Collection of measurable sets closed under countable unions implies existence of a set with maximal measure

Question

Let $(X,\mathcal{A}, \mu)$ be a finite measure space. Let $\mathcal{G} \subset \mathcal{A}$ be a subset, closed under countable unions, i.e., if $(G_n)_{n \in \mathbb{N}} \subset \mathcal{G}$, then $\cup_{n \in \mathbb{N}}G_n \in \mathcal{G}$. In this case, is it true that there exists a set $G \in \mathcal{G}$ with maximal $\mu$-measure, i.e.

\begin{align} \mu(G) \geq \mu(G'), \text{ for all } G' \in \mathcal{G}, \end{align} and if so, how do I see that this set exists?

Thanks a lot in advance!

Answer 1

Let $s$ be the supremum of the $\mu$-measures of members of $\mathcal G$. By definition of supremum, for each $n$, there is $G_n\in\mathcal G$ with $\mu(G_n)>s-1/n$. Letting $G=\bigcup_n G_n$, then $G\in \mathcal G$ since $\mathcal G$ is closed under countable unions, and $\mu(G)=s$, since it is at least $\sup_n\mu(G_n)$ but it is at most $s$ (by definition of $s$).

Answer 2

Yes it's a direct application of Zorn's lemma: https://en.wikipedia.org/wiki/Zorn%27s_lemma

Your order would be $A\geqslant B$ if $\mu(A)>\mu(B)$ or $B$ is included in $A$.

Then your set $\mathcal{G}$ is partially ordered, any chain has an upper bound, and the Zorn's lemma resulting maximal element is necessary maximal for the measure.