# Why does Fubini's theorem not hold/apply to this function?

Given the function $$f(x,y) = \begin{cases} e^{y-x}, & x > y \geq 0 \\ -e^{x-y}, & 0 \leq x \leq y \end{cases}$$ I have already determined that $$\int_0^\infty \left( \int_0^\infty f(x,y) \, dx \right) dy = 1$$ and $$\int_0^\infty \left( \int_0^\infty f(x,y) \, dy \right) dx = -1.$$ So it would appear that Fubini does not hold, however, $f(x,y)$ seems to be Lebesgue integrable on $R^2$ since it starts at $1$ or $-1$ on the diagonal and tends to zero away from the diagonal. What is the true issue in applying Fubini's here?

• Did you check if $\int_{\mathbb{R}^2} |f(x,y)| < \infty$? – Batman Apr 22 '15 at 2:14
• Words like "seems to be" may have a hint of where to look. – GEdgar Apr 22 '15 at 2:16

You should expect that Fubini's theorem might fail because the integral of $|f|$ is infinite. Here's a geometrically intuitive reason for that.
Considering $\varepsilon=1/2$, by continuity of $\exp$ there is a $\delta>0$ so that $|f(x,y)|>1/2$ for $(x,y)$ within $\delta$ of the diagonal. Now this $\delta$-band around the diagonal has infinite measure, and the integrand is at least $1/2$ inside it. Something very similar happens in the classic examples of failures of Fubini's theorem on $(\mathbb{N} \times \mathbb{N},c \times c)$.
• Is $c \times c$ in your last sentence the counting measure? Do you have any good literature on that? – mathjacks Apr 22 '15 at 11:11
• @mathjacks $c \times c$ is the product counting measure, which as it happens is also the counting measure on $\mathbb{N} \times \mathbb{N}$ itself. I think this is discussed in the Fubini/Tonelli section in Royden and Fitzpatrick, although it might have been confined to an exercise. – Ian Apr 22 '15 at 11:12
$$\int\limits_{\{(x,y): x>y>0\}} f(x,y)\,d(x,y) = \int_0^\infty \left( \int_y^\infty e^{y-x} \, dx\right) \, dy = \int_0^\infty 1\,dy = \infty.$$