How to show that the finite open intervals $(a, b)$ do not generate the Borel sets of the extended real line

Question

It seems that any set generated by the finite open intervals either contains $\{\infty, -\infty\}$ or none of these infinities separately, and one would actually need to take intevals of the form $(a, \infty]$ to generate the Borel sets of the extended real line, but how can I prove this?

Answer 1

Let $\mathcal B$ denote the collection of Borel sets on $\mathbb R$.

Then the collection $\mathcal B\cup\{B\cup\{-\infty,+\infty\}\mid B\in\mathcal B\}$ is a $\sigma$-algebra that contains all open intervals.

However it does not contain sets like $(a,+\infty]$.

Answer 2

Let $A$ be the set of finite open intervals and let $\mathcal T$ denote the subset of $\sigma(A)$ of all sets which contain either both $+\infty$ and $-\infty$, or neither. You can show that $\mathcal T$ is a sigma algebra, and it contains $A$. Therefore $\mathcal T=\sigma(A)$.