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I am trying to come up with a measurable function on $[0,1]^2$ which is not integrable, but such that the iterated integrals are defined and unequal.

Any help would be appreciated.

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No such counterexample exists. Since $[0,1]$ is $\sigma$-finite. – JSchlather Dec 8 '12 at 18:51
@JacobSchlather Nonsense. Just look up Fubini's theorem on wikipedia for a standard counterexample. – Erick Wong Dec 8 '12 at 19:16
up vote 5 down vote accepted

Consider the double integrals $$I:=\int_0^1\int_0^1{y-x\over(2-x-y)^3}\ dy\ dx\ ,\qquad J:=\int_0^1\int_0^1{y-x\over(2-x-y)^3}\ dx\ dy\ .$$ Then $$\int_0^1{y-x\over(2-x-y)^3}\ dy={y-1\over(2-x-y)^2}\Biggr|_{y=0}^1={1\over(2-x)^2}\ .$$ It follows that $$I=\int_0^1 {dx\over(2-x)^2}={1\over 2-x}\Biggr|_0^1={1\over2}\ .$$ Similarly you get $J=-{1\over2}\ne I$.

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Did you mean to integrate these over the unit square? Not that it makes a big difference in the result :). – Erick Wong Dec 8 '12 at 22:09

$$ \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dy\,dx \ne \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx\,dy $$

Obviously either of these is $-1$ times the other and if this function were absolutely integrable, then they would be equal, so their value would be $0$. But one is $\pi/2$ and the other is $-\pi/2$, as may be checked by freshman calculus methods.

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