Stipulate that the Axiom of Probabilistic Choice (APC) says that for every collection $\{ A_i : i \in I \}$ of non-empty sets, there is a function on $I$ that assigns to $i$ a finitely-additive probability measure $\mu_i$ on $A_i$.

Then Hahn-Banach (HB) implies APC, since HB is equivalent to the existence of a finitely-additive probability measure on every boolean algebra, and hence implies the existence of such a measure on the direct sum of the powerset algebras $P(A_i)$.

APC is non-trivial in that it implies the Banach-Tarski paradox and the existence of nonmeasurable sets (the proof Foreman and Wehrung use to show that HB implies Banach-Tarski works).

Question: Does APC imply HB?

(When I think about this, I find I keep on wanting to use Stone representation, but of course to do that would be to assume Boolean Prime Ideal, which is stronger than HB or APC.)

  • $\begingroup$ Well, using Bartle integrals you can get linear functionals on $\ell^\infty(A_i)$ which do not come from $\ell^1(A_i)$. And HB is equivalent to "Every Banach space has a nontrivial functional" (continuous or otherwise). So... $\endgroup$
    – Asaf Karagila
    Sep 14, 2017 at 15:20
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    $\begingroup$ ... sorry, I don't see the "so"... $\endgroup$ Sep 14, 2017 at 15:40
  • $\begingroup$ Oh, I wasn't implying it's trivial. I was hoping for someone to complete the proof. :) $\endgroup$
    – Asaf Karagila
    Sep 14, 2017 at 15:41
  • $\begingroup$ If I remember well, HB is equivalent to AC. $\endgroup$ Sep 14, 2017 at 16:33
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    $\begingroup$ @FedorPetrov Not at all. It is very much weaker. It follows from the ultrafilter lemma (and at the same time doesn't imply it). $\endgroup$
    – Wojowu
    Sep 14, 2017 at 16:37


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