In a well-written wikipedia article titled "Boolean Prime Ideal Theorem" (BPIT) the author states: "A not too well known application of the BPIT is the existence of non-measurable sets". It follows that the existence of non-measurable sets is strictly weaker than the axiom of choice (since the BPIT is strictly weaker than AC, as established by J. D. Halpern and Azriel Lévy). Does anybody know a reference to a direct proof of the existence on non-measurable sets from the BPIT?
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3You don't need this to conclude that the existence of non-measurable sets is weaker than AC - just have the first failure of AC occur high in the cumulative hierarchy. In general, no "rank-bounded" fact can imply full AC. – Noah Schweber Dec 28 '22 at 21:58
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The Boolean prime ideal theorem gives you an ultrafilter $U$ on the natural numbers $\mathbb{N}$, but any ultrafilter, viewed as a subset of Cantor space $2^{\mathbb{N}}$, is not measurable with respect to the coin-flipping measure.
I described one way to see this in my answer to this question, A remark of Connes:
We may regard $U$ as a subset of $2^\mathbb{N}$, which carries a natural probability measure. But a nonprincipal ultrafilter cannot be measurable there, since the full bit-flipping operation, which is measure-preserving, carries $U$ exactly to its complement, so $U$ would have to have measure $\frac12$, but $U$ is invariant by the operation of flipping any finite number of bits, and so must have measure $0$ or $1$ by Kolmogorov's zero-one law. (See also this article by Blackwell and Diaconis proving the same fact.)
Joel David Hamkins
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