Does anyone know how to do this exercise? I couldn't do it.
Let $(X,S,\mu)$ be a $\sigma$ -finite measure space and $g\in M(X,S)$ such that $gs\in\mathcal{L}_1$ for any simple function $s\in\mathcal{L}_p$, $p\in(1,\infty)$. Suppose there is $A\geq0$ such that $\left|\int gs\,d\mu\right|\leq A||s||_p$ for all $s$. Show that $g\in\mathcal{L}_p$ with $\frac1p+\frac1q=1$ and $||g||_q\leq A$.
Define $T$ on the space $M$ simple functions in $L^{p}$ by $Ts=\int gs d \mu$. It is a bounded linear functional on $M$ with norm at most $A$. Since $M$ is dense in $L^{p}$ it extends uniquely to a bounded linear functional on $L^{p}$ with the same norm. Since the dual of $L^{p}$ is $L^{q}$ there exists $h \in L^{q}$ such that $\int s gd \mu=Ts=\int hs d\mu$ for all simple functions $s$ and $\|h\|_q=\|T\|\leq A$ . For any set $E$ of finite measure this gives $\int_E g d\mu= \int_E h d\mu$. Can you use sigma finiteness now to prove that $g=h$ a.e..?