LEC CAL2 7

Partitions of unity

The support $\text{supp}(f)$ of a function $f$ is defined to be the closure of the set of all points at which $f$ is nonzero.

Lemma 198.1(Spivak (1965) Problem 1-22).

If $U$ is open and $C\subseteq U$ is compact, there exists a compact set $D$ such that $C\subseteq D^{\circ }$ and $D\subseteq U$.

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Lemma 198.2(Lemma).

If $\{U_1,\dots ,U_n\}$ is an open cover of a compact set $A$, then there exist compact sets $D_i\subseteq U_i$ such that $\{D_1^{\circ },\dots ,D_n^{\circ }\}$ covers $A$.

Proof.

Let

$$C_1:=A\setminus (U_2\cup U_3\cup \cdots \cup U_n).$$

$C_1\subseteq U_1$ is compact. By Lemma 1, there exists compact $D_1\subseteq U_1$ such that $C_1\subseteq D_1^{\circ }$.

Suppose compact $D_1,\dots ,D_k$ have been chosen so that $\{D_1^{\circ },\dots ,D_k^{\circ },U_{k+1},\dots ,U_n\}$ covers $A$. Let

$$C_{k+1}:=A\setminus (D_1^{\circ }\cup \cdots \cup D_k^{\circ }\cup U_{k+2}\cup \cdots \cup U_n).$$

$C_{k+1}\subseteq U_{k+1}$ is compact, so there exists compact $D_{k+1}\subseteq U_{k+1}$ such that $C_{k+1}\subseteq D_{k+1}^{\circ }$.□

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Lemma 198.3(Spivak (1965) Problem 2-26).

Suppose $A\subseteq {\mathbb{R}}^n$ is open, $C\subseteq A$ is compact. Then, there exists a $C^{\infty }$ function $f:A\to \mathbb{R}$ such that $f(x)=1$ for all $x\in C$ and $f$ is $0$ outside a closed set $B$ in $A$, that is, $C\subseteq \text{supp}(f)\subseteq A$.

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Theorem 198.4(Spivak (1965) 3-11).

Let $A\subseteq {\mathbb{R}}^n$ and let $\mathcal{O}$ be an open cover of $A$. Then there is a countable collection $\Phi$ of $C^{\infty }$ functions $\phi$ defined in an open set containing $A$, with the following properties:

1. For each $x\in A$ and $\phi \in \Phi$, we have $0\le \phi (x)\le 1$.
2. For each $x\in A$, there exists an open set $V\ni x$ such that all but finitely many $\phi \in \Phi$ are $0$ on $V$.
3. For each $x\in A$, we have ${\sum }_{\phi \in \Phi }\phi (x)=1$ (this is a finite sum by (2))
4. For each $\phi \in \Phi$, there exists $U\in \mathcal{O}$ such that $\text{supp}(\phi )\subseteq U$.

A collection $\Phi$ satisfying (1) to (3) is called a $C^{\infty }$ partition of unity for $A$. If $\Phi$ also satisfies (4), it is said to be subordinate to the cover $\mathcal{O}$.

Proof.

Case 1: $A$ is compact

Let $\{U_1,\dots ,U_n\}\subseteq \mathcal{O}$ be a finite subcover of $A$. It clearly suffices to construct a partition of unity subordinate to this cover. Let $\{D_1,\dots ,D_n\}$ be given by Lemma 2. By Lemma 3, we can choose non-negative $C^{\infty }$ functions ${\psi }_i$ with $D_i\subseteq \text{supp}({\psi }_i)\subseteq U_i$.

Let $U=D_1^{\circ }\cup \cdots \cup D_n^{\circ }$. It follows that ${\psi }_1(x)+\cdots +{\psi }_n(x)>0$ for all $x$ in $U$. On $U$ we can define

$${\phi }_i(x):=\frac{{\psi }_i(x)}{{\psi }_1(x)+\cdots +{\psi }_n(x)}.$$

By Lemma 3, there exists a $C^{\infty }$ function $f:U\to [0,1]$ which is $1$ on $A$ and $0$ outside of some closed set in $U$. $\Phi =\{f{\phi }_1,\dots ,f{\phi }_n\}$ is the desired partition of unity ¹.

