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

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.

• If you just want the result in the title of your question the simplest way is probably to make Choice fail but only fail way above the reals. See math.stackexchange.com/questions/343923/… Jan 17, 2020 at 1:17
• I am not sure what precisely you mean by "not introducing choice". If you add a nonprincipal ultrafilter on $\omega$ (via, say, forcing with $\mathcal P(\omega)/\mathrm{Fin}$) to a decent model of ZF, you add a nonmeasurable set. Jan 17, 2020 at 4:53
• @Andrés: Is there an equivalent condition (or a simple proof) that that forcing will not well order the continuum? Jan 17, 2020 at 20:38
• @Asaf It doesn't under AD. (This is the barren extension.) Don't know how it behaves without any such assumption. Jan 18, 2020 at 0:14

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.)

• Regarding your paper is the version of the paper on Asperó's website an earlier one or the same as the submitted one? Jan 17, 2020 at 10:53
• Alessandro, that is the earlier one (which is the submitted one), but the argument in the second point is in the revised version. We were requested to make one more significant change, and after that we will post the new version online. Jan 17, 2020 at 10:54
• I see, thanks! I'll wait for it then Jan 17, 2020 at 10:57
• I will probably edit this answer once we post it online, so I can ping you while I'm at it. Jan 17, 2020 at 10:57
• I'd appreciate that, thanks Jan 17, 2020 at 11:04