4

I cant find example in 7.32

every continuous function is Riemann integrable,continuity is certainly not necessary.

I dont know anything about measure.

jujuju
  • 269
  • 4
  • 10

1 Answers1

5

Define $f$ to be identically $0$ on $\Bbb{R}$, except that $f(0) = 1$. To see that this is Riemann integrable, note that the lower sums are all $0$ (suppose we're integrating on $[-1, 1]$, for clarity). But the upper sums can be made arbitrary small, by choosing small intervals around $0$.

More generally, $f$ is Riemann integrable if and only if the set of points of discontinuities has Lebesgue measure $0$; so something like Thomae's function, which is discontinuous on $\Bbb{Q}$ is still Riemann-integrable.