用户: Ruiqi chen/CoursesNotes/数学分析 I (实验班)/Nov 27 2024

定义 1. 称 $E$ 为 (Lebesgue) 可测集, 如果 $E$ 满足 Caratheodory 条件: $m^{*} (T) = m^{*} (T \cap E) + m^{*} (T \cap E^{c}), \forall T \in R^{n}$ (1) $R^{n}$ 中全体可测集记作 $U$ .

引理 2. 若 $E$ 为零测集, 即 $m^{*} (E) = 0$ , 则 $E$ 可测.

证明. 有次可加性, 只需证明:

m^{*} (T) = m^{*} (E) + m^{*} (T) \geq m^{*} (T \cap E) + m^{*} (T \cap E^{c})

(2)

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定义 3. 所有零测集都可测的测度空间称为完备的测度空间.

注 4. 依 Cantor 集零测, 零测集子集亦零测, 故 Lebesgue 可测集 $U$ 的基为 $2^{c}$ .

定理 5. 集合系 $U$ 是 $σ$ -代数, 且对不交 $E_{k}$ 有: $m^{*} (k ⋃ E_{k}) = k \sum m^{*} (E_{k})$ (3)

证明. 取补是平凡的, 若

E_{1}, E_{2}

可测, 此时:

m^{*} (T) = m^{*} (T \cap E_{1}) + m^{*} (T \cap E_{1}^{c}) = m^{*} (T \cap E_{1} \cap E_{2}) + m^{*} (T \cap E_{1} \cap E_{2}^{c}) + m^{*} (T \cap E_{1}^{c} \cap E_{2}) + m^{*} (T \cap E_{1}^{c} \cap E_{2}^{c}) \geq m^{*} (T \cap (E_{1} \cup E_{2})) + m^{*} (T \cap (E_{1} \cup E_{2})^{c})

若

E_{1} \cap E_{2} = \emptyset

, 则:

m^{*} (T \cap (E_{1} \cup E_{2})) = m^{*} (T \cap (E_{1} \cup E_{2}) \cap E_{1}) + m^{*} (T \cap (E_{1} \cup E_{2}) \cap E_{1}^{c}) = m^{*} (T \cap E_{1}) + m^{*} (T \cap E_{2})

利用有限个的操作只需证明

E_{k}

两两不交情形, 记:

E : = k = 1 ⋃ + \infty E_{k}, F_{k} : = i = 1 ⋃ k E_{i}

(4)此时

F_{k}

可测, 且:

m^{*} (T) m^{*} (T) = m^{*} (T \cap F_{k}) + m^{*} (T \cap F_{k}^{c}) = i = 1 \sum k m^{*} (T \cap E_{i}) + m^{*} (T \cap F_{k}^{c}) \geq i = 1 \sum k m^{*} (T \cap E_{i}) + m^{*} (T \cap E^{c}) \geq i = 1 \sum \infty m^{*} (T \cap E_{i}) + m^{*} (T \cap E^{c}) \geq m^{*} (T \cap E) + m^{*} (T \cap E^{c})

可数可加性带入

T = E

有:

m^{*} (E) \geq i = 1 \sum \infty m^{*} (E_{i}) \geq m^{*} (E)

(5)

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定理 6. $E_{k}$ 是单调上升的可测集合列, 则: $m (k \to \infty lim E_{k}) = k \to \infty lim m (E_{k})$ (6)

证明. 有

lim_{k \to + \infty} E_{k} = ⋃_{k = 1}^{\infty} E_{k} = ⋃_{k = 1}^{\infty} (E_{k} ∖ E_{k - 1})

故:

m (k \to \infty lim E_{k}) = i = 1 \sum \infty m (E_{i}) - m (E_{i - 1}) = k \to \infty lim m (E_{k})

(7)

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定理 7. 若 $E_{k}$ 单调降, 而某 $E_{k}$ 测度有限, 则: $m (k \to \infty lim E_{k}) = k \to \infty lim m (E_{k})$ (8)

证明. 不妨假设

E_{1}

测度有限, 有:

m (E_{1} ∖ k \to \infty lim E_{k}) = k = 1 \sum \infty m (E_{k} ∖ E_{k + 1}) < + \infty

(9)故级数收敛, 此时:

m (E_{k}) - m (k \to \infty lim E_{k}) = l = k \sum \infty m (E_{l} ∖ E_{l + 1}) \to 0

(10)

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定义 8 (上下极限集). 给定 $A_{k}$ , 上下极限集定义为: $k \to \infty lim sup A_{k} k \to \infty lim inf A_{k} : = n = 1 ⋂ \infty k = n ⋃ \infty A_{k} : = n = 1 ⋃ \infty k = n ⋂ \infty A_{k}$

定理 9 (集合的 Fatou 引理). $m (k \to \infty lim inf E_{k}) \leq k \to \infty lim inf m (E_{k}) m (k \to \infty lim sup E_{k}) \geq k \to \infty lim sup m (E_{k})$ 后者要求对某个 $k$ , $m (⋃_{l = k}^{\infty} E_{l})$ 有限.

证明. 利用单调情形.

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定理 10 (Borel–Cantelli 定理). 若 $E_{k}$ 可测而 $\sum_{k = 1}^{\infty} m (E_{k}) < + \infty$ , 则: $m (k \to \infty lim inf E_{k}) = m (k \to \infty lim sup E_{k}) = 0$ (11)

证明. Omitted.

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定理 11 (第二 Borel–Cantelli 定理). 事件 $E_{k}$ 相互独立, 并且概率: $k = 1 \sum \infty m (E_{k}) = + \infty$ (12)那么: $m (k \to \infty lim sup E_{k}) > 0$ (13)

定理 12. 方块可测, 开闭集可测, 故 Borel 集可测.

对任意 $ε$ 存在开集 $G$ , 闭集 $F$ 使得: $F \subseteq E \subseteq G, m (G ∖ F) < ε$ (14)故 $E$ 为 $G_{δ}$ 去除一零测集, 或 $F_{σ}$ 并上一零测集.

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用户: Ruiqi chen/CoursesNotes/数学分析 I (实验班)/Nov 27 2024