Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)

share|cite|improve this question

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

share|cite|improve this answer
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

It's is simpler to use the complementary set, which is a union of countable many intervals, so that the integral on this union will be the sum of integrals on the intervals. Then by complementation you can find the integral on you fractal set.

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 $$ a_{n,k} = \frac{1+2k}{2^{n+1}}-\frac{1}{2^{2n+3}} $$ $$ b_{n,k} = \frac{1+2k}{2^{n+1}}+\frac{1}{2^{2n+3}} $$

share|cite|improve this answer
Thanks for your reply. I am trying on your idea. I will comment again soon. – user61681 Feb 10 '13 at 11:15
The intervals that are removed after n steps in Smith Voltera Cantor set is difficult to express. Unfortunately I am not able to carry out this. I would be thankful if you can help for this. – user61681 Feb 10 '13 at 15:59
I added some computation for a general $f$. – Emanuele Paolini Feb 10 '13 at 18:16
Thank you for your answer. Can you give me your email ? I want to be in touch with you through email. Thank you very much for your answer. – user61681 Feb 12 '13 at 2:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.