# Negative part of the integrand in an iterated integral

Hi everyone: Suppose that $(X,\mathfrak{M},\mu)$ and $(Y,\mathfrak{N},\nu)$ are two measure spaces and $f(x,y)$ is an extended real valued measurable function on $X\times Y$. Suppose we can not apply Fubini to $f$ but we know that the two iterated integrals $$\int_{X} \int_{Y}fd\nu d\mu$$ and $$\int_{Y} \int_{X}fd\mu d\nu$$ exist in $(-\infty,+\infty]$. Can we conclude that the iterated integrals (or at least one of them) of $f^{-}$ is less that $+\infty$, i.e. $$\int_{X} \int_{Y}f^{-}d\nu d\mu<+\infty?$$ Thanks for your help.

• Thanks for your reply. Are you saying that the existence of the iterated integrals doesn't say anything about the integral of the positive and negative parts? Can we have a case where both iterated integrals are finite but the product integral is $\infty-\infty$? – user127474 Aug 6 '14 at 3:44
• You should have commented on my answer, so I'd be notified. As I wrote there, every time the iterated integrals are unequal, the $\infty-\infty$ situation occurs. For a concrete example with finite iterated integrals, see Wikipedia: Failure of Fubini's theorem for non-integrable functions. – user147263 Aug 8 '14 at 17:26

No, it's rather the opposite. Assuming the spaces are $\sigma$-finite and $f$ is measurable, Tonelli's theorem applies to both $f^+$ and $f^-$, and asserts the equalities $$\iint_{X\times Y} f^+=\int_X\int_Y f^+ =\int_Y\int_X f^+ \tag1$$ $$\iint_{X\times Y} f^-=\int_X\int_Y f^- =\int_Y\int_X f^-\tag2$$ (where the integrals could be $+\infty$).
So, how can equality fail for $f$? Only because of both (1) and (2) being infinite, so that their subtraction produces "$+\infty-\infty$" which can mean anything or nothing. If just one of the integrals in (1)-(2) is finite, we have the conclusion of Fubini's theorem already.
Here is a concrete example, taken from Wikipedia article Failure of Fubini's theorem for non-integrable functions: $$\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}$$ Both iterated integrals are finite and they are unequal. Both the positive and negative parts of the function integrate to $\infty$.