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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 $f_n(x):=c_n\mathcal{X}_{E_n}(x).$ Then $f_n$ converges in measure to zero, iff $c_n\to 0$, or $m(E_n)\to 0$ as $n\to \infty$.

My approach:

($\Rightarrow$) Suppose that $f_n\to 0$ in measure. For all $\epsilon >0$, let $E:=\{x: |f_n(x)|\geq \epsilon\}.$ I feel that it is obvious, but here is my reasoning anyway: as $n \to \infty$, $\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}c_n\mathcal{X}_{E_n}(x)$, now if $x\in E_n$ then clearly $m(E_n)\to 0$ as $n\to \infty$ and if $x\notin E_n$ then by assumption we will have $c_n\to 0.$

($\Leftarrow$) I just realized this direction is the obvious one! Isn't it? If $c_n\to 0$ or $m(E_n)\to 0$ as $n\to \infty$ then hmmm... I need help!

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2 Answers

up vote 1 down vote accepted

A quite detailed explanation follows. If you have any questions, do not hesitate to ask me. By definition a function converges in measure to zero iff for every $\epsilon>0$ $$m(x\in X : |f_n(x)|>\epsilon) \rightarrow 0 $$ as $n \rightarrow 0 $. Your function is positive for all $x\in X$. Hence you may "neglect" the absolute value signs.

Let $\epsilon>0$ be given. If $c_n \rightarrow 0$ as $n \rightarrow \infty$, then there is a $N \in \mathbb{N}$ such that $c_n<\epsilon$ for $n\geq N$. Hence if $n\geq N$ then $m(x\in X : f_n(x)>\epsilon) = 0 $. If on the other hand $m(E_n)\rightarrow 0$ as $n \rightarrow \infty$, choosing $N \in \mathbb {N}$ such that $m(E_n)<\epsilon$ for $n\geq N$ we have: $$m(x\in X : f_n(x)>\epsilon)\leq m(E_n)<\epsilon{}$$ for $n\geq N$

For the other direction assume that $f_n(x)$ converges in measure to zero. If $c_n$ does not converge to zero, then by assumption in the exercise $c_n\geq c>0$ all $n$. Choosing $\epsilon<c$, we get $m(x\in X : f_n(x)>\epsilon)=m(E_n) $ all $n$. This is a contradiction unless $m(E_n)\rightarrow 0$ as $n\rightarrow{} \infty$. Lastly assume that $m(E_n)$ does not converge to zero. Then there is a $\delta>0$ such that for all $N$ there is a $n\geq N$ such that $m(E_n)>\delta$. If $c_n\geq c>\epsilon>0$ all $n \in \mathbb{N}$, then $m(x\in X:f_n(x)>\epsilon)=m(E_n)$ all $n\in \mathbb{N}$, and hence for every $N \in \mathbb{N}$ there is a $n\geq N$, $m(x\in X:f_n(x)>\epsilon)>\delta$. Contradicting the convergence in measure of the $f_n$:s.

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I'll try to give a hint:

concerning $(\Rightarrow)$: So you suppose $f_n \to 0$ in measure. Your task is to prove $c_n \to 0$ or $m(E_n) \to 0$. You define a set $E$ (depending on some $n$, which?, and $\epsilon$) and don't use it anymore. I'd argue as follows: So we have to prove $c_n \to 0$ or $m(E_n) \to 0$. Supposing that $c_n \not\to 0$, we have by assumption $c_n \ge c > 0$. So $f_n \ge c\chi_{E_n}$, but $f_n \to 0$ in measure. We have that for every $\epsilon > 0$ $m(|f_n| > \epsilon) \to 0$. What $\epsilon$ may help (compared to $c$)? Try to picture the set $\{|f_n| > \epsilon\}$ for different $\epsilon$.

concering $(\Leftarrow)$: As you say, this direction is more obvious (IMO). Consider the cases $c_n \to 0$ and $c_n \ge c > 0$ and $m(E_n)\to 0$ separately. If $c_n \to 0$ we have $f_n \to 0$ uniformly, so ... otherwise if $c_n$ is bounded below, you can use your picture of $\{|f_n| > \epsilon\}$ from the other direction. Now use $m(E_n) \to 0$.

Hope that helps.

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Did you mean to say suppose $c_n$ does not converge to $0$ in your hint for forward direction? –  Lyapunov Mar 26 '12 at 9:49
    
Yes. We have to prove $c_n \to 0$ or $m(E_n) \to 0$. We consider to cases: If $c_n \to 0$ then we are done, otherwise $c_n \not\to 0$ and we can continue as above. –  martini Mar 26 '12 at 9:53
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