In measure theory on $\mathbb{R}^n$, whenever you need an integral, you induce it from the measure $m$ of the space you are working with using the Lebesgue-integral: $$\int_{M} f \,\mathrm{d}m.$$
Obviously, this brilliant concept was not going to work smoothly on curved manifolds, since we do not have a uniform coordinate system and what used to be a rectangle in one coordinate system, becomes a parallelogram in the other - complete mess! So we cannot just write $$\int_{M} 1 \,\mathrm{d}m = \operatorname{vol}(M)$$ since we would be getting different, coordinate-dependent volumes for the same figures.
The solution, as given in most differential geometry books, is to induce the measure from the tangent spaces via the anti-symmetric covariant tensor fields, aka differential n-norms, which give you a "signed length meter to each 1-dimensional subspace of each tangent space in a coordinate independent way". In other words, we get stuff like "the integral of the covector field $\omega$ over a curve $\gamma$": $$\int_{\gamma} \omega = \int_{[a,b]} \gamma^*\omega$$ or integrate the covector field $\omega$ over $U$ (well-definedly using coordinate charts) $$\int_{U} \omega = \int_{\varphi(U)} (\varphi^{-1})^* \omega$$
QUESTION At this point, I lost the connection between the original problem and the proposed solution.
- Our problem was that we cannot measure volumes because of the coordinate dependency. Is my understanding correct that we replace sets (whose volumes we want to measure) by expressing these sets as images of $\mathbb{R}^k$-valued functions and accordingly integrating $k$-forms over them?
- If that is so, is there an intuitive, non-rigorous way to explain why a real number that we get in the end for the integral is not coordinate-dependent?
- Any other comments about further conceptual differences between two approaches are welcomed.
The sets that you want to measure in $\mathbb{R}^n$ are those which are contained in the Borel $\sigma$-algebra, which is generated by open sets. Similarly, you can generate a $\sigma$-algebra on the topological space which underlies a smooth manifold $M$, and it will again be the Borel $\sigma$-algebra of that topological space. To introduce a measure on $\mathbb{R}^n$, you use the notion of length. For example, the measure of an interval $[a,b]\subset\mathbb{R}$ is the length of that interval. On a smooth manifold, there is no canonical notion of length, because there is no canonical metric. Once you have chosen a metric, you get a Riemannian manifold $(M,g)$, and this gives you a Riemannian volume form $d\text{Vol}_g\in\Omega^n(M)$, provided $M$ is orientable. Now, for all open subsets $U\subseteq M$, we may define $\mu(U)=\int_Ud\text{Vol}_g$, and this defines a Borel measure on $M$, just like the flat Euclidean metric defines a Borel measure (the Lebesgue measure) on $\mathbb{R}^n$.
So to conclude, the situation on a smooth manifold is just a more general version of the situation on $\mathbb{R}^n$. In the latter, we have a canonical choice of a metric, the flat Euclidean metric. It gives us a measure, for the Borel $\sigma$-algebra of the topological space $\mathbb{R}^n$. Likewise, any choice of Riemannian metric on a smooth manifold $M$ will give you a Borel measure for the Borel $\sigma$-algebra. Alternatively, you can choose a volume form and the same story applies. In either case, the measure theoretic integral is precisely the same as the differential geometric integral. It is simply the choice of metric/volume form which corresponds to choosing a measure for the Borel $\sigma$-algebra.
(You can also do something with densities when your manifold is not orientable, but I'm not familiar with this. See J. Lee's "Introduction to Smooth Manifolds".)