Is the space of probability measures on a compact set is compact w.r.t Wasserstein metric?

Solution 1:

Let $(\mu_n)_{n\ge1}$ be a sequence in $\Delta(C)$. It is obviously tight because $\mu_n(K)=1\ge1-\varepsilon$ for the compact $K:=C$ and for all $n\ge1$ and $\varepsilon>0$. By Prokhorov's theorem, we know that $(\mu_n)_{n\ge1}$ has some limit point $\mu$ w.r.t. the topology of weak convergence. By Skorokhod's representation theorem, there exist a probabiltiy space $(\Omega,\mathcal A,\mathbb P)$ and random variables $X,X_1,X_2,\ldots:\Omega\to C$ such that $X$ has law $\mu$, $X_n$ has law $\mu_n$ for every $n\ge1$ and $$d(X_n,X)\xrightarrow[n\to\infty]{}0\quad\text{$\mathbb P$-a.s.}$$ Now because $C$ is compact, there exists a constant $a>0$ such that $d(X_n,X)\le a$, thus $d(X_n,X)\le a^p$ for every $n\ge1$. Hence $$\mathbb E[d(X_n,X)^p]\xrightarrow[n\to\infty]{}0$$ by dominated convergence. Thus we have couplings $\gamma_n\in\Pi(\mu_n,\mu)$ such that $$\left(\int d(x,y) ^p\,\mathrm d\gamma_n(x,y)\right)^{\!\frac1p}\xrightarrow[n\to\infty]{}\:0.$$ Since $W_p(\mu_n,\mu)$ is the infimum of the left-hand side over $\Pi(\mu_n,\mu)$, we deduce that $$W_p(\mu_n,\mu)\xrightarrow[n\to\infty]{}0,$$ which means that $\mu$ is also a limit point of $(\mu_n)_{n\ge1}$ for the Wasserstein-$p$ metric. Hence $(\Delta(C),W_p)$ is compact.