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!

  • $\begingroup$ A related question would be to start with a signed measure. Would countable unions (and not complements or intersections or differences) still be enough to conclude a set with maximal measure? $\endgroup$ – GEdgar Jul 12 '18 at 16:06

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 $s$).

  • $\begingroup$ I wrapped my head around it one more time, as I had trouble to get how you freed yourself from Zorn's lemma. In fact, you use the axiom of countable choice... so you get away without Zorn but still need a weaker form. $\endgroup$ – Benoit Gaudeul Jul 14 '18 at 16:43
  • $\begingroup$ Yes, countable choice is enough, which is why I referred to the full version as an overkill. I suspect we cannot really remove all choice. $\endgroup$ – Andrés E. Caicedo Jul 14 '18 at 17:31

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.

  • $\begingroup$ Zorn's lemma is an overkill. Anyway, if you proceed this way, it seems to me the key thing is rather arguing that "any chain has an upper bound". $\endgroup$ – Andrés E. Caicedo Jul 12 '18 at 15:57
  • $\begingroup$ Indeed, but the key problem then is to find a way to get always bigger sets. For example, in your answer, how do you get the existence of a supremum? $\endgroup$ – Benoit Gaudeul Jul 12 '18 at 16:03
  • $\begingroup$ Ah OK, the set of the sets measures is a non-empty bounded part of R... $\endgroup$ – Benoit Gaudeul Jul 12 '18 at 16:04
  • $\begingroup$ Then the upper bound would just be the union of the sets in the chain right? $\endgroup$ – vaoy Jul 13 '18 at 16:11
  • $\begingroup$ In the special case of the chain chosen by Andrés, yes, but it is easy to find other chains (adding a point at a time for example) $\endgroup$ – Benoit Gaudeul Jul 14 '18 at 16:37

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