Proof of Couplings on a Hilbert Space: Tightness, Closedness under Weak Convergence, and Lower Semicontinuity of the Quadratic Cost

Each result cited below is universally quantified over the data in its own statement and is applied with the data named at the point of citation.

Throughout, $\pi_{1},\pi_{2}:X\times X\to X$ are the coordinate maps of Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §pairs. They are Borel by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §product-sigma, and they are continuous: by Properties of the Product of Two Real Inner Product Spaces §coordinates, applied with $E_{1}=E_{2}=X$, each $\pi_{i}$ is linear with $|\pi_{i}w|\le|w|$, so $d(\pi_{i}z,\pi_{i}z')=|\pi_{i}(z-z')|\le|z-z'|=d(z,z')$ for $z,z'\in X\times X$ (Real Inner Product Space §distance); thus $\pi_{i}$ is Lipschitz with constant $1$ (Lipschitz Map Between Metric Spaces) and continuous by A Lipschitz Map is Uniformly Continuous. Push-forwards are those of Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §pushforward, so $((\pi_{i})_{\#}\lambda)(A)=\lambda(\pi_{i}^{-1}(A))$ for a Borel measure $\lambda$ on $X\times X$ and $A\in\mathcal{B}(X)$. Let $\rho$ be the product metric on $X\times X$; by Properties of the Product of Two Real Inner Product Spaces §metric, applied with $E_{1}=E_{2}=X$, the metric spaces $(X\times X,d)$ and $(X\times X,\rho)$ have the same open subsets.

Step 1 (claim 1). Let $\mathcal{M},\mathcal{N}$ be as in claim 1 and let $\varepsilon>0$ be real. First choose, by tightness of $\mathcal{M}$ (Tight Family of Borel Measures on a Metric Space §tight, with $\varepsilon/2$ in place of $\varepsilon$), a set $K\subseteq X$ compact in $(X,d)$ with $\mu(X\setminus K)\le\varepsilon/2$ for every $\mu\in\mathcal{M}$; then choose likewise, by tightness of $\mathcal{N}$, a set $K'\subseteq X$ compact in $(X,d)$ with $\nu(X\setminus K')\le\varepsilon/2$ for every $\nu\in\mathcal{N}$. By A Product of Compact Subsets is Compact in the Product Metric, applied with $(X,d_{X})=(Y,d_{Y})=(X,d)$ and $L=K'$, the set $K\times K'$ is compact in $X\times X$ for the topology of open subsets of $(X\times X,\rho)$. This topology is the topology of open subsets of $(X\times X,d)$, as recalled above, and compactness of a subset depends only on the topology through the subspace topology (Compact Topological Space and Compact Subset); so $K\times K'$ is compact in $(X\times X,d)$.

Now let $\mu\in\mathcal{M}$, $\nu\in\mathcal{N}$ and $\pi\in\Pi(\mu,\nu)$. The sets $X\setminus K$ and $X\setminus K'$ belong to $\mathcal{B}(X)$ by Compact Subsets of a Metric Space are Closed and Borel §borel, and since $\pi_{1},\pi_{2}$ are Borel the sets $A_{1}=\pi_{1}^{-1}(X\setminus K)$ and $A_{2}=\pi_{2}^{-1}(X\setminus K')$ belong to $\mathcal{B}(X\times X)$. A pair $z$ lies outside $K\times K'$ exactly when $\pi_{1}z\notin K$ or $\pi_{2}z\notin K'$, so $(X\times X)\setminus(K\times K')=A_{1}\cup A_{2}$. By claim 4 of Basic Properties of a Measure, applied to the measure $\pi$ on $(X\times X,\mathcal{B}(X\times X))$ and to the sequence $A_{1},A_{2},\varnothing,\varnothing,\dots$, whose sum of measures is $\pi(A_{1})+\pi(A_{2})$ because $\pi(\varnothing)=0$ (Measure, Measure Space, and Probability Measure), and by the marginal conditions of Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §coupling,

$$\pi\bigl((X\times X)\setminus(K\times K')\bigr)\le\pi(A_{1})+\pi(A_{2})=((\pi_{1})_{\#}\pi)(X\setminus K)+((\pi_{2})_{\#}\pi)(X\setminus K')=\mu(X\setminus K)+\nu(X\setminus K')\le\varepsilon .$$

