An Analysis of the Henstock-Kurzweil Integral

Home , Henri Lebesgue

A Thesis Submitted to the Graduate School in Partial Fulﬁllment for the degree of Masters of Science

Joshua L. Turner

Advisor: Dr. Ahmed Mohammed Ball State University July 2015

1 ACKNOWLEDGEMENTS

Thank you to my thesis advisor, Dr. Ahmed Mohammed, my thesis committee, Dr. Ralph Bremigan

and Dr. Rich Stankewitz, and to Dr. Hanspeter Fischer for their guidance and their fortitude in

enduring my ineptitude.

2 §1 INTRODUCTION.

Since the inception of integration theory, mathematicians have sought to find an integral which would make integration and differentiation truly inverse processes. Today, one of the first theories of integration that most mathematicians learn about is the Riemann integral. The Riemann integral, although very powerful is not without its flaws. One of the first mathematicians to exploit the flaws in Riemann’s integral was the French mathematician Johann Peter Gustav Lejeune Dirichlet. Dirichlet constructed the following bounded function which is not Riemann integrable  1 if x ∈ Q f(x) =  0 if x ∈ R \ Q.

This function illustrates an inherent flaw in Riemann’s integral by showing that it cannot integrate functions with too many discontinuities. Mathematicians began to notice other problems with the Riemann’s integral as well. For example, they realized that the Riemann integral cannot integrate every derivative. Hence, as an operator, it is not a true inverse to the derivative. Moreover, in order to integrate functions over infinite intervals or functions with vertical asymptotes, one must introduce an improper Riemann integral. Subsequently, these flaws led mathematicians to see integration in a whole new light. Prior to this time, the mathematical community had rarely considered pathological functions such as Dirichlet’s. As a result, the very foundation of analysis seemed to be on unsteady ground. It was not until nearly one-hundred years later that a young mathematician named Henri Lebesgue developed a theory of integration that could handle most of the problems that plagued the Riemann integral. Lebesgue examined the effect of partitioning the range of a function rather than the domain thus allowing more control over small variations in the graph of the function. However, Lebesgue’s theory was much more complicated than Riemann’s and forced mathematicians to learn very deep mathematical concepts most of which were relatively unfamiliar and new to mathematics. After some time, as is their nature, mathematicians began finding flaws in Lebesgue’s integral. First of all, like the Riemann integral, the Lebesgue integral is not a true inverse to differentiation. In addition, in the most general definition of the Lebesgue integral, the integrability of a function f is contingent upon the integrability of |f|. Although Lebesgue’s integral would prove to be a valuable tool for the working mathematician, it was not the end of the story. Nearly twenty years after Lebesgue’s death, two mathematicians simultaneously developed an integral that integrates any derivative making it a true inverse to differentiation. Furthermore, this new integral generalizes both the Riemann and Lebesgue integral. It is this theory that we discuss today.

Let I = [a, b] ⊆ R be a compact interval. Let f : I → R be a real-valued function. We will investigate

3 the Henstock-Kurzweil or “HK” integral and compare it to the integrals of Riemann and Lebesgue. In the classic version of the Fundamental Theorem of Calculus, given a compact interval I = [a, b] ⊆ R and functions f : I → R and F : I → R with F 0(x) = f(x) for all x ∈ I, neither the Riemann, or “R” integral nor the Lebesgue, or “L” integrals guarantees that

Z b f = F (b) − F (a). a

However, the HK integral does guarantee this result. Furthermore, the HK integral is an non-absolute integral in the sense that a function f may be HK integrable without |f| being HK integrable. In addition, the space

HK(I) of Henstock-Kurzweil integrable functions defined on a compact interval I = [a, b] ⊆ R cannot be extended by adding on improper integrals in the sense that if a function f has an improper integral, then f is already HK integrable. In this paper we look at some of these results. We begin by investigating specific properties of the HK integral which illustrate how a relatively small change in the definition of the R integral can have far reaching consequences. We follow this by defining the HK integral and looking at some of its most powerful results. Next, we examine some of the differences between the HK integral and the integrals of Riemann and Lebesgue. In these sections, we look at various functions which are HK integrable and yet are neither Riemann nor Lebesgue integrable. One such function resembles the Volterra function, but with a much simpler construction. In particular, we show that

R(I) $ L(I) $ HK(I) where R(I) and L(I) denote the space of Riemann and Lebesgue integrable functions deﬁned on a compact interval I ⊆ R respectively. In addition, we look at the beneﬁt of being a non-absolute integral in the sense that a function f can be Henstock-Kurzweil integrable without |f| being Henstock-Kurzweil integrable. We exploit this property by showing that the HK integral can integrate any conditionally convergent series while the Lebesgue integral cannot. Finally, we discuss the consequences of using the HK integral in various areas of applied mathematics.

