Justfication of the summability of $XY$ where $X$ and $Y$ are summable independent random variables. 
Suppose $\Omega$ is a countable sample space, $P$ a probability measure on $\Omega$, and $X$ and $Y$ are independent random variables on $\Omega$ with $E(|X|)<\infty$, $E(|Y|)<\infty$. Show that $E(|XY|)<\infty$.

By definition,
$$E(X)=\sum_{\omega\in\Omega}X(\omega)P(\omega)\text{.}$$
It is well-known that if a series converges absolutely then every rearrangement of it converges to the same limit, and that a sequence diverges to $\infty$ if and only if every subsequence of it diverges to $\infty$. Hence it suffices to show that there exists arrangement of $\Omega=\{\omega_1,\omega_2,\omega_3,\dots\}$ and a grouping of the terms such that
$$[XY(\omega_1)P(\omega_1)+\cdots+XY(\omega_{N_1})P(\omega_{N_1})]+[XY(\omega_{N_1+1})P(\omega_{N_1}+1)+\cdots XY(\omega_{N_2})P(\omega_{N_2})]+\cdots<\infty\text{.}$$
Let $\{x_n\}$ and $\{y_n\}$ be the sequence of values taken by $X$ and $Y$, respectively.  Arrange $\Omega$ in a sequence $\{\omega_n\}$ so that ...
This is what I tried. I don't know how to proceed any further. For example how to arrange the points so that $\sum_n X(\omega_n)Y(\omega_n)P(\omega_n)=\sum_j\sum_k x_jy_k P(X=x_j,Y=y_k)$?
 A: As $X$ and $Y$ are independent, the joint distribution satisfies
$$
F_{X,Y}(x,y) = F_X(x)F_Y(y),\ x,y\in\mathbb R.
$$
It follows that $$\mathsf P_{X,Y} = \mathsf P_X\times \mathsf P_Y,$$ and as the product of $\sigma$-finite measures is again a $\sigma$-finite measure, from Tonelli's theorem it follows that (as $(x,y)\mapsto |xy|$ is a non-negative measurable map)
\begin{align}
\mathbb E[|XY|] &= \int_{\Omega^2} |X(\omega)Y(\omega)|\ \mathsf d\mathbb P(\omega)\\
&= \int_{\mathbb R^2} |xy|\ \mathsf dF_{X,Y}(x,y)\\
&= \int_{\mathbb R}{\int_\mathbb R}|xy|\ \mathsf d F_X(x)\ \mathsf d F_Y(y)\\
&= \int_{\mathbb R}|x|\mathsf dF_X(x) \int_{\mathbb R}|y|\ \mathsf dF_Y(y)\\
&= \mathbb E[|X|]\mathbb E[|Y|]\\
& < \infty.
\end{align}
A: Since $X$ and $Y$ are independent, we get that $|X|$ and $|Y|$ are independent (used in the middle step) therefore
\begin{align*}
\mathbb E[|XY|]&=\mathbb E[|X| \cdot|Y|]\\
&=\mathbb E[|X|]\cdot\mathbb E[|Y|]\\
&<\infty
\end{align*}

If $U$ and $V$ are independent discrete random variable, then $\mathbb E[UV]=\mathbb E[U]\cdot\mathbb E[V]$, indeed
\begin{align*}
\mathbb E[UV] &= \sum_{u,v} u v p_{U,V}(u,v)\\
&=\sum_{u,v} uv p_U(u) p_V(v)\\
&=\sum_u up_U(u) \sum_v vp_V(v)\\
&=\mathbb E[U]\cdot\mathbb E[V]
\end{align*}
This is true also when $U$ and $V$ are not discrete (see @Math1000 's answer).
