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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 ?

edited Jul 12 2010 at 8:50

Yemon Choi
8,437●1●19●44

asked Jul 12 2010 at 8:34

ron l winger
49●3

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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? – Yemon Choi Jul 12 2010 at 8:50

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Can you expand in what do you mean by "uses usual topology"? – Franklin Jul 12 2010 at 8:50

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I propose that the OP makes his question more precise. – Martin Brandenburg Jul 12 2010 at 9:23

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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 ? – ron l winger Jul 12 2010 at 10:42

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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. – Yemon Choi Jul 14 2010 at 8:09

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Gerald Edgar · Answer 1 · 2010-07-12 13:40:17Z UTC

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.

Oleg Reinov · Answer 2 · 2010-07-12 10:40:37Z UTC

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See Lorant Schwartz's "Analysis".

answered Jul 12 2010 at 10:40

Oleg Reinov
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Before instant voting, how about some calm thinking ? The issue is simple : most of the convergent sequences used in measure theory do NOT correspond to a Hausdorff-Kuratowsky-Bourbaki, that is, usual topology. Any such usual topology would have either many more convergent sequences, of quite a lot fewer. And then the question is : Is there some other concept of topology appropriate instances of which may MATCH various sets of convergent sequences in measure theory ? Kelly in his famous 1950s book on topology lists four criteria for such a match but only in the case of usual topologies. – Elemer E Rosinger Jul 12 2010 at 14:36

Why not usual topology in measure theory ?

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