Timeline for Can we change the Lebesgue measure by forcing?
Current License: CC BY-SA 3.0
7 events
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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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| Feb 16, 2012 at 8:20 | vote | accept | Asaf Karagila♦ | ||
| Feb 16, 2012 at 7:54 | comment | added | Goldstern | Instead of powers, you could also use multiples of the new real. Or, you can appeal to the Hewitt-Savage 01-1-law, which says that any measurable tail set must have measure 1 or measure 0. (A tail set in $X \subseteq 2^\omega$ is one that is closed under rational translations, i.e., $X+s=X$ for all finite sequences $s$.) The same argument shows that any filter extending the Frechet filter must have measure 0, or be non-measurable. | |
| Feb 16, 2012 at 1:30 | comment | added | Joel David Hamkins | The argument doesn't seem to use AC, since I'm not picking representatives here, but using only the transcendence of the new reals over the ground model field. So perhaps it works fine as-is? | |
| Feb 16, 2012 at 1:27 | comment | added | Asaf Karagila♦ | Ah, so simple! Seeing how this is a Vitali style argument, my next question would have to be obvious: what happens when we remove the axiom of choice so the Vitali argument fails? | |
| Feb 16, 2012 at 1:21 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 35 characters in body; added 4 characters in body; added 127 characters in body
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| Feb 16, 2012 at 1:15 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |