Cusp points in Alexandrov spaces

Question

Given a space of bounded integral curvature (by which I mean a topological surface with an intrinsic metric, such that the sum of excesses of any finite collection of non-overlapping simple triangles is bounded), we have the curvature measure $\omega$, defined in terms of these excesses.

A cusp point is when $\omega(\{p\})=2 \pi$. A result of Kapovitch says that the space of directions at any point of an Alexandrov space with curvature bounded below $X$, which can be approximated in the Gromov-Hausdorff topology by smooth Riemannian manifolds with curvature bounded below, must be the sphere $S^{n-1}$, were $n$ is the dimension of $X$.

Since 2D Alexandrov spaces are smoothable in this way, the space of directions in this case is always a circle.

Now my question: does anyone know of a proof that the space of directions at a cusp point of an Alexandrov space is not a sphere? In this way, it would follow that $2D$ Alexandrov surfaces do not contain any cusp points.

Answer 1

Theorem 8 on page 166 in "Intrinsic geometry of surfaces" by Alexandrov and Zalgaller implies that if $\omega(\{p\})=0$, then the total angle around $p$ vanishes. Therefore, the space of directions at $p$ is a single point which is not a circle.