The dual of $L^1$

Question

Let $(\Omega, \mathcal{A}, \mu) $ be a $\sigma$-finite measure space. I want to show that $\Phi: L^\infty (\Omega, \mathcal{A}, \mu) \ni g \mapsto [u \mapsto \int u g \mathrm{d} \mu] \in (L^1(\Omega, \mathcal A, \mu))^\ast$ is a bijective isometry. How do I show that this map is surjective, i.e. that for every $T \in (L^1 (\Omega, \mathcal{A}, \mu))^\ast$ exists a unique $g_T\in L^\infty (\Omega, \mathcal{A}, \mu)$ such that $T(u)= \int u g_T \mathrm{d} \mu$ ? And how do I show that $\Phi$ is an isometry?

Answer 1

This is a standard theorem and here is a sketch: Let $\nu (A)=T(I_A)$. Then $\nu$ is signed measure (if you are considering real function spaces and $\nu$ a complex measure if you are considering complex function spaces) and $\nu <<\mu$. Your function $g_T$ is nothing but $\frac {d\nu} {d\mu}$.