Inner regular measures and tight measures

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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

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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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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.

$\omega = \{0,1,2,3,4,5,...\} \;\;$ ()

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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