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Let $\alpha:[0,1] \to R$ be the Cantor function. Evaluate $$\int_{0}^{1}xd\alpha $$and $$\int_{0}^{1}x^2d\alpha.$$

I know that the Cantor function is continuous and monotone increasing, how can I evaluate the integral above using the properties of $\alpha$? Can someone help me solve this question?

Martin Sleziak
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python3
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  • It might help that the union of all sets on which $\alpha$ is constant has measure 1. – Mnifldz Apr 14 '15 at 02:05
  • This is the nondecreasing Cantor function, which is constant on the middle third intervals? If so, then maybe try calculating it as a limit of $\int_0^1 x d \alpha_n$, where $\alpha_n$ are absolutely continuous approximations of $\alpha$. Then for such things you can integrate by parts, etc. – Ian Apr 14 '15 at 02:05
  • Then,I tried your way,then I am stuck in integrating $\int\alpha dx$ – python3 Apr 14 '15 at 02:42
  • These are $E(X)$ and $E(X^2)$ where $$X=\sum_{n\geqslant1}\frac{U_n}{3^n}$$ and $(U_n)$ is i.i.d. with $P(U_n=0)=P(U_n=2)=\frac12$. For example, the (rather simple) facts that $E(U)=1$ and $E(U^2)=2$ (directly) yield $$E(X)=\sum_{n\geqslant1}\frac1{3^n}=\frac12,\qquad E(X^2)=\sum_{n\geqslant1}\frac2{9^n}+\sum_{n\ne m}\frac1{3^n3^m}=\frac58.$$ – Did Apr 14 '15 at 07:37
  • Also see http://math.stackexchange.com/q/57721/442 – GEdgar Apr 14 '15 at 12:00

1 Answers1

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For the first integral we have by parts$$\int_{0}^{1}xd\alpha\left(x\right)=\left.x\alpha\left(x\right)\right|_{0}^{1}-\int_{0}^{1}\alpha\left(x\right)dx=1-\int_{0}^{1}\alpha\left(x\right)dx $$ and since $\int_{0}^{1}\alpha\left(x\right)dx $ is the area of the Cantor function on $\left[0,1\right] $ we get $$\int_{0}^{1}xd\alpha\left(x\right)=\frac{1}{2}. $$ For the second integral see here.

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Marco Cantarini
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