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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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    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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    @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