# Understanding Equality of Hausdorff Measures in Global and Submanifold Riemannian Metrics

I am reading chapter 12 of Measure and Integration Theory of Micheal E.Taylor. I struggle with a point on the proof of the following proposition.

Proposition 12.7:

Let $$\Omega$$ be a $$C^1$$ manifold with a $$C^0$$ metric tensor, and let $$M$$ be a $$C^1$$ embedded submanifold of $$\Omega$$, with the induced metric tensor, so $$M$$ has two metrics, the metric $$d_{M}$$ obtained minimizing curve in $$M,$$ and the one obtained minimizing curves in $$\Omega:$$ please forgive the slang, I obsviously mean Riemmanian distances. If $$H_M$$ and $$H_\Omega$$ denote the respective $$r$$-dimensional Hausdorff measures, then, for any $$r \in \mathbb{R}^+$$,

$$\text{(12.33)} \quad S \subset M \ \ \text{Borel} \Rightarrow H_M(S) = H_\Omega(S).$$

I'll now present the proof given in the book.

Proof: It suffices to note that

$$\text{(12.34)} \quad d_M(p, q) = \varphi(p, q) \, d_{\Omega}(p, q), \quad \varphi(p,p) = 1,$$

with $$\varphi : M \times M \longrightarrow \mathbb{R}^+$$ continuous. Then we can apply proposition $$12.5$$: I did not reported proposition $$12.5$$ because this is not the unclear passage.

The piece that I am missing is the following: proving the existence of $$\phi.$$ The only good candidate is clearly $$\varphi(p,q) := \begin{cases} \dfrac{d_{M}(p,q)}{d_{\Omega}(p,q)} && p \neq q \\ 1 && p = q \end{cases},$$ but how can I prove that $$\lim_{(p,q) \to \Delta} \varphi(p,q) = 1?$$ Here, $$\Delta$$ denotes the diagonal of $$M \times M$$.

One thing we can say for sure is that $$\frac{d_{M}(p,q)}{d_{\Omega}(p,q)} \geq 1 \quad \forall (p,q) \in M \times M \setminus{\Delta}.$$ Consequently, we have $$\liminf_{(p,q) \to \Delta} \frac{d_{M}(p,q)}{d_{\Omega}(p,q)} \geq 1.$$ Now, I know I have to play around with the definition of Riemannian distance, but I get entangled between the $$\epsilon$$-s and $$\delta$$-s of the definitions; moreover, I do not know how to adopt a convenient system of coordinates!

• What is $\phi$? What are the assumptions or required properties of $\phi$? Commented Feb 10 at 20:02
• I suggest trying this first where $\Omega=\mathbb{R}^3$ and $M$ is a smoothly embedded surface. Maybe even start with $M$ equal to the standard unit sphere. Also, I think you have the inequality reversed. $\phi(p,q)$ is bounded from below, not above, by $1$. Commented Feb 10 at 20:07
• Yes, thank you. The function $\varphi$ is just a continuous functions on the product $M \times M.$ I will try to prove the case you suggested me. Commented Feb 10 at 20:26
• @Deane I think I got it. Indicate the two distances with $d_1$ and $d_{2}.$ Given an arbitrary positive $\epsilon$ there exists a curve $\gamma_2$ in $\Omega$ such that $\mathcal{l}(\gamma_2) < \epsilon + d_2.$ Indicate with $\bar{\gamma}_{2}$ the 'projection' of $\gamma_2$ on $M,$ then we have : $$\dfrac{d_1}{d_2+\epsilon} \leq \frac{l(\bar{\gamma}_2)}{l(\gamma_2)} \leq 1.$$ The last inequality hold because the length of the two curves differ at most for one component. I have to scale up to the general case. Could you please indicate the tools necessary for this last step? Commented Feb 10 at 20:44

Let $$M\subset \Omega$$ be an embedded manifold, and $$p, q\in M$$. We want to compare $$d_M(p, q)$$ and $$d_\Omega(p, q)$$. As in the post and the comments, it is easy to see that $$d_\Omega(p, q)\leq d_M(p, q),\tag{1}$$ since a path in $$M$$ is also in $$\Omega$$.

Now we try to find a sort of bound in the other way, and in particular prove that $$\lim_{q\in M, q\to p} \frac{d_M(p, q)}{d_\Omega(p, q)} = 1.\tag{2}$$

Let $$U\subset \Omega$$ be a neighborhood of $$p$$ in $$\Omega$$ with adapted submanifold coordinates $$(x^1,\dots,x^m,y^1,\dots,y^n)$$ such that $$M\cap U$$ is defined by $$y^j=0, 1\leq j\leq n$$ and $$p$$ is defined by furthermore $$x^i=0, 1\leq i\leq m$$. (So $$M$$ has dimension $$m$$ and codimension $$n$$.)

