It is given that (1) $\int_a^x g(s) \, ds \geq 0$ for $a \leq x \leq b$ and (2) $\int_a^b g(x) \, dx = 0$. I am to show that if $f$ is an increasing function then (3) $\int_a^b f(x)g(x)dx \leq 0$.
First of all I am not sure if it is true. I see if $g$ is increasing then (1) is true but not (2). Also if $g$ is increasing the strict inequality of (3) may not hold. So it must be that $g$ is not monotone. In that case I cannot argue that $fg \leq 0$ to conclude (3).
This is true and requires no other information about $g$ other than integrability.
With $G(x) = \int_a^x g(t) \, dt \geqslant 0$ and integrating by parts, we have
$$\tag{*}\int_a^bf(x) g(x) \, dx = \int_a^b f \, dG \\ = f(b)G(b)- f(a) G(a) - \int_a^b G \, df = - \int_a^b G \, df \leqslant 0$$
The first equality in (*) between the Riemann and Riemann-Stieltjes integrals follows from the Riemann integrability of both $f$ and $g$ and is proved here. The final inequality follows because $G$ is nonnegative and $f$ is increasing.
