The Borel $\sigma$-algebra on $\mathbb R$ is the $\sigma$-algebra B($\mathbb R$) generated by the $\pi$-system $\mathcal J$ of intervals $\ (a, b]$, where $\ a<b$ in $\mathbb R$ (We also allow the possibility that $\ a=-\infty\ or \ b=\infty$) Its elements are called Borel sets. For A $\in$ B($\mathbb R$), the $\sigma$-algebra $$B(A)= \{B \subseteq A: B \in B(\mathbb R)\}$$ of Borel subsets of A is termed the Borel $\sigma$-algebra on A.

I struggle with this part especially "generated by the $\pi$-system $\mathcal J$ of intervals (a, b]"

In addition could someone please provide an example of a Borel set, preferably some numerical interval :)

Also is $\mathbb R$ the type of numbers that the $\sigma$-algebra is acting on?


2 Answers 2


Ignore the phrase "$\pi$-system" for the time being : What you are given is a collection $\mathcal{J}$ of subsets of $\mathbb{R}$ and the $\sigma$-algebra you seek is the smallest $\sigma$-algebra that contains $\mathcal{J}$. This is the definition of the Borel $\sigma$-algebra. For example $\{1\}$ is a Borel set since $$ \{1\} = \bigcap_{n=1}^{\infty} (1-1/n,1] = \mathbb{R}\setminus \left(\bigcup_{n=1}^{\infty} \mathbb{R}\setminus (1-1/n,1]\right) $$ Does this help you understand what this $\sigma$-algebra can contain? It is not possible to list down all the elements in $B(\mathbb{R})$ though.

Now, the reason we choose this $\sigma$-algebra is simple : We want continuous functions to be measurable - a rather reasonably requirement which is often imposed when dealing with measure spaces that are also topological spaces.

  • $\begingroup$ Yes that made alot of sense! however what does it mean to say smallest $\sigma$-algebra? $\endgroup$
    – Brofessor
    Dec 27, 2014 at 19:58
  • $\begingroup$ You take the collection of all $\sigma$-algebras containing $\mathcal{J}$ (for instance, the power set $\mathcal{P}(\mathbb{R})$ is one such), and you take their intersection. You can check that this intersection is a $\sigma$-algebra. $\endgroup$ Dec 28, 2014 at 7:00

Borel $\sigma$-algebras turn op if we are working on topological spaces. If $X$ denotes a topological space and $\tau$ is its topology then the smallest $\sigma$-algebra that contains $\tau$ (in other words: the $\sigma$-algebra generated by $\tau$) is the Borel $\sigma$-algebra on that space. In special case $X=\mathbb R$ equipped with its usual topology it can be shown that the $\sigma$-algebra generated by the open sets (by definition the Borel $\sigma$-algebra) coincides with the $\sigma$-algebra generated by intervals $(a,b]$.


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