Integration on compact manifold

Integration on a nice enough manifold of a function $f:M \to \mathbb{R}$ is defined $$\int f = \sum_{ i \in I} \int_{U_i}\phi_i f$$ where $\phi_i$ is a partition of unity subordinate to the open cover $U_i$ of $M$.

The index set $I$ can be infinite in general.

If $M$ is compact (bounded), can I take $I$ to be a finite set so that the sum is a finite sum?

• Yes, if $M$ is compact, you can always have a finite partition of unity subordinate to any open cover. – Daniel Fischer Jan 12 '14 at 23:37

1 Answer

Yes, since every open cover has a finite subcover (Heine-Borel).

• Thanks, but why don't books emphasise that the sum can be finite?? – michael_carbon Jan 12 '14 at 23:39
• I am sure some do. Also, many manifolds are not compact, but they are generally locally compact, which means that every point lies in a finite number of the open sets (this is obviously important, since infinite sums are a big pain). – Igor Rivin Jan 12 '14 at 23:40
• @PedroTamaroff I actually think that the local finiteness (global finiteness less so) is crucial for the whole machine to work, and I am sure if you read the first partitions of unity papers, it is probably emphacized. – Igor Rivin Jan 12 '14 at 23:45