Suppose $M$ is a $\sigma$-finite measure space, and $\Gamma$ a set of positive-measure subsets of $M$ such that every point in $M$ belongs to only finitely many members of $\Gamma$. Must $\Gamma$ be countable? (And if the answer to this question is no for general σ-finite measures, is it also no for, say, Lebesgue measure?)
• I can see that if the conjecture that Γ has to be countable is true when $M$ has finite measure, it must be also be true when $M$ is σ-finite. (Just decompose $M$ into $M_0 ∪ M_1 ∪ \ldots$ where each $M_i$ is finite-measure. For each $i$, only countably many of Γ will have positive measure intersection with $M_i$; but every member of Γ must have positive-measure overlap with some $M_i$, so Γ must be countable.) So we can focus on the case where $M$ is finite measure.
• I can prove the following weaker result: suppose $M$ is finite-measure and Γ a set of positive-measure subsets of $M$ such that every point in $M$ belongs to at most $n$ members of Γ, then Γ must be countable. (For each ε>0, at most $nμ(M)/ε$ members of Γ have measure ≥ε. So we can enumerate all the members of Γ by first enumerating those with measure≥1, then those with measure ≥1/2, then those with measure ≥1/4, etc.)
• Returning to the original conjecture, I can see how to prove it if only I could take it for granted that for each $i$, the set $E_i$ of points in $M$ that belong to exactly $i$ members of Γ is measurable. In that case, I could first use the weaker result to show that for each $i$, at most countably many members of Γ have positive-measure overlap with $E_i$, and then argue that since $M = E_0 ∪ E_1 ∪ \ldots$ and the $E_i$ are measurable, every member of Γ must have positive-measure overlap with at least one of the $E_i$, establishing that Γ is itself countable. Unfortunately this doesn't work if the $E_i$ aren't measurable, and I don't see any way to show that they are.
(This question is a follow-up to Can an uncountable family of positive-measure sets be such that no point belongs to uncountably many of them?.)
Best Answer
Assume a finite measure. There is an $\epsilon > 0$ and countable collection $E_i$ with $\mu(E_i) > \epsilon$, because the alternative is that $\lbrace E: \mu(E) > \frac 1 n \rbrace$ is finite for all $n$, which would make $\lbrace E: \mu(E) > 0 \rbrace$ countable. Let $F_n = E_n \cup E_{n+1} \cup .... $. The sequence of indicators $1_{F_n} $ is monotone decreasing and converges to $0$ on the set where $x \in E_i$ finitely often, and $1$ if $x \in E_i$ infinitely often. It is also bounded by 1. By bounded convergence it converges in $ \mathbb L _1$. But since $\int 1_{F_n} d\mu > \epsilon $ the same must be true of the limit.