4 Contents

1 INTRODUCTION 3

2 BASIC DEFINITIONS 6

3 GAUGES 9

4 THE HK INTEGRAL 10

5 PROPERTIES OF THE HK INTEGRAL 11

6 THE FUNDAMENTAL THEOREM OF CALCULUS 13

7 FUNCTIONS WHICH ARE HK INTEGRABLE BUT NOT R INTEGRABLE 15

8 FUNCTIONS WHICH ARE HK INTEGRABLE BUT NOT L INTEGRABLE 26

9 APPLICATIONS OF THE HK INTEGRAL 34

10 FINAL COMMENTS 36

5 §2 BASIC DEFINITIONS.

We begin by introducing some preliminary deﬁnitions which serve to illustrate the prowess of the HK integral. Let I = [a, b] be a compact interval in R with a < b.

n A partition P is a ﬁnite collection of non-degenerate closed intervals {Ii}i=1 whose union is I. Here,

Ii = [xi−1, xi], where

a = x0 < x1 < . . . < xn = b.

We call each closed interval Ik = [xk−1, xk] a subinterval of the partition P.

· n A tagged partition P = {(Ii, ti)}i=1 is a ﬁnite set of ordered pairs where the closed intervals Ii form a partition of I and the numbers ti ∈ Ii are the corresponding tags.

The mesh kPk of a partition P is given by max{`([xi−1, xi]) : i = 1, 2, . . . , n} where `([xi−1, xi]) = xi − xi−1.

· n Given a tagged partition P = {(Ii, ti)}i=1 of I, and a function f : I → R, the real-number

n · X S(f; P) = f(ti)∆xi, i=1

· where ∆xi = xi − xi−1, is the Riemann Sum of f induced by P.

Remark: As we will see, in our approach to integration we view the integral as a limit of Riemann sums. By using tagged partitions, we allow ourselves some ﬂexibility with the placement of tags in that adding or · eliminating points from a partition P, often the tags themselves, if done with care, will not eﬀect the value of the Riemann sum.

· n For example, let P = {(Ii, ti)}i=1 be a tagged partition of I. Let tk be an interior point of the subinterval · · Ik = [xk−1, xk]. Now, let Q be the partition of I obtained from P by adding the new partition point ξ = tk, so that

a = x0 < . . . < xk−1 < ξ < xk < . . . < xn = b.

· We can now use ξ as the tag for both subintervals [xk−1, ξ] and [xk, ξ] of Q. Since tk = ξ, and

f(tk)(xk − xk−1) = f(ξ)(ξ − xk−1) + f(ξ)(xk − ξ),

6 · · S(f; P) = S(f; Q).

This process can also be reversed by merging two subintervals which share a tag ξ. In this case, the tag of the resulting subinterval will no longer be an endpoint. Thus, when using tagged partitions, we may assume any of the following:

• Every tag tk is an endpoint of a subinterval Ik.

• No point tk is a tag for two distinct subintervals Ik and Ik+1.

Let f : I → R be a real-valued function. We say that f is Riemann or R integrable on I if there is a · n real number A such that for every > 0 there is a number δ > 0 such that if P = {(Ii, ti)}i=1 is any tagged partition of I with kPk < δ, then

· S(f; P) − A < .

It can be shown that the number A is unique, and in this case,

Z b A = R f a or if it is clear that we are using a Riemann integral,

Z b A = f. a

Remark: If f is Riemann integrable on I then it is known that f is bounded on I. Using an alternate scheme, we can view the Riemann integral through the lens of the following terminology.

