How does using partitions of unity avoid the problem of double-counting? I've been thinking about integration on manifolds and am pondering the notion of partitions of unity.
Given a partition of unity $ \Phi$ of an open set $A \subset \mathbb{R}^n$ subordinate to an open cover $O$, and $f: A \to \mathbb{R}$ is locally bounded in $A$, and $\sum_{\phi \in \Phi} \int_A \phi \cdot \left| f \right|$ converges, we define
$$ \int_A f := \sum_{\phi \in \Phi} \int_A \phi \cdot f. $$
My question is very simple: How can we guarantee that in the overlap of any two open sets we are not counting them twice? I'm guessing this has something to do with the condition that $\sum_{\phi \in \Phi} \phi(x) = 1$ but would prefer to have it spelt out by someone who understands this topic better.
 A: Since $\sum_{\phi\in\Phi} \phi=1$, you have by linearity of integration
$$
\sum_\phi \int_A\phi\cdot f \,\textrm{d}x=\int_A\sum_\phi \phi\cdot f \,\textrm{d}x=\int_A 1 \cdot f \,\textrm{d}x=\int_A f\,\textrm{d}x.
$$
There is a technical subtlety in the first equality, because linearity of integration only allows finite sums. There are two ways to resolve that: you can use the fact that $\Phi$ is countable and all $\phi$ are non-negative and apply the Lebesgue dominated convergence theorem (to partial sums) or you can use the fact that the partition of unity is locally finite, in which case locally it makes perfect sense (the sum is finite), and globally an integral over a set is the sum of the integrals over its parts, even if there are infinitely many (provided the sum even makes sense, i.e. converges absolutely). (Note that a rigorous proof of the latter part would likely use the Lebesgue dominated convergence theorem or the monotone convergence theorem anyhow.)
A: Partition of unity are used to use the well-known definition of the integration in open subset of $R^n$, for every $U\in O$, $U$ is a chart which is diffeomorphic to an open subset of $R^n$.
If $U,V\in O$, $\int_{U\cap V}f=\int_{U\cap V}\sum_{\phi\in \Phi}f.\phi=\int_{U\cap V}f$, since $\sum_{\phi\in \Phi}\phi_{\mid U\cap V}=1$
