Advanced Calculus of Several Variables (1973)

Part IV. Multiple Integrals

Chapter 3. STEP FUNCTIONS AND RIEMANN SUMS

As a tool for studying the integral, we introduce in this section a class of very simple functions called step functions. We shall see first that the properties of step functions are easily established, and then that an arbitrary function is integrable if and only if it is, in a certain precise sense, closely approximated by step functions.

The function h : ⁿ → is called a step function if and only if h can be written as a linear combination of characteristic functions φ₁, . . . , φ_p of intervals I₁, . . . , I_p whose interiors are mutually disjoint, that is,

with coefficients . Here the intervals I₁, I₂·, . . . , I_p are not necessarily closed—each is simply a product of intervals in (the latter may be either open or closed or “half-open”).

Theorem 3.1If h is a step function, as above, then h is integrable with

PROOFIn fact h is admissible, since it is clearly continuous except possibly at the points of the negligible set .

Assume for simplicity that each a_i > 0. Then the ordinate set O_h contains the set

whose volume is clearly , and is contained in the union of A and the negligible set

It follows easily that

It follows immediately from Theorem 3.1 that, if h is a step function and , then ∫ ch = c ∫ h. The following theorem gives the other half of the linearity of the integral on step functions.

Theorem 3.2If h and k are step functions, then so is h + k, and ∫(h + k) = ∫ h + ∫ k.

PROOFIn order to make clear the idea of the proof, it will suffice to consider the simple special case

where φ and ψ are the characteristic functions of intervals I and J. If I and J have disjoint interiors, then the desired result follows from the previous theorem.

Otherwise is an interval I₀, and it is easily seen that I − I₀ and J − I₀ are disjoint unions of intervals (Fig. 4.14)

Figure 4.14

Then

so we have expressed h + k as a step function. Theorem 3.1 now gives

as desired. The proof for general step functions h and k follows from this construction by induction on the number of intervals involved.

Our main reason for interest in step functions lies in the following characterization of integrable functions.

Theorem 3.3Let f : ⁿ → be a bounded function with bounded support. Then f is integrable if and only if, given ε > 0, there exist step functions h and k such that

in which case .

PROOFSuppose first that, given ε > 0, step functions h and k exist as prescribed. Then the set

has volume equal to ∫(k − h) < ε. But S contains the set _f = ∂O_f − ⁿ (Fig. 4.15). Thus for every ε > 0, _f lies in a set of volume < ε. It follows easily that _f is a negligible set. If Q is a rectangle in ⁿ such that f = 0 outside Q, then and are both subsets of , so it follows that both are negligible. Therefore the ordinate sets and are both contented, so f is integrable. The fact that follows from Exercise 2.6 (without using Axioms I and II, which we have not yet proved).

Now suppose that f is integrable. Since this implies that both f⁺ and f⁻ are integrable, we may assume without loss that . Then the fact that f is integrable means by definition that O_f is contented. Hence, given ε > 0, there exist

Figure 4.15

nonoverlapping intervals I₁, . . . , I_q contained in O_f, and intervals J₁, . . . , J_p with , such that

see Fig. 4.16.

Given , define

if the vertical line in ⁿ⁺¹ through intersects some I_i (respectively, some J_j), and let h(x) = 0 (respectively, k(x) = 0) otherwise. Then h and k are step

Figure 4.16

functions such that . Since

the above inequalities imply that

as desired.

ExampleWe can apply Theorem 3.3 to prove again that continuous functions are integrable. Let f : ⁿ → be a continuous function having compact support. Since it follows that the nonnegative functions f⁺ and f⁻ are continuous, we may as well assume that f itself is nonnegative. Let be a closed interval such that f = 0 outside Q. Then f is uniformly continuous on Q (by Theorem 8.9 of Chapter I) so, given ε > 0, there exists δ > 0 such that

Now let = {Q₁, . . . , Q_l} be a partition of Q into nonoverlapping closed intervals, each of diameter less than δ. If

and then and

so Theorem 3.3 applies.

We are now prepared to verify Axioms I and II of the previous section. Given integrable functions f₁ and f₂, and real numbers a₁ and a₂, we want to prove that a₁f₁ + a₂f₂ is integrable with

We suppose for simplicity that a₁ > 0, a₂ > 0, the proof being similar in the other cases.

