1
$\begingroup$

Consider $f:\mathbb{R}^{d_1}\times\mathbb{R}^{d_2}\to\mathbb{R}$ , Fubini’s theorem says if $f$ is Lebesgue integrable than we can integrate first over $x$ and then over $y$ and get the same value as the integral of $f$: $$\int_{\mathbb{R}^{d_2}}\int_{\mathbb{R}^{d_1}}f(x, y)dxdy=\int f$$ where $x\in\mathbb{R}^{d_1}$ and $y\in\mathbb{R}^{d_2}$. My question is that if we know that $$\int_{\mathbb{R}^{d_2}}\int_{\mathbb{R}^{d_1}}f(x, y)dxdy<\infty$$ Can we say that $f$ must be Lebesgue integrable?

$\endgroup$
  • 1
    $\begingroup$ No, it doesn't have to be. I'm sure, just don't know the counterexamples of the top of my head $\endgroup$ – Jakobian Mar 21 at 23:22
3
$\begingroup$

It is very easy to give an example where $\int f(x,y)dx=0$ for all $y$ but $f$ is not integrable. Just take $d_1=d_2=1$, $f(x,y)=g(x)h(y)$ where $h$ is not integrable but $g$ is an odd integrable function. In this case $f$ is not integrable.

However, if $\int \int |f(x,y)|dxdy<\infty$ then $f$ is integrable and this is immediate from Tonelli's Theorem.

$\endgroup$
  • $\begingroup$ Similarly, $f(x,y) = \begin{cases} 1, & -1 < x-y < 0, \\ -1, & 0 < x-y < 1, \\ 0, & \mathrm{otherwise} \end{cases}$ forms a counterexample where both iterated integrals give 0 but the function is not integrable overall. $\endgroup$ – Daniel Schepler Mar 21 at 23:31
  • $\begingroup$ @DanielSchepler Good observation! $\endgroup$ – Kavi Rama Murthy Mar 21 at 23:32

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.