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.



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> 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.)
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Measure of the boundary of Alexandrov space
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.


