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There is a proposition that states the following: Assume $E$ has finite measure. Let {$f_n$} be a sequence of measurable functions that converges pointwise a.e. on $E$ to $f$ and $f$ is finite a.e. on $E$. Then {$f_n$} converges in measure to $f$ on $E$. How do you show that this fails if $E$ has infinite measure?

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By finding a counterexample. –  cardinal Mar 28 '11 at 22:59
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@cardinal: Do you mean like {$f_n$} = $\chi_{[n,n+1]}$ –  Libertron Mar 28 '11 at 23:02
    
@Sachin: Thank you for editing. I've deleted my no longer relevant comments. @Whomever: In case anyone is wondering about the close votes, they were from before the question was edited. –  Jonas Meyer Mar 28 '11 at 23:03
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@Sachin: That is a good one. –  Jonas Meyer Mar 28 '11 at 23:03
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No this sequence does not converge uniformly to zero. It only converges pointwise. In fact $\Vert f_{n} - 0\Vert_\infty = 1$ for all $n$, while $f_{n}(x) = 0$ for all $n \geq x$ (that's what Jonas had in mind in his answer below). As I explained in my answer to your previous question, uniform convergence implies convergence in measure. Cardinal explained why convergence in measure fails. –  t.b. Mar 28 '11 at 23:15

2 Answers 2

up vote 6 down vote accepted

To see that the proposition is not true in general if $E$ does not have finite measure, you just need a counterexample. My suggestion is to start with $E=\mathbb{R}$, and come up with a sequence of measurable functions $(f_n)_{n\geq 1}$ converging pointwise to $0$, while the measure of the set where $f_n(x)\geq1$ is always infinite. For instance, if you have $f_n(x)=0$ when $x<n$, then the sequence will converge to $0$ at each point regardless of what $f_n(x)$ is when $x\geq n$. This gives a lot of freedom for strange counterexamples.

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We make the comment of Sachin clearer. Choose $E=\mathbb{R}$ and $\mu$ is the Lebesgue measure on $E$. Consider the sequence of measurable functions $\{f_n(x)\}_{n\in \mathbb{N}}$ given by $$ f_n(x)=\begin{cases} 1 & n\leq x\leq n+1, \\ 0& \text{ortherwise}. \end{cases} $$ for all $x\in E$ and $n\in \mathbb{N}$. Then:

  • For every $x\in E$ we have $x\notin [n,n+1]$ or $f_n(x)=0$ for sufficiently large $n$ and so $$|f_n(x)-0|=|0-0|=0$$ for sufficiently large $n$. This implies that $$\displaystyle\lim_{n\rightarrow\infty}f_n(x)= 0$$ for all $x\in E$. Therefore the sequence $\{f_n(x)\}_{n\in \mathbb{N}}$ is pointwise convergent to $f=0$.

  • We observe that $$ \left\{x\in E: |f_n(x)-0|\geq \frac{1}{2}\right\}=[n, n+1]. $$ and so $$ \mu\left\{x\in E: |f_n(x)-0|\geq \frac{1}{2}\right\}=\mu([n, n+1])=1. $$ Hence $\{f_n(x)\}_{n\in\mathbb{N}}$ is not convergent in measure to $f=0$.

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