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Every known construction of the Vitali set relies on the Axiom of Choice. It happens to not be Lebesgue-measurable.

Must every set whose construction relies on the Axiom of Choice not be Lebesgue-measurable?

I'm not sure if "relies on the Axiom of Choice" is mathematically meaningful, in which case I'll rephrase the question as asking whether there are any sets for which there is an existence proof with AC but no known existence proof without AC, which are known to be Lebesgue-measurable.

This related question asks whether there exists a Lebesgue non-measurable set that can be constructed without AC. The answer to that question is no.

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    $\begingroup$ Some subset of the Cantor set presumably has this property. $\endgroup$
    – André Nicolas
    Jan 21, 2016 at 6:34
  • $\begingroup$ Sorry. I just woke up. Next time, it doesn't hurt to put your actual question in a block quote to make it stand out! $\endgroup$
    – Asaf Karagila ♦
    Jan 21, 2016 at 6:36
  • $\begingroup$ @AsafKaragila Thanks for tip, added block quote. $\endgroup$
    – dshin
    Jan 21, 2016 at 6:37
  • $\begingroup$ @AndreNicolas: Every subset of the Cantor set is a null set and thus Lebesgue measurable (but not necessarily Borel measurable). $\endgroup$
    – PhoemueX
    Jan 21, 2016 at 6:38
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    $\begingroup$ I have read the statement that Solovay (maybe written differently) constructed a model of ZF (without choice) in which every subset of the reals is Lebesgue measurable. Thus, without using the axiom of choice, you will not be able to construct such a set. $\endgroup$
    – PhoemueX
    Jan 21, 2016 at 6:42

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The answer is negative.

An easy example is subsets of the Cantor set which have null measure, but can be "very non measurable". But this is sort of cheating since the Cantor set can be endowed with its own measure which is isomorphic to the actual Lebesgue measure on the unit interval, and there these subsets of the Cantor set need not be measurable themselves.

But there is an alternative. It is a nice theorem that the continuous image of a closed set Lebesgue measurable. Such set need not be a Borel set, and it is called an analytic set. But it is also consistent without the axiom of choice that every set is a Borel set, in which case there are no analytic sets which are not Borel.

So an analytic set would be Lebesgue measurable, but you need the axiom of choice to prove there is one which is not Borel.

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  • $\begingroup$ Thanks for the answer. Is it meaningful to add the condition that every known constructible Lebesgue-measurable superset has positive Lebesgue-measure? If so, does that change the answer? I believe this condition would differentiate the Vitali set from a Cantor set subset. $\endgroup$
    – dshin
    Jan 21, 2016 at 7:43
  • $\begingroup$ No, it's not meaningful because every open set has a copy of the Cantor set inside which has a null measure. Or alternatively you can always take the union with an interval. $\endgroup$
    – Asaf Karagila ♦
    Jan 21, 2016 at 7:58
  • $\begingroup$ Sorry, just deleted my prior comment. I understand your response now. Thanks! $\endgroup$
    – dshin
    Jan 21, 2016 at 8:13

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