Measure Theory

Rectangles

The main idea behind calculating the measure of a subset $A\subset {\mathbb{R}}^d$ is approximating the set $A$ with a union of geometrically simple shapes with known measure such has balls or rectangles.

A (closed) rectangle $R\subset {\mathbb{R}}^d$ is given by

$$R=[a_1,b_1]\times [a_2,b_2]\times \cdots \times [a_d,b_d].$$

Then the volume of a rectangle $R$ is

$$|R|=\prod _{i=1}^d(b_i-a_i).$$

We say a union of rectangles is almost disjoint if the interiors of the rectangles are disjoint.

Lemma

If a rectangle is the almost disjoint union of finitely many other rectangles, say $R={\bigcup }_{k=1}^N{R}_k$, then $|R|={\sum }_{k=1}^N|R_k|.$

If the rectangles are not almost disjoint, then

$$|R|\le \sum _{k=1}^N|R_k|.$$

Thm

Every open subset $\mathcal{O}$ of $\mathbb{R}$ can be written uniquely as a countable union of disjoint open intervals.

Proof: For each $x\in \mathcal{O}$, set $I_x$ to be the largest open interval containing $x$ and contained in $\mathcal{O}$. Define

$$a_x=\inf \{a<x:(a,x)\subset \mathcal{O}\}\text{ and }{b}_x=\sup \{b>x:(x,b)\in \mathcal{O}\}.$$

Clearly, $a_x<x<b_x$ so if $I_x=(a_x,b_x)$ then $x\in I_x$ and $I_x\subset \mathcal{O}$. Thus,

$$\mathcal{O}=\bigcup _{x\in \mathcal{O}}{I}_x.$$

If $I_x$ and $I_y$ intersect, then their $I_x\cup I_y\subset \mathcal{O}$ contains $x$. But we said that $I_x$ is maximal, so $I_x\cup I_y\subset I_x$ and $I_x\cup I_y\subset I_y$. Thus, $I_x=I_y$. Therefore, the collection of intervals $\mathcal{I}=\{I_x{\}}_{x\in \mathcal{O}}$ is disjoint. We also know this collection is countable as every open interval $I_x$ must contain a rational number.

Q.E.D.

This result generalizes to higher dimensions.

The Exterior Measure

Let $E$ be a subset of ${\mathbb{R}}^d$. Then the exterior/outer measure of $E$ is

$$m_{\ast }(E)=\inf \sum _{j=1}^{\infty }|Q_j|$$

where the infimum is taken over all countable covers $E\subset {\bigcup }_{j=1}^{\infty }{Q}_j$ of closed cubes.

For example, let $E$ be a closed cube. Clearly $E$ covers itself so $m_{\ast }(E)\le |E|$. Let $E\subset {\bigcup }_{j=1}^{\infty }{Q}_j$ be an arbitrary covering of $E$ by cubes and let $ϵ>0$. For each $j$, we choose an open cube $S_j$ containing $Q_j$ such that $|S_j|\le (1+ϵ)|Q_j|$. This forms an open cover of $E$, a compact set. Thus, there is a finite subcover $E\subset {\bigcup }_{j=1}^N{S}_j$. So

$$|E|\le \sum _{j=1}^N|S_j|\le (1+ϵ)\sum _{j=1}^N|Q_j|.$$

Therefore, $|E|\le m_{\ast }(E)$.

