# What is the Lebesgue Integral of $f(x) = x$ over the Smith-Volterra-Cantor Set?

The integral of the characteristic function is simply the measure of the Smith-Volterra-Cantor Set which is $1/2$. Also any Lebesgue integral function over the Cantor set evaluates to $0$ as the Cantor set is of measure $0$. But what about integrating over the Smith-Volterra-Cantor set which is similar to the Cantor set but has non-zero measure.

I am trying to "approximate" the Smith-Volterra-Cantor Set by closed or open sets and then evaluating the Lebesgue integral over that closed or open set, but not getting the idea. The question is about integrating over "discrete" type of measurable sets (such as the Cantor or the Smith-Volterra-Cantor Set)

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If $A$ is the Smith–Volterra–Cantor set, then $\int_A xdx = \int_A (1-x)dx$, and therefore $\int_A xdx =\frac12\int_A 1dx=\frac12m(A)$.

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How we get the property ∫Axdx=∫A(1−x)dx? –  user61681 Feb 10 '13 at 11:16
@user61681: Because the map $x\mapsto 1-x$ maps $A$ bijectively onto itself. E.g., a simple function $f$ satisfies $f(x)\leq x$ if and only if $g(x)=f(1-x)$ satisfies $g(x)\leq 1-x$, and you could show directly that for such simple functions $\int_A f(x)dx=\int_A g(x)dx$. In slightly more detail, note that $\int_{A\cap[0,1/2]}xdx = \int_{A\cap[1/2,1]}(1-x)dx$ and $\int_{A\cap[1/2,1]}xdx = \int_{A\cap[0,1/2]}(1-x)dx$ –  Jonas Meyer Feb 10 '13 at 11:20
(I should have mentioned in my last comment that $x\mapsto 1-x$ is measure preserving.) –  Jonas Meyer Feb 10 '13 at 11:27
Thanks. I am trying to understand your answer. How can we evaluate the Lebesgue integral of f(x)= x*x on Smith Volterra Cantor set? –  user61681 Feb 10 '13 at 11:32
@user61681: Attempting the same trick isn't enough to evaluate $\int_A x^2dx$; it will depend on more about $A$ than just the fact that $x\mapsto 1-x$ is a measure preserving bijection. I don't know off hand of an approach other than trying to keep track of the integrals over the removed integrals as suggested in manu-fatto's answer, and hoping that a pattern emerges. –  Jonas Meyer Feb 10 '13 at 11:37
The complementary of the Smith-Volterra-Cantor set is the set composed by the union for $n=0,1,2,\dots$ of $2^n$ intervals centered at points $\frac{1+2k}{2^{n+1}}$ with $k=0,\dots,2^n-1$ with size $\frac{1}{2^{2n+2}}$. So the integral of a function $f(x)$ on the set $C$ is given by: $$\int_C f(x)\, dx = \int_0^1 f(x)\, dx - \sum_{n=0}^\infty \sum_{k=0}^{2^n-1} \int_{a_{n,k}}^{b_{n,k}} f(x)\, dx$$ with $$a_{n,k} = \frac{1+2k}{2^{n+1}}-\frac{1}{2^{2n+3}}$$ $$b_{n,k} = \frac{1+2k}{2^{n+1}}+\frac{1}{2^{2n+3}}$$
I added some computation for a general $f$. –  Emanuele Paolini Feb 10 '13 at 18:16