Let $(S,\mathcal B)$ be a compact metric space with the Borel-$\sigma$-Algebra. Let $\mathcal M$ be the space of signed Borel measures and $\mathcal P \subset \mathcal M$ the set of probability measures on $(S,\mathcal B)$.
Now we say a sequence of measures $(\mu_n)_{n\in\mathbb N}$ with $\mu_n\in \mathcal M$ for each $n$ converges in the weak*-topology to a measure $\mu \in \mathcal M$ if
(I) $\int fd\mu_n\overset{n\rightarrow\infty}{\rightarrow}\int f d\mu$ for each countinuous and bounded function $f$ from $S$ to $\mathbb R$ that vanishes at infinity
(since $S$ is compact there aren't unbounded continuous $f$ anyways, and also "vanishing at infinity" has no meaning, hence this convergence coincides with the weak convergence of measures).
I know that for non-locally-compact sets $S$ the $\mathcal P$ is not closed in this topology. But several times I stumbled over the restriction non-locally-compact which makes me wonder how to show closedness of $\mathcal P$ in the compact case, i.e.
Question: Why can't there be a $\mu \in \mathcal M\setminus\mathcal P$ and a sequence of probability measures $(\mu_n)_{n\in\mathbb N}$ such that (I) holds?
What I have tried:
If $\mu\not\in\mathcal P$ then either $\mu(S)\neq1$ or there is a set $A\in\mathcal B$ such that $\mu(A)<0$. In the former case choose $f\equiv1$ and see that (I) cannot be true.
In the latter case I think there has to be some open set $U$ such that $\mu(U)<0$ and then I can take a function $f$ that is strictly positive on $U$ and zero elsewhere, then (I) wouldn't hold either (left side would be zero or greater and right side negative). But I don't know if there really has to be such a set $U$.
I already asked a different question regarding the existence of such a set (Obtaining the measure of a set from a limit of measures of open sets), but my approach there doesn't work.
Best Answer
Let $A$ be any Borel set in $S$. By Lusin's theorem, we can choose a sequence of continuous functions $f_m$ with $f_m \to 1_A$ $\mu$-almost everywhere. Moreover, if we set $g_m = \max(\min(f_m, 1), 0)$ then $g_m$ is also continuous, $0 \le g_m \le 1$ and $g_m \to 1_A$ $\mu$-almost everywhere as well.
Since $\int g_m\,d\mu = \lim_{n \to \infty} \int g_m\,d\mu_n$, we have $\int g_m\,d\mu \ge 0$. And by dominated convergence, $\lim_{m \to \infty} \int g_m \,d\mu = \int \lim_{m \to \infty} g_m \,d\mu = \int \mathbf 1_A \,d\mu = \mu(A)$. So $\mu(A) \ge 0$, and we conclude that $\mu$ is a positive measure.