| Line 1: | Line 1: | ||
| − | 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). | ||
Revision as of 01:48, 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). $
