Let $(X;\mathcal{A}; \mu)$ be a measure space with $\mu (X)=1$. Let $T:X\rightarrow X$ be a measure preserving map which is ergodic. I.e. for all $A\in \mathcal{A}$ s.t. $T^{-1}(A)=A$, $\mu(A) =0$ or $1$. I want to show that for all $B\in \mathcal{A}$ with $\mu(B)>0$ $$\mu(\{x\in X: \exists k\geq 1~~\text{such that}~~T^kx\in A\})=1$$
I have somehow now idea where to start. I think it should not be really complicated so my first Idea was to explicitly use the definition. But I could't show that $T^{-1}(A)=A$ where $A:=\{x\in X: \exists i\geq 1~~\text{such that}~~T^kx\in A\}$, since in my opinion $\subset$ does not hold.
I also know that ergodicity implies that for all $B\in \mathcal{A}$ $$\mu\left(T^{-1}(B)\triangle B\right)=0\Rightarrow \mu(B)=0 ~~\text{or}~~1$$
I also thought about using Pointcaré's recurrence theorem, and I see that my statement above is somehow a generalisation of the theorem where we have an additional assumption, namely that $T$ is ergodic. But I don't see how to get from the recurrence theorem my above exercise. One idea was to say that since $$\{x\in A: \exists~ n\geq 1 ~~\text{s.t}~~T^n(x)\in A\}\subset \{x\in X: \exists~ n\geq 1 ~~\text{s.t}~~T^n(x)\in A\}$$ we know that $$1=\mu(\{x\in A: \exists~ n\geq 1 ~~\text{s.t}~~T^n(x)\in A\})\leq \mu(\{x\in X: \exists~ n\geq 1 ~~\text{s.t}~~T^n(x)\in A\})$$ I don't use ergodicity here but I don't see where it's needed, so I think it is wrong.
but also this does not help me further.
Can someone help me?
I think that a nice approach is to note that the set $$ \bigl\{x\in X: \exists k\geq 1 \text{ such that }T^kx\in B\bigr\}=\bigcup_{k\ge1}T^{-k}B=:C $$ is invariant (modulo sets of measure zero). Indeed, $T^{-1}C=\bigcup_{k\ge2}T^{-k}B\subset C$, but $$ 0\le\mu(C\setminus T^{-1}C)=\mu(C)-\mu(T^{-1}C)=0 $$ because the measure is invariant. On the other hand, $C$ has positive measure (it contains $T^{-1}B$, which has measure $\mu(T^{-1}B)=\mu(B)>0$). Since $\mu$ is ergodic, the set $C$ has full measure.