Under a measure preserving transformation of a probability space, I need to show the following. Given a measure preserving system (X,$\mathcal{B}$, $\mu$, T) where $\mu$ is a probability measure, I want to show that $\forall$ A $\in\mathcal{B}$ and $\forall \epsilon >$ 0, $\exists$ n $\in\mathbb{N}$ such that $\mu$(A $\cap$ T$^{n}$A) $>$ $\mu^{2}$(A) - $\epsilon$. 
I tried the way of contradiction which gives existence of an A $\in\mathcal{B}$ and $\epsilon_{0}$ $>$ 0 such that $\forall$ n $\in\mathbb{N}$ , $\mu$(A $\cap$ T$^{n}$A) $\leq$ $\mu^{2}$(A) - $\epsilon_{0}$ $<$ $\mu^{2}$(A) $<$ $\mu$(A). 
I don't know where to go from here. Suggestion for any other approach would be great too. Thanks
 A: Minor comment before proceeding: you probably mean to assume that $T$ is measurably invertible. Otherwise, $T^n A$ need not be measurable, and so you'd have a hard time defining $\mu(A \cap T^n A)$. Another fix is to replace $T^n A$ with $T^{-n} A$, which is always measurable in your setting. I'll pursue the latter fix in what follows.
Let's assume first that $\mu$ is ergodic. By the Birkhoff ergodic theorem, $S_N := \frac{1}{N} \sum_0^{N-1} \chi_A(T^n x)$ converges $\mu$-almost surely to $\mu(A)$. Applying the Bounded Convergence Theorem to the sequence of functions $\chi_A \cdot S_N$, we obtain
$$
(*)\, \, \,  \mu(A)^2 = \lim_{N \to \infty} \frac{1}{N} \sum_{n = 0}^{N-1} \mu(A \cap T^{-n} A) 
$$
We obtain that for all $\epsilon > 0$, there exists $N = N_{\epsilon}$ such that
$$
\mu(A)^2 - \epsilon < \frac{1}{N} \sum_{n = 0}^{N-1} \mu(A \cap T^{-n} A) \, .
$$
One concludes that for some $n \in \{ 0, \cdots, N-1\}$, we have $\mu(A)^2 - \epsilon < \mu(A \cap T^{-n} A)$.
If $\mu$ is not ergodic, then the analogue of the formula (*) is
$$
\int (\chi^*_A)^2 d \mu = \lim_{N \to \infty} \frac{1}{N} \sum_{n = 0}^{N-1} \mu(A \cap T^{-n} A) \, ,
$$
where $\chi_A^*$ is the conditional expectation of $\chi_A$ w.r.t. the sigma algebra $\mathcal I$ of $T$-invariant sets in $\mathcal B$. By Jensen's inequality applied to the convex function $z \mapsto z^2$,
$$
\int (\chi^*_A)^2 d \mu \geq \left( \int \chi^*_A d \mu \right)^2 = \mu(A)^2 \, , 
$$
hence we can replace $(*)$ with
$$
(**) \, \, \,  \mu(A)^2 \leq \lim_{N \to \infty} \frac{1}{N} \sum_{n  = 0}^{N-1} \mu(A \cap T^{-n} A)
$$
and proceed as in the ergodic case. 
Addendum: nonergodic case
By the Birkhoff Ergodic Theorem, we have that $S_N$ converges to $\chi_A^*$ $\mu$-almost surely, hence
$\chi_A \cdot S_N$ converges $\mu$-almost surely to $\chi_A \cdot \chi_A^*$. Taking a $\mu$ integral of both sides of this limit, we conclude
$$
\int \chi_A \cdot \chi_A^* d \mu = \lim_{N \to \infty} \frac{1}{N} \sum_{n = 0}^{N-1} \mu(A \cap T^{-n} A) \, .
$$
The left-hand side coincides with $\int(\chi_A^*)^2 d \mu$ by basic properties of conditional expectations.
