Let $c\_n$ be a sequence of real numbers with $\sum c\_n$ converging conditionally but not absolutely. Suppose $\delta\_n > 0$ is another sequence with $\delta\_n \to 0$, and $\sum c\_n \delta\_n$ converging also conditionally but not absolutely.


Does there exist, for every $L^1$ function $f: [0, 1] \to \mathbb R$, a bijection $\gamma: \mathbb N \to \mathbb N$, and a sequence of measurable sets $A\_n$ with $\mu(A\_n) = \delta\_n$ such that


$\sum c\_{\gamma(n)}1\_{A\_{\gamma(n)}} \to f$,


in $L^1$ and pointwise a.e?


*Note: Here $\mu$ denotes the Lebesgue measure.*



Riemann rearrangement theorem for $L^1$ functions
The problem is trickier than I initially thought, but with the corrected condition it can be done. I need to assume that $\sum c\_n \delta\_n$ is conditionally but not absolutely convergent, $0 < \delta\_n < 1$ with $\delta\_n \to 0$ and $c\_n \to 0$ (which is currently only implied by the conditional convergence of $\sum c\_n$).


Wlog. we can assume $f: [0,1] \to \mathbb{R}$ to be non-negative (otherwise split the series in two and use them to approximate positive and negative part separately) and non-increasing (by rearrangement, just to simplify notation).


Now proceed as following: Denote the current partial sum by $\tilde{f}\_k$. Take the next unused $n\in \mathbb{N}$ such that $c\_n > 0$ and consider the sets $$A\_\lambda := \{x \in [\lambda,1]: f(x)- \tilde{f}\_k(x) \geq c\_n \}$$
for $\lambda \in [0,1]$. If there is a $\lambda$ such that $|A\_\lambda| = \delta\_n$, then choose $\gamma(k) := n$, $A\_n := A\_\lambda$, add $c\_n 1\_{A\_n}$ to the partial sum and iterate. (The "positive process")


If not, take the next unused $m \in \mathbb{N}$ such that $c\_m < 0$, choose $\gamma(k) := m$ and find a set $A\_m \subset [0,2\max(\delta\_n,\delta\_m)]$ with $|A\_m| =\delta\_m$, for which the approximation is best, i.e. $x \notin A\_m$ implies $f(x)-\tilde{f}\_k(x) \geq f(y)- \tilde{f}\_k(y)$ for all $y \in A\_m$, add $c\_m 1\_{A\_m}$ to the partial sum and iterate. (The "negative process")


The positive process cannot continue indefinitely, since it increases the integral by $c\_n \delta\_n$ and $\tilde{f}\_k \leq f$. The negative process cannot continue indefinitely, since at some point it will achieve $f(x) - \tilde{f}\_k(x) \geq c\_n$ on $[0,\delta\_n]$, at which point the positive process will take over again. Thus we use all indices and have constructed a bijection.


Now fix $\epsilon > 0$. At some point, all future $\delta$ are smaller than $\epsilon/2$, so after that on the interval $[\epsilon,1]$ the approximation $\tilde{f}\_k$ is monotone increasing in $k$. Furthermore, whenever the negative process takes over, we know that $f(x) - \tilde{f}\_k < c\_n$ except for a set of measure $< \delta\_n$. Thus as $f$ is decreasing, $$\int\_\epsilon^1 f - \tilde{f}\_k dx < c\_n (1-\epsilon) + \delta\_n \sup\_{[\epsilon,1]} (f-\tilde{f}\_k).$$
Here, the first term converges to $0$ as $c\_n \to 0$, while for the second term $\sup\_{[\epsilon,1]} f \leq f(\epsilon)$ as $f$ is non-increasing and $\sup\_{[\epsilon,1]} (- \tilde{f}\_k)$ is bounded as there were only a finite number of negative $c\_m$ until we reached monotonicity on $[\epsilon,1]$. Then since $\delta\_n \to 0$, that term converges as well and we have that $\tilde{f}\_k \to f$ in norm on $[\epsilon,1]$, which together with monotonicity implies pointwise convergence a.e. on $[\epsilon,1]$ and as $\epsilon$ was arbitrary, we then have pointwise convergence a.e. on $[0,1]$.


Finally consider the negative process. We know that $|\{x\in [0,2 \max(\delta\_n,\delta\_m)]: \tilde{f}\_k(x) < -c\_n \}| <\delta\_n \leq \max(\delta\_n,\delta\_m)$, otherwise the positive process would already have taken over. But this means that for the set $A\_m$ we choose, we have $\tilde{f}\_k(x) \geq -c\_n$ for all $x\in A\_m$ and hence at all points modified, we will have $\tilde{f}\_{k+1} (x) \geq -c\_n +c\_m$. But since this is true for all negative steps, we have $\tilde{f}\_k \geq - 2\max\_{n\in \mathbb{N}} |c\_n|$ for all $k\in\mathbb{N}$. But than we have an integrable lower bound which allows us to conclude that the pointwise limit $f$ is also the $L^1$ limit of $\tilde{f}\_k$ by dominated convergence.


