Is the set of probability measures on a closed set closed in the weak topology?

Question

Let $X$ be a metric space and $A\subseteq X$ be a closed subset of $X$. Let $\mathcal{P}(X)$ denote the set of all probability measures on $X$ equipped with the weak topology. Is $\mathcal{P}(A)$ a closed set?

Answer 1

Sure, if you mean the space of probability measures supported on $A$. The function $g:X\to\mathbb{R}$ given by $g(x)=\min\{1,d(x,A)\}$ is bounded, continuous, and strictly positive exactly at the complement of $A$. So the set $$\mathcal{P}(A)=\bigg\{\mu\in\mathcal{P}(X)\bigg\vert \int g~\mathrm d\mu=0\bigg\}$$ is closed.

Answer 2

Over metric spaces, weak convergence is metrised by the Levy-Prokhorov metric, $$ \pi(\mu,\nu) := \inf\{\delta: \forall\textrm{Borel } E, \mu(E) \le \nu(E^\delta) + \delta, \nu(E) \le \mu(E^\delta) + \delta\},$$ where $E^\delta = \{x : d(x,E) \le \delta\}.$ Using this just makes the following a little neater, but you should be able to translate the argument into the language of convergence of expectation for bounded continuous functions.

Fix some $Q \in \mathcal{P}(A)^c,$ and let $Q(A) = 1-L$ for some $L > 0$ (which must hold, since otherwise $Q \in \mathcal{P}(A)$. Observe that $A^{1/n} \searrow A$ as $n \to \infty$ since $A$ is closed. This in turn means that $Q(A^{1/n}) \to Q(A) < 1$.

But now, let $N$ be such that $Q(A^{1/n}) < 1-L/2$ for $n > N$. Then for any $n \ge \max(N, 4/L),$ and any $\tilde{Q}: \pi(\tilde{Q},Q) < 1/n,$ we have $$ \tilde{Q}(A) \le Q(A^{1/n}) + 1/n \le 1- L/2 + L/4 < 1,$$ meaning that $\mathcal{P}(A)^c$ is open.