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My question is that if $\sigma_{n}(x)\to f(x)$ almost everywhere, where $f(x)$ is essentially bounded function and $|\sigma_{n}(x)|\leq K$ then how can we prove that $|f(x)|\leq K$

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    Kns last inequality you mean almost surely ?2012-06-15
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    yes i want to show that $|f(x)|\leq K$ a.e..2012-06-15
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    Just take the limit in the inequality $|\sigma_{n}(x)|\leq K$2012-06-15
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    So we get $|f(x)|\leq K$ a.e., it implies $f(x)$ is essentially bounded. does it mean $f(x)$ is bounded?2012-06-15
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    No see the William's answer. It is not possible, in general, to show that $f$ is bounded. Another very simple example can be constructed by using a Dirac delta measure.2012-06-15

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