I have given the following function

$$ f(x,y) = \begin{cases} 1 &, \ x \in \mathbb{Q} \\ 2y & , \text{ otherwise} \end{cases} $$

This is a measurable function in sense of Lebesgue. Now, I have to answer the question if the following integral exists in the sense of Lebesgue and if $f(x,y)$ is Lebesgue-integrable.

$$ \int_{[0,1]} \int_{[0,1]} f(x,y) \ \mathrm d \lambda(x)\mathrm d \lambda(y) \text{ or } \int_{[0,1]} \int_{[0,1]} f(x,y) \ \mathrm d \lambda(y)\mathrm d \lambda(x) $$

We had the theorem of Fubini for non-negative functions. And this function is on $[0,1]\times [0,1]$ non-negative. So this integral should exist and both are equal.

To show that $f(x,y)$ is Lebesgue-integrable I just have to calculate one of the integrals above

$$\int_{[0,1]\times[0,1]} f(x,y) \ \mathrm d \lambda(x,y) =\int_{[0,1]} \int_{[0,1]} f(x,y) \ \mathrm d \lambda(y)\mathrm d \lambda(x) \\ = \underbrace{ \int_{[0,1]\cap \mathbb{Q}} \int_{[0,1]} f(x,y) \ \mathrm d \lambda(y)\mathrm d \lambda(x)}_{=0 \text{ (null set) }}+\int_{(\mathbb{R}\setminus \mathbb{Q})\cap[0,1]} \int_{[0,1]} f(x,y) \ \mathrm d \lambda(y)\mathrm d \lambda(x) \\ = \underbrace{ \int_{(\mathbb{R}\setminus \mathbb{Q})\cap[0,1]}\underbrace{ \int_{[0,1]} 2y \ \mathrm d \lambda(y)}_{=1}\mathrm d \lambda(x)}_{\lambda((\mathbb{R}\setminus \mathbb{Q})\cap[0,1])=\lambda([0,1])=1} = 1 < \infty$$

So it follow that $f(x,y)$ is lebesgue-integrable. The answer is both integrals exists in the sense of lebesgue and the function is integrable.


Is my caluclation right or did I something wrong? Because my tutor said, this function is not lebesgue integrable, but did not say why.

  • 1
    $\begingroup$ If you haven't written down the wrong function, and the real one isn't integrable, then your tutor made a mistake. The function you have here is indeed Lebesgue-integrable. $\endgroup$ – Daniel Fischer Aug 16 '14 at 18:37
  • $\begingroup$ @DanielFischer: And is the argumentation right? $\endgroup$ – DerJFK Aug 16 '14 at 18:40
  • 1
    $\begingroup$ Yes, it's right. (Aside: "Satz", in the sense used here, is "theorem" or "proposition", depending on how important it is. Sentence is a linguistic term. And, "of", not "from". The theorem of Fubini; or Fubini's theorem.) $\endgroup$ – Daniel Fischer Aug 16 '14 at 18:44
  • $\begingroup$ Thank you! I just translated it directly and didn't think about, sorry. $\endgroup$ – DerJFK Aug 16 '14 at 18:47

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