Clarification on the two assumptions of Lebesgue integral?

Question

The Lebesgue measure has the following properties:

1. $\mu(0) = 0$;
2. $\mu( C) = \operatorname{vol} C$ for any $n$-cell $ C$;
3. if $\{M_1, M_2,\ldots \}$ is a collection of mutually disjoint sets in $\mathbb{M}$, then $\mu(M_1 \cup M_2 \cup\cdots) = \mu(M_1) + \mu(M_2) +\cdots$
4. the measure of any measurable set can be approximated from above by open sets; that is, for any measurable $M$, $\mu(M) = \inf\{\mu(O) : M \subset O, O \text{ is open}\};$
5. the measure of any measurable set can be approximated from below by compact sets that is, for any measurable $M$, $\mu(M) = \sup\{\mu(C) : C \subset M, C \text{ is compact}\}.$

Can someone clarify what property $\color{red}{4}$ and $\color{red}{5}$ means? What does it mean exactly for the measure to be approximated from above by open sets and below by compact sets?

Answer 1

We call property $4$ the outer measure (and denote it $\mu_o$) and property $5$ the inner measure (and denote it $\mu_i$). We say that a set $S$ is measurable iff $\mu_o(S) = \mu_i(S)$.

As Ramiro Guerreiro identifies correctly, Property 4 is called outer regularity and property 3 is called inner regularity. To make sense of why these properties are important, I will try to explain the idea of the inner measure and outer measure in the context of the Lebesgue measure. We will denote the outer measure $\mu_o(T) = \inf\{S: T\subseteq S$ and $S$ is $open\}$ and the inner measure $\mu_i(T) = \sup\{S: S \subseteq T$ and $S$ is $compact \}$.

For example, suppose we are working with the Lebesgue measure and we want to know if $[0,1]$ is Lebesgue measurable. Then we need to demonstrate that $\mu_o ([0,1]) = \mu_i([0,1])$. Since $[0,1]$ is compact, we get the inner measure of $S$ for free and conclude $\mu_i([0,1]) = 1$. To find the outer measure, we look at $\mu_o(S) = \inf\{T \subset \mathbb{R} :[0,1] \subset T$ and $T$ is open $\}$. By looking at the collection of sets $(0-\epsilon, 1 + \epsilon)$, we can conclude that $\mu_i([0,1]) = 1$. Since $\mu_i([0,1]) = \mu_o([0,1])=1$, we can conclude $[0,1]$ is measurable (as well as compute its measure).

All this means is that two different perspective on the same set (building the set inward v.s. building the set outward) yield the same result. However, not all sets are measurable (for instance the Vitali set). Sets which are unmeasurable look different when building them up from compact sets versus building them down with open sets.

Additionally, your property 3 is incorrect. As Ramiro Guerreiro also points out, you must make sure that each $M_i$ is itself measurable. Finally, it should be the case that measure is countably additive, not uncountably additive. Recall that singletons have measure zero, but when you measure the union of points in $[0,1]$, we have a set with measure one. For instance, $\mu(\bigcup_{i \in [0,1]} \{i\}) \neq \mu([0,1])$.

Answer 2

Actually, as you present, property 4 is called outer regularity, and property 5 is called inner regularity.

Generally, property 4 and 5 are not part of the definition of Lebesgue measure, but properties that we prove for the Lebesgue measure.

The way you defined the Lebesgue measure has flaws: it is not defined what is a mensurable set. Also, note that using the five properties you list, we don't have a way to know $\mu(O)$ for an arbitrary open set $O$, so we can not use 4. In a similar way, we don't have a way to know $\mu(C)$ for an arbitrary compact set $C$, so we can not use 5.

The Lebesgue measure has the five properties you listed, but it is NOT DEFINED by those five properties.

Moreover, there are other measures that share the same five properties (assuming for property 2 that $\operatorname{vol} C$ is the result of the measure applied to $C$). In fact, any (inner and outer) regular measure has the five properties you listed.

Remark: Regarding the property 3: since when we write a collection of sets as $\{M_1,M_2,M_3,\cdots\}$ it means the collection is countable, then you are clearly dealing only with countable collections. However, you should state that the sets are measurable.