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Let $K$ be a compact Hausdorff space. Denote by $ca_r(K)$ the set of all countably additive, signed Borel measures which are regular and of bounded variation. Let $(\mu_n)_{n\in\mathbb{N}}\subset ca_r(K)$ be a bounded sequence satisfying $\mu_n\geq 0$ for all $n\in\mathbb{N}$. Can we conclude that $(\mu_n)$ (or a subsequence) converges in the weak*-topology to some $\mu\in ca_r(K)$ with $\mu\geq 0$?

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If $K$ is a Polish space, this works since the weak*-topology on $K$ is then metrizable... –  Andy Teich Jul 9 '12 at 14:45
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up vote 4 down vote accepted

We cannot.

Let $K = \beta \mathbb{N}$ be the Stone-Cech compactification of $\mathbb{N}$, and let $\mu_n$ be a point mass at $n \in \mathbb{N} \subset K$. Suppose to the contrary $(\mu_n)$ has a weak-* convergent subsequence $\mu_{n_k}$. Define $f : \mathbb{N} \to \mathbb{R}$ by $f(n_k) = (-1)^k$, $f(n) = 0$ otherwise. Then $f$ has a continuous extension $\tilde{f} : K \to \mathbb{R}$. By weak-* convergence, the sequence $\left(\int \tilde{f} d\mu_{n_k}\right)$ should converge. But in fact $\int \tilde{f} d\mu_{n_k} = \tilde{f}(n_k) = (-1)^k$ which does not converge.

If $C(K)$ is separable, then the weak-* topology on the closed unit ball $B$ of $C(K)^* = ca_r(K)$ is metrizable. In particular it is sequentially compact and so in that case every bounded sequence of regular measures has a weak-* convergent subsequence. As Andy Teich points out, it is sufficient for $K$ to be a compact metrizable space. Also, since there is a natural embedding of $K$ into $B$, if $B$ is metrizable then so is $K$.

One might ask whether it is is possible for $B$ to be sequentially compact without being metrizable. I don't know the answer but I suspect it is not possible, i.e. that metrizability of $B$ (and hence $K$) is necessary for sequential compactness.

We do know (by Alaoglu's theorem) that closed balls in $C(K)^*$ are weak-* compact, so what we can conclude in general is that $\{\mu_n\}$ has at least one weak-* limit point. However, as the above example shows, this limit point need not be a subsequential limit.

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So, if I understood you correctly, together with your last argument, this implies that for a sequence $\{\mu_n\}$ with $\mu_n\geq 0$ we have at least one weak* limit point $\mu$ which also satisfies $\mu\geq 0$ –  Andy Teich Jul 9 '12 at 17:19
    
@AndyTeich: Yes, that is right. (It is clear that $\mu \ge 0$, since the nonnegative measures are weak-* closed in $ca_r(K)$.) –  Nate Eldredge Jul 9 '12 at 17:30
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