Partial converse of the fact: $f\in L^p , g\in L^q \Rightarrow fg\in L^1$

Question

Hölder's inequality says If $p^{-1}+q^{-1}=1$ and $ f\in L^p, g\in L^q$, then $fg\in L^1$.

Then how about the following converse:

Let $g$ be measurable. If for all $f\in L^p$ we have $fg \in L^1$, then is $g\in L^q$?

I guess this may not be true. But I guess there exist a condition that makes this true. Am I right? If there exist such condition, can you explain it?

Answer 1

The theory of $L^p$ spaces is usually developed over spaces with a $\sigma$-finite measure. In this case the result is true: see Discontinuous functionals on $L^p$

For general measures this is not true, as David C. Ullrich demonstrated by an example:

Let $X=\{0\}$, with the unique $\sigma$-algebra. Define $\mu$ by $\mu(X)=\infty$. Let $g(0)=1$. Then $gf\in L^1$ for every $f\in L^p$, simply because $f\in L^p$ implies $f=0$. But $g\notin L^q$.