# Countable convergence-determining class for weak convergence of probability measures

Suppose that $$E$$ is a Polish space.

Portmanteau theorem asserts that a sequence $$(\mu_n)$$ of Borel probability measures weakly converges to a Borel probability measure $$\mu$$ (shortly, $$\mu_n\overset{w}{\to\mu}$$) if and only if $$\limsup_n \mu_n(C)\le \mu(C)$$ for all closed set $$C\subset E$$. My question is whether there exists a countable convergence-determining class of closed sets. Namely, if there exists a countable collection $$\mathcal C$$ of closed subsets of $$E$$ such that $$\limsup_n \mu_n(C)\le \mu(C)$$ for all closed set $$C\in\mathcal C$$ implies that $$\mu_n\overset{w}{\to\mu}$$.

Let $$E$$ be any separable space. The assertion is equivalent to ($$*$$) $$\liminf_{n\to \infty} \mu_n(U) \geq \mu(U)$$ for any open $$U \subset E$$. Since $$E$$ is separable, there is a countable base $$\cal{U}$$, which is $$\cap$$- and $$\cup$$-stable, for the open sets in $$E$$. This $$\cal{U}$$ is "convergence-determining" for open sets. Let $$U$$ be an arbitrary open subset of $$E$$. Then there is a sequence $$U_k$$ in $$\cal{U}$$ with $$U_k \uparrow U$$. But then $$\liminf_n \mu_n(U) \geq \liminf_n \mu_n(U_k) \geq \mu(U_k)$$ for any $$k \in \mathbb{N}$$, since $$\mu_n(U) \geq \mu_n({U_k})$$. Since $$\mu(U_k) \uparrow \mu(U)$$ ($$*$$) follows.