单调收敛定理

单调级数的收敛性

定理

$\lim_{j\to\infty} \sum_k a_{j,k} = \sum_k \lim_{j\to\infty} a_{j,k}$

勒贝格单调收敛定理

定理

$\forall x\in X, k\in \mathbb{N}, f_k(x) \leq f_{k+1}(x)$

$\forall x\in X, f(x):= \lim_{k\to\infty} f_k(x)$

$\lim_{k\to\infty} \int f_k d\mu = \int f d\mu$

证明

$f^{-1}(I) = \{x\in X | f(x)\in I \}$

$f(x)\in I \Leftrightarrow f_k(x)\in I, ~ \forall k\in \mathbb{N}$

$\{x\in X | f(x)\in I\} = \bigcap_{k\in \mathbb{N}} \{x\in X | f_k(x)\in I\}$

$\int f d \mu = sup \{\int g d \mu | g \in SF, g\leq f \}$

$\left\{\int g d \mu | g \in SF, g\leq f_k \right\}\subseteq \left\{\int g d \mu | g \in SF, g\leq f \right\}.$

$\int f d \mu \leq \lim_k \int f_k d \mu.$

$\lim_k \int g_k d \mu = \int f d \mu.$

$\int g_k d \mu \leq \lim_j \int f_j d \mu$

$\lim_j f_j(x) \geq g_k(x)$

$\lim_j \int f_j d \mu \geq \int g_k d \mu.$

$B_n = \{x \in B: f_n(x) \geq 1 - \epsilon \}.$

$\mu(B_n) (1 - \epsilon) = \int (1 - \epsilon) 1_{B_n} d \mu \leq \int f_n d \mu$

$\bigcup_n B_n = B$

$\int g_k d\mu =\int 1_B d\mu = \mu(B) = \mu(\bigcup_n B_n) .$

$\mu(\bigcup_n B_n)=\lim_n \mu(B_n) \leq \lim_n (1 - \epsilon)^{-1} \int f_n d\mu$

$k \rightarrow \infty$，并利用这对任何正数$\epsilon$都正确的事实，定理便得证。

注释

1. ^ J Yeh. Real analysis. Theory of measure and integration. 2006.
2. ^ Erik Schechter. 21.38. Analysis and Its Foundations. 1997.