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Let $(X,\mathcal{B},\mu)$ be an arbitrary finitely additive probability measure space, let $a>0$ and let $(A_i)_{i\in I}$ be an uncountable family of subsets with measure $\geq a$.

Is there an uncountable set $J\subseteq I$ such that, for all $j,j'\in J$, $A_j\cap A_{j'}$ has positive measure?

And if not, are there any conditions on $(X,\mathcal{B},\mu)$ that imply a positive answer to the question? This question seems to give conditions for a stronger result than I am looking for in $\mathbb{R}$, but it uses more axioms apart from ZFC.

I am also interested in any bibliography that deals with uncountable families of measurable sets.

Edit: About the discussion in the comments, it seems for any family $(A_i)_{i\in I}$ of set of measure $\geq a$, where $\#I>2^{\aleph_0}$, there is an uncountable set $J\subseteq I$ such that for all $i,j\in J$, $\mu(A_i\cap A_j)\geq a^2$. To see why, consider the complete graph $G$ with vertex set $I$ and color the edge $(i,j)$ of color "$n$" if $\mu(A_i\cap A_j)\in\left[a-\frac{1}{n},a-\frac{1}{n+1}\right)$ and of color "$0$" if $\mu(A_i\cap A_j)\geq a^2$.

Then the Erdős–Rado theorem implies that there is a monochromatic clique of uncountable size. If this clique has color $n$ for some $n>0$ we reach a contradiction, so the clique is of color $0$, and we are done.

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    $\begingroup$ I'm sure you know about Bergelson's lemma which gets a countable $J \subseteq I$. $\endgroup$
    – mathworker21
    Commented Nov 1 at 23:17
  • $\begingroup$ Thanks, yes, I am aware of the intersectivity lemma for countably many sets $\endgroup$
    – Saúl RM
    Commented Nov 2 at 1:24
  • $\begingroup$ Do you know if the same conclusion but with "$|A_j \cap A_{j'}| \ge a$" is true? (I.e., there exists $J$ and $a$ such that $|A_j \cap A_{j'}| \ge a$ for all $j, j' \in J$) $\endgroup$
    – mathworker21
    Commented Nov 2 at 14:28
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    $\begingroup$ @mathworker21 it would be true with $|A_j\cap A_{j'}|\geq\lambda$ for any $\lambda<a^2$, for the same reasoning as the answer below and an argument using Cauchy-Schwartz (you cannot have infinitely many measure $a$ sets with pairwise intersections bounded above by a constant below $a^2$). For any $\lambda>a^2$ it is not true (let the sets $A_i$ be independent) $\endgroup$
    – Saúl RM
    Commented Nov 2 at 15:04
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    $\begingroup$ I wonder about the case $\lambda=a^2$ though. In the countable case it is false, but I see no immediate reason why it would be false in the uncountable case too. Perhaps one use transfinite induction to construct a measure space with uncountably many sets of measure $a$ but intersections of measure $<a^2$? $\endgroup$
    – Saúl RM
    Commented Nov 2 at 15:52

1 Answer 1

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By the Erdős–Dushnik–Miller theorem, if the index set $I$ is infinite, then either there is a subset $J\subseteq I$ of the same cardinality as $I$ such that $A_j\cap A_{j'}$ has positive measure for all $j,j'\in J$, or else there is an infinite set $J\subseteq I$ such that $A_j\cap A_{j'}$ has measure zero whenever $j,j'\in J$ and $j\ne j'$. The latter is impossible if $\mu$ is a probability measure and $\mu(A_i)\ge a\gt0$ for all $i\in I$.

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    $\begingroup$ That's very nice! $\endgroup$
    – Saúl RM
    Commented Nov 2 at 8:46
  • $\begingroup$ Why do you say "if the index set $I$ is infinite"? The question said $I$ is uncountable. $\endgroup$
    – mathworker21
    Commented Nov 2 at 14:27
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    $\begingroup$ @mathworker21 Mea culpa. I thought it was permissible to state a result that is slightly more general than needed to answer the question. First, for uncountable $I$, the conclusion "$J$ is uncountable" can be strengthened to "$J$ has the same cardinality as $I$"; but this improved formulation is also true when $I$ is countably infinite. $\endgroup$
    – bof
    Commented Nov 2 at 15:32

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