Proof of The Couplings of Two Probability Measures on Euclidean Space are Closed under Weak Convergence

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named where it is used.

By Probability Measures on Euclidean Space and Random Vectors: Standing Notation (iv) the push-forward $\mu'=(\mathrm{pr}_{1})_{\#}\pi$ belongs to $\mathcal{P}(\mathbb{R}^{m})$, and likewise $\nu'=(\mathrm{pr}_{2})_{\#}\pi$; by Probability Measures on Euclidean Space and Random Vectors: Standing Notation (i) each of $\mu$, $\mu'$, $\nu$, $\nu'$ is a finite Borel measure on the metric space $(\mathbb{R}^{m},d_{E})$. By Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost (i) it suffices to prove that $\mu'=\mu$ and $\nu'=\nu$.

Let $f:\mathbb{R}^{m}\to\mathbb{R}$ be bounded and Lipschitz as a map from $(\mathbb{R}^{m},d_{E})$ to $\mathbb{R}$ with the absolute-value metric. By A Lipschitz Map is Uniformly Continuous and Continuous and A Uniformly Continuous Map Between Metric Spaces Is Continuous the map $f$ is continuous on $\mathbb{R}^{m}$, hence Borel by claim 3 of Borel Measurability and Bounded Integration on a Metric Space. By Continuity of the Coordinate Projections and of the Quadratic Cost Function, and Passage of a Cost Bound to a Weak Limit (i) the projection $\mathrm{pr}_{1}$ is continuous on $\mathbb{R}^{m+m}$, so $f\circ\mathrm{pr}_{1}$ is continuous on $\mathbb{R}^{m+m}$ by claim 3 of Semicontinuity and Continuity Under Composition with a Continuous Map; it is bounded by any bound for $f$, and Borel by claim 3 of Borel Measurability and Bounded Integration on a Metric Space. Every bounded Borel map is integrable with respect to every probability measure by Probability Measures on Euclidean Space and Random Vectors: Standing Notation (iii).

The change-of-variables formula of Probability Measures on Euclidean Space and Random Vectors: Standing Notation (iv), applied to the Borel map $\mathrm{pr}_{1}$ and the bounded Borel map $f$, gives

$$\int_{\mathbb{R}^{m}}f\,d\mu'=\int_{\mathbb{R}^{m+m}}f\circ\mathrm{pr}_{1}\,d\pi, \qquad \int_{\mathbb{R}^{m+m}}f\circ\mathrm{pr}_{1}\,d\pi_{n}=\int_{\mathbb{R}^{m}}f\,d\bigl((\mathrm{pr}_{1})_{\#}\pi_{n}\bigr)=\int_{\mathbb{R}^{m}}f\,d\mu$$

for every $n\in\mathbb{N}$, the last equality because $(\mathrm{pr}_{1})_{\#}\pi_{n}=\mu$ by Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost (i).

Since $f\circ\mathrm{pr}_{1}$ is bounded and continuous and $(\pi_{n})_{n\in\mathbb{N}}$ converges weakly to $\pi$, Weak Convergence of Finite Borel Measures on a Metric Space gives that the sequence whose $n$th term is $\int_{\mathbb{R}^{m+m}}f\circ\mathrm{pr}_{1}\,d\pi_{n}$ converges to $\int_{\mathbb{R}^{m+m}}f\circ\mathrm{pr}_{1}\,d\pi$. By the display that sequence is the constant sequence with value $\int_{\mathbb{R}^{m}}f\,d\mu$, which converges to $\int_{\mathbb{R}^{m}}f\,d\mu$: for every positive real $\varepsilon$ the difference of a term and that value is $0$, whose absolute value is $0<\varepsilon$ by claim 1 of Properties of the Absolute Value in an Ordered Field. By Uniqueness of Limits and Boundedness of Convergent Real Sequences a sequence of real numbers has at most one limit, so

$$\int_{\mathbb{R}^{m}}f\,d\mu'=\int_{\mathbb{R}^{m+m}}f\circ\mathrm{pr}_{1}\,d\pi=\int_{\mathbb{R}^{m}}f\,d\mu .$$

This holds for every bounded Lipschitz $f:\mathbb{R}^{m}\to\mathbb{R}$, so claim 1 of Lipschitz Test Functions Determine a Finite Borel Measure, and Uniqueness of Weak Limits, applied to the metric space $(\mathbb{R}^{m},d_{E})$ and the finite measures $\mu'$ and $\mu$ on its Borel $\sigma$-algebra, gives $\mu'=\mu$.

The same argument with $\mathrm{pr}_{2}$ in place of $\mathrm{pr}_{1}$, using $(\mathrm{pr}_{2})_{\#}\pi_{n}=\nu$ from Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost (i) and the continuity of $\mathrm{pr}_{2}$ from Continuity of the Coordinate Projections and of the Quadratic Cost Function, and Passage of a Cost Bound to a Weak Limit (i), gives $\nu'=\nu$. Hence $\pi\in\Pi(\mu,\nu)$.