(Generalized Cantor Set) Constructing a closed set with measure $1-\epsilon$ Hello dear Mathematicians!
For every $\epsilon>0$ I want to construct a closed set $F_\epsilon$ on $[0,1]$ such that $F_\epsilon$ does not contain any closed interval $[c,d]$ and $mF_\epsilon\geq1-\epsilon$.
The cantor set is obtained by successively removing the "open middle third" from each closed interval in $K_n$ to obtain $K_{n+1}$, where $K_0=[0,1]$. Now I want to construct a generalized cantor set described above. Moreover, there is a theorem saying that If there is a closed set which is not open, then there is a nonempty closed set which does not contain any nonempty open set. But, I did not find any theorem about not containing any closed set.
I'll appreciate any hints that take me into the right direction.

You can make these sets arbitrarily close to measure $1$ in the following sense.
Define a geometric series $\sum_{n \in \mathbb N}ar^{n}<1$. As with the cantor set, remove an interval of length exactly $ar^n$ from each component in the $(n-1)^{th}$ step. Define
$$C:= \bigcap_{n \in \mathbb N}C_n$$
which is closed, while $m(C_n)=1-\sum_{k=1}^{n}ar^n>0$. All we need to do is make sure that $a \cdot \frac{r}{1-r}$ is sufficiently small.

For reference, the cantor set is pretty much given by $\sum_{n \in \mathbb N} \frac{1}{2}(2/3)^n=1$, and $m(1-1)=0$, so it has measure zero.

By the way, these are known as fat cantor sets
