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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 ? –  Leandro Jun 15 '12 at 7:30
    
yes i want to show that $|f(x)|\leq K$ a.e.. –  Kns Jun 15 '12 at 7:33
    
Just take the limit in the inequality $|\sigma_{n}(x)|\leq K$ –  Leandro Jun 15 '12 at 7:34
    
So we get $|f(x)|\leq K$ a.e., it implies $f(x)$ is essentially bounded. does it mean $f(x)$ is bounded? –  Kns Jun 15 '12 at 7:37
    
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. –  Leandro Jun 15 '12 at 7:51

1 Answer 1

Suppose that $E = \{x : f(x) > K\}$ is not measure zero. Then for all $x \in E$, $\sigma_n(x)$ does not converge to $f(x)$ since $f(x) > K$ but $\sigma_n(x) \leq K$. Hence $\sigma_n$ does not converge to $f$ almost everywhere since it does not converge on $E$ which is not measure zero. Contradiction!

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Upvoted, but with a reservation. The inequality $\sigma_n(x)\le K$ holds only a.e. When learning a.e. convergence, it is a good practice to give names to exceptional sets: for example, $E_n=\{x\colon |\sigma_n(x)|>K\}$, and $F=\{x\colon \sigma_n(x)\to f(x)\}^c$. Now let $E=F\cup\bigcup_n E_n$. Since $E$ is a countable union of null sets, it is a null set. For all $x\notin E$ we have $|f(x)|=\lim_{n\to\infty} |\sigma_n(x)|\le K$. QED. –  user31373 Jun 15 '12 at 17:38

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