# If $f_k \not\to 0$ a.e., does there exist a subsequence, a set of positive measure, and $c > 0$, on which $\liminf |f_{k_j}| > c$?

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 $x\in A$?

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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 '10 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 '10 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. –  Digital Gal Jul 20 '10 at 18:54

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