I want to 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. 



Hausdorff dimension of convex set in ${\bf R}^n$
1. Yes. The boundary even has a locally finite Haudsorff $(m-1)$-measure.
2. No. A convex function of 1 variable has increasing derivative, but this derivative can have a
dense set of jumps.


In general, the function describing the boundary is only Lipschitz (and differentiable almost
everywhere).


For all these facts, you may consult a nice book Hormander, Notions of convexity, Chap II.


On your other questions. Of course, there is no reason to consider Hausdorff measure of a convex
set: it is ordinary Lebesgue measure in the linear span of this set.
I guess the paper you mention considers Hausdorff measure on the BOUNDARY of a convex set.
As I said in 1, it has integer dimension. But so what?
It is not a smooth surface. What other measure you propose to consider
on it?


