How can we prove that if $\phi:\mathbb{R}^N\to\mathbb{R}$ is a $C^1(\mathbb{R}^N)$ function with the property that $\phi^{-1}(0)\neq \emptyset$ and $\nabla\phi(x)\neq 0,\ \forall\ x\in \phi^{-1}(0)$ then the set $\phi^{-1}(0)$ has Lebesgue Measure $0$, i.e.:
$$\mathcal{L}^N(\phi^{-1}(0))=0$$
Remarks
If $\phi^{-1}(0)$ contains an interior point then we have a ball $B$ on which $\phi(x)=0,\ \forall\ x\in B$. But this means that $\nabla\phi(x)=0,\ \forall\ x\in B$, contradicting our assumption.
However, there are closed sets (even compact) that have positive Lebesgue measure and have not an interior point (see: Cantor fat set for example). But beacause $\phi^{-1}(0)$ is locally a $C^1 $graph of a function the conclusion will be true.
I can't figure out how to complete this proof.
Moreover, I know that if $\phi^{-1}(0)$ is bounded (hence compact) then $\phi^{-1}(0)$ has a positive Hausdorff $N-1$ dimensional measure. If the last can be proved then the former is a consequence.
Best wishes and a happy new year!