# "Funny Integral" over the Cantor Set

I was thinking about integrals and how one might generalize them to be able to integrate over fractals rather than just over intervals. For example, consider the cantor set $$C$$. Let us assume that $$\int_C dx=1$$ for our funny, not-yet-well-defined integral-like operation. If we assume linearity of this "funny integral," then we may calculate the integral of $$xdx$$ over the Cantor set, because of its symmetry about $$x=1/2$$: $$\int_C xdx=\int_C (1-x)dx=1-\int_C xdx=\frac{1}{2}$$ We may also make use of the fact that the left half of the Cantor set, or $$C_1$$, is a contraction of $$C$$ by a factor of $$3$$, and the right half $$C_2$$ is also a contraction of $$C$$ by a factor of $$3$$. If $$f$$ is a function defined over the Cantor set, by extending another property of integrals to our "funny integral," we have that $$\int_C f(x)dx=\int_{C_1}f(x)dx+\int_{C_2}f(x)dx$$ However, if we wish to make the substitution $$x\to x/3$$, we should not replace $$dx$$ with $$dx/3$$, because shrinking $$C$$ by a factor of $$3$$ does not actually decrease its "size" by a factor of $$3$$, but rather by a factor of $$2$$ (this is also why the fractal dimension of $$C$$ is $$\log_3(2)$$). Thus, when we let $$x\to x/3$$, we must also let $$dx\to dx/2$$, giving us $$\int_C f(x)dx=\int_{C}\frac{f(x/3)+f(1-x/3)}{2}dx$$ This formula, derived by assuming some of the familiar properties of the classical integral for our "funny integral," allows one to compute the integrals of $$x^2,x^3,x^4,$$ and so on recursively.

My question is the following: Is there a "proper" way (a way already accepted and used by mathematicians, I mean) to integrate over a nasty fractal set like $$C$$, and if so, do my assumptions about the "funny integral" still hold? I would be very surprised if this sort of thing has not been formalized yet.

• It seems a Riemann-Stieltjes integral with respect to $df$, where $f$ is the Cantor-Vitali function. Nov 13, 2018 at 14:40
• you are just integrating against a singular (probability) measure, or you can view it as the Haar measure on Cantor set $C=(C_2)^\mathbb{N}$. Nov 13, 2018 at 14:42
• @Rigel I don't understand... that function isn't differentiable everywhere, is it? Its derivative is zero on the "plateaus;" does its derivative exist anywhere else? Nov 13, 2018 at 14:47
• Indeed the integral $\int_0^1 x \, df$, for example, is intended in the sense of Riemann-Stieltjes. Otherwise, you can see $Df$ as a measure (it is the derivative of a bounded variation function) and integrate w.r.t. this measure, that is supported on $C$. Nov 13, 2018 at 14:50
• @Rigel Oh, I see. I was not familiar before with the Riemann-Stieltjes integral. Nov 13, 2018 at 14:52

For the specific example of integration over the Cantor set, these manipulations can be rigorously interpreted by considering the random variable $$X=\sum_{i=1}^{\infty}B_i/3^i$$. Here $$B_i$$ are iid and take the values $$0$$ and $$2$$ with probability $$1/2$$. Then an "integral over the cantor set" can be seen as integration with respect to the distribution of $$X$$. In particular, we have $$X=^dB/3+X/3$$ where $$B=^d B_1$$ and $$B$$ and $$X$$ are independent.
For example, $$EX=EB/3+EX/3$$. Since $$EB=1$$, we conclude $$EX=1/2$$.
Similarly, $$EX^2=E(B^2)/9+2EBEX/9+E(X^2)/9$$ which can be solved for $$EX^2$$ as above, and we can obtain a recursive formula for expectations of higher powers by using the binomial theorem and the independence of $$B$$ and $$X$$