# Integrable in the extended sense: Question about Spivak “Calculus on Manifolds” definition.

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: $$$$X_C = 1 \text{ if }x\in C \text{ else } 0.$$$$ And use this concept for generalize the definition of integral for a region $$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: $$$$\sum_{\phi \in \Phi}\int_A\phi \cdot f$$$$ where $$\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?

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. $$$$\text{supp}(\varphi) := \overline{\{ x \in \mathbb{R^n}: \varphi(x)\neq 0\}}.$$$$

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 $$$$\int_{\text{supp}(\varphi)} \varphi \cdot |f| \qquad \text{and} \qquad \int_{\text{supp}(\varphi)} \varphi \cdot f$$$$ exist according to the old definition (the one involving characteristic functions). We define $$f$$ to be integrable on $$A$$, in the extended sense if $$$$\sum_{\varphi \in \Phi} \int_{\text{supp}(\varphi)} \varphi \cdot |f|$$$$ converges. In this case, we define $$$$(\text{extended}) \int_{A} f = \sum_{\varphi \in \Phi} \int_{\text{supp}(\varphi)} \varphi \cdot f$$$$

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 $$$$(\text{extended})\displaystyle \int_{A} \varphi \cdot |f| = (\text{old}) \displaystyle \int_{\text{supp}(\varphi)} \varphi \cdot |f|$$$$ But, from a logical standpoint, we should not use the symbol $$\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, $$$$\int_V \varphi \cdot f = \int_{\text{supp}(\varphi)} \varphi \cdot f$$$$

Spivak often skips many steps and you have to read every sentence leading up to a theorem carefully.

The discussion of the definition and convergence of the extended integral $$\sum_{\phi \in \Phi}\int_A\phi \cdot f$$ uses the fact that the integral $$\int_A \phi \cdot f$$ exists. This in turn is based on the assumptions stated in the first sentence on page 65:

An open cover $$\mathcal{O}$$ of an open set $$A \subset \mathbb{R}^n$$ is admissible if each $$U \in \mathcal{O}$$ is contained in $$A$$.

Note that $$A$$ is assumed to be open and $$\mathcal{O}$$ is assumed to be admissible, meaning that $$A$$ is covered by open subsets $$U \subset A$$.

Spivak goes on to say:

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

Implicit in this statement is the existence of $$\int_A \phi \cdot f$$ which implies the existence of $$\int_A \phi \cdot |f|$$.

Given that $$A$$ is open and $$\mathcal{O}$$ is admissible, we can proceed to prove the existence of the integral. Since $$\Phi$$ is subordinate to $$\mathcal{O}$$, for each $$\phi \in \Phi$$ there is some open set $$U \in \mathcal{O}$$ 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 $$\int_{A \setminus U} \phi \cdot f$$ exists regardless of the measure of the boundary of A. Also $$\phi \cdot f$$ vanishes on the boundary of $$U$$ and is continuous almost everywhere in $$U$$ (since $$\phi \in C^\infty$$ with compact support in $$U$$). Thus, $$\int_U \phi \cdot f$$ exists and, regardless of the Jordan-measurability of $$A$$, it follows that

$$\int_A \phi \cdot f = \int_U \phi \cdot f + \int_{A \setminus U} \phi \cdot f$$

Spivak then defines $$f$$ to be integrable in the extended sense if $$\sum_{\phi \in \Phi} \int_A \phi \cdot |f|$$ (with $$\phi$$ arranged in a sequence) converges. Since $$\left| \int_A \phi \cdot f\right| \leqslant \int_A \phi \cdot |f|$$, the series $$\sum_{\phi \in \Phi} \int_A \phi \cdot f$$ is absolutely convergent.