# Measure of the boundary of Alexandrov space

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.
• It seems to me that this argument only shows $dim_H \partial X \le n-1$. How can one prove $H_{n-1}(\partial X)>0$? Jun 7, 2021 at 10:32
• May I ask what is the definition of a regular point in $T_p(\partial X)$? As $\partial X$ may not be an Alexandrov space, the definition is unclear to me. Anyway, one can use the same method and prove by induction that any $m$-dimensional extremal set contains a "cube" which is the image of $I^m$ such that the opposite faces of the cube have positive distance. Jun 8, 2021 at 2:25
• @Totoro A point $p$ is a regular point of subset S if the tangent space $T_pS$ is Euclidean. Jun 8, 2021 at 2:35