Let $T$ be a measurepreserving 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.

4$\begingroup$ Perhaps you want $\mu(T^{n}A \cap B)>0$ rather than nonemptiness, since the former is more natural in a probability space. This is certainly not a silly question. In the positivemeasure form this condition implies weak mixing by Theorem 4.31 in Furstenberg's book "Recurrence in Ergodic Theory and Combinatorial Number Theory". In Parry's book "Topics in Ergodic Theory" (p.89) a transformation is discussed which is weak mixing but does not meet this condition. I suspect that the answer to your question is positive but it may not be widely known. $\endgroup$– Ian MorrisMar 22, 2013 at 10:42

1$\begingroup$ You're right, positive measure is definitely more natural than nonemptiness. But perhaps the requirements are in fact the same, since one is quantifying over all measurable sets. $\endgroup$– EtienneMar 22, 2013 at 16:25
2 Answers
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".

1$\begingroup$ What a fascinating paper! If I'm not wrong, when $T$ is invertible, Lemma 2.4 in that article shows that Étienne's condition (with positivemeasure rather than nonempty intersections) is precisely light mixing. $\endgroup$ Mar 22, 2013 at 14:17

1$\begingroup$ Actually now I am suspicious of Lemma $2.4$. In particular, the identity transformation satisfies the condition which is listed there as equivalent to lightly mixing. But the identity is definitely not lightly mixing. I haven't looked to see how this is involved with the details of the paper, but I'll see if I can find another source for a result like this. $\endgroup$ Mar 22, 2013 at 15:05

$\begingroup$ You're right: if $T$ is not lightly mixing then there is no reason why we should be able to find $E$ such that $\liminf_{n \to \infty}\mu(T^{n}E\cap E)=0$. I think that the authors err when they state that in order to check light mixing it is sufficient to check the case $A=B$: this is fine for weak, strong and probably mild mixing because in those cases the relevant expressions are linear in $\chi_A$ and $\chi_B$, but lim inf is of course not linear. $\endgroup$ Mar 22, 2013 at 15:13

1$\begingroup$ Here is a correct modification of Lemma 2.4: if $\mu(T^{n}A \cap B)$ is eventually nonzero whenever $\mu(A)$ and $\mu(B)$ are both nonzero then $T$ is light mixing. Proof: suppose that $T$ is not light mixing. Choose $A,B$ with $\mu(A),\mu(B)>0$ and $\liminf_{n \to \infty} \mu(T^{n}\cap B)=0$. Choose a strictly increasing sequence $(n_k)$ such that $\mu(T^{n_k}A \cap B)<3^{k}\mu(B)$ for all $k \geq 1$. Let $C:=B \setminus \bigcup_{k=1}^\infty \left(T^{n_k}A \cap B\right)$. Then $\mu(C)>\frac{1}{2}\mu(B)>0$ and $\mu(T^{n_k}A \cap C)=0$ for all $k$, a contradiction. $\endgroup$ Mar 22, 2013 at 15:22

1$\begingroup$ Many thanks for this answer and the reference! $\endgroup$– EtienneMar 22, 2013 at 15:58
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, $\lim_{k\to \infty} \mu (T^{n_k+\ell}A'\cap A')=0$.