Riemann integral of characteristic function of Cantor set 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!
 A: 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 $0$.
A: 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 $0$.
