In this page, we will deal with the standard -algebra making (in particular ) a measurable space, as well as measurable functions that has as its codomain. We will not assume any conditions more than a measurable space on the domain of .
Borel Sets
Recall the definition of a measurable space:
We call such elements of measurable.
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.
The Borel -algebra on , denoted as , is the smallest $\sigma$-algebra containing all open sets. The term “smallest” means that if is any -algebra that contains all open sets in , then necessarily . Elements of the Borel -algebra are called Borel sets.
e.g.
Every closed subset of is a Borel set because every closed subset of is the complement of an open subset of .
Every half-open interval (where ) is a Borel set because .
Measurable Functions
Recall the definition of measurable functions:
Suppose and are measurable spaces. A function is an measurable function if
We usually consider the case where is simply endowed with the Borel -algebra. In such case, we call -measurable.
e.g. If , then the only -measurable functions from to are the constant functions.
Characteristic Function
The characteristic function of a set is a function defined by
As its name indicates, we observe that
Proposition
is -measurable if and only if .
Proof This is obvious once we have written out preimage: for any Borel set ,
Proposition
Suppose is a measurable space. A function is -measurable, if and only if, or or or is -measurable for all .
Proof It suffices to prove for all implies measurability of . Suppose for all . Consider . We claim that the Borel -algebra is a subset of . To show this, we show that contains every open subset, and is indeed a -algebra. Clearly, because . Suppose , then , and thus , so . If , then for all , and thus , so . Therefore, is a -algebra. Finally, we show that every open subset of is in . It suffices to show that every open interval is in . But we have , and , and thus . Therefore, the Borel -algebra is a subset of , and thus is measurable.
Proposition
Suppose is a measurable space and are -measurable. Then
, , and are -measurable functions;
if for all , then is an -measurable function.
Proposition
Suppose is a measurable space and is a sequence of -measurable functions from to . Suppose exists for each . Define by , then is -measurable.
Proof Fix some . We claim that For any , we have . Choose such that , say , then there exists some such that for all . This proves that is in the RHS.
For the other direction, picks some , then there exists some and , such that for all . Note that , so , and thus . Therefore, the claim holds, and is -measurable.
Extended Real-Valued Measurable Functions
Extended Reals
We define the extended reals. It is a measurable space with -algebra defined by
This is (also) called the Borel -algebra on .
In other words, a set is a Borel set if and only if there exists a Borel set such that or or or .
Similarly, a function is measurable if and only if is measurable for every Borel set , and we have analogous conditions for a function to be measurable:
Proposition
Suppose is a measurable space and , then is -measurable if and only if or is -measurable for all .
Proposition
Suppose is a measurable space, and is a sequence of -measurable functions from to . Then are -measurable. In particular, if exists a.e., it is measurable.
Proof Let . Note that , and , and other cases are similar.
Remark
The above is NOT true if we have an uncountable family of measurable functions. For example, consider with the countable-cocountable $\sigma$-algebra. Then is measurable for each , but is not measurable.
Simple Function, Step Function
A simple function is a function that can be expressed as a finite linear combination of characteristic functions of measurable sets with finite measure: where are constants. In particular, if each is a rectangle, then is called a step function.