If X, Y are random variables and there exists a constant $c>0$ so that $P(|X| \geq x) + P(|Y| \geq x) \leq e^{-cx}$ for all x > 0, then $E[X^m Y^n]$ is integrable for all nonnegative integers m, n.
I tried coming up with some sort of argument by contradiction, supposing that $E[X^m Y^n] = \infty$ for some nonnegative pair of integers m, n; then by Cauchy-Schwarz I know that $E[X^m]E[Y^n] \geq E[X^m Y^n] = \infty$ as well, but I'm stuck on how to turn either $E[X^m Y^n]$ or $E[X^m]E[Y^n]$ into something involving the given inequality.
I think $P(|X| \geq x) + P(|Y| \geq x) = E[\chi_{|X| \geq x}] + E[\chi_{|Y| \geq x}]$, but I'm not sure whether that's actually right, and if so, whether or not it has any useful relationship to the product of expectations (or expectation of products) that I'm actually trying to bound.
First $|X^mY^n|\leqslant|X|^{2m}+|Y|^{2n}$ hence it is enough to show that, if $Z\geqslant0$ almost surely and if $$P(Z\geqslant z)\leqslant\mathrm e^{-cz},$$ for every nonnegative $z$ then $E(Z^k)$ is finite for every $k$. To prove this, note that, for every positive $b$, $$\mathrm e^{bz}=\sum\limits_{i\geqslant0}b^i\frac{z^i}{i!}\geqslant b^k\frac{z^k}{k!},$$ hence $Z^k\leqslant \frac{k!}{b^k}\mathrm e^{bZ}$ and it suffices to show that $E(\mathrm e^{bZ})$ is finite for some positive $b$. Now, $$E(\mathrm e^{bZ})=1+b\int_0^\infty\mathrm e^{bz}P(Z\geqslant z)\mathrm dz,$$ hence the bound on the CDF of $Z$ implies that $E(\mathrm e^{bZ})$ is indeed finite for every $b\lt c$.