3

Question:

Consider a measure space $(X, \mathcal{X}, \mu)$, with $\mu(X)< \infty$ and measurable functions $(f_n)_{n \geq 1}, f$ from $\mathcal{X}\to\mathbb{R}$.

If the following are satisfied,

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

(2) $\sup_{t>0} t \cdot \mu\left( \big\{x\ :\ \sup_{n\in\mathbb N} |f_n(x)-f(x)| >t\big\}\right) <\infty,$

does that imply $f_n$ converges to $f$ pointwise almost everywhere?

I know that (1) implies there exists a subsequence $(f_{n_j})$ converging to $f$ pointwise a.e. but I am interested in the convergence for the entire sequence $(f_n)_{n \geq 1}$.

Background:

My question was inspired by this post where someone suggested this may be the case because of Steins Maximal Principle, but I am really struggling to see how that can imply pointwise a.e. convergence in a case like this.

Here is an example where the commentor uses Stein to show something does not converge pointwise a.e. but I'm failing to see how to do the converse.

If someone has an alternative to (2) to make this work (besides monotonicity) that would also be highly appreciated.

Chad Brown
  • 145
  • 3
    If $f_n$ are bounded say by $a$ then $(2)$ is always satisfied as for $t>2a$ the LHS is $0$ and for $t\le 2a$ the LHS is less than or equal $2a\mu(X).$ The condition $f_n\to 0$ a.e. is equivalent to $f_{n,a}\to f_{a}$ a.e., where the subscript $a $ denotes the restriction to the values less than or equal $a.$ Thus the condition $(2)$ does not seem essential in proving a.e. convergence. – Ryszard Szwarc Jul 22 '22 at 20:58

0 Answers0