On wikipedia, there is the following :

A partition of unity can be used to define the integral (with respect to a volume form) of a function defined over a manifold: One first defines the integral of a function whose support is contained in a single coordinate patch of the manifold; then one uses a partition of unity to define the integral of an arbitrary function; finally one shows that the definition is independent of the chosen partition of unity.

My question is the following : do we really need to use a partition of unity to define the integral of a function defined over a manifold ?

Couldn't we just use a sequence $(U_n)$ of chart domains that cover the manifold, and then define the sequence of Borel sets $B_{n+1}:=U_{n+1}\backslash U_n$, and $B_0=U_0$, and then just define the integral of a function $f$ with respect to a volume form to be the sum of all integrals of $f$ over $B_n$ (since $B_n$ is in a chart domain, the integral can be computed through a chart).


Strictly speaking, you don't "need" partitions of unity to define integration over a manifold and you never use them anyway when you actually want to compute some integral, using instead something which pretty much reminds your idea.

Let us assume for simplicity that our manifold $M$ is compact. If I understand correctly, your idea of defining the integral of $\omega \in \Omega^{\text{top}}(M)$ is to split $M$ as a disjoint union of finitely many sets that are contained in coordinate charts and then pull back the differential form, integrate it on each set and add everything up. While this is how one usually computes the integral in practice, it requires much work to make a satisfying definition of an integral based on this idea (even more work if you only want to work with the Riemann integral). Let me point up some difficulties which must be addressed:

  1. Let us say that we managed to cover $M$ by three charts $(\varphi_1, \varphi_2, \varphi_3)$. The three charts might have non-trivial triple intersection and so if you compute the integral as $\int_{U_1} \varphi_1^{*}(\omega) + \int_{U_2 \setminus U_1} \varphi_2^{*}(\omega) + \int_{U_3 \setminus U_2} \varphi_3^{*}(\omega)$ because then you will be counting $\int_{V_1 \cap V_2 \cap V_3} \omega$ twice! Thus, it is better to work in advance with some partition of $M$ into a disjoint union of pieces that are contained in coordinate charts in order to avoid inclusion/exclusion.
  2. We surely want $\int_M \omega$ to be well-defined and finite but it is not obvious that $\int_{U} \varphi^{*}(\omega)$ is finite! If $\varphi \colon U \rightarrow M$ is an arbitrary chart, the set $U$ might be pretty ugly/unbounded and it is quite hard to justify in advance that $\int_{U} \varphi^{*}(\omega)$ should be finite. This can be handled if you don't use arbitrary charts but special charts (for example, in which $U$ is bounded and which can be extended into charts on a larger subset $V$ with $\overline{U} \subset \subset V$).
  3. You want the integral to be independent of the partition used to define it.

The bottom line is that the definition using partition of unity is much easier to work with technically and to prove using it various properties of the integral (the best example being Stoke's theorem) but it is not really practical in order to explicitly compute integrals. If you don't care about computing integrals explicitly and just want the theoretical framework, you can completely ignore the question of "how do I explicitly find a partition of unity in order to compute the integral" which is what theoretical books often do. If you do, you find out that you can't really compute a partition of unity and then prove that the integral (defined originally using a partition of unity) can be computed by dividing $M$ into disjoint pieces that sit in coordinate charts and satisfy appropriate conditions and compute everything on $\mathbb{R}^n$. For example, see Proposition 16.8 in Lee's "Introduction to Smooth Manifolds" book.

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  • $\begingroup$ Thank you for your answer. For the point 1, I should have said : $B_{n+1}:=U_{n+1}/B_n$ to avoid inclusion/exclusion. $\endgroup$ – Phil-W Nov 10 '16 at 1:18
  • $\begingroup$ For the point 2, it seems to me that if we defined first the integral for positive measurable functions valued in $[0,\infty]$, finiteness questions are not relevant. Once the integral (valued in $[0,\infty]$) is defined for such functions, we can define it for real valued functions by decomposing into positive and negative parts, and define the "integrable" functions as being those such that the integrals in the decomposition are finite. $\endgroup$ – Phil-W Nov 10 '16 at 1:26
  • $\begingroup$ For the point 3, the same thing has to be done for partition of unity, and I don't see why the proof of this independence for partition of unity is simpler than with $(B_n)$. $\endgroup$ – Phil-W Nov 10 '16 at 1:29
  • $\begingroup$ @Phil-W: I'm not saying that it can't be done, but I do think that it requires more technical details. Remember that you can't integrate a function on a manifold, only a top differential form (or density). You suggest to define $\int_M \omega$ as $\sum \int_{U_i} \omega$ where each $U_i$ is "simple". The partition of unity approach defines $\int_M \omega$ as $\sum \int_M \omega_i$ where each $\omega_i$ is simple. This is the same philosophy (only you split $\omega$ instead of $M$). Try to do it rigorously and see if you agree with me. $\endgroup$ – levap Nov 10 '16 at 3:56
  • $\begingroup$ Also, partitions of unity are useful in differential geometry for many other reasons and the partition of unity definition makes the proof of Stokes's theorem extremely elegant so that's another advantage. $\endgroup$ – levap Nov 10 '16 at 4:12

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