When are iterated limits of sets equal to a double limit? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:03:04Z http://mathoverflow.net/feeds/question/90415 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90415/when-are-iterated-limits-of-sets-equal-to-a-double-limit When are iterated limits of sets equal to a double limit? Neil Toronto 2012-03-07T01:00:33Z 2012-03-07T01:00:33Z <p>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?</p> <p>$$\lim_{n \rightarrow \infty}\ \lim_{m \rightarrow \infty}\ A_{n,m}\ =\ \lim_{n \rightarrow \infty}\ A_{n,n}$$</p> <p>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?</p> <p>I'm using the standard measure-theoretic definition for limits of sets; i.e. start with</p> <p>$$\liminf_{n \rightarrow \infty}\ A_n = \bigcup_{N &lt; \infty} \left(\bigcap_{n \ge N} A_n \right)$$</p> <p>$$\limsup_{n \rightarrow \infty}\ A_n = \bigcap_{N &lt; \infty} \left(\bigcup_{n \ge N} A_n \right)$$</p> <p>Then a sequence of measurable sets $A_1,A_2,...$ converges when it has a limit $$\lim_{n \rightarrow \infty}\ A_n = \liminf_{n \rightarrow \infty}\ A_n = \limsup_{n \rightarrow \infty}\ A_n$$</p> <p>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 $x \notin A_n$.</p>