Hypotheses in Fubini’s theorem cannot be dispensed with

In this post we proved Fubini’s theorem in the sense of Lebesgue measure, which makes it easier to evaluate multi variable integral. But these two classic counterexamples in this post prevent you from using Fubini’s theorem without enough consideration.

Counterexamples

So we said $f(x,y)$ has to be integrable. What if $f$ is not? First let’s see this function.

$f(x,y)=\frac{y^2-x^2}{(x^2+y^2)^2}=\frac{\partial^2}{\partial x \partial y}\arctan(\frac{y}{x})$

This function is not Lebesgue integrable on $[0,1] \times [0,1]$, since we have

$\begin{aligned} \iint\limits_{D} |f(x,y)|dxdy & = \int_\varepsilon^1 \int_0^{\frac{\pi}{2}}\frac{r^2|\cos{2\theta}|}{r^4}rd\theta{dr} \\ &=\int_\varepsilon^1\frac{1}{r}dr \\ &\to \infty \quad (\varepsilon \to 0)\end{aligned}$

where $D=\{(x,y):\varepsilon^2 \leq x^2+y^2 \leq 1, x \geq 0, y \geq 0\}$. Since $D \subset [0,1] \times [0,1]$, we see the integral of $|f(x,y)|$ is not finite. Fubini’s theorem fails then since

$\begin{aligned} \int_0^1 dx \int_0^1 f(x,y)dy &= \int_0^1 \frac{-1}{1+x^2}dx \\ &= -\frac{\pi}{4}\end{aligned}$

meanwhile

$\begin{aligned} \int_0^1 dy \int_0^1 f(x,y)dx &= \int_0^1\frac{1}{1+y^2}dy \\ &=\frac{\pi}{4}\end{aligned}$

See, everything messes up, and the equation disappears. This function is too ‘large’ for Fubini’s, so is the next one.

The following function is generated by series. First consider the sequence on $[0,1]$ generated by

$0= \delta_1<\delta_2<\cdots,\delta_n \to 1$

And a sequence of functions $g_n$ generated by $\int_0^1 g_ndx=1$ with supports in $(\delta_n,\delta_{n+1})$.

Define $f(x,y)$ on $[0,1] \times [0,1]$ such that

$f(x,y)=\sum_{n=1}^{\infty}[g_n(x)-g_{n+1}(x)]g_n(y)$

The right hand is convergent since for each point $(x,y)$ there is at least one term in this sum that is different from $0$.

An easy computation shows that

$\int_0^1\int_0^1 |f(x,y)|dxdy = \infty$

and

$\int_0^1 dx \int_0^1 f(x,y)dy = 1 \neq 0 = \int_0^1 dy \int_0^1 f(x,y)dx$