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As happens every so often, I find myself in a set-theoretical morass. Please help me out of it.

Let $\Omega$ be a set and $\mathcal{B}$ a Boolean algebra of subsets of $\Omega$. The question, broadly, is how to reconcile the following three observations.

(1) From this post, it seems that the existence of a merely finitely additive probability (i.e., a probability that is finitely but not countably additive) on $(\Omega, \mathcal{B})$ cannot be shown in ZF. The proof of this claim proceeds as follows. It is known that the existence of a merely finitely additive probability $\mu$ on $(\mathbb{N}, 2^\mathbb{N})$, such that $\mu(n)=0$ for all $n$, cannot be shown in ZF. If there exists a merely finitely additive probability $P$ on $(\Omega, \mathcal{B})$, then a simple construction shows that $P$ can be used to define a merely finitely additive probability $\mu$ on $(\mathbb{N}, 2^\mathbb{N})$ such that $\mu(n)=0$ (see the linked post for details). Since the existence of $\mu$ cannot be shown in ZF, neither can the existence of $P$.

(2) Now, a friend of mine denies this conclusion. He thinks that merely finitely additive probabilities on $(\Omega, \mathcal{B})$ can be shown to exist using only the ZF axioms. His argument is essentially the same as the one that appears in this post. He assumes that $\mathcal{B}$ is well-ordered and extends the filter of cofinite subsets of $\Omega$ to a non-principal ultrafilter using transfinite induction on the well-order of $\mathcal{B}$. This yields a non-principal ultrafilter which, restricted to $\mathcal{B}$, defines a merely finitely additive probability (details can be provided if need be). So long as defining the well order on $\mathcal{B}$ doesn't require any choice (for example, let $\mathcal{B}$ be countable), the argument does seem to demonstrate that choice is not needed to show the existence of a merely finitely additive $P$.

(3) But now, to complicate things further, there is Theorem 2 of this paper by Pincus and Solovay, which states:

It is consistent with ZF, DC, and the Hahn-Banach theorem that every ultrafilter on any set is principal.

This seems to contradict my friend's claim in (2), adding further evidence to the claim made in (1), that the existence of a merely finitely additive $P$ on $(\Omega, \mathcal{B})$ cannot be shown in ZF.

To sum up, I am left wondering

Does there exist a set $\Omega$ and Boolean algebra $\mathcal{B}$ of subsets of $\Omega$ such that the existence of a merely finitely additive probability $P$ on $(\Omega, \mathcal{B})$ can be shown using only the ZF axioms?

and

Is there a set $\Omega$ such that the existence of a non-principal ultrafilter of subsets of $\Omega$ can be shown using only the ZF axioms?

Added. Here is the proof of (2), which seems to contradict the result cited in (3). Perhaps someone can indicate where it goes wrong.

We assume that $\mathcal{B}$ can be well-ordered without using the axiom of choice. For example, assume that $\mathcal{B}$ is countable. Let $\mathcal{F}_0$ be the filter of cofinite subsets of $\Omega$. We extend $F_0$ to a non-principal, proper ultrafilter of subsets of $\Omega$ by transfinite induction on the well order of $\mathcal{B}$.

Any proper filter $\mathcal{F}$ that extends $\mathcal{F}_0$ is clearly non-principal. Let $B \in \mathcal{B}$. The members of the filter generated by $\mathcal{F}$ and $B$ are supersets of sets that are intersections of members of $\mathcal{F}$ and $B$. If $B \cap F_1 = \emptyset$ where $F_1 \in \mathcal{F}$ and $(\Omega - B) \cap F_2 = \emptyset$ where $F_2 \in \mathcal{F}$, then $F_1 \cap F_2 = \emptyset$, contradicting the assumption that $\mathcal{F}$ is a proper filter. This shows that at each stage in the induction, we can add a member $B \in \mathcal{B}$ or its complement. At limit ordinals, we consider the filter that is the union of the chain of filters and continue until we end up with a non-principal, proper ultrafilter.

