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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.

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  • $\begingroup$ 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 $\infty-\infty$? $\endgroup$ – user127474 Aug 6 '14 at 3:44
  • $\begingroup$ 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. $\endgroup$ – user147263 Aug 8 '14 at 17:26
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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.

Stated as a contrapositive: if the equality of iterated integrals fails (be they finite or infinite), all integrals in (1)-(2) are infinite.

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 $\infty$.

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  • $\begingroup$ Why is this strongly downvoted, but with no comments? Is it wrong? $\endgroup$ – Katie Aug 7 '14 at 1:34
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    $\begingroup$ @Katie Long story. The answer is correct. $\endgroup$ – user147263 Aug 7 '14 at 1:37
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    $\begingroup$ The downvotes are entirely in regards to your 'vigilance' mission? I don't understand how someone could disagree with deleting terrible questions(I likely have a few myself), even if they have attached good answers. Keep up the good work with both your answers and your efforts. $\endgroup$ – Katie Aug 7 '14 at 2:18

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