Converse of Fubini's theorem

Let $f \colon [0,1]^2 \to \mathbb R$ a Lebesgue measurable function such that for all measurable subsets $A, B \subseteq [0,1]$ the iterated integrals exist and are equal to each other: $$\int_A \int_B f(x,y) \,\textrm{d}y \,\textrm{d}x = \int_B \int_A f(x,y) \,\textrm{d}x \,\textrm{d}y.$$

Is $f$ necessarily $\lambda^2$-integrable over its domain, e.g. $[0,1]^2$?

Here is the explanation why I'm unsure about answer to the above question. SierpiĆski showed that there exists a set $A \subseteq [0,1]^2$ such that $A \not \in \mathfrak L^2$, but every section $A_x$ and $A^y$ is empty or consisting of just one point. Therefore the iterated integrals (again, from above) for $f = \chi_A$ exist, but $\chi_A$ isn't $\lambda^2$-integrable.

1 Answer

It is false that $f$ is necessarily integrable. Consider that if $f$ is non-negative, then Tonelli's Theorem applies, which yields the result of Fubini's Theorem for $f$ (see this Wikipedia page). But $f$ can fail to be integrable.

EDIT (to give counterexample): Write down $$f(x,y)=\frac1{({x^2+y^2})^{1/2}}.$$

By Tonelli's Theorem, the result of Fubini's applies. But a calculation gives $$\int\limits_0^1\int\limits_0^1|f(x,y)|\,dy\,dx\geq c\int\limits_0^{1/2}\frac1{r^2}r\,dr=c\ln r\Big|_0^{1/2}=+\infty,$$ where the first inequality holds since the integral of $f$ over the square $[0,1]^2$ is larger than or equal to the integral of $f$ over the upper quarter circle of radius $1/2$. So $f$ is not integrable.

• By "$\lambda^2$ integrable" I mean: integrable with respect to two-dimensional Lebesgue measure. – Santiago Aug 13 '16 at 15:56
• Please see the latest edit; I wrote down a counterexample. – Lentes Aug 13 '16 at 16:01
• What if I will add an assumption that the iterated integrals are finite? – Santiago Aug 13 '16 at 16:05
• Then I'm not sure in general. But, if $f$ is non-negative, then it is true. – Lentes Aug 13 '16 at 16:13
• I gave an (nonmeasurable) counterexample. – Santiago Aug 13 '16 at 16:29