Density of an $n$-dimensional embedded submanifold Let $ M $ be a n-dimensional embedded $ C^1 $ submanifold in $ R^{n+k} $. Let  $ H^n $ be the n-dimensional Haussdorf measure of $ R^{n+k} $. I want to prove that 
$$ \lim_{r \rightarrow 0} \frac{H^n(B_r(x)\cap M)}{\alpha(n)r^n}=1 \;\text{ for every } \;x \in M $$
where $ B_r(x) $ is the $(n+k)-$ dimensional ball of $ R^{n+k} $ and $ \alpha(n) $ is the n-dimensional Lebesgue measure of the unit ball of $ R^n $.
The most obvious approach is to use local charts and area formula. But i'm not able to conclude....
Moreover i'm looking for a proof without the use of area formula...
Thanks 
 A: Let $P$ be the tangent hyperplane to $M$ at $x$. Choose the coordinates on $\mathbb R^{n+k}$ so that $x=0$ and $P$ is the coordinate hyperplane of $x_1,\dots,x_n$. A neighborhood of $0$ in   $M$ can be written as the graph $x_{j}=f_j(x_1,\dots,x_n)$, $j=n+1,\dots,n+k$ where $\nabla f_j(0)=0$. 
Consider the orthogonal projection $\pi:M\to P$. It is a $1$-Lipschitz map, therefore it does not increase any Hausdorff measure. Also, the inverse of $\pi$ is 
$$(x_1,\dots,x_n)\mapsto (x_1,\dots,x_n,f_{n+1},\dots,f_{n+k})$$ 
which is $(1+\epsilon)$-Lipschitz, where $\epsilon$ can be made as small as you wish by choosing smaller neighborhood, due to $\nabla f_j=0$.
The above Lipschitz properties imply the following: 


*

*$(1+\epsilon)^{-n}\mathcal H^n(E)\le \mathcal H^n(\pi(E))\le \mathcal H^n(E)$ for any set $E\subset M$ contained in the aforementioned neighborhood of $0$.

*$B_{r/(1+\epsilon)}(0)\cap P \subseteq \pi(B_r(0)\cap M)\subseteq B_r(0)\cap P$ for sufficiently small $r$.


From 1 and 2 it follows that 
$$\lim_{r\to 0}\frac{\mathcal H^n(B_r(0)\cap M)}{\mathcal H^n(B_r(0)\cap P)}=1$$
