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

#### by tzy9393

**1. Introduction**

The key idea is to define a function which assigns a value , meaning “the size of ”, to a subset of a given set , so that

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

However, some issues arise in general since one cannot define a function which satisfies above or more generally

Therefore, we should restrict the domain to be a special class of subsets, so called the -algebra.

**2. Measureable Sets**

**2.1. Definition and Examples**

**Definition:** Let be an arbitrary set. An ** algebra**, , is a family of subsets of such that:

- If
- If for all from to , then
- If for all from to , then

**Definition:** Let be an arbitrary set. A –** algebra**, , is a family of subsets of such that:

- If
- If for all , then
- If for all , then

We call elements a –** measureable set**, and call the pair a

**.**

*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 to be a nonempty family of subsets of .

**Remark 2:** From this definition, we can see also: ; and for any finite integer ; hence a -algebra is an algebra but not vice versa.

**Examples:**

- is a -algebra.
- is a -algebra on .
- is a -algebra.
- Let be an infinite set, and let be the collection of all finite subsets of . Then is not even an algebra since .
- Let be an infinite set, and let be the collection of all subsets of such that either or is finite. Then is an algebra but not a -algebra.

We now consider ways of constructing -algebras, first starting with a proposition.

Proposition 1Let X be an arbitrary set, and let be a nonempty collection of -algebras on . Then is a -algebra on X.

*Proof:* Let . We need to check that fulfills the definiton. First, note that for all since each is a -algebra, so . Then if , for all by definition, thereby for all , which implies . Lastly, let be a sequence of elements of , then for all we have for all . So we know for all by definition of -algebra, hence .

Corollary 2Let X be an arbitrary set, and let be a family of subsets of X. Then there exists a smallest -algebra on X that includes .

*Proof:* Let be the collection of -algebras that includes and note since . Then by proposition 1, the intersection of -algebras in is desired.

We have this basic tool to construct -algebras. Consider any family of subsets of , . Then there exist a ** unique smallest **-algebra , i.e., . We call the -algebra generated by . Namely, any -algebra containing also contains . Now we turn our attention to an important -algebras.

**Remark3:** By corollary 2, we have an important relationship between -algebra and its subfamilies. If is a -algebra and , then which will be used in the proof of the following proposition.

\ell

**Definition:** The ** Borel -algebra** on is the -algebra generated by the collection/family of open subsets of , denoted by . We call elements of the

**of . In particular, if , one often writes .**

*Borel subsets*

Proposition 3is generated by each of the following collections of sets:

- the collection of all closed subsets of .
- the collection of all subintervals of of the form .
- the collection of all subintervals of of the form .

*Proof:* Let’s denote the corresponding -algebra generated by the family of sets in (1), (2), (3) by , , respectively. It then suffices to show and . First consider , the collection of all closed subsets of . Notice each any closed subset is just the complement of open subsets so every closed subsets of is an element of the -algebra , i.e., and by proposition 2 we know . Secondly, since each subintervals of the form is closed we have . Finally, intervals of the form can be written as

which gives us .

For the other direction we need to show . Notice that every open subset of is a countable union of open intervals and each open interval is a countable union of subintervals of the form hence .

Now we take a closer look at Borel sets on . Let’s denote the collection of open subsets by and the collection of closed subsets by . Then define

and

We have the following consequences.

Proposition 4Every open subset of is a and every closed subset of is a .

*Proof:* Let be a closed subset of and define

Clearly . For the other direction, note that by definition each point in is the limit of sequence of points in and is closed, so such limit should also be in ; therefore, .

Let be an open subset of then is closed. Thus by above is a , say where is a sequence of open subsets of . Notice is open for each , and the identity shows that U is a .

We end this section by a interesting proposition.

Proposition 5

**Definition:** Let and be measurable spaces. The ** product -algebra** on is defined by