# 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.

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So you're asking how you can show this without using Fubini's theorem? –  Deven Ware Dec 14 '12 at 5:33
No, it's an old exam question, and they ask why it doesn't contradict Fubini's theorem. According to my text, the only hypothesis on Fubini's theorem is that the function be measurable (which this clearly is) and the function be Lebesgue integrable. So my question is how to show the integral isn't finite. –  Clayton Dec 14 '12 at 5:36
Yes, so the contrapositive of Fubini's theorem says the function is not integrable. On the Wikipedia page for "Fubini's theorem" they also give an outline of showing this without Fubini. –  Deven Ware Dec 14 '12 at 5:41
@DevenWare: Thank you! I think that is exactly what I wanted, if I understand correctly. The discussion just underneath the portion "without Fubini's Theorem" actually evaluates the integral without regards to order. I appreciate it! –  Clayton Dec 14 '12 at 5:47
–  Nana Dec 14 '12 at 5:47

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.