Econometrics I #2: Borel sigma-algebra

Borel σ-algebra is a special type of σ-algebra defined over a metric space. Here we only concern the Borel σ-algebra over the real line.

Definition 1: The Borel σ-algebra over the reals, denoted as B, is the σ-algebra generated by all open sets. Elements of Borel σ-algebra are called Borel sets.

Definition 2: A subset E of the reals is an F_σ set if it is a countable union of closed sets; E is a G_δ set if it is a countable intersection of open sets.

Some consequences:

1. Every G_δ set is a Borel set, by property of a σ-algebra. Since the complement of a G_δ set is an F_σ set, an F_σ set is a Borel set.
2. Semi-open intervals of the form [a,b) (equivalently (a,b]) are both F_σ and G_δ, since [a,b)=∪[a,b-1/n]=∩(a-1/n,b). Hence [a,b) is a Borel set for any real number a and b.

An alternative definition 3 of Borel σ-algebra (Marcelo Moreira, Columbia): The Borel σ-algebra (over the reals) is the smallest σ-algebra containing sets of the form (-∞,x].

I want to show that the two definitions are in fact equivalent.

Proof:
First we need to observe the following lemma.


Lemma: 

Suppose A and B are two σ-algebras over the same space. Further, B is the σ-algebra generated by F, a set of subsets in that space. Then A contains B if and only if A contains its generators (i.e. A contains elements of F).

Main Proposition:

Let A denote the Borel σ-algebra in definition 1, i.e. A=σ(all open sets over reals). Let B denote the Borel σ-algebra in definition 3, i.e. B=σ((-∞,x]:x is real). Therefore, we need to prove A=B as sets and in fact we need to prove the two-way inclusions of the two sets.

First, to show B is a subset of A, by the lemma, we just need to show its generator (-∞,x]  is in A. But clearly (-∞,x] is G_δ, hence Borel in definition 1, a.k.a an element of A.

Then, to show A is a subset of B, similarly by the lemma, we need to show an arbitrary open interval (a,b) is in B. Notice the semi-open interval (a,b]=(-∞,b]∩{complement of (-∞,a]} for all real number a and b, hence it is in B. Then the open interval (a,b)=∪(a,b-1/n]. Since every (a,b-1/n] is in B, the countable union is in B, a.k.a (a,b) is in B for any real number a and b. With easy and similar treatment of the special cases (-∞,b), (a,+∞) and (-∞,+∞), we conclude that every open set is in B. Hence by lemma, A is a subset of B.

With the two inclusions, A=B and hence the two definitions are equivalent. QED