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.



A question on the monotonicity formula for minimal submanifolds
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 $\{ \nabla f = 0 \} \cap D\_r(x) \subset \Gamma$.


