I have a question about a function that is null outside a set which have measure zero being integrable or not. In the case it's integrable, I must prove that its integral is $0$. I know about this theorem that says:

If $f[a,b]→R$ is an integrable function and $g:[a,b]→R$ is bounded, then if $g$ coincides with $f$ except in a set that has measure zero, then g is integrable and $∫g=∫f$.

Can I say that the function $g(x) = 0$ coincides with $f$ except when the measure is $0$, then if $f$ is integrable, $g$ is and its integral is $0$? Now, for $f$ to be integrable, it must be bounded in the set where the measure is $0$, right? So the answer would be that the function is integrable if $f$ is bounded, but not if it's not bounded. And it's $0$ when integrable, because of the theorem.

Am I right?

  • $\begingroup$ Lebesgue integration right? $\endgroup$ – Jacky Chong Oct 5 '16 at 4:25
  • $\begingroup$ @JackyChong sorry, what? I didn't start with lebesgue yet, I've just seen Riemann and Darboux integration, and now this theorem about integrability $\endgroup$ – Guerlando OCs Oct 5 '16 at 4:35
  • $\begingroup$ @JackyChong did you understand what I wrote? Am I wrong? $\endgroup$ – Guerlando OCs Oct 5 '16 at 16:58
  • $\begingroup$ Seems correct to me. $\endgroup$ – Jacky Chong Oct 5 '16 at 17:47

Your Answer

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

Browse other questions tagged or ask your own question.