(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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| + | 2. Let <math>(X,\mathcal{F},\mu)</math> be a measure space and <math>f \in L(\mu)</math>. Show that <math>\{f \ne 0\}</math> is <math>\sigma</math>-finite. | ||
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| + | '''[[Solution]]''' | ||
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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 | 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_n.</math> | <math>\{f \ne 0\} = \cup_{n=1}^\infty F_n.</math> | ||
| − | Clearly all of the <math>F_n</math> must have finite measure, else <math>\int_X |f|d\mu</math> would be infinite. Thus, <math>\{f \ne 0\}</math> is <math>\sigma</math>-finite. | + | Clearly all of the <math>F_n</math> must have finite measure, else <math> \int_X |f|d\mu </math> would be infinite. Thus, <math>\{f \ne 0\}</math> is <math>\sigma</math>-finite. |
Revision as of 09:19, 6 July 2009
2. Let $ (X,\mathcal{F},\mu) $ be a measure space and $ f \in L(\mu) $. Show that $ \{f \ne 0\} $ is $ \sigma $-finite.
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.
