Based on our experience with metric spaces and Euclidean geometry, it is reasonable to think of the interval $(a,b)$ as having length $b-a$. In this section, using the tools we have developed, we will construct a measure on $\R$ that agrees with that suggestion.
What will the domain of the measure be? We know that measures are defined on σ-algebras, and we want all of the intervals $(a,b)$ to belong to the σ-algebra. Every σ-algebra that contains all of the intervals $(a,b)$ must contain the Borel σ-algebra $\borel(\R)$ on $\R$. We can therefore rephrase our goal as follows: construct a measure $\lambda$ on $\borel(\R)$ such that \[ \lambda((a,b)) = b-a \] for all $a \l b$. In this section we will construct such a measure. To do so we will introduce two new ideas: outer measures and Lebesgue measurable sets.
If a Borel set $B$ is covered by a union of intervals \[ B \subset \bigcup_{n=1}^\infty (a_n,b_n) \] then we must have \[ \lambda(B) \le \sum_{n=1}^\infty \lambda((a_n,b_n)) = \sum_{n=1}^\infty b_n - a_n \] by subadditivity. The idea behind the definition of outer measure is that we should assign a size to $B$ by finding the most efficient way of covering it by a union of intervals. We define \[ \lambda(B) = \inf \left\{ \sum_{n=1}^\infty b_n - a_n : B \subset \bigcup_{n=1}^\infty (a_n,b_n) \right\} \] for all sets $B \subset \R$.
The function $\lambda$ is defined on $\mathcal{P}(\R)$. It assigns a value in $[0,\infty]$ to every subset of $\R$. However, as we have seen already, it is not a measure on the σ-algebra $\mathcal{P}(\R)$ because it is not countably additive. However, it does have some properties in common with measures.
The map $\lambda : \mathcal{P}(\R) \to [0,\infty]$ has the following properties.
These properties are proved in the length section.▮
Quite generally, a mapping $\Xi$ from $\P(X)$ to $[0,\infty]$ satisfying the three properties above is called an outer measure.
The central step in constructing a measure from an outer measure is to identify as large a σ-algebra as possible on which $\lambda$ defines a measure. Identifying a criterion for membership in such a σ-algebra was one of Lebesgue's great insights. Suppose that we have such a σ-algebra to hand. It must contain the open intervals, and thus for any set $L$ in the σ-algebra we also have $I \cap L$ and $I \setminus L$ in the σ-algebra. Moreover, the measure of $I$ must be equal to the sum of the measures of $I \cap L$ and $I \setminus L$. Lebesgue defined a set to be measurable if, in that sense, it splits every open interval appropriately.
We will not proceed directly with Lebesgue's definition. Instead we will work with a stranger but more applicable definition due to Carathéodory in which not only the intervals, but all subsets of $\R$ must be split properly for a set to be measurable.
A subset $L$ of $\R$ is Lebesgue measurable if \[ \lambda(S) = \lambda(S \cap L) + \lambda(S \cap L^\mathsf{c}) \] for every set $S \subset \R$. Write $\leb(\R)$ for the collection of all Lebesgue measurable subsets of $\R$.
Since $\lambda$ is an outer measure we have \[ \lambda(S) \le \lambda(S \cap L) + \lambda(S \cap L^\mathsf{c}) \] for all $S,L \subset \R$. It therefore suffices, when aiming to check a set $L$ is Lebesgue measurable, that \[ \lambda(S) \ge \lambda(S \cap L) + \lambda(S \cap L^\mathsf{c}) \] for all sets $S \subset \R$.
The collection $\leb(\R)$ is a σ-algebra and the restriction of $\lambda$ to $\leb(\R)$ is a measure.
We know that $\lambda(\emptyset) = 0$. The empty set is therefore Lebesgue measurable because $S \cap \emptyset = \emptyset$ and $S \setminus \emptyset = S$ for every $S \subset \R$. Thus $\leb(\R)$ is non-empty.
The expression defining Lebesgue measurability is symmetric in terms of $L$ and $L^\mathsf{c}$ so it is immediate that $L \in \leb(\R)$ implies $L^\mathsf{c}$ is Lebesgue measurable.
Next we prove that if $L, M \subset \R$ are both Lebesgue measurable then so too is $L \cup M$. Fix $S \subset \R$. We have \[ \begin{aligned} \lambda(S) &= \lambda(S \cap L) + \lambda(S \cap L^\mathsf{c}) \\ &= \lambda(S \cap L \cap M) + \lambda(S \cap L \cap M^\mathsf{c}) + {}\\ & \qquad \lambda(S \cap L^\mathsf{c} \cap M) + \lambda(S \cap L^\mathsf{c} \cap M^\mathsf{c}) \\ &\ge \lambda(S \cap (L \cup M)) + \lambda(S \cap (L \cup M)^\mathsf{c}) \end{aligned} \] where we have used Lebesgue measurability of $L$ for the first equality and Lebesgue measurability of $M$ twice for the second equality. The inequality follows from subadditivity of outer measures. Since $S \subset \R$ was arbitrary we conclude that $L \cup M$ is Lebesgue measurable.
Now, if $L,M$ and Lebesgue measurable and disjoint, we can calculate \[ \begin{aligned} \lambda(L \cup M) &= \lambda((L \cup M) \cap L) + \lambda((L \cup M) \cap L^\mathsf{c}) \\ &= \lambda(L) + \lambda(M) \end{aligned} \] which proves $\lambda$ is additive on the σ-algebra $\leb(\R)$.
