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I am having a hard time finding a proof that $ZF+DC+\aleph_1\leq 2^{\aleph_0}$ implies that there are non-measurable sets. This fact can be found in Raisonnier's paper: https://link.springer.com/article/10.1007/BF02759764, however, it is it not exactly elementary. This was answered on math overflow but one of the links in the answer is dead: https://mathoverflow.net/questions/193465/some-questions-on-non-measurable-sets-without-ac?noredirect=1&lq=1

This is annoyingly left as an exercise in Herrlich Horst's "Axiom of Choice" so I am assuming there should be a somewhat simple proof. I would appreciate a reference or even a sketch of a proof.

Edit: I take back my comment. It seems that Hanul Jeon might be correct, which begs the question why it was left as an exercise in Horst's book. In Kharazishvili's "Set theoretical aspects of real analysis" they write:

"In this direction, the most impressive and beautiful result is due to Shelah and Raisonnier, which yields a strongly negative answer to the question posed by Luzin. As was already mentioned, Shelah and Raisonnier proved that in the theory $\textbf{ZF}+\textbf{DC}$ the existence of a subset of $\mathbb{R}$ with cardinality $\omega_1$ implies the existence of a [Lebsegue] nonmeasurable subset of $\mathbb{R}$. All the known proofs of this result are rather difficult."

The book was published in 2015 which suggests this is still open.

Giorgio Genovesi
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    I would not be surprised if Raisonnier's proof is the simplest one, as Horst also cited Raisonnier's paper in this exercise. – Hanul Jeon Apr 20 '25 at 13:35
  • Isn’t it always true that $\aleph_1 \leq 2^{\aleph_0}$? It seems superfluous to assert that as an axiom (unless I’m missing something basic?) – NikS Apr 21 '25 at 01:54
  • @NikS In general dependant choice is not enough to prove it. For example take Solovay's model of $\textbf{ZF}+\textbf{DC}+$ every set of reals is measurable, then the above result implies that in this model $\aleph_1\nleq 2^{\aleph_0}$. – Giorgio Genovesi Apr 21 '25 at 08:19
  • @GiorgioGenovesi : In such a model $\aleph_1 > 2^{\aleph_0}$? (that would seem like a contradiction) Or the ordering isn’t well-defined? – NikS Apr 21 '25 at 22:25
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    @NikS Cardinals need not be comparabile under choice. In fact, choice is equivalent over $\textbf{ZF}$ to cardinals being linearly ordered (Cardinals in the sense of equipotency classes) – Giorgio Genovesi Apr 22 '25 at 10:21

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