Let $(X,d)$ be a metric space and $\gamma:[a,b]\rightarrow X$ a curve. We define the length of $\gamma$ on $[a,b]$ as follows:
\begin{equation} \ell_a^b(\gamma):=\sup\left\{\ell_a^b(\gamma,P)\mid P\in\mathcal{P}([a,b])\right\}\end{equation}
where if $P:=\{t_i\}_{i=0}^{n}\in\mathcal{P}([a,b])$, then:
\begin{equation} \ell_a^b(\gamma,P):=\sum_{i=1}^{n}{d(\gamma(t_{i-1}),\gamma(t_i))}.\end{equation}
We say $\gamma$ is rectifiable if $\ell_a^b(\gamma)<+\infty$. In that case, let define $\ell:[a,b]\rightarrow\mathbb{R}$ as $\ell(t)=\ell_a^t(\gamma)$. I would like to prove that $\ell$ is continuous. I have the proof, but there are some things that I don't know how to prove it.
It is very ease to prove that $\ell$ is increasing. Therefore, there exist:
\begin{equation} \begin{split} \ell\left(t_0^+\right) & :=\lim_{t\to t_0^+}{\ell\left(t\right)}, \quad \forall t_0\in\left[a, b\right],\\ \ell\left(t_0^-\right) & :=\lim_{t\to t_0^-}{\ell\left(t\right)}, \quad \forall t_0\in\left(a, b\right). \end{split}\end{equation}
By contradiction, assume there exist $\delta>0$ and $t_0\in\left[a, b\right]$ such that $\ell\left(t_0^+\right)-\ell\left(t_0\right)>\delta$. Let $t_n\in\left(t_0, b\right)$. Then:
\begin{equation} \ell_{t_0}^{t_n}\left(\gamma\right)=\ell\left(t_n\right)-\ell\left(t_0\right)=\left(\ell\left(t_n\right)-\ell\left(t_0^+\right)\right)+\left(\ell\left(t_0^+\right)-\ell\left(t_0\right)\right)>0+\delta=\delta.\end{equation}
Thus, there exists $P:=\{s_i\}_{i=0}^m\in\mathcal{P}\left([t_0,t_n]\right)$ such that:
\begin{equation} \delta<\ell_{t_0}^{t_n}\left(\gamma,P\right)=\sum_{i=1}^{m}{d\left(\gamma\left(s_{i-1}\right),\gamma\left(s_i\right)\right)}=\lim_{s\to t_0^+}{d\left(\gamma\left(s\right),\gamma\left(s_1\right)\right)}+\sum_{i=2}^{m}{d\left(\gamma\left(s_{i-1}\right),\gamma\left(s_i\right)\right)}.\end{equation}
Let $t_{n-1}\in\left(t_0,s_1\right)$ be such that:
\begin{equation} \ell_{t_{n-1}}^{t_n}\left(\gamma\right)\geq\sum_{i=1}^{m-1}{d\left(\gamma\left(s_{i-1}\right),\gamma\left(s_i\right)\right)}+d\left(\gamma\left(s_{m-1}\right),\gamma\left(t_1\right)\right)>\delta.\end{equation}
Consequently $\ell_{t_{n-1}}^{t_n}\left(\gamma\right)>\delta$ (this is what I don't know how to prove).
Inductively, we find $P:=\{t_i\}_{i=0}^n\in\mathcal{P}\left([t_0,t_n]\right)$ such that $\ell_{t_{i-1}}^{t_i}\left(\gamma\right)>\delta$ for all $i\in\{1,...,n\}$. Therefore:
\begin{equation} \ell_a^b\left(\gamma\right)\geq\ell_{t_0}^{t_n}\left(\gamma\right)=\sum_{i=1}^{n}{\ell_{t_{i-1}}^{t_i}\left(\gamma\right)}>n\cdot\delta\stackrel{n\to\infty}{\rightarrow}+\infty.\end{equation}
We have reached a contradiction since $\gamma$ was a rectifiable curve.