Lebesgue Measurability

Measurable Sets

Lebesgue Measurable Set

A set $E\subseteq {\mathbb{R}}^d$ is said to be Lebesgue measurable if for all $ϵ>0$, there is an open $O\supset E$, such that $m_{\ast }(O\setminus E)\le ϵ.$If $E$ is Lebesgue measurable, we write $m(E)$ for the exterior measure of $E$.

We will see that there are several equivalent ways to define a Lebesgue measurable set in the end.

Proposition

If $m_{\ast }(E)=0$, then $E$ is Lebesgue measurable. Such sets are called null sets.

Proof For any $ϵ>0$, we can choose open set $O\supset E$ with $m_{\ast }(O)\le ϵ$. Then $m_{\ast }(O\setminus E)\le m_{\ast }(O)\le ϵ$. $\square$

e.g. $\text{Q}$, (standard middle-thirds) Cantor set are null sets in $\mathbb{R}$.

Proposition

Open sets in ${\mathbb{R}}^d$ are Lebesgue measurable.

Proof Straightforward from the definition. $\square$

Proposition

A countable union of measurable sets is measurable.

Proof Suppose $E_1,E_2,\cdots$ are measurable sets in ${\mathbb{R}}^d$. For all $ϵ>0$, choose open $O_i\supset E_i$ with $m_{\ast }(O_i\setminus E_i)\le ϵ/2^i$. Then ${\bigcup }_{i=1}^{\infty }{O}_i\supset {\bigcup }_{i=1}^{\infty }{E}_i,$and we have $m_{\ast }\left({\bigcup }_{i=1}^{\infty }{O}_i\setminus {\bigcup }_{i=1}^{\infty }{E}_i\right)\le {\sum }_{i=1}^{\infty }{m}_{\ast }(O_i\setminus E_i)\le ϵ.$ $\square$

Proposition

Closed sets in ${\mathbb{R}}^d$ are Lebesgue measurable.

Proof Suppose $F\subset {\mathbb{R}}^d$ is closed. Without loss of generality, we can assume $F$ is compact.

Lemma

If $F$ is closed, $K$ is compact, and these sets are disjoint, then $d(F,K)>0$.

Proposition

The complement of a measurable set is measurable.

Corollary

A countable intersection of measurable sets is measurable.

Proof The complement of a countable union is a countable intersection. $\square$

Theorem

If $E_1,E_2,\dots$ are disjoint measurable sets, and $E={\bigcup }_{j=1}^{\infty }{E}_j$, then $m(E)={\sum }_{j=1}^{\infty }m(E_j).$

Proof

Corollary

For any two measurable sets $E$ and $F$, we have $m(E\cup F)+m(E\cap F)=m(E)+m(F).$

Proof We add the following two equations together: $\begin{array}{r}m(E\cup F)=m(E)+m(F\setminus E)\\ m(F\setminus E)+m(E\cap F)=m(F)\end{array}$ $\square$

Corollary

Suppose $E_1,E_2,\dots$ are measurable subsets of ${\mathbb{R}}^d$.

1. If $E_k\nearrow E$, then $m(E)={\lim }_{N\to \infty }m(E_N)$.
2. If $E_k\searrow E$ and $m(E_k)<\infty$ for some $k$, then $m(E)={\lim }_{N\to \infty }m(E_N).$

Where if $E_1,E_2,\cdots$ is a countable collection of subsets of ${\mathbb{R}}^d$ that increases to $E$ in the sense that $E_k\subset E_{k+1}$ for all $k$, and $E={\bigcup }_{k=1}^{\infty }{E}_k$, then we write $E_k\nearrow E$. Similarly, if $E_1,E_2,\cdots$ is a countable collection of subsets of ${\mathbb{R}}^d$ that decreases to $E$ in the sense that $E_{k+1}\subset E_k$ for all $k$, and $E={\bigcap }_{k=1}^{\infty }{E}_k$, then we write $E_k\searrow E$.

Theorem

Suppose $E$ is a measurable subset of ${\mathbb{R}}^d$. Then, for every $ϵ>0$:

1. There exists an open set $\mathcal{O}$ with $E\subset \mathcal{O}$ and $m(\mathcal{O}\setminus E)\le ϵ$.
2. There exists a closed set $F$ with $F\subset E$ and $m(E\setminus F)\le ϵ$.
3. If $m(E)$ is finite, there exists a compact set $K$ with $K\subset E$ and $m(E\setminus K)\le ϵ$.
4. If $m(E)$ is finite, there exists a finite union $F={\bigcup }_{j=1}^N{Q}_j$ of closed cubes such that
   $m(E△F)\le ϵ.$

Invariance Properties

Translation Invariance

$E\subset {\mathbb{R}}^d$ is a measurable set iff for all $h\in {\mathbb{R}}^d$, the set $E+h:=\{x+h\mid x\in E\}$ is measurable, and $m(E+h)=m(E)$.

