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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.

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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.

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  • $\begingroup$ 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" $\endgroup$ – J.Guo Dec 5 '18 at 8:59
  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Dec 5 '18 at 9:20
  • $\begingroup$ @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. $\endgroup$ – Kavi Rama Murthy Dec 5 '18 at 9:23

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