Rectifiability implies continuity

Question

Consider a rectifiable curve $\gamma:[0,a] \rightarrow S$ on a compact and metric space S.

I wonder since the path length of $\gamma$ is finite, it implies that $\gamma$ is continuous?

Thanks in advance

Answer 1

Continuity is usually a requirement in the definition of a curve.

Let $x_0$ and $x_1$ be two different points in $S$ and let $\gamma\colon[0,1]\to S$ be given by $$ \gamma(t)=\begin{cases}x_0& \text{if }0\le t\le 1/2,\\x_1 &\text{if } 1/2<t\le1.\end{cases} $$ Then $\gamma$ is a discontinuous curve, but for any partition $0=t_0<t_1<\dots<t_n=1$ of $[0,1]$ $$ \sum_{k=1}^n d(\gamma(t_{k-1}),\gamma(t_{k}))= d(x_0,x_1). $$