# From pointwise summability to absolute summbability in $L^\infty$, approximately

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.

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 $$\sum_{i=0}^{\infty} \sum_{n=1}^{2^i} a_{i,n} \geq \sum_{i=0}^{\infty} (1 - \mu(Y)) = +\infty.$$