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 
$$
\int_{X \times Y}f(x,y)dxdy = \int_{X}dx\int_{Y}f(x,y)dy = \int_{Y}dy\int_{X}f(x,y)dx.
$$
 A: First, your desired statement is true, provided that the following conditions are fulfilled


*

*$f \geq 0$,

*$f$ is measurable (w.r.t. the product $\sigma$-algebra),

*$X,Y$ 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.
