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The other day I learned of a small error in the book Theory of Charges: A Study of Finitely Additive Measures. Example 11.4.1 goes as follows.

Let $\mu_0$ be a finitely additive probability measure defined on the powerset of $\mathbb N$ such that $\mu_0(A)=0$ if $A$ is finite (let us call such $\mu_0$ a diffuse measure). For $n \geq 1$, let $\mu_n(\{n\})=1$. Let $$\mu = \sum_{n=0}^\infty \frac{1}{2^{n+1}} \mu_n.$$

The book claims that $\mu$ does not take the value $1/2$, but this is not true in general. Using the Hahn-Banach theorem, one can find a diffuse measure $\mu_0$ and a set $A$ such that $\mu(A)=1/2$.

Moreover, I have been unable to see that it is possible to find a $\mu$ with range $[0,1/2) \cup (1/2, 1]$ without stronger assumptions.

For instance, if we assume that $\mu_0$ is a non-principal ultrafilter (that is, a 0-1 valued diffuse measure), then the claim in the example holds: $\mu$ does not take the value $1/2$. The existence of a non-principal ultrafilter is stronger in ZF+DC than the Hahn-Banach theorem, which in turn is stronger than the following "weak choice axiom":

PM$_\omega$: A diffuse measure exists.

So, this has me wondering: Is it consistent with ZF+DC+PM$_\omega$ that the range of every diffuse measure is closed?

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  • $\begingroup$ Is it even clear that Hahn-Banach implies there is a diffuse measure whose range is not $[0,1]$? $\endgroup$
    – Elliot Glazer
    Commented Nov 25 at 23:59
  • $\begingroup$ @ElliotGlazer No, not to me. Good point! $\endgroup$
    – aduh
    Commented Nov 26 at 0:21
  • $\begingroup$ The mere existence of a nonprincipal ultrafilter, even specifically on $\omega$ (as you apparently mean from the context), is a quite weak principle, and certainly does not imply the Hahn-Banach theorem. You are probably confusing it with the Ultrafiler Lemma, which states that every proper filter on every set extends to an ultrafilter. $\endgroup$
    – Emil Jeřábek
    Commented Nov 26 at 7:39
  • $\begingroup$ @EmilJeřábek Indeed, thanks for the correction. But does PM$_\omega$ imply that there exists a non-principal ultrafilter on $\omega$? Admittedly, I've been assuming it doesn't, although I don't have a reference for this off the top of my head. $\endgroup$
    – aduh
    Commented Nov 26 at 9:07
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    $\begingroup$ It does not. Even the Hahn-Banach theorem does not imply that $\omega$ (or any set) admits a non-principal ultrafilter. See Pincus and Solovay's Definability of Measures and Ultrafilters. I wouldn't be surprised if their model of HB with no np ultrafilters in fact has that all diffuse measures have range $[0,1].$ $\endgroup$
    – Elliot Glazer
    Commented Nov 26 at 10:24

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