Let $T$ be a measure-preserving transformation on a probability space $(\Omega, \mathcal{F}, P)$ and let $A \in \mathcal{F}$ such that $P(A) > 0$.
(i) Show that there exists $n \geq 1 $such that $P(A ∩ T ^{−n}A) > 0$.
(ii) Using part (a), show that for a.e. $\omega \in A$, $T ^n(\omega) \in A$ for infinitely many $n ≥ 1$.
It is given that $P(T^{-1}(A))= P(A) \ \forall A \in \mathcal{F}$. How do I proceed from here on?
For (i):
Suppose for a contradiction that $P(A\cap T^{-n}(A))=0$ for all $n\geq 1$, and consider the sets $A_n=T^{-n}(A)$. Since $P(A)>0$ and $T$ is a measure-preserving transformation, $P(A_n)=P(A)>0$. On the other hand, note that for $m<n$, \begin{align*} P(A_m\cap A_n)&=P(T^{-m}(A)\cap T^{-n}(A))=P(T^{-m}(A\cap T^{-(n-m)}(A)))\\ &=P(A\cap T^{-(n-m)}(A))=0. \end{align*}
Take $k\in\mathbb{N}$ large enough such that $k\cdot P(A)>1$. Then we have
\begin{align*} P\left(\bigcup_{n=1}^k A_n\right)=\sum_{n=1}^k P(A_n)=k\cdot P(A)>1 \end{align*}
which is a contradiction since we are working in a probability space.
For (ii):
Claim: For all $m\in\mathbb{N}$, the set $\displaystyle{B_m=\bigcap_{k\geq m}(A\setminus T^{-k}(A))}$ has probability zero.
Suppose $P(B_m)>0$ for some $m\in\mathbb{N}$. Note that since $T$ is a measure-preserving transformation, the function $S=T^m$ is also a measure-preserving transformation. Thus, by (i), there is $n\geq 1$ such that $$0<P(B_m\cap S^{-n}(B_m))=P(B_m\cap T^{-n\cdot m}(B_m)).\hspace{1cm} (*)$$
On the other hand, if $\omega\in B_m\cap T^{-n\cdot m}(B_m)$ we would have $\omega\in A$, $T^{n\cdot m}(\omega)\in A$ and $T^{k}(\omega)\not\in A$. This shows that $B_m\cap S^{-n}(B_m)=\emptyset$, which contradicts $(*)$.
Now, notice that
\begin{align*} A'&=\{\omega\in A:T^n(\omega)\in A \text{ only for finitely many $n\geq 1$}\}\\ &=\{\omega\in A: \text{there is some $m$ such that $T^k(\omega)\not\in A$ for all $k\geq m$}\}\\ &=\bigcup_{m\in\mathbb{N}} B_m \end{align*} and by the claim, $\displaystyle{P(A')\leq \sum_{m\geq 1}P(B_m)=0}$. Therefore, for almost all $\omega\in A$ we have that $T^n(\omega)\in A$ for infinitely many $n\geq 1$.