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<math>\int_{\{|f_n|>M\}}|f_n|\leq\int_{(0,1)}|f_n-f|+\int_{\{|f_n|>M\}}|f|</math>
 
<math>\int_{\{|f_n|>M\}}|f_n|\leq\int_{(0,1)}|f_n-f|+\int_{\{|f_n|>M\}}|f|</math>
  
<math>Since \int_{(0,1)}|f_n-f|\to0(n\to\infinity),</math>
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<math>Since \int_{(0,1)}|f_n-f|\to0(n\to\infty), it suffices to show\sup\int_{\{|f_n|>M\}}|f|\to0(M\to\infty)</math>

Revision as of 09:52, 2 July 2008

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$ \int_{\{|f_n|>M\}}|f_n|\leq\int_{(0,1)}|f_n-f|+\int_{\{|f_n|>M\}}|f| $

$ Since \int_{(0,1)}|f_n-f|\to0(n\to\infty), it suffices to show\sup\int_{\{|f_n|>M\}}|f|\to0(M\to\infty) $

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