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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 <math>E_n=\{x \in X : n-1 \leq |f| \leq n\}</math>, then | 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> \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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| + | Then 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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| + | If <math>f \in L^1</math>, then <math>\sum_{n = 1}^{\infty}(n-1)\mu(E_n) < \infty</math>. Then <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. | ||
Revision as of 01:57, 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) $.
Then if $ \sum_{n = 1}^{\infty}n\mu(E_n) < \infty $, then $ \int_X|f|\mbox{d}\mu < \infty, $, i.e. $ f \in L^1 $.
If $ f \in L^1 $, then $ \sum_{n = 1}^{\infty}(n-1)\mu(E_n) < \infty $. Then $ \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.
