Measure-dense subset of measurable set $E$

Question

Let $E\subset\mathbb{R}$ be measurable with $0<m(E)<\infty$. Construct a measurable set $G\subset E$ such that $m(G)=m(E)$, and the intersection of $E$ with any open interval centered at a point of $G$ has positive measure. That is, $$ m(E\cap(x-\alpha,x+\alpha))>0 $$ for all $x\in G$ and all $\alpha>0$.

So I have seen variations of this problem all over the place. However, they all assume that we are starting with the interval $[0,1]$, or some other closed interval. My question is, how can we do this for an arbitrary measurable set $E$? What if $E$ contains no intervals (like a fat cantor set)? I'd like to do something with rational intervals (as in the case with basically every other version of this problem I can find), but we cannot guarantee that any rational interval even lies in $E$. Also, how can we get $m(G)=m(E)$? This sounds like it is going to be a very specific construction. Any advice?

Answer 1

Let $E$ be any measurable subset of $\mathbb R$. Let $\mathcal I$ be the set of all rational open intervals $I$ such that $m(E\cap I)=0$, and let $G=E\setminus\bigcup\mathcal I$. Then $G$ is a measurable subset of $E$, and $m(E\setminus G)=0$, and the intersection of $E$ with any open set containing a point of $G$ has positive measure.