2
$\begingroup$

Let $(X,\Sigma,\mu)$ be a probability space and $(f_n)_{n \in \mathbb{N}}$ a sequence of measurable functions $f_n : X \to [0,1]$. Suppose that $\sum_n f_n$ converges pointwise, and let $\varepsilon > 0$ be given.

Does it follow that there is $A \in \Sigma$ with $\mu(A) > 1 - \varepsilon$ and such that $\sum_n \|f_n|_A\|_\infty < \infty$?

This looks like it should be a standard result (if it's true), but I haven't been able to find it anywhere. Here's part of what I know:

  • Clearly taking $A = X$ does not work: pointwise summability does not imply absolute summability.
  • It's not so hard to prove the statement if all the $f_n$ are $\{0,1\}$-valued.
  • One can apply Egorov's theorem to reduce to the case where $\sum_n f_n$ converges uniformly, but I don't see how this would help.
$\endgroup$

1 Answer 1

1
$\begingroup$

No. Consider the following sequence of functions on $[0,1]$ $$ f_{0,1}, f_{1,1}, f_{1,2}, \ldots, f_{i, 1}, \ldots, f_{i, 2^i}, f_{i+1, 1}, \ldots $$ where $$ f_{i, n} = \frac{1}{2^i} \mathbb{1}\left( x \in \left(\frac{n-1}{2^i}, \frac{n}{2^i} \right) \right). $$ I claim that if $\mu(A) > 0$, then $$ \sum_{i=0}^{\infty} \sum_{n=1}^{2^i} a_{i,n} = + \infty, $$ where $a_{i, n} = \| f_{i, n}|_A \|_{\infty}$.

Proof: take $A$ such that $\mu(A) > 0$, and take $Y = X \setminus A$. Note that $a_{i,n} = 0$ iff $\mu(Y \cap (\frac{n-1}{2^i}, \frac{n}{2^i})) = \frac{1}{2^i}$, and $a_{i,n}=\frac{1}{2^i}$ otherwise. Hence $$ \sum_{n=1}^{2^i} a_{i,n} \geq \left( \sum_{n=1}^{2^i} \frac{1}{2^i} \right) - \frac{1}{2^i} \frac{\mu(Y)}{1/2^i} = 1 -\mu(Y). $$ Hence $$ \sum_{i=0}^{\infty} \sum_{n=1}^{2^i} a_{i,n} \geq \sum_{i=0}^{\infty} (1 - \mu(Y)) = +\infty. $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .