Clarification on the definition of Lebesgue measure

Question

I'm doing some independent reading on the Lebesgue measure and I have the following questions:

1) on the definition of a Lebesgue measurable set: Let $E \subseteq \mathbb{R} $. On Wikipedia, $$\mu(E) = \mu^*(E) \iff \mu^*(A) = \mu^*(A\cap E) + \mu^*(A\cap E^c) \, \, \text{for every} \, \, A \subseteq \mathbb{R}.$$ However on another text, the definition of Lebesgue measurable is defined as $$ \mu(E) = \mu^*(E) \iff \mu^*(A) = \mu^*(A\cap E) + \mu^*(A ^c\cap E) \, \, \text{for every} \, \, A \subseteq \mathbb{R}.$$ The subtle, yet seemingly important, distinction is which intersection we're looking at. Is this distinction important? Are these two definitions equivalent? If so, could you please explain the compatibility?

2) on the motivation for such a definition: why is Lebesgue measurable defined in this way? Why not stop at the definition of $\mu^*(E)$? What is the motivation for using $E$ as a "partitioning" (for a lack of better words) set? And why do we require this for every $A \subseteq \mathbb{R}$?

3) on the existence of counter-examples: What are some common examples of non-Lebesgue measurable sets? Where does the importance of defining measurable in the above way(s) appear in these examples? I.e. what is an example of a set in which defining measurability differently fails to capture the "desired" property?

Answer 1

If we denote the Lebesgue outer measure by $m^{*}$, then you know $m^{*}$ is a function from $\mathcal{P}(\Bbb R)$ (the set of all subsets of $\Bbb R$) to $[0, \infty]$.

But $m^{*}$ is only an outer measure. In order for it to be a measure, we need to restrict its domain. So we won't be able to find the Lebesgue measure of every subset of $\Bbb R$. Then which subsets of $\Bbb R$ can we "measure" with Lebesgue measure? We can measure the sets that split every subset of $\Bbb R$. It turns out these sets form a $\sigma$-algebra, and $m^{*}$ satisfies the properties of a measure on this $\sigma$-algebra. We call the $\sigma$-algebra: the $\sigma$-algebra of Lebesgue measurable sets. We denote the restriction of $m^{*}$ to this $\sigma$-algebra by $m$.

Now, the point of your question is: what does it mean for an element of this $\sigma$-algebra to "split" every subset of $\Bbb R$? Well, we say $E$ is Lebesgue measurable if for every $A \subseteq \Bbb R$, $m^{*}(A) = m^{*}(A \cap E) + m^{*}(A \cap E^{c})$. And it makes intuitive sense why we use the word "split" -- we are dividing or splitting the elements of $A$ into those that are also in $E$ and those not in $E$, and the sum of the outer measures of each of these sets equals the measure of the whole set.

Answer 2

1. At least in Royden and Fitzpatrick, the first definition of a Lebesgue measurable set is used. I don't think they are equivalent; likely the text has a typo.
2. This definition of Lebesgue measure has the nice property of being countably additive: for any pairwise disjoint countable collection $\{E_n\}$ of sets in $\mathbb{R}$, $\mu(\cup_n E_n)=\sum_n \mu(E_n)$. whereas for outer measure, there exist disjoint sets $A$ and $B$ such that $\mu^*(A\cup B)<\mu^*(A)+\mu^*(B)$.
3. Vitali's Theorem: Any set $E$ of real numbers with positive outer measure contains a subset that fails to be measurable. You can find a free copy of Royden and Fitzpatrick's "Real Analysis" online in PDF form if you google it. In the 4th edition, on page 48, they show how to construct such nonmeasurable sets, although the construction they use (and every construction I have ever seen of nonmeasurable sets) relies on the Axiom of Choice.

Answer 3

I think of Lebesgue measurable sets as those subsets of $\mathbb{R}$ whose complement is sufficiently separated from itself. For example, $E = (0,2)$ and its complement are pretty easy to distinguish. Now, the outer measure $\mu^{\ast}$ of a given subset $A$ of the real line, is obtained by estimating the plot of $A$ on the number line by countably many straight lines. Each collection of straight lines must cover $A$ completely, and we take the infimum of the lengths of all possible covers.

Since $E$ and its complement can easily be separated on the real line, it is obvious that you can estimate the length of $A$ by just estimating the portion of it contained in $(0,2)$, and then estimating the portion of it outside $(0,2)$.

On the other hand, you wouldn't be able to do that with a nonmeasurable set (by definition). A nonmeasurable set $E$ would have to have such intricacy of points close together in infinitely many places, that we would not be able to separate the set and its complement nicely. The result would be that for some subset $A$ of the real line, there would always be a nontrivial overlap between countable open covers of $A \cap E$ and of $A \cap E^c$, giving you $$\mu^{\ast}(A \cap E) + \mu^{\ast}(A \cap E^c) > \mu^{\ast}(A)$$