10
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ It seems a Riemann-Stieltjes integral with respect to $df$, where $f$ is the Cantor-Vitali function. $\endgroup$
    – Rigel
    Nov 13, 2018 at 14:40
  • $\begingroup$ you are just integrating against a singular (probability) measure, or you can view it as the Haar measure on Cantor set $C=(C_2)^\mathbb{N}$. $\endgroup$ Nov 13, 2018 at 14:42
  • $\begingroup$ @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? $\endgroup$ Nov 13, 2018 at 14:47
  • $\begingroup$ 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 $C$. $\endgroup$
    – Rigel
    Nov 13, 2018 at 14:50
  • $\begingroup$ @Rigel Oh, I see. I was not familiar before with the Riemann-Stieltjes integral. $\endgroup$ Nov 13, 2018 at 14:52

1 Answer 1

9
$\begingroup$

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 $X$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .