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| − | \int_{\{|f_n|>M\}}|f_n|\leq\int_{(0,1)}|f_n-f| | + | # |
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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> | ||
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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> | ||
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| + | 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-\epsilon\}}|f|+\int_{\{|f|>M-\epsilon\}}|f|</math> | ||
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| + | <math>\leq\int_{\{|f-f_n|>\epsilon\}}|f|+\int_{|f|>M-\epsilon}|f|</math> | ||
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| + | <math>f_n\to f(L^1)=>f_n\to f(\mu),</math>so, by absolute continuity, <math>\int_{\{|f-f_n|>\epsilon\}}|f|\to0(n\to\infty)</math> | ||
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| + | So, <math>\sup\limits_n\int_{\{|f_n|>M\}}|f|\leq\int_{\{|f|>M-\epsilon\}}|f|\to0(M\to\infty)</math>,since <math>f\in L^1</math> | ||
Latest revision as of 09:27, 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\}}|f|+\int_{|f|>M-\epsilon}|f| $
$ f_n\to f(L^1)=>f_n\to f(\mu), $so, by absolute continuity, $ \int_{\{|f-f_n|>\epsilon\}}|f|\to0(n\to\infty) $
So, $ \sup\limits_n\int_{\{|f_n|>M\}}|f|\leq\int_{\{|f|>M-\epsilon\}}|f|\to0(M\to\infty) $,since $ f\in L^1 $
