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On page 65 of Calculus on Manifolds, right after defining a partition of unity $\Phi$, it states:

If $\Phi$ is subordinate to $\mathcal{O}$, $f:A \to \mathbb{R}$ is bounded in an open set around each point of $A$, and $\{x: f \text{ is discontinuous at }x \}$ has measure $0$, then each $\int_A \varphi \cdot |f|$ exists.

I don't see why this would be true unless it is assumed that the $\varphi$ have compact support, which I don't believe it is. Can someone help me parse this? Thanks

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  • $\begingroup$ Something here should have compact support, either $\varphi$ or $f$. The assumption is often on $f$. $\endgroup$ – user98602 Aug 9 '15 at 17:48
  • $\begingroup$ @MikeMiller Pretty sure I haven't missed this assumption. But it's clear that if you read on that he means $\varphi$ have compact support. Perhaps I'll leave the question here in case anyone else has the same question. $\endgroup$ – Eric Auld Aug 9 '15 at 18:19
  • $\begingroup$ Sure, I buy that. Paul's answer shows there are multiple notions of partitions of unity; I guess Spivak wants Paul's. $\endgroup$ – user98602 Aug 9 '15 at 18:20
  • $\begingroup$ @MikeMiller Actually, after reading it once more, it's not clear whether he wants $f$ or $\varphi$ to have compact support. Later he "recalls" for the reader that $\varphi \cdot f$ should have compact support, without ever assuming it, as far as I can tell. $\endgroup$ – Eric Auld Aug 9 '15 at 18:24
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    $\begingroup$ Does he sum over the entire partition and expect the final thing to be finite? If so, $f$ is the one with compact support. If it seems he's willing to believe it can be infinite, it's $\varphi$. $\endgroup$ – user98602 Aug 9 '15 at 18:25
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By definition of "partition of unity", $\varphi$ has compact support (at least, the definition I know - I can't say how your book defines it).

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    $\begingroup$ This definition, unless you allow one to associate multiple functions to each open set in the cover, means that not every open cover has a subordinate partition of unity - pick the cover whose only open set is $X$ itself. The definition I know doesn't allow that, but maybe yours allows a refinement of the cover. $\endgroup$ – user98602 Aug 9 '15 at 18:05
  • $\begingroup$ Sometimes people choose to define things differently. The definition I learned explicitly requires the partition functions $\varphi$ to have compact support. Now I am curious myself how Spivak handles things without this requirement. $\endgroup$ – Paul Sinclair Aug 9 '15 at 18:24
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From page 63 theorem 3-11. Let $A\subset \mathbb{R}^{n}$ and let $\mathcal{O}$ be an open cover of $A$. Then there is a collection $\Phi$ of $\mathcal{C}^{\infty}$ functions $\varphi$ defined in an open set containing $A$, with the following properties: (I am only going to write the most relevant one)

$(4)$ For each $\varphi\in\Phi$ there is an open set $U$ in $\mathcal{O}$ such that $\varphi=0$ outside of some closed set contained in $U$.

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    $\begingroup$ The closed set does not need to be compact. $\endgroup$ – user98602 Aug 9 '15 at 18:09

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