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On a Hausdorff topological space $X$ with a sigma algebra $ Σ$ at least as fine as the Borel sigma algebra,

  • a measure $\mu$ is said to be inner regular, if for every set $A \in Σ$, $\mu(A) = \sup \{ \mu(K) | \text{ compact }K \subseteq A \}$.
  • a measure is said to be tight, if for all $ε > 0$, there is some compact subset $K$ of $X$ such that $μ(X - K) < ε$.

Wikipedia says that a measure is inner regular iff it is tight. I was wondering why? Is it still true when the topological space $X$ is not necessarily Hausdorff? References are also appreciated!

Thanks and regards!

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2 Answers 2

up vote 3 down vote accepted

Define $\: X = \omega \:$ and $\: \mathcal{T} = 2^X \:$ and $\: \Sigma = 2^X \:$. $\;\;$ $\langle X,\mathcal{T}\hspace{.01 in}\rangle$ is a Hausdorff topological space.
Define $\: \mu_c : \Sigma \to [0,\scriptsize+\normalsize\infty] \:$ to be the counting measure.

$\operatorname{Borel}(\langle X,\mathcal{T}\hspace{.01 in}\rangle) = \operatorname{Borel}\left(\left\langle X,2^X\right\rangle\right) = 2^X = \Sigma \;\;\;$ and $\mu_c$ is inner regular but not tight.



$\omega = \{0,1,2,3,4,5,...\} \;\;$ (http://en.wikipedia.org/wiki/Ordinal_number)


For all finite subsets $S$ of $\omega$, $S$ is compact, which gives $\mu(S) \leq \operatorname{sup}(\{\mu_c(K) : (K \text{ is compact) and } (K\subseteq S)\}) = \mu(S) \;\;$.

For all compact subsets $K$ of $X$, $K$ is finite, so $\: X-K \:$ is infinite, so $\: 0 < 1 < \scriptsize+\normalsize\infty = \mu_c(X-K) \;\;$.


Yes. $\:$ azarel mentioned one direction, I don't know anything about the other.

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Thanks! (1) What is $\omega$ in $X=\omega$? (2) Why $μ_c$ is inner regular but not tight? (3) So are you saying even when $X$ is Hausdorff, it is not true that a measure is inner regular iff it is tight? Is only one direction true or both directions are false? –  Tim Jan 28 '12 at 7:04
    
If $\mu$ is a finite measure then inner regular clearly implies tight (you don't need the Hausdorff hypothesis) –  azarel Jan 28 '12 at 7:20
    
Thanks! If Wikipedia is wrong on "a measure is inner regular iff it is tight" for a Hausdorff topological space, what would you suggest to modify the conditions on the topological space and/or the measure so that it is true? –  Tim Jan 28 '12 at 7:49
    
@azarel: Thanks!(1) Under what some conditions, will tight implies inner regular? (2) Do you have some idea what Wiki is actually trying to say instead, since it is not correct? –  Tim Jan 28 '12 at 15:00
    
Well, $\omega$ is a certain ordinal number. Set theorists identify this with a certain set. So saying $X = \omega$ is an obscure way to say $X = \mathbb N$. Since Ricky used no special properties of $\omega$ or of $\mathbb N$, it may have been preferable to say that $X$ is any countably infinite set. –  GEdgar Aug 17 '12 at 14:48

Wikipedia is wrong because its definition of a tight measure as "for all ε>0, there is some compact subset K of X such that μ(X−K)<ε" fails to note that the measure must be a probability (or totally finite) measure. (Parthasarathy, 1967, Probability Measures on Metric Spaces, p. 26, fn 1.) Otherwise, a measure μ is said to be tight, if for every set A∈Σ, μ(A)=sup{μ(K)| compact K⊆A} and inner regular, if for every set A∈Σ, μ(A)=sup{μ(F)| F closed and F⊆A}.

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... and in non-Hausdorff spaces, a compact set need not be closed?? –  GEdgar Aug 17 '12 at 14:51

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