An equivalent condition for strong-mixing
For a measure-preserving (finite) system $(X,\mathcal{B},\mu,T)$, is it correct that the following are equivalent?
For every $A,B\in\mathcal{B}$ , $\displaystyle\lim_{n\rightarrow\infty}\mu(A\cap T^{-n}B)=\mu(A)\mu(B)$.
For every $A,B\in\mathcal{B}$ of positive measure, there is some $n_0\in\mathbb{N}$ such that for every $n>n_0$, $\mu(A\cap T^{-n}B)>0$.
Clearly 1 implies 2. Is the opposite direction also correct?
Solution 1:
Yes, I'm sure. The definition of lightly mixing is (2.1) in the Rank-one lightly mixing paper: $\liminf_{n\to \infty} \mu(T^nA\cap B)>0$ for $A,B$ of positive measure. This definition implies (2) above. Lemma 2.4 of the paper shows the lightly mixing definition is equivalent to (2) where B=A (as you observed!). See the proof on page 2.
The Rank-one lightly mixing paper gives an example of a measure preserving transformation that satisfies (2) but is not strongly mixing. This shows that (1) and (2) are not equivalent.
As a side note, this rank-one transformation is attributed to Chacon from a paper in 1969 "Weakly mixing transformations which are not strongly mixing". Possibly, Chacon already knew lightly mixing was distinct from strongly mixing? Also, Peter Walters may have been the first to introduce lightly mixing in print, but he called it intermixing (circa 1970). I don't have the reference right now, but presumably Walters knew lightly/intermixing mixing is different than strongly mixing. Ciao
Solution 2:
The answer is NO. The second condition is called lightly mixing, while the first condition is called strongly mixing. It was shown in a paper in 1991 that these conditions are not equivalent. There exists a lightly mixing measure preserving system which is not strongly mixing: http://link.springer.com/article/10.1007%2FBF02773841?LI=true.