We know that existence of a Lebesgue non-measurable set follows from the Axiom Of Choice. Is the converse true? That is, does the existence of a Lebesgue non-measurable set imply the Axiom Of Choice?

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    $\begingroup$ The existence of a non-measurable set is not only consistent with AC, it follows from it (assuming ZF). Also, the question "does the existence of a non-measurable set imply AC?" is not the converse of "non-measurable sets are consistent with AC." So in the first sentence of your question, you should've said "follows from" rather than "is consistent with." $\endgroup$ Aug 28 '11 at 21:00

No, the existence of a non-Lebesgue measurable set does not imply the axiom of choice. If ZF is consistent, then set-theorists can construct models of ZF having a non-Lebesgue measurable set, but still not satisfying AC.

This is quite reasonable, because the existence of a non-Lebesgue measurable set is a very local assertion, having to do only with sets of reals, and thus can be satisfied with a small example, by set-theoretic standards. The axiom of choice, in contrast, is a global assertion insisting that every set, even a very large set, has a well-order. So we don't expect to turn a mere non-measurable set into well-orderings of enormous sets, such as the power set $P(\mathbb{R})$.

And indeed, one can use forcing to produce a model $L(P(\mathbb{R})^{V[G]})$ which satisfies $ZF+\neg AC$, for similar reasons as in the usual $\neg AC$ models, but since it has the true $P(\mathbb{R})$, it will have all the same non-Lebesgue measurable sets as in the ambient ZFC universe $V[G]$.

(Finally, let me make a minor objection to the question: consistency is a symmetric relation, and so if $A$ is consistent with $B$, then $B$ would be consistent with $A$, and so one wouldn't ordinarily speak of a "converse". You seem instead to be refering to the implication that AC implies there is a non-measurable set, and this is how I took your question.)

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    $\begingroup$ Joel, how large of a fragment is still consistent with non-measurable sets? I mean, I know that DC should be enough, and probably ACC. Is it known about lower limits? $\endgroup$
    – Asaf Karagila
    Aug 28 '11 at 19:52
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    $\begingroup$ Asaf, I dont quite understand what you are asking. Full AC is consistent with non-measurable sets, since it implies that they exist. Perhaps you mean to ask how large a fragment is consistent with there being NO non-measurable sets, and for this it is at least DC, since this is what you get in Solovay's model, and I'm not sure how much more you can have. Perhaps you mean, what fragment of AC is sufficient to prove the existence of non-Lebesgue measurable sets, and for this I am unsure, and it seems unlikely to me that the optimal amount will be expressed directly as a choice principle. $\endgroup$ Aug 28 '11 at 20:46
  • $\begingroup$ Is the strong denial of AC consistent with ZF? That is, can it be that every infinite collection of sets does NOT have a choice function? $\endgroup$
    – Jeff Strom
    Aug 28 '11 at 22:42
  • $\begingroup$ Jeff, no, that is too strong. For any set $I$, no matter how large, we can make a family of singletons { {i} | i in I }, and this clearly has a choice function. There are also many other examples, such as canonical copies of the same set, etc., and these always have choice functions. $\endgroup$ Aug 28 '11 at 22:57
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    $\begingroup$ Following up one of Joel's comments, what is known is the following: ZF+DC+"every set of reals is measurable" is consistent (relative to the existence of an inaccessible), but ZF + DC + "there is a set of $\aleph_1$ reals" implies the existence of a non-measurable set of reals. I don't know if the boundary has been made tighter since Shelah's original paper, from whence the above statements have been lifted. $\endgroup$ Sep 29 '11 at 3:11

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