The sigma algebra would need to contain complements $[n,n+1]^c = (\infty, n) \cup (n+1, \infty) $, which can be made from countable union of open sets $(n,n+1), n\in \mathbb{Z}$. It also contains singletons $[n-1,n] \cap [n, n+1] = \{n\} $.
My question is: how do I make sure that's all there is to it (or is there more). I'm having trouble since the only way I know how to find out generated sigma algebra's is to 'guess' it. Is there a better way to do this?
Best Answer
Consider sets of the form $\bigcup_{n \in I} (n,n+1) \cup E$ where $I$ and $E$ are subsets of $\mathbb Z$. This class is clearly closed under countable unions. The complement of $\bigcup_{n \in I} (n,n+1) \cup E$ is $\bigcup_{n \notin I} (n,n+1) \cup (\mathbb Z \setminus E)$. It follows that this class is a sigma algebra. We can now easily see that it is the smallest sigma algebra containing the intervals $[n, n+1], n \in \mathbb Z$.
PS: The motivation for this comes from the fact that the sigma algebra generated by a countable partition $(E_n)$ is the class of all possible unions of the sets $E_n$.