Measurable Spaces and Functions

-Algebra and Measurable Spaces

-Algebra

Suppose is a set and is a set of subsets of . Then is called a -algebra on if it is closed under countable unions, countable intersections, and complements:

• ;
• if , then ;
• if , then .

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 -algebra . An element of is called an -measurable set, or just a measurable set.

Proposition

Suppose is a set and is a set of subsets of . Then the intersection of all -algebras on that contain is a -algebra on .

Proof

Measurable Functions

Measurable Function

Suppose and are measurable spaces. A function is an measurable function if

Essential Range

The essential range of a measurable function is the set where is a measure on .

Measure Spaces

Measure

Suppose is a measurable space. A measure on is a function such that: and for any countable collection of disjoint sets in . In this case, we call a measure space.

Proposition

For a measure defined on a measurable space , the following properties hold:

• Monotonicity: for ,
• Subadditivity: .

Proof Observe that for any , and are disjoint, and , thus For subadditivity, we can divide the union into disjoint sets. Suppose , let , and for . Then each is disjoint, and we have where the last inequality follows from monotonicity.

-Finite Measure

A measure space is -finite if there exists a countable collection of measurable sets such that and for all .

e.g.

• The Lebesgue measure on is -finite because .
• The counting measure on is not -finite, because one cannot decompose into a countable union of sets with finite cardinality.

Complete Measure

A measure defined on a measurable space is complete if for all that and implies .

Exterior Measure and Carathéodory Theorem

Exterior (Outer) Measure

Let be a set. The exterior measure on is defined on all subsets of to such that

• ;
• if ;
• .

Carathéodory Measurable

Suppose is an exterior measure. A set in is Carathéodory measurable or simply measurable if one has

Carathéodory Theorem

Given an exterior measure on a set , the collection of Carathéodory measurable sets forms a -algebra. Moreover, restricted to is a measure.

Proof Clearly, and belong to and ,

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 on a metric space is called a metric exterior measure if it satisfies

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

Theorem

If is a metric exterior measure on a metric space , then the Borel sets in are measurable. Hence restricted to the Borel sets is a measure.

The Extension Theorem

Boolean Algebra

Let be a set. A boolean algebra on is a nonempty collection of subsets satisfies

• ,
• If , then ,
• If and are elements of , then .

In other words, is closed under complements, finite unions, and finite intersections.

Premeasure

A premeasure on a boolean algebra over a set , is a function such that

• .
• If is a countable collection of disjoint sets in with , then

Premeasures give rise to exterior measures in a natural way:

Lemma

If is a premeasure on a boolean algebra over , define on any subset by Then is an exterior measure on satisfying for all , and all sets in are Carathéodory measurable.

Carathéodory’s Extension Theorem

Suppose that is a boolean algebra of sets in , is a premeasure on , and is the -algebra generated by (i.e., the smallest -algebra containing ). Then there exists a measure on that extends . Moreover, if is -finite, then it is unqiue.

Proof This is a direct consequence of the above lemma. induces an exterior measure , which is a measure on the -algebra of Carathéodory measurable sets. As all sets in are Carathéodory measurable, and is generated by , all sets in are also Carathéodory measurable. Therefore is a measure on as well, and we call it . To prove the uniqueness, we suppose is another measure defined on that extends . Let containing where each . Then there holds

Pick a set with . If for each , then so by taking the infimum over all such covers of . To prove the reverse inequality, note that

Assume is -finite. We claim that . Observe that since is -finite, we may write , where is a countable collection of disjoint sets in with . Then we have Let