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Consider a path $\pi : [a,b]\to C$ where $C$ is a curve. By definition, then length of the path is $$\ell(\pi)=\sup \sum_{i=0}^n\|\pi(t_i)-\pi(t_{i+1})\|,$$ where $\{t_0,...,t_{n+1}\}$ is a partition of $[a,b]$, and the sup is taken over all partition.

I know that if $\pi$ is rectifiable but not absolutely continuous, then $$\ell(\pi)\geq \int_a^b \|\pi '(t)\|dt= m(C),$$ where $m$ is the Lebesgue measure. So there are case where the measure of a path is in fact not the length of the path ? That looks weird...

Bernard
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Henri
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    That $m(C)$ thing makes no sense. You must mean that $m$ is one-dimensional Lebesgue measure. But if so then there's no such thing as $m(C)$, since $C$ is not a subset of $\Bbb R$. You meant to say something about $h_1(C)$, where $h_1$ is one-dimensional Hausdorff measure. With that change it makes sense, but no, $\int_a^b||\pi'||\ne h_1(C)$. – David C. Ullrich Sep 02 '18 at 14:31

2 Answers2

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This is indeed the case. The Cantor function $c$:

  • is continuous everywhere but has zero derivative almost everywhere. Hence $\displaystyle \int_0^1 \sqrt{1+\left(\dfrac{dc}{dt}\right)^2} \ dt$ is equal to $1$.
  • However its arc length is greater or equal than $2$. See Arc length of the Cantor function.
mathcounterexamples.net
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An example is the graph $\mathcal{G}(f)$ of the Cantor function $f$.

The Cantor function is a continuous, increasing function from $f:[0,1]\to[0,1]$ with $f(0)=0$, $f(1)=1$. Let $$\gamma(t)=(t,f(t)),\qquad t\in [0,1] $$ A simple argument shows that the length of the graph, i.e. the length of the curve $\gamma$ is larger than $\sqrt{2}$. For any partition $0=t_0<t_1<\dots <t_n=1$ we have \begin{align*}\ell(\mathcal{G}(f))\geq\sum_{i=0}^{n-1}\|\gamma(t_{i+1})-\gamma(t_i)\|&=\sum_{i=0}^{n-1}\sqrt{|t_{i+1}-t_i|^2+|f(t_{i+1})-f(t_i)|^2} \geq \\ \text(Cauchy-Schwarz)\quad &\geq \frac{1}{\sqrt{2}}\left[ \sum_{i=0}^{n-1}|t_{i+1}-t_i|+\sum_{i=0}^{n-1}\left|f(t_{i+1})-f(t_i)\right|\right]=\\ \text{(}f\textit{ is increasing)}\quad &=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\sum_{i=0}^{n-1}\left(f(t_{i+1})-f(t_i)\right) =\frac{1}{\sqrt{2}}\left[1+f(t_n)-f(t_0)\right]=\\ (f(0)=0,\;f(1)=1)\quad&=\frac{1}{\sqrt{2}}\left[1+f(1)+f(0)\right] =\sqrt{2} \end{align*} This could also be seen as an argument that any rectifiable curve connecting the points $(0,0)$ and $(1,1)$ cannot be shorter than $\sqrt{2}$, i.e. the length of the segment connecting those two points.

With the same computations but using the opposite of Cauchy-Schwarz inequality \begin{align*}\sum_{i=0}^{n-1}\|\gamma(t_{i+1})-\gamma(t_i)\|&=\sum_{i=0}^{n-1}\sqrt{|t_{i+1}-t_i|^2+|f(t_{i+1})-f(t_i)|^2} \leq \\ &\leq \sum_{i=0}^{n-1}|t_{i+1}-t_i|+\sum_{i=0}^{n-1}\left|f(t_{i+1})-f(t_i)\right|=2\end{align*} and taking the $\sup$ over all partitions of $[0,1]$, we get that $\mathcal{G}(f)$ is, indeed, a rectifiable curve, with $\ell(\mathcal{G}(f))\leq 2$.

However, we also $f'=0$ a.e. on $[0,1]$, and hence $$\ell (\mathcal{G}(f))\geq \sqrt{2} > 1 =\int_0^1dt=\int_0^1\sqrt{1^2+|f'(t)|^2}dt=\int_0^1\|\gamma'(t)\|dt$$

Lorenzo Q
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