# Having trouble understanding the concept of “mixing” in dynamical systems.

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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It just occurred to me that maybe the problem is that for a mixing system my choice of $A$ does not correspond to a measurable set. Is this the case? If so, how do we square that with the fact that the dynamics is measure-preserving? –  Tarun Apr 6 '12 at 20:09
I'm afraid your entire second paragraph doesn't make much sense. What does $\lim_{t \to \infty} \phi^t(B)$ mean? Why should $\mu(A) = \mu(B)$ imply that $\mu(A \cap B) = \mu(B \cap B)$? –  user83827 Apr 6 '12 at 20:37
$\lim_{t \rightarrow \infty} \phi^{t}(B)$ is the set of points to which $B$ evolves under the flow in the infinite time limit. –  Tarun Apr 6 '12 at 22:13
As for the second part, I agree that I've put it in an unnecessarily confusing manner. The point is that in the infinite limit, $\phi^{t}(B)$ and $A$ are the same set of points, so their intersection is just going to be that set. So I should have said $\lim_{t \rightarrow \infty} \mu(\phi^{t}(B) \cap A) = \mu(A) = \mu(B)$. The rest of the argument follows. Does this work? –  Tarun Apr 6 '12 at 22:19
Your definition of $\lim_{t \to \infty} \phi^t(B)$ still doesn't make sense, and is most likely the source of your misunderstanding. Given a point $x$, what does it mean for $x$ to be in $\lim_{t \to \infty} \phi^t(B)$? How about a concrete example: $B$ is the set of points in the unit circle subtended by central angle $\pi/2$ radians, and $\phi$ is the rotation of the circle by $1$ radian. What is $\lim_{t \to \infty} \phi^t(B)$? –  user83827 Apr 6 '12 at 22:42

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.

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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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