Find an integrable $g(x,y) \ge |e^{-xy}\sin x|$ I want to use Fubini theorem on $$\int_0^{A} \int_0^{\infty} e^{-xy}\sin x dy dx=\int_0^{\infty} \int_0^{A}e^{-xy}\sin x dx dy$$
Must verify that $\int_M |f|d(\mu \times \nu) < \infty$. I'm using the Lebesgue theorem, so far I've come up with $g(x,y)=e^{-y}$ but am not sure whether it's correct.
My argument is that if $x\in (0,1)$ then the $\sin x$ part is going to ensure that the inequality holds.

Try $g(x,y)=x\mathrm e^{-xy}$, then $|\mathrm e^{-xy}\sin x|\leqslant g(x,y)$ for every nonnegative $x$ and $y$. Furthermore, $\int\limits_0^\infty g(x,y)\mathrm dy=1$ for every $x\gt0$ hence $\int\limits_0^A\int\limits_0^\infty g(x,y)\mathrm dy\mathrm dx=A$, which is finite.
