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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