Suppose $\sum^{\infty}a_{i}1_{A_{i}}\geq \sum^{\infty}b_{i}1_{B_{i}}$, where $a_{i},b_{i}\geq 0$ and the sets possibly intersect i.e. $A_{i}\cap A_{j}\neq \varnothing $ and same with $B_{i}$.
Is it true that we can write $\sum^{\infty}a_{i}1_{A_{i}}-\sum^{\infty}b_{i}1_{B_{i}}=\sum^{\infty}c_{i}1_{C_{i}}$ for $c_{i}\geq 0$?
If we have disjointness i.e. $A_{i}\cap A_{j}=\varnothing $ and same for $B_{i}$ then yes.
My concern is that when I proved it the general case for finite order i.e. $\sum^{N}a_{i}1_{A_{i}}$ my proof again relies on rewriting them disjointly. There are sums $\sum^{\infty}a_{i}1_{A_{i}}$ that cannot be written disjointly without making them uncountable sums.
For example, $\sum a_{k}1_{A_{k}}$, where $A_{k}=\{x\in \mathbb{R}^{+}:\frac{d_{k}}{10^{k}}$ is in the decimal expansion of x $\}$ and so $x=\sum^{\infty} a_{k}1_{A_{k}}(x)$. Since $\mathbb{R}^{+}$ is uncountable, this sum cannot be written disjointly.
Proof for disjoint sets
We have from disjointness that $b_{i}\leq \inf(a_{j})$ for all j s.t. $B_{i}\cap A_{j}\neq \varnothing$. Therefore,
$\sum^{\infty}a_{i}1_{A_{i}}-\sum^{\infty}b_{i}1_{B_{i}}=\sum^{\infty}a_{i}1_{A_{i}\setminus \bigcup B_{j}}+\sum_{j}\sum^{\infty}a_{i}1_{A_{i}\cap B_{j}}-\sum^{\infty}b_{i}1_{B_{i}}$ $=\sum^{\infty}a_{i}1_{A_{i}\setminus \bigcup B_{j}}+\sum_{j}\sum^{\infty}(a_{i}-b_{j})1_{A_{i}\cap \bigcup B_{j}}.$
Here we used disjointess of $B_{i}$ to split the indicator.
Suggestions?
Attempt at proof for general case
For finite $\sum^{N}a_{i}1_{A_{i}}-b_{i}1_{B_{i}}$ the result follows because we can rewrite them disjointly.
So now maybe we can approximate $\sum^{\infty}a_{i}1_{A_{i}}-b_{i}1_{B_{i}}$ by $\sum^{N}c_{i}1_{C_{i}}$.
This shows up in proving DCT without having to use positive and negative parts.