Given a finite set $S$, in the space of strings $\Sigma=S^{(\omega)}$ equipped with the Bernoulli measure $\mu$, I want to know why the shift map $\sigma:\Sigma\rightarrow \Sigma $, define as $\sigma(\omega_1 \omega_2\omega_3\dots)=(\omega_2\omega_3\dots)$ where $\omega_1 \omega_2\omega_3\dots\in \Sigma$ is ergodic. That is, if $A$ is a subset of $\Sigma$ such that $\sigma^{-1}(A)=A$ then $\mu(A)$ is $0$ or $1$.
I am searching an elementary proof of this fact, since I do not need any other result of the ergodic theorem but this.
I am glad if you could help.
It is in general (very) hard to determine all the invariant sets of a transformation, and so one usually resorts to other means when we want to establish ergodicity.
In the case of the shift map the easiest way is to recall that for a measurable map $T \colon X \to X$ a $T$-invariant probability measure $\mu$ on $X$ is ergodic if and only if $$\tag1 \lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}\mu(T^{-k}A\cap B)=\mu(A)\mu(B) $$ for any measurable sets $A,B\subset X$. Actually, "for any measurable sets $A,B\subset X$" can be replaced by "for any measurable sets $A,B\subset X$ in some family of sets generating the $\sigma$-algebra".
So, it is sufficient to show that $(1)$ holds for any cylinder sets $A$ and $B$. For those you have $$ \mu(\sigma^{-n}A\cap B)=\mu(A)\mu(B) $$ for any sufficiently large $n\in\mathbb N$, which shows that $(1)$ holds.