3
$\begingroup$

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?

  • 1
    By finding a counterexample.2011-03-28
  • 2
    @cardinal: Do you mean like {$f_n$} = $\chi_{[n,n+1]}$2011-03-28
  • 0
    @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.2011-03-28
  • 2
    @Sachin: That is a good one.2011-03-28
  • 0
    @Jonas Meyer: How can I give a good explanation though? I guess it is clear that this particular sequence of functions converges uniformly to 0. What should I do about the convergence in measure?2011-03-28
  • 0
    $\mu(\{x:|f_n(x)| > \epsilon\}) = 1$ for all $\epsilon < 1$. What more do you need?2011-03-28
  • 1
    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.2011-03-28
  • 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 2

6

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

0

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 $f=0$.