Lemma 16.2.6. Let $(X, \cm)$ be a measurable space and $f \in \mathcal{L}^{+}(X, \cm)$, then there exists $\seq{f_n}\subset \Sigma^{+}(X, \cm)$ such that $f_{n} \upto f$ pointwise.
Proof. By Proposition 15.5.6, there exists $\seq{g_n}\subset \Sigma^{+}(X, \cm)$ such that $0 \le g_{n} \le f$ for each $n \in \natp$, and $g_{n} \to f$ pointwise. For each $n \in \natp$, let $f_{n} = \max_{1 \le k \le n}g_{k}$, then $f_{n} \in \Sigma^{+}(X, \cm)$, $0 \le f_{n} \le f$, and $f_{n} \upto f$ pointwise.$\square$