1
$\begingroup$

From the Stone-Weierstrass theorem, we see that $C(X\to \mathbb{R})$ is separable for compact metric space $X$, combining this with the Riesz representation theorem and the sequential Banach-Alaoglu theorem, we obtain $M(X)\equiv C_0(X\to\mathbb{R})^*$ is sequentially compact in weak$^*$ topology for compact metric space $X$, here $M(X)$ denotes the space of finite signed Radon measure on $X$.

Now I want to prove the non-compact case. Let $X$ be a locally compact metric space which is $\sigma$-compact, and let $\mu_n$ be a sequence of Borel probability measures. We assume that the sequence $\mu_n$ is tight, which means that for every $\varepsilon>0$ there exists a compact set $K$ such that $\mu_n(X\setminus K)\leq\varepsilon$ for all $n$. Show that there is a subsequence of $\mu_n$ which converges vaguely (i.e. in the weak$^* $ topology) to another Borel probability measure $\mu$.

It seems that I can use the diagonal argument, but I don't know how to get started.

$\endgroup$

1 Answer 1

0
$\begingroup$

Since the space is $\sigma$-compact, we can find a increasing sequence of compact sets $(K_j)_{j\geqslant 1}$ such that $\bigcup_{j\geqslant 1}K_j=X$. Using the assumption of local compactness, we may assume that $K_j$ is contained in the interior of $K_{j+1}$.

We construct a decreasing sequence of infinite sets of integers $(I_j)_{j\geqslant 1}$ such that the sequence $\left(\mu_{n\mid K_j}\right)_{n\in I_j}$ converges to some $\nu_j$ in $M(K_j)$ for each $j$. Now define $n_j$ as the $j$th element of $I_j$. In this way, we obtain that $(n_j)_{j\geqslant 1}$ is an increasing sequence of integers and $\left(\mu_{n_j\mid K_i}\right)_{j\geqslant 1}$ converges to $\nu_j$ in $M(K_i)$ for each $i$. It remains to check that $(\mu_{n_j})_{j\geqslant 1}$ converges in distribution to some measure $\mu$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .