Manifolds 43 | Integral is Well-Defined

Integration on a smooth manifold is only useful if the result doesn’t depend on arbitrary choices—especially how the measurable set is chopped into pieces. This installment proves that the integral of a volume form over a measurable set is well defined: any countable decomposition compatible with an atlas yields the same numerical value.

The setup starts with a smooth manifold M, an orientable atlas, and a measurable set A. Using a chosen countable atlas {U_k, H_k}, the set A is written as a disjoint union of measurable pieces A_k, each lying inside a single chart U_k. The integral is defined by transporting the volume form’s component function to R^n via the chart H_k and integrating over the corresponding image. Two conditions are required for the definition to make sense: (1) each chart-level integral exists, meaning the integral of the absolute value of the transported component function over the chart image is finite; and (2) the infinite sum of these chart-level integrals (without absolute value) converges to a finite number.

The core question is what happens if a different decomposition is chosen—equivalently, if a second atlas {Ũ_m, H̃_m} and a new disjoint measurable partition {Ã_m} are used. The proposition claims that the same two finiteness requirements still hold for the new atlas, and—most importantly—the resulting integral value matches the original one. In other words, the notation “∫_A Ω” is justified because it does not secretly depend on which atlas or partition was used.

The proof avoids heavy new machinery by combining the two decompositions into a third one. Intersecting the pieces from both partitions produces a countable collection of measurable sets of the form A_k ∩ Ã_m. Each such intersection lies inside both a chart domain U_k and a chart domain Ũ_m, so it can be evaluated in either coordinate system. From earlier results (the chart-independence of the integral on R^n under coordinate changes), the integral over each intersection agrees whether computed using H_k or H̃_m—both for the component function and for its absolute value.

Once equality holds piece-by-piece, the argument sums across all intersections. The proof first reorganizes the double sum by grouping terms: summing over m turns the collection of intersection images into the full image of A_k, and summing over k similarly reconstructs the full image of Ã_m. The finiteness assumptions ensure these rearrangements are legitimate, so the double-sum expression collapses to the same finite value for both atlases. The conclusion is that the integral over A is invariant under changing the measurable decomposition and the atlas, so the integral is genuinely well defined rather than a bookkeeping artifact.