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Mar 7, 2012 at 18:11 comment added Neil Toronto 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.)
Mar 7, 2012 at 1:18 comment added Anthony Quas 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 $n=m$.
Mar 7, 2012 at 1:00 history asked Neil Toronto CC BY-SA 3.0