Hausdorff dimension of convex set in ${\bf R}^n$ 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. 
 A: In fact one can say quite a lot of regularity of the boundary of a convex set.
Assume that $X\subset\mathbb{R}^n$ is a bounded convex set with non-empty interior.

*

*Convex functions are locally Lipschitz and therefore the boundary of a convex set is locally a graph of a Lipschitz functions defined on a ball in $\mathbb{R}^{n-1}$. It follows that $\mathcal{H}^{n-1}(\partial X)<\infty$.


*Locally Lipschitz functions are differentiable almost everywhere so convex functions are differnetiable almost everywhere. However, one can say much more.
The following beautiful result is due to Anderson and Klee [AK] and Zajíček [Z]:

Theorem. If $f:\mathbb{R}^{n-1}\to\mathbb{R}$ is convex, then there are counrably manly Lipschitz functions
$
g_i:\mathbb{R}^{n-2}\to\mathbb{R}^{n-1},
\quad
i=1,2,\ldots
$
such that $f$ is differentiable in
$$
D=\mathbb{R}^{n-1}\setminus \left(\bigcup_{i=1}^\infty g_i(\mathbb{R}^{n-2})\right).
$$
Moreover, $\nabla f:D\to\mathbb{R}^{n-1}$ is continuous.

In fact Anderson and Klee did not discuss points of non-differentiability of a function, but the above result follows from what they proved. On the other hand Zajíček proved a stronger result than the one above since he showed that the functions $g_i$ can be represented as a differennces of two convex fucntions, see also [H] for an easy to follow adaptation of the original proof due to Zajíček.
This theorem generalizes to convex functions defined on convex domains. As a corollary we obtain:

Corollary. Let $X\subset\mathbb{R}^n$, be a bounded convex domain with nonempty interior and let $B^{n-2}$ be the unit ball in $\mathbb{R}^{n-2}$. Then $\mathcal{H}^{n-1}(\partial X)<\infty$, and there are countably many Lipschitz functions $g_i:B^{n-2}\to\partial X$ such that $\partial X$ has a tangent space at every point of
$$
D=\partial X\setminus
\left(\bigcup_{i=1}^\infty g_i(B^{n-2})\right)
$$
and the tangent space changes continuously along $D$.

Therefore, in a sense, the boundary of $X$ is smooth away from a set of $\sigma$-finite Hausdorff measure $\mathcal{H}^{n-2}$.
Yet another result was proved in [AH].

Theorem. If $X\subset\mathbb{R}^n$ is a convex set with nonempty interior, then for any $\epsilon>0$ there is a convex set $X_\epsilon\subset X$ with the boundary of class $C^{1,1}$ such that
$\mathcal{H}^{n-1}(\partial X\setminus\partial X_\epsilon)<\epsilon$.

Here $C^{1,1}$ means the class of functions with Lipschitz derivative so
the boundary of class $C^{1,1}$ means that the boundary is locally a graph of a $C^{1,1}$ function.
[AH] D. Azagra, P. Hajłasz,
Lusin-type properties of convex functions and convex bodies.
J. Geom. Anal. 31 (2021), 11685–11701.
[AK] R. D. Anderson, V. L. Klee, Jr., Convex functions and upper semi-continuous collections. Duke Math. J. 19 (1952), 349–357.
[H] P. Hajłasz,
On an old theorem of Erdős about ambiguous locus.
Colloq. Math. 168 (2022), 249–256.
[Z] L. Zajíček,
On the differentiation of convex functions in finite and infinite dimensional spaces.
Czechoslovak Math. J. 29 (1979), 340–348.
A: *

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

*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?
