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<math>\sup\limits_n\int_{\{|f_n|>M\}}|f_n|\leq\sup\limits_n\int_{(0,1)}|f_n-f|+\sup\limits_n\int_{\{|f_n|>M\}}|f|</math> | <math>\sup\limits_n\int_{\{|f_n|>M\}}|f_n|\leq\sup\limits_n\int_{(0,1)}|f_n-f|+\sup\limits_n\int_{\{|f_n|>M\}}|f|</math> | ||
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<math>Since \int_{(0,1)}|f_n-f|\to0(n\to\infty), \sup\limits_n\int_{(0,1)}|f_n-f|=0</math> | <math>Since \int_{(0,1)}|f_n-f|\to0(n\to\infty), \sup\limits_n\int_{(0,1)}|f_n-f|=0</math> | ||
| − | + | Therefore, to show <math>\sup\limits_n\int_{\{|f_n|>M\}}|f_n|\to0(M\to\infty),</math>it suffices to show that <math>\sup\limits_n\int_{\{|f_n|>M\}}|f|\to0(M\to\infty)</math> | |
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| + | Actually, | ||
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| + | <math>\int_{\{f_n>M\}}|f|\leq\int_{\{|f_n|>M & |f|<M-\delta\}}|f|+\int_{\{|f|>M-\delta\}}|f|</math> | ||
Revision as of 09:06, 2 July 2008
$ \sup\limits_n\int_{\{|f_n|>M\}}|f_n|\leq\sup\limits_n\int_{(0,1)}|f_n-f|+\sup\limits_n\int_{\{|f_n|>M\}}|f| $
$ Since \int_{(0,1)}|f_n-f|\to0(n\to\infty), \sup\limits_n\int_{(0,1)}|f_n-f|=0 $
Therefore, to show $ \sup\limits_n\int_{\{|f_n|>M\}}|f_n|\to0(M\to\infty), $it suffices to show that $ \sup\limits_n\int_{\{|f_n|>M\}}|f|\to0(M\to\infty) $
Actually,
$ \int_{\{f_n>M\}}|f|\leq\int_{\{|f_n|>M & |f|<M-\delta\}}|f|+\int_{\{|f|>M-\delta\}}|f| $
