Prove that the length of a curve is given by integral I have found the following problem in "Introduction to Analysis" by Rosenlicht. I am not sure if my solution is correct and I highlighted my uncertainties.


First we have to show that the set $$S=\Bigg\{\sum_{i=1}^N d(f(x_{i-1}),f(x_i)):x_0,x_1,\ldots , x_N \mbox{ is a partition of } [a,b]\Bigg\}$$ is bounded from above and that the sequence of the above sums is increasing as the mesh of the partition goes to $0$, so this sequence necessarily converges to $\sup S$.

I'm not even sure if the above part is correct and how to formally show it. It is crucial for the later part.

Next, since $\sqrt{(f_1^{'}(x))^2+\ldots+(f_n^{'}(x))^2}$ is continuous on $[a,b]$, the integral $$\int_a^b \sqrt{(f_1^{'}(x))^2+\ldots+(f_n^{'}(x))^2} dx $$ exists.
Fix a partition $x_0,x_1,\ldots , x_N$ of $[a,b]$.
From the mean value theorem we know that for $j=1,\ldots,n$ if $x_{i-1}$ and $x_i$ are two points from $[a,b]$ such that $x_{i-1}<x_i$ then there exists some $\zeta_{i}^j$ between them and
$$f_j(x_i)-f_j(x_{i-1})=f'_j(\zeta_{i}^j)(x_i-x_{i-1})\mbox{.}$$
We have:
$$\sum_{i=1}^N d(f(x_{i-1}),f(x_i))=\sum_{i=1}^N \sqrt{(f_1(x_{i-1})-f_1(x_{i}))^2+\ldots+(f_n(x_{i-1})-f_n(x_{i}))^2}=$$
$$=\sum_{i=1}^N \sqrt{(f^{'}_1(\zeta_i^1))^2+\ldots+(f^{'}_n(\zeta_i^n))^2}(x_i-x_{i-1}) \mbox{.}$$

Intuition tells me that this is okay, but how can I formally show that it does not matter if $\zeta_i^1,\ldots, \zeta_i^n$ are different, as long as they are "trapped" between $x_{i-1}$ and $x_i$?

If we go with the mesh of the partition to $0$, then the above sum converges to the integral $$\int_a^b \sqrt{(f_1^{'}(x))^2+\ldots+(f_n^{'}(x))^2} dx $$ which is equal to $\sup S$.
 A: As for your first question: note that $f$ as a continuosly differentiable curve is Lipshitz continuous and the derivate (since it is continuous by assumption) is bounded, so $f$ is Lipschitz continuous with a bound on the Lipschitz constant $L$ given by a norm bound for the derivative. So your set is bounded by $L(b-a)$.
For your second question it is possible to show that, for continously differentiable $f$, the following is true: to $\varepsilon>0$ there is $\delta >0$ such that $|t-s|< \delta \Rightarrow$ 
$$ (\#)\quad\quad\quad\quad\quad\quad\quad ||\frac{f(t)-f(s)}{t-s} - f^\prime(t)||\le \varepsilon $$
(note the choice of the argument of $f^\prime$). If you can show this your second question will be resolved. $(\#)$ is a consequence of the mean value theorem the way you used it and uniform continuity of $f^\prime$.
Edit: more detailed explanation of the last remark.
To derive the result from $(\#)$ recall first that, for the Riemannian integral and the Riemann integrable function $||f^\prime(t)||$ and given $\varepsilon > 0$, for any sufficiently fine partition $a= t_0<t_1< \ldots < t_N=b$, e.g. for $|t_i-t_{i-1}|<\delta$, 
$$(*)\quad\quad \quad\quad\quad\left|\int_a^b ||f^\prime(t)|| dt - \sum\limits_{i=1}^N ||f^\prime(t_i)||(t_i-t_{i-1})\right| < \varepsilon $$
If the partition is fine enough we can apply $(\#)$ for adjacent values $t_k, t_{k-1}$ in this partition, use the triangle inequality and multiply by $t_k-t_{k-1}$ to get
$$(**) \quad\quad \quad \left|\, ||f(t_k)-f(t_{k-1})||-||f^\prime(t_k)|| (t_k-t_{k-1}) \right|\le \varepsilon(t_k-t_{k-1})$$
Now it's just a matter of putting things together: we look at the expression wich should become arbitrarily small, less than $\varepsilon^*$, say:
$$
(***)\quad\quad\quad \quad\quad\left|\sum_{i=1}^N ||f(t_i)-f(t_{i-1})|| - \int_a^b ||f^\prime(t)||\,dt\right|
$$
and add and subtract under the norm the term 
$$ \sum_{i=1}^N ||f^\prime(t_i)||(t_i-t_{i-1})$$
By the triangle inequality, $(***)$ is bounded from above by
$$\left|\sum_{i=1}^N ||f(t_i)-f(t_{i-1})|| -  \sum_{i=1}^N ||f^\prime(t_i)||(t_i-t_{i-1})\right|+\left| \sum_{i=1}^N ||f^\prime(t_i)||(t_i-t_{i-1})- \int_a^b ||f^\prime(t)||\,dt \right|$$
Here the second term is bounded by $\varepsilon$ according to $(*)$. The first one is bounded, according to $(**)$ by 
$$\sum_{i=1}^N \varepsilon (t_i-t_{i-1})= (b-a)\varepsilon$$
So by making $\varepsilon $ so small that $(b-a+1) \varepsilon <\varepsilon^*$ we may choose the partition fine enough and the result follows. 
(You may have noticed that I did not provide a proof of $(\#)$, only a hint on how to prove it, i.e. I left you some work to do on your own).
