Regular borel measures on metric spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:20:47Z http://mathoverflow.net/feeds/question/22174 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22174/regular-borel-measures-on-metric-spaces Regular borel measures on metric spaces Matthew Daws 2010-04-22T10:35:59Z 2012-12-02T18:46:54Z <p>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 <a href="http://en.wikipedia.org/wiki/Borel_hierarchy" rel="nofollow">Borel Hierarchy</a> and some transfinite induction. But, typically, I've lost the details.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Lp_space#Dense_subspaces" rel="nofollow">http://en.wikipedia.org/wiki/Lp_space#Dense_subspaces</a> 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.</p> <p>(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).</p> <p>To clarify, some definitions (thanks Bill!):</p> <ul> <li>I guess by "Borel" I mean: the sigma-algebra generated by the open sets.</li> <li>A measure $\mu$ is "outer regular" if <code>$\mu(B) = \inf\{\mu(U) : B\subseteq U \text{ is open}\}$</code> for any Borel B.</li> <li>A measure $\mu$ is "inner regular" if <code>$\mu(B) = \sup\{\mu(K) : B\supseteq K \text{ is compact}\}$</code> for any Borel B.</li> <li>A measure $\mu$ is "Radon" if it's inner regular and locally finite (that is, all points have a neighbourhood of finite measure).</li> </ul> <p>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.</p> http://mathoverflow.net/questions/22174/regular-borel-measures-on-metric-spaces/22175#22175 Answer by Bill Johnson for Regular borel measures on metric spaces Bill Johnson 2010-04-22T11:00:06Z 2010-04-22T11:00:06Z <p>I think you are asking about (finite) Radon measures, Matt. See</p> <p><a href="http://en.wikipedia.org/wiki/Radon_measure" rel="nofollow">http://en.wikipedia.org/wiki/Radon_measure</a></p> http://mathoverflow.net/questions/22174/regular-borel-measures-on-metric-spaces/22176#22176 Answer by Robin Chapman for Regular borel measures on metric spaces Robin Chapman 2010-04-22T11:08:40Z 2010-04-22T11:08:40Z <p>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.</p> http://mathoverflow.net/questions/22174/regular-borel-measures-on-metric-spaces/22177#22177 Answer by Ian Morris for Regular borel measures on metric spaces Ian Morris 2010-04-22T11:10:54Z 2010-04-22T11:24:17Z <p>The book <em>Probability measures on metric spaces</em> by K. R. Parthasarathy is my standard reference; it contains a large subset of the material in <em>Convergence of probability measures</em> 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.</p> <p>(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 <em>compact</em> subsets of $A$. Some other texts use "regular" to mean "regular and tight", so there is some room for confusion here.)</p> http://mathoverflow.net/questions/22174/regular-borel-measures-on-metric-spaces/22178#22178 Answer by Andrey Rekalo for Regular borel measures on metric spaces Andrey Rekalo 2010-04-22T11:18:27Z 2010-04-26T23:33:57Z <p>*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\ $F_ε$) &lt; 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\ $K_ε$) &lt; ε).*</p> <p>This result is proved on p. 70 in <em>"Measure Theory" vol. 2, Springer-Verlag, Berlin 2007, by V. I. Bogachev</em> (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> <p>P.S. Just a comment on the answer by Ian Morris: <em>tightness of a regular Borel measure on X may fail even if X is a separable metric space</em>. 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).</p> http://mathoverflow.net/questions/22174/regular-borel-measures-on-metric-spaces/22179#22179 Answer by Someone for Regular borel measures on metric spaces Someone 2010-04-22T11:21:05Z 2010-04-22T11:21:05Z <p>Every finite Borel measure defined on a <a href="http://en.wikipedia.org/wiki/Polish_space" rel="nofollow">Polish space</a> is regular, see e.g., Lemma 26.2 in <em>Heinz Bauer: Measure and Integration Theory</em>.</p> http://mathoverflow.net/questions/22174/regular-borel-measures-on-metric-spaces/22647#22647 Answer by M. A. Moshier for Regular borel measures on metric spaces M. A. Moshier 2010-04-26T20:45:14Z 2012-12-02T18:46:54Z <p>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.</p> http://mathoverflow.net/questions/22174/regular-borel-measures-on-metric-spaces/35298#35298 Answer by Ashutosh for Regular borel measures on metric spaces Ashutosh 2010-08-11T23:48:38Z 2010-08-11T23:48:38Z <p>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:</p> <p>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)).</p> <p>A proof can be found in Fremlin's Measure theory, volume 4, page 244.</p>