Theorem   Let $(X,\mathfrak{M},\mu)$ be a $\sigma$ -finite measure space and $V$ be the space of bounded linear maps $\Phi\colon L^\infty(\mu)\rightarrow\mathbb{R}$ satisfying bounded convergence. That is, if $|f_n|\le 1$ are in $L^\infty(\mu)$ and $f_n(x)\rightarrow 0$ for almost every $x\in X$ , then $\Phi(f_n)\rightarrow 0$ .

Then $g\mapsto\Phi_g$ gives an isometric isomorphism from $L^1(\mu)$ to $V$ .

Proof. First, the operator norm $\Vert\Phi_g\Vert$ is equal to the $L^1$ -norm of $g$ (see $L^p$ -norm is dual to $L^q$ ), so the map $g\mapsto\Phi_g$ gives an isometric embedding from $L^1$ into the dual of $L^\infty$ . Furthermore, dominated convergence implies that $\Phi_g$ satisfies bounded convergence so $\Phi_g\in V$ . We just need to show that $g\mapsto\Phi_g$ maps onto $V$ .

So, suppose that $\Phi\in V$ . It needs to be shows that $\Phi=\Phi_g$ for some $g\in L^1$ . Defining an additive set function $\nu\colon\mathfrak{M}\rightarrow\mathbb{R}$ by \begin{equation*} \nu(A)=\Phi(1_A) \end{equation*}for every set $A\in\mathfrak{M}$ , the bounded convergence property for $\Phi$ implies that $\nu$ is countably additive and is therefore a finite signed measure. So, the Radon-Nikodym theorem gives a $g\in L^1$ such that $\nu(A)=\int_A g\,d\mu$ for every $A\in\mathfrak{M}$ . Then, the equality \begin{equation*} \Phi(fh)= \int fg\,d\mu \end{equation*}is satisfied for $f=1_A$ with any $A\in\mathfrak{M}$ and the functional monotone class theorem extends this to any bounded and measurable $f\colon X\rightarrow\mathbb{C}$ , giving $\Phi_g=\Phi$ .