Regular borel measures on metric spaces - MathOverflow

Regular borel measures on metric spaces

2010-04-22T10:35:59Z

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 http://en.wikipedia.org/wiki/Lp_space#Dense_subspaces 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.

Answer by Bill Johnson for Regular borel measures on metric spaces

2010-04-22T11:00:06Z

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

http://en.wikipedia.org/wiki/Radon_measure

Answer by Robin Chapman for Regular borel measures on metric spaces

2010-04-22T11:08:40Z

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.

Answer by Ian Morris for Regular borel measures on metric spaces

2010-04-22T11:10:54Z

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.)

Answer by Andrey Rekalo for Regular borel measures on metric spaces

2010-04-22T11:18:27Z

*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).

Answer by Someone for Regular borel measures on metric spaces

2010-04-22T11:21:05Z

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

Answer by M. A. Moshier for Regular borel measures on metric spaces

2010-04-26T20:45:14Z

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.

Answer by Ashutosh for Regular borel measures on metric spaces

2010-08-11T23:48:38Z

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.