# 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?

-
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.
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 $\mu(A)=0,1$.