Proposition 12.1.8. Let $(X, \cm, \mu)$ be a measure space, $E$ be a Banach space over $K \in \RC$, $p \in [1, \infty)$, $\seq{f_n}\subset L^{p}(X; E)$, and $f \in L^{p}(X; E)$. If
$f_{n} \to f$ strongly pointwise.
There exists $g \in L^{p}(X) \cap L^{+}(X)$ such that $\norm{f_n}_{E} \le g$ for all $n \in \natp$.
then $f_{n} \to f$ in $L^{p}(X; E)$.
Proof. By assumptions $a$ and $b$, $\norm{f_n - f}_{E} \le 2g$ for all $n \in \natp$. Since $\seq{f_n}\subset L^{p}(X; E)$, $f \in L^{p}(X; E)$, and $g \in L^{p}(X)$, $\norm{f_n - f}_{E}^{p}, g^{p} \in L^{1}(X)$ for all $n \in \natp$. By the Dominated Convergence Theorem,
\[\limv{n}\norm{f_n - f}_{L^p(X; E)}^{p} = \limv{n}\int \norm{f_n - f}_{E} d\mu = 0\]
$\square$