STEP FUNCTIONS AND RIEMANN SUMS - Multiple Integrals - Advanced Calculus of Several Variables

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 : ImagenImage is called a step function if and only if h can be written as a linear combination of characteristic functions φ1, . . . , φp of intervals I1, . . . , Ip whose interiors are mutually disjoint, that is,

Image

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

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

Image

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

Assume for simplicity that each ai > 0. Then the ordinate set Oh contains the set

Image

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

Image

It follows easily that

Image

It follows immediately from Theorem 3.1 that, if h is a step function and Image, then ∫ ch = ch. 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

Image

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

Image

Image

Figure 4.14

Then

Image

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

Image

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

Image

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

Theorem 3.3Let f : ImagenImage 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

Image

in which case Image.

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

Image

has volume equal to ∫(k − h) < ε. But S contains the set Imagef = ∂OfImagen (Fig. 4.15). Thus for every ε > 0, Imagef lies in a set of volume < ε. It follows easily that Imagef is a negligible set. If Q is a rectangle in Imagen such that f = 0 outside Q, then Image and Image are both subsets of Image, so it follows that both are negligible. Therefore the ordinate sets Image and Image are both contented, so f is integrable. The fact that Image 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 Image. Then the fact that f is integrable means by definition that Of is contented. Hence, given ε > 0, there exist

Image

Figure 4.15

nonoverlapping intervals I1, . . . , Iq contained in Of, and intervals J1, . . . , Jp with Image, such that

Image

see Fig. 4.16.

Given Image, define

Image

if the vertical line in Imagen+1 through Image intersects some Ii (respectively, some Jj), and let h(x) = 0 (respectively, k(x) = 0) otherwise. Then h and k are step

Image

Figure 4.16

functions such that Image. Since

Image

the above inequalities imply that

Image

as desired.

Image

ExampleWe can apply Theorem 3.3 to prove again that continuous functions are integrable. Let f : ImagenImage 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 Image 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

Image

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

Image

and Image then Image and

Image

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

Image

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

Image

and ∫(kihi) < ε/2ai for i = 1, 2. Then

Image

where h and k are step functions such that

Image

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

At the same time it follows from (1) that

Image

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

Image

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

Image

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] → Image be a nonnegative function. Let the points a = x0 < x1 < · · · < xk = b subdivide the interval [a, b] into k subintervals [x0, x1], [x1, x2], . . . , [xk−1, xk]. For each i = 1, . . . , k, choose a point Image. (Fig. 4.17). Then f(xi*)(xixi−1) is the area of the rectangle of height f(xi*) whose base is the ith subinterval [xi−1, xi], so one suspects that the sum

Image

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

Image

where φi denotes the characteristic function of the interval [xi−1, xi].

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

Image

Figure 4.17

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

Image

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

Image

where φi denotes the characteristic function of Qi.

Theorem 3.4Suppose f : ImagenImage 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

Image

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

PROOFIf f is integrable, choose M > 0 such that Image for all x. By Theorem 3.3 there exist step functions h and k such that Image 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 Image0 = {Q1, . . . , Qs} of Q such that

Image

where Qi is the closure of the interval of which φi is the characteristic function, and likewise for ψi.

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

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

Image

are both between Image and Image so it follows that

Image

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

Since Image for all x, both

Image

lie between Image and Image, so it follows that

Image

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

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

Image

Let Q1, . . . , Qp be disjoint intervals (not closed) such that Imagei = Pi, i = 1, . . . , p, and denote by φi the characteristic function of Qi. If

Image

then

Image

are step functions such that Image.

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

Image

for each i. Then

Image

and similarly

Image

Then

Image

so it follows from Theorem 3.3 that f is integrable.

Image

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 : ImagenImage is an integrable function which vanishes outside the interval Q, and Image is a sequence of partitions of Q with associated selections Image such that limk→∞ (mesh of Imagek) = 0, then

Image

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

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

Image

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

Image

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

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

Image

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

Image

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

Image

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

Image

Figure 4.18

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

Image

and the function g : JImage by g(t) = ∫Af(x, t) dx. We ask whether

Image

that is, whether

Image

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 Image and ∫(k − h) < ε/M, where M is the maximum value of Imagek(x) + h(x)Image. Then prove that h2 and k2 are step functions with Image (we may assume that Imagesince f is integrable if and only if ImagefImage is—why?), and that ∫ (k2h2) < 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 : ImagenImage be a bounded function which vanishes outside the interval Q. Show that f is integrable with ∫ f = I if and only if

Image

for every sequence of partitions Image and associated selections Image such that limk→∞ (mesh of Imagek) = 0.

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

Image

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

Image

Hint: Note that Image.

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

Image

prove that

Image

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

Image

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

Image

Let

Image

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

Image

as desired.

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

Image

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

Image

and

Image

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 Jn(x) may be defined by

Image

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

Image

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 ∫ fn exists because Image is a Cauchy sequence of real numbers (why?).

Given ε > 0, choose N such that both ImageI − ∫ fNImage < ε/3 and Imagef(x) − fN(x)Image < ε/3v(Q) for all x. Then choose δ > 0 such that, for any partition Image of mesh < δ and any selection Image,

Image

Noting that

Image

also, conclude that

Image

as desired.

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

Image

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

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

Image

under appropriate hypotheses (state them).