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
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