(New page: Since <math> f \in L(\mu) </math>, we know that <math> \int_X |f|d\mu < \infty </math>. Let <math>F_{n} \equiv \{|f| > 1/n\}</math>. Then we have <math>\{f \ne 0\} = \cup_{n=1}^\infty F...) |
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Revision as of 09:12, 6 July 2009
Since $ f \in L(\mu) $, we know that $ \int_X |f|d\mu < \infty $. Let $ F_{n} \equiv \{|f| > 1/n\} $. Then we have $ \{f \ne 0\} = \cup_{n=1}^\infty F_n. $
Clearly all of the $ F_n $ must have finite measure, else $ \int_X |f|d\mu $ would be infinite. Thus, $ \{f \ne 0\} $ is $ \sigma $-finite.
