Measurable Spaces and Functions

$\sigma$-Algebra and Measurable Spaces

$\sigma$-Algebra

Suppose $X$ is a set and $\mathcal{S}$ is a set of subsets of $X$. Then $\mathcal{S}$ is called a $\sigma$-algebra on $X$ if it is closed under countable unions, countable intersections, and complements:

• $\varnothing \in \mathcal{S}$;
• if $Y\in \mathcal{S}$, then $X\setminus Y\in \mathcal{S}$;
• if $Y_1,Y_2,\cdots \in \mathcal{S}$, then ${\bigcup }_{i=1}^{\infty }{Y}_i\in \mathcal{S}$.

Note that the closedness on countable intersections comes freely from (2) and (3).

Measurable Space

A measurable space is an ordered pair of a set and an associated $\sigma$-algebra $(X,\mathcal{S})$. An element of $\mathcal{S}$ is called an $\mathcal{S}$-measurable set, or just a measurable set.

Proposition

Suppose $X$ is a set and $\mathcal{A}$ is a set of subsets of $X$. Then the intersection of all $\sigma$-algebras on $X$ that contain $\mathcal{A}$ is a $\sigma$-algebra on $X$.

Proof

Measurable Functions

Measurable Function

Suppose $(X,\mathcal{S})$ and $(Y,\mathcal{T})$ are measurable spaces. A function $f:X\to Y$ is an measurable function if $f^{-1}(E)\in \mathcal{S}\text{for all }E\in \mathcal{T}.$

Essential Range

The essential range of a measurable function $f:(X,\mathcal{S})\to (Y,\mathcal{T})$ is the set $\mathrm{ess}\mathrm{range}(f):=\{y\in Y\mid \forall U\in \mathcal{T}\text{ with }y\in U,\mu (f^{-1}(U))>0\},$where $\mu$ is a measure on $(X,\mathcal{S})$.

Measure Spaces

Measure

Suppose $(X,\mathcal{S})$ is a measurable space. A measure $\mu$ on $(X,\mathcal{S})$ is a function $\mu :\mathcal{S}\to [0,\infty ]$ such that: $\mu (\varnothing )=0$ and $\mu \left({\bigcup }_{k=1}^{\infty }{E}_k\right)={\sum }_{k=1}^{\infty }\mu (E_k)$for any countable collection $\{E_k{\}}_{k=1}^{\infty }$ of disjoint sets in $\mathcal{S}$. In this case, we call $(X,\mathcal{S},\mu )$ a measure space.

Proposition

For a measure $\mu$ defined on a measurable space $(X,\mathcal{S})$, the following properties hold:

• Monotonicity: $\mu (E)\le \mu (F)$ for $E\subset F$,
• Subadditivity: $\mu \left({\bigcup }_{i=1}^{\infty }{E}_i\right)\le {\sum }_i^{\infty }\mu (E_i)$.

Proof Observe that for any $E\subset F$, $E$ and $F\setminus E$ are disjoint, and $F=E\cup (F\setminus E)$, thus $\mu (F)=\mu (E)+\mu (F\setminus E)\ge \mu (E).$For subadditivity, we can divide the union into disjoint sets. Suppose $G_n={\bigcup }_{i=1}^n{E}_i$, let $F_1:=E_1$, and $F_i:=E_i\setminus G_{i-1}$ for $i\ge 2$. Then each $F_i$ is disjoint, and we have $\mu \left({\bigcup }_{i=1}^{\infty }{E}_i\right)=\mu \left({\bigcup }_{i=1}^{\infty }{F}_i\right)={\sum }_{i=1}^{\infty }\mu (F_i)\le {\sum }_{i=1}^{\infty }\mu (E_i),$where the last inequality follows from monotonicity. $\square$

$\sigma$-Finite Measure

A measure space $(X,\mathcal{S},\mu )$ is $\sigma$-finite if there exists a countable collection $\{A_i{\}}_{i=1}^{\infty }$ of measurable sets such that $X={\bigcup }_{i=1}^{\infty }{A}_i,$ and $\mu (A_i)<\infty$ for all $i$.

e.g.

• The Lebesgue measure on $\mathbb{R}$ is $\sigma$-finite because $\mathbb{R}={\bigcup }_{i=1}^{\infty }[-i,i]$.
• The counting measure on $\mathbb{R}$ is not $\sigma$-finite, because one cannot decompose $\mathbb{R}$ into a countable union of sets with finite cardinality.

Complete Measure

A measure $\mu$ defined on a measurable space $(X,\mathcal{S})$ is complete if for all $F\in \mathcal{S}$ that $\mu (F)=0$ and $E\subset F$ implies $E\in \mathcal{S}$.

Exterior Measure and Carathéodory Theorem

Exterior (Outer) Measure

Let $X$ be a set. The exterior measure ${\mu }_{\ast }$ on $X$ is defined on all subsets of $X$ to $[0,\infty ]$ such that

• ${\mu }_{\ast }(\varnothing )=0$;
• ${\mu }_{\ast }(A)\le {\mu }_{\ast }(B)$ if $A\subseteq B$;
• ${\mu }_{\ast }({\bigcup }_{i=1}^{\infty }{A}_i)\le {\sum }_{i=1}^{\infty }{\mu }_{\ast }(A_i)$.

Carathéodory Measurable

Suppose ${\mu }_{\ast }$ is an exterior measure. A set $E$ in $X$ is Carathéodory measurable or simply measurable if one has ${\mu }_{\ast }(A)={\mu }_{\ast }(E\cap A)+{\mu }_{\ast }(E^c\cap A)\text{ for every }A\subset X.$

Carathéodory Theorem

Given an exterior measure ${\mu }_{\ast }$ on a set $X$, the collection $\mathcal{M}$ of Carathéodory measurable sets forms a $\sigma$-algebra. Moreover, ${\mu }_{\ast }$ restricted to $\mathcal{M}$ is a measure.

