01 Probability theory

Lingfeng 2026-09-15

1. Measure Theory

Definition ( $\sigma$ Field)

$\emptyset \in \mathcal{F}$
$A \in \mathcal{F} \implies A^{c} \in \mathcal{F}$
$A_{i} \in \mathcal{F}, i=1,2,\dots \implies \bigcup_{i=1}^{\infty} A_{i} \in \mathcal{F}$

Definition (Measure)

A set function $\nu$ defined on $\mathcal{F}$ is called a measure if and only if it has the following properties.

$\nu(\emptyset)=0$
$\nu(A) \ge 0$ for all $A \in \mathcal{F}$
If $A_{i} \in \mathcal{F}, i=1,2,\dots$ and $A_{i}$ are disjoint, then
$\nu\left( \bigcup_{i=1}^{\infty} A_{i} \right) = \sum_{i=1}^{\infty} \nu(A_{i})$

Theorem (Properties of Measure Space)

(Monotonicity) $A \subset B \implies \nu(A) \le \nu(B)$
(Subadditivity) If $A_{i} \in \mathcal{F}, i=1,2,\dots$ , then
$\nu\left(\bigcup_{i=1}^\infty A_{i}\right) \le \sum_{i=1}^\infty\nu(A_i)$
(Continuity) If $A_{1} \subset A_{2} \subset \dots$ and $v(A_1) < \infty$ , then
$\nu \left( \lim_{n\to\infty} A_{n}\right) = \lim_{n\to\infty} \nu(A_{n})$ where $\lim_{n\to\infty} A_{n} = \bigcup_{n=1}^{\infty} A_{n}$

Proof.

$B = A \cup (B \setminus A)$ , disjoint, so $\nu(B) = \nu(A) + \nu(B \setminus A) \ge \nu(A)$
Let $B_{n} = A_{n} \setminus \bigcup_{i=1}^{n-1} A_{i}$ , then $B_{n}$ are disjoint and $\bigcup_{n=1}^{\infty} B_{n} = \bigcup_{n=1}^{\infty} A_{n}$ , so
$\nu\left(\bigcup_{n=1}^{\infty} A_{n}\right) = \sum_{n=1}^{\infty} \nu(B_{n}) \le \sum_{n=1}^{\infty} \nu(A_{n})$
Let $B_{n} = A_{n} \setminus A_{n-1}$ , then $B_{n}$ are disjoint and $\bigcup_{n=1}^{\infty} B_{n} = \bigcup_{n=1}^{\infty} A_{n}$ , so
$\nu\left(\bigcup_{n=1}^{\infty} A_{n}\right) = \sum_{n=1}^{\infty} \nu(B_{n}) = \lim_{n\to\infty} \sum_{i=1}^{n} \nu(B_{i}) = \lim_{n\to\infty} \nu(A_{n})$

Definition (CDF)

$F(x) = P((-\infty, x])$ where P is a probability measure on $\mathbb{R}$ .

Theorem (Properties of CDF)

Proof.

$F(-\infty) = \lim_{n\to\infty} F(-n) = \lim_{n\to\infty} P((-n, n]^{c}) = 0$ by continuity from above. Similarly for $+\infty$ .
If $x \le y$ , then $(-\infty, x] \subset (-\infty, y]$ , so $F(x) \le F(y)$ .
Let $x_{n} \downarrow x_{0}$ . Then $(-\infty, x_{n}] \downarrow (-\infty, x_{0}]$ , so by continuity from above, $F(x_{n}) \to F(x_{0})$ .