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
.