$\newcommand{\vol}{\operatorname{vol}}$
Let $(M^n, g)$ be a Riemannian manifold. We define $d\vol(g)$ as a canonical volume form and then integrate. It is clear that $d\vol(g)$ depends on the metric $g$ or the geometry and so it is not invariant under the change of metric (e.g. conformal deformation).
There is also another object named measure $m$, by means of which we may define $dm= e^f \ d\vol(g)$ ; $f$ is a density function.
There are some cases in which for two different metrics $g_1$ and $g_2$, $dm_1$ is considered to be equal to $dm_2$ (by setting $f_1$ and $f_2$). This is in violation with the property mentioned above.
My question is that if $dm$ is supposed to be a generalized form of $d\vol(g)$, how can it be invariant under the change of metric?
Thanks in advance.
Since the determinant bundle is 1-dimensional, any non-vanishing $n$-form $\omega$ can be written $\omega = \phi\ d\mathrm{vol}$ for some non-vanishing smooth function $\phi$. If you fix an orientation then this function will be positive, so it can be written $\phi = e^f$ for some other function $f$.
This means that if we fix an $n$-form $dm$, it can be written in the form $e^f d\mathrm{vol}(g)$ for any metric $g$. However, the function $f$ depends on the metric in general, so as an assignment of $n$-forms to metrics, $e^f d\mathrm{vol}(g)$ is not invariant under change of metric.