# Is this “inverse” statement of Fubini’s theorem true?

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?

• No, it doesn't have to be. I'm sure, just don't know the counterexamples of the top of my head – 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.
• 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. – Daniel Schepler Mar 21 at 23:31