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
 A: 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.
A: A Borel set is any set in a topological space that can be formed from open sets (or, equivalently, from closed sets) through the operations of countable union, countable intersection, and relative complement. Such sets are named Borel (measurable) sets.
For a topological space $X$, the collection of all Borel sets on $X$ forms a σ-algebra, known as the Borel $\sigma$-algebra. The Borel $\sigma$-algebra on $X$ is the smallest $\sigma$-algebra containing all open sets (or, equivalently, all closed sets).
As an example, the Borel $\sigma$-algebra on $\mathbb{R}$, written as $\mathcal{B}$, is the smallest $\sigma$-algebra generated by the open sets.
More formally, denote by $\mathcal{C}$ the collection of all open sets of $\mathbb{R}$. Also, denote by $\sigma(\mathcal{C})$ the $\sigma$-algebra generated by $\mathcal{C}$.
Then, $\mathcal{B} = \sigma(\mathcal{C})$.
