Let $C\subseteq \mathbb R^n$ be non-empty, convex and compact. For $v\in S^{n-1}$, let $H\_v$ be the supporting hyperplane in the direction of $v$ (i.e., $H\_v$ is the boundary of the smallest closed half-space with outward normal $v$ that contains $C$). Let $U\subseteq S^{n-1}$ be the set of directions $v$ such that $H\_v$ meets $C$ at exactly one point.


**Main question:** Does $S^{n-1}\backslash U$ have measure zero?


If not, then I have a second question: Is $U$ dense in $S^{n-1}$?


For $n=2$, it's easy to see that $S^{n-1}\backslash U$ is countable (otherwise there are uncountably many nondegenerate line segments in the boundary of $C$, and hence $C$ has infinite perimeter). But a cylinder shows that in general $S^{n-1}\backslash U$ need not be countable.



For most directions does the supporting hyperplane meeting a bounded convex set meet it in one point?
The support function of $C$, restricted to the unit sphere, is differentiable exactly at directions such that (the relevant) hyperplane normal to that direction has a single contact point with $C$.


Because $C$ is bounded, its support function is Lipschitz. Rademacher's theorem then says it is differentiable almost everywhere, giving a positive answer to the question.


