Iterated Integrals - "Counterexample" to Fubini's Theorem Showing the iterated integrals $$\int_{[0,1]}\left[\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}dx\right]dy\quad\text{and}\quad\int_{[0,1]}\left[\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}dy\right]dx$$ are different isn't too hard, so this certainly implies $f(x,y)\notin L^1([0,1]^2)$ (otherwise it would be a counterexample to Fubini's Theorem), so how do we show that $$\iint_{[0,1]^2}|f|\;dx\:dy=\infty?$$ I've been thinking about this, but my intuition always leads me back to iterated integrals. Any help would be appreciated in guiding me in which direction I should take.
 A: As Deven Ware says, this follows from the contrapositive of Fubini. But we can also show it directly. 
Let $D$ be the "pie wedge" shape $\{ (x,y) : x^2+y^2 \leq 1,\  x \geq y \}$. This is an eighth of a circle, contained in the unit square. We'll show that $\int_D \left| \frac{x^2-y^2}{(x^2+y^2)^2} \right| dx dy$ is infinite; the integral over the square will then be even bigger. On the domain $D$, we have $x \geq y$, so the absolute value sign has no effect. We now switch to polar coordinates:
$$\int_D \frac{x^2 - y^2}{(x^2+y^2)^2} dx dy = \int_{r=0}^1 \int_{\theta=0}^{\pi/4} \frac{r^2 (\cos^2 \theta - \sin^2 \theta)}{r^4} r dr d \theta.$$
(If your function was in $L^1$, then I believe this switching of coordinates would be legitimate. I must admit that it's been a long time since I took real analysis though, so check me.) 
So we want to compute
$$\int_0^1 \frac{dr}{r} \int_{\theta=0}^{\pi/4} \cos ( 2 \theta) d \theta.$$
The second factor is $1/2$, so you have $\frac{1}{2} \int_{r=0}^1 dr/r$, which is plainly infinite.
For another approach, see Wikipedia.
