I was working through a problem In a book on functional analysis. the problem is
Let X be a $ \sigma $ compact metric space such that the space of bounded real continuous functions on X $ C_b(X) $ is separable. If $ \mu_n $ is a sequence of finite signed Borel measures such that $ |\mu_n|(X) \leq 1 $ for all n. Then there exists a subsequence $\mu_{n_j}$ and a finite borel measure $\mu$ such that $$ \int_X g d\mu_{n_j} \rightarrow \int_X g d\mu$$ for all g $ \in C_b(X) $ It is obvious by using the separability of $ C_b(X) $ hat there exists a subsequence $\mu_{n_j} $ such that $$ \int_X g d\mu_{n_j} \rightarrow T(g) $$ $T \in C_b(X)^* $ However beyond this I can't figure out how to conclude the existence of $\mu$ in the general case without an assumption of local compactness or a tightness assumption on $ \mu_n $ Is the problem as stated false or am I being dense by overlooking something? Any help is appreceitated.
We know $C_b(X)$ is a separable Banach space by assumption. BTW, $X$ being metrisable plus this separability implies already that $X$ is compact, so I don't see why your text adds a superfluous assumption here..
The $\mu_n$ define a sequence in the unit ball of its dual, which we consider in the weak$^\ast$ topology. We define $j(\mu_n): C_b(X) \to \Bbb R$ by $j(\mu_n)(g)=\int_X g(x)d\mu_n$ and all $j(\mu_n)$ lie in the unit ball of the dual of $C_b(X)$.
By the sequential version of the Banach-Alaoglu theorem (see Wikipedia, e.g.) we have a convergent subsequence in that ball, so some $T \in C_b(X)^\ast$ and some subsequence $j(\mu_{n_k}) \to T$, and the last integral condition just states the weak$^\ast$ convergence of that subsequence.
Then Riesz' theorem (plus compactness of $X$, see above) finishes the argument.
I think that's all there is to it.