Find a nowhere dense closed subset of $\mathbb{R}$ with the following property Let $m$ be the Lebesgue measure on $\mathbb{R}$. Prove that for every $\epsilon>0$, there exists a nowhere dense closed subset $F$ of $\mathbb{R}$ such that for every interval $(a,b)$ of $\mathbb{R}$, $m(F \cap (a,b)) > b-a-\epsilon$.
To be honest I'm not really sure where to start with this. The statement of the problem reminds me of Lusin's theorem, but I don't know how to apply that here. I've also thought that I might need some version of the Cantor set.

Hint: It would be enough to get a nowhere dense closed set $F$ such that $m(\mathbb{R}\setminus F)<\epsilon$.  As a starting point, think about how to do this if you were on $[0,1]$ instead of $\mathbb{R}$: how could you construct such an $F\subset[0,1]$ with $m([0,1]\setminus F)<\epsilon$?  Then think about how you can combine such sets to work on all of $\mathbb{R}$.
Further hint for the last step:

 If you choose a closed nowhere dense subset $F_n\subset[n,n+1]$ for each $n\in\mathbb{Z}$, then $F=\bigcup_{n\in\mathbb{Z}}F_n$ will still be closed and nowhere dense.

