In particular, the claim I'm wanting to prove is that if $f_n:\mathbb{R}\to\mathbb{R}$ is lebesgue integrable, $|f_n|\searrow 0$ for a.e. $x$, and $\{f_n(x)\}$ is alternating for $a.e.$ $x$ then $$\int\sum f_n=\sum\int f_n.$$ I think I have a proof, but was hoping someone might be able to check it out, and see if it works. Thank you.
$\textbf{Proof:}$
First WLOG we may assume are $a.e.$ conditions hold for all $x$. Setting $P=\{x:f_1(x)> 0\}$, it follows from $|f_n(x)|\searrow 0$ and $f_n(x)$ alternating that if $$\sum_{n\in\mathbb{N}} f_n(x)=\sum_{n\in 2\mathbb{N}}(f_n(x)+f_{n+1}(x))>0,$$ then $x\in P,$ so since integration is infinitely additive on $L^+$ and finitely additive in general it follows that
$$\int\text{Pos}\left(\sum f_n\right)=\int_P\sum_{n\in\mathbb{N}}f_n=\sum_{n\in2\mathbb{N}}\int_P(f_n+f_{n+1})=\sum_{n\in\mathbb{N}}\int_P f_n.\text{ } (*^1)$$ Similarly setting $N=\{x:f_1(x)<0\}$ it follows that $$\int-\text{Neg}\left(\sum f_n\right)=\sum_{n\in\mathbb{N}}\int_N f_n.\text{ }(*^2)$$ So as $f_1(x)=0$ implies $f_n(x)=0$ for all $n$ (modulus is decreasing) which implies $ \sum f_n(x)=0$, we need only worry about integration over $P$ and $N$, so we can assume WLOG $\mathbb{R}=P\sqcup N.$
Notice that by Dominated convergence $|f_n|<f_1$ we have $$\int_Pf_n\to \int_P\lim f_n=0.\text{ }(*^3)$$ As $f_n$ is alternating, and $f_1$ is positive on $P$, $\int_P f_n$ is alternating, hence by $(*^3)$ and $(*^1)$the Alternating Series test $$\sum\int_P f_n=\int\text{Pos}\left(\sum f_n\right)<\infty,$$ so by $(*^1),(*^2)$ it follows $$\int\sum f_n=\int\text{Pos}\left(\sum f_n\right)-\int\text{Neg}\left(\sum f_n\right)=\sum\int_Pf_n+\sum\int_Nf_n=\sum\int f_n.$$ This completes the proof.$\blacksquare$
Thanks Masacroso for help completing the proof.