Let $T: X \to X$ be a continuos map defined on the compact space (Maybe Haussdorff) $X$. Denote by $M_T$ the space of $T$-invariant probability measures defined over the Sigma Algebra of Borel sets. Prove the following:
Show that $M_T$ is non-empty
Show that $M_T$ is compact and Haussdorff.
Important remark:
I know the result is true for spaces which are Compact and Metric.
I think that the result is false for general Compact and Haussdorff Topological Spaces. If the result is false, please give me a counterexample. I tried to find a counterexample but I couldn't.
At the moment I'm more interested to discover if $M_T$ is nonempty rather than the second proposition.
The fact that $M_T$ is compact Hausdorff (in weak * topology) has been pointed out in the comment. The proof is similar to the case in compact metric spaces. I learned this from Clinton Conley.
Proof: By definition of weak * convergence, $\mu_n \to \mu $ iff $\int f d\mu_n \to \int f d\mu \forall f\in C(X)$, in particular $\int f\circ T d\mu_n \to \int f \circ T d\mu \ \ \ \forall f\in C(X)$ so $\int f dT_*\mu_n \to \int f dT_*\mu \ \ \ \forall f\in C(X)$.
Consider an abelian group (notice the actions commute) generated by $K=\{\frac{1}{n} (1+ T_* + \cdots + T_*^{n-1}): n\in \mathbb{N} \}$. Since $M_T$ is compact Hausdorff, and also for any finite $s_i\in K, (s_1 \circ \cdots s_k)[M_T]\subset \bigcap_i s_i[M_T]$ so by finite intersection property. In particular there exists $\mu\in \bigcap_{s\in K} s[M_T]$.
Now it's left to show that $\mu$ is invariant. Suppose not, there exists continuous function $f\in C(X,[0,1])$ such that $\int f\circ T d\mu \neq \int f d\mu$. Let $\epsilon>0$ be smaller that the absolute value of their difference. Pick $n$ large such that $\frac{1}{n}<\epsilon$ and also $\nu\in M_T$ with $\mu=\frac{1}{n} (\nu+\cdots+ T_*^{n-1}\nu)$. Then $T_*\mu-\mu=\frac{1}{n} (T_*^n \nu- \nu)$. Then $\epsilon\leq |\int f\circ T d\mu - \int f d\mu|=|\frac{1}{n}\int fd(T_*\nu-\nu)|\leq \frac{1}{n}$, contradiction.