Countably representing all closed sets of positive measure

Question

This may be a naive question, but I don't see an immediate argument.

Question: Does there exist a sequence $\{C_m\}_{m=1}^\infty$ of Borel subsets of $[0,1]$ with positive Lebesgue measure $|C_m|>0$, such that every closed subset $C\subset[0,1]$ with positive Lebesgue measure $|C|>0$ contains at least one $C_m\subset C$?

It feels too strong to be true, but I need to know for sure. I was thinking of patterns like dyadic partitions of fat Cantor sets etc., but can't arrive at something useful.

Thank you.

Answer 1

Without the assumption that $C$ be closed the answer is no.

Indeed, suppose such $C_m$'s exist. Then without loss of generality $|C_m|\le1/3^m$ for all $m$. Let now $C:=\bigcap_{m=1}^\infty ([0,1]\setminus C_m)=[0,1]\setminus\bigcup_{m=1}^\infty C_m$.

Then $|C|\ge1/2>0$, but $C_m\not\subseteq C$ for any $m$.

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Using the hint by Aleksei Kulikov (why didn't I think of that? :-)), one can modify the above construction as follows, to get an unqualified no:

Suppose such $C_m$'s exist. Then without loss of generality $|C_m|\le1/3^m$ for all $m$. Let now $B:=\bigcap_{m=1}^\infty ([0,1]\setminus C_m)=[0,1]\setminus\bigcup_{m=1}^\infty C_m$.

Then $|B|\ge1/2>0$ and $C_m\not\subseteq B$ for any $m$. Let finally $C$ be a closed subset of $B$ with $|C|>0$; such a set $C$ exists by the regularity of the Lebesgue measure. Then $C_m\not\subseteq C$ for any $m$. $\quad\Box$

Answer 2

Suppose $\{C_m\}_{m=1}^\infty$ is a sequence of nonempty subsets of $[0,1]$. Choose $x_m\in C_m$ and let $X=\{x_m:m\in\mathbb N\}$. The set $X$, being countable, has measure zero, so $[0,1]\setminus X$ has positive measure and therefore contains a closed set $C$ of positive measure. Then $C_m\not\subseteq C$ since $x_m\notin C$.