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Assume ZF. Consider the claim:

(1) For any infinite set $\Omega$, there is a finitely additive probability measure $\mu:2^\Omega\to[0,1]$ with $\mu(A) = 0$ whenever $|A|<|\Omega|$.

Then (1) is implied by Hahn-Banach and the claim that the union of two sets of cardinality less than $|\Omega|$ has cardinality less than $|\Omega|$ when $\Omega$ is infinite. (Let $\scr B$ be the boolean algebra $2^\Omega$ and let $\scr I$ be the ideal of all subsets of $\Omega$ of smaller cardinality. HB is equivalent to the claim that every boolean algebra has a f.a. probability measure, so $\scr B/I$ has a f.a. probability measure, which lifts to a f.a. probability measure on $\scr B$ with the requisite property.)

[Deleted Question 1 as it was predicated on something false.]

Question 2: Is anything known about the strength of (1)?

A related claim is:

(2) Every set has a f.a. probability measure on its powerset which is not supported on any set of smaller cardinality (i.e., if $|A|<|\Omega|$ then $P(A)<1$).

Clearly (1) implies (2) (at least in the infinite case, but the finite case of (2) is trivial).

Question 3: Does (2) imply (1)?

Updates: The answers below so far show: (1) is equivalent to AC. ZF+DC is not strong enough to show (2). It's still open in this discussion whether (2) is equivalent to AC (I doubt it).

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Alexander, as I noted in a comment on my answer, your proof of (1) from Hahn Banach presumes that the collection of small subsets of $\Omega$ forms an ideal, but this is not true in ZF alone, and fails for example when there are infinite Dedekind finite sets. –  Joel David Hamkins Nov 13 '13 at 18:26
As Joel just explained, (1) does not work when $\Omega$ can be decomposed into two sets of smaller size, but (2) still makes sense in that case. When $\Omega$ is indecomposable, then (2) does imply (1). If $\mu$ is a f.a. measure as in (2), and $\mathcal{J}$ is an ideal then the directed limit $\lim_{A \in \mathcal{J}} \mu_A$ is a f.a. measure as in (1) where $\mu_A(X) = \mu_A(X - A)/(1-\mu(A))$. –  François G. Dorais Nov 13 '13 at 19:00
Oops. You're right of course that the proof doesn't work. I'll edit. –  Alexander Pruss Nov 13 '13 at 19:27
@Francois points out indecomposability. Note that the assertion "For every infinite $A$, $|A|=|A|+|A|$" is much weaker than the axiom of choice, and it does not prove - nor is proven by - countable choice, its various strengthened versions, and Boolean Prime Ideal theorem. –  Asaf Karagila Nov 13 '13 at 19:36
@Asaf: Does $|A|=|A|+|A|$ for all infinite sets imply indecomposability of all infinite sets? –  Alexander Pruss Nov 13 '13 at 19:55

4 Answers 4

up vote 7 down vote accepted

Assertion (1) is equivalent to the Axiom of Choice. This is because it implies that for every infinite set $\Omega$, the set $\{A \subseteq \Omega : |A| \lt |\Omega|\}$ is an ideal. It follows from this that every infinite cardinal number $\mathfrak{m}$ is indecomposable — there are no cardinals $\mathfrak{p},\mathfrak{q} \lt \mathfrak{m}$ such that $\mathfrak{p} + \mathfrak{q} = \mathfrak{m}$ — which is known to be equivalent to the Axiom of Choice. (Note that this implies that any two cardinals are comparable for if $\mathfrak{p}$ and $\mathfrak{q}$ are incomparable then $\mathfrak{p} + \mathfrak{q}$ must be greater than both. The assertion that cardinals are linearly ordered is well-known to be equivalent to the Axiom of Choice.)

Assertion (2) is more promising and (2) does imply (1) when $|\Omega|$ happens to be indecomposable. To see this, suppose $\mu$ is a finitely additive measure such that $\mu(A) \lt 1$ whenever $|A| \lt |\Omega|$. Since $\mathcal{J} = \{A \subseteq \Omega : |A| \lt |\Omega|\}$ is an ideal, the directed limit $$\nu(X) = \lim_{A \in \mathcal{J}} \frac{\mu(X-A)}{1-\mu(A)}$$ makes sense and the result is a finitely additive probability measure $\nu$ on $\Omega$ such that $\nu(A) = 0$ whenever $|A| \lt |\Omega|$.

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Great! . . . . . . –  Joel David Hamkins Nov 14 '13 at 3:23
That's lovely. Thank you. –  Alexander Pruss Nov 14 '13 at 14:36

Let me simply point out that (1) is not provable in ZF, assuming ZF is consistent. To see this, suppose with ZF that there is an infinite Dedekind finite set $A$. The existence of such sets is (relatively) consistent with ZF. It follows that every proper subset of $A$ has strictly smaller cardinality than $A$. So take any point $a\in A$, and consider the partition $A=(A-\{a\})\sqcup\{a\}$. According to your requirements, both pieces must have measure zero, forcing $A$ also to have measure zero also, a contradiction.

