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 $x\in A$?



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


