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Suppose that $(X, T, \mu)$ is an ergodic topological probability measure preserving dynamical system on a compact metric space $X.$ I want to show that almost every point in $X$ is $\mu$ generic.

Definition of a $\mu$-generic point $x$ is as follows:

A point $x$ is called $\mu$- generic if for every continuous function $f: X \to \mathbb R,$ we have that $$\lim_{n \to \infty} \frac{f(x) + f(T(x))+ \dots + f(T^n(x))}{n} = \int_X fd \mu$$

My thoughts were:

I feel like I can use that if $X$ is a compact metric space, then there exist a countable set of continuous functions $f_1: X \to \mathbb R, f_2: X \to \mathbb R, \dots $ such that for all $f: X \to \mathbb R$ continuous, there exists $\{f_{n_k}\}$ such that $f_{n_k}(x) \to f(x)$ for every $x \in X.$

But I am not sure if I am correct or no or how can I use it. Any help will be greatly appreciated!

Alp Uzman
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Emptymind
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1 Answers1

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Indeed, you are on the right track. Here are some ideas: For $\phi:X\to \mathbb{R}$ continuous, define $A(\phi)=A(\mu,T,\phi)$ to be the set of all those points along which the time average (relative to $T$) of $\phi$ converges to the space average (relative to $\mu$). Then the set of Birkhoff generic points for $(\mu,T)$ is

$$ \bigcap_{\phi}A(\phi). $$

By the Birkhoff Ergodic Theorem, each $A(\phi)$ has full measure, but the intersection is not necessarily countable, so we can not at this stage conclude.

We are however given that $X$ is compact metric, thus there is a sequence $\phi_\bullet$ of continuous functions on $X$ that is dense (see e.g. $C(X)$ is separable when $X$ is compact?).

Claim: $\bigcap_{\phi}A(\phi)=\bigcap_{m\geq 1}A(\phi_m)$.

Proof: The $(\subseteq)$ part is syntactic. For $(\supseteq)$, let $x\in \bigcap_{m\geq 1}A(\phi_m)$, $\phi:X\to\mathbb{R}$ be continuous and $\epsilon\in\mathbb{R}_{>0}$. By the density of $\phi_\bullet$, there is an $m$ such that $d_{C^0}(\phi,\phi_m)<\epsilon/3$. Since $x\in A(\phi_m)$, there is an $N\in\mathbb{Z}_{\geq1}$ such that for any $n\in\mathbb{Z}_{\geq N}$, one has

$$ \left|\dfrac{1}{n}\sum_{k=0}^{n-1}\phi_m\circ T^k(x)-\mathbb{E}_\mu(\phi_m)\right|<\epsilon/3. $$

Then

\begin{align*} \left|\dfrac{1}{n}\sum_{k=0}^{n-1}\phi\circ T^k(x)-\mathbb{E}_\mu(\phi)\right|& \leq \left|\dfrac{1}{n}\sum_{k=0}^{n-1}\phi\circ T^k(x)-\dfrac{1}{n}\sum_{k=0}^{n-1}\phi_m\circ T^k(x)\right|\\ &\quad+\left|\dfrac{1}{n}\sum_{k=0}^{n-1}\phi_m\circ T^k(x)-\mathbb{E}_\mu(\phi_m)\right|\\ &\quad+\left|\mathbb{E}_\mu(\phi_m)-\mathbb{E}_\mu(\phi)\right|\\ &<\epsilon. \end{align*}

Alp Uzman
  • 12,209