Constructing measures

Home | Assessment | Notes | Worksheets | Tutorials

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 $\mathbb{R}$ 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 $\mathsf{Bor}(\mathbb{R})$ on $\mathbb{R}$. We can therefore rephrase our goal as follows: construct a measure $\lambda$ on $\mathsf{Bor}(\mathbb{R})$ such that $$\lambda ((a,b))=b-a$$ for all $a<b$. In this section we will construct such a measure. To do so we will introduce two new ideas: outer measures and Lebesgue measurable sets.

Outer measure

If a Borel set $B$ is covered by a union of intervals $$B\subset \bigcup _{n=1}^{\mathrm{\infty }}(a_n,b_n)$$ then we must have $$\lambda (B)\le \sum _{n=1}^{\mathrm{\infty }}\lambda ((a_n,b_n))=\sum _{n=1}^{\mathrm{\infty }}{b}_n-a_n$$ 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 $$\lambda (B)=inf\{\sum _{n=1}^{\mathrm{\infty }}{b}_n-a_n:B\subset \bigcup _{n=1}^{\mathrm{\infty }}(a_n,b_n)\}$$ for all sets $B\subset \mathbb{R}$.

The function $\lambda$ is defined on $\mathcal{P}(\mathbb{R})$. It assigns a value in $[0,\mathrm{\infty }]$ to every subset of $\mathbb{R}$. However, as we have seen already, it is not a measure on the σ-algebra $\mathcal{P}(\mathbb{R})$ because it is not countably additive. However, it does have some properties in common with measures.

Theorem

The map $\lambda :\mathcal{P}(\mathbb{R})\to [0,\mathrm{\infty }]$ has the following properties.

1. $\lambda (\varnothing )=0$.
2. $\lambda (A)\le \lambda (B)$ whenever $A\subset B$.
3. $\lambda (A_1\cup A_2\cup \cdots )\le \lambda (A_1)+\lambda (A_2)+\lambda (A_3)+\cdots$ for all $A_1,A_2,A_3,\dots \subset \mathbb{R}$.

Proof:

These properties are proved in the length section.▮

Quite generally, a mapping $\mathrm{\Xi }$ from $\mathcal{P}(X)$ to $[0,\mathrm{\infty }]$ satisfying the three properties above is called an outer measure.

Lebesgue measurable sets

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 $\mathbb{R}$ must be split properly for a set to be measurable.

Definition

A subset $L$ of $\mathbb{R}$ is Lebesgue measurable if $$\lambda (S)=\lambda (S\cap L)+\lambda (S\cap L^{\mathsf{c}})$$ for every set $S\subset \mathbb{R}$. Write $\mathsf{Leb}(\mathbb{R})$ for the collection of all Lebesgue measurable subsets of $\mathbb{R}$.

Since $\lambda$ is an outer measure we have $$\lambda (S)\le \lambda (S\cap L)+\lambda (S\cap L^{\mathsf{c}})$$ for all $S,L\subset \mathbb{R}$. 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 $S\subset \mathbb{R}$.

Theorem

The collection $\mathsf{Leb}(\mathbb{R})$ is a σ-algebra and the restriction of $\lambda$ to $\mathsf{Leb}(\mathbb{R})$ is a measure.

Proof:

We know that $\lambda (\varnothing )=0$. The empty set is therefore Lebesgue measurable because $S\cap \varnothing =\varnothing$ and $S\setminus \varnothing =S$ for every $S\subset \mathbb{R}$. Thus $\mathsf{Leb}(\mathbb{R})$ is non-empty.

The expression defining Lebesgue measurability is symmetric in terms of $L$ and $L^{\mathsf{c}}$ so it is immediate that $L\in \mathsf{Leb}(\mathbb{R})$ implies $L^{\mathsf{c}}$ is Lebesgue measurable.

Next we prove that if $L,M\subset \mathbb{R}$ are both Lebesgue measurable then so too is $L\cup M$. Fix $S\subset \mathbb{R}$. We have $$\begin{array}{rl}\lambda (S)& =\lambda (S\cap L)+\lambda (S\cap L^{\mathsf{c}})\\ & =\lambda (S\cap L\cap M)+\lambda (S\cap L\cap M^{\mathsf{c}})+\\ & \lambda (S\cap L^{\mathsf{c}}\cap M)+\lambda (S\cap L^{\mathsf{c}}\cap M^{\mathsf{c}})\\ & \ge \lambda (S\cap (L\cup M))+\lambda (S\cap (L\cup M{)}^{\mathsf{c}})\end{array}$$ 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 $S\subset \mathbb{R}$ was arbitrary we conclude that $L\cup M$ is Lebesgue measurable.

Now, if $L,M$ and Lebesgue measurable and disjoint, we can calculate $$\begin{array}{rl}\lambda (L\cup M)& =\lambda ((L\cup M)\cap L)+\lambda ((L\cup M)\cap L^{\mathsf{c}})\\ & =\lambda (L)+\lambda (M)\end{array}$$ which proves $\lambda$ is additive on the σ-algebra $\mathsf{Leb}(\mathbb{R})$.

