Is sigma-additivity of Lebesgue measure deducible from ZF?

Question

Is sigma-additivity (countable additivity) of Lebesgue measure (say on measurable subsets of the real line) deducible from the Zermelo-Fraenkel set theory (without the axiom of choice)?

Note 1. Follow-up question: Jech's 1973 book on the axiom of choice seems to be cited as the source for the Feferman-Levy model. Can this be sourced in the work of Feferman and levy themselves? Are these S. Feferman and A. Levy?

katz, you may want to take a look at Chapter 56 of Fremlin's treatise on measure theory (56 is in volume 5, part 2), where he describes how one can develop Lebesgue measure without choice. You recover the basic properties as long as you work with "codes" rather than directly with the sets (which would be impossible as mentioned in the answers). — Andrés E. Caicedo, Nov 03 '13 at 16:20
Fremlin's book can be found here: http://www.essex.ac.uk/maths/people/fremlin/mt.htm — Asaf Karagila, Nov 03 '13 at 16:48
@AsafKaragila : Updated link for Fremlin's book: https://www1.essex.ac.uk/maths/people/fremlin/mt.htm (I don't know why institutions are so cavalier about breaking these, especially in such trivial ways, but they are). — Toby Bartels, Sep 13 '22 at 11:45

score 11 · Accepted Answer · answered Nov 03 '13 at 15:07

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This depends on exactly how you define Lebesgue measure since some definitions incorporate countable additivity. However, there is a model of ZF, the Feferman-Levy model, where $\mathbb{R}$ is a countable union of countable sets which ensures that any countably additive measure on $\mathbb{R}$ has to be trivial.

answered Nov 03 '13 at 15:07

François G. Dorais

43,723

François, do you recall the link where we briefly discussed how one can try to go around these issues? (Using Borel codes, etc?) Couldn't find it after a few minutes... – Andrés E. Caicedo Nov 03 '13 at 16:13
@AndresCaicedo, I see there is a related discussion here. – Mikhail Katz Nov 03 '13 at 16:25
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OK, found it! I was thinking of this thread, though Fremlin's book probably supersedes what we said. – Andrés E. Caicedo Nov 03 '13 at 16:29

Asaf Karagila · Answer 2 · 2013-11-03T15:44:29.583

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No, you can't have that. It is consistent that the real numbers are a countable union of countable sets, in which case you immediately have that there is no nontrivial measure which is countably additive on the real numbers.

There are other models, however, in which $\aleph_1$ is singular, the countable union of countable sets of real numbers is countable; but every set is Borel. In such models, I believe, you can't have a countably additive Lebesgue measure as well.

To your question, yes. These are Solomon Feferman and Azriel Levy. The result appears as an abstract in Notices of the AMS from 1964 (give or take a year, this is from memory).

edited Nov 03 '13 at 15:44

answered Nov 03 '13 at 15:07

Asaf Karagila

38,140

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18 seconds apart!!! – François G. Dorais Nov 03 '13 at 15:08
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@Francois: It's because I'm really hungover! ;-) – Asaf Karagila Nov 03 '13 at 15:10
@katz: It would have been faster, but as I said. Severely hungover. :-) – Asaf Karagila Nov 03 '13 at 15:19
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S. Feferman-A. Levy, "Independence results in set theory by Cohen's method II", Notices Amer Math Soc., 10, (1963) 593. The proof itself is probably best learned elsewhere. Jech's The axiom of choice, chapter 10, or Ioanna Dimitriou's master thesis. – Andrés E. Caicedo Nov 03 '13 at 16:17
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Thank you @Andres. Ioanna Dimitriou also used this as an example for a symmetric model in her Ph.D. thesis which I find better than her masters. – Asaf Karagila Nov 03 '13 at 16:19
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And her PhD thesis is here. – Andrés E. Caicedo Nov 03 '13 at 16:21

Is sigma-additivity of Lebesgue measure deducible from ZF?

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