0
$\begingroup$

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 $x\in A$?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ 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. $\endgroup$
    – Gerald Edgar
    Commented Apr 4, 2010 at 18:35
  • $\begingroup$ 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. $\endgroup$
    – Nicolò
    Commented Apr 5, 2010 at 15:14
  • $\begingroup$ 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. $\endgroup$
    – Mykie
    Commented Jul 20, 2010 at 18:54

1 Answer 1

Reset to default
3
$\begingroup$

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

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .