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What is the Lebesgue integral of $f(x)=x^2$ on $\bar{C}=[0,1]\setminus C$, where $C$ is a Cantor-set?

I already know that the measure of the Cantor-set is $0$ ($m(C)=0$), hence $\int_Cx^2 dm=0 $.

Would the same apply for $\bar{C}$, or does it mean that I can take the Riemann integral of $x^2$ on [0,1], meaning $\int_\bar{C}x^2 dm=\int_{0}^{1}x^2 dx$? If so, why?

Any help and suggestion is appreciated!

Forest
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Hint: If $A\subset B$, with $A$ and $B$ both measurable, then $\int_B f\,d\mu=\int_{B\setminus A} f\,d\mu+\int_A f\,d\mu$. In other words, we can split up our domain of integration into pieces, integrate over the pieces, and then add them together. It all works out because the integral is linear.

Plutoro
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  • Based on your hint, and the property of Lebesgue integral $\int_{A\cup B}f dm=\int_{A}f dm+\int_{B}f dm$ where A and B are disjunct, I could take $\int_{C\cup \bar{C}}f dm - \int_{C}f dm=\int_{\bar{C}}$ so $\frac{1}{3} - 0 = \frac{1}{3}$, hence the Lebesgue integral of $f$ on $\bar{C}$ is $\frac{1}{3}$? Based on another idea, can I say that the measure of $C$ is $0$, so $\bar{C}$ is continuous almost everywhere, hence I can take $\int_{[0,1]} x^2 dm$? – Forest Apr 04 '16 at 18:59
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    Your first method is right on. The other is not quite right. You should let $\tilde f=f\chi_{\overline C}$. Say that $\int_{\overline C}f,d\mu=\int_{[0,1]}\tilde f,d\mu$, and $\tilde f=f$ almost everywhere, so $\int_{[0,1]}\tilde f,d\mu=\int_{[0,1]}f,d\mu$. – Plutoro Apr 04 '16 at 19:05
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    Saying that a set is continuous almost everywhere nos not have meaning. – Plutoro Apr 04 '16 at 19:06
  • That's right, thank you. The answer became clear thanks to your hint! – Forest Apr 04 '16 at 19:09