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 $T$)?

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:
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. 