The compact set $K\times K'$ does not depend on $\mu$, $\nu$ or $\pi$, so by Tight Family of Borel Measures on a Metric Space §tight the set of all such $\pi$ is tight in $X\times X$.

For the final sentence of claim 1, let $\mu,\nu\in\mathcal{P}(X)$. 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, and $\mu$ is a Borel measure on it with $\mu(X)=1<\infty$, so $\{\mu\}$ is tight by Ulam's Theorem: a Finite Borel Measure on a Complete Separable Metric Space is Tight §tight; likewise $\{\nu\}$ is tight. The set of claim 1 formed with $\mathcal{M}=\{\mu\}$ and $\mathcal{N}=\{\nu\}$ is $\Pi(\mu,\nu)$, which is therefore tight in $X\times X$.

Step 2 (claim 2). Let $\mu_{j},\nu_{j},\mu,\nu,\pi_{j},\pi$ be as in claim 2. All $\pi_{j}$ and $\pi$ are Borel measures on $(X\times X,d)$ of total mass $1$. By Weak Convergence of Finite Borel Measures is Preserved by Continuous Maps between Metric Spaces §weak, applied with $(Y,d_{Y})=(X\times X,d)$, $(Z,d_{Z})=(X,d)$, the continuous map $T=\pi_{1}$ and the sequence $(\pi_{j})_{j\in\mathbb{N}}$ with limit $\pi$, we get $(\pi_{1})_{\#}\pi_{j}\Rightarrow(\pi_{1})_{\#}\pi$; the image measures there are the push-forwards used here, and $(\pi_{1})_{\#}\pi$ is a Borel measure on $(X,d)$ of total mass $\pi(X\times X)=1$. Since $\pi_{j}\in\Pi(\mu_{j},\nu_{j})$, Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §coupling gives $(\pi_{1})_{\#}\pi_{j}=\mu_{j}$, so the sequence $(\mu_{j})_{j\in\mathbb{N}}$ of finite measures on $(X,\mathcal{B}(X))$ converges weakly both to $(\pi_{1})_{\#}\pi$ and, by hypothesis, to $\mu$. The set $X$ is nonempty, as $0_{X}\in X$, so claim 2 of Lipschitz Test Functions Determine a Finite Borel Measure, and Uniqueness of Weak Limits, applied to the metric space $(X,d)$, gives $(\pi_{1})_{\#}\pi=\mu$. The same argument with $\pi_{2}$, $\nu_{j}$ and $\nu$ in place of $\pi_{1}$, $\mu_{j}$ and $\mu$ gives $(\pi_{2})_{\#}\pi=\nu$. As $\pi\in\mathcal{P}(X\times X)$, Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §coupling gives $\pi\in\Pi(\mu,\nu)$.

Step 3 (truncated costs). Let $c:X\times X\to\mathbb{R}$ be $c(z)=|\pi_{1}(z)-\pi_{2}(z)|^{2}$, a nonnegative function that is continuous on $(X\times X,d)$ and Borel, as recorded in Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §cost; by that definition $I(\lambda)=\int_{X\times X}c\,d\lambda$ for every coupling $\lambda$. For $N\in\mathbb{N}$ let $c_{N}(z)=\min\{c(z),N\}$. For real $a,b$ one has $|\min\{a,N\}-\min\{b,N\}|\le|a-b|$ (elementary, by cases). Hence, given $z\in X\times X$ and real $\varepsilon>0$, the $\delta>0$ provided for $c$ at $z$ by Continuous Map Between Metric Spaces also serves for $c_{N}$, so $c_{N}$ is continuous on $(X\times X,d)$; it is Borel, as a map into $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, by claims 2 and 3 of Borel Measurability and Bounded Integration on a Metric Space; and $0\le c_{N}(z)\le N$ for every $z$, so $c_{N}$ is bounded with bound $N$ (Bounded Real-Valued Function on a Set). Moreover $c_{N}\le c_{N+1}\le c$ pointwise, and for every $z$ there is $N\in\mathbb{N}$ with $N\ge c(z)$, for which $c_{N}(z)=c(z)$; hence $\sup_{N}c_{N}(z)=c(z)$ for every $z\in X\times X$.

