Can anyone tell me how to calculate the Riemann integral of the characteristic function of the Cantor set? It's probably obvious but I don't see how to write it down. Many thanks for your help!
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4$\begingroup$ You can prove that the upper Darboux integral $\leq (\frac{2}{3})^n$, $\forall n \in \mathbb{N}$ and the lower Darboux integral is non-negative and then can conclude that the Riemann integral makes sense and hence it has to be zero. $\endgroup$user17762– user177622011-01-21 23:28:53 +00:00Commented Jan 21, 2011 at 23:28
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2 Answers
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Let $C$ be the Cantor set, and let $C_n$ be the closed set left after $n$ steps of removing middle thirds from $[0,1]$, so $C_n$ is a disjoint union of $2^n$ closed intervals, and the sum of the lengths of these intervals is $\left(\frac{2}{3}\right)^n$, which converges to zero. The characteristic function $\chi_{C_n}$ of $C_n$ is a step function that dominates the characteristic function of $C$, so its integral, $\left(\frac{2}{3}\right)^n$, is an upper Riemann sum for $\chi_C$. Thus the infimum of the upper Riemann sums for $\chi_C$ is at most $\inf_n\left(\frac{2}{3}\right)^n=0$. The lower Riemann sums are all greater than or equal to $0$, so this shows that the Riemann integral exists and equals $0$.
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1$\begingroup$ @Moron: I think the OP wants to know if the cantor set in the first place is Riemann integrable. If so then we could invoke the fact that the Riemann integral is same as the Lebesgue Integral as you have done in your answer. $\endgroup$user17762– user177622011-01-21 23:38:42 +00:00Commented Jan 21, 2011 at 23:38
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1$\begingroup$ @Sivaram: The set of points of discontinuity of that function is of measure zero, that implies Riemann Integrability automatically, as it is bounded. $\endgroup$Aryabhata– Aryabhata2011-01-21 23:40:12 +00:00Commented Jan 21, 2011 at 23:40
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$\begingroup$ @Moron: True. I didn't see you updated answer. $\endgroup$user17762– user177622011-01-21 23:40:49 +00:00Commented Jan 21, 2011 at 23:40
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$\begingroup$ @Sivaram: Yeah, usually make some edits after posting once. Also, it is tagged measure theory :-) $\endgroup$Aryabhata– Aryabhata2011-01-21 23:43:14 +00:00Commented Jan 21, 2011 at 23:43
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$\begingroup$ In spite of the measure theory tag, I decided to write a sketch of an argument that doesn't mention measure theory. Moron's answer actually seems more natural to me. $\endgroup$Jonas Meyer– Jonas Meyer2011-01-21 23:43:36 +00:00Commented Jan 21, 2011 at 23:43
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I am presuming you are talking about the Cantor Set in $[0,1]$, where you remove the middle third.
Since the Cantor set is of measure zero, the Lebesgue integral of its characteristic function is $0$.
If it were Riemann integrable (which it is, as the points of discontinuity is of measure $0$), then the value of the Riemann integral would equal the Lebesgue integral and so would be $0$.
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$\begingroup$ Yes thanks, I know, but I would like to write down the Riemann sum of it. $\endgroup$Rudy the Reindeer– Rudy the Reindeer2011-01-22 09:12:03 +00:00Commented Jan 22, 2011 at 9:12
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2$\begingroup$ @Matt: You said you want to know "how to calculate the Riemann integral" of the function. You did not give any definition of "calculate", or any context to allow us to know that you intended to ask for a method dealing explicitly with Riemann sums. The tag "measure-theory" seems to put it in a context where it is natural to use basic measure theoretic facts to quickly do the calculation. $\endgroup$Jonas Meyer– Jonas Meyer2011-01-22 09:53:08 +00:00Commented Jan 22, 2011 at 9:53