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I have seen a proof showing that there are subsets of $\mathbb{R}$ which are not Lebesgue measurable. If I recall correctly it uses the axiom of choice. My first question is, are there sensible sets that are not measurable, i.e. something I can actually be given a description of and not just be shown to exist, or at least, has somebody found one? Do all such sets require the axiom of choice, and does the existence of such sets imply the axiom of choice?

Thanks for any ideas, or any good references (preferably online)!

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    Lebesgue measurability is a much more general notion than Borel measurability, so your comment "perhaps not even Borel measurable" is strange.2011-04-14
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    *Unmeasurable* is a strange word in this context!2011-04-14
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    There I go again, getting myself into trouble by using words I don't really know the definition of.2011-04-14

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you must use choice. one example is given by considering a collection of coset representatives for $\mathbb{R}/\mathbb{Q}$ ( http://en.wikipedia.org/wiki/Vitali_set ). another example are the sets involved in the banach tarski paradox. most real analysis books will have a discussion of this; folland has references at the end of chapter 1 of real analysis modern techniques and their applications.

here is a mathoverflow discussion of the topic (concerning choice): https://mathoverflow.net/questions/42215/does-constructing-non-measurable-sets-require-the-axiom-of-choice

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    Sure, the Vitali set. I am familiar with this construction. It is not however a very nice construction, and is often given as an example. Did you read the question? I know that such sets require the axiom of choice. I am more interested in other examples.2011-04-14
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    if you know it requires choice, why did you say "Do all such sets require the axiom of choice"2011-04-14
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    Oh boy. My question is, do all unmeasurable sets require the axiom of choice. Not just Vitali sets, which is a familiar class of unmeasurable sets.2011-04-14
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    @JBeardz: Yes, they all require the Axiom of Choice: Solovay constructed a set theory in which the Axiom of Choice is false, but where every subset of the real numbers is measurable.2011-04-14
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    Okay, thanks. Interesting. Don't tell anyone, but I secretly don't believe the axiom of choice!2011-04-14
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    @yoyo: Thanks, that link to MO is really helpful!2011-04-14
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    @JBeardz: your secret is safe with me. But by the way, not believing in AC gives you a wide array of options as to exactly what you do believe. *Only* believe in the axioms of Zermelo-Fraenkel set theory is not a very tenable position for those interested in many parts of mathematics. So you have to decide **how much choice** you want to have. For instance, there is known to be a weak form of AC which does not allow measurable sets but does allow reasonable set constructions analysts want to make (e.g. countable choice). This actually has some advantages over full belief...2011-04-14
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    ...Namely when you try to prove Fubini's Theorem, the real pain is to show that certain functions are measurable. With just the right amount of choice, all functions become measurable and Fubini's Theorem becomes much easier to prove! (Nevertheless, I confess that I believe in AC -- or at least, I know that it is equivalent to plenty of other results in algebra and topology that I can't live without, so I am definitely willing to use it. As a practicing mathematician who is not a set theorist, I've learned not to take my own "beliefs" about sets very seriously...)2011-04-14
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    @Pete L. Clark: the construction of a model where all functions are measurable is harder than just assuming the correct fragment of choice. Even if we only assume ZF (with no choice at all) this is still *consistent* with AC, so we can't hope to prove that every set of reals is measurable. To prove that, we have to explicitly assume M = "every set of reals is measurable" as an axiom, or something that implies M. But that leaves us with the problem of proving that the system ZF + M is consistent. This is essentially what Solovay's proof does; he showed that ZF + M + DC is consistent.2011-04-14
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    @Carl: sure, I don't disagree with anything you say (and I wouldn't dare, since you know this stuff much better than I do). My point was that there are some set theoretic axioms under which (i) every set is measurable and (ii) you have enough choice to do things you want to do in analysis. To get both of these, neither ZF nor ZFC fits the bill, but rather something with an intermediate amount of choice (and possibly including other axioms as well)...2011-04-14
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Here is a very readable article on the explicit construction and visualization of nonmeasureable sets on the torus.

Enjoy