The question is as stated in the title:

If $f$ is continuous almost everywhere on $[a,b]$, is it true that $f$ is Riemann integrable on $[a,b]$?

The natural thought is Lebesgue's criterion, which states that "A bounded function on a compact interval [a, b] is Riemann integrable if and only if it is continuous almost everywhere".

Hence, if we remove the condition "bounded", the statement becomes false right? I would see that the upper sum goes to infinity.

What would be a good concrete counterexample?


  • 4
    $\begingroup$ The function $f(x)=1/x$ for $x\in(0,1]$ and $0$ for $x=0$ is continuous almost everywhere. $\endgroup$ Aug 19, 2016 at 1:22

1 Answer 1


Think $\frac{1}{x}$ on $(0,1)$

  • 1
    $\begingroup$ I see. If I need a compact domain, can I take a function $f=\frac{1}{x}$ on (0,1], and $f(0)=0$. $\endgroup$
    – yoyostein
    Aug 19, 2016 at 1:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .