# Weak-* convergence of Borel probability measures on a nice space

Let $X$ be compact, metrizable space. (If it somehow changes the outcome of the question, I am mainly interested in $X=$ the Cantor set)

Let $E\subset X$ be a closed subset and let $\mu_n$ be a sequence of Borel probability measures on $X$ converging to $\mu$ in the weak-* topology. Moreover, suppose that $\mu_n(E)=0$ for all $n$.

Is it possible that $\mu(E)>0$?

• Yes, it is possible. Try measures with one-point support. Jul 22 '13 at 20:50
• On the other hand, if $E$ is open, then we will have $\mu(E) = 0$. Jul 22 '13 at 21:05
• Okay, so I completely forgot about this very trivial case, which I should have thought of myself. Is the answer also 'Yes' if I assume that $\mu$ has full support? Jul 22 '13 at 21:26
• Yes. For example, take $\mu=m+\delta_0$ on $[0,1]$, where $m$ is Lebesgue measure, $\mu_n=m+\delta_{1/n}$ and $E=\{ 0\}$. Jul 23 '13 at 14:48

Yes, as GEdgar suggested, you can have a one-point supported measures $\delta_n$ converge to a one-point measure $\delta$ at a point distinct from supports of the sequence.
More generally, in any separable metrisable space (in particular, any compact metrisable), finitely supported measures are dense among finite Borel measures (with weak-*-topology, which is also metrisable in this case). In particular, if you have a continuous measure $\mu$ (vanishing at points) you can pick a sequence $\mu_n$ of finitely supported measures converging to it and just take the universe minus the union of all supports of $\mu_n$'s which will have $\mu_n$ zero for each $n$, but of course will have countable (and hence null with respect to $\mu$) complement.
• Regarding your second paragraph, note that the set you construct (the complement of the union of the supports of the $\mu_n$s) is in general not closed. Jul 22 '13 at 21:04