Tag Archives: Borel sigma algebra

Proof: a representation of the Borel σ-algebra on reals

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 \mathcal{B}:=\mathcal{B}(\mathbb{R}) on the set \mathbb{R} of real numbers is given by \mathcal{B}=\sigma (\pi (\mathbb{R})), where \sigma(\pi(\mathbb{R})) denotes the sigma algebra on \mathbb{R} generated by \pi (\mathbb{R})=\{(-\infty, x]:x\in\mathbb{R}\}.

It is reminded that the Borel σ-algebra \mathcal{B}(S) on a set S is the σ-algebra \sigma(\mathcal{O}) generated by the family \mathcal{O} of open subsets of S, i.e. \mathcal{B}(S):=\sigma(\mathcal{O}).

Two lemmas are provided before proceeding with the proof.


Lemma 1

Let S be a set, and let \mathcal{C}_1,\mathcal{C}_2 be two collections of subsets of S. If \mathcal{C}_1\subseteq\sigma(\mathcal{C}_2), then \sigma(\mathcal{C}_1)\subseteq\sigma(\mathcal{C}_2).

Proof of lemma 1

\sigma(\mathcal{C}_2) is a σ-algebra on S that contains \mathcal{C}_1 and \sigma(\mathcal{C}_1) is the intersection of all σ-algebras on S that contain \mathcal{C}_1, so \sigma(\mathcal{C}_1)\subseteq\sigma(\mathcal{C}_2).


Lemma 2

Every open subset G of \mathbb{R} in the Euclidean metric space (\mathbb{R},\|\cdot\|) 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 G be an open subset of \mathcal{R} and x\in G. Introduce the sets \mathcal{A}_x:=\{a\in\mathbb{R}: a<x, (a,x)\subseteq G\} and \mathcal{B}_x:=\{b\in\mathbb{R}: x<b, (x,b)\subseteq G\}. Moreover, let a_x:=\mbox{inf}\mathcal{A}_x and b_x:=\mbox{sup}\mathcal{B}_x, allowing for a_x,b_x to be infinite. Consider also the interval I_x:=(a_x,b_x).

It will be shown that I_x\subseteq G. Let y\in I_x. Assume that (\forall a\in\mathcal{A}_x)y\le a, in which case y is a lower bound of \mathcal{A}_x, so y\le a_x, thus contradicting the assumption y\in I_x=(a_x, b_x). Hence, there exists a\in\mathcal{A}_x such that a<y. By a symmetric argument, there exists b\in\mathcal{B}_x such that y<b. So, y\in (a,b)\subseteq (a_x,b_x). If y=x, then obviously y\in G. If y\neq x, then y\in (a,x)\cup (x, b). Note that a\in\mathcal{A}_x means that (a,x)\subseteq G. So, if y\in (a,x), then y\in G. Similarly, if y\in (x,b), then y\in G. Thus, y\in I_x\Rightarrow y\in G, which means that I_x\subseteq G.

(\forall x\in G) I_x\subseteq G\Rightarrow\underset{x\in G}{\bigcup}I_x\subseteq G. Furthermore, let y\in G. As G=G^{o}, there exists r>0 such that (x-r,x+r)\subseteq G. It follows from y-r<y and (y-r,y)\subseteq (y-r,y+r)\subseteq G that y\in\mathcal{A}_y. In a similar way, y\in\mathcal{B}_y. So, a_y\le y\le b_y\Rightarrow y\in I_y\subseteq\underset{x\in G}{\bigcup}I_x. Hence, G=\underset{x\in G}{\bigcup}I_x.

Assume that I_x\cap I_y\neq\emptyset, in which case I_x\cup I_y is also an open interval. It is straightforward to confirm that if I is an open interval containing x and being contained in G, then I\subseteq I_x. So, I_x\cup I_y\subseteq I_x and I_x\cup I_y\subseteq I_y, which yields I_x=I_y. It has been shown that I_x\cap I_y\neq\emptyset\Rightarrow I_x=I_y. By contraposition, I_x\neq I_y\Rightarrow I_x\cap I_y=\emptyset, which means that the sets of the collection \{I_x:x\in G\} are pairwise disjoint.

