Let $X^n$ be an $n$-dimensional Alexandrov space with curvature bounded from 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.)

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.)

Let $X^n$ be an $n$-dimensional Alexandrov space with curvature bounded from 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 regulat points on convex hypersurfface should have full Hausdorff measure.)

Let $X^n$ be an $n$-dimensional Alexandrov space with curvature bounded from 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.)