Set of regular points in an Alexandrov space with curvature bounded below

Question

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.)

Answer 1

"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.]