Suppose $\mathcal{A}$ is a $\sigma$-algebra, and $A_{1,1},A_{1,2},...,A_{2,1},A_{2,2},... \in \mathcal{A}$ is a double sequence of measurable sets. Under what circumstances do we have the following?

$$\lim_{n \rightarrow \infty}\ \lim_{m \rightarrow \infty}\ A_{n,m}\ =\ \lim_{n \rightarrow \infty}\ A_{n,n}$$

For real limits, uniform convergence is sufficient. Is there a similar concept for limits of sets? Is something weaker sufficient? Or is this always true?

I'm using the standard measure-theoretic definition for limits of sets; i.e. start with

$$\liminf_{n \rightarrow \infty}\ A_n = \bigcup_{N < \infty} \left(\bigcap_{n \ge N} A_n \right)$$

$$\limsup_{n \rightarrow \infty}\ A_n = \bigcap_{N < \infty} \left(\bigcup_{n \ge N} A_n \right)$$

Then a sequence of measurable sets $A_1,A_2,...$ converges when it has a limit $$\lim_{n \rightarrow \infty}\ A_n = \liminf_{n \rightarrow \infty}\ A_n = \limsup_{n \rightarrow \infty}\ A_n$$

Another way to put it: if $A = \lim_{n \rightarrow \infty}\ A_n$, then $x \in A$ implies that eventually all $x \in A_n$, and $x \notin A$ implies that eventually all $x \notin A_n$.