Let $|A|$ denote the outer-measure of a set of real numbers $A\subseteq \Bbb R$. Let $\{I_n\}_{n\in\Bbb N}$ be a sequence of disjoint open intervals. I want to prove that $\left|\bigcup_{n=1}^\infty I_n\right| = \sum_{n=1}^\infty \ell(I_n)$.
I know that $|I|=\ell(I)$ for any individual interval, and I have monotonicity, and sub-additivity. This seems like it should be enough but I haven't been able to show it.
Of course $\{I_n\}$ is already a countable open interval cover of $\bigcup_{n=1}^\infty I_n$ and therefore $\sum_{n=1}^\infty \ell(I_n)$ is in the set of all sums of lengths. It then suffices to show that it's a lower-bound on the set of sums of lengths.
But if you take any open interval cover $\{J_n\}_{n\in\Bbb N}$, so that $\cup_{n=1}^\infty I_n \subseteq \cup_{n=1}^\infty J_n$, then we need to show
$$ \sum_{n=1}^\infty \ell(I_n) \le \sum_{n=1}^\infty \ell(J_n)$$
Monotonicity only gives
$$\left|\bigcup_{n=1}^\infty I_n\right|\le \left|\bigcup_{n=1}^\infty J_n\right|$$
and sub-additivity gives
$$ \left|\bigcup_{n=1}^\infty I_n\right|\le \sum_{n=1}^\infty |I_n| = \sum_{n=1}^\infty \ell(I_n)$$
and likewise for the $J$'s.
So it doesn't seem like I can handle this with theorems. I know $\cup J_n$ is an open set and therefore a countable union of open intervals. Maybe I can sort of replace $\{J_n\}$ with a sequence $\{K_n\}_{n\in\Bbb N}$ where the $K_n$'s are formed from taking unions of any overlapping $J_n$'s. I think that's not hard, maybe by defining an equivalence relation on $\{J_n\}$. But then how could you prove that $\sum \ell(K_n)\le \sum \ell (J_n)$? It's intuitively correct, but making a rigorous argument which is valid for infinitely many terms is not clear to me.
I think I found a solution. It's not perfect because it relies on something that seems a little too powerful to be "from scratch". But it's better than nothing:
Suppose that we already know $m^*(A\cup G)=m^*(A)+m^*(G)$ if $G$ is an open set and $A\cap G=\emptyset$. Therefore for empty sets we get finite additivity. Note that then if $\{I_n\}_{1\le n\le m}$ is a finite collection of disjoint open intervals then by monotonicity
$$\sum_{n=1}^m\ell(I_n) = m^*\left(\bigcup_{n=1}^m I_n\right) \le m^*\left(\bigcup_{n=1}^\infty I_n\right)$$
Since this is true for each $m\in\Bbb N$ we can take the limit as $m\to\infty$.