Measure theory was introduced in the early 1900s by Lebesgue, at the same time with Hausdorff introducing the usual concept of topology, and publishing it in his book just before World War I. Measure theory is full with convergence and limit properties of measures, functions and integrals. Yet none of them uses usual toplogy. Is there a well thought out reply to why such a thing happens ?
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2$\begingroup$ I would strongly dispute the penultimate sentence. The probabilists' usual notion of convergence of probability measures is a topological notion. Also, what is "usual topology" meant to be? $\endgroup$– Yemon ChoiCommented Jul 12, 2010 at 8:50
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2$\begingroup$ Can you expand in what do you mean by "uses usual topology"? $\endgroup$– O.R.Commented Jul 12, 2010 at 8:50
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4$\begingroup$ I propose that the OP makes his question more precise. $\endgroup$– Martin BrandenburgCommented Jul 12, 2010 at 9:23
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1$\begingroup$ Well, let us take one of the most important theorems, namely, the Lebesgue dominated convergence theorem for integrals. Where is the HKB, or in other words, usual topology which has as convergent sequences precisely those in that theorem, and none less, and none more ? $\endgroup$– ron l wingerCommented Jul 12, 2010 at 10:42
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1$\begingroup$ In any case, I don't quite follow why one should feel compelled to fit convergence a.e. in Procrustean fashion into what you and ron l winger call "usual topology". It is what it really is, to use your own phrase. Just as weak convergence of probability measures, a fairly standard and possibly important part of the theory is a functional-analytic notion and is precisely the convergence of a sequence in a certain weak topology. $\endgroup$– Yemon ChoiCommented Jul 14, 2010 at 8:09
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1 Answer
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It is true that convergence a.e. is not the convergence of a topology. If you want a topology you can go to convergence in measure. A theorem MORE GENERAL THAN the dominated convergence theorem: let $f_n \to f$ in measure on a set $E$ of finite measure, and suppose there is an integrable $g$ with $|f_n| \le g$ a.e. Then $\int_E f_n \to \int_E f$. SPECIAL CASE: $f_n \to f$ a.e.
answered Jul 12, 2010 at 13:40