Properties

Clearly by the definition, for every $ϵ>0$ there is a cover $E\subset {\bigcup }_{j=1}^{\infty }{Q}_j$ such that

$$\sum _{j=1}^{\infty }{m}_{\ast }(Q_j)\le m_{\ast }(E)+ϵ.$$

Some useful properties:

1. Monotonicity: If $E_1\subset E_2$, then $m_{\ast }(E_1)\le m_{\ast }(E_2)$.
2. Countable sub-additivity: If $E={\bigcup }_{j=1}^{\infty }{E}_j$, then $m_{\ast }(E)\le {\sum }_{j=1}^{\infty }{m}_{\ast }(E_j)$.
3. If $E\subset {\mathbb{R}}^d$, then $m_{\ast }(E)=\inf m_{\ast }(\mathcal{O})$ where the infimum is taken over all open sets $\mathcal{O}$ containing $E$.
4. If $E=E_1\cup E_2$ and ${\inf }_{e_1\in E_1,e_2\in E_2}d(e_1,e_2)>0$, then $m_{\ast }(E)=m_{\ast }(E_1)+m_{\ast }(E_2)$.
5. If $E$ is a countable union of almost disjoint cubes $E={\bigcup }_{j=1}^{\infty }{Q}_j$, then $m_{\ast }(E)={\sum }_{j=1}^{\infty }|Q_j|$.

The Lebesgue Measure

We say a subset $E\subset {\mathbb{R}}^d$ is Lebesgue measurable if for every $ϵ>0$ there exists an open set $\mathcal{O}$ with $E\subset \mathcal{O}$ and

$$m_{\ast }(\mathcal{O}-E)\le ϵ.$$

If $E$ is measurable, we define it’s Lebesgue measure to be $m(E)=m_{\ast }(E)$. In other words, the Lebesgue measure is simply the outer measure restricted to measurable sets. Clearly, properties 1-5 above still hold. Moreover, it is clear that every open set in ${\mathbb{R}}^d$ is measurable as $m_{\ast }(\mathcal{O}-\mathcal{O})=m_{\ast }(\varnothing )=0\le ϵ$.

Notice that for any set $E$ with $m_{\ast }(E)=0$, we get that $E$ is measurable. Indeed, for every $ϵ>0$ it follows by property 3 that we can choose an open set $\mathcal{O}\supset E$ with $m_{\ast }(\mathcal{O})\le ϵ$. Then by monotonicity,

$$m_{\ast }(\mathcal{O}-E)\le m_{\ast }(\mathcal{O})\le ϵ.$$

We also get that a countable union of measurable sets is measurable. If $E_j$ is a countable collection of measurable sets, then there is another countable collection ${\mathcal{O}}_j$ of open sets such that $E_j\subset {\mathcal{O}}_j$ and $m_{\ast }({\mathcal{O}}_j-E_j)\le ϵ/2^j$. Thus,

$$m_{\ast }(\bigcup _{j=1}^{\infty }{\mathcal{O}}_j-\bigcup _{j=1}^{\infty }{E}_j)\le \sum _{j=1}^n{m}_{\ast }({\mathcal{O}}_j-E_j)\le ϵ.$$

Thm

If $E_1,E_2,\dots$ are disjoint measurable sets, then

$$m\left(\bigcup _{j=1}^{\infty }{E}_j\right)=\sum _{j=1}^{\infty }m(E_j).$$

Notice that this result does not hold for the outer measure!

Some more properties:

• The complement of a measurable set is measurable
• Closed sets are measurable
• A countable intersection of measurable sets is measurable
• If $E_k\nearrow E$, then $m(E)={\lim }_{k\to \infty }m(E_k)$.
• If $E_k\searrow E$ and $m(E_k)<\infty$ for some $k$, then $m(E)={\lim }_{k\to \infty }m(E_k)$.

σ-algebras and Borel Sets

A $\sigma$-algebra on a set $X$ is a collection of subsets of $X$ that is closed under countable unions, countable intersections and complements. While the power set, $\mathcal{P}({\mathbb{R}}^d)$, is a $\sigma$-algebra we are more interested in the $\sigma$-algebra consisting of all measurable sets.

One special $\sigma$-algebra is the Borel $\sigma$-algebra, i.e., the smallest $\sigma$-algebra consisting of all open sets.

Abstract Measures

In the above, we focused on ${\mathbb{R}}^d$ and the Lebesgue measure. We know abstract our notion of measure to arbitrary sets.

