Let $C\subset \mathbb R^2$ be a simple closed curve.
Then, there exists continuous mapping $\gamma : [a,b]\to \mathbb R^2$ s.t. $\gamma([a,b])=C$,$\gamma(a)=\gamma(b), $ $\gamma(t_1)=\gamma(t_2) \Rightarrow t_1=t_2$ for $(t_1,t_2)\neq(a,b),(b,a)$. $(\gamma$ is called parametrization of $C.$)
Define the length of $C$ as $L(\gamma)=\int_a^b \| \gamma'(t) \| dt.$
Then, prove that for another parametrization $\delta : [c,d]\to \mathbb R^2$, $L(\gamma)=L(\delta)$ holds. (i.e., the length of $C$ is independent of parametrization.)
The proof is here.
We can choose $\theta : [c,d]\to[a,b]$ s.t. $\theta(c)=a, \theta (d)=b, \theta'(t)>0, \delta=\gamma \circ \theta.$
Then,
\begin{align*} L(\gamma)&=\int_a^b \| \gamma'(t)\| dt\\ &=_{t=\theta(s)} \int_c^d \| \gamma'( \theta(s))\| \theta'(s) ds \\ &=\int_c^d \left\| \dfrac{d}{ds}[(\gamma \circ \theta)(s)] \right\| ds \\ &=\int_c^d \| \delta'(s)\| ds \\ &=L(\delta). \end{align*}
I have a question about the existence of such $\theta$.
I tried to find such $\theta$ and found $\theta(t)=\dfrac{a-b}{c-d}t-\dfrac{ad-bc}{c-d}$.
This $\theta$ satisfies $\theta(c)=a, \theta (d)=b, \theta'(t)>0,$ but may not satisfy $\delta=\gamma \circ \theta.$
So, I think that such $\theta$ generally doesn't exist.
Is the existence of such $\theta$ just a supposition ?
or can we show the existence of such $\theta$ ?
I general your argument do not work as you did not assume that $\gamma '(t) \not = 0$. The existence of $\theta $ is juts a stupid nonsense argument $\gamma$ and $\delta$ ar bijection onto there image, so that $\theta =\gamma ^{-1} \circ \delta $ is well defined. But to prove differentiability you need something, and $\gamma '\not =0$ is enough.