For every set I_x in \{I_x :x\in G\}, pick a rational q_x\in I_x. The set S:=\{q_x: q_x\in I_x\} of selected rationals is obviously a subset of the set \mathbb{Q} of rationals. S\subseteq\mathbb{Q} implies the cardinality inequality |S|\le |Q|, so S is a countable set. Since  |\{I_x :x\in G\}|=|S|, it is deduced that the collection \{I_x :x\in G\} is also countable.

It has been proved so far that G 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 \{I_k : k\in K\} and \{J_{\ell} : \ell\in L\} be two collections of disjoint open intervals such that G=\underset{k\in K}{\bigcup}I_k=\underset{\ell\in L}{\bigcup}J_{\ell}.

For any k\in K, it holds that I_k=\underset{\ell\in L}\bigcup(I_k\cap J_{\ell}). Moreover, the sets of the collection \{I_k\cap J_{\ell}:\ell\in L\} are pairwise disjoint. Since I_k and J_{_\ell} are open, (I_k\cap J_{\ell})^{o}=I_k^{o}\cap J_{\ell}^{o}=I_k\cap J_{\ell}, which means that I_k\cap J_{\ell} is also open.

In the collection \{I_k\cap J_{\ell}:\ell\in L\}, at least one set is non-empty, otherwise trivially G=\emptyset.

In the Euclidean metric space (\mathbb{R},\|\cdot\|), every interval is a connected set. Since I_k is an interval, it is connected. Furthermore, a subset S of a metric space E is connected, if and only if there is no open cover \{A,B\} of S such that S\cap A\neq\emptyset, S\cap B\neq\emptyset and S\cap A\cap B=\emptyset. Hence, by also considering that I_k is connected, it is deduced that at most one set in \{I_k\cap J_{\ell}:\ell\in L\} is non-empty.

So, exactly one set in \{I_k\cap J_{\ell}:\ell\in L\} is non-empty, say I_k\cap J_{\ell}. Consequently, I_k=I_k\cap J_{\ell}, which in turns gives I_k\subseteq J_{\ell}.

In an analogous manner, starting from J_{\ell}=\underset{k\in K}\bigcup(J_{\ell}\cap I_k), it is concluded that J_{\ell}\subseteq I_k. So, I_k=J_{\ell}, therefore the collections \{I_k : k\in K\} and \{J_{\ell} : \ell\in L\} coincide.


Proof of \boldsymbol{\mathcal{B}(\mathbb{R})=\sigma(\pi(\mathbb{R}))}

Let \mathcal{O} be the family of open subsets in the Euclidean metric space (\mathbb{R},\|\cdot\|). It will be shown that the Borel σ-algebra \mathcal{B}(\mathbb{R})=\sigma(\mathcal{O}) on \mathbb{R} is equal to the σ-algebra \sigma(\pi(\mathbb{R})) on \mathbb{R} generated by \pi (\mathbb{R})=\{(-\infty, x]:x\in\mathbb{R}\}.

Firstly, it will be proved that \pi (\mathbb{R})\subseteq\sigma(\mathcal{O}). Consider a set (-\infty, x]\in\pi (\mathbb{R}). Note that (-\infty, x]=\underset{n\in\mathbb{N}}{\bigcap}(-\infty,x+1/n). Indeed, for y\in\underset{n\in\mathbb{N}}{\bigcap}(-\infty,x+1/n), it holds that (\forall n\in\mathbb{N})y<x+1/n. Now assume that x<y, in which case y-x>0, so by the Archimedean property, there exists n_o \in\mathbb{N} such that 0<1/n_o <y-x, whence x+1/n_o<y, which is a contradiction. Thus, y\le x, i.e. y\in(-\infty,x]. It has been established that \underset{n\in\mathbb{N}}{\bigcap}(-\infty,x+1/n)\subseteq (-\infty,x]. Obviously  (-\infty,x]\subseteq\underset{n\in\mathbb{N}}{\bigcap}(-\infty,x+1/n) holds too.

