# Subsequence which converges for compact metric space.

Let $$X$$ be a compact metric space and $$\mu_n$$ a sequence of finite Borel measures on $$X$$ with the property that

$$\sup_n \mu_n(X)<\infty.$$

Show for all $$f \in C(X)$$ there exists a subsequence $$n_j$$ such that $$\int f d \mu_{n_j}$$ converges.

Attempt:

We know as $$X$$ is a compact metric space, that $$C(X)$$ is separable. I.e., there exists a countable dense subset, call it $$A \subset C(X)$$ and so

$$A:=\{f_n: n \in \Bbb{N}, \text{f_n:X \to \Bbb{R} is continuous}\}.$$

and $$\bar{A}=X$$. I also know for each $$n$$, $$f_n(X)$$ attains a min and max. But I dont see how to produce this convergent subsequence. Also, since the sup of the $$\mu_n(X)$$ is finite and I have a sequence I cant say they have a convergent subsequence right cause they live in $$\Bbb{R}^{\geq 0}$$ which is not compact..

put $$A=\sup_n \mu_n(X)$$.Since $$X$$ is compact, $$f$$ is bounded, so there is some constants $$m,M$$ such that: $$\forall x\in X:m\leq f(x) \leq M$$,so $$mA\leq \int f d \mu_{n} \leq MA$$ \ This shows that $$u_{n}:=\int f d \mu_{n}$$ is a bounded sequence of real numbers,so it has a convergent subsequence $$\int f d \mu_{n_{j}}$$