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| − | Since <math> \int_X|f|\mbox{d}\mu = \sum_{n = 1}^{\infty}\int_{E_n}|f|\mbox{d}\mu </math>, and E_n={x \in X : n-1 \ | + | Since <math> \int_X|f|\mbox{d}\mu = \sum_{n = 1}^{\infty}\int_{E_n}|f|\mbox{d}\mu </math>, and <math>E_n=\{x \in X : n-1 \leq |f| \leq n\}</math>, then |
| + | <math> \sum_{n = 1}^{\infty}(n-1)\mu(E_n) \leq \sum_{n = 1}^{\infty}\int_{E_n}|f|\mbox{d}\mu \leq\sum_{n = 1}^{\infty}n\mu(E_n)</math>. | ||
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| + | <math>(\Leftarrow)</math> | ||
| + | If <math>\sum_{n = 1}^{\infty}n\mu(E_n) < \infty</math>, then <math> \int_X|f|\mbox{d}\mu < \infty, </math>, i.e. <math>f \in L^1</math>. | ||
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| + | <math>(\Rightarrow)</math> | ||
| + | If <math>f \in L^1</math>, then <math>\sum_{n = 1}^{\infty}(n-1)\mu(E_n) < \infty</math>. <math>\sum_{n = 1}^{\infty}n\mu(E_n) = \sum_{n = 1}^{\infty}(n-1)\mu(E_n) + \sum_{n = 1}^{\infty}\mu(E_n) = \sum_{n = 1}^{\infty}(n-1)\mu(E_n) + \mu(X) < \infty </math>, since <math>X</math> is a finite measure space. | ||
Latest revision as of 01:59, 10 July 2008
Since $ \int_X|f|\mbox{d}\mu = \sum_{n = 1}^{\infty}\int_{E_n}|f|\mbox{d}\mu $, and $ E_n=\{x \in X : n-1 \leq |f| \leq n\} $, then $ \sum_{n = 1}^{\infty}(n-1)\mu(E_n) \leq \sum_{n = 1}^{\infty}\int_{E_n}|f|\mbox{d}\mu \leq\sum_{n = 1}^{\infty}n\mu(E_n) $.
$ (\Leftarrow) $ If $ \sum_{n = 1}^{\infty}n\mu(E_n) < \infty $, then $ \int_X|f|\mbox{d}\mu < \infty, $, i.e. $ f \in L^1 $.
$ (\Rightarrow) $ If $ f \in L^1 $, then $ \sum_{n = 1}^{\infty}(n-1)\mu(E_n) < \infty $. $ \sum_{n = 1}^{\infty}n\mu(E_n) = \sum_{n = 1}^{\infty}(n-1)\mu(E_n) + \sum_{n = 1}^{\infty}\mu(E_n) = \sum_{n = 1}^{\infty}(n-1)\mu(E_n) + \mu(X) < \infty $, since $ X $ is a finite measure space.
