to $0_{X}$, that is, $|Q_{m}y|\to0$, and hence $a_{m}(y)=|Q_{m}y|^{2}\to0$. Since $|a_{m}(y)|\le q(y)$ for all $m$ and $y$, claim 3 of Dominated Convergence Theorem gives $\int_{X}a_{m}\,d\eta\to0$ as $m\to\infty$. Hence $\varepsilon_{m}\to0$.

Now let $\varepsilon>0$ be real and choose $m_{0}\in\mathbb{N}$ with $\varepsilon_{m}<\varepsilon^{2}$ for every $m\ge m_{0}$. By Step 4, applied with each $m\ge m_{0}$ in place of $n$, $W_{2}(\mu^{(m)},\mu)^{2}<\varepsilon^{2}$, and as $W_{2}(\mu^{(m)},\mu)\ge0$ this gives $W_{2}(\mu^{(m)},\mu)<\varepsilon$. Thus $\lim_{m}W_{2}(\mu^{(m)},\mu)=0$. Since $(\mu^{(m)})_{m\in\mathbb{N}}$ is a sequence in $\mathcal{P}_{2}(X)$ by claim 1 and $\mu\in\mathcal{P}_{2}(X)$, Wasserstein Convergence on a Hilbert Space: Weak Convergence, Integrals of Continuous Functions of Quadratic Growth, Convergence from Weak Convergence with Uniformly Integrable Second Moments, and Compactness §weak gives $\mu^{(m)}\Rightarrow\mu$. This proves claim 2.