Let $X, \mathcal{A}, \mu$ be a probability space and $T: X \rightarrow X$ a measure preserving measurable map (i.e. $\mu (T^{-1}(A)) = \mu (A)$ for all $A \in \mathcal{A}$). We say $T$ is mixing for sets $A, B \in \mathcal{A}$ if $$\lim_{n\rightarrow \infty}\mu \{T^{-n}(A) \cap B \} = \mu (A) \mu (B)$$

I'm trying to show that it suffices to check the mixing property on an algebra generating $\mathcal{A}$. In other words:

if $\mathcal{A}_0$ is an algebra that generates $\mathcal {A}$, then $T$ is mixing for all $A, B \in \mathcal {A}$ if $T$ is mixing for all $A, B \in \mathcal {A}_0$.

I figured the easiest thing to do would be start by showing that, for fixed $B$, the set

$$\Gamma _B := \{A : T \text{ is mixing for the pair } A,B\}$$

is a "monotone class" (i.e. closed under nested countable unions and intersections). This would show $\mathcal {A} \subset \Gamma _B$. Then try to make the same argument with $A$ and $B$ flipped. However, when I do this I run into a double limit that doesn't seem to commute.

I'm pretty sure that I can show (using the dominated convergence theorem) that $\Gamma _B$ is closed under countable disjoint unions as well as compliments. Therefore it would also suffice to show its closed under finite intersection. But I'm not sure how to do that either.

  • 1
    $\begingroup$ Would using this help? math.stackexchange.com/questions/228998/… $\endgroup$ – Evan Aug 13 '13 at 4:56
  • $\begingroup$ @Evan: Thanks! It certainly looks promising. $\endgroup$ – Tim kinsella Aug 13 '13 at 5:00
  • $\begingroup$ Oh! I noticed the last bullet point just now :). Guess it just remains to put it to use (useful for testing mixing). $\endgroup$ – Evan Aug 13 '13 at 5:03

This follows almost immediately from the statement Evan linked to in the comments (Thanks to @Evan and @DavideGiraudo !).

Let $A, B \in \mathcal{A}$ and $\epsilon >0$. Choose $A_0, B_0 \in \mathcal{A}_0$ such that $A \Delta A_0, B \Delta B_0 < \epsilon$. Then since $T$ is measure preserving, $T^{-n}(A) \Delta T^{-n}(A_0) < \epsilon$ and similarly for $B, B_0$. We therefore have $$|\mu \{T^{-n} A \cap B \} - \mu \{T^{-n} A_0 \cap B_0 \}| <2 \epsilon$$ for all $n$. Now, by hypothesis

$$\mu \{T^{-n} A_0 \cap B_0 \} \rightarrow \mu (A_0)\mu (B_0)$$ We therefore have,

$$| \limsup_n \mu \{T^{-n} A \cap B \} - \mu (A) \mu (B) | \leq \\ | \limsup_n \mu \{T^{-n} A \cap B \} - \mu (A_0)\mu (B_0) | + | \mu (A_0)\mu (B_0) - \mu (A) \mu (B) | \leq \\ 2 \epsilon + \epsilon \{\mu (A)+ \mu (B) \} +\epsilon ^2$$ And similarly for $\liminf$. Since $\epsilon$ was arbitrary, $T$ is mixing for $A, B$. $\square$

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