Let f : A → R be a bounded real-valued function and A a non-empty subset of R. We deﬁne the oscillation of f on A to be

ωf (A) = supf(x) − inf f(x). x∈A x∈A

Let f : I → R be a bounded real-valued function. Let P be a partition of I. Then deﬁne

Mj = sup f(x) , mj = inf f(x) , x∈[xj−1,xj ] x∈[xj−1,xj ]

7 n n X X U(P; f) = Mj∆xj and L(P; f) = mj∆xj j=1 j=1 where ∆xj = (xj − xj−1). The following criteria for Riemann integrability is useful: f is Riemann integrable if and only if there exists a partition P of I such that for each > 0,

U(P; f) − L(P; f) < .

We can rephrase the above criteria in terms of the oscillation since

n n X X U(P; f) − L(P; f) = Mj∆xj − mj∆xj j=1 j=1 n X = (Mj − mj)∆xj j=1 n X = ωf ([xj−1, xj])∆xj. j=1

Let A ⊆ R. The outer measure of A is deﬁned to be

( ∞ ∞ ) X [ inf `(Ik) A ⊆ Uk

k=1 k=1

∞ where {Uk}k=1 is a collection of nonempty, open, bounded intervals.

Let A ⊆ R and f : A → R a real-valued function.

We call f a null function if the set {x ∈ A : f(x) 6= 0} has outer measure zero.

A subset A of R is said to be a null set if its outer measure is zero.

Let f : I → R. The variation of f over I is given by

( n ) X V ar(f; I) = sup |f(xi) − f(xi−1)| P = {x0, x1, . . . , xn} is a partition of I .

i=1

If V ar(f; I) < ∞, we say that f is of bounded variation on I. We denote the collection of functions on I that

8 are of bounded variation by BV(I).

Let I = [a, b] ⊆ R be a compact interval. Let f : I → R be an HK integrable function (see section 4.) We say that f is absolutely integrable on I if |f| is also integrable on I. A function that is integrable on I but not absolutely integrable on I is said to be non-absolutely integrable on I.

§3 GAUGES.

We now discuss different methods of measuring the fineness of partitions. In the traditional approach to · · · the Riemann integral, one measures the fineness of a partition P with regard to the mesh kPk of P in the sense · · that the length of each subinterval Ik of P must be no greater than the length of kPk. In Lebesgue theory, the concept of fineness is developed from the theory of outer measure. In order to develop the Henstock-Kurzweil integral, we introduce the following terminology. Let I = [a, b] be a compact interval in R.

(3.1) Deﬁnition. A function δ : I → (0, ∞) is called a gauge. For each t ∈ I the interval around t controlled by the gauge δ is the interval B(t; δ(t)) = (t − δ(t), t + δ(t)).

· · n (3.2) Deﬁnition. Let P = {(Ii, ti)}i=1 be a tagged partition of I. Given a gauge δ on I, we say that P is δ-ﬁne if

Ii ⊆ (ti − δ(ti), ti + δ(ti)) for all i = 1, 2, . . . , n.

· · If a tagged partition P is δ-ﬁne we say that P is subordinate to δ.

(3.3) Deﬁnition. The Riemann integral relies on what is known as a constant gauge. Let I = [a, b] be a compact interval, and let δ0 be any positive real number. The function δ : I → (0, ∞) on I given by

δ(t) = δ0 for all t ∈ I

· n is a constant gauge on I. Any tagged partition P = {(Ii, ti)}i=1 on I will be subordinate to δ if and only if

Ii ⊆ (ti − δ0, ti + δ0) = B(ti; δ0) for all i = 1, 2, . . . , n.

9 One of the primary beneﬁts of using gauges lies in their ability to force a given point to be a tag for · · n a partition P. For example, let I = [a, b] ⊆ R and let P = {(Ii, ti)}i=1 be a tagged partition of I. Let δ : I → (0, ∞) be given by   1  4 if t = a δ(t) =  1  2 |t − a| if a < t ≤ b.

· Then, for any δ-ﬁne partition P = {a = x0 < x1 < . . . < xn = b} of [a, b] the tag t1 for [a, x1] is forced to be · t1 = a as we now show. To see this, since P is δ-ﬁne, we must have that

[a, x1] ⊆ (t1 − δ(t1), t1 + δ(t1)).

