Borel σ-algebra

By Dmitry Kabanov

September 14, 2019

When one studies probability theory, a notion of a Borel $σ$-algebra arises. Here we will look at what this is.

First thing is to understand what $σ$-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 $σ$-algebra.

math