Do Integrals over Fractals Exist?

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)

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