Timeline for Measurability and Axiom of choice
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 16, 2015 at 22:12 | comment | added | Avshalom | In economics an example of the usefulness of this mention might be found in measurable utility theory, pages 288-289: ma.huji.ac.il/raumann/pdf/Measurable%20Utility.pdf ; it is even argued that in the context of economic applications the relevant concept of set is that of a measurable set: ma.huji.ac.il/raumann/pdf/dp260.pdf | |
| Jul 14, 2015 at 22:07 | comment | added | Joel David Hamkins | @TimothyChow Good idea! I edited the answer to mention that the projective sets constitute a simple sufficient condition, assuming that there are large cardinals. | |
| Jul 14, 2015 at 22:06 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Added explanation about projective sets in reply to Timothy Chow
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| Jul 14, 2015 at 21:53 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Fixed reference to the Shelah/Woodin paper
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| Jul 14, 2015 at 21:50 | comment | added | Timothy Chow | Is there a weak form of these results that is usable by an analyst who doesn't want to try to understand what "absolute to $L(\mathbb{R})$" means and/or who doesn't want to assume large cardinals? In other words, is there some sort of general and easily checkable (without knowledge of set theory) sufficient condition for a definition to define a ZF-provably measurable set? | |
| Jul 14, 2015 at 21:33 | comment | added | Wojowu | I wish I could upvote this answer twice. | |
| Jul 14, 2015 at 20:19 | vote | accept | Matthias Ludewig | ||
| Jul 14, 2015 at 19:29 | comment | added | paul garrett | Very nice essay. | |
| Jul 14, 2015 at 18:51 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Fixed an issue about the complexity of the definition
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| Jul 14, 2015 at 15:48 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
deleted 40 characters in body
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| Jul 14, 2015 at 15:40 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Expanded and improved the exposition
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| Jul 14, 2015 at 13:12 | comment | added | Joel David Hamkins | The point is that even if we do not have the axiom of choice in our universe $V$, there is the definable inner universe $L$, which does have ZFC and which has definable sets that it thinks are not measurable. We do not need to use AC in order to define those sets in $V$ and they may still not be measurable in $V$; so this is a counterexample to the OP's bold statement. | |
| Jul 14, 2015 at 13:09 | comment | added | Joel David Hamkins | Sorry, I was using set-theoretic terminology. By $V$, we refer to the entire (current) set-theoretic universe, and $L$ is the universe of constructible sets, defined by Gödel (see en.wikipedia.org/wiki/Constructible_universe). | |
| Jul 14, 2015 at 13:03 | comment | added | Matthias Ludewig | Sorry, but what are $V$ and $L$? | |
| Jul 14, 2015 at 12:45 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 55 characters in body
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| Jul 14, 2015 at 12:34 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 190 characters in body
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| Jul 14, 2015 at 12:19 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |