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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. –  Andres Caicedo Apr 14 '11 at 2:28
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Unmeasurable is a strange word in this context! –  Mariano Suárez-Alvarez Apr 14 '11 at 3:11
    
There I go again, getting myself into trouble by using words I don't really know the definition of. –  Jon Beardsley Apr 14 '11 at 4:00

2 Answers 2

up vote 5 down vote accepted

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): http://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. –  Jon Beardsley Apr 14 '11 at 1:13
    
if you know it requires choice, why did you say "Do all such sets require the axiom of choice" –  yoyo Apr 14 '11 at 1:18
    
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. –  Jon Beardsley Apr 14 '11 at 1:50
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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. –  Arturo Magidin Apr 14 '11 at 2:02
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Okay, thanks. Interesting. Don't tell anyone, but I secretly don't believe the axiom of choice! –  Jon Beardsley Apr 14 '11 at 2:07

Here is a very readable article on the explicit construction and visualization of nonmeasureable sets on the torus.

Enjoy

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Thanks! That's cool. –  Jon Beardsley Apr 14 '11 at 2:17

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