# Change of variables between spaces with different dimensionality in using the Riemann integral

On wikipedia (https://en.wikipedia.org/wiki/Integration_by_substitution) I find that, $$\int_{\phi(U)} f(v)dv = \int_U f(\phi(u))|\det(D\phi(u))|du$$ for the Riemann integral under a number of conditions. I'm looking for a more general formulation, where $f:R^m\rightarrow R$, $M$ is a mapping $M:V \rightarrow U$, where $V\subset R^n$ is open and $U \subset R^m$ is some n dimensional surface, so that: $$\int_{V} f(M(v))dv = \int_U f(u) \cdot \text{something} \cdot du$$ Already in the same wikipedia article, conditions for doing this using Lebesgue integration is given, but I want to know when and how this can be done using the Riemann integral, because I'm writing a short text that should be easy to understand, so I don't want to make it too abstract. A literature reference would be appreciated.

• First of all you need to ask the question "What measure am I integrating with respect to on U." Note, $U$ is an $n$-dimensional surface living in $R^m$; we know that we don't want to integrate with respect to $m$ dimensional measure on $U$ because if $m>n$ then we would just trivially get $0$. This begs the question of whether there exists a "natural" analog to $n$ dimensional measure for surfaces imbedded in larger dimensional spaces. In the case of arbitrary Riemannian manifolds the answer is yes and the volume form can be defined via the square root of the determinant of the metric tensor. – Shalop Aug 8 '16 at 9:11
• Indeed, the question is "What measure am I integrating with respect to on U." And I appreciate @Shalops answer "In the case of arbitrary Riemannian manifolds the answer is yes and the volume form can be defined via the square root of the determinant of the metric tensor". I believe that the left hand side of the equation should define a specific metric tensor on the manifold. Can you develop a bit, or give a literature reference? – Axel Aug 8 '16 at 9:36
• I don't know any sources from my head because I don't have any formal training in differential geometry either. But you can see Wikipedia (en.wikipedia.org/wiki/…). Maybe someone can give a better source. – Shalop Aug 8 '16 at 9:49
• For a surface embedded in Euclidean Space, which is paramatrized by the map $M$ (as you call it), the metric tensor is given by $g = DM^tDM$. You can check that this $g$ is invariant under a change of parametrization. In particular, you get that $$\int_U f(u) \;\mu_g(du) = \int_V f(M(v)) \sqrt{\det (DM^t(v)DM(v))} dv$$ Note that this reduces to a familiar formula for arclength when $f\equiv 1$ and $\dim(U)=1$. Also note that it reduces to the usual change of variables formula when $m=n$. Thus it correctly "generalizes" the change of variables formula. – Shalop Aug 8 '16 at 9:54