Hence,

t1 − δ(t1) < a,

1 1 and this would imply t1 − 2 |t1 −a| < a or equivalently t1 −a < 2 (t1 −a) if a < t1 ≤ b, which is a contradiction.

Thus, a must be the tag for [a, x1]. From this a very natural question arises. Given a compact interval I and a gauge δ : I → (0, ∞), is · it always possible to ﬁnd a tagged partition P which is δ-ﬁne? The answer is yes, as given by the following theorem.

(3.4) Cousin’s Theorem. If I = [a, b] ⊆ R is a nondegenerate compact interval, and δ is a gauge on I, then there exists a tagged partition of [a, b] that is δ-ﬁne.

For a proof of Cousin’s Theorem, we refer the reader to [1]. Thus, given any compact interval [a, b] ⊆ R, and a gauge δ on [a, b], a δ-fine partition of [a, b] always exists. Now that we have some preliminary definitions and results out of the way, we define the “Henstock-Kurzweil” integral.

§4 THE HK INTEGRAL.

We begin by deﬁning the HK integral. Let I = [a, b] ⊆ R be a compact interval.

(4.1) Deﬁnition. Let f : I → R be a function. We say that f is Henstock-Kurzweil, HK, Gauge, or Gen- eralized Riemann Integrable on I if there is some real number A such that for every > 0 there is a gauge

10 · n δ : I → (0, ∞) such that if P = {(Ii, ti)}i=1 is any tagged partition of I that is δ-ﬁne, then

· S(f; P) − A < .

As with the Riemann integral, one can show that A is unique. In this case,

Z b A = HK f a or if it is clear that we are using an HK integral,

Z b A = f. a

In some cases, we can use an appropriately defined constant gauge δ0 and construct our partitions so that the length of the mesh is always smaller than δ0. In this way, it is easy to see that the R integral is a special case of the HK integral. Indeed, in the definition of the R integral we defined our limit in terms of the mesh · · of a partition. However, after some reconsideration, we actually see that the mesh kPk of a partition P is just a constant gauge. In fact, by using gauges, we are making only a small change to the classic definition of the Riemann integral. However, this small change will prove to have large consequences. Intuitively, whether using the Riemann approach or the Henstock-Kurzweil approach, we are using the fineness of the subintervals of the partition over which the function is defined to approximate the integral. The Riemann approach measures that fineness by taking the mesh of the partition. So, the lengths of subintervals are all less than or equal to some real number. By using a gauge, we allow more variation in the lengths of those subintervals. In this way, we get a better approximation of the area beneath a curve via Riemann sums. For intervals upon which the curve is changing rapidly, we use subintervals with sufficiently small length and for intervals where the change is slow, we use subintervals of larger length. This turns out to be one of the key advantages of the HK integral. In addition, a gauge allows us to enclose a countable set of points A within a collection of subintervals ∞ S∞ {Jk}j=1 so that k=1 Jk has very small outer measure thus nullifying the effect that A has on the Riemann sums. This is a handy tool when dealing with discontinuities. Furthermore, as we have already shown, a gauge can force certain “problem” points to be tags. This allows us to deal with asymptotes. Finally, as we will see, the use of gauges will give an improved version of the “Fundamental Theorem of Calculus.” We now look at some of the important results tied to the HK integral.

11 §5 PROPERTIES OF THE HK INTEGRAL.

(5.1) Theorem. Let I = [a, b] be a compact interval in R. Let A ⊆ I be a set of outer measure zero. Let ϕ be deﬁned as follows:

 1 if x ∈ A ϕ(x) =  0 if x ∈ I \ A.

Then, ϕ ∈ HK(I) and Z b φ = 0. a

∞ Proof: Let > 0 be given. Let {Jk}k=1 be a countable collection of open intervals such that

∞ ∞ [ X A ⊆ Jk and `(Jk) < . k=1 k=1

We deﬁne a gauge on I as follows. If t ∈ I \ A, put δ(t) = 1. If t ∈ A, let k(t) be the smallest index k such that t ∈ Jk and choose δ(t) > 0 such that (t − δ(t), t + δ(t)) ⊆ Jk(t).

· n Now, let P = {(Ii, t)}i=1 be a δ-ﬁne partition of I. If t ∈ I \ A, then ϕ(t) = 0 and the Riemann sum P · ϕ(ti)∆xi = 0. Since, P is δ-ﬁne, if t ∈ A, ti∈I\A

Jk(ti) ⊇ (t − δ(t), t + δ(t)) ⊃ Ii.

