# Can we construct a non-measurable set using only the axiom of countable choice? [duplicate]

I know that it's established that non-measurable sets cannot be constructed without the axiom of choice but I was wondering if countable choice was enough? Since the ones that I have seen use uncountable choice.

My intuition tells me no. Since I tend to think of measurable sets as being closed under "countable operations". Although if we're being precise the operations in question are unions, intersections and complements; so i was wondering if we can use countable choice in some way to construct a non-measurable set.

The axiom of countable choice is strictly weaker than the axiom of dependent choice, which is not strong enough to prove the existence of non-measurable sets. However, with dependent choice and $$\aleph_1\le 2^{\aleph_0}$$ you do get a non-measurable set. The axiom of choice for pairs is also sufficient.
It is not possible to construct a non-measurable set using only countable choice. Solovay showed that (assuming the existence of an inaccessible cardinal) it is consistent to have $$ZF+DC+$$every set of reals is Lebesgue measurable. Here $$DC$$ is dependent choice, which is strictly stronger than countable choice.