Say we have a globally continuous parametrized curve $c : \mathbb{R} \longrightarrow \mathbb{R}^k$ such that $s$ is nowhere differentiable. We define the length of a curve between a point $a$ and $b$ as $$l_{(a,b)}(c)=\sup _{a = t_0 < t_ 1 < \cdots < t_n = b} \sum_{i=0}^{n-1} \|c(t_i)-c(t_{i+1})\|$$ where the supremum is taken over all partitions of the real line restricted to $(a,b)$.

Is it possible for there to be an $\epsilon$-neighborhood on the real line $(x_0-\epsilon,x_0+\epsilon)$ such that $l_{(x_0-\epsilon,x_0+\epsilon)}(c) < \infty$ ? Could $k$ affect our answer?

  • $\begingroup$ Do you require $c$ to be surjective? $\endgroup$ Apr 27, 2014 at 2:02
  • $\begingroup$ @EricTowers No...good point $\endgroup$
    – Matt R
    Apr 27, 2014 at 2:13

1 Answer 1


Actually, any function f:[a,b] will have finite length iff each of the functions $f_1,..,f_k$ is of bounded variation. Now, bounded variation implies a.e differentiability....

  • $\begingroup$ Thanks you...do you have a reference or proof sketch for why this is true? $\endgroup$
    – Matt R
    Apr 27, 2014 at 2:15
  • $\begingroup$ Maybe the fact that $V_b^a (f_i)$ is the length of the graph along the i-th direction? When calculating the total variation, you're doing basically the same as when calculating the length of the curve, i.e. you're taking the $Sup |f(x_{i+1})-f(x_i)|$ over all partitions, and then you have BV if the sum converges. But let me see if I can find something more rigorous. $\endgroup$
    – user99680
    Apr 27, 2014 at 2:20
  • $\begingroup$ @MattR: I found a corroborating source here; please see note 1 in this document: people.math.osu.edu/costin.9/264H/Rectifiability.pdf $\endgroup$
    – user99680
    Apr 27, 2014 at 5:38

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