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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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this looks strange to me. Could you add the information how rectifiability and arc length are defined in that book? –  user20266 Mar 25 '12 at 12:51
    
I added it to the question statement. –  Pteromys Mar 26 '12 at 4:23
1  
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). –  yasmar Mar 26 '12 at 18:54
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