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!

  • $\begingroup$ What is $\phi$? What are the assumptions or required properties of $\phi$? $\endgroup$
    – Deane
    Commented Feb 10 at 20:02
  • $\begingroup$ 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 $1$. $\endgroup$
    – Deane
    Commented Feb 10 at 20:07
  • $\begingroup$ 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. $\endgroup$ Commented Feb 10 at 20:26
  • $\begingroup$ @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? $\endgroup$ Commented Feb 10 at 20:44

1 Answer 1


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) \begin{equation} |\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} \end{equation}

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).

  • $\begingroup$ 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 $q.$ $\endgroup$ Commented Feb 12 at 21:32
  • $\begingroup$ 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. $\endgroup$ Commented Feb 12 at 21:44
  • $\begingroup$ 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. $\endgroup$ Commented Feb 12 at 23:02
  • $\begingroup$ 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 $x'$. $\endgroup$ Commented Feb 13 at 19:06
  • $\begingroup$ 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! $\endgroup$ Commented Feb 13 at 20:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .