### A Mild Glimpse Into Measure Theory (1)

1. Introduction

The key idea is to define a function which assigns a value ${m(A)}$, meaning “the size of ${A}$”, to a subset ${A}$ of a given set ${E}$, so that

$\displaystyle m(A\cup B)=m(A)+m(B)\;\mathrm{if}\;A\cap B=\emptyset$

e.g. number of elements, length of an interval.

However, some issues arise in general since one cannot define a function ${m:\mathscr{P}(E)\longrightarrow[0,\infty]}$ which satisfies above or more generally

$\displaystyle m(\bigcup_{n\in\mathbb{N}}A_{n})=\sum_{n\in\mathbb{N}}m(A_{n})\;\; \mathrm{\;if\;\left\{A_{n}\right\}_{n\in\mathbb{N}}}\mathrm{\;is\;a\;family\;of\;disjoint\;sets}$

Therefore, we should restrict the domain to be a special class of subsets, so called the ${\sigma}$-algebra.

2. Measureable Sets

2.1. Definition and Examples

Definition: Let ${X}$ be an arbitrary set. An algebra, ${\mathscr{M}\subseteq\mathscr{P}(X)}$, is a family of subsets of ${X}$ such that:

1. ${X\in\mathscr{M}}$
2. If ${A\in\mathscr{M}, \mathrm{then}\;A^{c}\in\mathscr{M}}$
3. If ${A_{n}\in\mathscr{M}}$ for all ${n}$ from ${1}$ to ${N}$, then ${\bigcup_{n=1}^{N}A_{n}\in\mathscr{M}}$
4. If ${A_{n}\in\mathscr{M}}$ for all ${n}$ from ${1}$ to ${N}$, then ${\bigcap_{n=1}^{N}A_{n}\in\mathscr{M}}$

Definition: Let ${X}$ be an arbitrary set. A ${\sigma}$algebra, ${\mathscr{M}\subseteq\mathscr{P}(X)}$, is a family of subsets of ${X}$ such that:

1. ${X\in\mathscr{M}}$
2. If ${A\in\mathscr{M}, \mathrm{then}\;A^{c}\in\mathscr{M}}$
3. If ${A_{n}\in\mathscr{M}}$ for all ${n\in\mathbb{N}}$, then ${\bigcup_{n\in\mathbb{N}}A_{n}\in\mathscr{M}}$
4. If ${A_{n}\in\mathscr{M}}$ for all ${n\in\mathbb{N}}$, then ${\bigcap_{n\in\mathbb{N}}A_{n}\in\mathscr{M}}$

We call elements ${M\in\mathscr{M}}$ a ${\mathscr{M}}$measureable set, and call the pair ${(X,\mathscr{M})}$measurable space.

Remark 1: Note we can omit (3) or (4) in each definition by de Morgan laws and we can also replace condition (1) by restricting ${\mathscr{M}}$ to be a nonempty family of subsets of ${X}$.
Remark 2: From this definition, we can see also: ${(a)}$ ${\emptyset\in\mathscr{M}}$; ${(b)}$ ${\bigcap_{n\in\mathbb{N}}^{k} A_{n}\in\mathscr{M}}$ and ${\bigcup_{n\in\mathbb{N}}^{k} A_{n}\in\mathscr{M}}$ for any finite integer ${k}$; hence a ${\sigma}$-algebra is an algebra but not vice versa.

Examples:

• ${\mathscr{M}=\mathscr{P}(X)}$ is a ${\sigma}$-algebra.
• ${\mathscr{M}=\left\{\emptyset, X\right\}}$ is a ${\sigma}$-algebra on ${X}$.
• ${\mathscr{M}=\left\{A\in\mathscr{P}(X)|A\;\mathrm{or}\; A^c\;\mathrm{is}\;\mathrm{at}\;\mathrm{most}\;\mathrm{countable}\right\}}$ is a ${\sigma}$-algebra.
• Let ${X}$ be an infinite set, and let ${\mathscr{F}}$ be the collection of all finite subsets of ${X}$. Then ${\mathscr{F}}$ is not even an algebra since ${X\notin\mathscr{F}}$.
• Let ${X}$ be an infinite set, and let ${\mathscr{F}}$ be the collection of all subsets ${F}$ of ${X}$ such that either ${F}$ or ${F^{c}}$ is finite. Then ${\mathscr{F}}$ is an algebra but not a ${\sigma}$-algebra.

We now consider ways of constructing ${\sigma}$-algebras, first starting with a proposition.

Proposition 1 Let X be an arbitrary set, and let ${\left\{\mathscr{M}_{j}\right\}_{j\in J}}$ be a nonempty collection of ${\sigma}$-algebras on ${X}$. Then ${\bigcap_{j\in J}\mathscr{M}_{j}}$ is a ${\sigma}$-algebra on X.

Proof: Let ${\mathscr{A}=\bigcap_{j\in J}\mathscr{M}_{j}}$. We need to check that ${\mathscr{A}}$ fulfills the definiton. First, note that ${X\in\mathscr{M}_{j}}$ for all ${j}$ since each ${\mathscr{M}_{j}}$ is a ${\sigma}$-algebra, so ${X\in\mathscr{A}}$. Then if ${A\in\mathscr{A}}$, ${A\in\mathscr{M}_{j}}$ for all ${j}$ by definition, thereby ${A^{c}\in\mathscr{M}_{j}}$ for all ${j}$, which implies ${A^{c}\in\mathscr{A}}$. Lastly, let ${\left\{A_{n}\right\}_{n\in\mathbb{N}}}$ be a sequence of elements of ${\mathscr{A}}$, then for all ${n\in\mathbb{N}}$ we have ${A_{n}\in\mathscr{M}_{j}}$ for all ${j\in J}$. So we know ${\bigcup_{n\in\mathbb{N}}\in\mathscr{M}_{j}}$ for all ${j}$ by definition of ${\sigma}$-algebra, hence ${\bigcup_{n\in\mathbb{N}}\in\mathscr{A}}$.

$\Box$

Corollary 2 Let X be an arbitrary set, and let ${\mathscr{F}}$ be a family of subsets of X. Then there exists a smallest ${\sigma}$-algebra on X that includes ${\mathscr{F}}$.

Proof: Let ${\mathscr{C}}$ be the collection of ${\sigma}$-algebras that includes ${\mathscr{F}}$ and note ${\mathscr{C}\neq\emptyset}$ since ${\mathscr{P}(E)\in\mathscr{C}}$. Then by proposition 1, the intersection of ${\sigma}$-algebras in ${\mathscr{C}}$ is desired.

$\Box$

We have this basic tool to construct ${\sigma}$-algebras. Consider any family of subsets of ${E}$, ${\mathscr{C}\subseteq\mathscr{P}(E)}$. Then there exist a unique smallest ${\sigma}$-algebra ${\sigma(\mathscr{C})\supseteq\mathscr{C}}$, i.e., ${\forall C\in\mathscr{C}\;\;C\in\sigma(\mathscr{C})}$. We call ${\sigma(\mathscr{C})}$ the ${\sigma}$-algebra generated by ${\mathscr{C}}$. Namely, any ${\sigma}$-algebra containing ${\mathscr{C}}$ also contains ${\sigma(\mathscr{C})}$. Now we turn our attention to an important ${\sigma}$-algebras.

Remark3: By corollary 2, we have an important relationship between ${\sigma}$-algebra and its subfamilies. If ${\mathscr{M}}$ is a ${\sigma}$-algebra and ${\mathscr{A}\subseteq\mathscr{M}}$, then ${\sigma(\mathscr{A})\subseteq\mathscr{M}}$ which will be used in the proof of the following proposition.
\ell

Definition: The Borel ${\sigma}$-algebra on ${\mathbb{R}^{n}}$ is the ${\sigma}$-algebra generated by the collection/family of open subsets of ${\mathbb{R}^{n}}$, denoted by ${\mathscr{B}(\mathbb{R}^{n})}$. We call elements of ${\mathscr{B}(\mathbb{R}^{n})}$ the Borel subsets of ${\mathbb{R}^{n}}$. In particular, if ${n=1}$, one often writes ${\mathscr{B}(\mathbb{R})}$.

Proposition 3 ${\mathscr{B}(\mathbb{R})}$ is generated by each of the following collections of sets:

1. the collection of all closed subsets of ${\mathbb{R}}$.
2. the collection of all subintervals of ${\mathbb{R}}$ of the form ${(-\infty, b]}$.
3. the collection of all subintervals of ${\mathbb{R}}$ of the form ${(a, b]}$.

