# Lebesgue measure, Borel sets and Axiom of choice

I cannot proceed my study on measure theory since it seems my measure theory is really unstable. I desperately need someone to briefly answer below 3 questions...

**For convenience, i will write Lebesgue Measurable set to mean the usual Lebesgue Measurable set, and write Codable-Lebesgue Measurable set to mean the Lebesgue Measurable set defined by Codable-Borel sets.

$1$. It is well-known that "Existence of Non-Lebesgue measurable set is unprovable in ZF". WHAT Lebesgue measurable set?

My definition for Lebesgue measurable set is defined by using Riesz Representation Theorem (Whose existence is guranteed by Axiom of Countable Choice, hence it is undefinable without choice). However, I heard that the usual construction of Lebesgue measure is by using Caratheodory's existence theorem. In that way, can Lebesgue measurable set be definable without Axiom of Choice? That is, it really doesn't make sense to me, 'existence of non-Lebesgue Measurable set is unprovable in ZF' since Lebesgue measure cannot be defined without choice (in my way of construction of Lebesgue measure).

$2$. Can "Existence of a set that is Lebesgue measurable but non-Borel" be proved without choice? I saw the standard example (Luzin's example using continued fraction), but i'm not sure what Lebesgue measurable set stated there. Moreover, if it is not true in ZF, what about "Existence of a set that is Codable-Lebesgue Measurable but non-Borel"?

$3$. (This is closely related to 1) Why Lebesgue measure is unique? Assuming Axiom of Choice, it is a theorem, that "If $\mu_1$ and $\mu_2$ are translation-invariant measures on sigma algebras $\mathfrak{M}_1$ and $\mathfrak{M}_2$ repectively, containing all Borel sets, and $\mu_1(K), \mu_2(K) <\infty$ for every compact set $K$, then there exists a constant $c$ such that $\mu_1(E)=c\mu_2(E)$ for all the borel sets $E$". You can see that this theorem does not imply that $\mathfrak{M}_1=\mathfrak{M}_2$ in the hypothesis.

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Your "measure theory is really unstable"? You will find answers to 1. and 2. by searchin here and on mathoverflow. For the last result, modify the usual proof that two probability measures agreeing on a $\pi$-system coincide on the $\sigma$-algebra generated by the $\pi$-system. –  Michael Greinecker Jan 25 '13 at 13:40
@Michael I have searched both here and mathoverflow before i post this question, but i couldn't find it. Would you please tell me the link? –  Katlus Jan 25 '13 at 13:44
Here are some: math.stackexchange.com/questions/133999/… math.stackexchange.com/questions/74676/… See also the links there. –  Michael Greinecker Jan 25 '13 at 13:47
Wouldn't it be reasonable to learn this in two steps? First learn measure theory proper and then worry about the axiom of choice. It might be more efficient to see where you need which variant of the Axiom if you have some flexibility in the tools at your disposal and an altogether deeper understanding. –  Martin Jan 25 '13 at 13:54

For the first question, note that if we assume Dependent Choice then all $\sigma$-additivity arguments can be carried out perfectly. In such setting using the Caratheodory theorem makes perfect sense, and indeed the Lebesgue measure is the completion of the Borel measure.

It was proved by Solovay that $\mathsf{ZF+DC}$ is consistent with "Every set is Lebesgue measurable", relative to an inaccessible cardinal. This shows that assuming large cardinals are not inconsistent, we cannot prove the existence of a non-measurable set (Vitali sets, Bernstein sets, ultrafilters on $\omega$, etc.) without appealing to more than $\mathsf{ZF+DC}$.

Under the assumption of $\mathsf{DC}$ it is almost immediate that there are only continuum Borel sets, but if we agree to remove this assumption then it is consistent that every set is Borel, where the Borel sets are the sets in the $\sigma$-algebra generated by open intervals with rational endpoints. For example in models where the real numbers are a countable union of countable sets; but not only in such models. Do note that in such bizarre models the Borel measure is no longer $\sigma$-additive.

For sets with Borel codes the proof follows as in the usual proof in $\mathsf{ZFC}$. There are only $2^{\aleph_0}$ possible codes, but there are $2^{2^{\aleph_0}}$ subsets of the Cantor set, all of which are codible-Lebesgue measurable.

Lastly, we have a very good definition for the Lebesgue measure, it is simply the completion of the Borel measure. The Borel measure itself is unique, it is the Haar measure of the additive group of the real numbers. The Lebesgue measure is simply the completion of the Borel measure which is also unique by definition. This is similar to the case where the rational numbers could have two non-isomorphic algebraic closures, but there is always a canonical closure. Even if you can find two ways to complete a measure, "the completion" would usually refer to the definable one (adding all subsets of null sets).

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Isn't Axiom of countable choice necessary to prove the 'Existence of a completion of Borel measure'? If we cannot prove the existence of the lebesgue measure, how does that make sense to say "There exists a non-Lebesgue measurable set is unprovable in ZF"? –  Katlus Jan 25 '13 at 13:57
Katlus, yes. The term "Lebesgue measurable" is very very ill-defined. However if you observe that the measure algebra is simply the completion of the Borel sets with respect to the null idea, even if you can't really define a measure on that algebra, it's still the collection that we call "Lebesgue measurable sets". –  Asaf Karagila Jan 25 '13 at 13:59
Furthermore, if you cannot prove it from $\mathsf{ZF+DC}$ then you certainly can't prove it from $\mathsf{ZF}$. –  Asaf Karagila Jan 25 '13 at 14:00
It's very clear now. Thank you! –  Katlus Jan 25 '13 at 14:10
@Asaf The term measure algebra is usually used for what you get when you quotient out the null sets. –  Michael Greinecker Jan 25 '13 at 16:05