# Sequential Compactness of a set of Positive measures

Let $$X$$ be an open subset in $$\mathbb R$$. Let $$\mu_n$$ and $$\mu$$ be positive measures on $$X$$. We define the weak convergence in this way: We say $$\mu_n$$ weakly converges to $$\mu$$ and write $$\mu_n\to \mu$$ if for any function $$\varphi \in C_c(X)$$ we have $$\int \varphi d\mu_n \to\int\varphi d\mu .$$

Now there is a theorem says that, if $$\mu_n(K)$$ is uniformly bounded for every compact subset $$K$$ of $$X$$, then there exists a subsequence $$\mu_{n_i}$$ s.t. $$\mu_{n_i}\to \mu$$ for some positive measure $$\mu$$. My question is how to prove this theorem.

My attempt: Fix $$\varphi \in C_c(X)$$. Suppose we have a compact set $$K$$ s.t. $$\text{supp} \,\varphi \subset K\subset X$$. For this $$K$$, $$\{\mu_n(K)\}$$ is a bounded sequence of reals, so it has a convergent subsequence $$\mu_{n_i}(K)$$. By using the approximation of simple functions for measurable functions, we may conclude that $$\int \varphi d\mu_{n_i} \to\int\varphi d\mu .$$

But when $$\varphi$$ changes, our compact set $$K$$ changes too. And accordingly, the index set $$\{n_i\}$$ changes. How can we pick up an "uniform" index set $$\{ n_i\}$$?

Thanks for help. :)

## 1 Answer

Hint: Use the compact sets $$K_N=[-N,N]$$ and use a diagonal argument to get one subsequence which works for each $$N$$.

• Thanks. I will accept your answer after the ten-minutes constraint. May 10 '21 at 12:34