As $\leb(\R)$ is now known to be closed under complements and finite unions it is also closed under finite intersection. To prove it is a σ-algebra it therefore suffices to take pairwise disjoint sets $L_1,L_2,\dots$ in $\leb(\R)$ and prove that their union $M$ is also in $\leb(\R)$. Fix $S \subset \R$ and calculate for every $N \in \N$ that \[ \begin{aligned} & \lambda(S \cap (L_1 \cup \cdots \cup L_N)) \\ = {}& \lambda(S \cap (L_1 \cup \cdots \cup L_N) \cap L_N) + \lambda(S \cap (L_1 \cup \cdots \cup L_N) \cap L_N^\mathsf{c}) \\ = {}& \lambda(S \cap L_N) + \lambda(S \cap (L_1 \cup \cdots \cup L_{N-1})) \end{aligned} \] and therefore \[ \lambda(S \cap (L_1 \cup \cdots \cup L_N)) = \sum_{n=1}^N \lambda(S \cap L_n) \] by induction. We have for every $N \in \N$ that \[ \begin{aligned} \lambda(S) &{} \ge \lambda(S \cap (L_1 \cup \cdots \cup L_N)^\mathsf{c}) + \lambda(S \cap (L_1 \cup \cdots \cup L_N)) \\ &{} \ge \lambda(S \cap M^\mathsf{c}) + \sum_{n=1}^N \lambda(S \cap L_n) \end{aligned} \] and therefore \[ \begin{aligned} \lambda(S) &{} \ge \lambda(S \cap M^\mathsf{c}) + \sum_{n=1}^\infty \lambda(S \cap L_n) \\ & {}\ge \lambda(S \cap M^\mathsf{c}) + \lambda \left( S \cap \, \bigcup_{n=1}^\infty L_n \right) \ge \lambda(S) \end{aligned} \] from which we conclude all terms above are equal. In particular $M$ belongs to $\leb(\R)$ and taking $S = M$ gives \[ \lambda \left( \, \bigcup_{n=1}^\infty L_n \right) = \sum_{n=1}^\infty \lambda(L_n) \] which is countable additivity.▮
We find ourselves in the following situation: there is a σ-algebra $\leb(\R)$ on which $\lambda$ defines a measure. Membership in $\leb(\R)$ was defined by a bizarre property that made it possible to prove the above theorem. What sets actually have this bizarre property? It may be the case that $\leb(\R)$ is the trivial σ-algebra $\{\emptyset, \R \}$. To know that $(\R,\leb(\R))$ is an interesting measurable space, and to realize our ambition of defining a measure on a σ-algebra that contains the intervals and agrees with length, we conclude by checking that $\leb(\R)$ contains the Borel σ-algebra.
The Lebesgue σ-algebra $\leb(\R)$ contains the Borel σ-algebra $\borel(\R)$.
Since $\borel(\R)$ is generated by the open intervals $(a,b)$ with $a \l b$ it suffices to prove that each open interval $(a,b)$ belongs to $\leb(\R)$. Fix $S \subset \R$. We need to verify that \[ \lambda(S) \ge \lambda(S \cap I) + \lambda(S \cap I^\mathsf{c}) \] where $I = (a,b)$.
First we check that $\lambda(I) = b-a$. Certainly $\lambda(I) \l b-a$ because $I$ is an interval that covers $I$. For the converse fix $\epsilon > 0$. There is a sequence $n \mapsto (a_n,b_n)$ of intervals with \[ (a,b) \subset \bigcup_{n=1}^\infty (a_n,b_n) \] and \[ \sum_{n=1}^\infty b_n - a_n \le \lambda(I) + \epsilon \] as well. Now $n \mapsto (a_n,b_n)$ is an open cover of $[a + \epsilon, b - \epsilon]$ and we can therefore find $r(1),\dots,r(v)$ such that \[ (a_{r(1)},b_{r(1)}),\dots,(a_{r(v)},b_{r(v)}) \] is a finite subcover of $[a+\epsilon, b-\epsilon]$. It must then be the case that \[ \sum_{n=1}^v b_{r(n)} - a_{r(n)} \ge b - a - 2\epsilon \] whence \[ \lambda(I) \ge \left( \, \sum_{n=1}^\infty b_n - a_n \right) - \epsilon \ge b - a - 3\epsilon \] and we conclude, since $\epsilon > 0$ was arbitrary, that $\lambda(I) = b-a$.
If $\lambda(S) = \infty$ there is nothing to prove, so assume $\lambda(S)$ is finite. Fix $\epsilon > 0$ and let $n \mapsto J_n$ be a sequence of intervals with \[ \sum_{n=1}^\infty \lambda(J_n) \le \lambda(S) + \epsilon \] that covers $S$. (Here we have used that $\lambda$ assignes the correct lengths to intervals.) Now \[ \begin{aligned} \lambda(S) + \epsilon & {}\ge \sum_{n=1}^\infty \lambda(J_n) \\ & {}= \sum_{n=1}^\infty \lambda(J_n \cap I) + \sum_{n=1}^\infty \lambda(J_n \cap I^\mathsf{c}) \\ & {}\ge \lambda(S \cap I) + \lambda(S \cap I^\mathsf{c}) \end{aligned} \] for every $\epsilon > 0$. Thus $I$ belongs to $\leb(\R)$.▮