# Convergence of measure on clopen basis implies the existence of a limit measure?

Let $$X$$ be a compact metrizable space. Let $$B$$ be basis for the topology of $$X$$. Assume that all sets in $$B$$ are clopen.

Let $$(\mu_n)_{n=1}^{\infty}$$ be sequence of Borel regular probability measures on $$X$$. Assume that for every $$U\in B$$ the sequence $$(\mu_n(U))_{n=1}^{\infty}$$ converges.

Does this mean that there is a Borel regular probability measure $$\mu$$ on $$X$$ such that $$\lim_{n\rightarrow\infty}\mu_{n}(U)=\mu(U)$$ for all $$U\in B$$?

I read a little about weak-* converges and it seems to be the right notion for this, but I'm not sure.

My knowledge in measure theory is somewhat lacking, but my goal is to understand how to construct such a measure $$\mu$$ completely including the proofs of all required theorems (hence, elementary answers would be great).

Identifying regular Borel measures with bounded linear functionals on $$C(X)$$ we can consider $$(\mu_n)_n$$ to be a sequence in the unit ball of $$C(X)^*$$. By the Banach-Alaoglu theorem it has a weak$$^*$$-cluster point, say $$\mu$$, in the unit ball of $$C(X)^*$$. This means that there is a subsequence, say $$(\mu_{n_k})_k$$, such that $$\int f$$ $$d\mu_{n_k}\rightarrow \int f$$ $$d\mu$$ for all $$f\in C(X)$$. $$\mathcal{X}_U\in C(X)$$ for any clopen $$U\subset X$$, hence $$\mu_{n_k}(U)=\int \mathcal{X}_U$$ $$d\mu_{n_k}\rightarrow \int \mathcal{X}_U$$ $$d\mu=\mu(U)$$. Since $$\lim\limits_{n \rightarrow \infty} \mu_n(U)$$ is assumed to exist we must then have $$\lim\limits_{n \rightarrow \infty} \mu_n(U)=\mu(U)$$.