# Lebesgue-measurable sets requiring the Axiom of Choice to construct

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.

• Some subset of the Cantor set presumably has this property. – André Nicolas Jan 21 '16 at 6:34
• 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! – Asaf Karagila Jan 21 '16 at 6:36
• @AsafKaragila Thanks for tip, added block quote. – dshin Jan 21 '16 at 6:37
• @AndreNicolas: Every subset of the Cantor set is a null set and thus Lebesgue measurable (but not necessarily Borel measurable). – PhoemueX Jan 21 '16 at 6:38
• 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. – PhoemueX Jan 21 '16 at 6:42