# Verify that $R^d$ with the usual Lebesgue measure is separable

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.

• Any measurable set of finite measure in $R$ can be approximated by a finite disjoint union of half closed intervals. But we need to approximate $E$ by just one subset , we can not take the "union" – J.Guo Dec 5 '18 at 8:59
• The countable dense set I am suggesting is the collection of all sets which can be written as finite unions of sets of the type $[a_1,b_n)\times [a_2,b_2)\times\cdots \times [a_d,b_d)$ with $a_i$'s and $b_i$'s rational. – Kavi Rama Murthy Dec 5 '18 at 9:20
• @J.Guo In you argument for one-dimensional case you are considering just single intervals with rational end points. Why not consider finite unions of these. In that case $(0,1)\cup (2,3)$ is already in the countable family. – Kavi Rama Murthy Dec 5 '18 at 9:23