The Bernoulli measure of points with a dense orbit in $\Sigma(2)$ is $1$

Consider a probability measure $\nu$ defined on $\{0,1\}$.Let $\nu^{\mathbb{Z}}$ be the Bernoulli measure defined on $X=\{0,1\}^{\mathbb{Z}}$ and $\sigma : X \to X$ be the shift map. Show that

\begin{align} \nu^{\mathbb{Z}}(\{w \in X;\overline{O(\sigma,w)}=X\})=1 \end{align}

I have already proved these for the shift map:

1- $\sigma$ has $2^n$ periodic points with period $n$

2- The set of periodic points (Per($\sigma$) ) is dense in $X$.

3- There is a point in $X$ with a dense orbit.

4- $\sigma$ is ergodic with respect to Bernoulli measure $\nu^{\mathbb{Z}}$.

From the statements above I want to prove what the problem wants. I can say from 4 that for every $\sigma$-invariant subset $A$($\sigma^{-1}(A)=A)$, $\nu^{\mathbb{Z}}(A)$ is either $0$ or $1$ if the set $\{w \in X;\overline{O(\sigma,w)}=X\}$ is $\sigma$-invariant then the Bernoulli measure of that is either $0$ or $1$.

My question is how could I prove this ? is this a correct way to prove this question?

Solution 1:

I think I found an answer. In Peter Walter's Ergodic Theory book there is a theorem on page 29 which says If $X$ is a compact metric space and $\mathcal{B}(X)$ is the borel sigma algebra subsets of $X$ and $m$ be probability measure on this space such that $m(U) >0$ for every non-empty open set $U$ . Suppose $T: X \to X $ is a continuous transformation which preserves the measure $m$ and is ergodic.Then almost all points of $X$ have a dense orbit under $T$ i.e \begin{align} m(\{x \in X ;\overline{O(T,x)}=X\}=1 \end{align} The problem has all conditions above so problem solved.