Proof: Let’s denote the corresponding ${\sigma}$-algebra generated by the family of sets in (1), (2), (3) by ${\mathscr{B}_{1}}$, ${\mathscr{B}_{2}}$, ${\mathscr{B}_{3}}$ respectively. It then suffices to show ${\mathscr{B}(\mathbb{R})\supseteq\mathscr{B}_{1}\supseteq\mathscr{B}_{2}\supseteq\mathscr{B}_{3}}$ and ${\mathscr{B}_{3}\supseteq\mathscr{B}(\mathbb{R})}$. First consider ${\mathscr{C}}$, the collection of all closed subsets of ${\mathbb{R}}$. Notice each any closed subset is just the complement of open subsets so every closed subsets of ${\mathbb{R}}$ is an element of the ${\sigma}$-algebra ${\mathscr{B}(\mathbb{R})}$, i.e., ${\mathscr{C}\subseteq\mathscr{B}(\mathbb{R})}$ and by proposition 2 we know ${\mathscr{B}_{1}=\sigma(\mathscr{C})\subseteq\mathscr{B}(\mathbb{R})}$. Secondly, since each subintervals of the form ${(-\infty,b]}$ is closed we have ${\mathscr{B}_{2}(\mathbb{R})\subseteq\mathscr{B}_{1}(\mathbb{R})}$. Finally, intervals of the form ${(a,b]}$ can be written as

$\displaystyle (a,b]=(-\infty,b]\cap(-\infty,a]^{c}$

which gives us ${\mathscr{B}_{3}(\mathbb{R})\subseteq\mathscr{B}_{2}(\mathbb{R})}$.

For the other direction we need to show ${\mathscr{B}(\mathbb{R})\subseteq\mathscr{B}_{3}(\mathbb{R})}$. Notice that every open subset of ${\mathbb{R}}$ is a countable union of open intervals and each open interval is a countable union of subintervals of the form ${(a,b]}$ hence ${\mathscr{B}(\mathbb{R})\subseteq\mathscr{B}_{3}(\mathbb{R})}$.

$\Box$

Now we take a closer look at Borel sets on ${\mathbb{R}}$. Let’s denote the collection of open subsets by $\mathscr{G}$ and the collection of closed subsets by $\mathscr{F}$. Then define

$\displaystyle G_{\delta}:=\bigcap_{G\in\mathscr{G}} G$

and

$\displaystyle F_{\sigma}:=\bigcup_{F\in\mathscr{F}}F$

We have the following consequences.

Proposition 4 Every open subset of ${\mathbb{R}^{n}}$ is a ${F_{\sigma}}$ and every closed subset of ${\mathbb{R}^{n}}$ is a ${G_{\delta}}$.

Proof: Let ${F}$ be a closed subset of ${\mathbb{R}^{n}}$ and define

$\displaystyle U_{n}:=\left\{x\in\mathbb{R}^{n}|\left\|x-y\right\|<1/n\;\text{for some }y\in F\right\}$

Clearly ${F\subseteq \bigcap_{n}U_{n}}$. For the other direction, note that by definition each point in ${U_{n}}$ is the limit of sequence of points in ${F}$ and ${F}$ is closed, so such limit should also be in ${F}$; therefore, ${\bigcap_{n}U_{n}\subseteq F}$.

Let ${U}$ be an open subset of ${\mathbb{R}^{n}}$ then ${U_{c}}$ is closed. Thus by above ${U^{c}}$ is a ${G_{\delta}}$, say ${U^{c}=\bigcap_{n} U_{n}}$ where ${\left\{U_{n}\right\}}$ is a sequence of open subsets of ${\mathbb{R}^{n}}$. Notice ${U_{n}^{c}}$ is open for each ${n}$, and the identity ${U=(\bigcap_{n} U_{n})^{c}=\bigcup_{n} U_{n}^{c}}$ shows that U is a ${F_{\sigma}}$.

$\Box$

We end this section by a interesting proposition.

Proposition 5

Definition: Let ${(E_{1}, \mathscr{M}_{1})}$ and ${(E_{2}, \mathscr{M}_{2})}$ be measurable spaces. The product ${\sigma}$-algebra on ${E_{1}\times E_{2}}$ is defined by

$\displaystyle \mathscr{M}_{1}\otimes\mathscr{M}_{2}:=\sigma(\mathscr{M}_{1}\times\mathscr{M}_{2})$