One can use the de Rham theorem to define the Lebesgue integral without ever using any notion of measure theory. More precisely, the integral can be defined as the composition of the following sequence of maps: C^∞_cs(Dens(M))→H^n_cs,dR(M,Or(M))→H^n_cs(M,Or(M))→H_0(M)→H_0(∙)=R.C∞cs(Dens(M))→Hncs,dR(M,Or(M))→Hncs(M,Or(M))→H0(M)→H0(∙)=R.
Here C^∞_cs(Dens(M)) C∞cs(Dens(M)) denotes the space of all smooth densities with compact support. The space C^∞_cs(Dens(M)) C∞cs(Dens(M)) is mapped to H^n_cs,dR(M,Or(M)) Hncs,dR(M,Or(M)) (the nth de Rham cohomology of M with compact support twisted by the orientation sheaf of M) by the obivous map given by the definition of de Rham cohomology. The space H^n_cs,dR(M,Or(M)) Hncs,dR(M,Or(M)) is isomorphic to H^n_cs(M,Or(M)) Hncs(M,Or(M)) (the nth twisted ordinary cohomology of M with compact support) by the de Rham theorem. The space H^n_cs(M,Or(M)) Hncs(M,Or(M)) is isomorphic to H_0(M) H0(M) by Poincaré duality. Finally, H_0(M) H0(M) can be mapped to H_0(∙)=R H0(∙)=R by the usual pushforward map for homology.
More details are available in this answer: http://mathoverflow.net/questions/38439/integrals-from-a-non-analytic-point-of-view/38479#38479
Here is an easy application of the above definition: The easiest version of Stokes' theorem states that ∫dω=0, where ω∈Ω^{n-1}(M,Or(M)). ω∈Ωn-1(M,Or(M)). Proof: ∫ factors through the map to the de Rham cohomology. The form dω is a coboundary, hence its image vanishes in the de Rham cohomology and the integral equals zero.
