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‎content/.obsidian/workspace.json‎

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"state": {
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"type": "markdown",
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"state": {
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"file": "计算机/clash.md",
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"file": "数学分析/级数.md",
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"mode": "source",
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"source": false
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},
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"icon": "lucide-file",
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"title": "clash"
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"title": "级数"
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}
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}
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]
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"state": {
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"type": "backlink",
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"state": {
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"file": "计算机/clash.md",
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"file": "数学分析/级数.md",
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"collapseAll": false,
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"extraContext": false,
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"sortOrder": "alphabetical",
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"unlinkedCollapsed": true
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},
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"icon": "links-coming-in",
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"title": "Backlinks for clash"
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"title": "Backlinks for 级数"
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}
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},
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{
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},
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"active": "f1f064fa641803df",
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"lastOpenFiles": [
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"数学分析/real-number.md",
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"数学分析/index.md",
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"数学分析/derivate.md",
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"数学分析/funct-limit.md",
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"测度论与概率论/Lebesgue-integral.md",
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"测度论与概率论/index.md",
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"测度论与概率论/dependency_graph.md",
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"测度论与概率论/almost-everywhere.md",
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"计算机/clash.md",
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"计算机/blog.md",
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"计算机/bash.md",
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"计算机/automata.md",
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"故事/神圣叙事.md",
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"故事/纯粹理性批判.md",
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"故事/你变了.md",
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"故事/判断力批判.md",
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"故事/固执.md",
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"故事/安全.md",
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"故事/安全与道德.md",
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"故事/实践理性批判.md",
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"测度论与概率论/三大积分收敛定理.md",
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"测度论与概率论/Lebesgue-integral.md",
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"测度论与概率论/L-S-measure.md",
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"测度论与概率论/measurable-func.md",
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"测度论与概率论/measurable-mapping.md"
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"故事/判断力批判.md"
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]
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}

‎content/数学分析/级数.md‎

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@@ -48,6 +48,8 @@ $$S_N = \sum_{n=1}^N \left( f^+(g(n)) - f^-(g(n)) \right) = \sum_{n=1}^N f^+(g(n
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- $\lim_{N \to \infty} \sum_{n=1}^N f^+(g(n)) = \sum_{x \in X} f^+(x) = L^+$
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- $\lim_{N \to \infty} \sum_{n=1}^N f^-(g(n)) = \sum_{x \in X} f^-(x) = L^-$
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知:$S_N$极限存在。
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由极限的线性性质(有限实数减法法则):
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$$\lim_{N \to \infty} S_N = L^+ - L^- = \sum_{x \in X} f^+(x) - \sum_{x \in X} f^-(x)$$
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考察非负部分 $f^+$。由于 $g$ 是双射且 $\sum_{x \in X} f^+(x) = +\infty$,根据非负项级数的性质(其部分和序列单调递增):
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$$\lim_{N \to \infty} \sum_{n=1}^N f^+(g(n)) = \sup_{K \subset X, |K|<\infty} \sum_{x \in K} f^+(x) = +\infty$$
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知:$S_N$极限存在。
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利用扩展实数的运算性质,若 $A_N \to +\infty$ 且 $B_N \leq L$($L$ 为常数),则:
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$$\lim_{N \to \infty} S_N \geq \lim_{N \to \infty} \left( \sum_{n=1}^N f^+(g(n)) - L \right) = (+\infty) - L = +\infty$$
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同理可证 $\sum f^+ < \infty, \sum f^- = \infty$ 时极限为 $-\infty$。

‎content/测度论与概率论/Lebesgue-integral.md‎

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## 前置
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-
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- [[测度论与概率论/almost-everywhere]]
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## 非负简单可测函数的$L$积分
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设$f=\sum_{i=1}^{n} a_i I_{A_i} \in S^+(\Omega, \mathcal{F})$, 称广义实数$\sum_{i=1}^{n} a_i \mu(A_i)$为$f$的**$L$积分**,记作$\int f d\mu$或$\mu(f)$。
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设$f=\sum_{i=1}^{n} a_i I_{A_i} \in S^+(\Omega, \mathcal{F})$, 称广义实数$\sum_{i=1}^{n} a_i \mu(A_i)$为$f$的**L积分**,记作$\int f d\mu$或$\mu(f)$。
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$\int f d\mu$与$f$的表示形式无关。
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- $f$可积
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- $\sum_{i=1}^{\infty} n \mu(n \leq |f| < n+1) < +\infty$
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- $\sum_{i=1}^{\infty} \mu(|f| \geq n) < +\infty$
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15. 若设$f \in \overline{\mathcal{L}}(\Omega, \mathcal{F})$ 的积分存在,且恒非负(或恒非正),$c \in \overline{\mathbb{R}}$,则$cf$积分存在,且$\int cf=c\int f$
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16. $A \in \mathcal{F}, \int I_A=\mu(A)$
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---
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title: 符号测度与 Hahn-J ordan 分解
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---
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## 前置
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- [[测度论与概率论/Stieltjes_integral]]
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## 定义
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- 设 $f \in \overline{\mathcal{L}}(\Omega, \mathcal{F})$ 关于 $\mu$ 的积分存在, 称如下定义的广义实值集函数
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$$\nu(A) = \int_A f \mathrm{d}\mu, \quad A \in \mathcal{F}$$
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为 $f$ 关于 $\mu$ 的**不定积分** (indefinite integral), 记作 $\nu = f \cdot \mu$
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## 性质

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