TITLE: Closed under countable union
QUESTION [6 upvotes]: I am reading a tutorial on measure theory and it states: "Given an interval $E = [a, b]$ and a set $S$ of subsets of $E$ which is closed under countable unions, we deﬁne the following..."
I was wondering what 'closed under countable unions' meant in this case. Does it mean that $S$ contains every element of $E$?
So if $E = [0,5]$, $S$ could be something like $\{[0,2], [1,3], [2,3], [2,5]\}$ since the union of those would be $[0,5]$... but it would impossible for $S$ to be something like $\{[0,1], [1,4], [2,3]\}$ since the union of those would be $[0,4]$.
Am I correct? Thanks!

REPLY [11 votes]: That $S$ is closed under countable unions means that for every countable collection $(A_i)_{i\in\mathbb{N}}$ of sets from $S$, we have that $\bigcup_{i\in\mathbb{N}} A_i \in S$. In particular this must hold for every finite collection of sets, so in your case where $S=\{[0,2], [1,3],[2,3],[2,5]\}$, we have that $[1,3]\cup [2,5]=[1,5]$ which is not in $S$. So this $S$ is not closed under countable union.