Con(ZF) implies Con(ZF¬C+ "∃Non-measurable set of reals")

Question

I was running my mouth again and now someone wants a citation. I claimed that you can force a non-measurable set of reals into [some] model of ZF without introducing choice. I think I can write the proof, but that's not a citation.

Namely, you can do forcing on a model of ZF if you have choice in the ambient universe, right? So you just take a model of ZF¬C and add a random real, which forces there to be a non-measurable set.

Maybe it would be easier to start with a model of ZFC with such a random real, and kill Choice by building a permutation model whose set of atoms is uncountable and less that the size of the continuum?

At any rate, I'm excited to see what is already out there.

Answer 1

Forcing without choice is fine, but it is messy. There is a lot more finesse to things.

First of all, I will assume you model satisfies Dependent Choice. Without it measure is likely to fail.

Okay, so let's assume that $\Bbb R$ cannot be well-ordered and that $\sf DC$ holds. The easiest way to add a non-measurable set is by adding $\omega_1$ Cohen reals.

1. The forcing is well-ordered, so it cannot introduce a well-ordering of the reals.
2. The forcing is ccc (in a strong sense) and therefore does not collapse $\omega_1$ and preserves $\sf DC$.
3. Shelah proved that from ${\sf ZF+DC}+\aleph_1\leq2^{\aleph_0}$ we can prove there exists a non-measurable set.
4. Congratulate yourself on a job well-done, maybe have a beer.

(For the first fact, see the proof I gave here also mentioned in this paper, for the second one, the paper I have in mind is not online yet in the version holding this argument, but I can tell you it is this one.)