| Line 9: | Line 9: | ||
Actually, | Actually, | ||
| − | <math> | + | <math>\int_{\{f_n>M\}}|f|\leq\int_{\{|f_n|>M,|f|<M-\epsilon\}}|f|+\int_{\{|f|>M-\epsilon\}}|f|</math> |
| − | \int_{\{f_n>M\}}|f|\leq\int_{\{|f_n|>M,|f|<M-\epsilon\}}|f|+\int_{\{|f|>M-\epsilon\}}|f| | + | |
| − | + | <math>\leq\int_{\{|f-f_n|>\epsilon\}}+\int_{|f|>M-\epsilon}|f|</math> | |
| − | </math> | + | |
Revision as of 09:11, 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-\epsilon\}}|f|+\int_{\{|f|>M-\epsilon\}}|f| $
$ \leq\int_{\{|f-f_n|>\epsilon\}}+\int_{|f|>M-\epsilon}|f| $
