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I think I can solve the following exercise if $X$ is assumed to be separable, otherwise I can't.

Let $X$ be a (Hausdorff) locally compact space, $\pi\colon X \to Y$ a continuous map into a topological space $Y$ such that $Y$ is the union of a countable sequence of compact sets $(K_n)$, and such that $\pi^{-1} (K_n)$ is compact for each $n$. Let $\mu$ be a regular Borel probability measure on $Y$. Define the space $\mathcal{M}_\mu (X)_1$ consisting of all Borel probability measures $\nu$ on $X$ such that $\pi_* \nu = \mu$. Show that $\mathcal{M}_\mu(X)_1$ is compact with respect to the weak-* topology.

Now if $X$ is assumed to be separable, so is $C_0(X)$ and the same proof as in Helly-Bray's theorem works (mutatis mutandis). What if it's not?

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Isn't $\pi_*$ a continuous application from the space of Borel probability measures on $X$ to the set of Borel probability measures on $Y$, so that $\mathcal{M}_\mu(X)_1=(\pi_*)^{-1} (\mu)$ is closed ($\{\mu\}$ is closed because the weak-* topology is Hausdorff). Then since by Alaoglu's theorem the closed ball is compact with respect to the weak-* topology, we have the conclusion? What's wrong with this idea? – timofei Aug 22 '12 at 20:47
    
Just out of curiosity, where did this question come from? – Quinn Culver Aug 22 '12 at 23:29
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An exercise in Renato Feres' book about dynamics and semisimple Lie groups. – timofei Aug 23 '12 at 6:59

Here is what I got for a compact metric space:

We must show that any sequence $\mu_n$ in $\mathscr{M}(X)$ has a $\omega^*$-convergent subsequence (this shows that $\mathscr{M}(X)$ is actually sequentially compact).

Let $\{f_i\}_{i=1}^{\infty}$ be a countable dense subset of functions in $C(X)$. For any sequence $\mu_n$ in $\mathscr{M}(X)$ we have that \begin{equation*} |\int_X f_1 d\mu_n| \leq \|f_1\|_\infty \end{equation*} for all $n$, hence the sequence $\int_X f_1 d\mu_n$ is bounded and therefore has a convergent subsequence which we will denote by $\int_X f_1 d\mu_{n}^{1}$.

Now consider the sequence $\int_X f_2 d\mu_{n}^{1}$. It is again a bounded sequence of real numbers and so it has a convergent subsequence $\int_X f_2 d\mu_{n}^{2}$.

We continue in this fashion and obtain, for each $i \geq 1$, \begin{equation*} ... \subset \mu_{n}^{i} \subset \mu_{n}^{i-1} \subset ... \subset \mu_{n}^{1} \end{equation*} such that $\int_X f_j d\mu_{n}^{i}$ converges for $1 \leq j \leq i$. Now consider the sequence $\mu_{n}^{n}$. Since, for $n \geq i$, $\mu_{n}^{n}$ is a subsequence of $\mu_{n}^{i}$, $\int_X f_i d\mu_{n}^{n}$ converges for every $i \geq 1$. \ We can now use the fact that $\{f_i\}_{i=1}^{\infty}$ is dense to show that $\int_X f d\mu_n$ converges for all $f \in C(X)$. For any $\epsilon > 0$, choose $f_i$ such that $\|f-f_i\|_\infty \leq \epsilon$. Since $\int_X f_i d\mu_{n}^{n}$ converges, there exists $N$ such that if $n,m \geq N$ then \begin{equation*} \lvert \int_X f_i d\mu_{n}^{n} - \int_X f_i d\mu_{m}^{m} \rvert < \epsilon. \end{equation*}

Thus if $n, m \geq N$ we have \begin{equation*} | \int_X f d\mu_{n}^{n} - \int_X f d\mu_{m}^{m} | \leq \end{equation*} \begin{equation*}| \int_X f d\mu_{n}^{n} - \int_X f_i d\mu_{n}^{n} | + | \int_X f_i d\mu_{n}^{n} - \int_X f_i d\mu_{m}^{m} | + | \int_X f_i d\mu_{m}^{m} - \int_X f d\mu_{m}^{m} | \leq 3 \epsilon. \end{equation*}

So, $\int_X fd\mu_{n}^{n}$ converges, as required. To complete the proof, write \begin{equation*} \Lambda (f) = \lim_{n \rightarrow \infty} \int_X f d\mu_{n}^{n}. \end{equation*} It is easily verified that $\Lambda$ satisfies the hypothesis of the Riesz Representation Theorem. Therefore, there exists $\mu \in \mathscr{M}(X)$ such that \begin{equation*} \Lambda (f) = \int_X f d\mu.\end{equation*} We then have \begin{equation*} \int_X f d\mu_{n}^{n} \rightarrow \int_X f d\mu, \end{equation*} as $n \rightarrow \infty$, for all $f \in C(X)$, so we have proved that $\mu_{n}^{n}$ $\omega^*$-converges to $\mu$, as $n \rightarrow \infty$ and we are done. $\blacksquare$

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