Lebesgue Measure

It is quiet intuitive that intervals $(a,b),(a,b],[a,b),[a,b]$ have length $b - a$ . Now we want measure sets of real numbers more general than intervals, such as the set of all rational numbers.

In order to do so, we need a new model, a new concept of length. Before we show what Lebesgue Measure is, it is natural to have following question in mind:

**1. What do we mean when we talk about the length of a set? **
2. Does every set has a length?
3. What properties do the sets that have length satisfy?

One way to think of length of sets is regarding it as a function from sets to real numbers. Translating the above questions into the language of functions, we get

1. What is the function?
2. Especially, what is the domain of the function?
3. What are the typical properties of the function?

Before we delve into the details of constructing Lebesgue Measure, we shows some important goals of that construction (We forgo why we need such goals for the moment. Yet again, keep this question in mind):

Intervals like $(a,b],[a,b],[a,b),(a,b)$ are measurable and have length $b - a$ .
. If set $E$ is measurable, then its complement $\overline{E}$ is measurable.
. Countable unions of measurable sets are measurable.

One immediate corollary of goal (2) and (3) is that

4 . Countable intersections of measurable sets are measurable.

Actually, if we replace (3) with (4), we can derive (3) by combining (2) and (4). This is due to the symmetry inherent in set operation

\overline{E_1 \cup E_2} = \overline{E_1} \cap \overline{E_2}

Another corollary of (1), (2) and (3) is

5 . Countable intersections and unions of intervals are measurable.

Two Components for Lebesgue Measure

$m^*(E) \doteq \inf \{ \sum_n l(I_n) : I_n \ \text{is open interval and } E \subset \cup_n I_n\}$ , where $l(I_n)$ denotes the length of $I_n$ .
For any set A, a measurable set E must satisfy $m^*(A) \ge m^*(A \cap E) + m^*(A \cap \overline{E})$ .

Discussion: (1) is a quiet intuitive rule. It states the length of a set is the least upper bound(supermum) of the sum of length of open intervals whose union covers this set. But why don't we use the sum of lengths of intervals that constitute the set as the length of this set?
One possible explanation is that there are sets which are not countable unions or intersection intervals. So we have to use open intervals to approximate them. What do these set look like? Keep this question in mind.

(2) The second condition is required so that we find a set of sets that are close under complement, countable union and intersection operation with respect to the length definition of (1). Yet why (1) alone is not enough?. We will tough more about this issue when we come to non-Lebusgue measurable sets.

More formally, let

\begin{aligned} \mathfrak{L} = \{ &E: \forall A \subset R, \ m^*(A) \ge m^*(A \cap E) + m^*(A \cap \overline{E}), \\\\ & where \ m^*(E) \doteq \inf \{ \sum_n l(I_n) : I_n \ \text{is open interval and } E \subset \cup_n I_n\} \} \end{aligned}

we will show that

$[a,b], [a,b), (a,b], (a,b) \in \mathfrak{L}$ , and $m^*([a,b])=m^*([a,b))=m^*((a,b])=m^*((a,b))= b - a$ .
. $\forall E \in \mathfrak{L}, \overline{E} \in \mathfrak{L}.$
. $\{ E_n\}_{n \ge 1} \subset \mathfrak{L}, \cup_n E_n \in \mathfrak{L}$
. $\{ E_n\}_{n \ge 1} \subset \mathfrak{L}, \cap_n E_n \in \mathfrak{L}$

Before we move on, we notice the construction of $\mathfrak{L}$ so far involves (a) definition of length (b) constraint on the sets involved. Is it the only way of constructing $\mathfrak{L}?$ In other words, if we replace (a) or (b) with something else, we get a new set $\mathfrak{L'}$ . Will $\mathfrak{L'}$ satisfy the above four properties? Measure Theory studies these general topics.

Theorem

$\forall E \in \mathfrak{L}, \overline{E} \in \mathfrak{L}.$

Proof: Straightforward from definition of $\mathfrak{L}$ .

Lema

If $E \subset \cup_n E_n, \text{ then } m^*(E) \le \sum_n m^*(E_n)$ .

Proof: If $\sum_n m^*(E_n) = \infty$ then, this is obvious. Otherwise, for each n, $m^*(E_n) \le \infty$ , so there exist an set of open intervals $\{I_k^n \}$ , s.t.,

$E_n \subset \cup_k I_k^n \\ m^*(E_n) + {\epsilon \over 2^n}\ge \sum_k l(I_k^n)$

But $\{I_k^n : k \ge 1, n \ge 1\}$ is a cover of E, so

m^*(E) \le \sum_n \sum_k l(I_k^n) \le \sum_n (m^*(E_n) + {\epsilon \over 2^n}) = \sum_n m^*(E_n) + \epsilon

As $\epsilon$ takes arbitrary values, we have proved the inequality.

Corollary

$\text{If } E \subset E_1, \text{ then } m^*(E) \le m^*(E_1)$ .

Theorem

$\{ E_n\}_{n \ge 1} \subset \mathfrak{L}, \cup_n E_n \in \mathfrak{L}.$

Proof:

Let $F_n \doteq E_n - \cup_{i = 1}^{n - 1} E_i$ , so $\{ F_n \}_{n \ge 1}$ are disjoint and $\cup F_n = \cup E_n$ . In the following we denote $F \doteq \cup F_n$

By induction on n, we show that

\forall A \subset R, m^*(A) \ge \sum_{i = 1}^n m^*(A \cap F_i) + m^*(A \cap \overline{\cup_{i = 1}^n F_i}) \ \ \ \ \ \ \ \ \ \ (1)

when $n = 1$ , this clearly holds. Suppose the inequality above holds up to n, now we consider $n + 1$ . First of all, we notice that,

m^*(A) \ge m^*(A \cap F{n + 1}) + m^*(A \cap \overline{F{n + 1}})

by substituting A in (1) with $A \cap \overline{F_{n + 1}}$ ,

\begin{aligned} m^*(A \cap \overline{F{n + 1}}) \ge &\sum_{i = 1}^n m^*(A \cap \overline{F{n + 1}} \cap F_i) + m^*(A \cap \overline{F{n + 1}} \cap \overline{\cup_{i = 1}^n F_i}) \\ =& \sum_{i = 1}^n m^*(A \cap F_i) + m^*(A \cap \overline{\cup_{i = 1}^{n + 1} F_i}) \end{aligned}

so, $\forall n \in N$ ,

\begin{aligned} m^*(A) &\ge \sum_{i = 1}^n m^*(A \cap F_i) + m^*(A \cap \overline{\cup_{i = 1}^n F_i}) \\ & \ge m^*(A \cap (\cup_{i = 1}^n F_i)) + m^*(A \cap \overline{\cup_{i = 1}^n F_i}) \end{aligned}

let $n \rightarrow \infty$

\begin{aligned} m^*(A) \ge m^*(A \cap F) + m^*(A \cap \overline{F}) \end{aligned}