Let $f,g:A\subset\mathbb{R}^N\to[0,\infty)$, $f$ is Riemann integrable, and $g=f$ except in finite set, $A$ is a N-dimensional closed cell. Show that $g$ is Riemann integrable and $\int_Af(x)dx=\int_Ag(x)$

I tried to define $h(x)=g(x)-f(x)$, and prove that the integral exists and is zero. For that we could maybe use the lebesgue measure $\lambda(D_h)$, of the discontinuities of $h$, but I don't think it works, because if the difference between $g$ and $f$ was infinite and countable, the set would still have measure zero, but $g$ wouldn't be integrable.

Is there another way to prove it?


Let $B$ be the set of all $x \in A$ such that $f(x) \neq g(x)$. Then $B$ is finite by assumption and $B \subset A$. So $B$ has Lebesgue measure $0$. Note that $A = A\setminus B \cup B$; so the measure of $A$ is the measure of $A \setminus B$ plus the measure of $B$, and hence $A = A \setminus B$ in measure. Note that $A \setminus B$ is the set of all $x \in A$ such that $f(x) = g(x)$. Then $\int_{A}f = \int_{A\setminus B} f = \int_{A \setminus B} g = \int_{A}g$.

Note: The Riemann and Lebesgue integration consideration in the question was just made clear after this answer was posted.

  • $\begingroup$ But if $B$ was infinite and countable, Lebesgue measure would still be zero, but $g$ wouldn't be integrable. Although intuitive, I still can't see the difference between finite and infinite countable. $\endgroup$ – Bruno Mazeto Jun 20 '17 at 18:23
  • $\begingroup$ Could u come again? I did not get your current question. $\endgroup$ – Megadeth Jun 20 '17 at 18:25
  • $\begingroup$ @BrunoMazeto when you are talking about integrability you are referring to Riemann integrability right? $\endgroup$ – clark Jun 20 '17 at 18:31
  • $\begingroup$ Sorry, I thought that it would be clear, as I didn't learn Lebesgue integration. I'm talking about Riemann integration $\endgroup$ – Bruno Mazeto Jun 20 '17 at 18:35
  • $\begingroup$ @YngwieMalmsteen I put an example in the comments of the other answer. Just to be clear, it's Riemann integration. Sorry about the confusion. $\endgroup$ – Bruno Mazeto Jun 20 '17 at 18:38

Of course, $f=g$ almost everywhere (because finite sets have the Lebesgue measure zero). So, the set of discontinuity points of $g$ is also of measure zero (trivial observation based on discontinuities of $f$ and its integrability). So, $g$ is (Riemann) integrable.

  • $\begingroup$ I understand, but if I have $f(x) = 0$ and $g(x) = 0$ for $x$ irrational and $g(x) = 1$ for $x$ rational, the difference has still measure zero, but than $g$ is not integrable. And this argument doesn't work, does it? $\endgroup$ – Bruno Mazeto Jun 20 '17 at 18:15
  • $\begingroup$ I'm not sure continuty points are relevant since Lebesgue integrable functions can extremely discontinuous (even discontinuous everywhere). Plus it is never stated that $f$ has any continuity properties. $\endgroup$ – User8128 Jun 20 '17 at 18:15
  • $\begingroup$ @BrunoMazeto such $g$ is integrable in the Lebesgue sense and integrates to zero since it is zero a.e. $\endgroup$ – User8128 Jun 20 '17 at 18:16
  • $\begingroup$ @User8128 the question is regular Riemann integration. Should I add this to the question? $\endgroup$ – Bruno Mazeto Jun 20 '17 at 18:18
  • 1
    $\begingroup$ @BrunoMazeto Definitely! As of now, you added the lebesgue-measure tag to the question, which strongly suggests that you consider Lebesgue integratoin, not Riemann $\endgroup$ – Hagen von Eitzen Jun 20 '17 at 18:33

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.