I'm solving such task:
Given is $(X, d)$ and $(\mu_n)$ - a compact metric space and sequence of probability radon measures on $X$. We define set $K$ as follows: $$ K = X \smallsetminus\bigcup\mathscr{U} $$ where $$ \mathscr{U} = \lbrace U \subset X : U \texttt{ is opean and } \mu_n(U) \to 0 \rbrace $$ Show that there is a subsequence $(\mu_{k_i})$ if $(\mu_n)$ and probability measure $\nu$ such that of $a \in K$, then for all but countably many $r > 0$: $$ \mu_{k_i}(B(a, r)) \to \nu(B(a, r) \cap K) $$ where $B(a, r) = \lbrace x \in X : d(x, a) \leq r\rbrace$ is closed ball.
And I'm not sure how I can finish it. Below my attempt.
First things first, if $K = \emptyset$ then $\nu$ does not exist. Then I'm starting with proof that $K \neq \emptyset$.
Assume otherwise, then $\bigcup\mathscr{U} = X$, and as X is compact, we can choose finite family $\mathscr{U}' \subset \mathscr{U}$ such that $\bigcup\mathscr{U}' = X$.
As for every $U \in \mathscr{U}'$, $\mu_n(U) \to 0$ and $|\mathscr{U}'| < \infty$ it is true that $$ \lim_n \sum_{U \in \mathscr{U}'} \mu_n(U) = 0 $$ so for big enough $n$, $\sum_{U \in \mathscr{U}'} \mu_n(U) < \frac{1}{2}$, but for fixed $n$, as $\mu_n$ is probability radon measure (countably additive): $$ 1 = \mu_n(X) = \mu_n\left(\bigcup \mathscr{U}'\right) \leq \sum_{U \in \mathscr{U}'} \mu_n(U) < \frac{1}{2} $$ Contradiction. So $K \neq \emptyset$.
Now I want to move to main problem. As it is compact metric space, it is separable space, but after while it looks like compactness may be more important.
Obviously $\nu(K) = 1$.
However, I don't know what I can do with that. If $a \in K$, then there is no open set $U$ such that $\mu_n(U) \to 0$. And therefore, every fixed ball $B(a, r)$ with center in $a$ has positive limit $\limsup_n\mu_n(B(a, r))$. What can I do with that knowledge?
If I made mistake above, please let me know, otherwise, please help me with going further with the task.
Presumably you know that we can find a subsequence $(\mu_{k_i})$ converging weakly to some Radon measure, call it $\nu$. By the portmanteau theorem, this implies that $\mu_{k_i}(B(a,r)) \to \nu(B(a,r))$ whenever $\nu(\partial B(a,r)) = 0$. However, the sets $\partial B(a,r)$ are pairwise disjoint (each is contained in the sphere of radius $r$) so only countably many of them can have nonzero measure under $\nu$.
Thus if you can show $\nu(K) = 1$ you are done, for then $\nu(B(a,r)) = \nu(B(a,r) \cap K)$. I don't know that I'd say this is "obvious" but it is not too hard: suppose $E$ is any compact set disjoint from $K$. Cover $E$ with finitely many sets $U_1, \dots, U_n$ of $\mathscr{U}$; by another case of the portmanteau theorem, we have $\nu(U_i) = 0$ for each $i$, so $\nu(E)=0$. Now you can conclude using the fact that $\nu$ is Radon. (Incidentally, this also takes care of showing that $K$ is nonempty.)
I think the assumption $a \in K$ was unnecessary.