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Let $X=\{0,1\}^{\mathbb{N}}$, $T:X\to X$ the shift map, and $\mu$ a $T$-invariant probability measure on $X$. A point $x \in X$ is generic if $$ \lim\, \frac{1}{n}\sum_{i<n} \chi_{\sigma}(T^{i}(x)) $$ exists for every finite binary string $\sigma$ (so $\chi_{\sigma}$ is the characteristic function of the basic clopen cylinder set determined by $\sigma$). By Birkhoff's ergodic theorem, $\mu$-almost all $x$ are generic. Now when $x$ is generic, the set function $\mu_{x}$ defined on basic clopen cylinders by $$ \mu_{x}(\sigma) = \lim\, \frac{1}{n}\sum_{i<n} \chi_{\sigma}(T^{i}(x)) $$ is actually (extendable to) a $T$-invariant measure. This page claims that this measure is in fact ergodic.

Question: Why is $\mu_{x}$ ergodic when $x$ is generic?

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Let $\sigma$ a $T$-invariant set, i.e. $T^1(\sigma)=\sigma$. Then $T^{—j}(\sigma)=\sigma$ and $$\chi_{\sigma}(T^j(x))=\chi_{T^{-j}(\sigma)}(x)=\chi_{\sigma}(x).$$ Hence if $x$ is generic $$\mu_x(\sigma)=\chi_{\sigma}(x)\in\{0,1\},$$ which proves that $\mu_x$ is ergodic.

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