# definition of length of non-rectifiable curves

Following the usual definition of length of a curve, the curve have to be rectifiable to have finite length. However for example the curve

$\gamma \colon [0,1] \to \mathbb{R}, t \to t \cos^2(\pi/t)$

is not rectifiable. However I would say it makes perfectly sense to say it has length $1$.

So my question is, if there is a more general definition of length of a curve such that my example above has length $1$ (however such that for example the graph of the function $\gamma$ above has infinite length).

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Why would you say that? – Qiaochu Yuan Apr 29 '11 at 10:01
Since it is just a parametrization of the interval [0,1] – student Apr 29 '11 at 10:07
Check out the following rectifiable parametrizations: $\gamma \colon [0,1] \to \mathbb{R}, t \to a\cdot t$. Do they all have length $1$ ? – Raskolnikov Apr 29 '11 at 10:21
@Raskolnikov, They have length $|a|$, however (suppose $a > 0$) it is $\gamma([0,1]) = [0,a]$. So length $a$ makes sense. – student Apr 29 '11 at 10:38
@user3445: but in this example Raskolinikov is measuring the length of the range, whereas in yours you are measuring the length of the domain. The domain has length 1. That was the point of his question. – Ross Millikan Apr 29 '11 at 13:02

You have to distinguish between the ${\it curve}$, i.e., an equivalence class of maps $f:\ [0,1]\to {\mathbb R}^n$, and its "trace", i.e., the ${\it set}$ $f\bigl([0,1]\bigr)\subset{\mathbb R}^n$. If you declare your curve $\gamma$ to have length $1$ you would have to assign a twice surrounded circle the length $2\pi$. In this example the length of the "trace" is indeed still $2\pi$, but the length of the actual curve is $4\pi$.
Ok. How would one define the length of the trace $\Gamma$? May I say: if there is an injective, regular map $f\colon [0,1] \to \mathbb{R}^n$ such that $f([0,1]) = \Gamma$ the length of $\gamma$ is defined as the length of the curve $f$? – student Apr 29 '11 at 16:16