So, for each k ∈ N, the (nonoverlapping) intervals Ii with tags in A ∩ Jk have total length less than or equal P to `(Jk) by countable subadditivity. So for each k ∈ N, the Riemann sum ϕ(ti)∆xi ≤ `(Jk). Therefore, ti∈Ik

∞ ∞ X X X X 0 ≤ ϕ(ti)∆xi + ϕ(ti)∆xi ≤ `(Jk) ≤ `(Jk) < . ti∈I\A k=1 ti∈Ik k=1

Thus, ϕ ∈ HK(I) and Z b φ = 0. a

12 The next result, the proof of which can be found in [1] reveals one of the primary diﬀerences between the HK integral and the Lebesgue integral.

P∞ P∞ ∞ (5.2) Theorem. Let k=1 ak be any convergent series. Say k=1 ak = A. Let us deﬁne a sequence {cn}n=0 1 as cn = 1 − 2n . Now, let h : [0, 1] → R be deﬁned as follows:

 k 2 ak if x ∈ [ck−1, ck) k ∈ N h(x) = 0 if x = 1

In this case, h ∈ HK([0, 1]) and ∞ Z 1 X h = A = ak. 0 k=1 Another impressive result which comes from the HK integral is the so called “Hake’s Theorem.” One can ﬁnd the proof of this in [1]

(5.3) Hake’s Theorem. A function f ∈ HK([a, b]) if and only if f ∈ HK([a, c]) for every c ∈ [a, b) and lim R c f < ∞. In this case, c→b− a Z b Z c f = lim f. a c→b− a

Hake’s Theorem tells us that the space HK(I) cannot be extended by adjoining “improper integrals.” If such an integral exists, then the integrand must already belong to HK(I). This is not so for Riemann and Lebesgue integrals. We now examine one of the most powerful aspects of the HK integral - the Fundamental Theorem of Calculus.

§6 THE FUNDAMENTAL THEOREM OF CALCULUS.

Unlike the R and L integrals, the HK integral can integrate every derivative. In order to fully formulate the Fundamental Theorem of Calculus for HK integrals, we ﬁrst give some preliminary deﬁnitions. Let I =

[a, b] ⊆ R be a compact interval.

(6.1) Deﬁnition. Let F, f : I → R. We say that F is a Primitive of f on I if F 0(x) exists and F 0(x) = f(x) for all x ∈ I.

13 (6.2) Deﬁnition. Let F, f : I → R. We say that F is an a-primitive, c-primitive or f-primitive of f on I if F is continuous on I and there is a null, countable or ﬁnite set, respectively, E ⊆ I so that F 0 = f on I \ E. We call the set E the exceptional set for f.

(6.3) Deﬁnition. Suppose that f ∈ HK(I) and let u ∈ I. The function Fu : I → R given by

Z x Fu(x) = f u is called an indefinite integral of f with base point u. If the base point is the left endpoint of I or is well understood, then we may omit the subscript. Any function that differs from Fu by a constant is called an indefinite integral of f.

We can now begin our discussion of the Fundamental Theorem of Calculus for HK integrals. To un- derstand why we get an improved version of the Fundamental Theorem, we must ﬁrst consider the classic deﬁnition of the derivative.

(6.4) Definition. Given a differentiable function F :[a, b] → R, we say that F is differentiable at t ∈ [a, b], with derivative f(t), if for all > 0 there is a δ(t) > 0 such that if 0 < |x − t| < δ(t), where x ∈ [a, b], then

F (x) − F (t) − f(t) < . x − t

Notice that the number δ(t) can just as easily be thought of as a function of t and . Thus, if a function F is diﬀerentiable at a point t ∈ [a, b], there is a built-in gauge. It is this very idea that lies at the heart of the following result.

