14.1 Measures
Definition 14.1.1 (Measurable Space). Let $X$ be a set and $\cm \subset 2^{X}$ be a $\sigma$-algebra, then the pair $(X, \cm)$ is a measurable space.
Definition 14.1.2 (Measure). Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [0, \infty]$, then $\mu$ is a (countably-additive) measure if
$\mu(\emptyset) = 0$.
For any $\seq{E_n}\subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}E_n}= \sum_{n \in \natp}\mu(E_{n})$.
In which case, $(X, \cm, \mu)$ is a measure space.
If $\mu: \cm \to [0, \infty]$ instead satisfies (M1) and
For any $\seqf{E_j}\subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{j = 1}^n E_j}= \sum_{j = 1}^{n} \mu(E_{j})$.
then $\mu$ is a finitely-additive measure.
Definition 14.1.3 (Null Set). Let $(X, \cm, \mu)$ be a measure space, then $E \in \cm$ is $\mu$-null/null if $\mu(E) = 0$.
Definition 14.1.4 (Almost Everywhere). Let $(X, \cm, \mu)$ be a measure space. A statement holds $\mu$-almost everywhere/$\mu$-a.e./a.e. if it is false on a subset of a $\mu$-null set.
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),
Theorem 14.1.6 (Dynkin’s Uniqueness Theorem). Let $(X, \cm)$ be a measurable space, $\mathcal{P}\subset \cm$ be a $\pi$-syste, and $\mu, \nu: \cm \to [0, \infty]$ be measures. If
$\sigma(\mathcal{P}) = \cm$.
$\mu(E) = \nu(E)$ for all $E \in \mathcal{P}$.
There exists $\seq{E_n}\subset \mathcal{P}$ such that $E_{n} \upto X$ and $\mu(E_{n}) < \infty$ for all $n \in \natp$.
then $\mu = \nu$.
Proof. Let $F \in \mathcal{P}$ with $\mu(F) < \infty$ and
then $\alg(F) \supset \mathcal{P}$ by (b), and
For any $E, E' \in \alg(F)$ with $E \subset E'$ and $F \in \mathcal{P}$,
\begin{align*}\mu((E' \setminus E) \cap F)&= \mu(E' \cap F) - \mu(E \cap F) \\&= \nu(E' \cap F) - \nu(E \cap F) = \nu((E' \setminus E) \cap F)\end{align*}For any $\seq{E_n}\subset \alg$ with $E_{n} \subset E_{n+1}$ for all $n \in \natp$ and $F \in \mathcal{P}$,
\[\mu\paren{\bigcup_{n \in \nat}E_n \cap F}= \limv{n}\mu(E_{n} \cap F) = \limv{n}\nu(E_{n} \cap F) = \nu\paren{\bigcup_{n \in \nat}E_n \cap F}\]by continuity from below (Proposition 14.1.5).
so $\alg(F)$ is a $\lambda$-system. By (a) and Dynkin’s $\pi$-$\lambda$ theorem (Theorem 13.2.5), $\alg(F) = \cm$.
Let $\seq{E_n}$ as in assumption (c), then $\mu(E_{n} \cap F) = \mu(E_{n} \cap F)$ for all $n \in \natp$ and $F \in \cm$. Thus by continuity from below (Proposition 14.1.5),