$\int_{0}^{\infty}f(x)dx$ absolutely converges, $\int_{0}^{\infty}g(x)dx$ conditionally converges what can be said about $\int_{0}^{\infty}f(x)g(x)dx$?
$\int_{0}^{\infty}f(x)dx$ absolutely converges$\rightarrow$ $\int_{0}^{\infty}f(x)dx$ converges
$\int_{0}^{\infty}g(x)dx$ conditionally converges $\rightarrow$ $\int_{0}^{\infty}g(x)dx$ converges
Because both $\int_{0}^{\infty}f(x)dx$ and $\int_{0}^{\infty}g(x)dx$ converges ,so from linearity of the improper integral (Is there linearity?) so do $\int_{0}^{\infty}f(x)g(x)dx$ converges
Is it a valid proof? (it seems to be very trivial)
If $g$ is bounded, you actually get that the integral of $fg$ is absolutely convergent. (Why?) Thus you should look to unbounded $g$ to find a possible counter-example.
Let us pick $f(x) = x^{-\frac{1}{2}}\chi_{[0,1]}(x) = g(x)$, then $f(x)g(x) = x^{-1}\chi_{[0,1]}(x)$. Each of $f$ and $g$ are absolutely integrable (prove this yourself). What do you know about the integral of $fg$?