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This problem is related to Convergence in measure to zero with certain conditions.

(Let $\{f_n\}_{n\in \mathbb{N}}$ be a sequence of measurable functions on a measure space and $f$ measurable.

For $c_n>0$ such that either $\lim_{n\to \infty}c_n=0$, or $c_n\geq c>0$ for all $n$, and measurable sets $E_n$ with $m(E_n)>0$ consider the sequence $f_n(x):=c_n\mathcal{X}_{E_n}(x).$ )

For the same sequence(same assumptions!) $\{f_n\}_{n\in\mathbb{N}}$, $f_n$ converges almost uniformly to zero, iff $c_n\to 0$ as $n\to \infty$ or $m(\cup_{n\geq N}E_n)\to 0$ as $N\to \infty.$

My approach: ($\Rightarrow$) By definition of almost uniformly, we have that for all $\epsilon>0$ there exists $A_\epsilon $ such that $m(A_\epsilon)<\epsilon$ such that $f_n$ converges to uniformly to $0$ on $A_\epsilon^c$. Observe that we have $A_\epsilon \subset \cup_{n\geq N}E_n$, so $\epsilon\geq m(A_\epsilon)\leq m(\cup_{n\geq N}E_n)$. Now if $x\in A_\epsilon$ then clearly $m(\cup_{n\geq N}E_n)\to 0$ as $N\to \infty$ and if $x\notin A_\epsilon$ then again clearly as $f_n(x)=c_n\mathcal{X}_{E_n}(x)\to 0$, $c_n\to 0.$

($\Leftarrow$) I think to prove this direction is hard.

Any help, comments and suggestions will be appreciated.

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as in your other question I think it is a good idea to suppose for $(\Rightarrow)$ that we have $c_n \not\to 0$ (and therefore $c_n \ge c > 0$). I don't understand what you mean when you write "Now if $x \in A_\epsilon$ then clearly $m(\bigcup_{n\ge N} E_n) \to 0$ as ...". How can you conclude from $x \in A_\epsilon$ something about the measure of some set which doesn't depend on $x$. And why $A_\epsilon \subseteq \bigcup_{n\ge N} E_n$?

For $(\Rightarrow)$: Suppose $c_n \ge c > 0$. We want to prove $m(\bigcup_{n\ge N} E_n) \to 0$. For $\epsilon > 0$ choose $A_\epsilon$ as you did. Now $f_n \to 0$ uniformly on $A_\epsilon^c$. Now use again (as in the other task) that $f_n \ge c\chi_{E_n}$. So $\chi_{E_n\cap A_\epsilon^c} \to 0$ uniformly. Does this tell you something?

$(\Leftarrow)$ is IMO more direct. If $c_n \to 0$ use $|f_n| \le c_n$ to conclude uniform convergence, otherwise for some $\epsilon$ you can find a $N$ with $m(\bigcup_{n\ge N} E_n) < \epsilon$. Can you show convergence outside this set?

Hope this helps, otherwise feel free to ask more.

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Thanks for your help. ($\Rightarrow$): I think it tells me that for each $n\geq N$, $m(E_n\cap A_\epsilon^c)\to 0$? Then $m(\cup_{n\geq N}E_n \cap A_\epsilon)=m(\cup_{n\geq N}(E_n\cap A_\epsilon))\to 0$. –  Lyapunov Mar 26 '12 at 10:48
    
Not exactly. What is the range of $\chi_{E_n \cap A_\epsilon^c}$? What follows if such functions converge to $0$? –  martini Mar 26 '12 at 10:55
    
Range of $\chi_{E_n\cap A_\epsilon^c}$ is $0$ or $1$. Then $x\notin E_n\cap A_\epsilon^c$. –  Lyapunov Mar 26 '12 at 12:10
    
@SriPot Exactly. For which $x$? –  martini Mar 26 '12 at 12:30
    
for a.e. $x$? Is there a theorem I should be using now? Bounded Convergence Theorem? –  Lyapunov Mar 26 '12 at 13:34
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