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?](https://math.stackexchange.com/questions/319358/).)
['https://math.stackexchange.com/questions/332362', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/4909/']
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.
Here is a more detailed and more elementary version of mike's proof, using continuity rather than the bounded convergence theorem, which I'm writing up mostly for my own edification. 

Suppose that for some finite-measure $S$ and $ε>0$, there are infinitely many $E⊂Γ$ with $μ(S∩E)≥ε$. Let $E\_0, E\_1, E\_2\ldots$ be a countable sequence of such $E$. And let $F\_n = S ∩ (E\_n ∪ E\_{n+1} ∪ E\_{n+2} ∪ \cdots)$. Clearly $μ(S) ≥ μ(F\_i) ≥ μ(S∩E\_i) ≥ ε$ for each $i$, and $F\_i ⊂ F\_j$ whenever $j<i$. So by the fact that measures are continuous from above (see <http://en.wikipedia.org/wiki/Measure_(mathematics)#Properties>), $μ(\bigcap\_{i=1}^\infty F\_i) = \lim\_{i→∞}μ(F\_i) ≥ ε$. But for any point $p$ in $S$, $p∈\bigcap\_{i=1}^\infty F\_i$ if and only if p belongs to infinitely many of the $E\_i$. Therefore there are some points in $M$ (in fact, a positive-measure set of them) that belong to infinitely many of the $E\_i$, contradicting the stipulation that each point only belongs to finitely many members of Γ.

So we can conclude that for every finite-measure subset $S$ of $M$ and every $ε>0$, there are only finitely many $E∈Γ$ with $μ(S∩E)≥ε$. Then the set of $E∈Γ$ with $μ(S∩E)≥0$ is countable for every finite-measure $S$, since we can enumerate these $E$ by first listing those for which $μ(S∩E)≥1$, then those for which $μ(S∩E)≥1/2$, then those for which $μ(S∩E)≥1/3$, etc. But $M$ is σ-finite, so there is a countable collection of finite-measure sets $S\_0, S\_1, S\_2\ldots$ such that every positive-measure subset of $M$ has positive-measure overlap with at least one of $S\_0, S\_1, S\_2\ldots$. So Γ itself is countable, being the union of countably many countable sets.