The existence of ergodic measures is usually proved under the assumptions that the space $\Omega$ is compact metric and the transformation $T: \Omega \rightarrow \Omega$ is continuous, by using results on the existence of extreme points. Is it possible to establish the existence of ergodic measures if $T: \Omega \rightarrow \Omega$ is only a measurable transformation? ($\Omega$ can still be assumed compact metric.) If not, is there any extra condition that would be sufficient for this (of course, weaker than continuity of $T$)?

$\begingroup$ Measurability alone is not enough to guarantee existence of invariant (or ergodic) measures. See math.stackexchange.com/questions/94981/… and mathoverflow.net/questions/66669/… $\endgroup$ – Vaughn Climenhaga Aug 14 '13 at 19:06

1$\begingroup$ $x\mapsto x+1$ in $\mathbb{R}$ is the standard counterexample, if you want $\Omega$ to be compact you can take $[0,1]$ with $T\colon x\mapsto x/2$ for $x>0$ and $T(0)=1$. $\endgroup$ – Vaughn Climenhaga Aug 14 '13 at 19:07

$\begingroup$ Great, thanks. But would then a sufficient condition be the existence of an invariant measure? $\endgroup$ – John Learner Aug 14 '13 at 19:45

$\begingroup$ Yes, if $T$ is a measurable transformation and $\mu$ is a $T$invariant measure, then there must exist an ergodic measure. This is using the ergodic decomposition, see for example this question: mathoverflow.net/q/124066/5701 $\endgroup$ – Vaughn Climenhaga Aug 14 '13 at 20:05
As indicated in the comments, measurability alone is not enough, and there are easy counterexamples.
As for a condition between continuity and measurability that still does the trick, I'm not sure if there's a natural one. Using Lusin's theorem one can prove that Condition (C) below is equivalent to existence of an invariant measure, but it's not a particularly pleasant condition, as you'll see. Here we assume that $\Omega$ is a complete separable metric space (not necessarily compact), and $f\colon \Omega\to \Omega$ is measurable (nothing more yet...)
(C) There exist a sequence of points $x_k\in \Omega$ and times $n_k\in \mathbb{N}$ such that for every $\epsilon>0$ there is a compact set $K_\epsilon\subset \Omega$ satisfying: (1) $f_K$ is continuous, and (2) $\#\{1\leq j\leq n_k \mid f^j(x_k) \in K_\epsilon\} \geq (1\epsilon)k$ for all sufficiently large $k$.
If you have an invariant measure, you can use the ergodic decomposition to get an ergodic measure, and then the Birkhoff ergodic theorem together with Lusin's theorem to show that (C) holds. On the other hand, if (C) holds then by considering the empirical measures $\mu_k = \frac 1{n_k} \sum_{j=0}^{n_k} \delta_{f^jx_k}$ one can observe the following:
 $\mu_k(K_\epsilon)>1\epsilon$ for all $\epsilon$ and all large $k$, and hence the sequence $\mu_k$ is tight, so by Prohorov's theorem it has a convergent subsequence $\mu_{k_i}\to \mu$.
 If $D$ is a metric on the space of probability measures on $\Omega$ that induces the weak* topology, then $D(\mu, f_*\mu) \leq D(\mu,\mu_k) + D(\mu_k,f_*\mu_k) + D(f_*\mu_k,f_*\mu)$. The first term goes to zero along $k_i$ by the assumption on the subsequence. The second term goes to zero along $k$ because the empirical measures are averages over long orbit segments. The third term goes to zero because $f$ is continuous on $K_\epsilon$ and both $\mu$ and $\mu_k$ put nearly all their weight on $K_\epsilon$. Thus $D(\mu,f_*\mu)=0$ and so $\mu$ is $f$invariant.
So (C) is equivalent to the existence of an invariant measure. It's not a terribly nicelooking condition, though. At the very least you can probably use this to cover some reasonable classes of examples.