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
Best Answer
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.