$U$ is an open subset of $\mathbb R^n$, and $f: U \to \mathbb R$ is a $C^1 $ function such that $\nabla f \neq 0$ everywhere.
Define $A = \{x \in U: f(x) = 0\}$.
Show that $A$ and $\partial A$ have zero volume.
Here is what I did:
Suppose $A$ actually has volume. Then the interior $A^\circ$ is not empty. This means it contains some small closed ball $B \subset A^\circ \subseteq A$.
We have that $f(x) = 0$ on $B$, so $f$ is constant there, so the gradient is zero there, which is a contradiction, hence it can't be that $A$ has volume.
Is this correct? How would I prove that $\partial A$ has zero volume?
A possible idea:
take $a\in A$. Then $f(a)= 0$. Since $\nabla f(a) \neq 0$ we have that for some $i$ between $1$ and $n$, $\frac{\partial f}{\partial x_i}(a) \neq 0$. Without loss of generality, assume $i = n$
From the implicit function theorem, there is an open set $W$ containing $(a_1,a_2, \dots a_{n-1})$, and there is an open set $V$ containing $a_n$, and there is a $C^1$ function $g:W \to V$ such that $g(a_1,a_2, \dots, a_{n-1})=a_n$, and for all $w \in W$ we have $f(w,g(w)) = 0$.
Consider the graph of the function $g$: $\Gamma_g = \{(w,g(w)), w\in W\}$. This is a zero volume set, since it's a graph of a $C^1$ function, but also $\Gamma_g \subseteq A$.
Maybe we can cover $A$ by countably many $\Gamma_g$?