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Let $A$ be a definable subset of $\mathbb{R}$ in $\mathsf{ZF}$, and let $\mathcal{M},\mathcal{N}\models\mathsf{ZF}$ such that $A$ is lebesgue measurable in both models.

Is $\mu^\mathcal{M}(A^\mathcal{M})=\mu^\mathcal{N}(A^\mathcal{N})$?

Are there any conditions (applied to $A$ or $\mathcal{M},\mathcal{N}$) under which we know more about this question?

Andrés E. Caicedo
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Shay Ben Moshe
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    What exactly do you mean "definable" here? – Asaf Karagila Aug 04 '14 at 16:30
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    I meant in the sense of first-order formula with one free variable that defines $A$ as a subset of $\mathbb{R}$. I actually don't care about the definability, but I can't see how to ask about $A$ in different models otherwise. – Shay Ben Moshe Aug 04 '14 at 16:32
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    Not in the way you stated it. The usual counterexample is to define $ A $ to be empty or the whole line depending on whether $\mathsf {CH} $ holds. What you want instead is to have a "robust" description, in which case the answer is yes. The problem is to formalize this "robustness". For Borel sets and their continuous images, we can exhibit "codes" that describe how the set is made up starting with basic open sets. Beyond this, we do not have entirely satisfactory codes and the problem becomes set-theoretic. – Andrés E. Caicedo Aug 04 '14 at 16:33
  • Shay, definable in what language and so on, that's the question. If you really just want to ask about changing the measure, then this is in fact a duplicate of at least two questions. Let me find links. – Asaf Karagila Aug 04 '14 at 16:34
  • If you replace CH with Con(ZF) in the usual counterexample, wouldn’t it have a perfectly robust code as a closed set? – Emil Jeřábek Aug 04 '14 at 16:35
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    Here is one and here is another (which might be less duplicate, but also relevant). – Asaf Karagila Aug 04 '14 at 16:36
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    The strongest results we have require that we assume large cardinals, and restrict our attention to models and their forcing extensions, the codes being what we call term relations. You can read about this in the Feng-Magidor-Woodin paper on universally Bare sets, and in Steel's first paper on the derived model theorem. This is now part of the core model induction machinery. – Andrés E. Caicedo Aug 04 '14 at 16:38
  • Asaf, the language of set theory. Also the models should probably be well-founded or something similar. – Shay Ben Moshe Aug 04 '14 at 16:38
  • Do you want ZFC or something, to make sure Lebesgue measure is not trivial? ... What happens if, say, the real line is a countable union of countable sets? – Gerald Edgar Aug 04 '14 at 16:41
  • Asaf, I saw the links you sent, thanks, Also, Asaf and Andres, I would love it if you could elaborate more (and maybe answer rather than comment). – Shay Ben Moshe Aug 04 '14 at 16:42
  • @Emil I assume your comment was for me? No, Borel codes are more robust than this. One can see a quick description in Solovay's paper on all sets of reals being measurable. These codes generalize to projections of trees, and ultimately to universally Baire sets and the setting of term relations. – Andrés E. Caicedo Aug 04 '14 at 16:42
  • Gerald, if you have anything interesting to say in that case I would love to hear that. I have edited my question accordingly... – Shay Ben Moshe Aug 04 '14 at 16:45
  • (@Emil The point being that the code is simply a description of the construction of the set: Start with precisely these intervals with rational end points, take their union, form the complement of the result, and so on.) – Andrés E. Caicedo Aug 04 '14 at 16:46
  • Shay, I am writing from my phone, and suffering with auto-correct, so no chance for a while of writing an answer, but the references I suggested should give you all the details you need. – Andrés E. Caicedo Aug 04 '14 at 16:48
  • And I am too hungry and otherwise busy for writing an answer right now. If you're around the university tomorrow we can meet afternoon, or on Wednesday morning and talk about this. – Asaf Karagila Aug 04 '14 at 16:54
  • Andres, thanks, I'll read some of it later. Asaf, I am returning to the university in October. (finally!) and you made me hungry too. – Shay Ben Moshe Aug 04 '14 at 16:58
  • @Andres: My point was that if $\phi$ is an arithmetic sentence, you can write ${x\in\mathbb R:\phi}$ explicitly as a tree of intersection and unions (corresponding to the quantifiers in $\phi$) of intervals (whose endpoints are $\Delta^0_0$-definable, and correspond to evaluation of atomic formulas in $\phi$). I’m not in office now so I can’t check Solovay’s paper, but I would be very surprised if this can’t be expressed as a Borel code. It transpired in later comments that one needs to restrict attention to nice models (in particular, with absolute $\omega$), and I guess this is one reason. – Emil Jeřábek Aug 04 '14 at 18:55
  • @EmilJeřábek I see. Yes, everything I said is for transitive proper class models. There may be something interesting to say for $\omega$-models, but I have not thought much about it. – Andrés E. Caicedo Aug 04 '14 at 19:05
  • It's easy to show by induction that Borel sets defined by a code in the intersection of the models have absolute measure. – Monroe Eskew Aug 06 '14 at 17:18

1 Answers1

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Even when $\mathcal{M, N}$ are transitive models of $ZFC$ with same ordinals and cardinals, a nice counterexample is given by the set $L$ of constructible numbers which has a $\Sigma^1_2$ definition. One way to see this is by adding a Cohen real to a model of $V = L$.

Ashutosh
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  • Perhaps it should be added that the set goes from full measure in $L$ to null measure in $V$. It's what the expert probably don't call "Full to null" (because it doesn't quite rhyme in American English...) – Asaf Karagila Aug 04 '14 at 17:25
  • Yes, this is one of the reasons why we need large cardinals in our models, to avoid counterexamples of this sort. – Andrés E. Caicedo Aug 04 '14 at 17:48