(6.5) The Fundamental Theorem of Calculus I. Let I = [a, b] ⊆ R be a compact interval. Let f : I → R be a function. If f has a c-primitive F on I, then f ∈ HK(I) and

Z b f = F (b) − F (a). (1) a

Notice that in the hypothesis of the Fundamental Theorem of Calculus for the HK integral we need not assume that f ∈ HK(I). This is not the case for the R integral, and for the Lebesgue integral additional

14 constraints are necessary to make equation (1) valid. Consequently, this version of the Fundamental Theorem of Calculus makes diﬀerentiation and integration truly inverse processes. Furthermore, we have the following version of the Fundamental Theorem of Calculus II.

(6.6) The Fundamental Theorem of Calculus II. If f ∈ HK(I) then any indeﬁnite integral F is continuous on I and is an a-primitive of f on I. Thus, there exists a null set A ⊆ I such that

0 F (x) = f(x) for all x ∈ I \ A.

For the proofs of the above theorems, we refer the reader to [1].

Remark: A primary feature of the HK integral is its relationship to the c-primitive. The following statements are meant to add clarity to the subtle distinction between c-primitives and indeﬁnite integrals.

• An HK integrable function f always has an indeﬁnite integral and every indeﬁnite integral of a function in HK(I) is an a-primitive.

• An HK integrable function does not always have a c-primitive.

• If F is a c-primitive of f : I → R, then f ∈ HK(I) and F is an indeﬁnite integral of f.

• If F is an a-primitive of f ∈ HK(I), then F need not be an indeﬁnite integral of f.

Examples or counterexamples for each of these assertions can be found in [1].

Now that we have discussed some preliminary results, we consider the eﬀect of these results. We ﬁrst show that R(I) $ HK(I).

§7 FUNCTIONS WHICH ARE HK INTEGRABLE BUT NOT R INTEGRABLE.

The very ﬁrst integral that most mathematicians learn about is Riemann’s. While the importance of the Riemann integral to a general theory of integration cannot be overemphasized, it is not without its

15 drawbacks. In this section we examine some of those drawbacks. We ﬁrst show the power of the HK version of the Fundamental Theorem of Calculus.

(7.1) Example. Consider the function

 − 1 x 4 if x ∈ (0, 1] f(x) = 0 if x = 0.

3 4 4 Since f is not bounded, it fails to be Riemann integrable. Let F : [0, 1] → R be given by F (x) = 3 x . Then, F is continuous on [0, 1] and F 0(x) = f(x) for all x ∈ (0, 1] but F 0(0) does not exist. Thus, F is an f-primitive for f on [0, 1] with exceptional set E = {0}. Therefore, by the Fundamental Theorem of Calculus I for the HK integral (6.5), Z 1 4 f(x) dx = F (1) − F (0) = . 0 3

In practice, we write Z 1 − 1 x 4 dx 0 with the understanding that the integrand is zero at x = 0.

In any undergraduate real analysis course, students learn that the Riemann integral can handle any function with a ﬁnite number of discontinuities. This idea is later extended by saying that the set of discontinuities of a Riemann integrable function must have measure zero. Using this idea, we can create bounded functions which are not Riemann integrable. To show that such functions are actually HK integrable we construct a gauge which will enclose the points of discontinuity in a set with small outer measure making the contribution of said discontinuities to the Riemann sums negligible. To illustrate the full power of this technique, we will now construct such a function.

(7.2) Example. Let A be a countable, dense subset of [0, 1]. Examples include, but are not limited to, the algebraic numbers and the rational numbers. We claim that the following function is HK integrable but not R integrable.

 1 if x ∈ A f(x) =  0 if x ∈ [0, 1] \ A

16 1 n Proof: We begin by showing that f is not R integrable. To this end, let = 2 , and let P = {Ii}i=1 be a partition of [0, 1]. Since A and [0, 1] \ A are both dense in [0, 1], for each subinterval Ii induced by P,Ii contains both an element of A and an element of [0, 1] \ A. So, Mi = 1 and mi = 0 for i = 1, 2, . . . , n. Clearly, L(P; f) = 0, and

n X U(P; f) = Mi∆xj j=1 n X = 1∆xj j=1 n X = ∆xj j=1

= (1 − 0)

= 1.

Thus, U(P; f) − L(P; f) = 1 − 0 = 1.

Since P was arbitrary, we see that 1 U(P; f) − L(P; f) ≥ = 2 for all partitions P. Therefore, f is not Riemann integrable on [0, 1].

