Proof of Cauchy Sequences in the Quadratic Wasserstein Space and Weakly Convergent Sequences on a Hilbert Space are Tight

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

Throughout, $(X,d)$ is complete and separable by Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §space, every member of $\mathcal{P}(X)$ is a Borel measure of total mass $1<\infty$, and so every $\nu\in\mathcal{P}(X)$ satisfies the hypotheses of Ulam's Theorem: a Finite Borel Measure on a Complete Separable Metric Space is Tight §tight. Open balls are written $B(a,r)$ and closed balls $\bar{B}(a,r)=\{x\in X:d(x,a)\le r\}$, as in Closed Ball in a Metric Space. Measure-theoretic facts (monotonicity, $\nu(X\setminus E)=1-\nu(E)$ for $\nu\in\mathcal{P}(X)$, and countable subadditivity, applied to two sets by padding with empty sets) are Basic Properties of a Measure §monotone, Basic Properties of a Measure §differences and Basic Properties of a Measure §subadditivity. Finite unions of finite sets are finite by Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §finite-union, a compact subset $K$ of $(X,d)$ and its complement $X\setminus K$ are Borel by Compact Subsets of a Metric Space are Closed and Borel §borel, so that $\nu(K)$ and $\nu(X\setminus K)$ are defined for $\nu\in\mathcal{P}(X)$, and such a $K$ is totally bounded by A Compact Subset of a Metric Space is Totally Bounded, so that for every real $r>0$ there is a finite $F\subseteq X$ with $K\subseteq\bigcup_{a\in F}B(a,r)$ by Totally Bounded Subset of a Metric Space.

Step 0 (a compact set built from finite sets). Let $F_{k}\subseteq X$ be a finite set for every $k\in\mathbb{N}$, and put

$$C_{k}=\bigcup_{a\in F_{k}}\bar{B}(a,2/k),\qquad A=\bigcap_{k\in\mathbb{N}}C_{k}$$

(with $C_{k}=\varnothing$ if $F_{k}=\varnothing$). Each closed ball is closed by claim 3 of Elementary Properties of the Closed Ball in a Metric Space, so each $C_{k}$ is closed by claims 1 and 2 of Complements, Unions and Intersections of Closed Sets in a Topological Space, and $A$ is closed by claim 3 there; hence $C_{k}$ and $X\setminus C_{k}$ are Borel by claim 1 of Borel Measurability and Bounded Integration on a Metric Space. The set $A$ is totally bounded: given a real $r>0$, choose $k\in\mathbb{N}$ with $3/k<r$; then $A\subseteq C_{k}\subseteq\bigcup_{a\in F_{k}}B(a,r)$, because $d(x,a)\le2/k<r$, and $F_{k}$ is finite, so Totally Bounded Subset of a Metric Space applies. Since $(X,d)$ is complete, $A$ is compact by A Closed Totally Bounded Subset of a Complete Metric Space is Compact §compact. Finally $X\setminus A=\bigcup_{k}(X\setminus C_{k})$, so for every $\nu\in\mathcal{P}(X)$ countable subadditivity gives

$$\nu(X\setminus A)\le\sum_{k\in\mathbb{N}}\nu(X\setminus C_{k}).\tag{0}$$

Step 1 (clause 1: choices). Let $(\mu_{m})_{m\in\mathbb{N}}$ be a Cauchy sequence in $(\mathcal{P}_{2}(X),W_{2})$ and fix a real $\varepsilon>0$. For $j\in\mathbb{N}$ put $\delta_{j}=\varepsilon2^{-j-1}$. The choices are made in the following order, for each $j\in\mathbb{N}$: first $N_{j}$, then the compact sets $K_{j,i}$, then the finite sets $F_{j,i}$. By Cauchy Sequence in a Metric Space, applied with the positive real $j^{-1}\sqrt{\delta_{j}}$ (the square root of Existence and Uniqueness of the Nonnegative Square Root), choose $N_{j}\in\mathbb{N}$ with $W_{2}(\mu_{m},\mu_{\ell})<j^{-1}\sqrt{\delta_{j}}$ for all $m,\ell\ge N_{j}$. For each $i\in[N_{j}]$ choose, by Ulam's Theorem: a Finite Borel Measure on a Complete Separable Metric Space is Tight §tight, a compact $K_{j,i}\subseteq X$ with $\mu_{i}(X\setminus K_{j,i})\le\delta_{j}$, and then a finite $F_{j,i}\subseteq X$ with $K_{j,i}\subseteq\bigcup_{a\in F_{j,i}}B(a,1/j)$. Put $F_{j}=\bigcup_{i\in[N_{j}]}F_{j,i}$, a finite set, and let $C_{j}$ and $A$ be as in Step 0.

