Let $X$ be a compact $n$-dimensional Alexandrov space with curvature bounded below. Let $\partial X$ denote its boundary in the sense of the theory of Alexandrov spaces.
Is it true that if $\partial X\ne \emptyset$ then it has finite and positive $(n-1)$-Hausdorff measure? (The case $n=2$ is already interesting to me.)
It can be proved by induction on $n$. Base case $n=1$. The step follows since gradient exponent is locally Lipschitz and it maps $T_p(\partial X)=\mathrm{Cone}[\Sigma_p(\partial X)]$ to a neighborhood of $p$ in $\partial X$. It proves that $\partial X$ has finite $(n-1)$-Hausdorff measure.
The differential of gradient exponent at the origin is the identity map. It follows that a small cube $\square$ around a regular point in $T_p(\partial X)$ is mapped amost isometrically; in particular the opposite faces of $\square$ go to sets on positive distance. By Besicovitch inequality, the image of $\square$ has positive $(n-1)$-Hausdorff measure.
The same proof shows that an $m$-dimensional extremal subset in a compact Alexandrov space has finite positive $m$-dimesnional Hausdoeff measure.
