Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $x \notin A_n$.

share|improve this question
    
Presumably you want this almost everywhere? Otherwise the measurability is not doing anything for you. It's certainly not always true: Let $A_{n,m}$ be all of $\mathbb N$ for $n\ne m$ and $\{n+1,n+2\ldots\}$ for $n=m$. –  Anthony Quas Mar 7 '12 at 1:18
    
Ah, of course. Then I'm looking for an analogue of uniform convergence. I'd like to do this without a measure if I can, so I don't want this to hold almost everywhere just yet. I suppose I gave the $\sigma$-algebra just in case it could help somehow. (I doubt it can, though.) –  Neil Toronto Mar 7 '12 at 18:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.