Step 2 (clause 1: the early terms). Let $i\le N_{j}$. Every point of $K_{j,i}$ lies in some $B(a,1/j)\subseteq\bar{B}(a,2/j)$ with $a\in F_{j,i}\subseteq F_{j}$, so $K_{j,i}\subseteq C_{j}$ and $\mu_{i}(X\setminus C_{j})\le\mu_{i}(X\setminus K_{j,i})\le\delta_{j}$.

Step 3 (clause 1: the late terms). Let $n>N_{j}$ and write $K=K_{j,N_{j}}$. Both $\mu_{N_{j}}$ and $\mu_{n}$ lie in $\mathcal{P}_{2}(X)$, so by Existence of an Optimal Coupling of Two Borel Probability Measures with Finite Second Moment on a Hilbert Space §existence there is $\pi\in\Pi(\mu_{N_{j}},\mu_{n})$ with $I(\pi)=W_{2}(\mu_{N_{j}},\mu_{n})^{2}<\delta_{j}/j^{2}$, the inequality by Step 1 and claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. Let $\varphi(z)=|\pi_{1}(z)-\pi_{2}(z)|^{2}$, a nonnegative Borel function on $X\times X$ with $I(\pi)=\int\varphi\,d\pi$, by Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §cost. By The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §markov, applied on $(X\times X,\mathcal{B}(X\times X),\pi)$ with $f=\varphi$ and $t=1/j^{2}$, the set $E=\{z:1/j^{2}\le\varphi(z)\}$ is Borel and $\pi(E)\le j^{2}I(\pi)<\delta_{j}$.

We claim $\pi_{2}^{-1}(X\setminus C_{j})\subseteq\pi_{1}^{-1}(X\setminus K)\cup E$. Let $z\in X\times X$ with $\pi_{1}(z)\in K$ and $z\notin E$. Then $d(\pi_{1}(z),\pi_{2}(z))^{2}=\varphi(z)<1/j^{2}$, so $d(\pi_{1}(z),\pi_{2}(z))<1/j$ by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field; and $\pi_{1}(z)\in B(a,1/j)$ for some $a\in F_{j,N_{j}}\subseteq F_{j}$, by Step 1. By the triangle inequality of the metric $d$ (The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §metric), $d(a,\pi_{2}(z))<2/j$, so $\pi_{2}(z)\in C_{j}$. This proves the claim.

By the marginal conditions of Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §coupling, the claim, monotonicity and subadditivity,

$$\mu_{n}(X\setminus C_{j})=\pi\bigl(\pi_{2}^{-1}(X\setminus C_{j})\bigr)\le\pi\bigl(\pi_{1}^{-1}(X\setminus K)\bigr)+\pi(E)\le\mu_{N_{j}}(X\setminus K)+\delta_{j}\le2\delta_{j},$$

the last step by the choice of $K=K_{j,N_{j}}$ in Step 1.

Step 4 (clause 1: conclusion). By Steps 2 and 3, $\mu_{n}(X\setminus C_{j})\le2\delta_{j}=\varepsilon2^{-j}$ for all $n,j\in\mathbb{N}$. By Step 0, $A$ is compact and, by (0), $\mu_{n}(X\setminus A)\le\sum_{j}\varepsilon2^{-j}=\varepsilon$ for every $n\in\mathbb{N}$. As $\varepsilon>0$ was arbitrary, the set $\{\mu_{n}:n\in\mathbb{N}\}$ is tight by Tight Family of Borel Measures on a Metric Space §tight, that is, $(\mu_{n})$ is tight by Tight Family of Borel Measures on a Metric Space §sequence.

