TITLE: Integrable in the extended sense: Question about Spivak "Calculus on Manifolds" definition.
QUESTION [5 upvotes]: On Spivak "Calculus on Manifolds" he builds the concept of integration on an incremental fashion:

He starts by defining the integral $\int_R  f$ on a rectangle R;
Next he define the concept of characteristic function:
\begin{equation}
 X_C = 1 \text{ if }x\in C \text{ else } 0.
 \end{equation}
And use this concept for generalize the definition of integral for a 
region $C$, by defining $\int_C  f = \int_R  f \cdot X_C$ for $C$ 
contained in a rectangle $R$. This concept works for all the cases when C 
boundary has measure 0 and $X_C$ is integrable (see theorem 3-9 of the 
same book).
Then he defines partitions of the unit to generalize this concept even further. Using the concept of partition of the unit he defines the integral in the extended sense as:
\begin{equation}
\sum_{\phi \in \Phi}\int_A\phi \cdot f
\end{equation}
where $\Phi$ is a collection of functions such that $\phi \in \Phi$. Some properties of this functions are described next


Be $A$ a bonded region and $O$ and open cover to it, it can be proved
     that (see theorem 3-11 of the same book)  there exist a collection $\Phi$ of $C^\infty$ functions such:

$0 \le\phi(x) \le 1$
A finite number of $\phi(x)$ is different than zero in a open set containing $x \in A$
$\sum_{\phi \in \Phi} \phi(x) = 1$
For each $\phi \in \Phi$ there is an open set $U  \in O$ such that $\phi=0$ outside of some closed set contained in $U$. Let us call this closed set $C$.



So my question is: how can we prove $\int_A\phi \cdot f$ is integrable?

My understanding about the question is the following: From the above definition it follows that $\int_A\phi \cdot f = \int_C\phi \cdot f$. So if $C$ boundary has measure $0$ we could use the previous definition of integration to say this function is integrable in this region...But how can we prove that this is indeed the case?

REPLY [4 votes]: I studied from Calculus on Manifolds this year, and in this section, I found that his treatment was a little sloppy. First, there is a huge error in the entire section of partitions of unity: in property ($4$) of Theorem $3$-$11$, "... outside of some closed set contained in $U$", the word "closed" should be replaced with "compact". So, property (4) can be rephrased equivalently by requiring that the support of $\varphi$ be a compact subset of $U$, where the support is defined  as the topological closure of the set of points where $\varphi$ is non-zero.
\begin{equation}
\text{supp}(\varphi) := \overline{\{ x \in \mathbb{R^n}: \varphi(x)\neq 0\}}.
\end{equation}
Next, to define the extended integral, I think this is a better definition (it's almost the same, but there are a few subtle differences):

Definition/Proposition:
Let $A$ be an open subset of $\mathbb{R^n}$, $\mathcal{O}$ an admissible open cover for $A$, and $\Phi$ be a $\mathcal{C^0}$ partition of unity for $A$ subordinate to $\mathcal{O}$, with compact support. Let $f: A \to \mathbb{R}$ be a locally bounded function (every point has a neighbourhood on which $f$ is bounded) such that $\mathcal{D}_f$, the set of discontinuities of $f$ has measure zero. Then, for every $\varphi \in \Phi$, the integrals
\begin{equation}
\int_{\text{supp}(\varphi)} \varphi \cdot |f| \qquad  \text{and} \qquad \int_{\text{supp}(\varphi)} \varphi \cdot f
\end{equation}
exist according to the old definition (the one involving characteristic functions). We define $f$ to be integrable on $A$, in the extended sense if
\begin{equation}
\sum_{\varphi \in \Phi} \int_{\text{supp}(\varphi)} \varphi \cdot |f|
\end{equation}
converges. In this case, we define
\begin{equation}
(\text{extended}) \int_{A} f = \sum_{\varphi \in \Phi} \int_{\text{supp}(\varphi)} \varphi \cdot f
\end{equation}

