Timeline for Does constructing non-measurable sets require the axiom of choice?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| S May 6, 2018 at 8:55 | history | suggested | Mike Rosoft | CC BY-SA 4.0 |
Fix link to Wikipedia
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| May 6, 2018 at 7:57 | review | Suggested edits | |||
| S May 6, 2018 at 8:55 | |||||
| Oct 17, 2010 at 20:25 | comment | added | Andrés E. Caicedo | @Kevin: Clearly, Lebesgue measurable. There is a subtlety: There are (consistently) measures defined on all sets of reals that, when restricted to the Lebesgue measurable sets, coincide with Lebesgue measure. This is not an innocuous assumption: It contradicts CH, and implies the Lebesgue measurability of the sets that the descriptive-set theorists call $\Delta^1_3$. Neither of these facts holds without additional assumptions. | |
| Oct 17, 2010 at 18:57 | answer | added | Daniel Mehkeri | timeline score: 10 | |
| Oct 17, 2010 at 16:50 | comment | added | Kevin O'Bryant | Of course. Well, it depends on the measure. Did you mean to say "Lebesgue measurable"? | |
| Oct 15, 2010 at 9:16 | vote | accept | Mark Bell | ||
| Oct 15, 2010 at 8:46 | comment | added | Hany | There is an interesting discussion on: math.niu.edu/~rusin/known-math/99/AD_AC | |
| Oct 14, 2010 at 22:38 | answer | added | Andrés E. Caicedo | timeline score: 24 | |
| Oct 14, 2010 at 21:55 | answer | added | Andreas Blass | timeline score: 37 | |
| Oct 14, 2010 at 21:42 | answer | added | Bjørn Kjos-Hanssen | timeline score: 7 | |
| Oct 14, 2010 at 21:37 | comment | added | Pietro | In a sense, no. This is a very popular question on the mathematical internet. Look up en.wikipedia.org/wiki/Solovay%27s_model | |
| Oct 14, 2010 at 21:30 | history | asked | Mark Bell | CC BY-SA 2.5 |