I have a question about Lebesgue integral.
Let $(S,\Sigma,m)$ be a $\sigma$-finite measure space. Let $f$ be a $\Sigma$-measurable real-valued function.
If $f$ satisfies that $\forall g \in L^{2}(S;m), fg \in L^{1}(E;m)$ then $f \in L^{2}(E;m)$ is hold?
I'm looking for a counterexample.
Thank you in advance.
This can be proved using the Baire category theorem. Here is a sketch:
Define $$E_k = \left\{ g \in L^2(S;m) : \int_S |fg| \, dm \le k \right\}.$$ According to your hypothesis, $L^2(S;m) = \cup_k E_k$. It is a straightforward exercise to show that each $E_k$ is closed in $L^2(S;m)$. The Baire category theorem implies that at least one $E_k$ has an interior point.
This means there exists $h \in L^2(S;m)$ and $\epsilon > 0$ with the property that $$\|f-h\|_{L^2} \le \epsilon \implies \int_S |fg| \, dm \le k.$$
If $f \in L^2(S;m)$ and $\|f\|_{L^2} \le 1$, then $\|(\epsilon f + h) - h\|_{L^2} \le \epsilon$ so that $$\int_S |(\epsilon f + h)g| \, dm \le k.$$ From here you get $$ \sup_{\|f\|_{L^2} \le 1} \int_S |fg| \, dm \le \frac{2k}{\epsilon}$$ which implies $g \in L^2(S;m)$.