Let $X$ be a metric space. If it is necessary, one can assume that it is complete and separable, or even compact. Consider a sequence of Borel probability measures $(\mu_n)$ on $X$ that weakly converges to a Borel probability measure $\mu$ on $X$. It is well known that there are equivalent characterizations for this convergence, namely, that
\begin{equation*}
\limsup\limits_{n\to +\infty} \mu_n(F)\leq \mu(F)
\end{equation*}
for every closed set $F\subseteq X$ and that
\begin{equation*}
\liminf\limits_{n\to +\infty} \mu_n(G)\geq \mu(G)
\end{equation*}
for every open set $G\subseteq X$. My question, roughly speaking, if these estimates, after passing to a subsequence of $(\mu_n)$, can be made uniform over all Borel subsets of $X$?
Let me explain what I mean. Suppose that $\mu$ is fully supported on $X$, that is, $\mu(G)>0$ for every nonempty open set $G\subseteq X$, fix $\varepsilon>0$. Can I find a subsequence $(\mu_{n_k})$ such that for any Borel set $V\subseteq \mathrm{X}$ and any $k\in \mathbb{N}$ one has
\begin{equation*}
\mu_{n_k}(V)\leq (1+\varepsilon)\mu(V)?
\end{equation*}
If $\mu$ is not fully supported, one can consider arbitrary fully supported probability measure $\nu $ on $X$ and replace $(1+\varepsilon) \mu(V)$ with $\mu(V)+\varepsilon \nu(V)$ in the expression above. Are there some sufficient conditions for this property to hold? Or it fails dramatically?
The only thoughts on this I have are as follows. Suppose we are given a countable family $\mathcal{V}$ of closed subsets of $X$ such that $\mu(F)>0$ for each $F\in \mathcal{V}$. As it seems to me, standard Cantor's diagonal argument and the equivalent characterization of the weak convergence above imply that there is a subsequence $(\mu_{n_k})$ such that the desired property holds for all sets from $\mathcal{V}$. Explicitly,
\begin{equation*}
\mu_{n_k}(F)\leq (1+\varepsilon)\mu(F)
\end{equation*}
for every $F\in \mathcal{V}$ and for every $k\in \mathbb{N}$. Does this imply the inequality for each Borel subset of $X$, if we take an appropriate family $\mathcal{V}$, for instance, a generating set for the Borel sigma-algebra on $X$ in the case of its countable generatedness? Or there are some other arguments?
Will be greatful for any help!
Best Answer
No, this is not the case. Consider $X = [0,1]$ and $\mu_n = \frac 1{2^n} \sum_{i=1}^{2^n} 1_{i/2^n}$. Observe that $(\mu_n)$ converges weakly to the Lebesgue measure $\mu$, which is fully supported. Let $V = \{1/2\}$. Then $\mu(V) = 0$, but $\mu_n(V) > 0$ for all $n$, so for any $\varepsilon > 0$ we have $\mu_n(V) > (1+\varepsilon)\mu(V)$. In fact, one can take $V = \{i/2^n : n \in \mathbb{N}, i \in \mathbb{N} \cap [1,2^n]\}$ to get $\mu(V) = 0$ (because $V$ is countable), but $\mu_n(V) = 1$ for all $n$.