is measure $m$ invariant under diffeomorohisms of a manifold?

is measure $m$ invariant under diffeomorohisms of a manifold?

$\newcommand{\vol}{\operatorname{vol}}$
Let $(M^n, g)$ be a Riemannian manifold. We can define $d\vol(g)$ as a
canonical volume form and do integration by that. It is clear that
$d\vol(g)$ depends on metric $g$ or our geometry and so it is not
invariant under change of metric (active diffeomorphisms).
On the other hand there is another object, named measure $m$, which by
means of that we may define $dm= e^f \, d\vol(g)$ that $f$ is a density
function. So it seems we may use $dm$ for integration instead of
$d\vol(g)$.
It has been seen in some cases $dm_1$ has been considered equal to $dm_2$
for two different metric $g_1$ and $g_2$, while this is in violation of
property of $d\vol(g)$ which has been mentioned at the first paragraph,
i.e: $d\vol(g_1)\neq d\vol(g_2)$ .my question is if $dm$ is supposed to be
a generalized form of $d\vol(g)$, why their "invariance under the change
of metric (active diffeomorphism)" is different?
Thanks in advance.