Advanced Calculus of Several Variables (1973)

Part IV. Multiple Integrals

Chapter 2. VOLUME AND THE n-DIMENSIONAL INTEGRAL

In Section 1 we used the properties of area (without verification) to study the integral of a function of one variable. This was done mainly for purposes of review and motivation. We now start anew, with definitions of volume for subsets of ⁿ, and of the integral for functions on ⁿ.

These definitions are the obvious generalizations to higher dimensions of those given in Section 1. Recall that a closed interval in ⁿ is a set I = I₁ × I₂ × · · · × I_n, where . Thus a closed interval I in ⁿ is simply a Cartesian product of “ordinary” closed intervals. The volume of I is, by definition, v(I) = (b₁ − a₁)(b₂ − a₂) · · · (b_n − a_n). Now let A be a bounded subset of ⁿ. Then we say that A is a contented set with volume v(A) if and only if, given ε > 0, there exist

(a)nonoverlapping closed intervals such that , and

(b)closed intervals J₁, . . . , J_q such that

We have seen (Example 1 of Section 1) that not all sets are contented. Consequently we need an effective means of detecting those that are. This will be given in terms of the boundary of a set. Recall that the boundary ∂A of the set is the set of all those points of ⁿ that are limit points of both A and ⁿ − A.

The contented set A is called negligible if and only if v(A) = 0. Referring to the definition of volume, we see that A is negligible if and only if, given ε > 0, there exist intervals J₁, . . . , J_q such that and . It follows immediately that the union of a finite number of negligible sets is negligible, as is any subset of a negligible set.

Theorem 2.1The bounded set A is contented if and only if its boundary is negligible.

PROOFLet R be a closed interval containing A. By a partition of R we shall mean a finite collection of nonoverlapping closed intervals whose union is R.

First suppose that A is contented. Given ε > 0, choose closed intervals I₁, . . . , I_p and J₁, . . . , J_q as in conditions (a) and (b) of the definition of v(A). Let be a partition of R such that each I_i and each J_j is a union of closed intervals of . Let R₁, . . . , R_k be those intervals of which are contained in , and R_k+1, . . . , R_k+l the additional intervals of which are contained in . Then

and

Now suppose that ∂A is negligible. Note first that, in order to prove that A is contented, it suffices to show that, given ε > 0, there exist intervals J₁, . . . , J_q containing A and nonoverlapping intervals I₁, . . . , I_p contained in A such that

for then (the greatest lower and least upper bounds for all such collections of intervals) satisfies the definition of volume for A.

To show this, let R₁, . . . , R_k be intervals covering ∂A with , and let be a partition of R such that each R_i is a union of elements of . If I₁, . . . , I_p are the intervals of which are contained in A, and J₁, . . . , J_qare the intervals of which are contained in , then , and . Since the intervals I₁, . . . , I_p are included among the intervals J₁, . . . , J_q, it follows that

as desired.

Corollary The intersection, union, or difference of two contented sets is contented.

This follows immediately from the theorem, the fact that a subset of a negligible set is negligible, and the fact that , and ∂(A − B) are all subsets of .

The utility of Theorem 2.1 lies in the fact that negligible sets are often easily recognizable as such. For instance, we will show in Corollary 2.3 that, if f: A → is a continuous function on the contented set , then the graph of f is a negligible subset of ⁿ.

ExampleLet Bⁿ be the unit ball in ⁿ, with ∂Bⁿ = Sⁿ⁻¹, the unit (n − 1)-sphere. Assume by induction that Bⁿ⁻¹ is a contented subset of . Then Sⁿ⁻¹ is the union of the graphs of the continuous functions

and

Hence Sⁿ⁻¹ is negligible by the above remark, so Bⁿ is contented by Theorem 2.1.

We turn now to the definition of the integral. Given a nonnegative function f : ⁿ → , we consider the ordinate set

In order to ensure that the set O_f is bounded, we must require that f be bounded and have bounded support. We say that f : ⁿ → is bounded if there exists such that for all , and has bounded supportif there exists a closed interval such that f(x) = 0 if (Fig. 4.9). We will define the integral only for bounded functions on ⁿ that have bounded support; it will turn out that this restriction entails no loss of generality.

Figure 4.9

Given f: Rⁿ → R, we define the positive and negative parts f⁺ and f⁻ of f by

Then f = f⁺ − f⁻ (see Fig. 4.10). That is, f⁺(x) = f(x) if f(x) > 0, and f⁻(x) = −f(x) if f(x) < 0, and each is 0 otherwise.

Suppose now that f: ⁿ → is bounded and has bounded support. We then say that f is (Riemann) integrable (and write ) if and only if the ordinate sets and are both contented, in which case we define

Thus ∫ f is, by definition, the volume of the region in ⁿ⁺¹ above ⁿ and below the graph of f, minus the volume of the region below ⁿ and above the graph of f (just as before, in the 1-dimensional case).

Although f: ⁿ → has bounded support, we may think of ∫f as the “integral of f over ⁿ.” In order to define the integral of f over the set , thinking of the volume of the set in ⁿ⁺¹ which lies “under the graph of f and over the set A,” we introduce the characteristic function φ_A of A, defined by

Figure 4.10

We then define

provided that the product fφ_A is integrable; otherwise ∫_A f is not defined, and f is not integrable on A (see Fig. 4.11).

Figure 4.11

The basic properties of the integral are the following four “axioms” (which of course must be established as theorems).

Axiom I The set of integrable functions is a vector space.

Axiom II The mapping ∫ : → is linear.

Axiom III If everywhere, then .

