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?
 A: 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.
