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I am studying Evans-Gariepy book and in corollary 1 of section 3.1.2, he prove that if $f:\mathbb{R}^N\to\mathbb{R}^M$ is locally Lipschitz and $$Z=\{x:\ f(x)=0\},$$

then $Df(x)=0$ a.e. $x\in Z$. He prove it by contradiction, however, I think I can prove it directly, but I am stuck in the following.

Let $f:\mathbb{R}^N\to\mathbb{R}$ be a measurable function (with respect to Lebesgue measure) and $$E=\{x\in \mathbb{R}^N:\ f(x)=0\}.$$

Suppose that there is $x\in E$ with $$\lim_{r\to 0}\frac{|E\cap B(x,r)|}{|B(x,r)|}=1, $$

where $|\cdot|$ stands for Lebesgue measure. Fix $v\in \partial B(0,1)$. Can I conclude that there exist two sequences, $v_n\in\partial B(0,1)$ and $t_n>0$ such that $$v_n\to v,\ t_n\to 0\ \mbox{and}\ f(x+t_nv_n)=0.$$

If we assume by contradiction that this is not true then, for each sequence $v_n\to v$ and $t_n\to 0$, we will have without loss of generality that $f(x+t_nv_n)\neq 0$ for big $n$. Now I am stuck here, because it seems to me that such assumption will led to a set of points with positive density in $E$, contradicting the assumption. Any idea is appreciated.

Remark: The answer below provided a affirmative answer to my question. So, to prove corollary 1 with it, take $v_n,t_n$ as above and note that $$0=\frac{f(x+t_nv_n)-f(x)}{t_n}=f'(x)(v_n)+\frac{o(t_n)}{t_n},$$

hence, by taking the limit, we conclude that $$f'(x)(v)=0,\ \forall v.$$

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Let's take $x=0$ to simplify things. Suppose there is $v\in \partial B(0,1)$ for which this fails. Then there is $r>0$ and $\epsilon>0$ such that for every $y\in E$ with $0<|y|<r$ we have $|y/|y| - v|>\epsilon$. But this means that $E$ is disjoint from a cone with vertex at $0$. This contradicts the assumption that $E$ has Lebesgue density $1$ at $0$. Indeed, the intersection of this cone with a ball $\{|y|<\rho\}$ has volume $c\rho^n$ with $c=c(\epsilon)>0$.

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