Case 2: $A=A_1\cup A_2\cup \dots$, where each $A_i$ is compact and $A_i\subseteq A_{i+1}^{\circ }$

For each $i$ let ${\mathcal{O}}_i$ consist of all $U\cap (A_{i+1}^{\circ }\setminus A_{i-2})$ for $U\in \mathcal{O}$ ². Then ${\mathcal{O}}_i$ is an open cover of the compact set $B_i=A_i\setminus A_{i-1}^{\circ }$. By case 1, there is a partition of unity ${\Phi }_i$ for $B_i$ subordinate to ${\mathcal{O}}_i$. For each $x\in A$ the sum

$$\sigma (x)=\sum _{\phi \in {\Phi }_i,i\in \mathbb{N}}\phi (x)$$

is a finite sum in some open set containing $x$, since if $x\in A_i$, then $x$ must have a neighborhood in $A_{i+1}$, so we have $\phi (x)=0$ for $\phi \in {\Phi }_j$ with $j\ge i+3$. For each $\phi$ in each ${\Phi }_i$, define ${\phi }^{\prime }(x)=\phi (x)/\sigma (x)$. The collection of all ${\phi }^{\prime }$ is the desired partition of unity.

Case 3: $A$ is open

Let

$$A_i=\{x\in A:|x|\le i\text{ and distance from }x\text{ to }\partial A\ge 1/i\}$$

and note that $A=A_1\cup A_2\cup \dots$. Apply case 2.

Case 4: $A$ is arbitrary

Let $B$ be the union of all in . By case 3 there is a partition of unity for $B$; this is also a partition of unity for $A$.□

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Note:

• The partitions of unity we constructed were countable in every case.
• Each $\phi \in \Phi$ is compactly supported.

Integration on arbitrary open sets

So far we have defined the integrability of bounded functions on closed rectangles in ${\mathbb{R}}^n$, and then extended this notion to bounded functions on Jordan-measurable subsets of ${\mathbb{R}}^n$.

Not every open set is Jordan-measurable; this could be merely due to it being unbounded, but examples of bounded open sets which are not Jordan measurable exist (Ramadas, n.d.). We now define the class of absolutely integrable functions on an arbitrary open set $A\subseteq {\mathbb{R}}^n$. In case $A$ is bounded and Jordan measurable, this new definition will agree with the earlier one.

Definition 198.5(Definition).

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

Let $A\subseteq {\mathbb{R}}^n$ be open and $\mathcal{O}$ be an admissible open cover. By Theorem 4, there exists a countable family $\Phi$ subordinate to $\mathcal{O}$ of nonnegative continuous functions $\phi$ with compact support such that locally finitely many are non-zero. Let $f:A\to \mathbb{R}$ be such that

1. $f$ is bounded in some open set around each point of $A$ (so $f$ is bounded on any closed rectangle in $A$), and
2. the set of discontinuities of $f$ has measure zero.

Then, ${\int }_A\phi |f|$ exists³ for each $\phi \in \Phi$. We define $f$ to be integrable if

$$\sum _{\phi \in \Phi }{\int }_A\phi |f|<\infty .$$

Since the $\phi$‘s are nonnegative, this implies ${\sum }_{\phi \in \Phi }\left|{\int }_A\phi f\right|<\infty$, which implies the absolute convergence of ${\sum }_{\phi \in \Phi }{\int }_A\phi f$, which we define to be ${\tilde{\int }}_{A}f$ ⁴

$${\tilde{\int }}_{A}f:=\sum _{\phi \in \Phi }{\int }_A\phi \cdot f.$$

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The following theorem proves soundness and backward compatibility of Definition 5.

Theorem 198.6(Spivak (1965) 3-12).

1. Definition 5 does not depend on $\mathcal{O}$ or $\Phi$. That is, if $\Psi$ is another partition of unity subordinate to an admissible open cover $\mathcal{U}$ of $A$, then ${\sum }_{\psi \in \Psi }{\int }_A\psi |f|$ also converges, and

$$\sum _{\phi \in \Phi }{\int }_A\phi \cdot f=\sum _{\psi \in \Psi }{\int }_A\psi \cdot f.$$

Proof.