Axiom IV If the set A is contented, then

Axiom III follows immediately from the definition of the integral. Axiom IV is almost as easy. For if I₁, . . . , I_q and J₁, . . . , J_p are intervals in ⁿ such that

and

then

So it follows that

Axioms I and II, which assert that, if and , then and ∫(af + bg) = a ∫f + b ∫ g, will be verified in the following section.

In lieu of giving necessary and sufficient conditions for integrability, we next define a large class of integrable functions which includes most functions of frequent occurrence in practice.

The function f is called admissible if and only if

(a) f is bounded,

(b) f has bounded support, and

(c) f is continuous except on a negligible set.

Condition (c) means simply that, if D is the set of all those points at which f is not continuous, then D is negligible. It is easily verified that the set of all admissible functions on ⁿ is a vector space (Exercise 2.1).

Theorem 2.2Every admissible function is integrable.

PROOF Let f be an admissible function on ⁿ, and R an interval outside of which f vanishes. Since f⁺ and f⁻ are admissible (Exercise 2.2), and is a vector space, we may assume that f is nonnegative.

Choose M such that for all x, and let D denote the negligible set of points at which f is not continuous.

Given ε > 0, let Q₁, . . . , Q_k be a collection of closed intervals in ⁿ such that

By expanding the Q_i's slightly if necessary, we may assume that .

Then f is continuous on the set . Since Q is closed and bounded, it follows from Theorem 8.9 of Chapter I that f is uniformly continuous on Q, so there exists δ > 0 such that

if and x − y < δ.

Now let be a partition of R such that each Q_i is a union of intervals of , and each interval of has diameter <δ. If R₁, . . . , R_q are the intervals of that are contained in Q, and

Figure 4.12

then b_i − a_i < ε/2v(R) for i = 1, . . . , q. Finally let, as in Fig. 4.12,

Let . Since O_f = O_f* − R, and the interval R is contented, it suffices to see that O_f^* is contented (Fig. 4.13). But I₁, . . . , I_q is a collection of

Figure 4.13

nonoverlapping intervals contained in O_f^*, and J₁, . . . , J_q+k is a collection of intervals containing O_f^*, while

Hence it follows that O_f^* is contented as desired.

We can now establish the previous assertion about graphs of continuous functions.

Corollary 2.3 If f : A → is continuous, and is contented, then the graph G_f of f is a negligible set in ⁿ⁺¹.

PROOF If we extend f to ⁿ by f(x) = 0 for , then f is admissible because its discontinuities lie in ∂A, which is negligible because A is contented. Hence O_f is contented by the Theorem, so ∂O_f is negligible. But G_f is a subset of ∂O_f.

It follows easily from see Corollary 2.3 that, if f : A → is a nonnegative continuous function on a contented set , then the ordinate set of f is a contented subset of ⁿ⁺¹ (Exercise 2.7). This is the analogue for volume of Property E in Section 1.

The following proposition shows that ∫_A f is defined if f : ⁿ → is admissible and is contented. This fact serves to justify our concentration of attention (in the study of integration) on the class of admissible functions—every function that we will have “practical” need to integrate will, in fact, be an admissible function on a contented set.

Proposition 2.4 If f is admissible and A is contented, then fφ_A is admissible (and, therefore, is integrable by Theorem 2.2).

PROOFIf D is the negligible set of points at which f is not continuous, then fφ_A is continuous at every point not in the negligible set (see Exercise 2.3).

Proposition 2.5 If f and g are admissible functions on ⁿ with everywhere, and A is a contented set, then

PROOF everywhere, so Axiom III gives . Then Axiom II gives

It follows easily from see Proposition 2.5 that, if A and B are contented sets in ⁿ, with , then (Exercise 2.4). This is the analog for volume of Property A of Section 1.

Proposition 2.6 If the admissible function f satisfies for all , and A is contented, then

In particular, ∫_Af = 0 if A is negligible.

PROOFSince , this follows immediately from Proposition 2.5 and Axioms II and IV.

Proposition 2.7 If A and B are contented sets with negligible, and f is admissible, then

PROOFFirst consider the special case . Then , so

If is negligible, then by Proposition 2.6. Then, applying the special case, we obtain

It follows easily from see Proposition 2.7 that, if A and B are contented sets with negligible, then is contented, and (Exercise 2.5). This is the analog for volume of Property B of Section 1.

Theorem 2.8Let A be contented, and suppose the admissible functions f and g agree except on the negligible set D. Then

PROOFLet h = f − g, so h = 0 except on D. We need to show that ∫_A h = 0. But

by Proposition 2.6, because is negligible.

This theorem indicates why sets with volume zero are called “negligible”—insofar as integration is concerned, they do not matter.

Exercises

2.1If f and g are admissible functions on ⁿ and , show that f + g and cf are admissible.

2.2Show that the positive and negative parts of an admissible function are admissible.

2.3If D is the set of discontinuities of f : ⁿ → , show that the set of discontinuities of fφ_A is contained in .

2.4If A and B are contented sets with , apply Proposition 2.5 with f = φ_A = φ_Bφ_A, g = φ_B to show that

2.5Let A and B be contented sets with negligible. Apply Proposition 2.7 with to show that is contented with

2.6If f and g are integrable functions with for all x, prove that without using Axioms I and II. Hint: and and .

2.7If A is a contented subset of ⁿ and f : A → is continuous and nonnegative, apply Corollary 2.3 to show that O_f is contented. Hint: Note that the boundary of O_f is the union of , the graph of f, and the ordinate set of the restriction f : ∂A → . Conclude that ∂O_f is negligible, and apply Theorem 2.1.