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I'm trying to understand the concept of mixing in dynamical systems theory, especially when the system in question has a measure-preserving flow. Here's how the condition is expressed mathematically: If $\mu$ is the measure and $\phi$ is the flow, then for all subsets $A$ and $B$ of positive measure, $\lim_{t \rightarrow \infty}\mu(\phi^{t}(B) \cap A) = \mu(B) \times \mu(A)$.

Now suppose $B$ is an arbitrary set with measure greater than 0 and less than 1. If the flow is measure preserving, then for all $t$, $\mu(\phi^{t}(B)) = \mu(B)$. Pick $A = \lim_{t \rightarrow \infty}\phi^{t}(B)$. Then, $\mu(A) = \mu(B)$. It follows that $\lim_{t \rightarrow \infty}\mu(\phi^{t}(B) \cap A) = \mu(A) = \mu(B)$.

So if the dynamics is mixing, then we will have $\mu(B) = \mu(B) \times \mu(B)$. But this is only possible if $\mu(B)$ is 0 or 1, contradicting our initial assumption.

Isn't this a problem with the definition of mixing? Is the definition in my source wrong? Or am I doing something wrong?

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    OK, I see your point. Now that I think about, it's obvious that the limit won't exist for any bounded state space if the flow is measure-preserving. Thanks for pointing out my mistake!2012-04-07

3 Answers 3

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There seems to be suspicious reasoning as pointed out in the comments when you define $A$ and then conclude that since $\mu(A)=\mu(B)$ then $\lim_{t\rightarrow\infty}\mu(\phi^t(B)\cap A)=\mu(A)$.

Have you encountered ergodic theory? Your mixing condition is more formally referred to as "strong mixing" which implies ergodicity, in that $\phi_t$ must be ergodic. By definition a flow is ergodic when $A=\phi_{-t}(A)$ implies $\mu(A)=0,1$.

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You may want to think about the notion of independent events in probability theory: two events are independent if $Pr(A \cap B) = Pr(A)\cdot Pr(B).$
So the definition you give says that in the large time limit, the events of a point being in $\phi^t(B)$ and in $A$ are independent. So however $A$ and $B$ are positioned with respect to one another, after a long time $t$, the position of $\phi^t(B)$ is completely independent of the position of $A$.

Intuitvely, the points in $B$ are being completely mixed throughout the set, independently of where they were originally positioned. Hence the term mixing.