Proposition 12.1.8. Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty)$, then $\Sigma(X, \cm; E) \cap L^{p}(X; E)$ is dense in $L^{p}(X; E)$.
Proof. Let $f \in L^{p}(X; E)$. By Definition 17.1.1, there exists $\seq{f_n}\subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_{E} \le \norm{f}_{E}$ and $\norm{f_n - f}_{E} \to 0$ pointwise as $n \to \infty$.
For each $n \in \nat$, $\norm{f_n}_{E} \le \norm{f}_{E}$, so $\norm{f_n}_{L^p(X; E)}\le \norm{f}_{L^p(X; E)}< \infty$, and $\norm{f_n - f}_{E} \le 2\norm{f}_{E}$. By the Dominated Convergence Theorem,
\[\limv{n}\int \norm{f_n - f}_{E}^{p} d\mu = \int \limv{n}\norm{f_n - f}_{E}^{p} d\mu = 0\]
Therefore $\norm{f_n - f}_{L^p(X; E)}\to 0$ as $n \to \infty$.$\square$