Countable convergence-determining class for weak convergence of probability measures

Question

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 $\mu_n\overset{w}{\to\mu}$.

Answer 1

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.