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Timeline for Is Lebesgue/Borel non-measurability actually caused by non-uniqueness?

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Jun 22, 2022 at 7:16 history edited CommunityBot
replaced http://front.math.ucdavis.edu/ with https://arxiv.org/abs/
Jun 15, 2020 at 7:27 history edited CommunityBot
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Jul 5, 2011 at 4:27 history edited Ali Enayat CC BY-SA 3.0
Fixed the description of Krivine's model
Jul 5, 2011 at 4:21 comment added Ali Enayat @Andreas: And sorry for misspelling your name!
Jul 5, 2011 at 4:02 comment added Andreas Blass Ali, I think that the result of Krivine that you mention gets only that all OD sets are Lebesgue measurable. For OD$(\mathbb R)$ sets you really need a Solovay-style argument, using an inaccessible.
Jul 5, 2011 at 1:04 comment added Neil Toronto I am in awe at how deep the answer to this question is. Thanks!
Jul 5, 2011 at 0:52 vote accept Neil Toronto
Jul 4, 2011 at 17:13 history edited Ali Enayat CC BY-SA 3.0
Relected comments and other answers
Jul 4, 2011 at 16:28 comment added Ali Enayat @Martin: you are right, Solovay's result already does the job, and Friedman's extends it further. I will edit to clarify this point.
Jul 4, 2011 at 12:32 comment added Goldstern I am a bit confused. I think that already Solovay's theorem provides a negative answer to the original question (if you read "identifies a unique set" as "defines a set").
Jul 4, 2011 at 1:28 comment added Joel David Hamkins Ali, thank you for mentioning my paper with Linetsky and Reitz. But it should also be mentioned that Ali himself has done important work on exactly the same topic (as we mention in our paper). See, for example, academic2.american.edu/~enayat/DO.pdf, and perhaps Ali can post other suitable links.
Jul 3, 2011 at 22:44 history answered Ali Enayat CC BY-SA 3.0