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I have the following claim which is used in several proofs concerning Borel or Radon measures, but I don't see its triviality. Can someone help me?

We have a $\sigma$-compact space $(X,d)$ and want to show that there exist a sequence of compact non-empty subsets $X_n$ with

$X=\cup_n X_n $ and

$dist(X\setminus X_{n+1} , X_n) > 0$.

tubmaster
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  • Sorry, I'm a bit confused. Are you asking for a proof of the monotonicity of measures? – Matthew Cassell Nov 29 '16 at 13:09
  • Thanks for your answer! I am asking for a formal proof. I think that here has not so much to do with monotonicity of measures... – tubmaster Nov 29 '16 at 14:31
  • That can’t be right: $\operatorname{dist}(X\setminus X_{n-1},X_n)>0$ implies that $X_n\subseteq X_{n-1}$ and hence that $X=\bigcup_nX_n=X_0$ is compact. – Brian M. Scott Nov 29 '16 at 20:27
  • SORRYYYY, I mistyped, EDIT now ;) – tubmaster Nov 29 '16 at 21:55
  • That requires that $X$ is also locally compact. And if $X$ is locally compact and $\sigma$-compact, take a sequence of compact sets such that each is contained in the interior of the next. – Daniel Fischer Nov 29 '16 at 21:58
  • Ok well, that cant be the answer because in our notes they want to deduce afterwards that $X_n$ is contained in the interior of $X_{n+1}$ (with the convention $dist({},A)=\infty$ for nonempty set $A$. – tubmaster Nov 29 '16 at 22:08
  • And I think locally compactness is really not required – tubmaster Nov 29 '16 at 22:09
  • For a compact $K$, $\operatorname{dist}(X\setminus A, K) > 0$ is equivalent to $K \subset \overset{\Large\circ}{A}$. And since $X_{n+1}$ is compact, it follows that all $x\in X_n$ have a compact neighbourhood. Since $X$ is the union of the $X_n$, it follows that $X$ is locally compact. – Daniel Fischer Nov 29 '16 at 22:12
  • And you can find a counterexample here: http://math.stackexchange.com/questions/49707/sigma-compact-and-locally-compact-metric-space –  Nov 29 '16 at 22:14
  • sorry but I dont get your point... Do you want to say that local compactness comes automatically? And I think that in the given counterexample, the requirememnt are others – tubmaster Nov 29 '16 at 22:40
  • The point is: such a sequence exists if and only if $X$ is locally compact and $\sigma$-compact. If you have only $\sigma$-compactness, such a sequence need not exist. – Daniel Fischer Nov 29 '16 at 23:25
  • ok well, I looked up our definition of sigma compactness and it includes two things: locally compactness AND countable union of compact sets – tubmaster Nov 30 '16 at 08:23

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