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When teaching Measure Theory last year, I convinced myself that a finite measure defined on the Borel subsets of a (compact; separable complete?) metric space was automatically regular. I used the Borel Hierarchy and some transfinite induction. But, typically, I've lost the details.

So: is this true? Are related questions true? What are some good sources for this sort of questions? As motivation, a student pointed me to where it's claimed (without reference) that (up to a slight change of definition) the result is true for finite Borel measures on any metric space.

(I'm normally only interested in Locally Compact Hausdorff spaces, for which, e.g. Rudin's "Real and Complex Analysis" answers such questions to my satisfaction. But here I'm asking more about metric spaces).

To clarify, some definitions (thanks Bill!):

  • I guess by "Borel" I mean: the sigma-algebra generated by the open sets.
  • A measure $\mu$ is "outer regular" if $\mu(B) = \inf\{\mu(U) : B\subseteq U \text{ is open}\}$ for any Borel B.
  • A measure $\mu$ is "inner regular" if $\mu(B) = \sup\{\mu(K) : B\supseteq K \text{ is compact}\}$ for any Borel B.
  • A measure $\mu$ is "Radon" if it's inner regular and locally finite (that is, all points have a neighbourhood of finite measure).

So I don't think I'm quite interested in Radon measures (well, I am, but that doesn't completely answer my question): in particular, the original link to Wikipedia (about L^p spaces) seems to claim that any finite Borel measure on a metric space is automatically outer regular, and inner regular in the weaker sense with K being only closed.

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up vote 9 down vote accepted

The book Probability measures on metric spaces by K. R. Parthasarathy is my standard reference; it contains a large subset of the material in Convergence of probability measures by Billingsley, but is much cheaper! Parthasarathy shows that every finite Borel measure on a metric space is regular (p.27), and every finite Borel measure on a complete separable metric space, or on any Borel subset thereof, is tight (p.29). Tightness tends to fail when separability is removed, although I don't know any examples offhand.

(Definitions used in Parthasarathy's book: $\mu$ is regular if for every measurable set $A$, $\mu(A)$ equals the supremum of the measures of closed subsets of $A$ and the infimum of open supersets of $A$. We call $\mu$ tight if $\mu(A)$ is always equal to the supremum of the measures of compact subsets of $A$. Some other texts use "regular" to mean "regular and tight", so there is some room for confusion here.)

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I'm accepting this, as it was first, and Parthasarathy's book (once I got it out of our library's stack!) is very nice. – Matthew Daws Apr 22 '10 at 11:32

Every discrete space is a metric space. If we consider a measurable cardinal $\kappa$ as a discrete space, then it has an ultrafilter $\mathcal{F}$ in which the intersection of fewer than $\kappa$ elements of $\mathcal{F}$ lies in $\mathcal{F}$. There is a measure on $\kappa$ such that $\mu(A)=1$ or $0$ according to wheher or not $A\in\mathcal{F}$. For every finite set $\mu(A)=0$. But every compact set is finite so $\mu(A)=0$ for every compact $A$. But $\mu(\kappa)=1$ and so $\mu$ is not inner regular.

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Nice! Although one might quibble about the existence of measurable cardinals... – Matthew Daws Apr 22 '10 at 11:35
Of course, it's debatable whether measurable cardinals exist, but this example at least shows that some hypothesis is needed to prove this version of regularity. – Robin Chapman Apr 22 '10 at 11:41

Here's a reason why it appears hard to come up with an example of a non-tight probability measure on a complete metric space:

Theorem: Let X be a complete metric space. Denote by w(X) the smallest cardinality of a basis for the topology on X. Then there is a non-tight probability measure on the class of borel subsets of X iff w(X) is a measurable cardinal (i.e. there is a non-atomic measure on the power set of w(X)).

A proof can be found in Fremlin's Measure theory, volume 4, page 244.

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Let X be a metric space. Then every Borel measure μ on X is regular (i.e. for every Borel set B and every ε > 0, there exists a closed set $F_ε$ such that $F_ε ⊂ B$ and μ(B\ $F_ε$) < ε). If X is complete and separable, then the measure μ is Radon (i.e. for every Borel set B and ε > 0, there exists a compact set $K_ε$ ⊂ B such that μ(B\ $K_ε$) < ε).

This result is proved on p. 70 in "Measure Theory" vol. 2, Springer-Verlag, Berlin 2007, by V. I. Bogachev (Theorem 7.1.7.) An example of a regular Borel measure which is not tight is provided on the same page (Example 7.1.6).

P.S. Just a comment on the answer by Ian Morris: tightness of a regular Borel measure on X may fail even if X is a separable metric space. For example, we may take a restriction of the standard Lebesgue measure to a nonmeasurable subset of the interval $[0, 1]$ with zero inner measure and unit outer measure (endowed with the usual metric).

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You must mean another regular, or what happens to the measurable cardinal example? Inner regular by closed, instead of compact? – Henno Brandsma Apr 26 '10 at 21:06
Good point. According to Bogachev, a Borel measure $\mu$ is called - inner regular if $\mu(B)=sup\{\mu(K): K\subset B, K \mbox{is closed}\}$ for any Borel $B$, and - tight if $\mu(B)=sup\{\mu(K): K\subset B, K \mbox{is compact}\}$ for any Borel $B$. – Andrey Rekalo Apr 26 '10 at 21:45

I think you are asking about (finite) Radon measures, Matt. See

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Every finite Borel measure defined on a Polish space is regular, see e.g., Lemma 26.2 in Heinz Bauer: Measure and Integration Theory.

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A closely related question: I am reading a thesis that claims that the property "All measures on metric spaces are $\tau$-smooth" is independent of ZFC. A measure on Borel sets is $\tau$-smooth iff for any directed family of opens $U_t$, we have $\mu(\bigcup U_t) = \sup_t \mu(U_t)$. Robin Chapman's answer above tells us why the claim is plausible (non-inner regular measures exist when measurable cardinals do), but the author does not give a usable citation (only an out of print textbook, instead of a primary source). Does anyone have a pointer into the literature.

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You may have more luck getting an answer if you post this as a separate question of your own, with a link back to Robin Cooper's answer. – Yemon Choi Apr 26 '10 at 20:50
This would be better as a question so it gets all the attention it deserves. – François G. Dorais Apr 26 '10 at 20:50
@Choi: Chapman, not Cooper. – Henno Brandsma Apr 26 '10 at 21:08
@Henno: Sorry, I just copied the text above without thinking (am currently a couple of coffees short of functioning properly) – Yemon Choi Apr 26 '10 at 21:17

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