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Is axiom of choice required to show the existence of non-measurable sets? Is there a Lebesgue non-measurable set that can be constructed without axiom of choice?

Related question on MO says it is consistent: https://mathoverflow.net/questions/73902/axiom-of-choice-and-non-measurable-set

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    Solovay proved that there is a model of Set Theory without Choice where every subset of the real numbers is measurable. In that model, the Axiom of Dependent Choice is true. So you need at least something stronger than DC in order to establish the existence of non-measurable sets.2012-04-19
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    Yes, you can construct non-measurable sets using non-principal ultrafilters and the existence of such a thing is strictly weaker than AC: http://terrytao.wordpress.com/2008/10/14/non-measurable-sets-via-non-standard-analysis/2012-04-19
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    That's very interesting. I had thought that the existence of ultrafilters was equivalent to AC. Thanks!2012-04-19
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    @Neal: AC is equivalent to the statement "Every lattice has an ultrafilter." It is strictly stronger than "There is a non-principal ultrafilter over $\mathbb{N}$." This is perhaps the source of your confusion.2012-04-19
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    @Neal: It is possible to have a model in which the ultrafilter lemma holds but there is a countable family of countable sets whose product is empty.2012-04-19
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    @Cameron: I actually believe that the main source of confusion is the fact that people are usually ignorant to how much choice is *really* needed, and they simply use Zorn's lemma to do things. This gives the impression that many things require full choice while some require none.2012-04-19
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    True. Choice certainly makes proofs easier, but unnecessary choice is misleading.2012-04-19
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    @Neal, too add even more it is possible that every infinite set *has* a free ultrafilter but not every filter can be extended to ultrafilters. So the existence of free ultrafilters is not enough to even ensure the extension of filters to ultrafilters!2012-04-19
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    Related mathoverflow question: http://mathoverflow.net/questions/73902/axiom-of-choice-and-non-measurable-set2012-06-04

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The answer is, you cannot.

It is consistent with ZF that the real numbers are a countable union of countable sets, this implies that every set of reals is Borel and therefore measurable. Of course, in such model it is nearly impossible to develop the analysis we know.

However it is consistent relative to an inaccessible cardinal that there is a model of ZF+DC where all the sets of real numbers are Lebesgue measurable, and DC allows us to do most of classical analysis too.

Non-measurable sets can be generated by free ultrafilters over $\mathbb N$ too, which as remarked is a strictly weaker assumption that the axiom of choice. If there are $\aleph_1$ many real numbers and DC holds then there is an non-measurable set as well, which implies that ZF+DC($\aleph_1$) also implies the existence of non-measurable sets of real numbers - however this is not enough to imply the existence of free ultrafilters over the natural numbers!

Several other ways to generate non-measurable sets of real numbers:

  1. The axiom of choice for families of pairs;
  2. Hahn-Banach theorem;
  3. The existence of a Hamel basis for $\mathbb R$ over $\mathbb Q$.

There are several other ways as well, but none are quite close to the full power of the axiom of choice.

One important remark is that we can ensure that the axiom of choice holds for the real numbers as usual, but breaks in many many severe ways much much further in the universe (that is counterexamples will be sets generated much later than the real numbers in the von Neumann hierarchy). This means that the axiom of choice is severely negated - but the real numbers still behave as we know them.

The above constructions and to further read about ways to construct non-measurable sets cf. Horst Herrlich, Axiom of Choice, Lecture Notes in Mathematics v. 1876, Springer-Verlag (2006).

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    Can you provide a source of the Hahn-Banach way to generate a non-measurable?2012-04-19
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    @leo: The HB theorem is enough to prove the Banach-Tarski paradox.2012-04-19