Showing that a curve is not rectifiable if its arc length is not a continuous function

Question

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 $L(C) <\infty$.

Answer 1

I am not sure this is what the author intended, but the claim that holds "by definition" can be shown as follows:

1. Take any $t_1<t_0$. Then $L(C|[t_1,t_0])\ge \epsilon_0$, so there exists a partition $t_1=s_0 < \cdots < s_n=t_0$ such that $$\sum_{j=1}^n |\phi(s_j)-\phi(s_{j-1})|> 3 \epsilon_0/4.$$ We may assume $|\phi(s_n)-\phi(s_{n-1})|< \epsilon_0/4$, since $\phi$ is continuous.

   (Otherwise there exists $\hat s_n \in (s_{n-1},t_0)$ with $|t_0-\hat s_n|<1/4 \epsilon_0$. Take $\hat s_j = s_j$ for $j=1,\ldots n-1$, and $\hat s_{n+1}=t_0$. Then $$\sum_{j=1}^{n+1} |\phi(\hat s_j)-\phi(\hat s_{j-1})| \ge \sum_{j=1}^n |\phi(s_j)-\phi(s_{j-1})|> 3 \epsilon_0/4$$ by triangle inequality.)

   Hence $L(C|[t_1,s_{n-1}])>\epsilon_0/2$. Set $t_2=s_{n-1}$.

2. We still have $L(C|[t_2,t_0])\ge\epsilon_0$, so we can start over to get $t_3, t_4,\ldots$ with $L(C|[t_j,t_{j+1}]) > \epsilon_0/2$ by induction.

(Not sure why we need $\tilde t_j$, but if we want to, we can take any $\tilde t_j \in (t_j,t_0)$ and then proceed in step 2 with $\tilde t_j$ instead of $t_j$.)