Silly question about mixing Let $T$ be a measure-preserving transformation on a probability space $(\Omega,\mathcal B,\mu)$. Assume that for any pair of measurable sets $A,B\in\mathcal B$ with $\mu(A), \mu(B)>0$, one can find $N$ such that $T^{-n}(A)\cap B\neq\emptyset$ for all $n\geq N$. Is $T$ mixing with respect to $\mu$? This is likely to be a simple exercise. So, apologies, and thanks in advance.
 A: There is a condition known to be intermediate between the one you mention and mixing.  A transformation is lightly mixing if $\liminf_{n\to\infty} \mu(T^{-n}(A)\cap B) > 0$ for all $A$ and $B$ of positive measure.  For a transformation which is lightly mixing but not mixing, see for example Friedman and King's paper "Rank One Lightly Mixing".
A: The authors left out that $T$ is ergodic in Lemma 2.4.  If $T$ is ergodic and $\liminf_{n\to \infty}\mu (T^nA\cap A) > 0$ for all sets $A$ of positive measure, then $T$ is lightly mixing. In other words, $\liminf_{n\to \infty}\mu (T^nA\cap B) > 0$ for all sets $A,B$ of positive measure. Proof: Suppose there exists sets $A,B$ of positive measure such that $\liminf_{n\to \infty}\mu (T^nA\cap B)=0$. There exists a sequence $n_k$ such that $\lim_{k\to \infty} \mu (T^{n_k}A\cap B)=0$. Since $T$ is ergodic choose, $\ell$ such that $\mu (A\cap T^{\ell}B)>0$. Let $A'=A\cap T^{\ell}B$. Thus, $\mu (T^{n_k+\ell}A'\cap A')\leq \mu (T^{n_k+\ell}A\cap T^{\ell}B) = \mu (T^{n_k}A\cap B)$. The last equality holds by measure preservance. Therefore, $\lim_{k\to \infty} \mu (T^{n_k+\ell}A'\cap A')=0$.
