# Integrability

• Nov 18th 2010, 03:59 AM
JG89
Integrability
Question:

Let $Q$ be a rectangle in $\mathbb{R}^n$. Let $S \subset Q$. Consider the characteristic function of S on Q given by $f_s (x) = 1$ if $x \in S$, and $f_s (x) = 0$ otherwise. Prove that $f_s$ 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 $f_s$ 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?
• Nov 18th 2010, 04:56 AM
Ackbeet
I don't think your counterexample holds up, because I don't think $\text{bd}(S)=\mathbb{Q}\cap[0,1].$ I think $\text{bd}(S)=[0,1].$ 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.