# Measure-dense subset of measurable set $E$

Let $$E\subset\mathbb{R}$$ be measurable with $$0. Construct a measurable set $$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?

• The set of all Lebesgue points of $\chi_E$ work?
– r9m
Commented Apr 30, 2020 at 0:02

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.