Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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. $\blacksquare$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.