Can anyone give an example of a closed set contains no interval but with finite non-zero Lebesgue measure? Can anyone give an example of a closed set $F$ of $\Bbb{R}$ such that $0<|F|<+\infty$ and $F$ contains no open interval? Thank you!
 A: Because $\mathbb {Q}$ has measure $0,$ there is an open set $U$ containing $\mathbb {Q}$ such that $m(U) < 1/2.$ Then $[0,1]\setminus U$ is closed (in fact compact), and contains no interval (because it contains no rational). We have $m([0,1]\setminus U) > 1/2,$ so we're done.
A: Pick a strictly decreasing sequence $(a_n)_n$ with positive limit. Construct a sequence of sets $S_0,S_1, S_2, \ldots$ such that 


*

*$S_n\subset S_{n-1}$

*$S_n$ is the disjoint union of $2^n$ closed intervals of equal length

*$\mu(S_n)=a_n$


recursively: Just cut away the middle $\frac{a_{n-1}-a_n}{2^n}$ from each of the $2^n$ intervals making $S_{n-1}$ in order to obtain $S_n$.
Then $S:=\bigcap S_n$ is closed and has measure $\lim a_n$. It does not contain any open interval because the parts of $S_n$ have length $ \frac{a_n}{2^n}$, which gets arbitrarily small.
A: Enumerate the rational points of $[0,1]$ as a sequence $(r_n)_{n\in\mathbb N}$. Then, choose a sequence of positive numbers $(\varepsilon_n)$ such that $\sum_1^\infty 2\varepsilon_n<1$ and set $K:=[0,1]\setminus \bigcup_1^\infty ]r_n-\varepsilon_n, r_n+\varepsilon_n[$. This is a closed set with finite measure; it contains no non-trivial interval because the sequence $(r_n)$ is dense in $[0,1]$, and $\vert K\vert$ is at least $1-\sum_1^\infty 2\varepsilon_n>0$.
A: We have:
$$ \prod_{n=2}^{+\infty}\left(1-\frac{1}{n^2}\right) = \frac{1}{2}, $$
so we may consider the unit interval $I_0=[0,1]$ and remove its middle fourth to get:
$$ I_1 = \left[0,\frac{3}{8}\right]\cup\left[\frac{5}{8},1\right].$$
Then, from any connected component of $I_n$ we remove a central open interval of relative measure $\frac{1}{(n+2)^2}$ to get $I_{n+1}$.
$$ \lim_{n\to +\infty} I_n = \bigcap_{n\in\mathbb{N}} I_n$$
is a fat Cantor set having Lebesgue measure $\frac{1}{2}$ and containing no interval, since the length of any connected component of $I_n$ is:
$$ \frac{1}{2^n}\prod_{k=1}^{n}\left(1-\frac{1}{(k+1)^2}\right) \xrightarrow[n\to +\infty]{} 0.$$
