Lemma 26.2.3.label Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^{+}(X, \cm)$.
- (1)
For each $\phi \in \Sigma^{+}(X, \cm)$, denote $\phi \le_{u} f$ if there exists $\delta > 0$ such that $\phi + \delta \le f$ on $\bracs{\phi > 0}$, then
\[\int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi \le_u f}\] - (2)
If $\mu$ is semifinite, then
\[\int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm) \cap L^1(X; \real), \phi \le_u f}\]
Proof. (1): Let $\phi \in \Sigma^{+}(X, \cm)$ with $\phi \le f$ and $\alpha \in (0, 1)$. Since $\phi(X) \setminus \bracs{0}\subset (0, \infty)$ is finite, $\delta = \min_{y \in \phi(X) \setminus \bracs{0}}(1 - \alpha)y > 0$. Thus $\alpha \phi + \delta \le \phi \le f$ on $\bracs{\phi > 0}= \bracs{\alpha \phi > 0}$, and $\alpha \phi \le_{u} f$.
By linearity on simple functions,
\[\int \phi d\mu = \sup_{\alpha \in (0, 1)}\alpha \int \phi d\mu = \sup_{\alpha \in (0, 1)}\int \alpha \phi d\mu\]
Thus
\[\int \phi d\mu \le \sup\bracs{\int \psi d\mu \bigg | \psi \in \Sigma^+(X, \cm), \psi \le_u f}\]
As the above holds for all $\phi \in \Sigma^{+}(X, \cm)$,
\[\int f d\mu = \sup\bracs{\int \psi d\mu \bigg | \psi \in \Sigma^+(X, \cm), \psi \le_u f}\]
$\square$
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