Denote by $\mathcal{M}(X)$ the set of Borel probability measures of a space $X$. This set is equipped with weak$^*$ topology defined by the convergence $\mu_n\to \mu$ if and only if $\int \phi d\mu_n\to \int \phi d\mu$ for every continuous map with compact support $\phi:X\to \mathbb{R}$.
Let $\varphi:G\times X\to X$ be a continuous action, $G$ be a finitely generated group and $X$ be a compact metric space.
A measure $\mu\in \mathcal{M}(X)$ is called an invariant measure under $\varphi$ if $\mu(\varphi(g, A)= \mu(A)$, for every $g\in G$ and every measurable set $A$. Let $\mathcal{A}\subseteq \mathcal{M}(X)$ be a closed, convex and $\varphi_*$-invariant (this means that if $\mu\in\mathcal{A}$ , then $\varphi_{g,*}(\mu)\in\mathcal{A}$, where $\varphi_{g,*}(\mu)(A)= \mu(\varphi(g, A)$).
I claim that $\mathcal{A}$ has an invariant measure of $\varphi$. Since $G$ is a finitely generated group, hence it is countable, let $G=\{g_i\}_{i\in\mathbb{N}}$. For every $n\in \mathbb{N}$, take $\nu_n=\frac{1}{n}\sum_{i=1}^n \varphi_{g_i,*}(\mu)$. Since $\mathcal{A}$ is comact, the sequence $\nu_n$ has a limit measure $\mu$. Since $\mathcal{A}$ is convex and $\nu_n\in\mathcal{A}$, it follows that $\mu\in\mathcal{A}$ and also an invariant measure of $\varphi$.
Please help me to know that this proof is true.
Thanks