I am trying to prove that if I have 2 paramterizations of the same curve $\gamma$ and $\sigma$ (i.e. there is continuous bijective map $\phi$ such that $\sigma = \gamma \circ \phi$) then if the curve is rectifiable $L(\gamma) = L(\sigma)$ but without using the integral formula since I do not know if $\gamma'$ exists.
I am not sure how to use the supremum definition. Could someone explain how to use that in this context?
On
$L(\gamma)=L(\sigma)$ is obvious. The difficult thing is to prove that these suprema can be written as an integral.
Nevertheless, here is why one has $L(\gamma)=L(\sigma)$:
Both $L(\gamma)$ and $L(\sigma)$ are the sup of the same set, namely the set of all sums of the form $$\sum_{k=1}^N |\gamma(t_k)-\gamma(t_{k-1})|$$ with $a=t_0<t_1<\ldots<t_N=b$ for some $N\geq1$. It's just that in the computation of $L(\sigma)$ these same sums appear in a different disguise.
Hint: Let $[a, b]$ and $[\alpha, \beta]$ denote the domains of $\sigma$ and $\gamma$, respectively, so that $\phi:[a, b] \to [\alpha, \beta]$ is a continuous bijection.
For each partition $\{t_{i}\}_{i=0}^{n}$ of $[a, b]$, there is an associated partition $\{\phi(t_{i})\}$ of $[\alpha, \beta]$. Using this partition to approximate the arc length of $\gamma$ tells you something about the arc length of $\gamma$ in terms of the arc length of $\sigma$. Reversing the roles of the intervals easily finishes the proof.