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.

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    $\begingroup$ Note that, for uncountable $I$, the inequality is not true. $\endgroup$ – Martin Argerami Jan 12 '19 at 4:33

(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 $$ m^*(A) \leq \sum_{n=1}^\infty m^*(A_n). $$

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    $\begingroup$ You should note that, in the question, the union was not necessarily assumed to be countable. $\endgroup$ – Theo Bendit Jan 12 '19 at 2:11
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    $\begingroup$ @DavidC.Ullrich I thought about that, but sets with zero outer measure need not be empty. I don't see how the $\varepsilon / 2^n$ trick works in those circumstances. $\endgroup$ – Theo Bendit Jan 12 '19 at 2:47
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    $\begingroup$ When the union is uncountable the inequality is not true, so I'm not sure what you expect to prove. $\endgroup$ – Martin Argerami Jan 12 '19 at 3:19
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    $\begingroup$ Yes. $[0,1]=\bigcup_{t\in[0,1]}\{t\}$. $\endgroup$ – Martin Argerami Jan 12 '19 at 3:28
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    $\begingroup$ @MartinArgerami Ugh, obviously. :-( $\endgroup$ – Theo Bendit Jan 12 '19 at 4:11

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