So I know that the Borel $\sigma$-algebra of $\mathbb{R}$ is the $\sigma$-algebra generated by open sets. I have been able to prove that this Borel $\sigma$-algebra is also generated by the family of open intervals of the form $(a,b), a,b \in \mathbb{R}$ Now I want to show that the family of open intervals $(a,\infty)$ also generate the Borel $\sigma$-algebra. So what I have done is first show that
$(a,b) = \bigcup_{q \in \mathbb{Q}, q<b} ((a, \infty)-(q,\infty))$. This shows that every open interval is a countable union of intervals of the form $(a,\infty)$ and thus $\sigma$-algebra generated by $(a,b)$ is contained in the $\sigma$-algebra generated by $(a, \infty)$ But how would I show the other way around, i.e. that every interval $(a, \infty)$ is in the Borel $\sigma$- algebra?
(Can we just say that every interval $(a, \infty)$ is infact an open interval in itself and thus belongs to the $\sigma$-algebra generated by $(a,b)$?)
Best Answer
Denote $\mathcal{B}=\sigma\left(\tau\right)=\sigma\left(\left\{ \left(a,b\right)\mid a,b\in\mathbb{R}\right\} \right)$ (this equality was found out by you allready) where $\tau$ denotes the topology and $\mathcal{B}_{1}=\sigma\left(\left\{ \left(a,\infty\right)\mid a\in\mathbb{R}\right\} \right)$.
Then $\left\{ \left(a,\infty\right)\mid a\in\mathbb{R}\right\} \subset\tau$ so that $\mathcal{B}_{1}\subseteq\mathcal{B}$.
To prove the converse inclusion it is enough to show that $\left\{ \left(a,b\right)\mid a,b\in\mathbb{R}\right\} \subset\mathcal{B}_{1}$.
We have $\left(a,c\right]=\left(a,\infty\right)-\left(c,\infty\right)\in\mathcal{B}_{1}$ for each $c$, and consequently $\left(a,b\right)=\bigcup_{n\in\mathbb{N}}\left(a,b-\frac{1}{n}\right]\in\mathcal{B}_{1}$.