Why complete measure spaces? [duplicate]

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This is a longstanding confusion of mine without a clear answer. Why do we complete measure spaces? Certainly, it is nice to have the property that if some set is of measure zero, then a smaller set also should be so. However, when I looked up the mathematical literature, I am unable to find a single theorem that works better for complete measures. So, the only reasons in support of completing the measure seem to be: 1)., it agrees with our intuition of what can be neglected, 2)., the Caratheodory extension process automatically gives a complete measure.

I was also surprised to realize that in probablity theory, very little use is made of the Lebesgue $\sigma$-algebra. Almost always the Borel $\sigma$-algebra is used. This prominence of Borel $\sigma$-algebras also seems to be the case in ergodic theory, where one considers for instance the space of all probability measures on a given measurable space(usually equipped with a Borel $\sigma$-algebra), and the null sets might differ from one measure to other(for example, consider Dirac measures concentrated at different points).

So, what are some better mathematical reasons to argue for complete measures? Are there some theorems that I do not know, which are true only for complete measures? There ought to be some, to give credence to all W. Rudin's going-on about the process of completing a measure being as significant in analysis, if not more than, the process of completing the rationals into the real numbers. One can readily find a number of theorems in analysis that would break without the least upper bound property of real numbers. Such does not seem to be the case with completing the measure space, with a superficial look.

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marked as duplicate by 900 sit-ups a day, Steven Taschuk, Lost1, T. Bongers, G.T.RJul 2 at 19:29

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See Why do we essentially need complete measure space? for a discussion of a few of the points you raise. Also related the MO thread Why do probabilists take random variables to be Borel (and not Lebesgue) measurable? –  t.b. Aug 30 '12 at 21:01
I'm under the impression that it's mostly a matter of convenience. If one wants to show something is true almost everywhere, then this is often achieved by proving that the set of exceptions has outer measure zero. By considering the completion of a measure you don't need to worry about measurability of such a set, which makes things easier. –  Sam Aug 30 '12 at 21:47

1 Answer

In the theory of stochastic processes it is nearly always assumed that the filtrations we are working with are the ones satisfying the so called usual conditions, i.e. they are right continuous and their initial (and thus every) sigma-field is complete.

While it is hard to give some good philosophical explanation of that state of affairs, one thing is certain: the usual conditions are really needed from the technical point of view, they are necessary to perform many constructions and prove theorems in the theory of stochastic processes.

The general theory of stochastic integration is developed with respect to filtrations satisfying usual conditions and complete probability spaces.

If you would like to see the stochastic integration in its most general form and many deep theorems requiring the usual conditions assumption about filtrations and the completeness assumption about probability spaces see "Probabilities and potentials" by Dellacherie and Meyer and "Semimartingales" by Metivier.

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One can do stochastic integration without the usual conditions and sometimes they are totally inappropriate (e.g. in connection with Girsanov's theorem). However, there is one important result, the debut theorem, whose full generality requires the usual conditions. –  Jochen Aug 31 '12 at 7:00
@Jochen Is it really possible to develope the stochastic integration theory with respect to arbitrary (not necessarily continous!) martingales without the usual conditions assumption? –  Godot Aug 31 '12 at 13:03