Let $M$ be a smooth oriented manifold. Does there exist a smooth measure $m$ on $M$ which is not induced by the volume form of some Riemannian metric $g$ on $M$? I would say that the set of volume forms induced by Riemannian metrics is strictly contained in the set of all smooth measures on $M$...My interest would be to have some criteria for deciding whether a given measure on $M$ is induced by a Riemannian metric or not

The only possible meaning for smoothness that comes to mind is the following: A measure on ${\mathbb R}^n$ is smooth, if it has a smooth density against the Lebesguemeasure. On a manifold, a measure is smooth if it transforms to smooth measures on every smooth chart. The question, whether a given smooth measure comes from a metric is equivalent to the question whether its density has a zero or not. If it has no zero, simply choose any Riemannian metric. Then your given measure has a nowhere vanishing density against the measure coming from the metric. Simply multiply the metric with a suitable power of the density to get a metric that induces the given measure. Since RadonNikodym densities are uniquely determined, this is an if and only if criterion. 


I think the answer to your question is given at least within the class of k\"ahlerian manifolds. I would refer you to standard texts of complex geometry for the preliminaries. But in the case of k\"hlerian manifolds, any volume form, that is, a measure, whose RadonNikodym density with respect to a fixed reference metric $g$ is smooth enough, and has the same total volume as $d Vol_g$ can be realised as the volume form of a k\"ahlerian metric $g^{\phi}$ in the same cohomology class, ie. $g_{i \bar{j}}^{\phi} = g_{i \bar{j}} + \partial_{i} \partial_{\bar{j}}$. This was Calabi's volume conjecture. Also, you can realise an reasonably arbitrary volume forms by conformal change of the metric. However, in the conformal case the total volume can change. 

