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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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marked as duplicate by François G. Dorais Feb 28 '13 at 17:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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
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2 Answers 2

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.

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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. –  Andres 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 $\mathsf{AD}$.) –  Andres 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
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
@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. –  Andres Caicedo Feb 22 '13 at 21:58
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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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Thanks Ali! I have now closed this question as a duplicate. –  François G. Dorais Feb 28 '13 at 17:49
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