This is a (translated) proof from a textbook of the fact that arc length of a rectifiable curve is a continuous function.
Let $\phi:[T_0,T_1]\rightarrow\mathbb C$ be a function whose real part and imaginary part are continuous, and $C$ be a curve represented by $\phi$. Suppose $C$ is rectifiable. Define $f:[T_0,T_1]\rightarrow\mathbb R$ by $f(t) = L(C|[T_0,t])$, where $L(C|I)$ is the arc length of $C$ restricted to the interval $I$. To show by contradiction that $f$ is continuous, assume $f$ is not continuous at $t_0\in [T_0, T_1]$. Since $f$ is monotonously increasing, either of $ \lim_{t\rightarrow t_0-0} f(t) < f(t_0) $ or $\lim_{t\rightarrow t_0+0} f(t) > f(t_0)$ holds. WLOG we may assume the former holds. Here $t_0 > t$. Let $\epsilon_0 = f(t_0) - \lim_{t\rightarrow t_0-0}f(t)$. By definition, there exists an infinite number of $t_j < \tilde{t_j} < t_{j+1} < \tilde {t}_{j+1}\quad(j = 1,2,\dots)$ s.t. $L(C|[t_j,\tilde{t}_j])>\epsilon_0/2$. Then we have $L(C) = +\infty$. Contradiction.
I can't figure out why the sentence that begin with "By definition" is true. Why are there such $t_j$'s?
EDIT: In the book, $L(C)$ is defined to be the supremum (possibly $+\infty$) of $\sum_{j=1}^{n}|\phi(s_j)-\phi(s_{j-1})|$ for any partition $T_0 = s_0 < s_1 < \dots < s_n = T_1$. $C$ is rectifiable iff $L(C) <\infty$.