Let $f$ be a continuous, real-valued function on $I=[0, \infty)$ with $\mathrm{sup}_{x\in I} | \int_0^x f|<\infty$ and $g$ be a differentiable, real-valued function on $I$ sith $\lim_{x\to\infty}g(x)=0$. Prove that if $g'\le0$ and $g'$ is integrable on each compact interval in $I$, then $\int_0^\infty fg$ converges.
My attempt: Let $F(x)=\int_0^x f$ and by integration by parts, $\int_0^\infty fg=Fg|_0^\infty - \int_0^\infty Fg'=-\int_0^\infty Fg'$ since $F$ is bounded, $F(0)=0$, $g$ goes to 0 as $x\to\infty$.
Since $F$ is bounded, I guess this would help to prove the convergence, but I cannot proceed further. What should I do?
Hint. Note that, since $F$ is bounded and $g'\leq 0$ then $$\int_0^\infty |F(x)(-g'(x))|dx=\int_0^\infty |F(x)|(-g'(x))dx\leq M\int_0^\infty (-g'(x))dx\\=M[-g(x)]_0^{\infty}=Mg(0).$$