Is it possible to get fonction $$f\,:\,P\to [0,1]$$ where $$(\Omega,P)$$ is a sigma algebra, such that $$f$$ is additive but not sigma additive, without using ultra filter axiom?

Note that if $$F=\mathcal P(\mathbb N)$$ and $$f$$ is $$1$$ on every element of some maximal ultrafilter, and $$0$$ anywere else, then $$f$$ does the job.

Another idea is, on the same $$P$$, to give the Abel density for any element of $$P$$ that has an Abel density and try to extend it using Hann-Banach Theorem.

• Hahn-Banach theorem is known to be weaker than the ultrafilter lemma. (I am not certain that they are equivalent though.) – Hanul Jeon Jan 8 at 12:09
• @Hanul Jeon : thank you, I edited the part mentionning Choice – jcdornano Jan 9 at 17:44
• @HanulJeon It's strictly weaker – user435571 Jan 9 at 17:45

You have to use the Hahn–Banach theorem at some point.

The following theorem is due to Luxemburg:

The following are equivalent (in ZF):

1. Every Boolean-algebra has a finite additive real-valued measure.
2. The Hahn–Banach theorem.

This is proved in

Luxemburg, W. A. J., Reduced powers of the real number system and equivalents of the Hahn-Banach extension theorem, Appl. Model Theory Algebra, Anal., Probab., Proc. Int. Sympos. Calif. Inst. Technol. 1967, 123-137 (1969). ZBL0181.40101.

As remarked in the comments, this is weaker than the ultrafilter lemma, which in turn is weaker than the axiom of choice.

• Thank you very much Asaf Karagila. It answer to the question by the negative (because Hahn-Banach is strictly weaker, provided that the use i do of Hahn-Banach to extend the Abel density is correct) But you precise the answer and claim that not ultrafilter lemma , but HB has got to be used, that's really a great answer! thank you again. Can I ask you some details to help me understand better your answer? (sorry in advance for my weakness in the domain) – jcdornano Jan 12 at 14:23
• I don't have this off hand, but a first place to check would be Eric Schechter's book Handbook of Analysis and its Foundations. – Asaf Karagila Jan 12 at 15:54
• Thank you for the reference i will get it soon. The point that i don't get is why the necessity of hann banach to get the result for any boolean algebra implies the necessity for some boolean (sigma) algebra... (another point i don't get is but I think is that if by "finite additive" you mean that it is NOT sigma additive, like what we are looking for, but I think we can get one from the other by some quotient, so that it is not important to precise...) – jcdornano Jan 14 at 10:37
• First of all, you did not specify "some sigma algebra", and to me that reads "given any sigma algebra ...", if you ask imprecise questions, you get confusing answers. Secondly, by finitely additive I mean it is not countably additive, yes. Because on a Boolean algebra which is not a sigma algebra you might not have a countably additive measure, but you still might have a finitely additive one. I am not sure what you mean by that last sentence, but it is important to be precise. Precisely for that reason. – Asaf Karagila Jan 14 at 10:40
• My question in the main post was (i think it is obvious) "Does there exist a sigma algebra that has a finite real valued mesure of weight 1, that is not sigma additive, in a model of ZF where ultr filter lemma is not true?" I'm now asking 1) the same question replacing Ultrafilter lemma by HB theorem 2) if what you said in the answer leads to the answer of 1) – jcdornano Jan 14 at 18:59