Proof This is easy to see by replacing $O$ with $O\pm h$. $\square$

Dilation Invariance

Suppose $\lambda >0$, and denote by $\lambda E$ the set $\{\lambda x\mid x\in E\}$, then $\lambda E$ is measurable iff $E$ is measurable, and $m(\lambda E)={\lambda }^{d}m(E)$.

Proof This is easy to see by replacing $O$ with $\lambda O$, and dilation of an open set remains open. Moreover, for every cube $Q$, we have $|\lambda Q|={\lambda }^d|Q|$, so $m(\lambda E)={\lambda }^{d}m(E)$. $\square$

Reflection Invariance

$E$ is measurable iff $-E$ is measurable, and $m(-E)=m(E)$.

Proof This is easy to see by replacing $O$ with $-O$. $\square$

Sigma Algebra and Borel Sets

Lebesgue $\sigma$-Algebra

The Lebesgue $\sigma$-algebra on ${\mathbb{R}}^d$ is the [[Measurable Spaces and Functions#^60a516|$\sigma$-algebra]] generated by all Lebesgue measurable sets.

Borel $\sigma$-Algebra

The Borel $\sigma$-algebra on ${\mathbb{R}}^d$, denoted as ${\mathcal{B}}_{{\mathbb{R}}^d}$, is the smallest $\sigma$-algebra containing all open sets. The term “smallest” means that if $\mathcal{S}$ is any $\sigma$-algebra that contains all open sets in ${\mathbb{R}}^d$, then necessarily ${\mathcal{B}}_{{\mathbb{R}}^d}\subset \mathcal{S}$. Elements of the Borel $\sigma$-algebra are called Borel sets.

Corollary

A subset $E$ of ${\mathbb{R}}^d$ is measurable

1. if and only if $E$ differs from a $G_{\delta }$ by a set of measure zero, i.e., $E\subseteq G_{\delta }$, and $m(G_{\delta }\setminus E)=0$.
2. if and only if $E$ differs from an $F_{\sigma }$ by a set of measure zero. i.e., $E\supseteq F_{\sigma }$, and $m(E\setminus F_{\sigma })=0$.

Where a $G_{\delta }$ set is a countable intersection of open sets, and an $F_{\sigma }$ set is a countable union of closed sets.

Here is a summary of the equivalence of the definitions of a Lebesgue measurable set:

Equivalences for Lebesgue Measurability

Suppose $A\subseteq {\mathbb{R}}^d$. Then the following are equivalent:

1. $A$ is Lebesgue measurable.
2. For each $ϵ>0$, there exists a closed set $F\subseteq A$ with $m_{\ast }(A\setminus F)<ϵ$.
3. There exist closed sets $F_1,F_2,\dots$ contained in $A$ such that
   $m_{\ast }\left(A\setminus {\bigcup }_{k=1}^{\infty }{F}_k\right)=0.$
4. There exists a Borel set $B\subseteq A$ such that $m_{\ast }(A\setminus B)=0$.
5. For each $ϵ>0$, there exists an open set $G\supseteq A$ such that $m_{\ast }(G\setminus A)<ϵ$.
6. There exist open sets $G_1,G_2,\dots$ containing $A$ such that
   $m_{\ast }\left({\bigcap }_{k=1}^{\infty }{G}_k\setminus A\right)=0.$
7. There exists a Borel set $B\supseteq A$ such that $m_{\ast }(B\setminus A)=0$.

Non Measurable Sets

We now explicitly construct a non-measurable set.

Proposition

Consider $[0,1]/\text{Q}$, we identify a set $\mathcal{N}\subset [0,1]$ of representatives of each equivalence class in $[0,1]/\text{Q}$. Then $\mathcal{N}$ is a non-measurable set.

Proof We shall prove by contradiction. Suppose $\mathcal{N}$ is measurable.

Axiom of Choice

The construction of such a set relies on the axiom of choice, and the existence of non-measurable sets is independent of the standard axioms of set theory.