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?
Convergence in measure
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0@Sachin: It does not converge uniformly to $0$. As Theo pointed out in an answer to a question you asked a little while ago, uniform convergence would imply convergence in measure. It does converge pointwise to zero, and it does not converge in measure to 0 for the reason cardinal gave. – 2011-03-28
2 Answers
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
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 $f=0$.