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My general question was is it consistent that any uncountable family of less than $\mathrm{non}(\mathcal{N})$ sets, each with positive measure, has an uncountable subfamily $\mathcal{F}$ s.t. $\bigcap \mathcal{F}$ has positive measure?

Here, $\mathrm{non}(\mathcal{N})$ denotes the uniformity number of the ideal of null sets, i.e. the smallest cardinality of a set that is not of measure zero.

The original question I was thinking about was:

Is it true that every uncountable family of sets of positive measure has an uncountable subfamily with an intersection of positive measure?

It is not difficult to see that under CH, the answer is strongly negative, namely if $\mathfrak{c} = \omega_1$, then there exists an uncountable family $\mathcal{X}$ of sets of even full measure such that every uncountable subfamily of $\mathcal{X}$ has even an empty intersection. To see this, order the reals in an $\omega_1$-sequence $\{x_\alpha\}_{\alpha < \omega_1}$ and for each $\alpha < \omega_1$ define a tail of the reals above $x_\alpha$, i.e. let $X_\alpha = \{x_\beta: \alpha < \beta < \omega_1\}$. Then, the complement of each $X_\alpha$ is a null set, so all the sets in $\mathcal{X} = \{X_\alpha\}_{\alpha < \omega_1}$ are of full measure, yet every real $x$ belongs to at most countably many sets $X_\alpha$, so it cannot be a member of any intersection of uncountably many elements of $\mathcal{X}$.

Now, observe that the same reasoning seems to work in any model of ZFC, where $\mathrm{non}(\mathcal{N}) = \aleph_1$. If this is the case, then, by definition there exists a set $A \subseteq \mathbb{R}$ with $\mu^\ast(A) > 0$ and $|A| = \aleph_1$ (of course, if CH does not hold, $A$ cannot be measurable, but it does not matter here). Apply the same reasoning to $A$ as we applied above to the set reals.

Furthermore, this actually shows that there always exists a family of size $\mathrm{non}(\mathcal{N})$ sets of positive outer measure such that any intersection of at least $\mathrm{non}(\mathcal{N})$ many elements of this family is empty. We just take $A \subseteq \mathbb{R}$ of size $\mathrm{non}(\mathcal{N})$ and $\mu^\ast(A) >0$, and repeat the argument with $A = \{x_\alpha: \alpha < \mathrm{non}(\mathcal{N})\}$, and the sets $X_\alpha = \{x_\beta: \alpha < \beta < \mathrm{non}(\mathcal{N})\}$, where $\alpha < \mathrm{non}(\mathcal{N})$ - for each element $x$ of $A$ the number of sets $X_\alpha$ with $x \in X_\alpha$ is bounded in $\mathrm{non}(\mathcal{N})$.

So, it is true that there is always a family of at least $\mathrm{non}(\mathcal{N})$ sets of positive outer measure such that each intersection of $\mathrm{non}(\mathcal{N})$ of its elements is actually empty.

Two questions, then:

[reformulated in light of the comment of Wojowu referring to Łuzin-Sierpiński construction that actually answers the questions if we do not restrict to measurable sets]

(1) Is it consistent that each uncountable family of strictly less than $\mathrm{non}(\mathcal{N})$ measurable sets of positive measure has an uncountable subfamily (a) of same size, or (b) smaller, yet uncountable, with an intersection of positive measure?

(2) Is it consistent that each uncountable family at least $\mathrm{non}(\mathcal{N})$ measurable sets of positive measure has a strictly smaller, yet uncountable subfamily with an intersection of positive measure?

An additional, loosely related, question out of curiosity - it is known that if we take any family of less than $\mathrm{add}(\mathcal{N})$ (which denotes the additivity number, i.e. the minimal cardinality of null sets such that their sum is not null) sets of full measure, then their intersection also has the full measure. It is obvious that this statement does not hold if "full" is replaced with "positive (outer)" in the above - simply take the sets $[k, k+\epsilon]$ for some fixed small $\epsilon$, and for all integers $k$ - the intersection of this countable family is empty. But what about uncountable collections of strictly less than $ \mathrm{add}(\mathcal{N})$ sets of positive (outer) measure: can there be such families with intersection of every uncountable subfamily being null?

And even more loosely related question - it is obvious that e.g. the uniformity number would not make much sense in the context in which we restricted the attention to measurable sets of reals (namely, each measurable set of positive Lebesgue measure has to have cardinality continuum). But, it is not obvious to me if the same holds for the additivity number - if we restrict attention to measurable sets of reals, would it also hold that this measurable-additivity would be equal to continuum?

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  • $\begingroup$ This answer mentions a continuum-sized family of pairwise disjoint sets such that each has full outer measure. I haven't read your question fully so I'm not sure if this family (or some uncountable subfamily of it) leads to a complete answer. $\endgroup$ – Wojowu Feb 26 at 16:44
  • $\begingroup$ Thanks! Actually, the Łuzin-Sierpiński construction answers the question w.r.t. the families of sets of positive outer measure. I added the restriction to measurable sets. $\endgroup$ – mtg Feb 26 at 17:12
  • $\begingroup$ Of possible related interest is this 13 May 2005 sci.math post which gives of survey of ZFC results inspired by the following: Let $E_1,$ $E_2,$ $\ldots$ be Lebesgue measurable subsets of $[0,1],$ and let $C>0.$ Does there exist a subsequence ${E_{n_k}}$ such that $\bigcap_{k=1}^{\infty}E_{n_k}$ is nonempty? (Yes, by Fatou's lemma for appropriate characteristic functions.) Does there exist a subsequence ${E_{n_k}}$ such that $\bigcap_{k=1}^{\infty}E_{n_k}$ has positive measure? (No, and a common counterexample (continued) $\endgroup$ – Dave L Renfro Feb 26 at 19:03
  • $\begingroup$ is to let $E_n$ be those numbers having a nonzero $n$'th decimal digit—each $E_n$ has measure $0.9,$ and the intersection of any $k$-many of these sets has measure $(0.9)^k,$ and thus the measure of the intersection of any infinite subcollection of these sets is zero.) $\endgroup$ – Dave L Renfro Feb 26 at 19:03
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The answer to both questions (1) and (2) is yes.

It is consistent, and in fact it follows from Martin's Axiom for $\aleph_1$, that:

For every uncountable family $\mathcal F$ of positive-measure subsets of $\mathbb R$, there is an uncountable subfamily $\mathcal G \subseteq \mathcal F$ such that $\bigcap \mathcal G$ (is measurable and) has positive measure.

This is Exercise 27 in chapter 2 in Kunen's 1980 set theory book.

So if $\mathsf{MA}$ holds and $\mathfrak{c} = \aleph_2$, then what you've asked about in 1(a) is consistent. 1(b) and 2 follow from $\mathsf{MA}(\aleph_1)$.

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