Timeline for Is a random subset of the real numbers non-measurable? Is the set of measurable sets measurable?
Current License: CC BY-SA 3.0
13 events
when toggle format | what | by | license | comment | |
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Nov 29, 2012 at 22:06 | history | edited | Gene S. Kopp | CC BY-SA 3.0 |
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Jul 17, 2012 at 0:12 | answer | added | Michael Greinecker | timeline score: 9 | |
Jul 16, 2012 at 23:33 | vote | accept | Gene S. Kopp | ||
Jul 16, 2012 at 21:50 | answer | added | Gerald Edgar | timeline score: 5 | |
Jul 16, 2012 at 21:23 | answer | added | Will Sawin | timeline score: 19 | |
Jul 16, 2012 at 21:17 | comment | added | George Lowther | ...because the only subset of $\Sigma$ which is measurable with respect to the product sigma-algebra is the empty set, and the only superset of $\Sigma$ which is measurable in this sense is the whole of $2^\mathbb{R}$. This is because measurability of a subset of $2^\mathbb{R}$ can only depend on sampling at a countable set of points in $\mathbb{R}$. Determining whether a set $S$ is in $\Sigma$ requires sampling it at uncountably many points. | |
Jul 16, 2012 at 21:03 | comment | added | George Lowther | $\Sigma$ has to have inner measure 0 and outer measure 1, so is not measurable. | |
Jul 16, 2012 at 21:02 | answer | added | Andreas Blass | timeline score: 23 | |
Jul 16, 2012 at 21:02 | comment | added | Asaf Karagila♦ | Right, thanks Ricky. I keep confusing those finite numbers! :-P | |
Jul 16, 2012 at 20:56 | comment | added | user5810 | @Asaf: $\:$ No, in that case the dimension of the vector space is $0$. $\;\;$ | |
Jul 16, 2012 at 20:44 | comment | added | Qiaochu Yuan | Relevant: terrytao.wordpress.com/2008/10/14/… | |
Jul 16, 2012 at 20:30 | comment | added | Asaf Karagila♦ | Allow me to be the wiseguy that brings it up, in a model of ZF without the axiom of choice, in which all sets of real numbers are measurable the answer is: Yes, $\Sigma$ is measurable and the dimension of this vector space is $1$. :-) | |
Jul 16, 2012 at 20:21 | history | asked | Gene S. Kopp | CC BY-SA 3.0 |