Can we construct the set $(\alpha, \infty]$ only out of open sets $(\alpha, \beta)$?

Question

Let $(\alpha, \beta) \in [-\infty, \infty]$ be open sets, and let $\mathscr{B}$ be the set of Borel sets generated by such open sets. Is it possible to construct the set $(\alpha, \infty]$ out of open sets? In other words is $(\alpha, \infty] \in \mathscr{B}$? Or, do we need to define the open sets of $[-\infty, \infty]$ as $(\alpha, \infty]$?

Answer 1

Consider the collection of all sets $A$ in the your sigma algebra $\mathcal B$ such that $I_A(\infty)=I_A(-\infty)$. It is easy to show that this is a sigma algebra contaning the intervals $(\alpha, \beta)$. Conclusion: for every $A$ in this sigma algebra $\infty \in A$ iff $-\infty \in A$. In particular $\{\infty\}$ is not in this sigma algebra. It follows that $(\alpha, \infty]$ cannot be in $\mathcal B$ for any $\alpha$.