Since (-\infty,x+1/n)\in\mathcal{O} and \sigma(\mathcal{O}) is a σ-algebra on \mathbb{R}, the countable intersection (-\infty, x]=\underset{n\in\mathbb{N}}{\bigcap}(-\infty,x+1/n) belongs to \sigma(\mathcal{O}).

With the help of lemma 1, \pi (\mathbb{R})\subseteq\sigma(\mathcal{O}) yields \sigma(\pi (\mathbb{R}))\subseteq\sigma(\mathcal{O})=\mathcal{B}(\mathbb{R}).

To prove the converse set inequality \sigma(\mathcal{O}) \subseteq\sigma(\pi (\mathbb{R})), start by proving that \mathcal{O}\subseteq\sigma(\pi (\mathbb{R})). Along these lines, consider an open set G\in\mathcal{O}. It is known from lemma 2 that X can be written as a countable union of open intervals.

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

  1. a\in\mathbb{R}, b\in\mathbb{R}. Set (a, b)=\underset{n\in\mathbb{N}}{\bigcup}(a,b-\epsilon/n], where \epsilon=(b-a)/2. This case will be treated in detail.
  2. a=-\infty, b\in\mathbb{R}. Set (-\infty, b)=\underset{n\in\mathbb{N}}{\bigcup}(-\infty,b-\epsilon/n], where \epsilon=(b-a)/2.
  3. a\in\mathbb{R}, b=\infty. Set (a,\infty)=\underset{n\in\mathbb{N}}{\bigcup}(a,n].
  4. a=\infty, b=\infty. Set (-\infty, \infty)=(-\infty,c]\cup(c,\infty) for any real c, which leads back to case 3.

Case 1 is now treated in detail. The equality (a, b)=\underset{n\in\mathbb{N}}{\bigcup}(a,b-\epsilon/n] is proved by observing that y\in (a,b)\Leftrightarrow (\exists n_o \in\mathbb{N}) a<y\le b-\epsilon/n_o\Leftrightarrow y\in\underset{n\in\mathbb{N}}{\bigcup}(a,b-\epsilon/n]. Leaving aside the remaining trivial deductions among these equivalences, it will be clarified that y\in (a,b)\Rightarrow (\exists n_o \in\mathbb{N}) a<y\le b-\epsilon/n_o. So, y<b means 0 < (b-y)/\epsilon. By the Archimedean property, there exists n_o\in\mathbb{N} such that 0 < 1/n_o < (b-y)/\epsilon, whence y<b-\epsilon/n_o.

A set of the form (a, u], a<u, u\in\mathbb{R}, can be written as (a,u]=(-\infty, u]\cap(-\infty, a]^{c}. So, the sets (a,b-\epsilon/n], n\in\mathbb{N}, appearing in (a, b)=\underset{n\in\mathbb{N}}{\bigcup}(a,b-\epsilon/n] can be expressed as (a,b-\epsilon/n]=(-\infty,b-\epsilon/n]\cap(-\infty, a]^{c}. Thereby, (-\infty,b-\epsilon/n]\in\pi(\mathbb{R}) and (-\infty,a]\in\pi(\mathbb{R}) along with the fact that \sigma (\pi(\mathbb{R})) is a σ-algebra lead to (a,b-\epsilon/n]\in\sigma (\pi(\mathbb{R})). So, (a,b)\in\sigma (\pi(\mathbb{R})), which implies \mathcal{O}\subseteq\sigma(\pi (\mathbb{R})). Finally, \sigma(\mathcal{O}) \subseteq\sigma(\pi (\mathbb{R})) by application of lemma 1, so \mathcal{B}(\mathbb{R})=\sigma(\mathcal{O}) =\sigma(\pi (\mathbb{R})).