Ergodic (equivalent characterization) Let $(\Omega,\mathcal{B},\mu,T)$ be a measuretheoretical dynamical system. Then this system is called ergodic if
$$
B\in\mathcal{B}, T^{-1}(B)=B\implies \mu(B)=0\text{ or }\mu(B^C)=0.
$$


Show: $(\Omega,\mathcal{B},\mu,T)$ is ergodic $\Leftrightarrow$ For every set $A$ that is invariant almost surely (i.e. $\mu(T^{-1}(A)\Delta A)=0$) it is $\mu(A)=0$ or $\mu(A^C)=0$.


Proof:
I think the direction "$\Leftarrow$" is easy. Because consider any invariant $B\in\mathcal{B}$ then it is invariant almost surely of course, too and so $\mu(B)=0$ or $\mu(B^C)=0$, what means, that the dynamical system is ergodic.
But I do not know how to prove "$\Rightarrow$". Let the dynamical system be ergodic. Consider any set $A$ that is invariant almost surely, i.e.
$$
\mu(T^{-1}(A)\Delta A)=0.
$$
This implies that
$$
\mu(T^{-1}(A)\setminus A)=\mu(A\setminus T^{-1}(A))=0.
$$
My first idea was to use
$$
A=\underbrace{[A\setminus\bigcup_{k\geqslant 0}T^{-k}(A)]}_{\subset A\setminus T^{-1}(A)}\cup[A\cap\bigcup_{k\geqslant 0}T^{-k}(A)]
$$
and so
$$
\mu(A)=\mu(A\cap\bigcup_{k\geqslant 0}T^{-k}(A)),
$$
If now $A\cap\bigcup_{k\geqslant 0}T^{-k}(A)$ were a measurable invariant set, then I could use the ergodicity... but I think it isn't.
Do not know how to make it...
 A: This answer is taken from Walters' An Introduction to Ergodic Theory.
Let $B \in \mathcal{B}$ and $\mu(T^{-1}B \triangle B)=0$. For each $n\geq 0$ we have $\mu(T^{-n}B \triangle B) =0$ because $T^{-n}B \triangle B \subset \bigcup_{i=0}^{n-1}T^{-(i+1)}B \triangle T^{-i}B = \bigcup_{i=0}^{n-1}T^{-i}(T^{-1}B\triangle B)$ and hence $\mu(T^{-n}B \triangle B) \leq n \mu(T^{-1}B \triangle B)$. Let $B_{\infty} = \bigcap_{n=0}^{\infty}\bigcup_{i=n}^{\infty}T^{-i}B$. By the above we know $\mu(B \triangle \bigcup_{i=n}^{\infty}T^{-i}B) \leq \sum_{i=n}^{\infty} \mu(B \triangle T^{-i}B) =0$ for each $n \geq 0$. Since the sets $\bigcup_{i=n}^{\infty} T^{-i} B$ decrease with $n$ and each has measure equal to $B$ we have $\mu(B_{\infty} \triangle B)=0$ and hence $\mu(B_{\infty}) = \mu(B)$. Also $T^{-1}B_{\infty} = \bigcap_{n=0}^{\infty} \bigcup_{i=n}^{\infty}T^{-(i+1)}B = \bigcap_{n=0}^{\infty} \bigcup_{i=n+1}^{\infty} T^{-i}B = B_{\infty}$. Therefore we have obtained a set $B_{\infty}$ with $T^{-1}B_{\infty} = B_{\infty}$ and $\mu(B_{\infty} \triangle B)=0$. By ergodicity we must have $\mu(B_{\infty})=0$ or $1$ and hence $\mu(B)=0$ or $1$.
