Is the definition of open interval used when discussing Borel $\sigma$-algebra different from the normal definition of open interval?

Question

I encounter the following problem when doing a problem on measure theory.

Every open subset of $\mathbb{R} = (-\infty, \infty)$, the real line, is a countable union of open intervals. Use this fact to show that $\mathcal{B}(\mathbb{R})$ is generated by the collection of all open intervals.

Certainly, the complement of (a, b) is not an open interval in the normal definition of interval since it is $(-\infty, a] \cup [b, \infty)$. What am I misunderstanding here? Is this considered an interval in measure theory?

Answer 1

This is not really an answer to your question, but what they mean is that - if every open interval is an element of a $\sigma$-algebra $\mathcal A$ - then automatically every countable union of open intervals will be an element of $\mathcal A$, since a $\sigma$-algebra is closed under countable unions. In case of $\mathbb R$ that means that every open set will be an element of $\mathcal A$ simply because every open set in $\mathbb R$ is a countable union of open intervals.

Consequently $\mathcal B=\sigma(\tau)\subseteq\mathcal A$.

Answer 2

For a collection of sets $S\subseteq 2^{\mathbb{R}}$, $\sigma(S)$ is the smallest $\sigma$-algebra containing $S$. We also refer to $\sigma(S)$ as the $\sigma$-algebra generated by $S$. The Borel $\sigma$-algebra, denoted by $\mathcal{B}(\mathbb{R})$ is the $\sigma$-algebra generated by the open subsets of $\mathbb{R}$; that is $\mathcal{B}(\mathbb{R})=\sigma(\mathcal{O})$ where $\mathcal{O}$ is the collection of all open subsets of $\mathbb{R}$. The question is essentially asking you to show the following: If $\mathcal{I}$ is the collection of all open intervals of $\mathbb{R}$, then $\sigma(\mathcal{I})=\mathcal{B}(\mathbb{R})$ as well. One containment is obvious. For the other, it suffices to show that $\mathcal{O}\subseteq\sigma(\mathcal{I})$. Since $\sigma(\mathcal{I})$ contains all countable unions of open intervals, it suffices to show that every element of $\mathcal{O}$ can be written as a countable union of open intervals. Hopefully this clarifies what they're asking.