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.
Best Answer
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 $m$.
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 $m$. $\quad\Box$