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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?

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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}
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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.

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