Proposition 14.1.5. Let $(X, \cm, \mu)$ be a measure space, then:
For any $E, F \in \cm$ with $E \subset F$, $\mu(E) \le \mu(F)$.
For any $\seq{E_n}\subset \cm$, $\mu\paren{\bigcup_{n \in \natp}E_n}\le \sum_{n \in \natp}\mu(E_{n})$.
For any $\seq{E_n}\subset \cm$ with $E_{n} \subset E_{n+1}$ for all $n \in \nat$, $\mu\paren{\bigcup_{n \in \nat}E_n}= \limv{n}\mu(E_{n})$.
For any $\seq{E_n}\subset \cm$ with $E_{n} \supset E_{n+1}$ for all $n \in \nat$ and $\mu(E_{1}) < \infty$, $\mu(\bigcap_{n \in \natp}E_{n}) = \limv{n}\mu(E_{n})$.
For any $\seq{E_n}\subset \cm$,
\[\mu\paren{\liminf_{n \to \infty}E_n}\le \liminf_{n \to \infty}\mu(E_{n})\]For any $\seq{E_n}\subset \cm$ with $\mu\paren{\bigcup_{n \in \nat}E_n}< \infty$,
\[\mu\paren{\limsup_{n \to \infty}E_n}\ge \limsup_{n \to \infty}\mu(E_{n})\]
Proof. (1): $\mu(F) = \mu(E) + \mu(F \setminus E) \ge \mu(E)$.
(2): For each $n \in \natp$, let $F_{n} = E_{n} \setminus \bigcup_{k = 1}^{n-1}E_{k}$, then $\seq{F_n}$ is pairwise disjoint with $\bigsqcup_{n \in \natp}F_{n} = \bigcup_{n \in \natp}E_{n}$. In which case,
(3): Denote $E_{0} = \emptyset$. For each $n \in \natp$, let $F_{n} = E_{n} \setminus E_{n - 1}$, then $\seq{F_n}$ is pairwise disjoint with $\bigsqcup_{n \in \natp}F_{n} = \bigcup_{n \in \natp}E_{n}$. In which case,
(4): Since $\mu(E_{1}) < \infty$,
by (3).
(5): Using (1) and (3),
(6): Using (1) and (4),