This article explicitly constructs a Cantor set $C\subseteq[0,1]$ of measure $1/2$. The same construction can clearly be used to build a Cantor set $C(a,b)\subseteq[a,b]$ of measure $(b-a)/2$ for any $a,b\in\Bbb R$ with $a<b$.
Let $D_0=C_0=C(0,1)$. Let $\{(a_0(n),b_0(n)):n\in\Bbb N\}$ be an enumeration of the maximal open subintervals of $[0,1]\setminus D_0$, let $$C_1=\bigcup_{n\in\Bbb N}C\Big(a_0(n),b_0(n)\Big)\;,$$ and let $D_1=D_0\cup C_1$.
In general, given $C_k$ and $D_k$ for some $k\in\Bbb N$, let $\{(a_k(n),b_k(n):n\in\Bbb N\}$ be an enumeration of the maximal open subintervals of $[0,1]\setminus D_k$, let $$C_{k+1}=\bigcup_{n\in\Bbb N}C\Big(a_k(n),b_k(n)\Big)\;,$$ and let $D_{k+1}=D_k\cup C_{k+1}$.
Now let $A=\bigcup_{k\in\Bbb N}C_{2k}$ and $B=\bigcup_{k\in\Bbb N}C_{2k+1}$. $A$ and $B$ are both measure dense in $[0,1]$, since any non-empty open set in $[0,1]$ contains intervals $\big(a_{2k}(n),b_{2k}(n)\big)$ and $\big(a_{2k+1}(m),b_{2k+1}(m)\big)$ for some $k,m,n\in\Bbb N$. Moreover, $|A\cap B|$ is countable and therefore a null set so for every open $U\subseteq[0,1]$ we have $$m(U\setminus A)\ge m\big(U\cap (B\setminus A)\big)=m(U\cap B)>0\;,$$ and hence $[0,1]\setminus A$ is measure dense in $[0,1]$ as well.
Now just extend this to $\Bbb R$ by replacing $A$ by $\bigcup\limits_{n\in\Bbb Z}(A+n)$, the union of its integer translates.
Because the Cantor set includes numbers which are not the endpoints of any intervals removed. For example, the number $\frac{1}{4}$ (0.02020202020... in ternary) belongs to the Cantor set, but is not an endpoint of any interval removed.
Best Answer
Let $C$ be the Cantor set and $X=[0,1] \setminus C$ its complement in the unit interval. Then $m(X)=m([0,1])-m(C)=1$, so the complement has measure $1$. Furthermore the Cantor set is nowhere dense, so its complement must be dense and thus $\overline{X}=[0,1]$.