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For any positive integer n, let E(n) denote n-dimensional Euclidean Space and let L(n) denote n-dimensional Lebesgue Measure on E(n). Take n=2 for simplicity. There are uncountably many convex subsets of E(2) which are neither open nor closed in E(2). For example, if the subset D of E(2) is an open disk, then the union of D and any subset of its boundary (with respect to E(2)) is convex. My first question is "Are all convex subsets of the Euclidean plane L(2)-measurable, if we allow L(2) to be infinite?". My second question is "If S is any closed non-convex subset of E(2) and if C(S) is the smallest convex subset of E(2) that contains S as a subset, is C(S) necessarily closed in E(2)?". I suspect that both questions are well known to have "Yes" answers, although I have been unable to find any mention of this in the literature to which I have access.

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    $\begingroup$ Is this question really on-topic here? $\endgroup$ – TaQ Aug 18 '13 at 15:46
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The first question has an affirmative answer; see this question on math.se: https://math.stackexchange.com/questions/207609/the-measurability-of-convex-sets

(This was the first Google hit for "convex set Lebesgue measurable").

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  • $\begingroup$ Thanks, Nate. I knew this had been answered previously but did not find it on MO after a quick search--now I see why. $\endgroup$ – Bill Johnson Aug 18 '13 at 15:00
  • $\begingroup$ Many thanks for your complete and illuminating answers. The answer to the question about measurability is very interesting. Since there are continuum many points in the boundary of the open disk D, there are more than continuum many convex subsets of E(2). But there are only continuum many analytic subsets of E(2). Hence there are $\endgroup$ – Garabed Gulbenkian Aug 19 '13 at 18:44
  • $\begingroup$ Many thanks for your complete and illuminating answers. The answer to the question about measurability is very interesting. Since there are continuum many points in the boundary of the open disk D, there are more than continuum many convex subsets of E(2). But there are only continuum many analytic subsets of E(2). Hence there are very many convex subsets of E(2) which are not even analytic. Nevertheless, it seems that they are all Lebesgue measurable. Incidentally, I feel stupid to have failed to see even one of those nice counter-examples to a "yes" answer for my second question. $\endgroup$ – Garabed Gulbenkian Aug 19 '13 at 19:21
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The answer to your second question is "no". Consider the union of $\{0\}$ with $\{(x, 1/x) : x >0\}$.

However, if you change "closed" to "compact", the answer is "yes" (use Caratheodory's theorem).

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  • $\begingroup$ The union of the $x$-axis and the point $(0,1)$ is a counter-example as well. The convex hull is the strip between the lines $y=0$ and $y=1$ including the $x$-axis, of course, but no point other than $(0,1)$ on the line $y=1$. $\endgroup$ – Wlodek Kuperberg Aug 18 '13 at 23:51

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