Let $X^n$ be an $n$-dimensional Alexandrov space with curvature bounded below. A point $x\in X$ is called regular if the space of directions $\Sigma\_x$ is isometric to the standard sphere $S^{n-1}$.


**QUESTION 1. Is it true that the set of regular points has full Hausdorff measure?**


(Rmk: Theorem 10.9.13 in the Burago-Burago-Ivanov book claims a weaker property: this set is everywhere dense, and moreover is a countable intersection of open everywhere dense subsets.)


If the answer is yes, a reference would be helpful.


**QUESTION 2.** Let now $X^n$ be a convex hypersurface in the Euclidean space $\mathbb{R}^{n+1}$. Let $x\in X$ be a smooth point of $X$, i.e. there is a unique supporting hyperplane at $x$. **Is it true that $x$ is regular in the above sense?**


(Rmk: if this is the case then the set of regular points on convex hypersurface should have full Hausdorff measure since the set of smooth points has full measure.)



Set of regular points in an Alexandrov space with curvature bounded below
"Yes" to both questions.


For the second, take the projection to the tangent plane and note that its bi-Lipschitz in a small neighborhood of $x$ with constants as close to 1 as you want.


For the first one, see in 10.6 in "Alekandrov's Space with Curvature bounded from below" by Burago, Gromov and Perelman.


[In fact you can say bit more about regular set;
it is convex and the complement is countably $(n-1)$-rectifiable;
that is it lies in the images of countable collection of Lipschitz maps $\mathbb {R}^{n-1}\to X^n$.
Moreover if there is no boundary then it is is countably $(n-2)$-rectifiable.
One can say yet more --- in some sense all you know about singularities of convex surfaces is known for Alexandrov spaces.]


