Let $A = \bigcup_{n \in I} A_n \subset \Bbb{R}^k$ where $I$ is an arbitrary index set. Define the Lebesgue outer measure by
$m^*(A) := \inf \ \{ \sum_{n} \text{vol}(I_n) : I_n, n \geq 1$ are each any type of interval and $A \subset \bigcup_n I_n \}$.
Then how can we prove $m^*(A) \leq \sum_n m^*(A_n)$ elegantly. The proof in my book is kind of hand-wavy and very complicated for something intuitively obvious.
Also this is the first property of such objects other than $m^*$ is monotonic: $A \subset B \implies m^*(A) \leq m^*(B)$. First properties should be easily proven or something is wrong with the proof technique! It has to be fixed.
So I am on the search for an elegant proof. Maybe Galois connections?
The idea is that if I am successful, I can apply the proof technique to other such over-complicated examples I encounter.
(This answer assumes that $I$ is countable)
This is a case for the $\epsilon/2^n$ trick, in combination with the "give yourself an epsilon of room" trick. Once we've internalized these tricks the proof seems straightforward.
Let $\epsilon > 0$. For each positive integer $n$, let $\{I_{nj}\}$ be a countable collection of intervals such that $A_n \subset \cup_j I_{nj}$ and $$\sum_{j} |I_{nj}| \leq m^*(A_n) + \frac{\epsilon}{2^n}. $$ Then
$$ A \subset \cup_{n,j} I_{nj} $$ and
\begin{align} m^*(A) &\leq \sum_{n,j} |I_{nj}| \\ &= \sum_{n=1}^\infty \sum_{j=1}^\infty |I_{nj}| \\ &\leq \sum_{n=1}^\infty m^*(A_n) + \frac{\epsilon}{2^n} \\ &= \sum_{n=1}^\infty m^*(A_n) + \underbrace{\sum_{n=1}^\infty \frac{\epsilon}{2^n}}_\epsilon. \end{align}
This shows that $$ m^*(A) \leq \sum_{n=1}^\infty m^*(A_n) + \epsilon $$ for any $\epsilon > 0$. It follows that $$ m^*(A) \leq \sum_{n=1}^\infty m^*(A_n). $$