# CH vs AC and non lebesgue measurable sets

According to this discussion https://mathoverflow.net/questions/42215/does-constructing-non-measurable-sets-require-the-axiom-of-choice, in order to construct a non lebesgue measurable set this requires the axiom of choice (via results of Solovay and Shelah). This is my understanding.

However, a friend of mine just showed me a proof of a non lebesgue measurable set assuming only CH, not AC. The same proof my friend showed me is discussed here: Lebesgue nonmeasurable sets

Also here is a comment of someone claiming to be able to make a non lebesgue measurable set using something weaker than AC: https://math.stackexchange.com/q/3791000

I know GCH implies AC and hence non lebesgue measurable sets. However is CH alone enough to give you non lebesgue measurable sets?

• Yes. CH (in its strong form $2^{\aleph_0}=\aleph_1$, not the weak version "every uncountable set of reals has size $2^{\aleph_0}$") implies that the reals are well-orderable, and this is enough choice for the construction of a non-measurable set to go through. Commented Nov 2, 2022 at 23:58

Without the Axiom of Choice the question becomes what do you mean by $$\sf CH$$. In Solovay's model the perfect set property holds, namely every uncountable set of reals contains a copy of a Cantor set, and in particular, any uncountable set of reals has size continuum. So in that sense, the Continuum Hypothesis holds.
On the other hand, requiring that $$|\Bbb R|=\aleph_1$$ means that the real numbers can be well-ordered, pretty much by definition. This provides us with all the choice we need to repeat any of the usual constructions of non measurable sets. From Vitali sets, to ultrafilters on $$\Bbb N$$, to Hahn–Banach and Banach–Tarski.
What Shelah showed is that in fact just requiring that $$\aleph_1\leq|\Bbb R|$$ is enough to construct a non measurable set. Although it might not be enough to get some of the aforementioned "traditional" examples.