Let the metric on $$\Omega$$ be presented in these coordinates as $$\begin{pmatrix} g & h\\ h^T & k \end{pmatrix},$$ where $$g_{ij}=\langle \frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}\rangle$$, $$h_{ij}=\langle \frac{\partial}{\partial x^i}, \frac{\partial}{\partial y^j}\rangle$$, and $$k_{ij}=\langle \frac{\partial}{\partial y^i}, \frac{\partial}{\partial y^j}\rangle$$.

Let $$q\in M\cap U$$. Note that distance is locally achieved by geodesics, since we can require $$U$$ to be a normal neighborhood of $$p$$. Let $$\epsilon=d_\Omega(p, q)$$, and $$\alpha: [0, \epsilon]\to U\subset \Omega;\ \alpha(0)=p, \alpha(\epsilon)=q$$ be the unit-speed geodesic in $$\Omega$$ realizing the distance.

In the coordinates, write $$\alpha(t)$$ out as $$(x^1(t),\dots,x^m(t),y^1(t),\dots,y^n(t)).$$ Let $$\beta:[0,\epsilon]\to M\cap U$$ be defined by $$(x^1(t),\dots,x^m(t),0,\dots,0).$$

By the form of the metric, we have (with obvious notation) $$$$|\alpha'(t)|_\Omega^2 = |\beta'(t)|_M^2 + 2 \sum_{ij} h_{ij}{x^i}'(t){y^j}'(t) + \sum_{ij} k_{ij}{y^i}'(t){y^j}'(t). \tag{3}$$$$

Here is the key point to me. Since $$p, q\in M$$, we have $$y^j(0)=y^j(\epsilon)=0,\quad 1\leq j\leq n.$$ Therefore, by the Rolle's theorem, $$\exists \xi_j\in (0, \epsilon)$$ such that $$(y^j)'(\xi_j)=0$$.

On $$U$$, we have that the Christoffel symbols are bounded, and that the $$(x^i)'(t)$$ and $$(y^j)'(t)$$ are bounded. ($$\alpha$$ has unit speed.) By the geodesic equation, we see that the $$(y^j)''(t)$$ (and also the $$(x^i)''(t)$$ ) are also bounded.

Therefore, $$\exists C>0$$ such that $$\forall t\in [0, \epsilon]$$ $$|(y^j)'(t)|\leq |(y^j)'(\xi_j) + (y^j)''(c) (t-\xi_j)|\leq C\epsilon.$$ Then (3) would mean $$|\beta'(t)|_M\leq |\alpha'(t)|_\Omega + C\epsilon,$$ for a different constant $$C$$.

After integration, we get $$d_M(p, q)\leq \int_0^\epsilon |\beta'(t)|_M\,dt\leq \int_0^\epsilon |\alpha'(t)|_\Omega\,dt + C\epsilon^2= \epsilon+C\epsilon^2.$$ Noting that $$d_\Omega(p, q)=\epsilon$$, we get by diving by it that $$\frac{d_M(p, q)}{d_\Omega(p, q)}\leq 1 + C{d_\Omega(p, q)}.$$ Therefore, $$\limsup_{d_\Omega(p, q)\to 0} \frac{d_M(p, q)}{d_\Omega(p, q)} \leq 1.$$

Combining this with the other direction (1), we have proved the assertion (2).

• Very insightful answer! Drawing closer to the surface $M$ implies that the growth rate of coordinates along the directions that would lead me away from $M$ also decreases! It decrease at least as fast as the distance between the two points $p$ and $q.$ Commented Feb 12 at 21:32
• I have only one objection the original question considered a C^1 manifold; thus I should not be using second derivatives. Anyway, this is a minor issue, but it seemed correct to point it out. Commented Feb 12 at 21:44
• Yes, I noticed that. I am a $C^\infty$ guy, but maybe there is a proof more in the spirit of analysis and continuity that does not require so much regularity. Commented Feb 12 at 23:02
• Thanks for the bounty. It is fun to work this out. I do want to say that I doubt a purely continuous proof exists. Then you don't have Christoffel symbols, and no geodesics. I thought the way I did it was rather tight, and in particular it involves the geodesic equation to relate $y''$ with $y'$ and $x'$. Commented Feb 13 at 19:06
• Yes, now that I think of it without C^2 you do not even have this neat trick of using geodesics. I had a rough idea in mind. Instead of a geodesic you can take a curve $\gamma_n$ that has length at least $\epsilon -1/n.$ The curves $\gamma_n$ approach the constant curve, and then I could conclude if having a $C^1$ manifold implied the uniform convergence of the curve and of it's velocities. Anyway, I think the exercise has become a little too time consuming. I should focus on other stuff. Moreover, there is certainly an approximation trick that avoids all the dull calculations! Thanks! Commented Feb 13 at 20:10