One Question about the Fubini's Theorem

The Fubini's Theorem says:

If function $f:X \times Y \rightarrow R$ is integrable over $X \times Y$, then $$ \int_{X \times Y}f(x,y)dxdy = \int_{X}dx\int_{Y}f(x,y)dy = \int_{Y}dy\int_{X}f(x,y)dx. $$

My question is: If $\int_{X}dx\int_{Y}f(x,y)dy = \int_{Y}dy\int_{X}f(x,y)dx$, could we say $f:X \times Y \rightarrow R$ is integrable over $X \times Y$?

If so, I can get: If $\int_{X}dx\int_{Y}f(x,y)dy = \int_{Y}dy\int_{X}f(x,y)dx$, then $$ \int_{X \times Y}f(x,y)dxdy = \int_{X}dx\int_{Y}f(x,y)dy = \int_{Y}dy\int_{X}f(x,y)dx. $$

Solution 1:

First, your desired statement is true, provided that the following conditions are fulfilled

1. $f \geq 0$,
2. $f$ is measurable (w.r.t. the product $\sigma$-algebra),
3. $X,Y$ are $\sigma$-finite measure spaces.

The exact statemement (sometimes called Tonelli's theorem, or the Fubini-Tonelli-theorem) is that if

$$ \int \int f(x,y) \, dx\, dy < \infty, $$

then $f$ is integrable w.r.t. the product measure and

$$ \int \int f(x,y) \, dx \, dy = \int \int f(x,y) \, dy \, dx. $$

If one of first two conditions fail, one can construct counterexamples. This is probably also true in the non $\sigma$-finite case, but I will have to think about that.

Regarding the first condition, consider (this is a classical counterexample)

$$ f : \Bbb{R} \times \Bbb{R} \to \Bbb{R}, (x,y) \mapsto \frac{xy}{(x^2 + y^2)^2}, $$

where both spaces $\Bbb{R}$ are equipped with the usual Lebesgue measure.

It is then easy to see that for each $x \neq 0$, we have

$$ \int f(x,y) \, dy = 0 \text{ (in particular, the integral exists)}. $$

This is based on the fact that easy estimates show the existence of the integral and that $f(x, -y) = - f(x,y)$.

Likewise,

$$ \int f(x,y) \, dx = 0 \text{ for each fixed } y \neq 0. $$

Together, this shows

$$ \int \int f(x,y) \, dx \, dy = 0= \int \int f(x,y) \, dy \, dx, $$

but $f$ is not integrable over $\Bbb{R}^2$, because introduction of polar coordinates shows

\begin{eqnarray*} \int_{\mathbb{R}^{2}}\left|f\left(x,y\right)\right|\, d\left(\begin{matrix}x\\ y \end{matrix}\right) & = & \int_{0}^{2\pi}\int_{0}^{\infty}r\cdot\frac{\left|r\cdot\cos\left(\varphi\right)\cdot r\cdot\sin\left(\varphi\right)\right|}{\left(r^{2}+r^{2}\right)^{2}}\,{\rm d}r\,{\rm d}\varphi\\ & = & \underbrace{\int_{0}^{2\pi}\left|\cos\left(\varphi\right)\sin\left(\varphi\right)\right|\,{\rm d}\varphi}_{>0\text{ because integrand continuous and not identically }0}\cdot\underbrace{\int_{0}^{\infty}\frac{r^{3}}{4r^{4}}\,{\rm d}r}_{=\infty}=\infty. \end{eqnarray*}

For the second condition (measurabiliy w.r.t. the product $\sigma$-algebra), there is a striking example constructed by Sierpinski (see https://eudml.org/doc/212592, but the paper is in french), which shows the following (quote from Folland's "Real Analysis"):

Using the axiom of choice, but not the continuum hypothesis, Sierpinski has proved the existence of a Lebesgue nonmeasurable subset of $\Bbb{R}^2$, whose intersection with any straight line contains at most two points.

Let us call this set $M$. The properties mentioned above imply that $\chi_M (x,y) = 0$ for each fixed $x$ and all but (at most) two points $y$ (depending on $x$). A similar statement holds for each fixed $y$. Here, $\chi_M$ is the characteristic function/indicator function of the set $M$. This yields

$$ \int \int \chi_M \, dx \, dy = 0 = \int \int \chi_M \, dy \, dx, $$

but $\chi_M$ is not integrable w.r.t. the product measure, because it is not even measurable w.r.t. the product (even the Lebesgue) $\sigma$-algebra.