Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer

$x$ can be generic by your definition while $\mu_x$ is not ergodic:

A generic point for a non-ergodic measure

The page you linked is only considering points that are generic for ergodic measures.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.