Show that $\{\cup_{n\in K} (n, n+1]: K \subset \mathbb{Z}\}$ is a $\sigma$-algebra on $\mathbb{R}$

Solution 1:

Note that $\mathbb{R} = \bigcup\limits_{n \in \mathbb{Z}} (n, n + 1]$, so the purported $\sigma$-algebra has the whole space as an element. In fact, note that every element of $\mathbb{R}$ is in exactly one of the sets of the form $(n, n + 1]$ for $n \in \mathbb{Z}$.

It follows that the complement of the set $\bigcup\limits_{k \in K} (k, k + 1]$ is the set $\bigcup\limits_{k \in \mathbb{Z} - K} (k, k + 1]$. Thus, the purported $\sigma$-algebra is closed under complements.

Finally, if we take $\bigcup\limits_{i = 1}^\infty \bigcup\limits_{k \in K_i} (k, k + 1]$, this is clearly equal to $\bigcup\limits_{k \in K} (k, k + 1]$ where we define $K = \bigcup\limits_{i = 1}^\infty K_i$. So the purported $\sigma$-algebra is closed under countable union.

Thus, the purported $\sigma$-algebra actually is a $\sigma$-algebra.