By Fubini-Tonelli theorem, if one of $\int(\int |f(x,y)| dy)dx$, $\int(\int |f(x,y)| dx)dy$, $\int\int |f(x,y)| dxdy$ is finite , then $\int(\int f(x,y) dy)dx = \int(\int f(x,y) dx)dy=\int \int f(x,y) dxdy$. Does this imply that the following cases are impossible,

(i)Both of the iterated integrals exist and agree but the double integral does not exist.

(ii)At least one of the iterated integrals is finite but they are not equal.

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    $\begingroup$ If one of those things on your first line is finite, then both your cases are impossible. It doesn't say what happens when those things on your first line are infinite. $\endgroup$ – GEdgar Oct 31 '15 at 21:51
  • $\begingroup$ @GEdgar So if one of the iterated integral is finite then the double integral has to exist? What about $\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}$? $\endgroup$ – user136592 Oct 31 '15 at 21:57
  • $\begingroup$ @user136592 The integral is not absolutely convergent. To see, calculate it e.g. over the upper half square $\int_0^1\int_0^y...dxdy$. $\endgroup$ – A.Γ. Oct 31 '15 at 22:33
  • $\begingroup$ @A.G. But the iterated integrals exist and agree in this case right? $\endgroup$ – user136592 Oct 31 '15 at 22:36
  • $\begingroup$ @user136592 Even iterated integrals are not agree here: they have the opposite signs $\int\int...dydx=\pi/4$, $\int\int...dxdy=-\pi/4$. $\endgroup$ – A.Γ. Oct 31 '15 at 22:39

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