# 測度

## 測度の定義

* $S$ is a set.
* $O$ is a 完全加法族
* $\mu$ is a function that filles following conditions.
$\mu:O \mapsto [ 0, \infty]$
then $\mu$ is called "$O$上の測度"(measure over $O$) if it fills following conditions.
1. $\mu(\emptyset) = 0$
2. $\{F_n|F_n \in O\}, F_m \cap F_n = \emptyset (\forall m,n \in \mathbb{N},m \neq n)$ then $\mu(\cup_{i = 0}^{\infty} F_i) = \Sigma_{i = 0}^{\infty} \mu(F_i)$

## 測度空間の定義

* $S$ is a set.
* $O$ is a 完全加法族 over $S$
* $\mu$ is a measure over $O$
$(S, O, \mu)$ is called a measure space(測度空間) over $S$

## 完備(complete)な測度空間の定義

* $(S, O, \mu)$ is a measure space(測度空間) over $S$
* Let $E \in O, \mu(E) = 0$ then $(F \subset E \Rightarrow F \in O)$
$(S, O, \mu)$ is called a complete measure space(完備な測度空間) over $S$
(注意) 一般に測度空間は完備とは限らない

## ジョルダン測度の定義

n次元ユークリッド空間 $R^n$ の部分集合$C \subset R^n$を
$$C = [a_1,b_1] \times \dots \times [a_n, b_n]$$
とする時、ジョルダン測度 $m(C)$ は
$$m(C) = (b_1 - a_1)(b_2 - a_2) \dots (b_n - a_n)$$
と定義する

## Definition カラテオドリの外測度

* Let $X$ is a set and $(X\neq \emptyset)$
* Let $\mu^*: 2^X \rightarrow \mathbb{R}$
and
1. $\mu^*(\emptyset) = 0$ and $\mu(A) \in [0,\infty] \quad (A \in 2^X)$
2. $A,B \in 2^X, A \subset B \Rightarrow \mu^*(A) \le \mu^*(B)$
3. $\mu^*(\cup_{i=1}^\infty A_i) \le \Sigma_{i=1}^\infty \mu^*(A_i) \quad (A_i \in 2^X)$
then
$\mu^*$ is called Caratheodoryの外測度 or simply 外測度


## Theorem

Let
$$m^*(A) = \inf\{\sum_{j=1}^\infty m(E_j) \} \quad E_j \in \mathcal{R}(R^d), A \subset \bigcup_{j=1}^\infty E_j$$ then
$m^*$ is 外測度 over $\mathcal{P}(R^d)$

### Proof

(1.)
* $m^*(\emptyset) = 0$ は明らか
* $\forall E \in \mathcal{R}(R^d), m(E) \ge 0 \therefore m^*(A) \ge 0$

(2.)
* $\because$ $A\subset B$より$B$の被覆は$A$の被覆

(3.)

$\sum_{j=1}^\infty \mu^*(A_j) \lt \infty$と仮定して証明すれば十分

$\forall \epsilon \gt 0, \forall j \in \mathbb{N}, \exists \{E_{j,k} | k=1,2,\cdots\} \subset \mathcal{R}(R^d)$

s.t. $$A_j \subset \bigcup_{k=1}^\infty E_{j,k}, \quad \sum_{k=1}^\infty m(E_{j,k}) \lt m^*(A_j) + \frac{\epsilon}{2^j}$$ then $$\bigcup_{j=1}^\infty A_j \subset \bigcup_{j=1,k=1}^\infty E_{j,k}$$ and

$m^*$ の定義より
$$\begin{eqnarray} m^*(\bigcup_{j=1}^\infty A_j) & \le & \sum_{j=1,k=1}^\infty m(E_{j,k}) \\ & = & \sum_{j=1}^\infty \sum_{k=1}^\infty m(E_{j,k})\\ & \lt & \sum_{j=1}^\infty (m^*(A_j) + \frac{\epsilon}{2^j})\\ & = & \sum_{j=1}^\infty m^*(A_j) + \epsilon \end{eqnarray}$$ $\epsilon \gt 0$ は任意でよいので $$m^*(\bigcup_{j=1}^\infty A_j) \le \sum_{j=1}^\infty m^*(A_j)$$

## Proposition

$$m^*(\mathbb{Q}) = 0$$

### Proof

$\forall r \in \mathbb{R}, m^*(r) = 0$

$m^*(\mathbb{Q}) = \sum_{i=1}^\infty m^*(r_i) \quad (r_i \in \mathbb{Q})$

$\because |\mathbb{Q}| = \aleph_0$

## Definition 零集合

* A set $E$ is 零集合 if and only if $m^*(E) = 0$

## ルベーグ測度

$$\tilde m = m^*|_{\mathcal{M}(R^d)}$$ $\tilde m$ is called ルベーグ測度

## ルベーグ外測度の定義

Let * $E \subset \mathbb{R}^n$ * $E \subset \cup_{n=1}^\infty I_n$ $$Lm^*(E) = \inf \sum_{n=1}^\infty |I_n|$$