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.
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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 positive-measure 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 MorrisCommented Mar 22, 2013 at 10:42
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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. $\endgroup$– EtienneCommented Mar 22, 2013 at 16:25
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2 Answers
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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".
answered Mar 22, 2013 at 13:00
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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 positive-measure rather than nonempty intersections) is precisely light mixing. $\endgroup$ Commented Mar 22, 2013 at 14:17
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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. $\endgroup$ Commented Mar 22, 2013 at 15:05
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$\begingroup$ 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. $\endgroup$ Commented Mar 22, 2013 at 15:13
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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. $\endgroup$ Commented Mar 22, 2013 at 15:22
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1$\begingroup$ Many thanks for this answer and the reference! $\endgroup$– EtienneCommented Mar 22, 2013 at 15:58
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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, $\lim_{k\to \infty} \mu (T^{n_k+\ell}A'\cap A')=0$.
