In his "Calculus on manifolds" Spivak first defines $n$-dimensional (Riemann-) integral over rectancles, then over Jordan measurable subsets of rectangles and finally extends it to open sets using partitions of unity. It's this final part I'm having some problems with.

He defines a partition of unity for a set $A\subset \mathbb{R}^n$ to be family $\Phi$ of $C^{\infty}$ functions satisfying the following three axioms:

  1. $0\leq\phi(x)\leq 1$ $\forall \phi\in\Phi$ $\forall x\in A$

  2. For every $x\in A$ there exists an open set $V$ s.t. only finitely many $\phi\in\Phi$ are nonzero in $V$

  3. $\sum_{\phi\in\Phi} \phi(x)=1$ $\forall x\in A$

Moreover, $\Phi$ is subordinate to an open cover $\mathcal{O}$ of $A$ if for each $\phi\in\Phi$ we can find an open set $U\in\mathcal{O}$ s.t. $\phi=0$ outside some closed set contained in $U$.

Suppose now that $A$ is open and we have an open cover $\mathcal{O}$ for $A$ s.t. $U\subseteq A$ for each $U\in\mathcal{O}$. We shall call such a cover admissible. Let $f:A\to R$ be locally bounded (i.e. every point has a neighbourhood $V$ s.t. $f$ bounded in $V$) and continuous almost everywhere (i.e. the set of discontinuties has measure $0$). Furthermore let $\Phi$ be a countable partition of unity for $A$ subordinate to $\mathcal{O}$. We say that $f$ is integrable over $A$ if $\sum_{\phi\in\Phi} \int_A \phi |f|$ converges.

Spivak asserts that the integrals $\int_A \phi|f|$ all exist but I don't see why because we have thus far only defined integrals over bounded sets. Obviously we could assume that the open cover is composed of bounded sets or maybe instead that the functions $\phi$ have compact support but neither appears to be assumed in the book.

So do we need some extra assumptions here or am I missing something?


If you look at the proof in Spivak's book you'll see the following as the first statement: "since $\varphi\cdot f = 0$ except on some compact set $C$...". So, quite obviously, he assumes the $\varphi$ to have compact support and forgot to make this explicit.

  • $\begingroup$ Ok, thanks. Wanted to make sure I'm not missing something. $\endgroup$ – KPK Jul 20 '12 at 21:43

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