# Without AOC, every subset of $\mathbb{R}^n$ is measurable [duplicate]

My analysis prof constructed a non-measurable subset of $$\mathbb{R}$$ today. A student noticed he used axiom of choice and asked for a construction that doesn't use it. Then, another student responded that there exist models of ZF where every subset of $$\mathbb{R}$$ is measurable--which blows my mind. Can anybody show me a proof of this? Also, ZF is just a set of axioms, what did my classmate mean by "models of ZF"?

• mathoverflow.net/a/42220/158937 Nov 10, 2021 at 1:21
• If you don't know what models are, I'd suggest learning some logic before diving into this question. Nov 10, 2021 at 1:21
• See the Solovay Model But I second the comment that you should first look up what models are.
– lulu
Nov 10, 2021 at 1:22
• Without $AC$ you can't construct a non-measurable set. But you can't prove that they don't exist, either. So your title is not quite right. Nov 10, 2021 at 1:26
• @Ian "ZF-C is equiconsistent with ZF-C+axiom of determinacy which does get you that" That's not true: $\mathsf{ZF+AD}$ proves the consistency of $\mathsf{ZF}$ (this is nontrivial, but basically $\mathsf{ZF+AD}\vdash$ "$L_{\omega_1}\models\mathsf{ZFC}$"), and so has strictly greater consistency strength. (In fact it has much greater consistency strength.) Nov 10, 2021 at 1:42