Lemma 16.2.3. Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^{+}(X, \cm)$, then
\[\int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}, \phi \le f}\]
Proof. Let $\phi \in \Sigma^{+}(X, \cm)$ with $\phi \le f$, then for any $\alpha \in (0, 1)$, $\alpha \phi|_{\bracs{f > 0}}< f|_{\bracs{f > 0}}$. Since
\[\int \phi d\mu = \sup_{\alpha \in (0, 1)}\alpha \int \phi d\mu = \sup_{\alpha \in (0, 1)}\int \alpha \phi d\mu\]
the two sides are equal.$\square$