Exchanging the order of integration in improper double integrals My question arise from  Improper Riemann Integrals by Ioannis M. Roussos


Why the improper double integral is absolutely convergent,then we can exchange order of integration in improper Riemann sense? I really don't understand that it said  $\textbf{Condition II}\ \Longleftrightarrow \textbf{Condition III} \Longleftrightarrow \textbf{Condition IV}$ and each one of them can ensure the equalities
$$ \iint_{\mathbf{R}^2} f(x,y) \, \mathrm{d}x \, \mathrm{d}y
= \int_{-\infty}^\infty \left[ \int_{-\infty}^\infty f(x,y) \, \mathrm{d}x \right] \, \mathrm{d}y
= \int_{-\infty}^\infty \left[ \int_{-\infty}^\infty f(x,y) \, \mathrm{d}y \right] \, \mathrm{d}x $$
hold. This has been bothering me for a long time！Any help you can provide would be greatly appreciated!

I got a counterexample function $f(x,y)$ as following which is discontinuous on $[0,1]\times[0,1].$
$f(x,y)$ is defined on $\left [0,1  \right ] \times\left [ 0,1 \right ] $
$$f(x,y)=\begin{cases}
  2^n,\quad x=\frac{2m-1}{2^n}\text{ and }0<y\le\frac{1}{2^n}\\
\quad \left ( n=1,2,\ldots;m=1,2,\ldots,2^{n-1} \right ) ,\\
  0,  \text{ otherwise}.
\end{cases}$$
$\textbf(1).$ For any $\varepsilon>0,$ $$\iint_{[0,1]\times[\varepsilon ,1]} f(x,y)\,dx\,dy=0 \Rightarrow $$$$\iint_{[0,1]\times[0 ,1]} f(x,y) \, dx\, dy = \lim_{\varepsilon \to 0} \iint_{[0,1]\times[\varepsilon ,1]} f(x,y) \, dx \, dy = 0.$$
$\textbf(2).$ For any fixed $y\in\left [ 0,1 \right ],$ $$ \int_0^1 f(x,y) \, dx=0\Rightarrow \int_0^1 dy \int_0^1 f(x,y) \, dx = 0.$$
$\textbf(3).$
When $x\in[0,1]$ but $x\ne\frac{2m-1}{2^n}\left ( n=1,2,\ldots;m=1,2,\ldots,2^{n-1} \right ) ,$ we get $\int_0^1 f(x,y) \, dy = 0;$
When $x\in[0,1]$ and $x=\frac{2m-1}{2^n}\left ( n=1,2,\ldots;m=1,2,\ldots,2^{n-1} \right ) ,$ we get $$\int_0^1 f(x,y) \, dy = \int_0^{\frac{1}{2^n}}f(x,y) \, dy=1.$$
So $$ \int_0^1 dx\int_0^1 f(x,y) \, dy$$ does not exist in Riemann sense!

I am definitely not disagreeing with Fubini's Theorem. Actually,I want find a counterexample/example  $f(x,y)$ is bounded or unbounded,continuous  on some region $\mathcal{R}$ which is bounde or unbounded and not necessarily closed or open and the function $f(x,y)$  satisfies the $\textbf{Condition IV}$,
but
$$ \text{ the improper double integral } \iint_{\mathbf{R}^2}  \lvert f(x,y) \rvert \, \mathrm{d}x \, \mathrm{d}y < \infty \nRightarrow \int_{-\infty}^\infty \left[ \int_{-\infty}^\infty \lvert f(x,y) \rvert \, \mathrm{d}x \right] \mathrm{d}y < \infty ;$$
and $$\text{ the improper double integral } \iint_{\mathbf{R}^2}  \lvert f(x,y) \rvert \, \mathrm{d}x \, \mathrm{d}y < \infty \nRightarrow \int_{-\infty}^\infty \left[ \int_{-\infty}^\infty \lvert f(x,y) \rvert \, \mathrm{d}y \right] \, \mathrm{d}x < \infty.$$
further,the the equalities
$$ \iint_{\mathbf{R}^2} f(x,y) \, \mathrm{d}x \, \mathrm{d}y
= \int_{-\infty}^\infty \left[ \int_{-\infty}^\infty f(x,y) \, \mathrm{d}x \right] \, \mathrm{d}y
= \int_{-\infty}^\infty \left[ \int_{-\infty}^\infty f(x,y) \, \mathrm{d}y \right] \, \mathrm{d}x $$ can not hold.
 A: Further clarification here is warranted.
Note that
$$f(x,y) = (2xy-2x^3y^3)e^{-x^2y^2} = \frac{\partial}{\partial x}x^2ye^{-x^2y^2} = \frac{\partial}{\partial y}xy^2e^{-x^2y^2}$$
Hence,
$$\int_0^\infty f(x,y)\, dx = \left.x^2ye^{-x^2y^2} \right|_{x=0}^{x\to\infty} = 0, \quad \int_0^1 f(x,y)\, dy = \left.xy^2e^{-x^2y^2} \right|_{y=0}^{y=1} = xe^{-x^2},$$
and
$$\int_0^1 \left(\int_0^\infty f(x,y)\, dx \right)  dy=0, \quad \int_0^\infty \left(\int_0^1 f(x,y)\, dy \right)  dx=\int_0^\infty xe^{-x^2} \, dx= \frac{1}{2}$$
Thus, $f$ is continuous on $[0,\infty] \times [0,1]$, all improper integrals converge, but the order of integration cannot be switched. However, the condition $\int_0^\infty \int_0^1 |f(x,y)| \, d(x,y) < \infty$ does not hold so this is not a counterexample to the revised question.
Also, you have given an example (and it is not difficult to find others)  where the improper double Riemann integral of $|f|$ converges but the iterated integrals are not equal (because one does not exist as an improper Riemann integral).  However, the function $f$ is discontinuous in these cases.
There remains the question of finding a counterexample where $f$ is absolutely integrable, $f$ is continuous, and the iterated integrals are unequal and/or do not exist as improper Riemann integrals.  If it is true in the sense of the improper Riemann integral, that
$$\int \int_{\mathbb{R}^2}|f(x,y)| \, d(x,y)< \infty,$$
and in addition $f$ is continuous, then  both Riemann and Lebesgue integrals exist and coincide over finite regions. It also follows that the double integral over $\mathbb{R}^2$ exists in the sense of Lebesgue  and Fubini’s theorem implies equality of the iterated integrals of $f$ also in the sense of Lebesgue. The question then reduces to whether or not
$$\int_{-\infty}^\infty \left( \int_{-\infty}^\infty f(x,y) \, dy \right)dx, \quad\int_{-\infty}^\infty \left( \int_{-\infty}^\infty f(x,y) \, dx \right)dy$$
can fail to exist as iterated improper Riemann integrals (or exist and be unequal).  I suspect -- at this moment -- when $f$ is both continuous and absolutely integrable, that it can be shown using the DCT that this cannot be true.
