
Integrability
Question:
Let $\displaystyle Q $ be a rectangle in $\displaystyle \mathbb{R}^n $. Let $\displaystyle S \subset Q $. Consider the characteristic function of S on Q given by $\displaystyle f_s (x) = 1 $ if $\displaystyle x \in S $, and $\displaystyle f_s (x) = 0 $ otherwise. Prove that $\displaystyle f_s $ is integrable if and only if bd(S) has measure 0.
I don't see how this can be true. Take Q to be the unit interval [0,1], and let S be the set of irrationals in [0,1]. bd(S) is the set of rationals in [0,1], which has measure 0, but the integral of $\displaystyle f_s $ doesn't exist, because for any partition P the upper sum is 1 and the lower sum is 0. Am I correct in my thinking?

I don't think your counterexample holds up, because I don't think $\displaystyle \text{bd}(S)=\mathbb{Q}\cap[0,1].$ I think $\displaystyle \text{bd}(S)=[0,1].$ At least as I understand it, the set of boundary points of a set are all those points which, if you take any neighborhood around one of them, that neighborhood always contains points of S and the compliment of S. The irrationals work for this definition just as well as the rationals. Thus, you get the entire interval.
So how would you go about proving your main result?