Are all convex sets Borel sets? 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?
 A: 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.
A: 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.
