Wikipedia defines a regular measure as a measure which, given on a topological space, satisfies: $$\mu(A)=\sup\{\mu(F):F\subseteq A,F\text{ compact and measurable}\},$$ and that is inner regular, or: $$\mu(A)=\inf\{\mu(U):U\supseteq A,U\text{ open and measurable}\},$$ and that is inner regular. Both conditions must hold for all measurable subsets $A\subseteq X$, $X$ being the space. My thesis professor told me a regular measure on a manifold is the measure associated to a volume form, i.e.: $$\mu(A)=\int_A\omega,$$ $\omega$ being a volume form. What is the relationship between these notions? Does either imply another? Wikipedia lists examples of measures that are inner but not outer regular and viceversa: are there conditions where inner and outer regularity coincide?

  • $\begingroup$ Lebesgue measure on $R^n$ is both inner and outer regular. I have seen different definitions of inner regular, not all equivalent. $\endgroup$ – DanielWainfleet Sep 27 '15 at 22:45
  • $\begingroup$ Point masses are regular in the inner/outer sense but are not induced by volume forms. I think the best measure-theoretic characterization of "coming from a volume form" is something like "absolutely continuous with smooth Radon-Nikodym derivative". This needs a choice of volume form to compare to, but is independent of the choice. $\endgroup$ – Anthony Carapetis Sep 28 '15 at 3:45
  • $\begingroup$ @anthonycarapetis "point masses [...] do not come from a volume form", well, unless you're on a 0-manifold, right :)? I can see how, in $\mathbb R^n$, form measures have a smooth density w.r.t. the Lebesgue measure, but on a manifold which measure do I use for absolute continuity? And can I prove that, at least in $\mathbb R^n$, smooth density iff form measure? $\endgroup$ – MickG Sep 28 '15 at 7:36
  • $\begingroup$ Also, density w.r.t. Lebesgue which is inner and outer regular implies the measure with density also is, doesn't it? $\endgroup$ – MickG Sep 28 '15 at 7:46

Volume forms produce measures that are inner and outer regular, but the converse is not true: if $n>0$ then point masses are (inner/outer) regular measures that are not induced by volume forms. (If $n=0$ then you're talking about a discrete space, so every measure is regular in both senses.)

Let's assume $M$ is orientable so that volume forms exist, and refer to the measures induced by them as smooth measures. The bundle $\Omega^n(M)$ of top forms is thus trivial and rank-1; i.e. top forms $\simeq$ smooth functions. Thus any two volume forms ($\simeq$ non-vanishing functions) $\omega, \sigma$ are related by $\omega = \rho \sigma$ for some non-vanishing smooth function $\rho$. Thus smooth measures have non-vanishing smooth densities with respect to any other smooth measure. This also implies $\int_X \omega = 0 \iff \int_X \sigma = 0$, so we don't need a distinguished volume form to define absolute continuity of an arbitrary measure: $\mu$ is absolutely continuous if $\int_X \omega = 0 \implies \mu(X) = 0$ for some/every (doesn't matter!) $\omega$.

If $\mu$ is absolutely continuous, then the Radon-Nikodym theorem provides a measurable density $f$ s.t. $\mu = f \omega$. If this $f$ is smooth and positive then we can interpret $f \omega$ as a multiple of a smooth function and a volume form, so $\mu$ is smooth.

This isn't a very powerful observation in terms of recognizing smooth measures: after all, it characterizes smooth measures in terms of... smooth measures. But it does tell you a lot about the structure of the set of all smooth measures - once you know one, you know all of them. I suspect the multiplicity of smooth structures makes it impossible to find a general characterization using only measure-theoretic and topological data.

| cite | improve this answer | |
  • $\begingroup$ How do I show smooth measures are inner/outer regular? $\endgroup$ – MickG Sep 28 '15 at 8:34
  • 1
    $\begingroup$ In local coordinates we have $\mu = f \lambda$ for $\lambda$ the Lebesgue measure, so with the right hypotheses it follows very easily from the regularity of $\lambda$. Falls out in a few lines if you assume $f$ is bounded, anyway - I'm not actually sure whether this kind of assumption is necessary. Might not be true in general. I'll think about it. $\endgroup$ – Anthony Carapetis Sep 28 '15 at 9:19
  • $\begingroup$ It certainly works out fine for precompact test sets. I think $\sigma$-compactness should then be enough for any continuous multiple of a regular ($\sigma$-additive) measure to be regular. Maybe have a look through some textbooks for results of this nature or try to put the proof together yourself. $\endgroup$ – Anthony Carapetis Sep 28 '15 at 9:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.