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Let us agree to say that $\mu$ is a Radon measure on a metric space $X$ if it is a Borel measure which is finite on compact subsets and is such that:

  • Every measurable subset $A$ is outer regular, meaning that $$\mu(A)=\inf\{\mu(V)\ |\ A\subset V,\ V\ \text{is open}\};$$
  • Every open subset $U$ is inner regular, meaning that $$\mu(U)=\sup\{\mu(K)\ |\ K\subset U,\ K\ \text{is compact}\}.$$

The present question regards inner regularity. Indeed, as I read in Evans-Gariepy's Measure theory and fine properties of functions, Theorem 4, Chapter 1 (*), if $X=\mathbb{R}^n$ then every measurable subset is automatically inner regular. Is this a property of $\mathbb{R}^n$ alone? Formally:

Question. Which metric spaces have the property that for any Radon measure every measurable subset is inner regular?


(*) Notation and conventions in this book are a bit different from the ones of the present post.

EDIT. Specifically, in Evans-Gariepy's book a measure is a extended-real valued set function which is monotone and subadditive (usually, this is called a outer measure). A measurable set is one which satisfies Caratheodory's criterion: $$E\ \text{is measurable} \iff \forall T\subset X,\ \mu(T)=\mu(T\cap E)+ \mu(T\cap E^c).$$ A Radon measure is a (outer) measure which is:

  1. Borel regular, meaning that every Borel set is measurable and every set (even nonmeasurable ones) is contained in a Borel set of the same (outer) measure;
  2. Finite on compact subsets.

The aforementioned Theorem 4 of Chapter 1 says that, given a Radon (outer) measure on $\mathbb{R}^n$, every set (measurable or not) is outer regular and every measurable set is inner regular.

Remark 1. If $\mu$ is a Borel regular measure on $X$ such that every Borel set is inner regular, then every measurable subset of finite measure is inner regular. Indeed, if $M\subset X$ is measurable and has finite measure, then by applying two times the Borel regularity property we can get a Borel set $M'$ which is contained in $M$ and has the same measure as $M$.

Remark 2. In particular, if $X$ is locally compact and is expanding union of compact sets, as in Micheal's kind answer below, and if $\mu$ is a Borel regular Radon (outer) measure, I believe that every (Caratheodory) measurable subset is inner regular. Indeed let $M\subset X$ be measurable. If $\mu(M)<\infty$ we are done. If $\mu(M)=\infty$ then we can write it as an expanding union of sets of finite measure: $M=\cup_1^\infty M_j$. Every $M_j$ contains a compact $K_j$ such that $\mu(K_j)\ge \mu(M_j)-1$. Letting $j\to \infty$, $\mu(M_j)\to \mu(M)=\infty$ and so $\mu(K_j)\to \infty$ too. This proves the claim.

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    What kind of spaces are you interested in? Do you care about *large* spaces (not $\sigma$-compact)? Would you mind adding completeness? Have you considered the example of Lebesgue measure times counting measure on the reals times the discrete reals (it has two incarnations: one is inner regular but not outer regular and the other is outer regular but not inner regular)?2012-12-19
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    I didn't work it out in full detail, but I think that one can use [the example](http://math.stackexchange.com/a/215261/5363) I alluded to above to show that on every non-separable complete metric space without isolated points there is an outer measure as in Evans-Gariepy, but that the associated measure (from Carathéodory) will not be tight while satisfying all your conditions. On the other hand, one can always construct a tight version of it which will fail to be outer regular.2012-12-19
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    @t.b.: Actually I am exploring some measure theory because of a course in fractal geometry which I am attending. So the spaces I am most interested in usually are $\sigma$-compact. For those spaces, if I understand well Micheal's answer below, all Radon measures are automatically tight (meaning that any measurable set is inner regular). By all means it would be nice to construct your counterexample so as to put a period to this question. I'll think about it. Thank you!2012-12-20
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    I see. Here's the idea: The example of a non-tight Radon measure on an uncountable disjoint union of $\mathbb{R}$ I linked you to can be modified to use the Cantor set with the "coin-flipping measure" instead of $\mathbb{R}$. Thus the task is essentially reduced to find a closed subspace of a metric space which is homeomorphic to an uncountable disjoint union of Cantor sets. This can be done using [this construction](http://math.stackexchange.com/q/68396) and a locally finite disjoint family of open sets (and some effort). I'm not so sure if the actual construction is all that enlightening.2012-12-20
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    You understand Michael's answer correctly. You can find proofs in many places, for example Theorem 3.2 in Parthasarathy's *Probability Measures on Metric Spaces* or Theorem 17.11 of Kechris's *Classical Descriptive Set Theory*.2012-12-20
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    This concept of Radon topological space is exactly the same thing this question is about: https://en.wikipedia.org/wiki/Radon_space2016-02-25

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