two definitions of arc length Let $f:[a,b]\rightarrow\mathbb{R}$ be an absolute continuous function with $-\infty <a <b <\infty. $ Define the length $L $ be the total variation of the graph $g (x)=(x,f (x)) $ on $[a,b] $. Show that $L=\displaystyle\int^b_a \sqrt {1+f'^2(x)} dx. $
I think $L\le\displaystyle\int^b_a \sqrt {1+f'^2(x)} dx $ is easily provable but the converse doesn't seem easy.. can anyone help me?
 A: Yes, one half is trivial. Since $f$ is AC it follows that $g$ is AC, hence $g$ is the integral of its derivative. So $$|g(y)-g(x)|
=\left|\int_x^yg'(t)\,dt\right|\le\int_x^y|g'(t)|\,dt=\int_x^y(1+f'(t)^2)^{1/2}\,dt.$$This shows that $$L\le\int_a^b(1+f'^2)^{1/2}.$$
The other inequality follows from a certain lemma. I'll let you consider why the result follows from the lemma (take $F=g'$).
Lemma If $F\in L^1([0,1])$ then $$\lim_{n\to\infty}\sum_{j=1}^n\left|\int_{(j-1)/n}^{j/n}F\right|=\int_0^1|F|.$$ 
Proof. Say the left side is $\lim_{n\to\infty}I_n(F)$. Note that $|I_n(F)-I_n(G)|\le||F-G||_1$. Let $\epsilon>0$. Choose $G\in C([0,1])$ with $||F-G||_1<\epsilon$.
The lemma holds for continuous functions, hence with $G$ in place of $F$ (another exercise). So if $n$ is large enough we have $$|I_n(F)-||F||_1|
\le|I_n(F)-I_n(G)|+|I_n(G)-||G||_1|+|||G||_1-||F||_1|<\epsilon+\epsilon+\epsilon.$$
A: Let $\Gamma=\{g(x)=(x,f(x))\mid x\in[a,b]\}$
$$L=\int_{\Gamma} d\gamma=\int_a^b\|g'(x)\|dx=\int_a^b\sqrt{(f'(x))^2+1}dx.$$
