I am stuck on the following problem, which is a previous step for proving existence of invariant measures, and also is a previous step for proving the existence of a Ergodic System.

Let $G$ be an abelian group, $\Omega$ a Hausdorff compact space, and $\alpha:G\times\Omega \to \Omega$ a continuous action on $\Omega$. A probability measure $\mu \in M(\Omega)$ is called $\alpha$-invariant if $\mu(\alpha_t(E)) = \mu(E) \ \forall E$ Borel set, and for all $t \in G$. Let $M_{\alpha}$ be the set of $\alpha$-invariant measures. Prove that $M_\alpha$ is convex and w$*$-closed.

I've already proved that $M(\Omega)$ is w$*$-compact, using Banach-Alaoglu and some tricks. Hence if I prove that $M_\alpha$ is w$*$-closed I finish. To prove this I take a convergence net $\mu_i\to \mu$ and I've tried to prove that $\mu$ belongs to $M_{\alpha}$. The best I can do was that:

Since $\mu_i$ is invariant, it can be proved that $$\int_{\Omega}fd\mu_i = \int_{\Omega}f\alpha_t^{-1}d\mu_i,$$ now taking limits the left side converges to $\int_{\Omega}fd\mu$ and the right to $\int_{\Omega}f\alpha_t^{-1}d\mu$. The right member converges to this because $f\alpha_t^{-1} \in C(\Omega)$.

I would appreciate any ideas.

EDIT 1: By probability measure I mean regular borel measures on $\Omega$. In this case, we have that $C(\Omega)^*\cong M(\Omega) $ where $M(\Omega)$ are the sets of all complex regular borel measures on $\Omega$. We consider the set of probability measures meaning the subset of $M(\Omega)$ which are probability measures. We have a natural topology here that is the weak star topology of $M(\Omega)$ restricted to the sets of probability measures. Here a net $\mu_i$ of probability measures converges to a probability measures $\mu$ if and only if $$\int fd\mu_i \to \int f d\mu \ \ \forall f\in C(\Omega)$$

  • 1
    $\begingroup$ You're a bit vague on what you mean by probability measure. If you consider arbitrary measures, the natural weak-$^*$ topology would be that of $\mathcal{L}^\infty(X)^*$ (the space of bounded measurable functions), and then I don't see a reason why the action of $\mathrm{Homeo}(X)$ on the set of measures would be continuous (even for a fixed self-homeomorphism). But in this context it's more common and natural to consider the space of Radon measures, and the weak-$^*$ would be that of $C(X)^*$. Then the action is continuous and hence the set of fixed points is closed. $\endgroup$ – YCor Nov 8 at 15:17
  • $\begingroup$ You are right! I am going to make the clarification $\endgroup$ – HFKy Nov 8 at 16:11
  • $\begingroup$ Ive added the clarifications! Thanks for the advice $\endgroup$ – HFKy Nov 8 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.