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Let $(X,d)$ be an $n (\geq2)$ dim Alexandrov space with curvature $\geq k$. $B(x,r)$ is an open ball in $X$. Let $M_{k,n}$ be the $n$ dim space form of constant curvature $k$. $B_k(r)$ is an open ball with radius $r$ in $M_{k,n}$.

Question: It is true that $H^{n-1}(\partial B(x,r)) \leq H^{n-1}(\partial B_k(r)) $? Here $H^{n-1}$ means the $(n-1)$ dim Hausdroff volume.

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Yes, it is true.

In fact there is is a distance non-contracting map $\ell\colon\partial B(x,r)\to \partial B_k(r)$.

If $x$ is a regular point then $\ell$ is the $k$-logarithm --- it sends point $z$ to the point $\tilde z$ such that geodesics $[xz]$ and $[\tilde x\tilde z]$ go in the same direction for some identification of tangent spaces $\mathrm{T}_xX=\mathrm{T}_{\tilde x}\mathbb{M}^n[k]$.

If $x$ is a singular point, approximate it by a regular points apply the construction as above and pass to the limit.

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