is it consistent with AC that every set is measurable?

Question

Does every set is measurable follow from AC, or from it's negation? I think that by Vitali's construction from the AC follows that some set is not masurable. But here in the 1st comment they claim that it may be consistent with ZFC that every set is measurable. How is it?

Answer 1

It doesn't follow from AC or from its negation.

In fact it is refuted by AC, as you pointed out, e.g. the Vitali set.

It doesn't follow from the negation, because a section of $\mathbb{R}\to\mathbb{R/Q}$ is enough to perform the Vitali construction and certainly AC doesn't follow from such a section existing.

What is true is that under certain large cardinal hypotheses, it is consistent with ZF + Dependent choice that every set is measurable. And under no hypotheses at all it is consistent with ZF that every set is Borel (so measurable): indeed it is consistent with ZF that $\mathbb{R}$ is a countable union of countable sets, and if this happens then so is every subset of $\mathbb{R}$, and so every set is Borel as countable sets are Borel.

Finally, that's not what the linked 1st comment claims at all. It actually gives another proof that AC refutes said proposition : if every set were measurable, then the Banach Tarski paradox would actually be a paradox : it would be contradictory because the unit ball doesn't have measure $0$ or $\infty$ and isometries preserve measure