Proof, [Section 3.3, Rya02]. Let $T: F \to L^{1}(X; K)$ be defined by $T\phi = \phi \circ f$. To see that $T$ is continuous, it is sufficient to apply the Closed Graph Theorem.

Let $\seq{\phi_n}\subset F$, $\phi \in F$, $\seq{g_n}\subset L^{1}(X; K)$, and $g \in L^{1}(X; K)$ such that

1. (i)

   $g_{n} = \phi_{n} \circ f$ almost everywhere for all $n \in \natp$.

2. (ii)

   $\phi_{n} \to \phi$ and $g_{n} \to g$ as $n \to \infty$.

By passing through a subsequence using Theorem 26.7.4, assume further without loss of generality that $g_{n} \to g$ almost everywhere. In which case, $\phi_{n} \circ f \to g$ almost everywhere as well, and $\phi \circ f = g$ almost everywhere. Thus $T$ is continuous.

Now, let $T^{*}: L^{1}(X; K)^{*} \to F^{*}$ be the adjoint of $T$. For each $A \in \cm$, the mapping $\Phi_{A}: L^{1}(X; K) \to K$ defined by $g \mapsto \int_{A} g d\mu$ is a continuous linear functional on $L^{1}(X; K)$. As such, for each $\phi \in F$,

Therefore $T^{*}\Phi_{A} = \int_{A}^{*} f d\mu$ is the desired Dunford $\lambda$-integral of $f$ over $A$.$\square$