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| − | <math>\sup\int_{\{|f_n|>M\}}|f_n|\leq\sup\int_{(0,1)}|f_n-f|+\sup\int_{\{|f_n|>M\}}|f|</math> | + | <math>\sup\limits_n\int_{\{|f_n|>M\}}|f_n|\leq\sup\int_{(0,1)}|f_n-f|+\sup\int_{\{|f_n|>M\}}|f|</math> |
<math>Since \int_{(0,1)}|f_n-f|\to0(n\to\infty), \sup\int_{(0,1)}|f_n-f|=0</math> | <math>Since \int_{(0,1)}|f_n-f|\to0(n\to\infty), \sup\int_{(0,1)}|f_n-f|=0</math> | ||
To show <math>\sup\int_{\{|f_n|>M\}}|f_n|\to0(M\to\infty),</math>it suffices to show that <math>\sup\int_{\{|f_n|>M\}}|f|\to0(M\to\infty)</math> | To show <math>\sup\int_{\{|f_n|>M\}}|f_n|\to0(M\to\infty),</math>it suffices to show that <math>\sup\int_{\{|f_n|>M\}}|f|\to0(M\to\infty)</math> | ||
Revision as of 08:59, 2 July 2008
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$ \sup\limits_n\int_{\{|f_n|>M\}}|f_n|\leq\sup\int_{(0,1)}|f_n-f|+\sup\int_{\{|f_n|>M\}}|f| $
$ Since \int_{(0,1)}|f_n-f|\to0(n\to\infty), \sup\int_{(0,1)}|f_n-f|=0 $
To show $ \sup\int_{\{|f_n|>M\}}|f_n|\to0(M\to\infty), $it suffices to show that $ \sup\int_{\{|f_n|>M\}}|f|\to0(M\to\infty) $
