# Lebesgue theory and axiom of choice

I have been told that the existence of non-Lebesgue-measurable sets on $\mathbb R$ is impossible without axiom of choice. Do any other well-known results in Lebesgue theory depend on the axiom of choice?

I have been told of Radon-Nikodym theorem in particular as a theorem which probably needs the Axiom of Choice. If this is indeed the case, I would like a reference.

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Countable union of countable sets are countable requires some form of the axiom of choice. This fact is used a quite a bit in measure theory. –  William Aug 30 '12 at 16:50
I suppose you are the same Doubting Thomas as this one and that one. It would be nice if you registered your latest account to keep the ability of commenting on your earlier questions and accept answers. –  t.b. Aug 30 '12 at 17:01
@t.b.: The same one. Unfortunately I used a fictitious e-mail account all the time, and I have no control over registration. If I register using another address, will the moderators be able to merge them? –  Doubting Thomas Aug 30 '12 at 17:04
@DoubtingThomas: Yes. Even without registering they are able to merge them, but there is no guarantee that it won't split again if you don't register. –  Asaf Karagila Aug 30 '12 at 17:05
@DoubtingThomas: I have merged the accounts. –  Eric Naslund Aug 31 '12 at 1:33

You can find a lot in Fremlin's book which has an extensive chapter about measure theory without the axiom of choice. See t.b.'s comment below for the exact reference.

Some basic things which may break without the axiom of choice:

1. In Solovay's model all sets are Lebesgue measurable, Dependent Choice holds and most analysis goes through as usual.
2. The continuum is a countable union of countable sets (cf. Feferman-Levy model) and every set is Borel; the Lebesgue measure is not sigma-additive. Many theorems which require even minor uses of the axiom of choice fails (e.g., sequences defined by induction, where each step is chosen arbitrarily from a set of possible elements is likely to fail).
3. It is possible that there is a proper subset of the real numbers which is uncountable, but is a countable union of countable sets. In such model measure theory should probably behave strangely, but it might still be developed to some extent if one is careful enough.

It is important to observe that this sort of set is Borel (in fact $\Sigma^0_4$ or so), and if we require the measure to be sigma-additive then it has to have measure zero. How many peculiar sets like this exist will be dependent on the exact model, and might have strange effects on measure theory. On the other hand it is possible that there is essentially one set like this (by this I mean up to cardinality, of course) and subsuming it into the ideal of null sets might be harmless.

Note that the second and third possibilities are models in which the axiom of countable choice breaks down. In the first option measure theory is immediately damaged because the entire space is "broken", but in the second it might be possible to rescue quite a bit of theory.

In either case there is a possibility, discussed in Fremlin's book, to discuss only measures and sets that we can construct effectively in which case the theory holds up relatively well -- although some changes are bound to happen.

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Fremlin's Volume 5.II chapter 56, to be precise... There's a Borel codable version of the Radon-Nikodym theorem in 564L to answer that part of the question. –  t.b. Aug 30 '12 at 16:55
@t.b.: Thanks for the exact reference! –  Asaf Karagila Aug 30 '12 at 16:58