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Simon (Geometric Measure Theory) says:

If $X$ is a locally compact, separable metric space and $\mu(K) < \infty$ for all $K$ compact, then $X=\cup_{i=1}^\infty U_i$ where $U_i$ are open and $\mu(U_i) < \infty$.

I tried showing this by using separability $X=\cup_{i=1}^\infty B(x_i,r)$ for some $r>0$. However, I am unable to prove that each of the open balls $B(x_i,r)$ is contained in a finite union of compact sets (using the local compactness property). This would ensure that the measures of the open balls are finite and so the assertion will be ture. In $\mathbb{R}^n$ this can be done.

I am not sure if this is true in general metric spaces. Any help on how to prove this assertion generally is much appreciated.

jpv
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1 Answers1

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Hint: Since $X$ is locally compact and separable, there exists a countable basis $\{U_i\}_{i=1}^{\infty}$ for the topology such that $\overline{U}_i$ is compact for every $i$...

Tom
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  • Tom: In this case, I am not sure how to show that $X=\cup_{i=1}^\infty U_i$. – jpv Jun 25 '14 at 14:18
  • @jpv You're right! It needn't be the case. I'll edit my response (along the same lines, but without this oversight). – Tom Jun 25 '14 at 14:30
  • Tom: Thanks for the edit. I think this proves. I hope I will be able to get a proof of this somewhere. – jpv Jun 25 '14 at 15:34
  • Tom: I was unable to find a proof of this. Could you provide a reference? – jpv Jun 26 '14 at 06:24
  • I found the way: A separable metric space is Lindelof. Now, $X$ can be covered by a union of balls centered at every point of $X$. These balls are taken from the local compactness of $X$. This way, we get a countable cover of $X$ by the balls and hence by the compact sets. – jpv Jun 29 '14 at 10:14