Show that if $\mu$ is semifinite, and $\mu(E)=\infty,$ then for all $C>0$ there exists $F\subseteq E$ with $C<\mu(F)<\infty$

Question

So I'm trying to prove that claim from Folland's Real Analysis. The techniques I was using on the other problems weren't working, so I thought I would try something else. I was wondering if someone could verify my proof, or tell me what I did wrong. Thank you.

To motivate what's going on, I am planning to use Zorn's Lemma.

$\textbf{Proof:}$

Suppose the claim doesn't hold. Choose a $C>0$ such that exists no exists no $F\subseteq E$ with $C<\mu(F)<\infty.$ Let $$S=\{F\subseteq E:0<\mu(F)<\infty\}$$ be indexed by inclusion. Since semifinite we see $S\not=\emptyset.$ Let $\{F_i\}_{i\in I\subseteq\mathbb{N}}$ a chain. If $I$ if finite, then by finite additivity $\bigcup_{i\in I}F_i\in S$ is an upper bound for our chain, so WLOG $I=\mathbb{N}.$ By continuity from below it follows that $$\mu\left(\bigcup_{i\in\mathbb{N}}F_i \right)=\lim_{n\to\infty}\mu(F_i)\leq C.$$ Then $\bigcup_{i\in\mathbb{N}}F_i\in S$ is an upper bound for our chain. So, every chain has an upper bound in $S$. By Zorn's Lemma choose $F\in S$ maximal with respect to inclusion.

Note that $$\mu(E)=\mu(F)+\mu(E\backslash F),$$ so $\mu(E\backslash F)=\infty.$ Since $\mu$ is semifinite we can and do choose $F'\subseteq E\backslash F$ with $0<\mu(F')<\infty.$ Since $F,F'$ are disjoint it follows that $$\mu(F\cup F')=\mu(F)+\mu(F')<\infty,$$ hence $F\cup F'\in S.$ Since $F$ is maximal it follows that $F'=\emptyset$ and $\mu(\emptyset)>0.$ A contradiction, hence there exists $F\subseteq E$ with $C<\mu(F)<\infty$ as desired. $\blacksquare$

Answer 1

No, that's not right. To apply Zorn's lemma you have to show every chain has an upper bound, not just every countable chain.

(Simple counterexample to the version of Zorn's lemma you're applying here: Let $A$ be an uncountable set and let $X$ be the class of countable subsets of $A$, ordered by inclusion. Then every countable chain in $X$ has an upper bound, but $X$ does not have a maximal element.)

One could fix this with a little fiddling. Happily, the sort of constructions that you need to fix your argument actually give a direct proof, without any Zorn's lemma:

Let $$\alpha=\sup\{\mu(F):F\subset E, \mu(F)<\infty\}.$$We need to show that $\alpha=\infty$. Suppose that $\alpha<\infty$. There exists a sequence of sets $F_n\subset E$ with $\mu(F_n)<\infty$ and $$\mu(F_n)\to\alpha.$$ Let $$S_n=F_1\cup\dots\cup F_n.$$Since $F_n\subset S_n$ and $\mu(S_n)<\infty$ the definition of $\alpha$ shows that $$\mu(F_n)\le\mu(S_n)\le\alpha.$$Hence $$\mu(S_n)\to\alpha;$$since $S_n\subset S_{n+1}$ this shows that if $$S=\bigcup S_n$$then $$\mu(S)=\alpha.$$Now (as above) choose $F\subset E\setminus S$ with $$0<\mu(F)<\infty;$$then if $S'=S\cup F$ we have $$\alpha<\mu(S')<\infty,$$contradiction.