Calculate $\int_0^1 f(x) dx$, where $f(x)= \begin{cases} 0 & x= x\in C \\ \frac{2}{7^n} & x= x\not \in C \end{cases} $, with $C$ the Cantor set.

Question

I am looking back at notes, and problems from the semester, and I came across this problem that I am having trouble solving.

Let $$f(x)= \begin{cases} 0 & x= x\in C \\ \dfrac{2}{7^n} & x= x\not \in C \end{cases} $$

Where $C$ is the cantor set, and $n$ stands for $x$ being removed in the $n$th time from the Cantor set.

where $f:[0,1]\to \mathbb{R}$. Find $\int_0^1 f(x) dx$.

I am having a tough time Calculating this. Any help would be appreciated. Mainly I having trouble understanding how to do this, because the Cantor set gives me trouble.

Answer 1

At stage $n$ the measure of the remaining set (to generate the Cantor set) is $$\mu(T_n) = \left( \frac{2}{3} \right)^n$$ So the portion of the complement that will integrate with value $2/7^n$ is $$\mu(T_{n-1}) - \mu(T_n)= \left( \frac{2}{3} \right)^{n-1} - \left( \frac{2}{3} \right)^{n} = \frac{2^{n-1}}{3^n} $$ So your integral (over the complement of the Cantor set) is just $$\sum_{n=1}^\infty\left( \frac{2}{7^n} \right)\frac{2^{n-1}}{3^n} = \sum_{n=1}^\infty\frac{2^n}{3^n7^n} = \frac{1}{1 - 2/21} - 1 = \frac{2}{19} $$