TITLE: A question on the monotonicity formula for minimal submanifolds
QUESTION [5 upvotes]: I'm reading the proof of monotonicity formula from A Course in Minimal Surfaces by Colding-Minicozzi. The theorem says

Suppose $\Sigma^k \subset \mathbb{R}^n$ is a minimal submanifold and $x_0\in\mathbb{R}^n$; then for all $0<s<t$,
$$
\frac{\mathrm{Vol}(B_t(x_0)\cap\Sigma)}{t^k} - \frac{\mathrm{Vol}(B_s(x_0)\cap\Sigma)}{s^k} \\
= \int_{(B_t(x_0)\setminus B_s(x_0))\cap\Sigma} \frac{|(x-x_0)^N|^2}{|x-x_0|^{k+2}},
$$
where $B_t(x_0)$ is the ball in $\mathbb{R}^n$ with center $x_0$ and radius $t$, and $(\cdot)^N$ the projection onto the normal space of $\Sigma$.

In the proof, they define $f(x)=|x-x_0|$ and then argue that
$$
\mathrm{Vol}(\{f\le s\}\cap\Sigma) = \int_{\{f\le s\}\cap\Sigma} |\nabla_\Sigma f|^{-1} |\nabla_\Sigma f| d\mathcal{H}^k = \int_0^s \int_{f=\tau} |\nabla_\Sigma f|^{-1} d\mathcal{H}^{k-1} d\tau
$$
My question is about the first equality. Here the authors implicitly assumed that
$$
|\nabla_\Sigma f|>0 \quad \mathcal{H}^k\mathrm{-a.e.} \text{ on } \Sigma,\label{1}\tag{$*$}
$$
but didn't give a proof.

How do I prove \eqref{1}?

My idea: Since $\Delta_\Sigma f^2 = 2k>0$, $f^2$ is a subharmonic function on $\Sigma$. It suffices to show that the set of critical points of a smooth subharmonic function has $\mathcal{H}^k$ measure $0$. But I can't find a reference for the latter.

REPLY [4 votes]: A stronger property is true: the critical set of a strictly subharmonic function $f: \Sigma \to \mathbf{R}$ is locally contained inside a codimension one submanifold. Explicitly: for every critical point $x \in \Sigma$ there is $r > 0$ and a $(k-1)$-dimensional submanifold $\Gamma \subset \Sigma \cap D_r(x)$ so that $\{ \nabla f = 0 \} \cap D_r(x) \subset \Gamma$. (Here $D_r(x)$ is the ball of radius $r$ in the intrinsic metric on the surface.)
Let $x \in \Sigma$ be such a critical point. In local normal coordinates one can identify $\nabla^2 f(x) = (\partial^2_{ij} f(x))$. This matrix has positive trace because $\Delta f(x) > 0$. Therefore at least one partial derivative $\nabla \partial_i f(x) \neq 0$, and by the implicit function theorem the zero set $\{ \partial_i f = 0 \}$ is a $C^1$ embedded surface in a small enough neighbourhood of $x$. We conclude by setting $\Gamma = \{ \partial_i f = 0 \} \cap D_r(x)$ and observing that $\{ \nabla f = 0 \} \cap D_r(x) \subset \Gamma$.