I am trying to prove the following statement:
Let $\mathscr{L}(x)$ denote the length of $f$ on $[a,x]$. If $f$ is rectifiable on $[a,x]$ for every $x\in[a,b)$ and $$\lim_{x\to b^-} \mathscr{L}(x)=L$$ then $f$ is rectifiable on $[a,b]$ and $\mathscr{L}(b)=L$
in order to show that a semi-circle is rectifiable and its length is given by the improper integral $\int_0^1\frac{1}{\sqrt{1-t^2}}dt$. I am able to do this if I assume that $f$ is continuous on $[a,b]$, which is the case here, but it got me wondering if this statement was true in cases where $f$ is rectifiable but not continuous. Can a function be rectifiable on an interval without being continuous?
For context, I am using the following definition of rectifibiality of a function on a closed interval:
Let $P$ be a partition of $[a,b]$ and let $\ell(f,P)$ denote the length of the chord-polygonal inscribed in $f$ i.e. $$\ell (f,P):=\sum_{k=1}^{n}\sqrt{[x_{k}-x_{k-1}]^{2}+[f(x_{k}-f(x_{k-1}))]^{2}}$$If $\{\ell (f,P)|P \text{ is a partition of }[a,b]\}$ is bounded above, the curve is said to be rectifiable and the length of $f$ on $[a,b]$ is defined as $$\mathscr{L}:=\sup \{\ell (f,P)|P \text{ is a partition of }[a,b]\} $$
