A measure space $(X,\mu)$ is sepearable if there is a countable family of measurable subsets $\{E_k \}_{k=1}^\infty $ so that if $E$ is any measurable set of finite measure , then $$\mu(E \triangle E_{n_k}) \to 0 \,\,\,\,\,\,\,as \,k\to0$$ for an appropriate subsequence $\{n_k \}$ which depends on $E$ . Verify that $R^d$ with the usual Lebesgue measure is separable.
I want to show that $E_k$ is a collection of open subset centered at rational numbers with rational radius. However , let $R^d =R$ , and $E=(0,1) \cup (2,3)$ , the collection of $E_k$ defines above fails. Then I want to let $M$ denote the collection of all the subset of countable union of $E_k$ , But it is obvious $M$ is not countable.
The following basic approximation results will give the answer.
1) Any set $E$ of finite measure in $\mathbb R^{d}$ can be approximated by a finite disjoint union of measurable rectangles.
2) Any measurable set of finite measure in $\mathbb R$ can be approximated by a finite disjoint union of half- closed intervals.
3) Any half closed interval can be approximated by a half closed interval with rational end points.
Here approximating $A$ by $B$ means making $m(A\Delta B)$ small.