It is clear, by Theorem (5.1), that f ∈ HK([0, 1]). We will now examine this assertion in further detail by using an explicit gauge. Since A is a countable set, it is enumerable. Let A = {r1, r2,...} be an enumeration of A. Let > 0 be given. Consider the following gauge on [0, 1].

 1 if t ∈ [0, 1] \ A δ(t) =   2k+1 if t = rk ∈ A.

· Let P be a δ-ﬁne, tagged partition of [0, 1]. Since f = 0 on [0, 1] \ A, the tags in [0, 1] \ A contribute nothing · · to the Riemann sum of f induced by P. So, since P is δ-ﬁne, we need only to consider the following sum:

∞ · X S(f; P) ≤ = . 2k k=1

Therefore, f ∈ HK([0, 1]) and Z 1 f = 0. 0

In the above example, we constructed a gauge which allowed us to nullify the eﬀect that the points of

A had on the Riemann sums. By doing this, we have eﬀectively made f identically zero on [0, 1]. However, this is not surprising since, alternatively, we could have shown that f ∈ HK([0, 1]) using Theorem (6.5) and considering the function F (x) = 0 for all x ∈ [0, 1]. Notice that F is a c-primitive for f on [0, 1] with exceptional set A. Hence, by the HK version of the Fundamental Theorem of Calculus,

Z 1 f(x)dx = F (0) − F (1) = 0. 0

We have now shown that R(I) $ HK(I). However, the above functions, or variations therein, are so well known that in some sense they are trivial examples. We now construct one ﬁnal example which will show the power of the Fundamental Theorem of Calculus for HK integrals. The following is a bounded function which is the derivative of another function, yet which is not Riemann integrable. The initial construction of this function was suggested by Goﬀmann.[2]

(7.3) Example. We now construct an oscillating function on a set of positive outer measure by using a method similar in spirit to the construction of the classic Volterra function.

1 We begin with the unit interval, which we denote by K1 = [0, 1]. Consider x1 = 2 , the midpoint of K1. 3 5 1 First, we delete the interval G1 = I1 = ( 8 , 8 ) from K1. Note that I1 is symmetric about x1 and `(I1) = 4 . 3 5 We are now left with two closed intervals, K2 = [0, 8 ] and K3 = [ 8 , 1]. Let

L1 = K2 ∪ K3.

3 13 Let x2 = 16 and x3 = 16 , the midpoints of K2 and K3 respectively. We now delete the open intervals 5 7 25 27 3 5 I2 = ( 32 , 32 ) and I3 = ( 32 , 32 ), which are symmetric about x2 and x3, from K2 = [0, 8 ] and K3 = [ 8 , 1] respectively. Note that 1 `(I ) = = [`(I )]2 for k = 2, 3. k 16 1

Let 5 7 25 27 G = I ∪ I = , ∪ , . 2 2 3 32 32 32 32

18 This leaves us with the four closed intervals

5 7 3 5 25 27 K = 0, ,K = , ,K = , and K = , 1 . 4 32 5 32 8 6 8 32 7 32

Let

L2 = K4 ∪ K5 ∪ K6 ∪ K7.

5 19 45 59 Let x4 = 64 , x5 = 64 , x6 = 64 and x7 = 64 which are the midpoints of K4,K5,K6, and K7 respectively. Next, we delete the open intervals

9 11 37 39 89 91 117 119 I = , ,I = , ,I = , and I = , 4 128 128 5 128 128 6 128 128 7 128 128

5 7 3 5 25 27 from K4 = 0, 32 ,K5 = 32 , 8 ,K6 = 8 , 32 and K7 = 32 , 1 respectively. Note that

1 `(I ) = = [`(I )]3 for k = 4, 5, 6, 7. k 64 0

Let 9 11 37 39 89 91 117 119 G = I ∪ I ∪ I ∪ I = , ∪ , ∪ , ∪ , . 3 4 5 6 7 128 128 128 128 128 128 128 128

If we continue this process inductively, we obtain the following construction:

• G1 = I1.

• Gn = I2n−1 ∪ I2n−1+1 ∪ ... ∪ I2n−1+(2n−1−1) for n ≥ 2, where n ∈ N.

n 1 n n−1 n−1 n−1 n−1 • `(Ik) = [`(I1)] = 4 where k = 2 , 2 + 1 ..., 2 + (2 − 1) for each n ∈ N.