As $\mathsf{Leb}(\mathbb{R})$ 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 $\mathsf{Leb}(\mathbb{R})$ and prove that their union $M$ is also in $\mathsf{Leb}(\mathbb{R})$. Fix $S\subset \mathbb{R}$ and calculate for every $N\in \mathbb{N}$ that $$\begin{array}{rl}& \lambda (S\cap (L_1\cup \cdots \cup L_N))\\ =& \lambda (S\cap (L_1\cup \cdots \cup L_N)\cap L_N)+\lambda (S\cap (L_1\cup \cdots \cup L_N)\cap L_N^{\mathsf{c}})\\ =& \lambda (S\cap L_N)+\lambda (S\cap (L_1\cup \cdots \cup L_{N-1}))\end{array}$$ 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 $N\in \mathbb{N}$ that $$\begin{array}{rl}\lambda (S)& \ge \lambda (S\cap (L_1\cup \cdots \cup L_N{)}^{\mathsf{c}})+\lambda (S\cap (L_1\cup \cdots \cup L_N))\\ & \ge \lambda (S\cap M^{\mathsf{c}})+\sum _{n=1}^N\lambda (S\cap L_n)\end{array}$$ and therefore $$\begin{array}{rl}\lambda (S)& \ge \lambda (S\cap M^{\mathsf{c}})+\sum _{n=1}^{\mathrm{\infty }}\lambda (S\cap L_n)\\ & \ge \lambda (S\cap M^{\mathsf{c}})+\lambda (S\cap \bigcup _{n=1}^{\mathrm{\infty }}{L}_n)\ge \lambda (S)\end{array}$$ from which we conclude all terms above are equal. In particular $M$ belongs to $\mathsf{Leb}(\mathbb{R})$ and taking $S=M$ gives $$\lambda \left(\bigcup _{n=1}^{\mathrm{\infty }}{L}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\lambda (L_n)$$ which is countable additivity.▮

We find ourselves in the following situation: there is a σ-algebra $\mathsf{Leb}(\mathbb{R})$ on which $\lambda$ defines a measure. Membership in $\mathsf{Leb}(\mathbb{R})$ 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 $\mathsf{Leb}(\mathbb{R})$ is the trivial σ-algebra $\{\varnothing ,\mathbb{R}\}$. To know that $(\mathbb{R},\mathsf{Leb}(\mathbb{R}))$ 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 $\mathsf{Leb}(\mathbb{R})$ contains the Borel σ-algebra.

Theorem

The Lebesgue σ-algebra $\mathsf{Leb}(\mathbb{R})$ contains the Borel σ-algebra $\mathsf{Bor}(\mathbb{R})$.

Proof:

Since $\mathsf{Bor}(\mathbb{R})$ is generated by the open intervals $(a,b)$ with $a<b$ it suffices to prove that each open interval $(a,b)$ belongs to $\mathsf{Leb}(\mathbb{R})$. Fix $S\subset \mathbb{R}$. 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 $\lambda (I)<b-a$ because $I$ is an interval that covers $I$. For the converse fix $ϵ>0$. There is a sequence $n\mapsto (a_n,b_n)$ of intervals with $$(a,b)\subset \bigcup _{n=1}^{\mathrm{\infty }}(a_n,b_n)$$ and $$\sum _{n=1}^{\mathrm{\infty }}{b}_n-a_n\le \lambda (I)+ϵ$$ as well. Now $n\mapsto (a_n,b_n)$ is an open cover of $[a+ϵ,b-ϵ]$ 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 $[a+ϵ,b-ϵ]$. It must then be the case that $$\sum _{n=1}^v{b}_{r(n)}-a_{r(n)}\ge b-a-2ϵ$$ whence $$\lambda (I)\ge (\sum _{n=1}^{\mathrm{\infty }}{b}_n-a_n)-ϵ\ge b-a-3ϵ$$ and we conclude, since $ϵ>0$ was arbitrary, that $\lambda (I)=b-a$.

If $\lambda (S)=\mathrm{\infty }$ there is nothing to prove, so assume $\lambda (S)$ is finite. Fix $ϵ>0$ and let $n\mapsto J_n$ be a sequence of intervals with $$\sum _{n=1}^{\mathrm{\infty }}\lambda (J_n)\le \lambda (S)+ϵ$$ that covers $S$. (Here we have used that $\lambda$ assignes the correct lengths to intervals.) Now $$\begin{array}{rl}\lambda (S)+ϵ& \ge \sum _{n=1}^{\mathrm{\infty }}\lambda (J_n)\\ & =\sum _{n=1}^{\mathrm{\infty }}\lambda (J_n\cap I)+\sum _{n=1}^{\mathrm{\infty }}\lambda (J_n\cap I^{\mathsf{c}})\\ & \ge \lambda (S\cap I)+\lambda (S\cap I^{\mathsf{c}})\end{array}$$ for every $ϵ>0$. Thus $I$ belongs to $\mathsf{Leb}(\mathbb{R})$.▮