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.

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