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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.

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

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    $\begingroup$ 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 $\bar A$. $\endgroup$ – user251257 Oct 21 '15 at 4:00
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    $\begingroup$ @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). $\endgroup$ – Ian Oct 21 '15 at 4:21
  • $\begingroup$ @user251257 Edited my comment accordingly. $\endgroup$ – Ian Oct 21 '15 at 4:22
  • $\begingroup$ @user251257 I edited my answer as well, is it clearer? $\endgroup$ – Ian Oct 21 '15 at 4:25
  • $\begingroup$ @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. $\endgroup$ – Ian Oct 21 '15 at 4:33

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