Given ε > 0, Theorem 3.3 provides step functions h₁, h₂, k₁, k₂ such that

and ∫(k_i − h_i) < ε/2a_i for i = 1, 2. Then

where h and k are step functions such that

So it follows from Theorem 3.3 that a₁f₁ + a₂f₂ is integrable.

At the same time it follows from (1) that

(by Exercise 2.6), and similarly from (2) that

Since ∫ k − ∫ h < ε, it follows that ∫(a₁f₁ + a₂f₂) and a₁ ∫f₁ + a₂∫f₂ differ by less than ε. This being true for every ε > 0, we conclude that

We now relate our definition of the integral to the “Riemann sum” definition of the single-variable integral which the student may have seen in introductory calculus. The motivation for this latter approach is as follows. Let f : [a, b] → be a nonnegative function. Let the points a = x₀ < x₁ < · · · < x_k = b subdivide the interval [a, b] into k subintervals [x₀, x₁], [x₁, x₂], . . . , [x_k−1, x_k]. For each i = 1, . . . , k, choose a point . (Fig. 4.17). Then f(x_i^*)(x_i − x_i−1) is the area of the rectangle of height f(x_i^*) whose base is the ith subinterval [x_i−1, x_i], so one suspects that the sum

should be a “good approximation” to the area of O_f if the subintervals are sufficiently small. Notice that the Riemann sum R is simply the integral of the step function

where φ_i denotes the characteristic function of the interval [x_i−1, x_i].

Recall that a partition of the interval Q is a collection = {Q₁, . . . , Q_k} of closed intervals, with disjoint interiors, such that . By the mesh of

Figure 4.17

is meant the maximum of the diameters of the Q_i. A selection for is a set = {x₁, . . . , x_k} of points such that for each i. If f : ⁿ → is afunction such that f = 0 outside of Q, then the Riemann sum for f corresponding to the partition and selection is

Notice that, by Theorem 3.1, the Riemann sum R(f, , ) is simply the integral of the step function

where φ_i denotes the characteristic function of Q_i.

Theorem 3.4Suppose f : ⁿ → is bounded and vanishes outside the interval Q. Then f is integrable with ∫f = I if and only if, given ε > 0, there exists δ > 0 such that

whenever is a partition of Q with mesh <δ and is a selection for .

PROOFIf f is integrable, choose M > 0 such that for all x. By Theorem 3.3 there exist step functions h and k such that and ∫ (k − h) < ε/2. By the construction of the proof of Theorem 3.2, we may assume that h and k are linear combinations of the characteristic functions of intervals whose closures are the same. That is, there is a partition ₀ = {Q₁, . . . , Q_s} of Q such that

where Q_i is the closure of the interval of which φ_i is the characteristic function, and likewise for ψ_i.

Now let int Q_i. Then v(A) = 0, so there exists δ > 0 such that, if is a partition of Q with mesh <δ, then the sum of the volumes of those intervals P₁, . . . , P_k of which intersect A is less than ε/4M. Let P_k+1, . . . , P_l be the remaining intervals of , that is, those which lie interior to the Q_i.

If = {x₁, . . . , x_l} is a selection for , then if i = k + 1, . . . , l, so that

are both between and so it follows that

because ∫_Q(k − h) < ε/2 by assumption.

Since for all x, both

lie between and , so it follows that

Since , (3) and (4) finally imply by the triangle inequality that I − R(f, , ) < ε as desired.

Conversely, suppose = {P₁, . . . , P_p} is a partition of Q such that, given any selection for , we have

Let Q₁, . . . , Q_p be disjoint intervals (not closed) such that _i = P_i, i = 1, . . . , p, and denote by φ_i the characteristic function of Q_i. If

then

are step functions such that .

Choose selections ′ = {x₁′, . . . , x_p′} and ″ = {x₁″, . . . , x_p″} for such that

for each i. Then

and similarly

Then

so it follows from Theorem 3.3 that f is integrable.

Theorem 3.4 makes it clear that the operation of integration is a limit process. This is even more apparent in Exercise 3.2, which asserts that, if f : ⁿ → is an integrable function which vanishes outside the interval Q, and is a sequence of partitions of Q with associated selections such that lim_k→∞ (mesh of _k) = 0, then

This observation leads to the formulation of several natural and important integration questions as interchange of limit operations questions.

