# Uncountable collections of distinct subsets of an interval (existence)

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?

• 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. Commented May 28 at 2:42
• @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. Commented May 28 at 2:49
• The question need be reformulated: for an arbitrary subset the measure $\mu$ is not defined. You probably mean "Borel subset" or "measurable subset".
– YCor
Commented May 28 at 19:59
• @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". Commented May 28 at 20:12
• Does the word "distinct" in the original question mean "disjoint", i,e, having empty intersection? Commented May 28 at 22:23

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 $$B_n$$.

• 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: Commented May 28 at 8:40
• 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 $I \subseteq J$, $$\mu\left(\bigcap_{i \in I} U_i \cap \bigcap_{j \in J \setminus I} U^c_j\right) = 0. \tag{1'}$$ Commented May 28 at 8:41
• (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...) Commented May 28 at 8:43
• 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 Commented May 28 at 10:45

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

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

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.

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

• 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. Commented May 28 at 19:59