Let $\psi$ be measurable function such that for all $ f \in L^{1}(\mu)$, $ f\psi \in L^{1}(\mu)$. Then $\psi$ is bounded almost everywhere(in sense of measure $\mu$).
My attempt: Use contraposition: if $\psi $ isn't bounded everywhere then $\forall M>0$, there exists a set $A$ such that $\mu (A)>0 $ and $ \psi(x) > M , \forall x \in A$. Now i need to prove that there is an $f \in L^{1} $ such that $ f\psi \not\in L^{1} $. But I don't know how.
Any help or hint would be appreciated.Thanks in advance.
We can prove this statement using Baire Category argument. But here, I will argue by contrapositive showing that $$ \psi \notin L^\infty(\mu) \implies \exists f\in L^1(\mu) \ : \ f\psi\notin L^1(\mu). $$ Let $E_j =\{ 2^j\le |\psi|<2^{j+1}\}$ and $c_j = \mu (E_j).$ Note that assumption implies $c_j>0$ for infinitely many $j$'s. Define $J=\{j:c_j>0\}$ and $$ f(x) = \sum_{j\in J} \frac{1}{c_j \sqrt{2}^{j}} 1_{E_j}(x). $$ Then $\int |f|=\int f =\sum_{j\in J} 2^{-j/2}<\infty$. However, $$ |\psi(x)||f(x)| \ge 2^j\frac{1}{c_j \sqrt{2}^{j}}=\frac{\sqrt{2}^{j}}{c_j} $$ on each $E_j, \ j\in J$, which implies $$ \int |\psi f|=\sum_{j\in J} \int_{E_j} |\psi f| \ge \sum_{j\in J}\int_{E_j}\frac{\sqrt{2}^{j}}{c_j}=\sum_{j\in J} \sqrt{2}^j =\infty. $$ Thus $\psi f\notin L^1(\mu)$ and the claim is proved.