Product of two random variables How can one show that the product $X \cdot Y$ of two real-valued random variables $X,Y$ is again a random variable?
We can fix some set generating the Borel sigma algebra on the real line, then take for instance an arbitrary open interval, and consider $(X \cdot Y)^{-1}((a,b))$. We need to show it belongs to the sigma algebra on the underlying space $\Omega$.
We could take any $c$ in the interval, write $c = f \cdot \frac{c}{f}$, and consider $X^{-1}(f) \cap Y^{-1}(c/f)$, then take a union over $f$, and then again over $c$. But these are uncountable unions, so the argument doesn't work.
 A: Assuming that you already know that sums and constant multiples of random variables are again random variables, then all we really need to know is that the square of a random variable is a random variable, using the fact that $$XY=\frac14\bigl((X+Y)^2-(X-Y)^2\bigr).$$
As for how to prove that, suppose that $X$ is a random variable on a set $S$. For $\alpha\in\Bbb R,$ we have $$(X^2)^{-1}\bigl((\alpha,+\infty)\bigr)=\begin{cases}S & \text{if }\alpha<0\\X^{-1}\bigl((\sqrt\alpha,+\infty)\bigr)\cup(-X)^{-1}\bigl((\sqrt\alpha,+\infty)\bigr) & \text{if }\alpha\ge 0.\end{cases}$$ Then $(X^2)^{-1}\bigl((\alpha,+\infty)\bigr)$ is measurable for all $\alpha\in\Bbb R,$ and so $X^2$ is a random variable on $S$ whenever $X$ is.
A: If you already know that if $X$ and $Y$ are random variables and $r$ a real number, then $X+Y$ and $rX$ are random variables too, then you can do the following: Show that if $X$ is a random variable, so is $X^2$. Then you can use the fact that $XY=1/4[(f+g)^2-(f-g)^2]$ to get the result.
There is a slightly more messy but less sneaky approach: Let $X$ and $Y$ be both nonnegative random variables and note that if $X(\omega)Y(\omega)<c$, then there are
nonnegative rational numbers $r_1,r_2$ with $X(\omega)\leq r_1$, $Y(\omega)\leq r_2$ and $r_1r_2<c$. Let $R_c$ be the set of all pairs of nonnegative rational numbers with product smaller than $c$ and note that $R_c$ is countable. Then $$(XY)^{-1}\big((-\infty,c)\big)=\bigcup_{(r_1,r_2)\in R_c}X^{-1}\big((-\infty,r_1]\big)\cap Y^{-1}\big((-\infty,r_2]\big).$$
A similar approach works with general random variables, but you have to take care of the signs. 
