# Two notions of regularity for measures

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?

• Lebesgue measure on $R^n$ is both inner and outer regular. I have seen different definitions of inner regular, not all equivalent. – DanielWainfleet Sep 27 '15 at 22:45
• 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. – Anthony Carapetis Sep 28 '15 at 3:45
• @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? – MickG Sep 28 '15 at 7:36
• Also, density w.r.t. Lebesgue which is inner and outer regular implies the measure with density also is, doesn't it? – 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.
• 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. – Anthony Carapetis Sep 28 '15 at 9:19
• 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. – Anthony Carapetis Sep 28 '15 at 9:33