Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 if $C_0(X)$ and the same proof as in Helly-Bray's theorem works (mutadis mutandis). What if it's not?

share|improve this question
2  
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

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.