Topology + implicit function theorem Question

Question

Let $f:\mathbb{R}^n \to \mathbb{R} \in C^1$ (continuously differentiable)

Let $E=\{x|f(x)=0\}$, We know that $Df(x)\neq 0\ \forall x\in E$ and that $lim_{|x|\to\infty}f(x)=\infty$

Prove that E is of volume zero

My work: Let $y\in E$, because $Df(y)\neq 0 \ \exists 1\leq i \leq n$ s.t $\nabla f_{x_i}(y)\neq 0$. WLOG Lets assume that its $x_1$. So by the implicit function theorem there is a neighbourhood of y, $U_y \ s.t \ \forall x=(x_1,...,x_n)\in U_y, x_1=g(x_2,...,x_n) $ for some $g\in C^1$ which means x is a graph of a smooth function, which means $U_y$ is of volume zero. Now I'm stuck on proving that E is a finite union of $U_y$'s. Can anyone help?

Answer 1

Measure is countably additive, so you need only proof that it is a countable union, which follows since $\mathbb{R}^n$ is second countable

alternatively: by assumption $E\subset B(0,N)$ for some large enough $N$. This is compact. $E$ is closed in $B(0,N)$, so $E$ is also compact. Hence you can cover $E$ by finitely many of the $E_y$.