# What is the most elegant known proof of $m^*(A) \leq \sum_{n} m^*(A_n)$ when $A = \bigcup_n A_n$?

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.

• Note that, for uncountable $I$, the inequality is not true. – 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 $$m^*(A) \leq \sum_{n=1}^\infty m^*(A_n).$$

• You should note that, in the question, the union was not necessarily assumed to be countable. – Theo Bendit Jan 12 '19 at 2:11
• @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. – Theo Bendit Jan 12 '19 at 2:47
• When the union is uncountable the inequality is not true, so I'm not sure what you expect to prove. – Martin Argerami Jan 12 '19 at 3:19
• Yes. $[0,1]=\bigcup_{t\in[0,1]}\{t\}$. – Martin Argerami Jan 12 '19 at 3:28
• @MartinArgerami Ugh, obviously. :-( – Theo Bendit Jan 12 '19 at 4:11