This post elaborates on the proof of a proposition in page 17 of the book “Probability with martingales” by David Williams. The relevant proposition is first stated; the Borel σ-algebra on the set of real numbers is given by , where denotes the sigma algebra on generated by .

It is reminded that the Borel σ-algebra on a set is the σ-algebra generated by the family of open subsets of , i.e. .

Two lemmas are provided before proceeding with the proof.

**Lemma 1**

Let be a set, and let be two collections of subsets of . If , then .

**Proof of lemma 1**

is a σ-algebra on that contains and is the intersection of all σ-algebras on that contain , so .

**Lemma 2**

Every open subset of in the Euclidean metric space can be represented uniquely as a countable union of disjoint open intervals.

**Proof of lemma 2**

This lemma appears also as theorem 1.3, p. 6, in the book “Real analysis: measure theory, integration and Hilbert spaces” by Elias M. Stein and Rami Shakarchi, where a partial proof is provided as uniqueness is not established. In what follows, the proof from Stein’s and Shakarchi’s book is reproduced in more detail, plus uniqueness is proved.

Let be an open subset of and . Introduce the sets and . Moreover, let and , allowing for to be infinite. Consider also the interval .

It will be shown that . Let . Assume that , in which case is a lower bound of , so , thus contradicting the assumption . Hence, there exists such that . By a symmetric argument, there exists such that . So, . If , then obviously . If , then . Note that means that . So, if , then . Similarly, if , then . Thus, , which means that .

. Furthermore, let . As , there exists such that . It follows from and that . In a similar way, . So, . Hence, .

Assume that , in which case is also an open interval. It is straightforward to confirm that if is an open interval containing and being contained in , then . So, and , which yields . It has been shown that . By contraposition, , which means that the sets of the collection are pairwise disjoint.

For every set in , pick a rational . The set of selected rationals is obviously a subset of the set of rationals. implies the cardinality inequality , so is a countable set. Since , it is deduced that the collection is also countable.

It has been proved so far that can be expressed as a countable union of disjoint open intervals. Although not pertinent to the subsequent proof, it will be shown that such representation is unique for the sake of completeness of exposition.

Let and be two collections of disjoint open intervals such that .

For any , it holds that . Moreover, the sets of the collection are pairwise disjoint. Since and are open, , which means that is also open.

In the collection , at least one set is non-empty, otherwise trivially .

In the Euclidean metric space , every interval is a connected set. Since is an interval, it is connected. Furthermore, a subset of a metric space is connected, if and only if there is no open cover of such that , and . Hence, by also considering that is connected, it is deduced that at most one set in is non-empty.

So, exactly one set in is non-empty, say . Consequently, , which in turns gives .

In an analogous manner, starting from , it is concluded that . So, , therefore the collections and coincide.

**Proof of **

Let be the family of open subsets in the Euclidean metric space . It will be shown that the Borel σ-algebra on is equal to the σ-algebra on generated by .

Firstly, it will be proved that . Consider a set . Note that . Indeed, for , it holds that . Now assume that , in which case , so by the Archimedean property, there exists such that , whence , which is a contradiction. Thus, , i.e. . It has been established that . Obviously holds too.

Since and is a σ-algebra on , the countable intersection belongs to .

With the help of lemma 1, yields .

To prove the converse set inequality , start by proving that . Along these lines, consider an open set . It is known from lemma 2 that can be written as a countable union of open intervals.

Let be one of these open intervals in the countable union. Recall from the proof of lemma 2 that can be infinite. Consequently, four cases are distinguished:

- . Set , where . This case will be treated in detail.
- . Set , where .
- . Set .
- . Set for any real , which leads back to case 3.

Case 1 is now treated in detail. The equality is proved by observing that . Leaving aside the remaining trivial deductions among these equivalences, it will be clarified that . So, means . By the Archimedean property, there exists such that , whence .

A set of the form , can be written as . So, the sets , appearing in can be expressed as . Thereby, and along with the fact that is a σ-algebra lead to . So, , which implies . Finally, by application of lemma 1, so .