*** A set $E\subseteq \Bbb R^n$ will be called negligible iff for every $\varepsilon >0$ there exists a countable set (possibly finite) of n-dimetional rectangles $\{Q_n\}_{n\in \Bbb N}$ s.t. $E\subseteq \bigcup_{n} Q_n$ and $\sum_{n} Vol(Q_n) <\varepsilon$
I'm having a bit of trouble with this Question although it does not seem that difficult, here is my approach:
for each $x\in f^{-1}(0)$ we have $\nabla f(x)\neq 0$ so by the implicit function theorem a neighborhood $U_x$ of $x$ contained in $f^{-1}(0)$ is (up to permutation of coordinates) a negligible set as a graph of a $C^1$ function.
Hence if we could cover $f^{-1}(0)$ with only countably many $U_x$'s we would have that $f^{-1}(0)$ is negligible. However, I cannot justify this last step.
Edit:
since $E$ is open we could cover it with a countable collection of compact sets $\{K_i\}_i$. in each of them, the above argument would let us conclude that $(f|_{K_i})^{-1}(0) = f^{-1}(0)\cap K_i$ is negligable (we could choose finitely many $U_x$'s to cover $f^{-1}(0)\cap K_i$ since it is a compact set as a closed subset ($f^{-1}(0)$ is closed since $f$ is continuous) of a compact set) and hence:
$$f^{-1}(0) = \bigcup_i (f|_{K_i})^{-1}(0) = \bigcup_i f^{-1}(0)\cap K_i$$
and so $f^{-1}(0)$ is a countable union of negligable sets.
Is this solution correct?