In Folland's Real Analysis, Section 1.6, it says
Our characterization of the $\sigma$-algebra$\mathcal{M}(\mathcal{E})$ generated by a family $\mathcal{E} \subset \mathcal{P}(X)$ is nonconstructive, and one might ask how to obtain$\mathcal{M}(\mathcal{E})$ explicitly from $\mathcal{E}$. The answeris rather complicated. One can begin as follows: Let $\mathcal{E}_1 = \mathcal{E} \cup \{E^c : E \in \mathcal{E}\}$, and for $j > 1$ define$\mathcal{E}_j$ to be the collection of all sets that are countableunions of sets in $\mathcal{E}_{j-1}$ or complements of such. Let$\mathcal{E}_\omega = \bigcup_{j=1}^\infty \mathcal{E}_j$; is$\mathcal{E}_\omega = \mathcal{M}(\mathcal{E})$? In general, no.$\mathcal{E}_\omega$ is closed under complements, but if $E_j \in \mathcal{E}_j \setminus \mathcal{E}_{j-1}$ for each $j$, there is noreason for $\bigcup_{j=1}^\infty E_j$ to be in $\mathcal{E}_\omega$.So one must start all over again.
Is there a simple example of a family of sets $\mathcal{E}$ such that the resulting sets $(\mathcal{E_j})_{j \in \mathbb{N}}$ (as defined above) are strictly increasing? If $\mathcal{E}$ is the family of open intervals (or closed / half open - whatever is easier), are the sets $\mathcal{E_j}$ strictly increasing? If so what's an example of a set $E \in \mathcal{E_j} \setminus \mathcal{E_{j-1}}$ for any $j \in \mathbb{N}$?