Borel σ-algebra

When one studies probability theory, a notion of a Borel \(\sigma\)-algebra arises. Here we will look at what this is.

First thing is to understand what \(\sigma\)-algebra means.

Let \(X\) be a set. Then if we have another set \(\mathcal B\), which we call σ-algebra over \(X\), if it has the following properties:

• it consists of subsets of \(X\),

• \(X \in \mathcal B\),

• for any \(b \in \mathcal B\), it follows that \(b^\C \in \mathcal B\), where superscript \(\C\) denotes complement,

• for any set of sets \(b_n \subset \mathcal B\), \(n=1, 2, \dots\), their union is also in \(\mathcal B\):

  \begin{equation} \bigcup_{n=1}^{\infty} b_n \subset \mathcal B, \end{equation}

• for any set of sets \(b_n \subset \mathcal B\), \(n=1, 2, \dots\), their intersection is also in \(\mathcal B\):

  \begin{equation} \bigcap_{n=1}^{\infty} b_n \subset \mathcal B. \end{equation}

It also follows that as \(X \in \mathcal B\), and \(X^\C = \emptyset\), then \(\emptyset \in \mathcal B\).

Consider collection \(\mathcal O\) of open subsets of \(\R\). This collection \(\mathcal O\) is not a σ-algebra of subsets of \(\R\). Why? For any open subset \(A \subseteq \R\), its complement \(A^\C\) in \(\R\) is closed. For example, if \(A=(0, 1)\), then \(A^\C=(-\infty, 0] \cup [1, \infty)\). Therefore, if \(A \in \mathcal O\), then \(A^\C \not \subset \mathcal O\). That explains, why \(\mathcal O\) is not a \(\sigma\)-algebra.

• Math