I was reading Morgan's "GMT: a beginner's guide" and I stucked on a very simple fact. I was not very familiar with Hausdorff measures, hence I have some troubles.
The topic is the same of this question. Consider the cone $C:=\{x^2+y^2=z^2\} \subset \mathbb R^3$ and suppose $0 \ne x \in C$. Then I want to prove that $$ \vartheta^{2}(C,x) :=\lim_{r \to 0} \frac{\mathscr{H}^2(C \cap B^{3}(x,r))}{\pi r^2} = 1 $$
where $B^3(x,r)$ is the three dimensional ball of radius $r$ and center $x$. Is there a simple explanation? I can't manage to prove it, I cannot evaluate the Hausdorff measure of the intersection... Could you provide any hint, please?
Thanks.
Let $P$ be the tangent plane to $C$ at the point $x$. Choose a coordinate system $(u,v,w)$ so that $x$ is its origin, $u,v$ axes are parallel to $P$, and $w$-axis is orthogonal to it. In this coordinate system, $C$ is given by the equation $w = f(u,v)$ where $f$ is a smooth function such that $\nabla f(0,0)=0$. Note that this is nothing special about the cone; the above holds for any submanifold of $\mathbb R^3$, pretty much by the definition of a submanifold.
For every $\epsilon>0$, there is a neighborhood of $(0,0)$ in which the map $(u,v)\mapsto (u,v,f(w))$ satisfies the bi-Lipschitz condition with constants $1\pm \epsilon$. Indeed, its derivative at $(0,0)$ is a matrix of norm $1$, which implies it's $(1+\epsilon)$-Lipschitz in some neighborhood. That the inverse map $(u,v,f(w))\mapsto (u,v)$ is $1$-Lipschitz is easy to see directly.
So, the $\mathcal H^2$ measure of $C\cap B^3(x,r)$ is asymptotic to the area of the orthogonal projection of $C\cap B^3(x,r)$ onto $P$. This projection is contained in a disk of radius $r$, which gives an upper bound on $\mathcal H^2$. On the other hand, $C\cap B^3(x,r)$ is contained in the slab $|w|\le \sup_{|(u,v)|<r} |f| <\delta r $ where $\delta\to 0$ as $r\to 0$. Looking at the intersection of $C$ with the boundary of $B^3(x,r)$, you will conclude that the projection of $C\cap B^3(x,r)$ onto $P$ contains the disk of radius $\sqrt{r^2-(\delta r)^2}$. This gives a lower bound on $\mathcal H^2$.