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Consider a measure space $(X, \mathcal{X}, \mu)$, with $\mu(X)< \infty$. For measurable functions $(f_n)_{n \geq 1}, f$ we know that

$\Vert f_n -f \Vert_{L_1(\mu)}:=\int_X|f_n(x)-f(d)|\mu(dx)\to_{n \to \infty}0$

entails that there exists a subsequence $(f_{n_j})$ converging to $f$ almost uniformly and, hence, pointwise almost everywhere. Under which conditions such a statement can be extended to the entire sequence $(f_n)_{n \geq 1}$?

I know that a relatively strong additional condition that does the job is $$ \sum_{n \geq 1}\Vert f_n -f \Vert_{L_1(\mu)} <\infty. $$

Is there anything milder, concerning for example continuity or nonnegativity of $f_n$ and $f$? In particular, what if, for example, $X=[0,1]^d$, for some $d \in \mathbb{N}_+$, and $\mu$ is the Lebesgue measure? In this case could we combine Theorem 5 in

https://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/

and possibly some relation between convergence in (Lebesgue) measure with pointwise a.e. convergence?

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    $\begingroup$ The criterion is called Stein's maximal principle. $\endgroup$
    – Giuseppe Negro
    Jul 23, 2019 at 8:45
  • $\begingroup$ Actually, the operator formulation of Stein's maximal principle makes it a bit hard to digest to me. Would you mind giving some hints in a $(f_n)$ sequence formulation, as considered above? $\endgroup$
    – Jack London
    Jul 23, 2019 at 9:12
  • $\begingroup$ This is an interesting question. The "fast L1 convergence" of Exercise 5 of the linked Tao's blog page gives almost uniform convergence, which is more than you want. I conjecture that the corresponding criterion for a.e. convergence is the one in my community wiki answer, but I have not proven that and I am not even sure it is correct. $\endgroup$
    – Giuseppe Negro
    Jul 23, 2019 at 14:23
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    $\begingroup$ Since in my question I'm only considering a finite measure space, by Egorov’s theorem a.e. pointwise convergence and almost uniform convergence are equivalent (e.g. Theorem 2 in terrytao.wordpress.com/2010/10/02/…). Outside a finite measure framework, of course, almost uniform convergence is more than I want. $\endgroup$
    – Jack London
    Jul 23, 2019 at 15:32

1 Answer 1

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WARNING: I am not sure this is true, I am just conjecturing.

If $\|f_n\|_{L^1}\to 0$, the maximal principle of Stein makes me think that a necessary and sufficient condition for a.e.-pointwise convergence $f_n(x)\to 0$ is $$ \sup_{t>0} t \left\lvert \{x\ :\ f^\star(x)>t\}\right\rvert <\infty, $$ where $\lvert\cdot\rvert$ denotes the measure of a set, and $$ f^\star(x):=\sup_{n\in\mathbb N} |f_n(x)|.$$

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