I have that $(X,\Omega,\mu)$ is a sigma finite space, and I have that $g$ is a measurable function. Assume that $fg\in L^p$ for all $1\leq p\leq \infty$. I want to show that $g\in L^\infty$.
My idea was to assume that $g$ is not essentially bounded. Then for all $n\in \mathbb{Z}^+$ we have that $A_n:=\{x\mid |g(x)|>n\}$ has positive measure. Since $X$ is sigma-finite, write $X=\cup_{i=1}^\infty X_i$. For any $n$, let us assume wlog that $0<\mu(A_n)<\infty$ because otherwise consider $A_n\cap X_i$ for some $i$. Then, $$\|\chi_{A_n}g\|_p\geq \|\chi_{A_n}\|_p n$$Hence, we see that $M_g$ (this will turn out not to be an operator because the result will be that there is an $f$ s.t. $M_g(f)\notin L^p$, but for sake of contradiction say it is an operator) is an operator with infinite norm.
I can't however construct a function such that $M_g(f)$ fails to be in $L^p$. I was thinking perhaps arguing with some sort of convergence theorem that the $f_n$ go to some function $f$ and that their $p$-norms converge, but I have not been able to.
Any hints?