Simple question on Mean value theorem Let $\phi,\psi\in C^1[a,b]$ of bounded variation, and $C$ a curve given by $x=\phi(t),\ y=\psi(t)$. Let $\Gamma=\{t_i\}_{i=0,\cdots,m}$ be a partition of $[a,b]$. By $C$ rectifiable we mean $$ L=\sup_\Gamma S_\Gamma=\sup_\Gamma \sum_{i=1}^m \left( (\phi(t_i)-\phi(t_{i-1}))^{2}+ (\psi(t_i)-\psi(t_{i-1}))^{2}\right)^{1/2}<\infty$$
Here because of bounded variation condition holds, we have $C$ rectifiable. The problem is asking me to prove that $$L=\int_a^b \left( \phi'(t)^{2}+ \psi'(t)^{2}\right)^{1/2}\ dt\ \ \  (*)$$
Since $\phi',\psi'$ are continuous, I know that
$$L=\lim_{|\Gamma|\to 0} S_\Gamma $$
So it seems natural to use, $$\phi(t_i)-\phi(t_{i-1})=\phi'(\xi_i)(t_i-t_{i-1}) \ \mathrm{and}\  \psi(t_i)-\psi(t_{i-1})=\psi'(\rho_i)(t_i-t_{i-1})$$ for some $t_{i-1}\leq \xi_i,\rho_i\leq t_i$ and for all $i=1,\cdots,m$, to form a Riemann sum for $(*)$ that converges to $L$, indeed starting from $S_\Gamma$ and replacing we get
$$S_\Gamma = \sum_{i=1}^m \left( \phi'(\xi_i)^{2}+ \psi'(\rho_i)^{2}\right)^{1/2}(t_i-t_{i-1}) \ \ (**)$$
My question starts here:
Intuitively, if $|\Gamma|\to 0$, is expected that $t_{i-1},\xi_i,\rho_i,t_{i}$ will be close enough, so $(**)$ is very close to a Riemann sum for $(*)$ and we are done. But, is it just like that I mean, can I take $\left( \phi'(\xi_i)^{2}+ \psi'(\rho_i)^{2}\right)^{1/2}$ as if it were $\left( \phi'(\xi_i)^{2}+ \psi'(\xi_i)^{2}\right)^{1/2}$? what is the formal justification for this?
 A: Unless by chance it so happens that $\rho_i = \xi_i$ for all $i$, $(\ast\ast)$ will not be a Riemann sum for $(\ast)$. Thus we have a little more work to do. But, since $\phi$ and $\psi$ are assumed to be continuously differentiable, fortunately not much more work.
By the uniform continuity of $\psi'$, $(\ast\ast)$ differs very little from a Riemann sum for $(\ast)$. Using the triangle inequality for the Euclidean norm, we have
$$\bigl\lvert \bigl(\phi'(\xi_i)^2 + \psi'(\rho_i)^2\bigr)^{1/2} - \bigl(\phi'(\xi_i)^2 - \psi'(\xi_i)^2\bigr)^{1/2}\bigr\rvert \leqslant \lvert \psi'(\rho_i) - \psi'(\xi_i)\rvert\,. \tag{1}$$
And by the uniform continuity of $\psi'$, for every $\varepsilon > 0$ there is a $\delta > 0$ such that
$$\lvert \psi'(\rho_i) - \psi'(\xi_i)\rvert \leqslant \varepsilon$$
for all $i$ when $\lvert \Gamma\rvert \leqslant \delta$. Thus, writing
$$\tilde{S}_{\Gamma} = \sum_{i = 1}^{m} \bigl(\phi'(\xi_i)^2 + \psi'(\xi_i)^2\bigr)^{1/2}(t_i - t_{i-1})$$
we have
$$\lvert S_{\Gamma} - \tilde{S}_{\Gamma}\rvert \leqslant \sum_{i = 1}^{m} \lvert \psi'(\rho_i) - \psi'(\xi_i)\rvert (t_i - t_{i-1}) \leqslant \sum_{i = 1}^{m} \varepsilon\cdot (t_i - t_{i-1}) = \varepsilon\cdot (b-a)$$
whenever $\lvert \Gamma\rvert \leqslant \delta$.
Now $\tilde{S}_{\Gamma}$ is a Riemann sum for $(\ast)$, and thus it follows that
$$\Biggl\lvert\, L - \int_{a}^{b} \bigl(\phi'(t)^2 + \psi'(t)^2\bigr)^{1/2}\,dt \Biggr\rvert \leqslant \varepsilon \cdot (b-a)$$
holds for every $\varepsilon > 0$. And that implies
$$L = \int_{a}^{b} \bigl(\phi'(t)^2 + \psi'(t)^2\bigr)^{1/2}\,dt\,.$$
We have used the continuity of only one of the derivatives ($\psi$ here, we could obviously swap the rôles), of the other we only use Riemann integrability (which ensures that indeed $\tilde{S}_{\Gamma}$ tends to the integral as $\lvert \Gamma\rvert \to 0$). The conclusion that the arc-length of the curve is given by the integral holds under still weaker assumptions, but then more work must be put into the proof.
A: Note that
$$S_\Gamma = \sum_{i=1}^m \left( \phi'(\xi_i)^{2}+ \psi'(\xi_i)^{2}\right)^{1/2}(t_i-t_{i-1}) + \sum_{i=1}^m (f(\xi_i,\rho_i) - f(\xi_i,\xi_i))(t_i-t_{i-1}),$$
where $f:(s,t) \mapsto  \sqrt{\phi'(s)^2 + \psi'(t)^2}$ is uniformly continuous on $[a,b]\times[a,b]$.
For any $\epsilon > 0$, we have for all sufficiently fine partitions,
$$|f(\xi_i,\rho_i) - f(\xi_i,\xi_i)| < \epsilon,$$
and,
$$|S_\Gamma - I| \leqslant \left|\sum_{i=1}^m \left( \phi'(\xi_i)^{2}+ \psi'(\xi_i)^{2}\right)^{1/2}(t_i-t_{i-1})-I \right|+ \sum_{i=1}^m \epsilon (t_i - t_{i-1}) \\ = \left|\sum_{i=1}^m \left( \phi'(\xi_i)^{2}+ \psi'(\xi_i)^{2}\right)^{1/2}(t_i-t_{i-1})-I \right|+ \epsilon(b-a),$$
where $I = \int_a^b \left( \phi'(t)^{2}+ \psi'(t)^{2}\right)^{1/2}\ dt$.
