# 单调收敛定理

## 单调级数的收敛性

### 定理

${\displaystyle \lim _{j\to \infty }\sum _{k}a_{j,k}=\sum _{k}\lim _{j\to \infty }a_{j,k}}$

## 勒贝格单调收敛定理

### 定理

${\displaystyle 0\leq f_{k}(x)\leq f_{k+1}(x)}$

${\displaystyle f(x):=\lim _{k\to \infty }f_{k}(x)}$

${\displaystyle f}$ 會是 ${\displaystyle \mu }$-可测函數，且：

${\displaystyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu =\int _{X}f\,d\mu }$。（参见[2]第21.38节）

### 证明

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

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

${\displaystyle \{x\in X|f(x)\in I\}=\bigcap _{k\in \mathbb {N} }\{x\in X|f_{k}(x)\in I\}}$

${\displaystyle \int fd\mu =sup\{\int gd\mu |g\in SF,g\leq f\}}$

${\displaystyle \left\{\int gd\mu |g\in SF,g\leq f_{k}\right\}\subseteq \left\{\int gd\mu |g\in SF,g\leq f\right\}.}$

${\displaystyle \int fd\mu \leq \lim _{k}\int f_{k}d\mu .}$

${\displaystyle \lim _{k}\int g_{k}d\mu =\int fd\mu .}$

${\displaystyle \int g_{k}d\mu \leq \lim _{j}\int f_{j}d\mu }$

${\displaystyle \lim _{j}f_{j}(x)\geq g_{k}(x)}$

${\displaystyle \lim _{j}\int f_{j}d\mu \geq \int g_{k}d\mu .}$

${\displaystyle B_{n}=\{x\in B:f_{n}(x)\geq 1-\epsilon \}.}$

${\displaystyle \mu (B_{n})(1-\epsilon )=\int (1-\epsilon )1_{B_{n}}d\mu \leq \int f_{n}d\mu }$

${\displaystyle \bigcup _{n}B_{n}=B}$

${\displaystyle \int g_{k}d\mu =\int 1_{B}d\mu =\mu (B)=\mu (\bigcup _{n}B_{n}).}$

${\displaystyle \mu (\bigcup _{n}B_{n})=\lim _{n}\mu (B_{n})\leq \lim _{n}(1-\epsilon )^{-1}\int f_{n}d\mu }$

${\displaystyle k\rightarrow \infty }$，并利用这对任何正数${\displaystyle \epsilon }$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.