# Do Integrals over Fractals Exist?

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 –  Nikolaj K. Jan 31 '12 at 14:01
@NickKidman related, but not quite what I was looking for. Are there examples, where the integral is, for example, taken over Koch's Snowflake? –  draks ... Jan 31 '12 at 15:22

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? –  BBz Jan 15 '13 at 4:02