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.
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 $A, B$. $\square$