If two functions/random variables are integrable and independent, then their product is integrable. What if they are not independent? What is an example?
What I tried:
Let $X, Y \in \mathscr L^{1}(\Omega, \mathscr F, \mathbb P)$.
Consider $X$ and $X - Y$ w/ $X$ having an infinite second moment but finite first moment.
Then
$$E[X(X-Y)] = E[X^2] - E[XY] = \infty$$
assuming $-\infty \le E[XY] < \infty$ and indeterminate otherwise. In either case, $X(X-Y)$ is not integrable.
An example is $X$ having a student-t distribution with two degrees of freedom and $Y$ can be anything integrable, I guess.
If that's right, what are examples are better/simpler (in your opinion) or more well-known (from what you've observed)?
If that's wrong, which part is wrong, why, and please provide examples.
Let $U$ be a random variable which is uniform on $(0,1]$. Let $X=Y=U^{-1/2}$. Then $E[X]=E[Y]=2$ but $E[XY]=+\infty$.
You can make lots of examples of functions like this, which are not integrable but their square root is integrable, because it diverges "more slowly".
On an infinite measure space you can also have the opposite phenomenon with tailing: $x^{-2}$ is integrable over $[1,\infty)$ but its square root is not, because it decays more slowly. But this is impossible on a finite measure space.