Prove that the bounded function $f$ defined by $f(x)=0$ if $x$ is irrational and $f(x)=1$ if $x$ is rational is not Riemann integrable on $[0,1]$.

I was given the hint to use the inverse definition of Riemann integrable and consider the cases of the partition being all rationals, and all irrationals between $[0,1]$, but I'm not too sure how to go about it.


1 Answer 1


Hint: Every subinterval of $[0,1]$ contains both rational numbers and irrational numbers.

Look at the definition of a function being Riemann integrable on the interval $[a,b]$ (in this case $[0,1]$). This definition must fail for this function.


Well, since every subinterval contains irrational numbers, the infimum of the function values over any subinterval is $0$. Similarly, since every sub-interval contains rational numbers, the supremum of the function values over any subinterval is $1$.

What does this tell you about the upper and lower sums over any partition?

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    $\begingroup$ Wouldn't that make the upper integral equal to 1 and the bottom equal to 0, but since they don't equal it's not Riemann integrable? In my real analysis class we never went over upper or lower sums so I had to look it up on the net. How would I write this down as a proof? Also sorry it took me a while to respond, had to make dinner for little ones. $\endgroup$
    – xCanaan
    Nov 29, 2015 at 4:02
  • $\begingroup$ @xCanaan That's exactly right. And that's also how you would write the proof out. State that for any arbitrary partition $0 = x_{0} < x_{1} < \dots < x_{n - 1} < x_{n} = 1$, by the argument above the upper integral is 1 and and the lower integral is $0$, and since they aren't equal, $f$ is not Riemann integrable. $\endgroup$
    – layman
    Nov 29, 2015 at 4:11
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    $\begingroup$ @xCanaan Remember that proofs don't have some specific structure or language. A proof should be a story -- a list of logical steps or deductions where each one follows from the last, and it should be a story of how you understand the proof so that the reader can understand it too. Feel free to put the proof in your own words, as long as the ideas are there. $\endgroup$
    – layman
    Nov 29, 2015 at 4:12
  • $\begingroup$ @xCanaan I feel it is important to bring out how this shows a failure of convergence of Riemann sums. It is not the fact that for every partition the upper and lower Riemann sums are different, is is that they do not converge to one another as the partition becomes finer. You may have understood this, but, it wasn't made super explicit. $\endgroup$
    – James
    Nov 29, 2015 at 4:44

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