20.2.1 Vector Measure Version
Definition 20.2.3 (Bochner Integral). Let $(X, \cm)$ be a measurable space, $E, F$ be normed spaces over $K \in \RC$, $G$ be a Banach space over $K$,
be a bounded bilinear map, and $\mu: \cm \to F$ be a vector measure, then there exists a unique $I_{\lambda} \in L(L^{1}(X, |\mu|; E); G)$ such that:
For any $x \in E$ and $A \in \cm$, $I_{\lambda}(x \cdot \one_{A}) = x \mu(A)$.
For any $f \in L^{1}(X, |\mu|; E)$, $\normn{I_\lambda f}_{G}\le \norm{\lambda}_{L^2(E, F; G)}\cdot \norm{f}_{L^1(X, |\mu|; E)}$.
For any $f \in L^{1}(X; E)$, $I_{\lambda} f = \int \lambda(f, d\mu)$ is the Bochner integral of $f$ with respect to $\mu$ and $\lambda$.
Proof. Same as Definition 20.2.1.$\square$
Theorem 20.2.4 (Dominated Convergence Theorem). Let $(X, \cm, \mu)$ be a measure space, $E, F$ be normed spaces over $K \in \RC$, $G$ be a Banach space over $K$,
be a bounded bilinear map, $\mu: \cm \to F$ be a vector measure, $\seq{f_n}\subset L^{1}(X, |\mu|; E)$, and $f \in L^{1}(X, |\mu|; E)$. If
$f_{n} \to f$ strongly pointwise.
There exists $g \in L^{1}(X) \cap L^{+}(X)$ such that $\norm{f_n}_{E} \le g$ for all $n \in \natp$.
then $\int \lambda(f , d\mu) = \limv{n}\int \lambda(f_{n}, d\mu)$.
Proof. By the classical Dominated Convergence Theorem, $f_{n} \to f$ in $L^{1}(X, |\mu|; E)$. Since $h \mapsto \int \lambda(f , d\mu)$ is a bounded linear operator, $\int \lambda(f , d\mu) = \limv{n}\int \lambda(f_{n}, d\mu)$.$\square$