How to prove the measurability of convex sets in $R^n$ ? I have seen a proof, but too long and not very intuitive.If you have seen any, please post it here.
The measurability of convex sets
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real-analysis
convex-analysis
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0What do you know about the topological properties of convex sets and the Lebesgue Measure on $\mathbb{R}^n$? – 2012-10-05
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1There is the following way to prove it, which is probably not the most elementary. One can assume that the convex set has non empty interior. Then the projection onto the closure can be shown to be Lipschitz and differentiable everywhere, except at the boundary. By Rademacher's theorem, it implies that the boundary is Lebesgue negligible, and the measurability follows easily. – 2012-10-05
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0I suspect that if the interior is empty, then you can show that the interior of the affine hull is empty, and from this show that the set is contained in a hyperplane, and hence has measure zero. – 2012-10-05
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0If you can show that the boundary of your set $\partial C$ has measure zero by squeezing it and using convexity, then $\partial C \cap C$ is measurable by completeness of the Lebesgue measure. You can also reduce the problem to the bounded case by cutting off your set with a countable collection of larger and larger balls. – 2012-10-05
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7Note that the answer depends on what you mean by "measurable". A convex set need not be Borel measurable. (Take the open unit ball together with a non-Borel subset of the unit sphere.) – 2012-10-05