It is reasonable to think of the interval $(a,b)$ as having length $b-a$. In this section 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 $a < b$.
We already know that $\Lambda$ is an outer measure on $\P(\R)$. Our goal is to prove that the restriction $\lambda = \Lambda | \borel(\R)$ of $\Lambda$ to the Borel subsets of $\R$ is a measure. We will not do this directly, as it is difficult to work with arbitrary Borel sets.
$\lahResCounter{equation}$$\lahResCounter{equation}$$\lahResCounter{equation}$$\lahResCounter{equation}$$\lahResCounter{equation}$5.1 Lebesgue measurable sets
The central step in constructing a measure $\mu$ from an outer measure $M$ on $\P(X)$ is to identify as large a σ-algebra as possible on which $\lambda$ defines a measure.
Identifying – in the special case $X = \R$ and $M = \Lambda$ – 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 correctly.
5.2 Carathéodory's construction
We will not proceed using with Lebesgue's definition. Instead we will work with a stranger and more abstract – but as a consequence more applicable – definition due to Carathéodory in which measurable sets are those that correctly split all sets.
Definition 5.2.1
Fix an outer measure $M$ on a set $X$. A set $L \subset X$ is Carathéodory measurable for $M$ if \[ M(S) = M(L \cap S) + M(L^\mathsf{c} \cap S) \] for every $S \subset X$. Write $\mathsf{Car}(M)$ for the collection of all subsets of $X$ that are Carathéodory measurable for $M$.
Of great importance is the special case $X = \R$ and $M = \Lambda$.
Definition 5.2.2
A subset $L$ of $\R$ is Lebesgue measurable if \[ \Lambda(S) = \Lambda(L \cap S) + \Lambda(L^\mathsf{c} \cap S) \] for every set $S \subset \R$. Write $\leb(\R)$ for the collection of all Lebesgue measurable subsets of $\R$.
Whenever $M$ is an outer measure on $X$ we have \[ M(S) \le M(L \cap S) + M(L^\mathsf{c} \cap S) \] for all $S,L \subset X$. It therefore suffices, when aiming to check a set $L$ is Carathéodory measurable, to verify \[ M(S) \ge M(L \cap S) + \lambda(L^\mathsf{c} \cap S) \] for all sets $S \subset X$.
Theorem 5.2.3
Fix an outer measure $M$ on a set $X$. The collection $\mathsf{Car}(M)$ is a σ-algebra and the restriction of $M$ to $\mathsf{Car}(M)$ is a measure.
Proof:
We know that $M(\emptyset) = 0$. The empty set is therefore Carathéodory measurable because $S \cap \emptyset = \emptyset$ and $S \setminus \emptyset = S$ for every $S \subset X$. Thus $\mathsf{Car}(M)$ is non-empty.
The expression defining Carathéodory measurability is symmetric in terms of $L$ and $L^\mathsf{c}$ so it is immediate that $L \in \mathsf{Car}(M)$ implies $L^\mathsf{c}$ is Carathéodory measurable.
Next we prove that if $L, K \subset X$ are both Carathéodory measurable then so too is $L \cup K$. Fix $S \subset X$. We have \[ \begin{aligned} M(S) & = M(L \cap S) + M(L^\mathsf{c} \cap S) \\ & = M(K \cap (L \cap S)) + M(K^\mathsf{c} \cap (L \cap S)) + {} \\ & \qquad M(K \cap (L^\mathsf{c} \cap S)) + M(K^\mathsf{c} \cap (L^\mathsf{c} \cap S)) \\ & \ge M(S \cap (L \cup K)) + M(S \cap (L \cup K)^\mathsf{c}) \end{aligned} \] where we have used Carathéodory measurability of $L$ for the first equality and Carathéodory measurability of $K$ twice for the second equality. The inequality follows from subadditivity of outer measures. Since $S \subset X$ was arbitrary we conclude that $L \cup K$ is Carathéodory measurable.
Now, if $L,K$ are Carathéodory measurable and disjoint, we can calculate \begin{align*} M(L \cup K) & = M(L \cap (L \cup K)) + M(L^\mathsf{c} \cap (L \cup K)) \\& = M(L) + M(K) \end{align*} which proves $M$ is additive on $\mathsf{Car}(M)$.
