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.


1 Answer 1


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.

  • $\begingroup$ OK thank you (Now that you gave the counterexample, it seems so obvious that I'm a bit embarrassed having asked the question in the first place) $\endgroup$
    – Florian
    Mar 20, 2012 at 13:11
  • $\begingroup$ I was trying to think of some extra condition which would save the result; but I don't see one... $\endgroup$ Mar 20, 2012 at 13:12

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