Counterexample to: $f$ is Riemann Integrable if and only if its discontinuous set is Jordan measure zero

Asked Feb 20 '21 at 18:20

Active Feb 20 '21 at 18:37

Viewed 152 times

The forward direction is covered in some textbooks. But the reverse I am only able to find sparingly. Is this a counterexample?:

$f=\begin{cases} 0 & (x\not\in\mathbb{Q})\\ 1/n & (x\in\mathbb{Q}) \end{cases}$

where if $x\in\mathbb{Q}$ we write $x=m/n$ such that $n\not| m$. This is the pathological "continuous at irrational points but discontinuous at rational points" which integrates to $0$.

Edit: "some textbooks", maybe not "most"

edited Feb 20 '21 at 18:37

asked Feb 20 '21 at 18:20

William Loman

Which textbooks? The theorem that I am aware of is that a bounded function $f\colon[a,b]\longrightarrow\Bbb R$ is Riemann-integrable if and only if the Lebesgue measure of the set of its points of discontinuity is $0$. This is known as the Lebesgue's criterion for Riemann integrability. – José Carlos Santos Feb 20 '21 at 18:24
I believe Munkres covers it. – William Loman Feb 20 '21 at 18:39
See my answer to Jordan measure zero discontinuities a necessary condition for integrability. – Dave L. Renfro Feb 20 '21 at 19:17

Counterexample to: $f$ is Riemann Integrable if and only if its discontinuous set is Jordan measure zero

0 Answers0