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Lebesgue Measurability and Weak CH

I have studied a little set theory and I found that Solovay constructed a model of ZF+DC+"All set of reals are Lebesgue measurable" and I found that the same model satisfies the "perfect set property" which implies that every uncountable subset of $\mathbb{R}$ is bijectable with $\mathbb{R}$. In other words, the Continuum Hypothesis (CH) is true in that model. I also found that the Axiom of Determinacy (AD) implies that every set of reals is Lebesgue measurable and it also implies the perfect set property.

Therefore, I have a question: Does ZF+DC+"All set of reals are Lebesgue measurable" always imply CH?

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Iván Ongay Valverde
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    ZFC is inconsistent with "All sets are Lebesgue measurable". Did you mean ZF maybe, or ZF+DC (so the Lebesgue measure have some context to it)? – Asaf Karagila Feb 21 '13 at 19:16

2 Answers2

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The question seems to be an open research problem; it was posed in 2011 on MO (and has remained unanswered), see:

Lebesgue Measurability and Weak CH

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Ali Enayat
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Some remarks:

First of all it is known since 1906 that the axiom of choice implies the existence of non-measurable sets. What Solovay have shown is that it is consistent (relative to the existence of an inaccessible cardinal) that $\small\sf{ZF+DC+CH+}\text{"Every set of reals is Lebesgue measurable"}$ is consistent, for brevity we abbreviate the last axiom of $\small\sf LM$. The axiom of determinacy ($\small\sf AD$) requires a lot more than just one inaccessible cardinal, and it indeed implies the same properties as Solovay's model does (except $\small\sf DC$ which is true in some models of determinacy, but not in others; it is true that $\small\sf AD$ implies the axiom of countable choice for sets of real numbers, which is weaker than $\small\sf DC$).

$\small\sf DC$ is the best you can hope for in terms of choice principles and measurability for two reasons:

  1. $\small\sf DC_{\aleph_1}$ implies that there is a non-measurable set.
  2. The ultrafilter lemma implies that there is a non-measurable set.

It is also partially required. If we remove this axiom, then measure theory becomes hard to develop because the measure need not be countably additive. We could ask about sets being Borel sets instead, and it is indeed it is consistent that $\small\sf ZF+CH+\text{"Every set of reals is Borel"}$ this is proved in a work of Truss which showed that if we repeat a similar construction as Solovay's model then we may end up with a universe where $\small\sf CH$ holds, but $\aleph_1$ is singular and every set is Borel; but also $\small\sf ZF+\lnot CH+\text{"Every set of reals is Borel"}$ which is true in the Feferman-Levy model which is somewhat of a prelude to Solovay's work, and in this model the real numbers are a countable union of countable sets (still uncountable though) so every set is Borel, but as it turns out there is an intermediate cardinality.

Asaf Karagila
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  • My latest pet peeve is that in the absence of appropriate amounts of choice, rather than Borel sets, we should work with Borel codes. With this move, we can develop Lebesgue measure even in ZF. – Andrés E. Caicedo Feb 21 '13 at 19:55
  • (Anyway, the question should be under $\mathsf{ZF}+ \mathsf{DC}_{\mathbb R}$, which also avoids (most of) the issue regarding $\mathsf{AD}$.) – Andrés E. Caicedo Feb 21 '13 at 19:57
  • Andres, does $\small\sf DC_\Bbb R$ prove that $\aleph_1$ is regular? – Asaf Karagila Feb 21 '13 at 20:04
  • Countable choice for sets of reals suffices: Pick $f:\omega\to \omega_1$ cofinal, and for each $n$, pick a real that codes a well-ordering of type $f(n)$, by countable choice. Put them together to get a real of type $\omega_1$. – Andrés E. Caicedo Feb 21 '13 at 20:13
  • D'oh :-)$\vphantom{}$ – Asaf Karagila Feb 21 '13 at 20:23
  • @Andres, I'm totally unfamiliar with what happens with that development of Lebesgue measure, so this is probably a silly question: is the statement "in the Feferman-Levy model, every set of reals is measurable" sensible, and if so, what is the answer? – Noah Schweber Feb 21 '13 at 21:47
  • @Noah I think the question is sensible, and the answer is negative (not every set has a Borel code). – Andrés E. Caicedo Feb 21 '13 at 21:53
  • @Noah, I actually have asked a related question here. http://mathoverflow.net/questions/111124/existence-of-non-borel-sets-in-models-of-all-sets-measurable if you want to learn more on Borel coded measure theory, Fremlin's Measure Theory has a lot of information, vol 5 chapter 6. – Asaf Karagila Feb 21 '13 at 22:12
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    Andres, I think Noah was asking about measurability in the Lebesgue sense rather than the Borel sense. The answer is not as clear in that case, even for the most restrictive definition of Lebesgue measurable I can think of. – François G. Dorais Feb 22 '13 at 20:58
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    @François Yes, I understand. I do not know how to define Lebesgue measure in any meaningful way beyond sets with Borel codes in ZF without any choice-like assumptions. – Andrés E. Caicedo Feb 22 '13 at 21:58
  • @Francois: yes, that's right. – Noah Schweber Feb 22 '13 at 22:38
  • Definition 1: Define the outer measure $\mu^(X) = \inf \sum_{i \in I} (b_i - a_i)$ over all collections of intervals such that $X \subseteq \bigcup_{i \in I} (a_i,b_i)$. Then, following Carathéodory, we can define $X$ to be measurable if $\mu^(A) = \mu^(A \cap X) + \mu^(A \setminus X)$ for all (bounded) $A \subseteq \mathbb{R}$. Note that this is always a nontrivial notion since open intervals $(a,b)$ are measurable and $\mu^*(a,b) = b-a$. – François G. Dorais Feb 22 '13 at 23:17
  • Definition 2: Define $X \subseteq \mathbb{R}$ to be null if there is a sequence $(U_n)_{n\lt\omega}$ of open sets containing $X$ such that $\mu(U_n) \lt 2^{-n}$ for each $n$. (As above, the measure of an open set is the sum of the lengths of its connected components, which are all open intervals.) Define $X$ to be measurable if there is a coded Borel set $A$ such that the symmetric difference $(X \setminus A)\cup(A \setminus X)$ is null. – François G. Dorais Feb 22 '13 at 23:22
  • Definition 3: Define $X \subseteq \mathbb{R}$ to be measurable if there is a coded Borel set $A$ such that $\mu^((X\setminus A)\cup(A \setminus X)) = 0$, with $\mu^$ as in Definition 1. – François G. Dorais Feb 22 '13 at 23:28
  • Francois, those definitions are interesting and they make a lot of sense. I will point out that what I dislike in the Borel code requirement is that we would expect that if the real numbers are the countable union of countable sets, then every set is measurable. But if we restrict to Borel codes I'm not sure this is the case, in fact I am fairly certain it would not be the case. – Asaf Karagila Feb 22 '13 at 23:44