Please check my arguments.
Let $f$ be continuous on $[a,b)$, $g$ be integrable on $[a,b)$ s.t. $g>0$ and $g' \geq 0$. Also Define $F(x) = \int_{a}^{x} f(t) dt$ with $F$ being bounded on $[a,b)$ and suppose $\lim_{x \to b^{-}} g(x) = \infty$.
From the assumptions we have
$$-L \leq F(x) \leq L, \forall x \in [a,b) \iff -L \leq \int_{a}^{x} f(t)dt \leq L$$
for some $L>0$. Since $g>0$ and $g$ integrable on $[a,b)$, by MVT for integrals $\exists c \in [a,b)$ s.t.
$$\int_{a}^{x} f(t)g(t)dt = f(c) \int_{a}^{x} g(t)dt$$
$\forall x \in [a,b)$. Now since $\lim_{x \to b^{-}} g(x) = \infty$,
$$\frac{1}{[g(x)]^{p}} \to 0 \text{ as } x \to b^{-}.$$
for $p > 1$. Thus
$$\lim_{x \to b^{-}} \frac{1}{[g(x)]^p} \int_{a}^{x} f(t)g(t)dt = \lim_{x \to b^{-}} \frac{1}{[g(x)]^p} \lim_{x \to b^{-}} \int_{a}^{x} f(t)g(t)dt = \lim_{x \to b^{-}} \frac{1}{[g(x)]^p} \lim_{x \to b^{-}} f(c) \int_{a}^{x} g(t)dt$$ $$ = f(c) \lim_{x \to b^{-}} \frac{1}{[g(x)]^p} \lim_{x \to b^{-}} \int_{a}^{x} g(t)dt$$
We know for a fact that $\lim_{x \to b^{-}} \frac{1}{[g(x)]^p} = 0$> Also since $g$ is integrable on $[a,b)$, $\lim_{x \to b^{-}} \int_{a}^{x} g(t)dt$ is finite. Now I have a feeling that $f(c)$ must also be finite but I'm unsure how to justify it (if its true). On top of that, I'm unsure on why the boundedness of $F(x)$ is a sufficient condition for the problem.
Thanks!
I will assume that $g$ is continuously differentiable on $[a,b)$. [The precise hypothesis on $g$ is not clearly stated]. It is not clear that $f(c)$ does not tend to $\infty$. So instead of MVT for integrals just integrate by parts. $\int_a^{x}f(t)g(t)dt =F(x)g(x)-\int_a^{x} F(t)g'(t)dt$. The first term divided by $|g(x)|^{p}$ tends to $0$ becuase $F$ is bounded and $p>1$. In the second term note that $|\int_a^{x} F(t)g'(t)dt| \leq L (g(x)-g(a))$ so second term divided by $|g(x)|^{p}$ also tends to $0$.