Welcome to MathOverflow

Question

0

Here you are another question in basic measure theory...

Let $f_k$ be a measurable sequence of functions on $(X,M,\mu)$ measure space. Suppose that $f_k$ does not go to 0 a.e.. Can I then find a set $A\subseteq X$ with positive measure and a subsequence $f_{k_j}$ and an $\varepsilon > 0$ such that $\liminf_j |f_{k_j}(x)| > \varepsilon$ foreach $x\in A$?

edited Apr 4 2010 at 16:34

asked Apr 4 2010 at 16:24

Nicolò
512●3●17

2

Your measure theory textbook surely has a discussion of this. Perhaps called "convergence in measure". The "no" answer below shows that convergence in measure does not imply a.e. convergence. This same example is probably in your textbook. – Gerald Edgar Apr 4 2010 at 18:35

Thanks for your comment. I know that convergence in measure does not imply a.e. convergence (I use Folland's Real Analysis, and there there is a good section on convergence in measure), but my question was more about the inverse, I think. – Nicolò Apr 5 2010 at 15:14

Convergence a.e. does not imply convergence in measure. Consider f_{n} defined as f_{n}(x)=1 x>n and f_{n}(x)=0 x<=n. – Myke Jul 20 2010 at 18:54

Nicolò · Accepted Answer · 2010-04-04 16:42:39Z UTC

That's not true. For example, in $(0,1)$ take

$f_1 =1$,

$f_2=1_{(0,1/2)}$, $f_3= 1_{(1/2,1)}$

$f_4=1_{(0,1/3)}$, $f_5= 1_{(1/3,2/3)}$, $f_6= 1_{(2/3,1)}$

and so on. $f_k(x)$ does not go to 0 a.e. (the limit does not exist, for each x), but we can't find any succession that satisfies the statement, because $m(supp f_k)$ goes to zero

If f_k does not go to 0 a.e., there exists a subsequence, a set of positive measure, and c > 0, on which liminf |f_k_j| > c ?| >

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.