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Given, for example, a line integral like $$ \int_\gamma f \; ds $$ with $f$ not further defined, yet.

  • What happens, if the contour $\gamma$ happens to be a fractal curve?

Since all fractal curves (to my knowledge) are not differentiable and have an infinite length, we should run into some trouble. Further questions of interest are:

  • What happens if the integral is taken over a Julia Set?
  • What happens in the case of a space-filling curve? Do we get a surface integral then?

I'd be glad, if anybody could wrap her/his 2.79-Hausdorff-dimensional brain surface around it!

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Lebesgue integration en.wikipedia.org/wiki/Lebesgue_integration can integrate very nasty domains I believe –  Thomas Rot Jan 31 '12 at 11:43
@ThomasRot: Do you know example or an application of those integrals? –  draks ... Jan 31 '12 at 11:45
Any Riemann integrable function is Lebesgue integrable with the same value, so all integrals you know are computable in Lebesgue sense. Applications where the Riemann integral is insufficient are abundant. It often comes up in statistics/stochastics/probability theory I believe –  Thomas Rot Jan 31 '12 at 11:57
related –  NikolajK Jan 31 '12 at 14:01
As for applications: people do solve the Laplace equation or the wave equation in the interior of the Koch snowflake, and to do that they need to compute integrals on the boundary curve and so on. This is motivated for concrete applications. –  Mariano Suárez-Alvarez Jan 31 '12 at 17:47

2 Answers 2

up vote 19 down vote accepted

Re @Thomas Rot's comment: Lebesgue integration is just the thing here. However, if the fractal in question is a subset of some nice measure space, e.g. $\mathbb{R}^n,$ then the measure one will use in the Lebesgue integration shouldn't be the inherited Lebesgue measure. Otherwise, one gets dumb results: the Lebesgue measure of a subset of $\mathbb{R}^n$ of strictly fractional Hausdorff dimension is, almost by definition, 0.

Instead, one must use the Hausdorff measure as specified in, say, Stein and Shakarchi's book Real Analysis, on pages 325--327. The definition is that the exterior $\alpha$-dimensional Hausdorff measure of a subset $E \subset \mathbb{R}^d$ is $$m_\alpha^\ast(E) = \lim_{\delta \to 0} \inf \left\{\sum_k(\mbox{diam }F_k)^\alpha\ \left|\ E \subset \bigcup_{k=1}^\infty F_k,\ \mbox{diam }F_k \leq \delta\ \forall k\right.\right\}.$$ One can show that this is in fact an outer measure in the sense of Caratheodory, and restricting it to Borel sets yields a measure. The Hausdorff dimension of a Borel subset $E\subset \mathbb{R}^n$ is the infimum over all $\alpha$ with $m_\alpha^\ast(E) = 0.$ Equivalently it is the supremum over all $\alpha$ with $m_\alpha^\ast(E) = \infty.$

With this measure one can, presumably, do Lebesgue integration and differentiation as usual. (One might still get boring results: say the set has Hausdorff dimension $\delta.$ It might be the case that the set's $\delta$-dimensional Hausdorff measure is 0.)

This is just general measure theory, though. As far as I know, oriented integration theory (e.g. line integrals and Stokes's theorem) hasn't yet been generalized to fractals, though there is a notion of Laplacian on some fractals, as explained in an article of Strichartz.

If one is content to remain with general measure theory, one gets the following eerie result, Theorem 3.1 in Chapter 7 of Stein and Shakarchi:

There exists a curve $\mathcal{P}:[0,1] \to [0,1]\times [0,1]$ such that

  • $\mathcal{P}$ is continuous and surjective.
  • The image under $\mathcal{P}$ of any subinterval $[a,b] \subset [0,1]$ is a compact subset of the square of (two-dimensional) Lebesgue measure $b - a.$
  • There are subsets $Z_1 \subset [0,1]$ and $Z_2 \subset [0,1]\times[0,1],$ each of measure zero, such that $\mathcal{P}|_{[0,1]\setminus Z_1}:[0,1]\setminus Z_1 \to [0,1]\times [0,1] \setminus Z_2$ is bijective and measure preserving!

(cue eerie music)

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+1 for the link ;-)... and the answer of course. –  draks ... Jan 31 '12 at 21:39
Could you help answering this question? –  draks ... Apr 5 '12 at 18:12
Is there a geometric explanation or a layman's analogy that would explain how a fractal is bijective and measure-preserving? –  Atav32 Jan 15 '13 at 4:02

Sure, there are several ways to integrate over fractals. Jenny Harrison has worked out a lot using chains, Boris Katz has some papers you could look up... there is this using fractal measures http://arxiv.org/pdf/chao-dyn/9804006.pdf, and http://www.minet.uni-jena.de/geometrie/publicat/pdffiles/iwrtff1_ptrf98.pdf using fractional calculus. I have been formulating methods using the hyperreal number system, where, for instance, the limit case of the Cantor middle thirds set is a set of closed infinitesimal length intervals. An integral over infinitesimal elements can be defined, then the integrals summed over the whole fractal. See also "Analysis on Fractals" by Robert Strichartz.

John Starrett

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Thanks for the links. I like Zyczkowski's way of writing! –  draks ... Nov 23 '13 at 22:03
@user110946, do you have references for the papers by by Harrison and Katz? –  user72694 Nov 25 '13 at 12:15
My paper on the technique will be appearing in the American Mathematical Monthly later this year. It is called Hyperreal Fractals and Integration. –  user110946 Feb 25 '14 at 18:12

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