Step 5 (clause 2: choices). Let $(\mu_{j})_{j\in\mathbb{N}}$ in $\mathcal{P}(X)$ converge weakly to $\mu\in\mathcal{P}(X)$; we index the sequence by $n$, that is, we write $\mu_{n}$ for the $n$-th term, so that the sequence is $(\mu_{n})_{n\in\mathbb{N}}$ (the letter $j$ is not used for it below). Fix a real $\varepsilon>0$ and put $\delta_{k}=\varepsilon2^{-k}$ for $k\in\mathbb{N}$. For each $k\in\mathbb{N}$ the choices are made in the order: $K$, $F$, $n_{k}$, then $K_{k,n}$ and $F_{k,n}$ for $n<n_{k}$. By Ulam's Theorem: a Finite Borel Measure on a Complete Separable Metric Space is Tight §tight choose a compact $K\subseteq X$ with $\mu(X\setminus K)\le\delta_{k}/2$, and then a finite $F\subseteq X$ with $K\subseteq G$, where $G=\bigcup_{a\in F}B(a,1/k)$. Each open ball is open by Open Ball in a Metric Space is Open, so $G$ is open, the open sets forming a topology by Metric Open Sets Form a Topology, hence Borel by claim 1 of Borel Measurability and Bounded Integration on a Metric Space. Then $\mu(G)\ge\mu(K)=1-\mu(X\setminus K)>1-\delta_{k}$. The set $X$ is nonempty, and members of $\mathcal{P}(X)$ are probability measures on $(X,\mathcal{B}(X))$, so claim 3 of Portmanteau Theorem on a Metric Space gives $1-\delta_{k}<\mu(G)\le\liminf_{n}\mu_{n}(G)$, the sequence $(\mu_{n}(G))$ being bounded by $1$. By Limit Inferior of a Bounded Sequence of Real Numbers this limit inferior is the least upper bound of the numbers $\inf\{\mu_{m}(G):m\ge n\}$, so $1-\delta_{k}$ is not an upper bound of them: choose $n_{k}\in\mathbb{N}$ with $\inf\{\mu_{m}(G):m\ge n_{k}\}>1-\delta_{k}$. Hence $\mu_{n}(X\setminus G)=1-\mu_{n}(G)<\delta_{k}$ for every $n\ge n_{k}$. For each of the finitely many $n\in\mathbb{N}$ with $n<n_{k}$, choose by Ulam's Theorem: a Finite Borel Measure on a Complete Separable Metric Space is Tight §tight a compact $K_{k,n}$ with $\mu_{n}(X\setminus K_{k,n})\le\delta_{k}$, and then a finite $F_{k,n}$ with $K_{k,n}\subseteq\bigcup_{a\in F_{k,n}}B(a,1/k)$. Let $F_{k}$ be the union of $F$ and of the sets $F_{k,n}$, $n<n_{k}$, a finite set, and let $C_{k}$ and $A$ be as in Step 0.

Step 6 (clause 2: conclusion). Since $B(a,1/k)\subseteq\bar{B}(a,2/k)$, one has $G\subseteq C_{k}$ and $K_{k,n}\subseteq C_{k}$ for $n<n_{k}$. Hence, by monotonicity and Step 5, $\mu_{n}(X\setminus C_{k})\le\mu_{n}(X\setminus G)<\delta_{k}$ for $n\ge n_{k}$, and $\mu_{n}(X\setminus C_{k})\le\mu_{n}(X\setminus K_{k,n})\le\delta_{k}$ for $n<n_{k}$. By Step 0, $A$ is compact and, by (0), $\mu_{n}(X\setminus A)\le\sum_{k}\varepsilon2^{-k}=\varepsilon$ for every $n\in\mathbb{N}$. As in Step 4, $(\mu_{n})$ is tight by Tight Family of Borel Measures on a Metric Space §tight and Tight Family of Borel Measures on a Metric Space §sequence.