# Borel $\sigma$ -algebra

So I have a proposition which states the following:

Each of the following families of sets generate the Borel $\sigma$-algebra:

1)The family of all open intervals $(a,b)$, $a,b, \in \mathbb{R}$.

2) The family of all closed intervals $[a,b]$, $a,b, \in \mathbb{R}$.

3) The family of open intervals $(a,\infty)$, $a\in \mathbb{R}$.

4) The family of closed intervals $[a,\infty)$, $a\in \mathbb{R}$.

The proof of part 1 involved us showing that every open set is a countable union of open intervals and then since every open set is contained in this $\sigma$- algebra therefore it contains the borel $\sigma$-algebra but I am not sure how to show the other way around i.e. to show that the $\sigma$-algebra generated by the family of open intervals is contained in the borel $\sigma$-algebra

• Simply, every open interval is a borel set, so the $\sigma$ lagebra generated by these open intervals, which is the smallest $\sigma$ algebra conatianing them, is in the borel $\sigma$ algebra . – Nizar Apr 16 '15 at 9:11

The Borel $\sigma$-algebra is generated by all the open sets. A $\sigma$-algebra generated by a subset of the open sets must therefore be a subset of the Borel $\sigma$-algebra.
Similarly, the set of closed intervals is a subset of the Borel $\sigma$-algebra, so the $\sigma$-algebra generated by those intervals will be contained within the Borel $\sigma$-algebra. You should be able to make this work for all your other cases too.
• So this is what I have understood so far; Every open set in $\mathbb{R}$ is of the form $(a,b)$, $a,b \in \mathbb{R}$ and a countable union of open sets is open and a finite intersection of open sets is also open. So every open set in $\mathbb{R}$ is a countable union of open intervals as described above which means that every open set is contained in the $\sigma$- algebra generated by these open intervals and hence it contains the Borel $\sigma$-algebra and for the other way around since every open set is an open interval itself we can show the other inclusion as well. Is this correct? – user1314 Apr 16 '15 at 9:34
• Ahh yes that makes sense now, it is because every open interval is an open set that we can say the $\sigma$-algebra generated by open intervals is contained in the Borel $\sigma$-algebra. So for part 2 and others, again we can indeed show that every open interval is a countable union of those sets and therefore show that the sigma algebra generated is contained in the Borel algebra but how about showing the reverse direction for part 2 ? – user1314 Apr 16 '15 at 9:55
For a topological space X, the collection of all Borel sets on X forms a σ-algebra, known as the Borel algebra or Borel σ-algebra. The Borel algebra on X is the smallest σ-algebra containing all open sets (or, equivalently, all closed sets) for example be have The Borel algebra on the reals is the smallest σ-algebra on R which contains all the intervals. \begin{lemma} let $C = {(a,b) :a<b}$ then $\sigma(C)=B_{R}$ IS the Borel field generated by the family of all open intervals C. \end{lemma}