This question is about Durrett Edition 5 Example 1.1.8, in which he claims that
Let $\Omega=\mathbb{R}$, and $\mathcal{S}=\mathcal{S}_{1}$ then $\overline{\mathcal{S}}_{1}=$the empty set put all sets of the form $\bigcup_{k=1}^{n}(a_{k}, b_{k}]$, where $-\infty\leq a_{i}<b_{i}\leq\infty$ is an algebra.
He did not specify what is $\mathcal{S}_{1}$, so I assumed it is a set containing empty set and the intervals of the form $(a,b]$ where $-\infty\leq a<b\leq\infty.$
It is clear that $\overline{\mathcal{S}}_{1}$ is the set of finite disjoint union of sets in $\mathcal{S}_{1}$. Thus, $\overline{\mathcal{S}}_{1}$ being an algebra follows immediately if we prove that $\mathcal{S}_{1}$ is a semi-algebra.
Indeed, for $(a_{1}, b_{1}], (a_{2}, b_{2}]\in\mathcal{S}_{1}$, the intersection of them is of the form $(x_{1},y_{2}]$ where $x_{1}\in \{a_{1}, a_{2}\}$ and $y_{2}\in\{b_{1}, b_{2}\}$, and thus $\mathcal{S}_{1}$ is closed under finite intersection.
However, I don't know how to show for any $(a,b]\in\mathcal{S}_{1}$, $(a,b]^{c}$ can be expressed as a finite disjoint union of elements in $\mathcal{S}_{1}$. I don't even this this is possible, so the definition of $\mathcal{S}_{1}$ must be something else.
Any idea? I think perhaps I assumed the wrong definition. Thank you!
Edit:
Okay I figured it out. The definition of $\mathcal{S}_{1}$ is not wrong, Durrett actually gave a general example in example 1.1.5 of $\mathcal{S}_{d}$, with respect to $\mathcal{R}^{d}$.
The point here is that the end points of the interval can be $\pm\infty$.
I will answer my own question.
If for $(a, b]\in\mathcal{S}_{1}$, then $(a,b]^{c}=(-\infty, a]\sqcup (b,+\infty]$ which is a finite disjoint union of sets in $\mathcal{S}_{1}$.
As I said in my edit, the point here is that you can have $\pm\infty$, so it is immediate.