Let $A$ be a bounded open set in $\mathbb{R}^n$. Give an example where $\int_{\bar{A}} f$ exists but $\int_A f$ does not.

Integration in the Riemann sense.

Let $A$ be a bounded open set in $\mathbb{R}^n$; let $f: \mathbb{R}^n\to \mathbb{R}$ be a bounded continuous function. Give an example where $\int_{\bar{A}} f$ exists but $\int_A f$ does not.

I cannot think of any such example. I would greatly appreciate it if anyone can provide me with some insight.

• In which sense are you defining integration, Lebesgue or Riemann? In the Lebesgue sense this result is nonsensical, since $f 1_A$ will necessarily also be integrable if $f$ is. – Ian Oct 21 '15 at 2:59
• Sorry should've made it clear, I mean Riemann. – nomadicmathematician Oct 21 '15 at 3:14
• @Ian: this might be silly, but is it obvious that $1_A$ is measurable if $1_{\overline{A}}$ is? – Giovanni Oct 21 '15 at 3:14
• @Giovanni $A$ is open and $\overline{A}$ is closed, so the indicator functions are definitely measurable. – Ian Oct 21 '15 at 3:15
• @takecare How do you define Riemann integration on an open set? (All notions of Riemann integration that I understand depend on compactness.) – Ian Oct 21 '15 at 3:16

I don't think this statement makes sense as formulated. In particular, if $f$ is integrable on $\overline{A}$, then its set of discontinuity points has measure zero. The restriction of $f$ to $A$ will not add any discontinuity points (because it can be given by the composition of $f$ with the inclusion map, and the inclusion map is continuous). So the set of discontinuity points of the restriction will also have measure zero, so $f$ will be integrable on $A$ as well (if that notion even makes sense in the Riemann setting).
A reformulation (assuming we've somehow sensibly defined Riemann integration on an open set) is as follows. Let $C$ be a fat Cantor set, let $A=[0,1] \setminus C$, and then consider $f=1_A$. Then $f$ is actually continuous on $A$, because it is constant there. But no matter what, $f$ cannot be Riemann integrable on $\overline{A}$, because its set of discontinuity points is $C$ which has positive measure.
• I don't really understand your example. $f$ is supposed to be continuous on $\mathbb R^n$, not only on $A$. Beside $f$ is supposed to be integrable on $\bar A$. – user251257 Oct 21 '15 at 4:00
• @user251257 I agree, I had to change some of the original hypotheses somewhat because as formulated they do not make sense. In particular, if $f$ is integrable on $\overline{A}$, then no discontinuities are introduced by restricting to $A$, so by the Lebesgue criterion, $f$ is also integrable on $A$ (whatever that even means in the Riemann case). – Ian Oct 21 '15 at 4:21
• @user251257 Er, what? $A$ is a bounded open set. $f=1$ will be integrable on both $\overline{A}$ and $A$. There is no way to satisfy the OP's condition by having the integrals blow up. – Ian Oct 21 '15 at 4:33