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What would be an example of $f$ s.t. $\int_0^1 \int_0^1 f dx dy = \int_0^1 \int_0^1 f dx dy < \infty$ but the double integral $$\int_{[0,1]^2} f \ d(x,y) = \infty$$

Or maybe there doesn't exist such $f$ (which I doubt)? The integrals are Lebesgue.

user16015
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  • an $f$ with (large) discontinuities inside the unit rectangle maybe? – Nikos M. Sep 12 '15 at 00:48
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    What if $\int_0^1 f dx$ is actually always zero but only gets that way by "catastrophic cancellation". For example $f(x,y)=\operatorname{sign}(x-1/2)/(y-1/2)$. Or do you specifically want the double integral to be $+\infty$ rather than indeterminate? That much I suspect is impossible, since if the positive or negative part has finite integral, then I think you can "remove" it and then be able to apply Tonelli (and then Tonelli lets you apply Fubini). – Ian Sep 12 '15 at 00:56
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    I want the double integral to be $\infty$... Actually I would like to know if the two iterated integrals agree and are finite does that imply that the double integral is also finite? – user16015 Sep 12 '15 at 01:03
  • @user16015: No, it does not. It can still happen that the double integral does not exist at all. But if you insist on the double integral having a definite value (even $\pm\infty$), then the iterated integrals will have the same value as (a slight modification) of my answer shows. – PhoemueX Sep 12 '15 at 03:13

1 Answers1

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I will assume that $f$ is measurable (otherwise the double integral does not make sense).

In this case, there is no counterexample, since your assumption on the double integral implies $\int_{[0,1]^2} f_+ d(x,y)=\infty$ and $\int_{[0,1]^2} f_- d(x,y)<\infty$ (where $f_+$ and $f_-$ denote the positive/negative parts of $f$).

But now, the Fubini-Tonelli theorem yields

\begin{eqnarray*} \int_{[0,1]}\int_{[0,1]} f_+ (x,y) \,dx\,dy =\infty\\ \int_{[0,1]}\int_{[0,1]} f_-(x,y) \,dx\,dy <\infty \end{eqnarray*} and analogously for the other ordering of integrals. In particular, the "inner" integral over $f_-$ is finite for almost all $y$.

Adding these two identities shows $$ \int_{[0,1]}\int_{[0,1]} f(x,y) \, dy\, dx=\infty $$ and likewise for the other ordering.

PhoemueX
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