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Let's say we construct measure theory without the axiom of choice in our pocket. Since we don't have to worry about unmeasurable sets anymore (see this thread), is there any good reason one might define a measure on a $\sigma$-algebra other than the power set? If not, would $\sigma$-algebras still serve any purpose at all? Or are they used in analysis solely to sidestep problems of unmeasurability?

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The failure of the Axiom of Choice does not imply that all sets of reals are Lebesgue measurable. The question you link to only indicates that it is consistent with the failure of the Axiom of Choice that all sets of reals are Lebegsue measurable. (Also, Asaf appears to explicitly mention that the Axiom of Choice could fail in general, but still apply to "small enough" sets so that the usual construction of, say, a Vitali set can go through.) –  Arthur Fischer Jan 7 '13 at 8:20
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You don't mean "without the axiom of choice" -- you mean "without the axiom of choice and assuming the axiom that all subsets of $\mathbb{R}$ are Lebesgue measurable". –  Hurkyl Jan 7 '13 at 8:20
    
Yes, you're both right. Sorry for the sloppy phrasing. –  GMB Jan 7 '13 at 8:24
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2 Answers 2

Non-measurability is not a problem per se, in other words, it is not a priori desirable to make every subset measurable, even when one can. In fact, conditioning (on non trivial sigma-algebras) is seen by some as the first and foremost distinctive tool used in probability. One wants to condition on sigma-algebras G which are neither the full power set nor the trivial sigma-algebra, just like, in $L^2$ spaces, one wants to consider projections which are neither the identity nor the mean. Then, subsets not in the sigma-algebra G are, even for a while, even in finite probability spaces, non-measurable.

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Does conditioning really change the $\sigma$-algebra? Conditioning will change the measure we work with, but I can't imagine it giving an unmeasurable set a probability or making a measurable set have no defined probability. Also, unrelatedly: I've been asking a ton of questions here lately and you've answered basically all of them. Thanks mate, I appreciate all the help. –  GMB Jan 7 '13 at 8:35
    
Conditioning does replace the probability of measurable sets by something different, namely random variables. But my main point here is that in natural situations one must juggle with several different sigma-algebras, and that one often wants to, and that this runs contrary to viewing as a defect the fact that (due to AC or whatever) one cannot always pick the full power set as sigma-algebra. –  Did Jan 7 '13 at 9:12
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The answer is a mixture of yes and no. $\sigma$-algebras serve the purpose of defining the collection of sets you want to measure. In measure theory with AC one may naively wish to measure all subsets (say of $\mathbb R$) but that is impossible and thus one must consider a $\sigma$-algebra strictly smaller than $\mathcal P (X)$. However, while it is true that in the absence of AC it may be the case that all subsets of $\mathbb R$ are measurable it still does not mean that you will necessarily want such a measure. For instance, if you are interested in Borel measurability then you don't really need the more general Lebesgue measure, let alone a measure in which all subsets are measurable.

So, much like anything at all, it all depends on what you want to do. Rarely are you interested in setting up a measure on as many sets as possible. Rather you want the measure to measure just what you need. Tailoring the measure to the purpose at hand is the point and if the most natural $\sigma$-algebra is strictly smaller than what is actually possible then so be it. This can be seen as a case of Polya's principle of the right amount of generality. There is no point of generalizing for the sake of generalization.

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