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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. :)

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

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  • $\begingroup$ Thanks. I will accept your answer after the ten-minutes constraint. $\endgroup$
    – Sam Wong
    May 10 '21 at 12:34

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