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!
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).