2
$\begingroup$

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

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.