$\textbf{Example 4.10:}$ For a family $\mathcal{F}$ of open intervals of real numbers, the fact that the endpoints of the intervals in $\mathcal{F}$ are dense in $\mathbb{R}$ does not imply that $\sigma(\mathcal{F}) = \mathcal{B}(\mathbb{R})$.
$\textbf{Proof:}$ Recall tha the positive rationals are countable and let $\{ r_n \ ; \ n \in \mathbb{N} \}$ be an enumeration of $\mathbb{Q} \cap (0, \infty)$. Now, let $\mathcal{F} = \{ (- r_n, r_n) \ ; \ n \in \mathbb{N} \}$. Note that the endpoints of the intervals in $\mathcal{F}$ are dense in $\mathbb{R}$. Obviously, $\sigma(\mathcal{F})$ contains only Borel subsets of $\mathbb{R}$ since $\sigma(\mathcal{F})$ is generated by a family of open sets. However, note that $\sigma(\mathcal{F})$ does not contain the Borel set $[0,1]$. Thus, $\sigma(\mathcal{F})$ is a proper subset of $\mathcal{B}(\mathbb{R})$. $\square$
I'm trying understand this counterexample, but I'm stuck in understand why $[0,1] \notin \sigma(\mathcal{F})$. It's clear for me that $[0,1]$ can't be generated by elements of $\mathcal{F}$ since such elements generate subsets of $\mathbb{R}$ which are symmetric with respect to $0$, but how can I ensure that doesn't exist elements in $\sigma(\mathcal{F}) \ \backslash \ \mathcal{F}$ that generate $[0,1]$?
The collection $\cal S$ of subsets of $\Bbb R$ which are symmetric about $0$ (i.e. have the property that $a\in A$ implies $-a\in A$) is a sigma-algebra. As $\cal{F}\subseteq S$ then $\sigma(\cal{F})\subseteq S$ (since $\sigma(\cal F)$ is the intersection of all sigma-algebras containing $\cal F$). But $[0,1]\notin\cal S$.