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Timeline for Measurability and Axiom of choice

Current License: CC BY-SA 3.0

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Jul 14, 2015 at 20:19 vote accept Matthias Ludewig
Jul 14, 2015 at 20:01 history edited Benoît Kloeckner CC BY-SA 3.0
Changed a "Borel" by a "Lebesgue".
Jul 14, 2015 at 12:19 answer added Joel David Hamkins timeline score: 45
Jul 14, 2015 at 12:18 history edited Matthias Ludewig CC BY-SA 3.0
added 3 characters in body
Jul 14, 2015 at 12:12 answer added Gerald Edgar timeline score: 6
Jul 14, 2015 at 12:04 comment added Andrés E. Caicedo (To see that some restrictions are needed, either by requiring the definition to be "sufficiently simple", or by having to assume a theory beyond $\mathsf{ZFC}$, consider that well-orderings of $\mathbb R$, seen as subsets of $\mathbb R^2$, are not measurable. This is a classical result due to Sierpinski. On the other hand, it is consistent with $\mathsf{ZFC}$ that $V=L$, which implies that there is a (simply, though not quite "very" simply) definable well-ordering of the reals.)
Jul 14, 2015 at 12:00 comment added Simon Henry Isn't the general form of this argument just wrong ?: it only proves that it is not possible to prove in ZF that the function is non-measurable, it does not prove that it is actually possible to prove that it is measurable, and I don't even think it shows that ZFC will not prove that the function is non measurable.
Jul 14, 2015 at 11:59 comment added Andrés E. Caicedo Yes, this can be made rigorous. If the definition is simple enough (measured in descriptive set theoretic terms), this follows from classical theorems on so-called analytic sets. For much more generous notions of definability, this is a theorem, but not of $\mathsf{ZFC}$ alone. Rather, of the theory resulting from extending $\mathsf{ZFC}$ with appropriate large cardinals. For an example of what I mean here, see this paper; stronger results are possible.
Jul 14, 2015 at 11:42 history edited Asaf Karagila
edited tags
Jul 14, 2015 at 11:39 history asked Matthias Ludewig CC BY-SA 3.0