aduh
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    Your friend assumes $\mathcal{B}$ is well-orderable, which is not generally hold unless we have the axiom of choice. – Hanul Jeon Mar 19 '18 at 11:40
  • @HanulJeon Yes, I realize that, and should have pointed it out. I will edit. I don't think that this observation dissolves my confusion though because choice isn't always needed to find a well order. For example, let $\mathcal{B}$ be countable. – aduh Mar 19 '18 at 11:44
  • Yes, we do not need the choice if $\mathcal{B}$ is countable. But I don't understand why my observation does not solve your question, because you don't make any assumption on either $\Omega$ or $\mathcal{B}$. – Hanul Jeon Mar 19 '18 at 11:51
  • To answer your second question, the existence of a nonprincipal ultrafilter over $\Omega$ is known as the ultrafilter theorem, which is known not provable by ZF alone. – Hanul Jeon Mar 19 '18 at 11:52
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    @HanulJeon: The issue here is an apparent contradiction between two apparent facts: we have an explicit way to construct a nonprincipal ultrafilter, and its impossible to prove any nonprincipal ultrafilters exist. –  Mar 19 '18 at 11:55
  • @HanulJeon Let me try rewording the last two questions to see if that helps. I think Hurkyl has understood me. – aduh Mar 19 '18 at 11:58
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    How can you be left wondering about the second question? You clearly cite a paper where they answer that to the negative. – Asaf Karagila Mar 19 '18 at 12:01
  • @AsafKaragila Because my friend is a very good mathematician and he has me doubting my understanding of the cited paper! Perhaps I should just post his proof of (2) so that someone can point out the error, which I've been unable to find. – aduh Mar 19 '18 at 12:05
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    Is he better than Solovay? I doubt that. But even without an appeal to authority. Give him the paper, and ask him to point out exactly where the mistake lies. Once a theorem has been accepted by the "mathematical cannon", it's his responsibility to point out the mistake. Just saying "I don't believe it" is not enough. – Asaf Karagila Mar 19 '18 at 12:07
  • @AsafKaragila Agreed. I am assuming the Solovay result is correct. What I'm unsure of is whether I understand it correctly. Surely I'm misunderstanding something as I seem to be believe a contradiction at the moment. The question is supposed to be about addressing my misunderstanding, not verifying Solovay. – aduh Mar 19 '18 at 12:27
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    @aduh in general, when people talk about ultrafilters on a set, they are implicitly referring to ultrafilters on the Boolean Algebra $\mathcal{P}(\Omega)$ for some infinite set $\Omega$. This is a very specific restriction. – Not Mike Mar 19 '18 at 12:38
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    I think https://math.stackexchange.com/questions/2271493/all-ultrafilters-are-principal-consistent-with-zf/ is relevant here. Possibly even a duplicate. – Asaf Karagila Mar 19 '18 at 12:42
  • @aduh it would be worth your time to show that under the restriction I mentioned in the previous comment, that no non-principal ultrafilter can be countably generated (that is, every subset generating the full filter must be uncountable) – Not Mike Mar 19 '18 at 12:44
  • @NotMike I think your first comment gets to the heart of what was confusing me, thanks. The point is that Solovay's result is about ultrafilters over power sets, whereas my friend is not requiring that the Boolean algebra be a powerset. – aduh Mar 19 '18 at 13:24
  • @aduh correct. glad to have helped. – Not Mike Mar 19 '18 at 13:45

1 Answers1

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Answer for this question

Does there exist a set $\Omega$ and Boolean algebra $\mathcal B$ of subsets of $\Omega$ such that the existence of a merely finitely additive probability $P$ on $(\Omega,\mathcal B)$ can be shown using only the ZF axioms?

(1) Existence of a finitely additive, not sigma additive, probability measure on a sigma-algebra cannot be proved in ZF. An example of a sigma-algebra is the power set of some set. Most of the discussion of the OP is for this case: sigma-algebra, not merely Boolean algebra. We cannot prove in ZF that there exists an infinite well-orderable sigma-algebra...

(2) Existence of a finitely additive, not sigma additive, probability measure on a Boolean algebra can easily be proved in ZF.

Proof for (2). The set is $\mathbb N$. The Boolean algebra $\mathcal B$ consists of the finite subsets of $\mathbb N$ and the complements of finite sets. The measure $\mu$ is: $\mu(A) = 0$ if $A$ is finite, $\mu(A) = 1$ if the complement of $A$ is finite.

GEdgar
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  • Wasn't the task to prove that every Boolean algebra has a finitely-additive probability measure, not just to exhibit a particular Boolean algebra that has one? – hmakholm left over Monica Mar 19 '18 at 12:37
  • @HenningMakholm ... I clarified which question I am answering – GEdgar Mar 19 '18 at 12:41
  • Ok, seems I can't read. – hmakholm left over Monica Mar 19 '18 at 12:48
  • Thanks. So there must be a problem with the argument in (1) of my post. The point there is that since your $\mu$ is not countably additive Then there must be a disjoint sequence $ A_n$ in $\mathcal{B}$ such that $$\sum_n \mu(A_n)<\mu\bigg(\bigcup_n A_n\bigg).$$ Now define the function $\nu$ on $2^\mathbb{N}$ by letting $$\nu(F)=\frac{\mu\big(\bigcup_{n\in F} A_n\big)-\sum_{n\in F}\mu(A_n)}{\mu\big(\bigcup_n A_n\big)-\sum_n\mu(A_n)}.$$But $\nu$ shouldn't be constructible in ZF. – aduh Mar 19 '18 at 13:04
  • If $\mu$ is defined on a Boolean algebra $\mathcal B$ which is not a sigme-algebra, then for many $F \subset \mathbb N$, the set $\bigcup_{n\in F} A_n$ is not in $\mathcal B$, so $\mu$ is not defined for it. – GEdgar Mar 19 '18 at 13:08
  • Ah, of course. Stupid of me to not see that. – aduh Mar 19 '18 at 13:12