If we equip a manifold with an inner product (i.e. we have a Riemannian Manifold) then we get a canonical volume form on that manifold (please mentally insert the prefix "pseudo" into my question whenever you feel it necessary).
Whenever we have a submanifold we can pull back the metric along the inclusion map to get an inner product on the submanifold. We then get a volume form on the submanifold.
I'm wondering if there are any converse results to this. I'm looking for some result of the form:
Suppose $\mathcal{M}$ is a manifold with a volume form assigned to every submanifold, such that these volume forms satisfy (some consistency condition). Then $\mathcal{M}$ is Riemannian, and the volume forms arise as above.
My only progress so far is to notice that you can define some kind of "norm" on such an $\mathcal{M}$ by, for any vector $v$, picking a curve $\gamma:(-1,1)\rightarrow\mathcal{M}$ with $\gamma'(0)=v$, letting $\mu$ be the measure on the submanifold which is the image of this curve and then defining $$\lVert v \rVert=\lim_{\varepsilon\rightarrow0}\frac{\mu(\gamma((0,\varepsilon))}{\varepsilon}.$$ But I don't see what properties I need to put on the volume forms in order to get the triangle inequality or the parallelogram law.