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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> | ||
Revision as of 09:01, 2 July 2008
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$ \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 $
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) $
