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Throughout, $\mu$ is just the Lebesgue measure.

Question: does there exist an uncountable family of distinct subsets of $[-1, 1]$, denoted by $(U_j)_{j \in [-1, 1]}$, with $\mu(U_j) > 0$ for each $j \in [-1, 1]$, and such that the following holds?

For any $k \in (0, 1]$ and any non-empty $I \subseteq [-k, k]$,

$$\ \ \mu\left(\bigcap_{i \in I} U_i \cap \bigcap_{j \in [-k, k]\smallsetminus I} U^\mathsf{c}_j \right) = 0. \tag{1}\label{472113_1}$$

Thoughts: this would be easy if e.g. we only required \eqref{472113_1} to hold for $k =1$. For then $U_x = \{y: \frac{1}{2}(x - 1) \leq y \leq \frac{1}{2}(x +1)\}$, $x \in [-1, 1]$ would be one such collection. But how to handle the case in question?

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    $\begingroup$ Just an idea to show that the answer cannot be negative: If the continuum hypothesis held, then we could give a well order $\prec$ to $\mathbb{R}$ isomorphic to the first uncountable ordinal. And then for each $j\in[-1,1]$ we could let $U_j=\{i\in[-1,1];j\prec i\}$ (so $\mu(U_j)=1$ for all $j$). Then for any $k\in(0,1]$ and any nonempty $I\subseteq[-k,k]$, if $I$ is uncountable then $\bigcap_{i\in I}U_i$ is empty. And if not, then there is some $j\in[-k,k]\setminus I$, and $U_j^c$ is countable. $\endgroup$
    – Saúl RM
    Commented May 28 at 2:42
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    $\begingroup$ @SaúlRM That is a very nice observation, and I encourage you to post it as an answer. Meanwhile, I believe we will be able to weaken the CH assumption to something about cardinal characteristics. $\endgroup$
    – Joel David Hamkins
    Commented May 28 at 2:49
  • $\begingroup$ The question need be reformulated: for an arbitrary subset the measure $\mu$ is not defined. You probably mean "Borel subset" or "measurable subset". $\endgroup$
    – YCor
    Commented May 28 at 19:59
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    $\begingroup$ @YCor Even if the sets themselves were assumed to be Borel, the uncountable intersection of them still would not have to be such. I think the equality means "the set inside is Lebesgue measurable and its measure equals zero". $\endgroup$
    – Adayah
    Commented May 28 at 20:12
  • $\begingroup$ Does the word "distinct" in the original question mean "disjoint", i,e, having empty intersection? $\endgroup$
    – Daniel Asimov
    Commented May 28 at 22:23

2 Answers 2

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My comment reposted as an answer:

If the continuum hypothesis holds, then we can give a well order $\prec$ to $\mathbb{R}$ isomorphic to the first uncountable ordinal. And then for each $j\in[-1,1]$ we can let $U_j=\{i\in[-1,1];j\prec i\}$ (so $\mu(U_j)=1$ for all $j$). Then for any $k\in(0,1]$ and any nonempty $I\subseteq[-k,k]$, if $I$ is uncountable then $\bigcap_{i\in I}U_i$ is empty. If not, there is some $j\in[-k,k]\setminus I$, and $U_j^c$ is countable.

And now, a construction without CH:

Let $U_x = \{y: \frac{1}{2}(x - 1) \leq y \leq \frac{1}{2}(x +1)\}$ as in the question. Note that If $D$ is any dense subset of $[-1,1]$, then for any nonempty $I\subseteq D$ we have

$$\ \ \mu\left(\bigcap_{i \in I} U_i \cap \bigcap_{j \in D\smallsetminus I} U^\mathsf{c}_j \right) = 0. \tag{1}$$

This is clear if $I=D$, and if not consider some point $x\in\overline{I}\cap\overline{D\setminus I}$ and note that when $i,j$ are close to $x$, $U_i\cap U_j^c$ is small.

So we can consider a bijection $f:[-1,1]\to[-1,1]$ such that $f([-1/n,1/n])$ is dense in $[-1,1]$ for all $n$ (see below for a construction), and define $V_x=U_{f(x)}$ for all $x\in[-1,1]$. Then the collection of sets $(V_x)_{x\in[-1,1]}$ satisfies what you want.

To construct the bijection $f:[-1,1]\to[-1,1]$, let $(A_n)_n$, $(B_n)_n$ be pairwise disjoint sequences of countable sets (the sets $A_i$ are pairwise disjoint, the sets $B_i$ are pairwise disjoint and the $A_i$ are disjoint with the $B_i$), such that $A_n\subseteq[-1/n,1/n]$ and $B_n$ is dense in $[-1,1]$ for all $n$.

