Proof for the length of plane curves Given that the plane curve is a parametric curve with equations $x=f(t)$ and $y=g(t)$ and we are finding the length of the curve in the interval (a,b) and so we divide the given plane curve into smaller arcs and write the length of the plane curve as an approximation of this sum
$$\sum_{k=1}^n \sqrt {\Delta x_k^2 + \Delta y_k^2}$$
$$\sum_{k=1}^n \sqrt {(f(t_k)-f(t_{k-1}))^2 + (g(t_k)-g(t_{k-1}))^2}$$
As per the mean value theorem ther are numbers such that  $t_k^*$ and $t_k^{**} \in [t_{k-1},t_{k}] $
$$\Delta x_k = f(t_k)-f(t_{k-1})=f^|(t_k^*)\Delta t_k$$ 
$$ \Delta y_k = g(t_k)-g(t_{k-1})=g^|(t_k^{**})\Delta t_k$$
$$\sum_{k=1}^n \sqrt {f^|(t_k^*)^2 + g^|(t_k^{**})^2}\Delta t_k$$
My question
The above equation is not a riemann sum but a theorem in advanced calculus guarantees its limit as the norm of the partition tends to zero (This is what I read in the Thomas Calculus) and I want to know the proof(How does the above sum conerges to below integral) of this and any book to such advanced topics and any links to such.
As the norm of the partition tends to zero the sum conerges to the integral 
$$\int_a^b \sqrt {[f^|(t)]^2+[g^|(t)]^2}dt$$
Any type of help is appreciated
 A: You need to assume some good behaviour of $f'$ and $g'$: let's say, that $f'$ and $g'$ are continuous on $[a,b]$ (and therefore, by the Heine-Cantor theorem, uniformly continuous on that interval).   
Let $\epsilon > 0$ be given, and take $\delta$ such that for all $s,t \in [a,b]$ with $|s - t| < \delta$, $|f'(s) - f'(t)| < \epsilon$ and $|g'(s)-g'(t)| < \epsilon$.  Let $a = t_0 < t_1 < \ldots < t_n = b$ such that
$t_{j+1} - t_j < \delta$, and let $x_j = f(t_j)$ and $y_j = g(t_j)$.
We write $\Delta x_j = x_{j+1} - x_j$, and similarly for $\Delta y_j$ and $\Delta t_j$.
By the Mean Value Theorem, for each $j$ we have 
$ \Delta x_j = f'(c) \Delta t_j$ with $c \in [t_j, t_{j+1}]$. 
Now $|f'(c) - f'(t_j)| < \epsilon$ so
$$  |\Delta x_j - f'(t_j)\Delta t_j| < \epsilon \Delta t_j$$
and similarly 
$$  |\Delta y_j - g'(t_j)\Delta t_j| < \epsilon \Delta t_j$$ 
By the Triangle Inequality,
$$ \eqalign{\left|\sqrt{(\Delta x_j)^2 + (\Delta y_j)^2} - \sqrt{f'(t_j)^2 + g'(t_j)^2)} \Delta t_j \right| &\le \sqrt{(\Delta x_j - f'(t_j)\Delta t_j)^2 + 
(\Delta y_j - g'(t_j)\Delta t_j)^2}\cr &< \sqrt{2} \epsilon \Delta t_j}$$
so that
$$ \left|\sum_{j=0}^{n-1} \sqrt{(\Delta x_j)^2 + (\Delta y_j)^2} - \sum_{j=0}^{n-1} \sqrt{f'(t_j)^2 + g'(t_j)^2)} \Delta t_j \right| < \sqrt{2} \epsilon \sum_{j=0}^{n-1} \Delta t_j = \sqrt{2} (b-a) \epsilon$$
As $n \to \infty$ with the mesh size $\to 0$, $$\sum_{j=0}^{n-1} \sqrt{f'(t_j)^2 + g'(t_j)^2)} \Delta t_j  \to L = \int_a^b \sqrt{f'(t)^2 + g'(t)^2}\; dt$$ i.e. if $n$ is sufficiently large and $\delta$ is sufficiently small, 
$$ \left| \sum_{j=0}^{n-1} \sqrt{f'(t_j)^2 + g'(t_j)^2)} \Delta t_j - L \right| < \epsilon $$
so that 
$$  \left|\sum_{j=0}^{n-1} \sqrt{(\Delta x_j)^2 + (\Delta y_j)^2} - L \right| < (\sqrt{2} (b-a) + 1) \epsilon $$
