Is S a $\sigma$-alegrba?

Question

If $S=\{ \cup_{n \in A} [n,n+1): A \subset \mathbb{Z} \}$, then is $S$ a $\sigma$-algebra? I know that I need to show that: $\emptyset \in S$, if $E \in S$ then $X \backslash E \in S$, if $E_1,E_2,...$ is a sequence of elements of $S$ then $\cup_{k=1}^{\infty}E_k \in S$. However, I don't seem to be getting anywhere with any of these. Any help would be appreciated, thanks in advance.

Answer 1

You can take $\emptyset\subset\Bbb Z$ and then $\cup_{n\in\emptyset}[n,n+1)=\emptyset$, so $\emptyset\in S$.

If $E\in S$ then there is some $A\subset Z$ such that $E=\cup_{n\in A}[n,n+1)$. Note that $x\notin E\Leftrightarrow x\notin[n,n+1)\quad\forall \, n\in A\Leftrightarrow x\in[m,m+1)$ for some $m\notin A\Leftrightarrow x\in[m,m+1)$ for some $m\in \Bbb Z\setminus A$; this is due to $\Bbb R$ being the disjoint union $\Bbb R=\cup_{n\in\Bbb Z}[n,n+1)$.

Then we have $E^\mathsf{c}=\cup_{m\in \Bbb Z \setminus A}[m,m+1)$, so $E^\mathsf{c}\in S$ considering $\Bbb Z\setminus A\subset Z$.

Consider $\{E_i\}_{i\in I}$ a countable collection of sets in $S$ (hence $I$ is countable). For each $E_i$ there is some $A_i\subset\Bbb Z$ such that $E_i=\cup_{n\in A_i}[n,n+1)$. Now note that $x\in\cup_{i\in I}E_i\Leftrightarrow x\in E_{i_0}$ for some $i_0\in I\Leftrightarrow x\in[n,n+1)$ for some $n\in A_{i_0}$ for some $i_0\in I\Leftrightarrow x\in[n,n+1)$ for some $n\in\cup_{i\in I}A_i\Leftrightarrow x\in\cup_{n\in\cup_{i\in I}A_i}[n,n+1)$.

Then we have $\cup_{i\in I}E_i=\cup_{n\in\cup_{i\in I}}[n,n+1)$, so $\cup_{i\in I}E_i\in S$ considering $\cup_{i\in I}A_i\subset\Bbb Z$.

We conclude $S$ is a $\sigma$-algebra.

Answer 2

A general result: Let $E_1,E_2,..$ be a partition of $\mathbb R$. [This means the sets are pair-wise disjoint and their union is $\mathbb R$]. Then all possible unions of the sets $E_i$ forms a sigma algebra. Proof: The complement of any such unions is exactly the union of the remaining $E_i$'s. It is obvious that union of any sequence of sets in this family also belongs to the family. QED.