Finally, define $f$ as the identity in $[-1,1]\setminus\bigcup_n(A_n\cup B_n)$, and let $f(A_n)=B_n$ and $f(B_n)=A_n$ for all $n$, using bijections between $A_n$ and $B_n$.

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    $\begingroup$ Thanks very much, this is great! A quick follow-up, and in case you don't find it entirely trivial, I'd be happy to ask a new question: $\endgroup$
    – Stepan Plyushkin
    Commented May 28 at 8:40
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    $\begingroup$ Follow-up question: Can we modify either one of the two methods to make $(U_j)_{j \in J}$ satisfy the following two properties? For any $i, j \in [-1, 1]$ with $i \neq j$, $$\mu((U_i \setminus U_j) \cup (U_j \setminus U_i)) > 0. \tag{2}$$ For any interval $J \subseteq [-1, 1]$ and any non-empty $I \subseteq J$, $$ \mu\left(\bigcap_{i \in I} U_i \cap \bigcap_{j \in J \setminus I} U^c_j\right) = 0. \tag{1'} $$ $\endgroup$
    – Stepan Plyushkin
    Commented May 28 at 8:41
  • $\begingroup$ (I reckon that the CH based approach gives us (1') but doesn't give us (2), and the other way around for the second approach...) $\endgroup$
    – Stepan Plyushkin
    Commented May 28 at 8:43
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    $\begingroup$ The approach without using CH should work: instead of the intervals $[-1/n,1/n]$ use the intervals $[q,q']$, where $q<q'$ are rationals in $[-1,1]$, thus obtaining a bijection $f:[-1,1]\to[-1,1]$ such that the image of every nontrivial interval is dense $\endgroup$
    – Saúl RM
    Commented May 28 at 10:45
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Take any sequence $(A_n)$ of measure independent subsets of $[-1, 1]$ of measure $1$, e.g.

$$\begin{align*} A_1 & = [-1, 0), \\ A_2 & = \left[ -1, -\frac{1}{2} \right) \cup \left[ 0, \frac{1}{2} \right), \\[1ex] A_3 & = \left[ -1, -\frac{3}{4} \right) \cup \left[ -\frac{1}{2}, -\frac{1}{4} \right) \cup \left[ 0, \frac{1}{4} \right) \cup \left[ \frac{1}{2}, \frac{3}{4} \right), \\ & \vdots \end{align*}$$

Now define $U_j$, $j \in [-1, 1]$ such that $U_{1/n} = A_n$ and $U_j$ is anything otherwise.${}^{\dagger}$

Given $m \in \mathbb{N}$, let $\mathbb{N}_m = \{ n \in \mathbb{N} : n \geqslant m \}$. Note that when $N$ is any subset of $\mathbb{N}_m$, by the independence of $U_{1/n}$ we have

$$\begin{equation} \tag{1} \mu \left( \bigcap_{n \in N} U_{1/n} \cap \bigcap_{n \in \mathbb{N}_m \setminus N} U_{1/n}^c \right) = 0. \end{equation}$$

As a consequence, when $k \in (0, 1]$ and $I$ is any subset of $[-k, k]$ (empty or not), we have

$$\mu \left( \bigcap_{i \in I} U_{i} \cap \bigcap_{j \in [-k, k] \setminus I} U_{j}^c \right) = 0,$$

because this set can be obtained from the intersection of the form $(1)$ by adding more sets to the intersection (where $m$ is such that $\frac{1}{m} \leqslant k$).

$\dagger$ There are many ways to define the missing $U_j$ in such a way that all $U_j$ are distinct and have positive measure. For the sake of concreteness we can take the ones that you proposed, i.e.

$$U_x = \left[ \frac{1}{2}(x-1), \frac{1}{2}(x+1) \right] \quad \text{for } x \notin \left\{ \frac{1}{n} : n \in \mathbb{N} \right\}.$$

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    $\begingroup$ Regarding the extra conditions from the discussion under Saúl RM's answer: the condition $\mu( U_i \Delta U_j ) > 0$ for $i \neq j$ can be satisfied by rejecting $A_1$ (which is almost identical to $U_{-1}$). The zero measure condition where $J \subseteq [-1, 1]$ is any interval (rather than $[-k, k]$ for some $k \in (0, 1]$) can be met by setting $U_j$ equal to $A_n$ with $j$ ranging over a countable dense subset of $[-1, 1]$ rather than $\left\{ \frac{1}{n} : n \in \mathbb{N} \right\}$. Explicitly, $U_{q_n} = A_n$ where $\left< q_n : n \in \mathbb{N} \right>$ is an enumeration of rationals. $\endgroup$
    – Adayah
    Commented May 28 at 19:59

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