Proposition 16.2.7. Let $(X, \cm, \mu)$ be a measure space, then
For any $f, g \in \mathcal{L}^{+}(X, \cm)$ and $\alpha \ge 0$, $\int \alpha f + g d\mu = \alpha \int f d\mu + \int g d\mu$.
For any $\seq{f_n}\subset \mathcal{L}^{+}(X, \cm)$, $\sum_{n \in \natp}\int f_{n}d\mu = \int \sum_{n \in \natp}f_{n} d\mu$.
For any $f, g \in \mathcal{L}^{+}(X, \cm)$ with $f \le g$, $\int f d\mu \le \int g d\mu$.
For any $f \in \mathcal{L}^{+}(X, \cm)$ with $\int f d\mu < \infty$, $\mu(\bracs{f = \infty}) = 0$.
For any $f \in \mathcal{L}^{+}(X, \cm)$ with $\int f d\mu < \infty$, $\bracs{f > 0}$ is $\sigma$-finite.
For any $f \in \mathcal{L}^{+}(X, \cm)$ with $\int f d\mu = 0$, $f = 0$ $\mu$-almost everywhere.
Proof. (1): By Lemma 16.2.6, there exists $\seq{f_n}, \seq{g_n}\subset \Sigma^{+}(X, \cm)$ with $0 \le f_{n} \le f$ and $0 \le g_{n} \le g$ for each $n \in \natp$, $f_{n} \upto f$, and $g_{n} \upto g$. By Proposition 16.1.3 and the Monotone Convergence Theorem,
(2): By (1) and the Monotone Convergence Theorem,
(3): By Definition 16.2.2.
(4): Since $\int f d\mu \ge \int \infty \cdot \one_{\bracs{f = \infty}}d\mu = \infty \cdot \mu(\bracs{f = \infty})$, $\int f d\mu < \infty$ implies that $\mu(\bracs{f = \infty}) = 0$.
(5): For each $\eps > 0$,
so $\int f d\mu < \infty$ implies that $\mu(\bracs{f \ge \eps}) < \infty$. Since $\bracs{f > 0}= \bigcup_{n \in \natp}\bracs{f_n \ge 1/n}$, $\bracs{f > 0}$ is $\sigma$-finite.
(6): As in the proof of (5), $\mu(\bracs{f \ge \eps}) = 0$ for all $\eps > 0$. Thus $\bracs{f > 0}= \bigcup_{n \in \natp}\bracs{f_n \ge 1/n}$ is a $\mu$-null set, so $f = 0$ $\mu$-almost everywhere.$\square$