Hausdorff dimension of convex set in ${\bf R}^n$

I wanto know the smoothness of convex set in ${\bf R}^n$. Recall the following definition. Definition : $X$ is a bounded closed convex set in ${\bf R}^n$ if for $x$, $y\in X$, the any $d$-minimizing geodesic from $x$ to $y$ lies in $X$ where $d$ is a distance function of $X$.

That is, if $Y= S^{n-1}(1)$ and $X$ is convex then for $a$, $b\in X$, then $\frac{sa + (1-s)b}{|sa + (1-s)b |}$ is in $X$ for $0< s<1$

Question 1) Does the boundary of $m$-dimensional bounded convex set has dimension $m-1$ ?

Question 2) Is the following opinion is right ?

$(\ast)$ My thought : Let $m\geq 2$. A $(m-1)$-dimensional boundary of a $m$-dimensional bounded closed convex set $X$ is smooth except some $(m-2)$-dimensional set.

The motivation of this is as follows: In some paper, the Hausdorff measure of convex set in $S^{n-1}(1)$ is considered.

That is, in my thought convex set may be a set of noninteger Hausdorff dimension. Am I right ?

If $\ast$ is right, then why does one consider the Hausdorff measure of convex set ?

Thank you in advance.

[paper's content]-----------------------------------------------------

3.1 Proposition : $X$ is a closed convex set in $S^{n-1}(1)$ and $u$ is a point in $X$ Then area$(X\cap {\bf H}_u) \geq \frac{1}{2}$area$(X)$ where ${\bf H}_u = \{ p\in S^{n-1}(1) | p\cdot u \geq 0\}$

3.2 Note : If $X \subset S^{n-1}(1)$ is a convex $spherical$ set of Hausdorff dimension $d$, then $H^d(X\cap {\bf H}_u) \geq \frac{1}{2} H^d(X)$ where $H^d$ is the $d$-dimensional Hausdorff measure.

Here there is the word "spherical". I think that if we omit the word, then it is also fine.

• If $m< n$, the boundary of an $m$-dimensional closed set (convex or otherwise) is the set itself, as it has empty interior. And what is $S^{n-1}(1)$? A sphere has no convex subset with more than one point. In any case, a convex set has the same Hausdorff dimension as its affine hull, which is an integer. – Emil Jeřábek Nov 23 '12 at 16:34
• Perhaps this discussion would have more point if the OP told us which paper he was looking at ?! – Igor Rivin Nov 23 '12 at 20:14
• A geodesic ball in $S^{n-1}(1)$ is not convex according to the definition you’ve given. If $x,y\in X\subseteq S^{n-1}(1)$, $x\ne y$, then every point $tx+(t-1)y$ with $0< t< 1$ has norm strictly less than $1$, and thus is not an element of $X$. – Emil Jeřábek Nov 26 '12 at 12:39
• I see that the paper is using a different definition: they say that an $X\subseteq S^{n-1}$ is convex if any two points of $X$ can be joined by a distance-minimizing geodesic which lies in $X$. – Emil Jeřábek Nov 26 '12 at 12:44
• This question formulation was highly unclear to me. – Włodzimierz Holsztyński Apr 15 '14 at 21:56

1. Yes. The boundary even has a locally finite Haudsorff $(m-1)$-measure.