A measure space consists of a set $X$ with

1. A $\sigma$-algebra $\mathcal{M}$ of measurable sets; and
2. A measure $\mu :\mathcal{M}\to [0,\infty ]$ such that if $E_1,E_2,\dots$ is a countable family of disjoint sets in $\mathcal{M}$ then

$$\mu \left(\sum _{n=1}^{\infty }{E}_n\right)=\sum _{n=1}^{\infty }\mu (E_n).$$

One may denote a measure space by $(X,\mathcal{M},\mu )$, but it is typically abbreviated as $(X,\mu )$. As we saw above, $({\mathbb{R}}^d,\mathcal{M},m)$ is a measure space where $\mathcal{M}$ is the set of Lebesgue measurable sets. Another simple example is the counting measure, $(X,\mathcal{P}(X),\mu )$ where $\mu (E)=|E|$ the cardinality of $E$.

Caratheodory’s Extension Theorem

Let $X$ be a set. An exterior measure ${\mu }_{\ast }$ on $X$ is a function ${\mu }_{\ast }:\mathcal{P}(X)\to [0,\infty ]$ that satisfies the following properties:

1. ${\mu }_{\ast }(\varnothing )=0$.
2. Monotonicity: If $E_1\subset E_2$, then ${\mu }_{\ast }(E_1)\le m_{\ast }(E_2)$.
3. Countable sub-additivity: If $E_1,E_2,\dots$ is a countable family of sets, then

$${\mu }_{\ast }\left(\bigcup _{j=1}^{\infty }{E}_j\right)\le \sum _{j=1}^{\infty }{\mu }_{\ast }(E_j).$$

Given an exterior measure ${\mu }_{\ast }$, we say a set $E$ in $X$ is Caratheodory measurable if for every $A\subset X$,

$${\mu }_{\ast }(A)={\mu }_{\ast }(E\cap A)+{\mu }_{\ast }(E^c\cap A).$$

This is sometimes called the separation condition.

Prp

Lebesgue measurable sets are Caratheodory measurable.

Proof: Let $E$ Lebesgue measurable, i.e., for all $ϵ>0$ there is an open set $O_{ϵ}$ such that $E\subset O_{ϵ}$ and $m_{\ast }(O_{ϵ}-E)\le ϵ$. Let $A\subset {\mathbb{R}}^d$. Notice that

$$\begin{array}{rl}{\mu }_{\ast }(A)& ={\mu }_{\ast }((E\cap A)\cup (E^c\cap A))\\ & \le {\mu }_{\ast }(E\cap A)+{\mu }_{\ast }(E^c\cap A)\\ & \le {\mu }_{\ast }(O_{ϵ}\cap A)+{\mu }_{\ast }((E^c\cap O_{ϵ}\cap A)\cup (O_{ϵ}^c\cap A))\\ & \le {\mu }_{\ast }(O_{ϵ}\cap A)+{\mu }_{\ast }(E^c\cap O_{ϵ}\cap A)+{\mu }_{\ast }(O_{ϵ}^c\cap A)\\ & \le {\mu }_{\ast }(A)+ϵ.\end{array}$$

Let $ϵ\to 0$.

Caratheodory Extension Theorem

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

Measurable Functions

Let $(X,{\mu }_X)$ and $(Y,{\mu }_Y)$ be measurable spaces. A function $f:X\to Y$ is said to be measurable if for any ${\mu }_Y$-measurable set $E\subset Y$, the pre-image

$$f^{-1}(E)=\{x\in X\mid f(x)\in E\}$$

is ${\mu }_X$-measurable.

Some properties:

1. The sum and product of complex valued measurable functions is measurable.
2. The composition of measurable functions is measurable.
3. If $Y$ is a metric space, then the pointwise limit of a sequence of measurable functions $f_n:X\to Y$ is measurable.