Does there exist a if and only if condition so that arc length of a convergent sequence of functions to converge to the arc length of the limit. Is there a necessary and sufficient condition so that the arc length of a convergent sequence of functions converges to the arc length of the limit of the function?
I know that if $f'_n$ is continuously differentiable on the interval $[a,b]$ and   $f$ converges for one $x \in [a,b]$. If $f'_n$ converges uniformly, then $f'_n$ converges to $f'$.
So for a normal Cartesian coordinate system the arc length is $\int\sqrt{1+(f')^{2}}$ $dx$ and we have that $f'_n$ converges uniformly to $f'$ and if the arc length of $f_n$ is exists for all $n$ and the arc length of $f$ exists, then the arc length of $f_n$ converges to $f$.
But, I know that its possible for a convergent sequence of piecewise differentiable functions $f_n$ still have it's arc length converges to $f$ when $f_n$ converges to $f$.
So does there exist some kind of most weak condition that gives the arc length of a convergent sequence of functions converges to the arc length of the limit and does there exists a converse condition that will give an if and only if statement?
 A: A sufficient condition is that $f_n'\to f'$ in $L^1$, that is $\int_a^b |f_n'-f'|\to 0$. The justification for this (unexciting) observation is that the function $t\mapsto \sqrt{1+t^2}$ has derivative bounded by $1$, hence $|\sqrt{1+t^2}-\sqrt{1+s^2}|\le |t-s|$ for all $t,s\in\mathbb R$, which implies 
$$
\left| \int_a^b \sqrt{1+{f_n'}^2} - \int_a^b \sqrt{1+{f'}^2}\right| 
\le \int_a^b |f_n'-f'|
$$

Within some classes of functions one can give weaker sufficient conditions. For example, if every $f_n$ is convex, and $f_n\to f$ uniformly, then the lengths of graphs converge.  

Another relevant fact is that mere pointwise convergence $f_n\to f$ suffices to conclude that $$\int_a^b \sqrt{1+{f'}^2}\le\liminf_{n\to\infty} \int_a^b \sqrt{1+{f_n'}^2}$$
(the upper semicontinuity of length). To see this, pick a partition $t_j$ of $[a,b]$ that almost realizes the length of the graph of $f$. For sufficiently large $n$, the differences $|f_n(t_j)-f(t_j)|$ are as small as we want.  Since the length of the graph of $f_n$ between $(t_{j-1}, f_{n}(t_{j-1}))$ and $(t_j, f_n(t_j))$ is no less than the length of line segment connecting  these points, the claim follows.

All that said, I don't think there is any non-tautological necessary and sufficient condition for the general case. 
