I'm trying to solve this exercise but I need some hints because my teacher didn't give me the theory necessary to solve it.
Let $(X, M, \mu)$ be a measurement space such that $\mu (X) = 1$. Suppose that $T: X → X$ is measurable and $\mu (T^{−1} (E)) = \mu (E)$ for all $E\in M$. Prove that
For all $E ∈ M$ such that $\mu (E)> 0$ there exists a natural $n$ such that $\mu (E \cap T^n (E))> 0$. Here $T^0$ is the identity map in $X$ and $T^n = T \circ T^{n − 1}$ for $n \geq 1$.
For every $E$ the set of points for which there is a natural $n_0$ such that $T^n(x)$ $\notin$ $E$ for all $n \geq n_0$ has measure $0$.
Let $\mu(E) = \alpha > 0$. Consider the sets $E, T^{-1}E, T^{-2}E, \dots$. This is a countable list of sets, each with measure $\alpha$. Therefore, since $\mu(X) = 1$, some pair of them must overlap in a set of positive measure, i.e. there exists $n > m$ such that $\mu(T^{-n}E \cap T^{-m}E) > 0$. But this is also equal to $\mu(T^{-n}(E \cap T^{n-m}E)) = \mu(E \cap T^{n-m}E)$.
We need to show that $\mu$-a.e. $x \in E$ satisfies the statement: "for all $k$ there exists $n>k$ such that $T^n x \in E$." Define $E' = \bigcup_{n=0}^{\infty} T^{-n} E$. Fix an arbitrary $k$ and note that $T^{-k}E' = \bigcup_{n=k}^{\infty} T^{-n}E$, so we have $$ E' = (E \cup T^{-1}E \cup \dots \cup T^{-k+1}E) \cup T^{-k}E'. $$ Because $T$ is measure preserving we must have $\mu(E') = \mu(T^{-k}E')$, so the above implies that $(E \cup T^{-1}E \cup \dots \cup T^{-k+1}E) \setminus T^{-k}E'$ has measure zero. In particular, almost every $x \in E$ is also in $T^{-k}E'$. This shows that almost every $x \in E$ satisfies $T^n x \in E$ for some $n\geq k$. Thus we've proved that for each $k$ there is a set of full measure in $E$ with the desired property. To finish, take a countable intersection over all $k$ to obtain a fixed set of full measure in $E$ that has the desired property for all $k$, as required.