The following argument is from my class notes. "Suppose $\gamma(s): I\rightarrow \mathbb R^3$ is a regular curve. Fix $t_0$ and define $s(t) = \int_{t_0}^t \|\gamma'(t)\|dt$. That is, $s(t)$ is the length of the curve from $t_0$ to $t$. Then $s(t)$ is continuous." Why is $s(t)$ continuous, please?
My guess is that $\|\gamma'(t)\|$ is continuous and its integral is also continuous. Is this correct, please? if so, why is $\|\gamma'(t)\|$ continuous? Thank you!
If put $M = \max |f'(t)|$ on $I = [0, 1]$, then you can show $|s(b) - s(a)| < M(b - a)$. From this you conclude $s(t)$ is continuous.