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?

  • 3
    $\begingroup$ Largely in response to your second-to-last paragraph: If I had to give a one word description of the entire field of mathematical analysis, that one word would be "approximation." The actual objects which we want to study are hard (e.g. measurable functions), so we approximate them by more easily understood objects (e.g. characteristic functions or simple functions). See, for example, Littlewood's three principals of analysis. $\endgroup$ – Xander Henderson Mar 10 '18 at 0:40

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,

For every $\delta > 0$, there exist some $U \in \mathcal{U}$, such that, $ M \le m^\ast(U) < M - \delta$

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.

  • $\begingroup$ Thank you. I'll leave the question open for a some time, and definitely come back to close the loop. The answer is difficult, but so is the topic. There is one sentence at the beginning: "Since every open subset of R is the union of such a countable sequence." I see the word sequence, but if intervals can be open subsets, they are uncountable in R. Can you briefly explain? $\endgroup$ – Antoni Parellada Mar 10 '18 at 4:29
  • $\begingroup$ @mathkoan every open subset of the reals can be expressed as the union of countably many disjoint open intervals. The sequence I was referring to was the sequence of intervals $I_k$ $\endgroup$ – Not Mike Mar 10 '18 at 4:32

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