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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)? –  t.b. Dec 19 '12 at 8:00
    
I didn't work it out in full detail, but I think that one can use the example 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. –  t.b. Dec 19 '12 at 10:27
    
@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! –  Giuseppe Negro Dec 20 '12 at 2:58
    
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 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. –  t.b. Dec 20 '12 at 4:31
    
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. –  t.b. Dec 20 '12 at 4:39
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1 Answer 1

Every finite measure on a topological space is outer regular if and only if it is inner regular, so compactness would be sufficient (and every finite measure on metric space is automatically both inner and outer regular). Another sufficient condition is that the space is the union of an increasing sequence of compact sets, which covers the case of $\mathbb{R}^n$.

One can look this up in: Aliprantis & Border 2006, "Infinite Dimensional Analysis" in section 12.1.

Edit: Lurker has pointed out that aliprantis and Border use the weaker notion of inner regularity of apprximating sets from below by closed sets. What is called inner regularity here, they call tightness. It is automatic for finite measures on Polish spaces but not all metric spaces.

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A finite measure on metric space is both inner and outer regular for Borel sets, but does this extend to all measurable sets? If yes, how would you do it? –  Thomas E. Dec 14 '12 at 9:43
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@ThomasE: The question was about Borel sets. Inner regularity is straightforward since every measurable set can be written as the union of a Borel set and a null-set. If a set can be approximated from within by compact sets, it can (in a Hausdorff space) also be approximated from within by closed sets. So applying the argument to the complement of a measurable set gives you outer regulariy. –  Michael Greinecker Dec 14 '12 at 9:57
    
Note that Aliprantis & Border define Borel measure differently than Evans-Gariepy. In the first, a Borel measure is a measure such that the measurable sets are exactly the Borel sets, and in the latter, Borel measure is an outer-measure so that the collection of measurable sets contains Borel sets. So the question was about a broader collection than just Borel sets. And thanks for providing an argument, I will try to follow it in a second :-) –  Thomas E. Dec 14 '12 at 10:08
    
@ThomasE. Giuseppe wrote in the first sentence "Let us agree to say that $\mu$ is a Radon measure on a metric space $X$ if it is a Borel measure which[...]" and the last sentence is "Notation and conventions in this book are a bit different from the ones of the present post." So I did not focus on the definition in the book. –  Michael Greinecker Dec 14 '12 at 10:10
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If you regard a Borel measure to be only on the Borel sets, as the OP has now clarified, then I can see your point. But in general when Borel measures are defined on a broader $\sigma$-algebra that contains the Borel sets, then Radon and Borel-regularity are two properties that neither imply the other. In that case I don't see immediately how completions are related, but I will do my best to look into it. Thanks for providing the clarification. –  Thomas E. Dec 14 '12 at 12:24
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