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This is a (translated) proof from a textbook of the fact that arc length of a rectifiable curve is a continuous function.

Let $phi:[T_0,T_1]rightarrowmathbb C$ be a function whose real part and imaginary part are continuous, and $C$ be a curve represented by $phi$. Suppose $C$ is rectifiable. Define $f:[T_0,T_1]rightarrowmathbb R$ 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 $t_0in [T_0, T_1]$. Since $f$ is monotonously increasing, either of
$
lim_{trightarrow t_0-0} f(t) < f(t_0)
$ or $lim_{trightarrow t_0+0} f(t) > f(t_0)$ holds. WLOG we may assume the former holds. Here $t_0 > t$. Let $epsilon_0 = f(t_0) - lim_{trightarrow t_0-0}f(t)$. 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 $L(C) <infty$.

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. "e still have $L(C|[t_2,t_0])geepsilon_0$, 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 $t_j$.)