vague and weak topology on separable Hilbert space

Let $\mu_n$ and $\mu$ be Borel probability measures on a separable Hilbert space $X$. Assume that $\mu_n\to\mu$ vaguely. Is it necessarily true that $\mu_n\to\mu$ weakly?

I know that this is true if $X$ is a locally compact Polish space (because enlargements of compacta stay compact and we can get tightness, see below) . I have only seen the vague topology used on locally compact spaces so I suspect that things behave badly outside this setting.

A related question in case the answer is "no": denote the set of Borel probability measures on $X$ by $P(X)$. Is it at least true that the Borel sigma algebras generated from the vague and weak topologies on $P(X)$ are equal? I wanted to use something like the Lusin-Suslin theorem (11.4 here) but the vague topology is not complete when restricted to probability measures.

Proof when $X=\mathbb R^d$: let $K_1$ compact such that $\mu(K_1)>1-\epsilon$. Let $K_2=\{x:\|x - K_1\|\le \epsilon\}$, which is also compact. There exists a continuous function $0\le f\le 1$ with $f\equiv1$ on $K_1$ and $f\equiv0$ outside $K_2$. Then $$\mu_n(K_2) \ge \int fd\mu_n \to \int fd\mu \ge \mu(K_1)$$ is larger then $1-2\epsilon$ for $n$ large. Since $K_2$ is compact $(\mu_n)$ is tight, has a weak limit, and that limit has to be $\mu$. This is easily extended to any locally compact space instead of $\mathbb R^d$ with a more painful construction of the set $K_2$.

The answer to both questions is no. In fact, for any $\mu_n,\mu\in P(X)$, $\mu_n\to\mu$ vaguely. This comes from the fact that $C_0(X)=0$.
To see this, assume $f\in C_0(X)$. For each $n$, there exists a compact set $K_n$ such that $|f|\le\frac1n$ outside $K_n$. Recall that compact sets in infinite dimensional Hilbert spaces have empty interior, and hence are nowhere dense. Defining $K:=\bigcup_nK_n$, we have that $K$ is nowhere dense by the Baire category theorem. But $f=0$ outside of $K$, and hence $f\equiv0$.
• The conclusion is correct, but the argument is slightly wrong. Baire category says that $K$ has empty interior, but it need not be nowhere dense ($K$ need not be closed, and its closure could have nontrivial interior). But anyway you don't need Baire here: for each $n$ the set $\{|f| > 1/n\}$ is both open and contained in $K_n$, so it must be empty because, as you said, $K_n$ has empty interior. Hence you have $|f| \le 1/n$ everywhere, and $n$ was arbitrary. – Nate Eldredge Sep 20 '17 at 21:58