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In a paper by J.-L. Krivine, Modèles de ZF+AC dans lesquels tout ensemble de réels définissable en termes d'ordinaux est mesurable-Lebesgue [C. R. Acad. Sci. Paris Sér. A-B 269 (1969), A549–A552, MR0253894], he presents a model of ZFC in which every set of reals definable from ordinals is Lebesgue measurable.

My question is:

  1. He seems not to assume the existence of an inaccessible cardinal in the ground model, is this correct?

  2. In Theorem 3 of that paper, he concludes that every set of reals definable from ordinals is Lebesgue measurable. Does this continue to hold if we replace definable from ordinals by definable from a countable sequence of ordinals?

  3. Does Theorem 3 of that paper still hold if we replace "Lebesgue measurable" by "having the property of Baire"?

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1 Answer 1

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  1. Yes, that is correct. Krivine's observation is that by collapsing the continuum (or any larger cardinal) to $\aleph_0$ using finite conditions then any set of reals definable from ground model parameters (e.g. ordinals) is Lebesgue measurable in the extension.
  2. No, not at all countable sequences of ordinals are in the ground model.
  3. Yes, the argument works just as well replacing the use of random forcing by Cohen forcing at the end and the conclusion then is regarding to the Baire property instead of Lebesgue measurability.
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    $\begingroup$ A point should be made that the collection of sets obtained this way is generally quite small. $\endgroup$ Commented Mar 5, 2014 at 15:53
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    $\begingroup$ About item 2: Note also that measurability of projective sets requires large cardinal assumptions, while Krivine's theorem does not. Just on consistency strength grounds we see that the answer is no. $\endgroup$ Commented Mar 5, 2014 at 16:35

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