Is there any example which can show that the change of the order of integration is not plausible when the improper integral exists? Let $f(x,y)$ be continuous on $Q=\{(x,y)\mid x>0,y>0\}$ and $\iint_Q |f(x,y)|$ converge.
 $$\int_{0}^{\infty}\int_{0}^{\infty}f(x,y)\, \mathrm{d}x\, \mathrm{d}y \stackrel{?}{=}\int_{0}^{\infty}\int_{0}^{\infty}f(x,y)\, \mathrm{d}y\, \mathrm{d}x$$
Is there any example to show that the equality is wrong?
 A: No, because under the assumptions you gave, you are allowed to interchange the order of integration by Fubini-Tonelli's theorem and the equality will always be true.
If an integral exists absolutely, you are allowed to interchange the order of the integrals as you want. See here: https://en.wikipedia.org/wiki/Fubini%27s_theorem#Fubini%E2%80%93Tonelli_theorem
Edit: It does not matter in this case whether the integrals are understood as Lebesgue-integrals or improper Riemann-integrals. As $\iint_Q |f(x,y)|$ converges, we know that Fubini holds and 
$$\int_{[0,\infty)}\int_{[0,\infty)}f(x,y)\, \mathrm{d}x\, \mathrm{d}y =\int_{[0,\infty)}\int_{[0,\infty)}f(x,y)\, \mathrm{d}y\, \mathrm{d}x$$
as Lebesgue-integrals. But if an integral exists in the sense of Lebesgue, then the corresponding improper Riemann integral exists too and they are equal. You can prove this for example by using dominated convergency and $|f(x,y)|$ as the dominating fuction:
$$\int_{0}^{\infty}\int_{0}^{\infty}f(x,y)\, \mathrm{d}x\, \mathrm{d}y = \lim_{m \to \infty}\int_{0}^m \lim_{n \to \infty}\int_{0}^n f(x,y) dx dy = \lim_{m \to \infty}\int_{[0,m]} \lim_{n \to \infty}\int_{[0,n]} f(x,y) dx dy = \lim_{m \to \infty}\int_{[0,\infty)} \lim_{n \to \infty}\int_{[0,\infty)} f_{m,n}(x,y)dx dy = \int_{[0,\infty)} \int_{[0,\infty)} \lim_{m \to \infty} \lim_{n \to \infty} f_{m,n}(x,y)dx dy = \int_{[0,\infty)} \int_{[0,\infty)} f(x,y)dxdy$$
where 
$$f_{m,n}(x,y) =
\begin{cases}
 f(x,y), \ \ x \leq n, y \leq m\\
 0, \ \ \ \ \ \ \ \ \ \ \ otherwise
\end{cases}. $$
The Lebesgue-integrals are the ones where the range is denoted as a set.
