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.
 A: 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.
In view of the Lebesgue criterion for Riemann integrability, pretty much any example will be more or less of this type, unless somehow your definition of integration on an open set does not respect the Lebesgue criterion. Indeed it seems the entire discussion hinges on this definition, which is not at all obvious to me.
