If $S$ is a convex subset of $\mathbb R^d$, then the Hausdorff dimension of the boundary of $S$ is at most $d-1$

Question

Let $S$ be a subset of $\mathbb R^d$. Let $H^k$ be the $k$-dimensional Hausdorff measure and $\dim_H (S)$ the Hausdorff dimension of $S$, i.e., $$ \dim_H (S) := \inf \{k \in [0, +\infty) \mid H^k (S) = 0\}. $$

Could you elaborate on the proof or provide a reference of below theorem?

Theorem: If $S$ is convex, then $\dim_H (\partial S) \le d-1$.

Answer 1

HINT:

1. Reduce to $S$ convex and compact ( approximate $S$ by intersecting with larger and larger cubes)

2. For $S$ convex and compact, surround it with a sphere $S_1$. Now the map $S_1\to \partial S$, $x\mapsto $ closest point in $S$ to $x$ is Lipschitz with constant $1$ and surjective ( I have an answer on this site for this)

3. Reduce to the case $S$ is a ball, so $\partial S$ is a sphere. Now $\partial S$ is a smooth compact manifold of dimension $n-1$, so the Hausdorff dimension of it is exactly $n-1$.

Obs: In fact the Hausdorff dimension of $\partial S$ equals $n-1$ for any convex compact $S$ in $\mathbb{R}^n$ with non-void interior. Simply reverse the argument with the closest point, with a ball inside of $S$.