Is the Integrability of a Function Determined by the Measure of Its Boundary?

[JG89]

Homework Statement

Let Q be a rectangle in $$\mathbb{R}^n$$. Let S be a subset of Q. Consider the characteristic function of S on Q given by $$f_s (x) = 1$$ if x is in S, and 0 otherwise.
Prove that $$f_s$$ is integrable if and only if bd(S) has measure 0.

Homework Equations

The Attempt at a Solution

I don't see how this statement can be true. For example, let S be the set of irrationals in Q = [0,1]. bd(S) is the rationals in [0,1], which has measure 0. So the characteristic function of S $$f_s$$ should be integrable on [0,1], but it's obviously not, because for any partition P, the upper sums equal 1 and the lower sums equal 0.

Is my thinking correct?

 

[HallsofIvy]

Homework Statement

Let Q be a rectangle in $$\mathbb{R}^n$$. Let S be a subset of Q. Consider the characteristic function of S on Q given by $$f_s (x) = 1$$ if x is in S, and 0 otherwise.
Prove that $$f_s$$ is integrable if and only if bd(S) has measure 0.

Homework Equations

The Attempt at a Solution

I don't see how this statement can be true. For example, let S be the set of irrationals in Q = [0,1]. bd(S) is the rationals in [0,1], which has measure 0. So the characteristic function of S $$f_s$$ should be integrable on [0,1], but it's obviously not, because for any partition P, the upper sums equal 1 and the lower sums equal 0.

Is my thinking correct?

No, it isn't. Q has measure 0 but its boundary is the set of all real numbers in [0, 1] and has measure 1.