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Let $d\geq1$ and let $\mu$ be a regular Borel measure on $\mathbb{R}^d$. Show that $L^2(\mu)$ is separable.

I seem to have reduced the problem to showing that indicator functions of cubes with rational coordinates can be made arbitrarily close to indicator functions of bounded measurable sets, but I am unable to proceed further.

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Let $D$ be the collection of finites unions of cubes with rational coordinates. This collection is countable, and we will show that it is dense.

Due to the regularity of $\mu$, it suffices to approximate the indicator function of an open set. An open set is a countable union of open cubes with rational coordinates. Let $O$ be an open set of finite measure and let $\left(R_n\right)_{n\in\mathbb N}$ be a sequence of open cubes with rational coordinates such that $O=\bigcup_{n\in\mathbb N}R_n$. Let $O_N:=\bigcup_{n=1}^NR_n$. Then $O_N$ belongs to $D$ and the indicator of $O_N$ converges almost everywhere to that of $O$. Conclude by dominated convergence.

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  • $\begingroup$ But these open cubes with rational coordinates won't be disjoint, right? So we wouldn't have $\mu(O)=\Sigma\mu(C_i)$. How can we do the approximation then? $\endgroup$
    – Manan
    May 22, 2017 at 5:34
  • $\begingroup$ You can approximate by finite unions of cubes with rational coordinates. $\endgroup$ May 22, 2017 at 8:24
  • $\begingroup$ But how do I show that it will converge to the indicator function of the open set. Can you please elaborate? $\endgroup$
    – Manan
    May 22, 2017 at 8:28
  • $\begingroup$ I have edited. ${}{}$ $\endgroup$ May 22, 2017 at 8:40

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