For a Borel measure $\lambda$ on $X\times X$ with $\lambda(X\times X)=1$, claims 6(b) and 6(c) of Borel Measurability and Bounded Integration on a Metric Space, applied to the metric space $(X\times X,d)$, the measure $\lambda$, $f=c_{N}$ and $M=N$, show that $c_{N}$ is integrable with respect to $\lambda$, that $|\int c_{N}\,d\lambda|\le N$, and that its integral as a nonnegative measurable function is the same real number; we write $J_{N}(\lambda)$ for it. Since $c_{N}$ and $c$ are measurable functions into $[0,\infty]$ in the sense of Measure Spaces and the Lebesgue Integral: Standing Notation §measurable (for the measure space $(X\times X,\mathcal{B}(X\times X),\lambda)$, in force by Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §background) and $c_{N}\le c$, claim 1 of Linearity and Monotonicity of the Lebesgue Integral gives $J_{N}(\lambda)\le\int_{X\times X}c\,d\lambda$ in $[0,\infty]$.

Step 4 (claim 3). In the situation of claim 3, $\pi\in\Pi(\mu,\nu)$ by Step 2. Put $a_{j}=I(\pi_{j})$, a bounded sequence of real numbers by hypothesis, and $L=\liminf_{j}a_{j}$ as in Limit Inferior of a Bounded Sequence of Real Numbers. Fix $N\in\mathbb{N}$ and put $b_{j}=J_{N}(\pi_{j})$. By Step 3, $|b_{j}|\le N$ for every $j$, so $(b_{j})_{j\in\mathbb{N}}$ is bounded (Bounded Sequence of Real Numbers, with bound $N$, which is positive), and $b_{j}\le a_{j}$ for every $j$. Since $c_{N}$ is bounded and continuous on $(X\times X,d)$ and $\pi_{j}\Rightarrow\pi$, Weak Convergence of Finite Borel Measures on a Metric Space gives that $(b_{j})_{j\in\mathbb{N}}$ converges to $J_{N}(\pi)$. By claim 5 of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence, $\liminf_{j}b_{j}=J_{N}(\pi)$, and by claim 2 of that lemma, $\liminf_{j}b_{j}\le\liminf_{j}a_{j}$. Hence

$$J_{N}(\pi)\le L\qquad\text{for every }N\in\mathbb{N}.$$

By Step 3 the functions $c_{N}:X\times X\to[0,\infty]$ are measurable, nondecreasing in $N$ at every point, with pointwise supremum $c$. By Monotone Convergence Theorem, applied to the measure space $(X\times X,\mathcal{B}(X\times X),\pi)$ and the sequence $(c_{N})_{N\in\mathbb{N}}$,

$$I(\pi)=\int_{X\times X}c\,d\pi=\sup_{N}J_{N}(\pi).$$

Since $L$ is a real upper bound of $\{J_{N}(\pi):N\in\mathbb{N}\}$, the least upper bound in $[0,\infty]$ satisfies $\sup_{N}J_{N}(\pi)\le L$. Therefore $I(\pi)\le L=\liminf_{j}I(\pi_{j})$, and in particular $I(\pi)<\infty$. This proves claim 3.