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"?

  • $\begingroup$ mathoverflow.net/a/42220/158937 $\endgroup$ Nov 10, 2021 at 1:21
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    $\begingroup$ If you don't know what models are, I'd suggest learning some logic before diving into this question. $\endgroup$ Nov 10, 2021 at 1:21
  • $\begingroup$ See the Solovay Model But I second the comment that you should first look up what models are. $\endgroup$
    – lulu
    Nov 10, 2021 at 1:22
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    $\begingroup$ 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. $\endgroup$
    – TonyK
    Nov 10, 2021 at 1:26
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    $\begingroup$ @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.) $\endgroup$ Nov 10, 2021 at 1:42