Why this function is not Riemann integrable on [0, 1] ? \begin{equation} f\left(x\right)=\begin{cases} x & x\in\mathbb{Q}\\ 0 & x\notin\mathbb{Q} \end{cases} \end{equation} We can calculate the upper integral and lower integral, $\overline{\int}f\left(x\right)dx=0.5$, $\underline{\int}f\left(x\right)dx=0$, therefore, the Riemann's criterion does not hold. So the function is not Riemann inegrable.

But if you use the Lebesgue theorem, bounded function on [a, b] is Riemann integrable if and only if the set of discontinuities of $f\left(x\right)$ has measure zero.

We know that (1) every finite set has measure zero, and (2) every countable subset of $\mathbb{R}$ has measure zero.

The rational numbers $\mathbb{Q}$ is a countable subset of $\mathbb{R}$, and the rational numbers $\mathbb{R}\setminus\mathbb{Q}$ is a countable subset of $\mathbb{R}$.

Hence, the set of discontinuities of $f\left(x\right)$ on [0, 1] has measure zero. Therefore $f\left(x\right)$ is Riemann integrable.

What are the mistakes here? Thank you very much.


well, your discontinuities are a lot more than $\mathbb{Q}\cap [0,1]$ they are in fact all of $[0,1]$, since it is the boundary that matters. or show me one point apart from $0$ where your function is continuous!

  • $\begingroup$ The function is continuous at $x=0$ $\endgroup$ – JavaMan Oct 31 '18 at 14:01
  • $\begingroup$ Fair point! I apologize $\endgroup$ – Enkidu Oct 31 '18 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.