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All convex subsets of $\mathbb{R}^n$ are Lebesgue measurable. But not all Lebesgue measurable sets are Borel sets. So my question is, are all convex subsets of $\mathbb{R}^n$ Borel sets?

Keshav Srinivasan
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2 Answers2

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The answer is no (except in the trivial case $n=1$), by a counting argument.

Let $C$ be the closed unit ball centered at the origin, and let $O$ be the open unit ball centered at the origin. Any set $X$ with $$O\subseteq X\subseteq C$$ is convex; this is just because no line can meet the surface of a ball in three points.

But since the surface of the ball (as long as $n>1$!) has $2^{2^{\aleph_0}}$-many subsets, we get $2^{2^{\aleph_0}}$-many sets of this form, most of which won't be Borel (since there are only $2^{\aleph_0}$-many Borel sets).

We can rephrase this as: if $A,B$ are distinct subsets of the unit sphere then they have distinct convex closures. The point is that the sphere gives us a kind of "convexly independent set" of sufficiently large cardinality.

Noah Schweber
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    Alternatively, if we know that there is a non-Borel set $Y$ in $\mathbb R^{n-1}$, then by homeomorphism there is a non-Borel set $h(Y)$ in the unit sphere $C\setminus O$, so $X=O\cup Y$ is a non-Borel convex set. But I guess you'd use the same cardinality argument to show that there's a non-Borel set in $\mathbb R^{n-1}$. – bof Oct 26 '18 at 23:22
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Let the convex set be the unit ball in $\mathbb R^n$ with some, but not all, of its boundary points. Then it is convex. And we can choose the subset of the boundary that we include to be a non-Borel set. (Provided we know that there is a non-Borel set in $\mathbb R^{n-1}$, this will be easy, since the boundary of the ball, without one pont, is homeomorphic to $\mathbb R^{n-1}$.)

For fun, note that all convex sets in $\mathbb R^n$ are Lebesgue measurable.

GEdgar
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