Prove: If g is strictly positive then $\int_{2}^{x}o(g)\,dt = o\left(\int_{2}^{x}g\,dt\right)$.
I understand this question to mean: prove that $f=o(g)$ implies $F = o(G)$ and conversely. Please advise if this is not correct. If it is, then
Suppose $\lim\limits_{x \to \infty}{\dfrac{f}{g}} = 0 $ as $ x \to \infty $. Since $g \gt 0$, then $\lim\limits_{x \to \infty}{G} = \infty.$
These conditions allow application of l'Hôpital, as given in Rudin (3rd), p. 113. We can conclude that
$\lim_{x \to \infty} \frac{F}{G} = 0.$
Now suppose that $F=o(G)$. Again we have that $g \gt 0$ and so $\lim_{x\to\infty}{G} = \infty$.
If we can say that $F$ is at least piecewise continuous and differentiable then I suppose l'Hôpital applies and $f = o(G)$. But I can't state precisely why $F=o(G)$ and $ G \to \infty $ give that $F$ satisfies the requirements for application of L'Hôpital.
For example, why might we conclude that $\lim\limits_{x \to \infty} F = \infty$?
Thanks for any suggestions.
This problem is from a comment of Gerry Myerson. Please do not award me points for it.