integrability of $f^3$ for some Lebesgue measurable function

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I'm trying to solve the following problem from an old qualifying exam, but nothing I've tried has been successful, so any help would be greatly appreciated.

Suppose $f: \mathbb{R} \rightarrow (0, \infty)$ is Lebesgue measurable, and that $\int f^2gdx < \infty$ for every Lebesgue measurable function $g: \mathbb{R} \rightarrow \mathbb{R}$ with $\int g^3 dx < \infty$. Prove that $\int f^3 dx < \infty$

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My idea is inspired by Holder's inequality, so I'll pose it in that framework. Let $1<p<\infty,\frac{1}{p}+\frac{1}{q}=1$. (This works at the endpoints too, but the details are different, so I don't want to do it.) Try to argue that

$$\| h \|_{L^p} = \sup_{g \in L^q,\| g \|_{L^q}=1} \int hg dx.$$

For your problem, consider $q=3$, hence $p=3/2$, and take $h=f^2$. From the above you get $f^2 \in L^{3/2}$, hence $f \in L^3$, which is what you want.

Idea for the proof: try to take $g$ essentially equal to $\frac{h^{p-1}}{\| h^{p-1} \|_{L^q}}$, so that $g$ has $L^q$ norm $1$ and $hg$ is essentially proportional to $h^p$. You cannot directly do this: $(p-1)q=p$, so this implicitly assumes that $h \in L^p$ already. So instead approximate this by a sequence of functions that you know are in $L^q$. To do this, cut off the domain and range: you can cut off the domain to $[-n,n]$ and the range to $[0,n]$.