As $\mathsf{Car}(M)$ 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 $\mathsf{Car}(M)$ and prove that their union $K$ is also in $\mathsf{Car}(M)$. Fix $S \subset X$ and calculate for every $N \in \N$ that \begin{align*}& M(S \cap (L_1 \cup \cdots \cup L_N)) \\ = {}& M(S \cap (L_1 \cup \cdots \cup L_N) \cap L_N) + M(S \cap (L_1 \cup \cdots \cup L_N) \cap L_N^\mathsf{c}) \\ = {}& M(S \cap L_N) + M(S \cap (L_1 \cup \cdots \cup L_{N-1})) \end{align*} and therefore \[ M(S \cap (L_1 \cup \cdots \cup L_N)) = \sum_{n=1}^N M(S \cap L_n) \] by induction. We have for every $N \in \N$ that \begin{align*} M(S) &{}\ge M(S \cap (L_1 \cup \cdots \cup L_N)^\mathsf{c}) + M(S \cap (L_1 \cup \cdots \cup L_N)) \\&{}\ge M(S \cap K^\mathsf{c}) + \sum_{n=1}^N M(S \cap L_n) \end{align*} and therefore conclude \begin{align*} M(S) &{}\ge M(S \cap K^\mathsf{c}) + \sum_{n=1}^\infty M(S \cap L_n) \\&{}\ge M(S \cap K^\mathsf{c}) + M \left( S \cap \, \bigcup_{n=1}^\infty L_n \right) \ge M(S) \end{align*} from which we conclude all terms above are equal. In particular $K$ belongs to $\mathsf{Car}(M)$ and taking $S = K$ gives \[ M \left( \, \bigcup_{n=1}^\infty L_n \right) = \sum_{n=1}^\infty M(L_n) \] which is countable additivity. ▮
The above theorem is only useful if $\mathsf{Car}(M)$ is rich. We address this issue next.
5.3 Real Borel sets are Lebesgue measurable
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 check that $\leb(\R)$ contains the Borel σ-algebra on $\R$.
Theorem 5.3.1
The Lebesgue σ-algebra $\leb(\R)$ contains the Borel σ-algebra $\borel(\R)$.
Proof:
Since $\borel(\R)$ is generated by the open intervals $(a,b)$ with $a < b$ it suffices to prove that each open interval $(a,b)$ belongs to $\leb(\R)$. Fix $S \subset \R$. We need to verify that \[ \Lambda(S) \ge \Lambda(I \cap S) + \Lambda(I^\mathsf{c} \cap S) \] where $I = (a,b)$.
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 \[ \sum_{n=1}^\infty \mathsf{Length}(J_n) \le \Lambda(S) + \epsilon \] that covers $S$. Now \begin{align*} J_n = {}& (I \cap J_n) \cup (I^\mathsf{c} \cap J_n) \\ = {}&\Big( (a,b) \cap J_n \Big) \cup \Big( (-\infty,a] \cap J_n \Big) \cup \Big( [b,\infty) \cap J_n \Big) \end{align*} so \begin{align*}\mathsf{Length}(J_n) + \dfrac{2\epsilon}{2^n}{}&\ge \mathsf{Length}(I \cap J_n) \\&\quad {}+ \mathsf{Length}\Big( (-\infty,a+\tfrac{\epsilon}{2^n}) \cap J_n \Big) \\&\quad\quad {}+ \mathsf{Length}\Big( (b-\tfrac{\epsilon}{2^n},\infty) \cap J_n \Big) \end{align*} and we may conclude that \[ \begin{aligned} \Lambda(S) + 3\epsilon &{} \ge \sum_{n=1}^\infty \mathsf{Length}(J_n) + \dfrac{2\epsilon}{2^n} \\ &{} = \sum_{n=1}^\infty \mathsf{Length}(I \cap J_n) \\ & \qquad +{} \sum_{n=1}^\infty \mathsf{Length}\Big( (-\infty,a+\tfrac{\epsilon}{2^n}) \cap J_n \Big) \\ & \qquad\qquad +{} \sum_{n=1}^\infty \mathsf{Length}\Big( (b-\tfrac{\epsilon}{2^n},\infty) \cap J_n \Big) \\ &{} \ge \Lambda(I \cap S) + \Lambda(I^\mathsf{c} \cap S) \end{aligned} \] for every $\epsilon > 0$ because each $I \cap J_n$ is an open interval and together these intervals cover $S$ while \[ \Big( (b-\tfrac{\epsilon}{2^n},\infty) \cap J_n \Big), \Big( (-\infty,a+\tfrac{\epsilon}{2^n}) \cap J_n \Big) \] is a sequence of open intervals that cover $I^\mathsf{c} \cap S$. ▮