I have to prove that: $$\lim_{r \to 0} \frac {\mathcal H ^{n-1}(M \cap B(x,r))}{\omega_{n-1} r^{n-1}}=1, $$ where $\mathcal H ^{n-1}$ is the ($n-1$)-dimensional Hausdorff measure, $M$ is a $C^1$ hypersurface s.t. exists a function $\nu_E: \mathbb R ^n \to \mathbb R ^ {n-1}$, with $|\nu_E|=1$ and $\nu_E^\bot(x)=T_xM$, $B(x,r)$ is a $n$-dimensional sphere of centre $x$ and radius $r$, $\omega_{n}$ the volume of the$ $n-dimensional ball of radius $1$.
On the book, Sets of finite perimeter and variational problems by F.Maggi page 181, it is written that the equality follows from the proposition:
If $u: \mathbb R^{n-1} \to \mathbb R $ is a Lipschitz function, then for every $G \subset \mathbb R ^{n-1} $:
$$ \mathcal H^{n-1}(\Gamma(z,G))=\int_G \sqrt{1+|\nabla u(z)|^2} dz, $$
where $\Gamma(z,G)=\{z\in \mathbb R^n | z=(x,u(x)) \text{ where } x \in G \}$.
I don't understand how the equality follows from this, any help?
Fix $x^*\in M$. Choose a coordinate system in which $x^*$ is the origin and the tangent hyperplane to $M$ at $x^*$ is the plane $x_n=0$. In this coordinate system, a piece of $M$ near $x^*$ is represented as $\{z\in \mathbb{R}^n: z = (y,u(y))\}$ where $u$ is a $C^1$ function such that $u(0)=0$ and $\nabla u(0)=0$. Note that $\sqrt{1+|\nabla u(y)|^2}\to 1$ as $y\to 0$.
For small $r$, the intersection of $M$ with $B(x^*,r)$ projects to a subset $G$ of $\mathbb{R}^{n-1}$ that is contained in $(n-1)$-dimensional ball of radius $r$, and contains a ball of radius $(1-\epsilon)r$, where $\epsilon$ is small when $r$ is small.
So, the area of $M$, given by the integral of $\sqrt{1+|\nabla u|^2}$ over $G$, is $\omega_{n-1}r^{n-1}$ up to an error term that is $o(r^{n-1})$.