The two differences are: I only required $\Phi$ to be $\mathcal{C^0}$, not $\mathcal{C^{\infty}}$, and second, I put $\displaystyle \int_{\text{supp}(\varphi)} \varphi \cdot |f|$ rather than $\displaystyle \int_{A} \varphi \cdot |f|$. The reason I made the second change is because the purpose of this definition is to define integration on an open set (which may be unbounded), so writing $\displaystyle \int_{A} \varphi \cdot |f|$ isn't even defined based on all the old definitions. However, this isn't a huge deal, because later on we can show that
\begin{equation}
(\text{extended})\displaystyle \int_{A} \varphi \cdot |f| = (\text{old}) \displaystyle \int_{\text{supp}(\varphi)} \varphi \cdot |f|
\end{equation}
But, from a logical standpoint, we should not use the symbol $\displaystyle \int_A \varphi \cdot f$ in a definition where we're trying to define the meaning of integration on $A$
(note that we have to use another partition of unity $\Psi$ to make sense of the LHS above).

Proof $\displaystyle \int_{\text{supp}(\varphi)} \varphi \cdot f $ exists according to old definition:
To prove this, we need to show that $\varphi f$ is bounded on a rectangle $R$ containing supp$(\varphi)$, and that $\varphi f \cdot \chi_{\text{supp}(\varphi)}$ is integrable on $R$. To prove boundedness, note that for each $x \in \text{supp}(\varphi)$, since $f$ is locally bounded, there is an open neighbourhood $V_x$ of $x$, and a number $M_x > 0$ such that $|f| \leq M_x$ on $V_x$. The collection of all such $V_x$ forms an open cover of $\text{supp}(\varphi)$, hence by compactness, there is a finite subcover, say by $V_{x_1}, \dots, V_{x_k}$. Then $f$ is bounded by $M = \max \{M_{x_i} \}_{i=1}^k$ on supp($\varphi$). Since $\varphi = 0$ outside $\text{supp}(\varphi)$, it follows that $\varphi \cdot f$ is bounded everywhere (by $M$).
Next, let $R$ be a closed rectangle containing $\text{supp}(\varphi)$. It is easy to verify that
\begin{align}
\varphi f \cdot \chi_{\text{supp}(\varphi)} = \varphi f  \tag{*}
\end{align}
(because outside the support, both sides are $0$). Also, since $f$ has a discontinuity set of measure zero, and since $\varphi$ is continuous, it follows that $\varphi f \cdot \chi_{\text{supp}(\varphi)} = \varphi f$ also has a discontinuity set of measure zero; hence $\varphi f \cdot \chi_{\text{supp}(\varphi)}$ is integrable on $R$ according to the very first definition. This proves $\displaystyle \int_{\text{supp}(\varphi)} \varphi \cdot f$
exists according to the old definition. By replacing $f$ with $|f|$ everywhere, you can see that $\displaystyle \int_{\text{supp}(\varphi)} \varphi \cdot |f|$ also exists according to the old definition.

Remarks:

Notice that because of (*),  it doesn't matter whether or not the boundary of $\text{supp}(\varphi)$ has measure zero. $\varphi \cdot f$ is integrable on $\text{supp}(\varphi)$ anyway.
Notice that by definition, $\text{supp}(\varphi)$ is the closure of a set and hence closed. But this is not good enough, we need it to be compact so that the boundedness argument above works.
If $V \subset A$ is a bounded open set containing $\text{supp}(\varphi)$, then $\displaystyle \int_V \varphi \cdot f$ exists according to the old definition; this should be immediate since $\displaystyle \int_{\text{supp}(\varphi)} \varphi \cdot f$ has already been shown to exist. In this case,
\begin{equation}
\int_V \varphi \cdot f = \int_{\text{supp}(\varphi)} \varphi \cdot f
\end{equation}