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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 $\phi:[T_0,T_1]\rightarrow\mathbb C$ 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 $L(C) <\infty$.

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  • $\begingroup$ this looks strange to me. Could you add the information how rectifiability and arc length are defined in that book? $\endgroup$ – user20266 Mar 25 '12 at 12:51
  • $\begingroup$ I added it to the question statement. $\endgroup$ – Pteromys Mar 26 '12 at 4:23
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    $\begingroup$ I don't see the reasoning either, maybe there is some more context that is missing. Anyway, continuity is not difficult to demonstrate by other means. Burago, Burago, and Ivanov do it nicely in a few lines in "A Course in Metric Geometry", p. 35 Proposition 2.3.4 (you can find it online). $\endgroup$ – yasmar Mar 26 '12 at 18:54
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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 $t_j$.)

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