Let $\Omega$ be a measure space (which can be assumed to be an interval with Lebesgue measure).


It is well known that for a sequence $(f\_n)$ in $L^1(\Omega)$ which converges to zero (in $L^1(\Omega)$, that is, $\|f\_n\|\_{L^1(\Omega)}\to 0$) then a subsequence of $f\_n$ converges to zero pointwise almost everywhere. My question is: Is there a uniform version of this theorem in the following sense: For a countable set $A$ of sequences $(f\_n)$ in $L^1(\Omega)$ and
$$\sup\_{(f\_n)\in A} \|f\_n\|\_{L^1(\Omega)}\to 0$$
is it true that there exists a set of measure zero $N\subset \Omega$ and an increasing sequence of indices $(k\_n)$ such that for $x\notin N$,
$$\sup\_{(f\_n)\in A} |f\_{k\_n}(x)|\to 0$$
as $n\to \infty$?


Edit: "uniform convergence pointwise a.e." is obviously not such a great way to describe this kind of convergence, so if someone knows how this should be called, this might be of help too.



Uniform $L_1$ convergence implies uniform convergence pointwise a.e.
No. For each $n$ let $(A\_{n,m})$ be a sequence of subsets of $\Omega$ each with measure less than $1/n$, but with $\bigcup\_m A\_{n,m} = \Omega$ (certainly you can do this if $\Omega=[0,1]$ with Lebesgue measure).


Now set 


$f^{(m)}\_n = \chi\_{A\_{n,m}}$. Then


$$ \lim\_n \ \sup\_m \|f^{(m)}\_n\|\_1 = \lim\_n \ \sup\_m |A\_{n,m}| < \lim\_n \frac{1}{n} = 0. $$


However, for any sequence $(k\_n)$ and any $x\in\Omega$,$ \sup\_m |f^{(m)}\_{k\_n}(x)| = \sup\_m \chi\_{A\_{k\_n,m}}(x) = \chi\_{\bigcup\_m A\_{k\_n,m}}(x)
= 1.$


Thus your conclusion cannot hold.