Proof Clearly, $\varnothing$ and $X$ belong to $\mathcal{M}$ and ${\mu }_{\ast }(\varnothing )=0$,

One of the most important examples of exterior measure is the exterior measure on metric spaces, which is defined as follows:

Metric Exterior Measure

An exterior measure ${\mu }_{\ast }$ on a metric space $(X,d)$ is called a metric exterior measure if it satisfies ${\mu }_{\ast }(A\cup B)={\mu }_{\ast }(A)+{\mu }_{\ast }(B),\text{whenever }d(A,B)>0.$

This property plays a crucial role in the case of exterior Lebesgue measure.

Theorem

If ${\mu }_{\ast }$ is a metric exterior measure on a metric space $(X,d)$, then the Borel sets in $X$ are measurable. Hence ${\mu }_{\ast }$ restricted to the Borel sets is a measure.

The Extension Theorem

Boolean Algebra

Let $X$ be a set. A boolean algebra $\mathcal{A}$ on $X$ is a nonempty collection of subsets satisfies

• $\varnothing \in X$,
• If $E\in \mathcal{A}$, then $X\setminus E=E^c\in A$,
• If $E$ and $F$ are elements of $\mathcal{A}$, then $E\cup F\in \mathcal{A}$.

In other words, $\mathcal{A}$ is closed under complements, finite unions, and finite intersections.

Premeasure

A premeasure on a boolean algebra $\mathcal{A}$ over a set $X$, is a function $\mu :\mathcal{A}\to [0,\infty ]$ such that

• $\mu (\varnothing )=0$.
• If $E_1,E_2,\dots$ is a countable collection of disjoint sets in $\mathcal{A}$ with ${\bigcup }_{i=1}^{\infty }{E}_i\in \mathcal{A}$, then $\mu \left({\bigcup }_{i=1}^{\infty }{E}_i\right)={\sum }_{i=1}^{\infty }\mu (E_i).$

Premeasures give rise to exterior measures in a natural way:

Lemma

If $\mu$ is a premeasure on a boolean algebra $\mathcal{A}$ over $X$, define ${\mu }_{\ast }$ on any subset $E\subseteq X$ by ${\mu }_{\ast }(E)=\inf \left\{{\sum }_{j=1}^{\infty }\mu (E_j):E\subset {\bigcup }_{j=1}^{\infty }{E}_j,\text{where }{E}_j\in \mathcal{A}\text{ for all }j\in \mathbb{N}\right\}$ Then ${\mu }_{\ast }$ is an exterior measure on $X$ satisfying ${\mu }_{\ast }(E)=\mu (E)$ for all $E\in \mathcal{A}$, and all sets in $\mathcal{A}$ are Carathéodory measurable.

Carathéodory’s Extension Theorem

Suppose that $\mathcal{A}$ is a boolean algebra of sets in $X$, ${\mu }_0$ is a premeasure on $\mathcal{A}$, and $\mathcal{M}$ is the $\sigma$-algebra generated by $\mathcal{A}$ (i.e., the smallest $\sigma$-algebra containing $\mathcal{A}$). Then there exists a measure $\mu$ on $\mathcal{M}$ that extends ${\mu }_0$. Moreover, if $\mu$ is $\sigma$-finite, then it is unqiue.

Proof This is a direct consequence of the above lemma. ${\mu }_0$ induces an exterior measure ${\mu }_{\ast }$, which is a measure on the $\sigma$-algebra of Carathéodory measurable sets. As all sets in $\mathcal{A}$ are Carathéodory measurable, and $\mathcal{M}$ is generated by $\mathcal{A}$, all sets in $\mathcal{M}$ are also Carathéodory measurable. Therefore ${\mu }_{\ast }$ is a measure on $\mathcal{M}$ as well, and we call it $\mu$. To prove the uniqueness, we suppose $\nu$ is another measure defined on $\mathcal{M}$ that extends ${\mu }_0$. Let $E={\bigcup }_{i=1}^{\infty }{E}_i$ containing where each $E_i\in \mathcal{A}$. Then there holds $\nu (E)={\lim }_{n\to \infty }\nu \left({\bigcup }_{i=1}^n{E}_i\right)={\lim }_{n\to \infty }\mu \left({\bigcup }_{i=1}^n{E}_i\right)=\mu (E).$

Pick a set $F\in \mathcal{M}$ with $\mu (F)<\infty$. If $F\subset {\bigcup }_i^{\infty }{E}_i$ for each $E_i\in \mathcal{A}$, then $\nu (F)\le {\sum }_{i=1}^{\infty }\nu (E_i)={\sum }_{i=1}^{\infty }{\mu }_0(E_i),$so $\nu (F)\le \mu (F)$ by taking the infimum over all such covers of $F$. To prove the reverse inequality, note that $\mu (F)={\sum }_{j=1}^{\infty }\mu (F\cap G_j)={\sum }_{j=1}^{\infty }\nu (F\cap G_j)=\nu (F).$

Assume $\mu$ is $\sigma$-finite. We claim that $\mu (F)=\nu (F)$. Observe that since $\mu$ is $\sigma$-finite, we may write $X={\bigcup }_i{G}_i$, where $G_1,G_2,\dots$ is a countable collection of disjoint sets in $\mathcal{A}$ with $\mu (E_i)<\infty$. Then we have $F\subset {\bigcup }_i^{\infty }(F\cap G_i).$Let $E_F\cap G_i$