The Lebesgue outer measure of of a set $E$ is denoted as $m^*(E),$ and defined as
$$m^*(E)=\inf\Bigg\{\sum_{k=1}^\infty \ell(I_k)\;\Bigg\vert \;E\subseteq \bigcup_{k=1}^\infty I_k \Bigg\}$$
where $\{I_k\}$ are collections of open intervals.
I understand that the idea is to extend the idea of measure to sets that are not all that intuitive, but I'm having problems picturing the "process" even in straightforward sets.
I have found that there isn't a single $\{I_k\}_{k=1}^\infty=\bigcup_{k=1}^\infty I_k ,$ but a collection: $$\Big\{\{I_k^1\}_{k=1}^\infty,\{I_k^2\}_{k=1}^\infty,\dots\Big\}$$
and for each one of them (i.e. covers) the calculation
$$\sum_{k=1}^\infty \ell(I_k)$$ is carried out. Eventually the infimum or greatest lower bound is calculated.
What is the need for these multiple collections of open intervals – at first sight it sounds like a process of approximation, where some unions of open intervals might be extremely redundant.
What is the process involved in this search for the infimum?
Best Answer
In the context of $\mathbb R$, explicitly naming the sequence of open intervals is not really necessary. Since every open subset of $\mathbb R$ is the union of such a countable sequence, you can restate the definition of outer measure as follows, first given an arbitrary open set $U = \cup \{ I_k: k \in \mathbb{N} \} \subset \mathbb R$ define, $$ m^\ast(U) = \sum_{k=0}^\infty \ell(I_k)$$ then, extend the definition of $m^\ast$ to arbitrary subsets $E \subset \mathbb R$ by asserting $m^\ast(E) = $ $\inf \{ m^\ast(U): E \subset U \text{ and } U \text{ is open } \}$.
What does this simplification/restatement accomplish? It enables a cleaner analysis of the process by which the open sets $\mathcal{U} = \{ U \subset \mathbb{R}: U \text{ is open and } E\subset U\}$ are used to find the outer measure of $E$.
Recalling that the infimum $M = \inf \{ m^\ast(U): U \in \mathcal{U}\} = m^\ast(E)$ must satisfy,
we can choose/find/fix a sequence of open sets $U_0, \ldots, U_n, \ldots \in \mathcal{U}$ such that, for each $n\ge 0$,
$$ M \le m^\ast(U_n) < M - \frac{1}{2^n}. $$
Letting $V_n = U_0 \cap \ldots \cap U_n$, we get, $V_n \in \mathcal{U}$ (since $E \subset U_0 \cap \ldots \cap U_n = V_n$ and $V_n$ is open.) Moreover, applying the definition of $m^\ast$ for open sets, we have $V_{n+1} \subset V_n \implies$ $m^\ast(V_{n+1}) \le m^\ast(V_n)$,and
$$ M \le m^\ast(V_n) \le \min \{ m^\ast(U_0), \ldots, m^\ast(U_n) \} < M - \frac{1}{2^n}.$$
Putting this all together, we've extracted a descending sequence $V_n$ of open sets containing $E$, whose measures converge to the value of the outer measure $M = m^\ast(E)$ of $E$.
You can think of the sets $V_k$ as approximating $E$ by shrinking and progressively discarding non-essential segments. Moreover, capturing that processing is the whole idea behind how the outer measure is defined.