(1)Let and be contented sets, and f : A × B → a continuous function. Define g : A → by

where f_x(y) = f(x, y). Is g continuous on A? This is the question as to whether

for each . According to Exercise 3.3, this is true if f is uniformly continuous.

(2)Let be a sequence of integrable functions on the contented set A, which converges (pointwise) to the integrable function f : A → . We ask whether

According to Exercise 3.4, this is true if the sequence converges uniformly to f on A. This means that, given ε > 0, there exists N such that

for all . The following example shows that the hypothesis of uniform convergence is necessary. Let f_n be the function on [0, 1 ] whose graph is pictured in Fig. 4.18. Then

so . However . It is evident that the convergence of is not uniform (why?).

Figure 4.18

(3)Differentiating under the integral. Let f : A × J → be a continuous function, where is contented and is an open interval. Define the partial derivative D₂f : A × J → by

and the function g : J → by g(t) = ∫_Af(x, t) dx. We ask whether

that is, whether

According to Exercise 3.5, this is true if D₂f is uniformly continuous on A × J. Since differentiation is also a limit operation, this is another “interchange of limit operations” result.

Exercises

3.1(a) If f is integrable, prove that f² is integrable. Hint: Given ε > 0, let h and k be step functions such that and ∫(k − h) < ε/M, where M is the maximum value of k(x) + h(x). Then prove that h² and k² are step functions with (we may assume that since f is integrable if and only if f is—why?), and that ∫ (k² − h²) < ε. Then apply Theorem 3.3.

(b) If f and g are integrable, prove that fg is integrable. Hint: Express fg in terms of (f + g)² and (f − g)²; then apply (a).

3.2Let f : ⁿ → be a bounded function which vanishes outside the interval Q. Show that f is integrable with ∫ f = I if and only if

for every sequence of partitions and associated selections such that lim_k→∞ (mesh of _k) = 0.

3.3Let A and B be contented sets, and f : A × B → a uniformly continuous function. If g : A → is defined by

prove that g is continuous. Hint: Write g(x) − g(a) = ∫_B [f(x, y) − f(a, y)] dy and apply the uniform continuity of f.

3.4Let be a sequence of integrable functions which converges uniformly on the contented set A to the integrable function f. Then prove that

Hint: Note that .

3.5Let f : A × J → be a continuous function, where is contented and is an open interval. Suppose that is uniformly continuous on A × J. If

prove that

Outline: It suffices to prove that, if is a sequence of distinct points of J converging to a J, then

For each fixed J converging to x A, the mean value theorem gives such that

Let

Now the hypothesis that D₂f is uniformly continuous implies that the sequence converges uniformly on A to the function D₂ f(x, a) of x (explain why). Therefore, by the previous problem, we have

as desired.

3.6Let f : [a, b] × [c, d] → be a continuous function. Prove that

by computing the derivatives of the two functions g, h : [a, b] → defined by

and

using Exercise 3.5 and the fundamental theorem of calculus. This is still another interchange of limit operations.

3.7For each positive integer n, the Bessel function J_n(x) may be defined by

Prove that J_n(x) satisfies Bessel's differential equation

3.8Establish the conclusion of Exercise 3.4 without the hypothesis that the limit function f is integrable. Hint: Let Q be a rectangle containing A, and use Theorem 3.4 as follows to prove that f must be integrable. Note first that I = lim ∫ f_n exists because is a Cauchy sequence of real numbers (why?).

Given ε > 0, choose N such that both I − ∫ f_N < ε/3 and f(x) − f_N(x) < ε/3v(Q) for all x. Then choose δ > 0 such that, for any partition of mesh < δ and any selection ,

Noting that

also, conclude that

as desired.

The following example shows that the uniform convergence of is necessary. Let be the set of all rational numbers in the unit interval [0, 1]. If f_n denotes the characteristic function of the set {r₁, . . . , r_n}, then for each n = 1, 2, . . . , but the limit function

is not integrable. The point is that the convergence of this sequence is not uniform.

3.9If , apply Exercise 3.5, the fundamental theorem of calculus, and the chain rule to prove Leibniz' rule:

under appropriate hypotheses (state them).