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!
Riemann integral of characteristic function of Cantor set
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3You 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. – 2011-01-21
2 Answers
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@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. – 2011-01-21
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1@Sivaram: The set of points of discontinuity of that function is of measure zero, that implies Riemann Integrability automatically, as it is bounded. – 2011-01-21
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0@Moron: True. I didn't see you updated answer. – 2011-01-21
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0@Sivaram: Yeah, usually make some edits after posting once. Also, it is tagged measure theory :-) – 2011-01-21
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0In 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. – 2011-01-21
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0@Jonas: I like your answer though :-) – 2011-01-21
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0@Jonas: Actually your answer seems more natural to me. Since the OP wants to know about the Riemann integral, I would actually prefer going from the definitions of Riemann integral than to do the Lebesgue integral and then from there conclude that this is Riemann integrable and hence the integral is zero. However as Moron mentioned since this is in measure theory tag the OP probably wants to use the facts learnt in measure theory to conclude the answer. Either way $0=0$ :) – 2011-01-21
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0@Sivaram: It is quite inefficient to go by the definition each time. In this case, it is much easier (at least to me) to see that the points of discontinuity are of measure zero and that the lebesgue integral is 0. Of course, this is probably homework/self study, so... Also, apologies to Jonas for the pings. – 2011-01-22
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0I am not bothered by pings. – 2011-01-22
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0@Jonas: I see. If $S^+ (C_n) = \sum_{k=1}^n \frac{1}{n} \chi_{C_n} (x) \leq (\frac{2}{3})^n $ denotes the upper sum, I don't need to explicitly write down $S^- (C_n)$ because $S^- (C_n) \geq 0$ and $lim_{n -> \infty} S^+ (C_n) = 0$ and $S^- \leq R \leq S^+$, where $R$ is the Riemann integral I'm looking for. Thank you, that answered my question! – 2011-01-22
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0Don't know why I was trying to write down $S^-$ explicitly : / – 2011-01-22
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0@Matt: The sum you wrote is incorrect, and I'm not sure what it is supposed to mean. E.g., what is $x$ there, why are you dividing by $n$, and why are you summing from $1$ to $n$? $C_n$ is a union of disjoint intervals, say $I_1,I_2,\ldots,I_{2^n}$. The endpoints of these intervals are in $C$, and $C\subset C_n$, so the integral of $\chi_{C_n}$ is an upper Riemann sum for $\chi_C$, namely $\displaystyle{\sum_{k=1}^{2^n}1\cdot\operatorname{length}(I_k)}$ (the contribution of the other intervals being $0$). – 2011-01-22
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0@Matt:It is also straightforward to show that the lower sums must all be zero. It follows from the fact that the complement of the Cantor set is dense, so every interval contains a point where the characteristic function of the Cantor set must be $0$. But you're correct that in this case you don't need to show that explicitly. The idea is useful for some other examples, e.g. for showing that characteristic functions of fat Cantor sets are not Riemann integrable. – 2011-01-22
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0@Matt: By the way, you're welcome. I hope the more explicit description of the sum 2 comments up helps. I guess in your sum you were thinking of breaking up $[0,1]$ into $n$ even intervals, but (1) then you should have $C$ there instead of $C_n$ (2) then you should have $x_n$ there instead of $x$, where $x_n$ is some point in the $n^\text{th}$ interval. Since it is supposed to be an upper sum, you technically should be taking a supremum. (3) I recommend not breaking up $[0,1]$ that way, because it makes it harder. – 2011-01-22
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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0Yes thanks, I know, but I would like to write down the Riemann sum of it. – 2011-01-22
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1@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. – 2011-01-22