Proof [Proposition 7.12, Fol99]. Assume without loss of generality that $\int f d\mu < \infty$. By Proposition 4.22.2, $f$ is Borel measurable, so $f \ge f_{\alpha}$ for all $\alpha \in A$ implies that

Let $\phi \in \Sigma^{+}(X, \cm)$ with $0 \le \phi < f$ and $\beta < \int \phi d\mu$. Let $\seqf{a_j}\subset (0, \infty)$ and $\seqf{E_j}\subset \cb_{X}$ such that $\phi = \sum_{j = 1}^{n} a_{j} \one_{E_j}$. Since $\int \phi d\mu < \infty$, for each $1 \le j \le n$, $\mu(E_{j}) < \infty$ by Markov’s Inequality. By Proposition 17.1.3, there exists compact sets $\seqf{K_j}\subset 2^{X}$ such that $K_{j} \subset E_{j}$ for each $1 \le j \le n$ and $\int \sum_{j = 1}^{n} a_{j} \one_{K_j}d\mu > \beta$.

Let $\psi = \sum_{j = 1}^{n} a_{j} \one_{K_j}$ and $K = \bigcup_{j = 1}^{n} K_{j}$, then $K$ is compact by Proposition 4.16.2 and $-\psi$ is lower semicontinuous by Proposition 4.22.2.

For any $x_{0} \in K$, since $f_{\alpha} \upto f$ and $\psi \le \phi < f$, there exists $\alpha(x_{0}) \in A$ such that $f_{\alpha(x_0)}(x_{0}) > \psi(x_{0})$. By Proposition 4.22.2, $f_{\alpha(x_0)}- \psi$ is lower semicontinuous, so

is an open cover of $K$. Let $\bracs{x_j}_{1}^{N} \subset K$ such that $K \subset \bigcup_{j = 1}^{N} \bracsn{f_{\alpha(x_j)} > \psi}$. Since $f_{\alpha} \upto f$, there exists $\alpha_{0} \in A$ such that

so

As such an $\alpha_{0} \in A$ exists for all $\beta < \int \phi d\mu$,

Since the above holds for all $\phi \in \Sigma^{+}(X, \cm)$ with $0 \le \phi < f$,

by Lemma 19.2.3.$\square$