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?