Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Spivak's "Calculus on Manifolds", Spivak first defines integration over rectangles, then bounded Jordan-measurable sets (for functions whose discontinuities form a Lebesgue null set).

He then uses partitions of unity to define integration over arbitrary open sets (top of p.65). Used in the proof of this assertion is the claim that if:

i) $\Phi$ is subordinate to an admissible cover $J$ of our open set $A$.

ii) $f:A\rightarrow \mathbb{R}$ is locally bounded in $A$.

iii) The set of discontinuities of $f$ is Lebesgue-null.

then each $\int_A \phi\cdot|f|$ exists.

I cannot see how this statement makes sense, seeing as the integral is thus far only defined for bounded Jordan-measurable sets. Perhaps I am missing something simple here?

share|cite|improve this question
possible duplicate of Is this definition missing some assumptions? – Martin Argerami Nov 29 '12 at 14:07
up vote 2 down vote accepted

I may well be misinterpreting something here (your notation is a bit unclear on a few points, and I don't have Spivak's book available), but provided the cover consists of bounded, Jordan measurable sets, and you already know how to integrate over such sets, the rest should be easy: if $J = \{ A_i \}_{i \in I}$, and the partition of unity consists of continuous functions $\varphi_i: A \to \mathbb{R}$ with $\varphi_i$ supported in $A_i$ for each $i \in I$, then each $$ \int_{A_i} \varphi_i f $$ exists (the integrand is supported in $A_i$, a bounded Jordan measurable set, the set of discontinuities has measure $0$, so we know how to integrate it). Then you simply define $$ \int_A f := \sum_{i \in I} \int_{A_i} \varphi_i f $$ and hope that this is independent of the particular cover $J$ (and it usually is).

share|cite|improve this answer
Thank you for your answer. Whilst this certainly makes sense to me, he nowhere states that the sets of the cover are bounded (or even Jordan measurable!). I also feel that we need this to be true (or for every $\phi\in\Phi$ to be compactly supported, not just supported inside a closed subset of some $U\in J$) to make use of the local-boundedness property. This interpretation seems to fit the theme of using partitions of unity to "piece together" local properties but it still seems weird to me that the nature of this cover is not specified explicitly. – Sean Gomes Apr 3 '12 at 22:49
@SeanGomes: That seems strange. I guess I'll have to pick up a copy of Spivak's book at my university's library when I get back (sometime next week), and have a look for myself (unless, of course, someone else is able to shed some more light on it before then). – Martin Wanvik Apr 4 '12 at 8:05
No need to go out of your way, I am proceeding using Munkres' treatment of partitions of unity. I think I understand it well enough to persevere with Spivak, the key is that the $\phi$ may all be chosen to be compactly supported. – Sean Gomes Apr 4 '12 at 10:55
@SeanGomes: Not trying to go out of my way either - I just happen to be a bit curious about Spivak's treatment myself. – Martin Wanvik Apr 4 '12 at 15:07

This issue is also discussed in this post.

I think that the key is at the begining of the proof of item ($1$) in Theorem $3$-$12$, where Spivak says "Since $\varphi\cdot f=0$ except on some compact set $C$..." This compact set seems to depend on $\varphi$ therefore the statement suggests that Spivak may have had in mind a slightly different version of item ($4$) in Theorem $3$-$11$.

It seems then that the word "closed" in item ($4$) should be changed to "compact" and hence the sentence in the proof "If $f:U\rightarrow [0,1]$ is a $C^{\infty}$ function which is $1$ on $A$ and $0$ outside of some closed set in $U$,..." (p.64) should have the word "closed" changed to "compact".

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.