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?

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    $\begingroup$ 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

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.

  • $\begingroup$ 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

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