Let $f: (1,\infty) \to (0,\infty)$ be a continuous function which is bounded away from $0$ on compact sets. I'm looking for a statement of the following form:
If $g$ is a function satisfying *******, then $$\int_{(1,\infty)} \frac{1}{f(s)} g(s) \, ds = \infty \tag{1}$$ implies $$\int_{(1,\infty)} f(s) \, ds < \infty. \tag{2}$$
From my point of view the most natural choice would be $g(s) = 1/s^{2+\epsilon}$ for some $\epsilon>0$; if
$$\int_{(1,\infty)} \frac{1}{f(s) s^{1+\epsilon/2}} \frac{1}{s^{1+\epsilon/2}} \, ds = \infty \tag{3}$$
then it is to be expected that $\frac{1}{f(s) s^{1+\epsilon/2}}$ cannot be too close to $0$ for large $s$ (otherwise the integral $(3)$ would be finite). If $f$ is monotone, then this is pretty straight-forward, but I would like to avoid this assumption. Any references or ideas?
Counterexample for the choice $1/s^{2+\epsilon}$: Let $g(s) = s^{-3}.$ We can find $f$ such that $f(s) = 1/n^3$ on the interval $[n,n+1/3], n = 2,3,\dots,$ and $f=1$ on $[n+1/2,n+2/3], n = 2,3,\dots $ You'll have $(1)$ holding because of what's happening on the first set of intervals, yet $\int_1^\infty f = \infty$ because of what's happening on the second set of intervals.