A Mild Glimpse Into Measure Theory (1)

by tzy9393

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})

 

 

Advertisements