Let $(X, \mathcal{F},\mu, T)$ be a dynamical system.
I have tried to prove the following:
Theorem 1: Let function $T:X\rightarrow X$, preserves the probabilistic measure $\mu$. For any set $A$, such $\mu(A)>0$, we have: $$\mu(\{x\in A: \forall_{N\in \mathbb{N}}\exists_{n \ge N}T^n(x)\in A\})=\mu(A)$$
"Proof" Let $B(N) = \{x\in A: \forall_{n \ge N}T^n(x)\notin A\}$. The thesis of theorem is equivalent to: $$\mu(\{x\in A: \exists_{N\in \mathbb{N}}\forall_{n \ge N}T^n(x)\notin A\})=0$$ But i notice that: $$\mu(\{x\in A: \exists_{N\in \mathbb{N}}\forall_{n \ge N}T^n(x)\notin A\}) = \mu(\bigcup_{N\in \mathbb{N}}B(N))\le \sum_{N\in \mathbb{N}}\mu(B(N))$$ So it is sufficient to prove, that for every $N$ we have $\mu(B(N))=0$
Here is how to do this:
Let $n\ge N$ and $x\in B(N)$, we have: $T^{n}(x)\notin B(N)$, and $x\notin T^{-n}(B(N))$, hence $B(N)\cap T^{-n}(B(N))=\emptyset$
Now $$\emptyset =T^{-k}(\emptyset)= T^{-k}(T^{-n}(B(N))\cap B(N)) = T^{-k-n}(B(N))\cap T^{-k}(B(N))$$
I can see now, that sets of the form $T^{-n}(B(N))$ are pairwise disjoint and $\mu(T^{-n}(B(N)))=\mu(B(N))$
Now assume that $\mu(B(N))>0$, then
$$\mu(\bigcup_{n\ge N}T^{-n}(B(N)))=\sum_{n\ge N}\mu(T^{-n}(B(N)))=\sum_{n\ge N}\mu(B(N))=\infty$$
A contradiction, since $\mu(X)=1$
Did i do this proof correctly or no? If no, then please point me where i did something wrong.
Regards