Let real number $a > 0$ and let $S_a \subset \mathbb{R}$ that satisfies:
- $S_a$ is finite;
- If $b > a$, then $S_b \subseteq S_a$.
I am wondering whether the set $S = \bigcup_{a > 0} S_a$ has measure zero or not. It seems to me that $S$ might be uncountable in some cases, and it's obvious that each $S_a$ has measure $0$. But does the union of all of the $S_a$ still has measure zero?
For the background, I am actually unfamiliar with the definition of Lebesgue Measure in general. I know the definition of a set in real number being "measure zero" is that it can be covered by countably many open intervals with the sum of their length being arbitrarily small.
$S=\bigcup_{a>0}S_a$ is actually countable. In fact, let $x\in S$. Then, there is some $a>0$ such that $x\in S_a$ and therefore, by considering some $q\in\Bbb Q\cap (0,a)$, there is some $q\in\Bbb Q_{>0}$ such that $x\in S_q$. Therefore $S\subseteq \bigcup_{q\in\Bbb Q_{>0}}S_q$, which is union of countably many finite sets.