Let $T:X \to X$ be a continuous function on the compact metric space $X$. We say that $\mu$ is $T-$ invarint if $\mu(T^{-1}(A))=\mu(A) \hspace{0.2cm} \forall A \in \beta.$
We denote by $\mathcal{M}(X,T)$ the set of $T-$invariant Borel probability measures.
For $\mu \in \mathcal{M}(X, T),$ the set $G_{\mu}$ of $\mu-$generic points is defined by $$ G_{\mu}:=\left\{x \in X: \frac{1}{n} \sum_{j=0}^{n-1} \delta_{T^{j} x} \rightarrow \mu \text { in the weak }^{*} \text { topology as } n \rightarrow \infty\right\} $$ where $\delta_{y}$ denotes the probability measure whose support is the single point $y$.
Let $\mu$ be an ergodic measure. By Birkhoff ergodic theorem, $\mu(G_{\mu})=1$.
$\textit{Question:}$ Let $\mu$ be an ergodic measure and $x$ be a $\mu-$generic point. Why is $\mu(B(x, \frac{1}{h}))>0$ for every $h \in \mathbb{N}$, where $B(x, \frac{1}{h})$ is the ball of radius $\frac{1}{h}$ centered at $x.$?
I think one should consider $\chi_{B(x, 1/h)}$ in Birkhoff ergodic theorem, but I don't know how to get the sum is positive.
This claim seems false to me. Consider $X=\{0,1\}$ with $T(0)=T(1)=1$. Then $0$ is a generic point for the measure $\delta_1$, but is not contained in the support of the measure.