Since $\phi f=0$ except on some compact set $C$, and there are only finitely many $\psi$ which are non-zero on $C$ ⁵, we can write

$$\sum _{\phi \in \Phi }{\int }_A\phi \cdot f=\sum _{\phi \in \Phi }{\int }_A\sum _{\psi \in \Psi }\psi \cdot \phi \cdot f=\sum _{\phi \in \Phi }\sum _{\psi \in \Psi }{\int }_A\psi \cdot \phi \cdot f.$$E1

This result, applied to $|f|$, show the convergence of ${\sum }_{\phi \in \Psi }{\sum }_{\psi \in \Psi }{\int }_A\psi \phi |f|$, and hence of ${\sum }_{\phi \in \Psi }{\sum }_{\psi \in \Psi }\left|{\int }_A\psi \phi f\right|$. This absolute convergence justifies interchanging the order of summation in (E1); the resulting expression becomes ⁶

$$\sum _{\psi \in \Psi }{\int }_A\psi \cdot f.$$

This result applied to $|f|$ proves convergence of ${\sum }_{\psi \in \Psi }{\int }_A\psi \cdot |f|$.□

2. If $A$ and $f$ are bounded, then ${\tilde{\int }}_{A}f$ exists.

Proof.

If $A$ is contained in the closed rectangle $B$ and $|f(x)|\le M$ for $x\in A$, and $F\subseteq \Phi$ is finite, then

$$\sum _{\phi \in F}{\int }_A\phi \cdot |f|\le \sum _{\phi \in F}M{\int }_A\phi =M{\int }_A\sum _{\phi \in F}\phi \le M\cdot \text{Vol}(B),$$

since ${\sum }_{\phi \in F}\phi \le 1$ on $A$.□

3. If $A$ is Jordan-measurable and $f$ is bounded, Definition 5 agrees with Definition 196.2, that is, ${\tilde{\int }}_{A}f={\int }_{A}f$.

Proof.

If $ϵ>0$, there is a compact Jordan-measurable $C\subseteq A$ such that ${\int }_{A}1-{\int }_{C}1<ϵ$. Again, there are only finitely many $\phi \in \Phi$ which are nonzero on $C$. If $F\subseteq \Phi$ is any finite collection which includes these, then

$$\begin{array}{rl}\left|{\int }_{A}f-\sum _{\phi \in F}{\int }_A\phi \cdot f\right|& \le {\int }_A\left|f-\sum _{\phi \in F}\phi \cdot f\right|\\ & \le M{\int }_A\left(1-\sum _{\phi \in F}\phi \right)\\ & \le M\left({\int }_{A}1-{\int }_C\sum _{\phi \in F}\phi \right)\\ & =M\left({\int }_{A}1-{\int }_{C}1\right)\\ & <Mϵ.\end{array}$$ □

Footnotes

1. Note that the ${\psi }_i$‘s (and hence the ${\phi }_i$‘s) bleed out of $U$; we introduce $f$ only to ensure that the support of each function in $\Phi$ is contained in $U$. ↩

2. Take $A_0,A_{-1}$ to be $\varnothing$. ↩

3. As previously noted, we cannot rely on $A$ being Jordan measurable here. Recall that $\phi$ has compact support $C$, which must be contained in $A$ since $\mathcal{O}$ is admissible. We can cover $C$ by a collection $\mathcal{R}$ of finitely many disjoint closed rectangles. Thus, the integral ${\int }_A\phi |f|$ becomes ${\int }_C\phi |f|={\sum }_{R\in \mathcal{R}}{\int }_R\phi |f|$, which exists thanks to our hypotheses regarding $f$. ↩

4. Integrals using the new definition will be embellished with ~ in this article. ↩

5. For each $x\in C$, let $V_x$ be a neighborhood of $x$ such that only finitely many $\phi$‘s are nonzero on $V_x$. $\{V_x:x\in C\}$ is an open cover of $C$, of which there exists a finite subcover. ↩

6. For each fixed $\psi$, only finitely many of the $\phi$‘s are relevant, which allows us to swap the sum and integral. ↩

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References

Ramadas, T. R. (n.d.). (MULTIDIMENSIONAL INTEGRAL) CALCULUS.

Spivak, M. (1965). Calculus on Manifolds: A Modern Approach to Classical Theorems of Advanced Calculus. Addison-Wesley publ.