Theorem: 假設 $\{f_n\}\subset L^1$ 使得 $\sum_{n=1}^\infty\int |f_n| <\infty$。則 (a) $\sum_{n=1}^\infty f_n$ almost everywhere 收斂 。 (b) $\int \sum_{n=1}^\infty f_n = \sum_{n=1}^\infty \int f_n$。 Proof: (a) 令$\{f_n\}\subset L^1$ 使得 $\sum_{n=1}^\infty\int |f_n| <\infty$,則我們有 \[ \int \sum_{n=1}^\infty |f_n| = \sum_{n=1}^\infty\int |f_n| <\infty \]故我們知道 $ \sum_{n=1}^\infty |f_n| \in L^1$ 且 $\sum_{n=1}^\infty |f_n|<\infty$ almost everywhere。此表明對 almost every $x$, $\sum_{n=1}^\infty f_n(x) $ (絕對)收斂。令極限 $$ f(x):=\lim_{N\to \infty} \sum_{n=1}^N f_n(x) $$ 故我們僅需證明此 $f \in L^1$。注意到 對任意 $N$, $\lim_N \sum_{n=1}^N f_n \leq \lim_N \sum_{n=1}^\infty |f_n|$ a.e. 故 $f \leq \lim_N \sum_{n=1}^N |f_n|$ a.e. 由於 $\lim_N \sum_{n=1}^N |f_n| \in L^1$ 故 $f \in L^1$。 Proof (b) 要證明 $\int \sum_{n=1}^\infty f_n = \sum_{n=1}^\infty \int f_n$。首先觀察 \[ \int \sum_{n=1}^N f_n = \sum_{n=1}^N \int f_n \;\;\;\; (*) \]由於 $\sum_{n=1}^N f_n \leq \sum_{n=1}^\infty |f_n| \in L^1$ 且 $\sum_{n=1}^N f_n \to
If you can’t solve a problem, then there is an easier problem you can solve: find it. -George Polya