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In page 65 of his book, Spivak is saying $\int_A \varphi . |f|$ exists if

1.$\Phi$ is a partition of unity subordinate to an open cover $O$ of $A \subset \mathbf{R}^n$, $\varphi \in O$

2.Discontinuity of $f:A \rightarrow \mathbf{R} $ is measure $0$

3.$f$ is bounded in some open set around each point of $A$.

But I can't understand why it exists.

I know in his proof of existance of partition of unity, he actually proved each $\varphi \in O$ has compact support, so we can think above integral as integration on a subset of a rectangle $\prod^n_{i=1}[a_i,b_i] $. but I think still $A$ should be Jordan measurable since otherwise the integral may not exist.

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One more condition of the theorem is that the open cover $O$ is admissible, meaning each open set $U \in O$ is contained in $A$.

Since $\Phi$ is subordinate to $O$, for each $\phi \in \Phi$ there is some open set $U \in O$ and some closed set $F$ such that $F \subset U \subset A$ and $\phi = 0$ outside of $F$.

Hence, $\phi \,$ vanishes in $A \setminus U,$ and $\int_{A \setminus U} \phi \, |f|$ exists regardless of the measure of the boundary of A. Also $\phi \, |f|$ vanishes on the boundary of $U$ and is continuous almost everywhere in $U$ (since $\phi \in C^\infty$ with compact support in $U$). Thus, $\int_U \phi \, |f|$ exists and, regardless of the Jordan-measurability of $A$, it follows that

$$\int_A \phi \, |f| = \int_U \phi \, |f| + \int_{A \setminus U} \phi \, |f|.$$

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  • $\begingroup$ Thank you for your answer. But A should be open in R^n to exist such admissible cover. I see no assumption A is open in the book. $\endgroup$ – quicksilver Mar 25 '17 at 3:17
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    $\begingroup$ It appears you are trying to understand the argument on page 65 of Spivak. The first sentence reads: "An open cover $O$ of an OPEN set $A$ is admissible if each $U \in O$ is contained in $A$." $\endgroup$ – RRL Mar 25 '17 at 4:01
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    $\begingroup$ The proofs in Spivak are sometimes very terse -- big steps are skipped and key assumptions appear sometimes before the statement of the proof. I'm looking at the 1998 twenty-seventh printing reissued by Westview Press. $\endgroup$ – RRL Mar 25 '17 at 4:04
  • $\begingroup$ I have a question: in order for the integrals to exist, two conditions must be met. One is continuity almost everywhere, which we have by hypothesis. The other one is boundedness. Do we have any guarantee of the latter? We assume that f is locally bounded, but we have no guarantee that the cover is such that f is bounded on every element of the cover, do we? $\endgroup$ – Robly18 Mar 6 at 21:22
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    $\begingroup$ @Robly18: For a partition of unity the function $\phi$ has compact support in $A$. So it vanishes outside a compact set $F \subset A$. Now for every $x \in F$, there is an open set $G_x$ containing $x$ where $f$ is bounded. The collection of sets $G_x$ is an open cover of $F$, and, thus $F$ is contained in a finite subcollection $G_{x_1}, \ldots, G_{x_n}$. Since $f$ is bounded in each of the finitely many sets, it is bounded in $F$ and so $\int_F \phi |f|$ exists. But $\phi$ vanishes outside $F$ so the integral over $A$ exists. As I said Spivak skips many steps. $\endgroup$ – RRL Mar 7 at 7:56

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