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!

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:
$$ \frac{1}{2^n}\prod_{k=1}^{n}\left(1-\frac{1}{(k+1)^2}\right) \xrightarrow[n\to +\infty]{} 0.$$