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Nice, thanks! I wouldn't have expected (1) to be provable in ZF, but it's nice to know. I also wouldn't expect (1) to be provable in ZF + Countable Choice. –  Alexander Pruss Nov 13 '13 at 15:12
Joel, @Alexander: The existence of infinite Dedekind-finite sets is consistent with the Boolean Prime Ideal theorem which implies the Hahn-Banach theorem. Ergo, somewhere in this thread there exists a mistake. My money is on the conditions on the measure. –  Asaf Karagila Nov 13 '13 at 17:15
The problem seems to be: the "ideal" of small sets is not necessarily an ideal if there is a Dedekind finite set. For example, in my example, the OP's proof would take the quotient by all proper subsets, but this isn't a Boolean algebra. –  Joel David Hamkins Nov 13 '13 at 18:25

Your definition, as Joel points out, is slightly amiss. It should be that the measure vanishes on finite sets. In that case, it is consistent with the existence of Dedekind-finite sets, e.g. in Cohen's first model there exists an infinite Dedekind-finite set of real numbers, but the Hahn-Banach theorem holds and therefore we can define a measure on its power set.

Visiting the "Consequences of the Axiom of Choice" website, one can find that the [classical] Hahn-Banach theorem is Form 52 and the existence of non-principal measures on $2^A$, for every infinite $A$, is Form 222.

Indeed Hahn-Banach implies the existence of measures, but not vice versa. I suppose that one should pick up the actual "Consequences of the Axiom of Choice" book and search out the reference, which I don't remember off hand.

Comparing Form 222 to Form 0 (which is just "$\sf ZF$", or even without foundation) the result is also that 222 is unprovable, and again I am unfamiliar with a reference off the top of my head.

Searching a bit through MathSciNet I came across the following review:

Luxemburg, W. A. J., "Reduced powers of the real number system and equivalents of the Hahn-Banach extension theorem." 1969 Applications of Model Theory to Algebra, Analysis, and Probability (Internat. Sympos., Pasadena, Calif., 1967) pp. 123–137

From its review I quote the last paragraph:

The author shows that a modified form of the Hahn-Banach theorem is valid in set theory without the axiom of choice. The modification consists in allowing the extended functional to take its values in a reduced power of the reals. Using this result the author proves the main theorem of the paper which states that the Hahn-Banach theorem is equivalent to the statement that every Boolean algebra admits a nontrivial measure.

It seems that the above is incompatible with the top part of this answer. Lacking access to either the Consequences book as well to this paper it is difficult to judge the results. I will try to find more next week when I am in the university again and can flip through the Consequences book (and maybe the proceedings book as well if it exists there).

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I actually did mean for the measure to vanish on all subsets of smaller cardinality rather than just to vanish on all finite subsets. Is a non-principal measure one that vanishes on all singletons? –  Alexander Pruss Nov 13 '13 at 19:38
Yes, yes it does. –  Asaf Karagila Nov 13 '13 at 19:54
So my (1) is at least as strong than the existence of a non-principal measure. Wonder if it's stronger. –  Alexander Pruss Nov 13 '13 at 19:57
The model for form 222 is from Pincus & Solovay Definability of measures and ultrafilters (J. Symbolic Logic 42 (1977), 179-190; MR0480028). After a quick glance, the model in question has a non-principal finitely additive measure on $\omega$ and therefore a countably supported non-principal finitely additive measure on every Dedekind infinite set. In other words, this does not answer @Alexander's question for uncountable sets. –  François G. Dorais Nov 13 '13 at 22:23
Thanks @Francois. –  Asaf Karagila Nov 14 '13 at 18:02

A quick note that (2) doesn't follow from ZF (assuming ZF is consistent) or even ZF+DC.

(2)+Countable Choice for Finite Sets (CC(fin)) implies that every uncountable set has a non-principal (finitely additive, non-trivial) measure. Let $\Omega$ be uncountable. Let $\mu$ be as in (2). Let $A = \{ x : \mu(\{ x \}) > 0 \}$. The standard proof that a finite measure has only countable many atoms only uses CC(fin) (there are at most $n$ elements $x$ such that $\mu(\{ x \}) \in [1/n, 1/(n-1))$). So $A$ is countable. So $\mu(A)<1$. Then $\nu(B) = \mu(B-A)$ defines a non-principal measure and is non-trivial since $\nu(\Omega)=1-\mu(A)>0$.

But Blass gave a model of ZF+DC with no non-principal measures ("A model without ultrafilters", Bull. Acad. Polon. Sci. 25 (1977), 329–331; I haven't read the paper, but Pincus and Solovay say that it proves this). Since DC implies CC(fin), (2) is false in that model.

A different way to see that (2) doesn't follow from ZF is this. Take a model where there is an infinite Dedekind-finite set but CC(fin) still holds (the Consequences of AC website lists a bunch of models where Form 10 is true and Form 9 is false).

Now if $\Omega$ is an infinite Dedekind-finite set and $\mu$ satisfies (2), then every point of $\Omega$ will have to be an atom of $\mu$ (or else $\mu(\Omega - \{ x \}) = \mu(\Omega)$). But by CC(fin), the set of atoms is countable, which is impossible for a Dedekind-finite set.

I'm curious if (2) implies CC(fin).

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