• Ij ∩ Ik = ∅ for all j 6= k

• Gj ∩ Gk = ∅ for all j 6= k.

∗ n−1 For each n ∈ N, we deﬁne λ (Gn) to be the outer measure of Gn. Since Gn is the union of 2 disjoint open

19 n 1 n intervals, each of length [`(I1)] = 4 , by countable additivity,

∗ n−1 n λ (Gn) = 2 [`(I1)] 1n = 2n−1 4 1n+1 = . 2

Now, let us deﬁne the set ∞ [ G = Gk. k=1 By countable additivity,

∞ ! ∗ ∗ [ λ (G) = λ Gk k=1 ∞ X ∗ = λ (Gk) k=1 ∞ k+1 X 1 = 2 k=1 1 = . 2

n Notice that after the nth iteration in the construction of G we are left with a set Ln, a union of 2 closed intervals each of which we denoted by Kk for some k ∈ N. Then,

∞ \ K = Ln = [0, 1] \ G n=1

∗ 1 is a fat Cantor set with outer measure λ (K) = 2 . Since the Cantor set is known to be totally disconnected, its only connected sets are singletons. Thus, given any point x ∈ [0, 1]\G, and any open interval U = (x−, x+) containing x, U ∩ G 6= ∅. Hence, x ∈ cl(G) and [0, 1] ⊆ cl(G). Clearly, [0, 1] ⊇ cl(G). So, [0, 1] = cl(G) and G is dense in [0, 1]. Thus, G has the following properties:

• G is a union of pairwise disjoint open intervals.

• G is dense in [0, 1].

∗ 1 • G has outer measure λ (G) = 2 .

20 We can now begin constructing a bounded function f which is the derivative of another function, but

3 5 1 which is not Riemann integrable. We begin with the open interval I1 = ( 8 , 8 ) of length 4 , and the point 1 15 17 1 1 2 2 x1 = 2 which is the midpoint of I1. Let J1 = [ 32 , 32 ] be the closed interval of length 16 = 4 = [`(I1)] 1 1 15 17 which is symmetric about x1 = 2 . Define f( 2 ) = 1, f( 32 ) = f( 32 ) = 0. Let f be linear and continuous 15 1 1 17 15 1 1 17 on the open intervals ( 32 , 2 ) and ( 2 , 32 ) connecting f( 32 ) to f( 2 ) and f( 2 ) to f( 32 ) respectively. Define 5 7 25 27 1 f = 0 on I1 \ J1. Next, consider the open intervals I2 = ( 32 , 32 ) and I3 = ( 32 , 32 ), each of length 16 , which 3 13 are symmetric about x2 = 16 and x3 = 16 respectively. From I2 and I3, we construct the closed intervals 23 25 103 105 1 1 2 2 J2 = [ 128 , 128 ] and J3 = [ 128 , 128 ] of length 64 = 16 = [`(Ik)] , for k = 2, 3, symmetrically about x2 3 13 23 25 103 105 and x3, respectively. Define f( 16 ) = f( 16 ) = 1, f( 128 ) = f( 128 ) = f( 128 ) = f( 128 ) = 0. Let f be linear 23 3 3 25 103 13 13 105 23 3 and continuous on the open intervals ( 128 , 16 ), ( 16 , 128 ), ( 128 , 16 ), and ( 16 , 128 ), connecting f( 128 ) to f( 16 ), 3 25 103 13 13 105 f( 16 ) to f( 128 ), f( 128 ) to f( 16 ), and f( 16 ) to f( 128 ). Finally, define f to be identically zero on Ik \ Jk where k = 2, 3. Continuing this process inductively, we obtain the following construction:

1 n Let In be the nth open interval, of length 4 in the construction of G. Let Jn = [an, bn] ⊆ In be a 2 1 n2 closed interval which is symmetric about the midpoint xn of In so that `(Jn) = [`(In)] = 4 for each n ∈ N. Deﬁne the function f : [0, 1] → [0, 1] follows:

Let f(xn) = 1 and f(an) = f(bn) = 0 for each n. Let f be linear and continuous on the open intervals

(an, xn) and (xn, bn) connecting f(an) = 0 to f(xn) = 1 and f(xn) = 1 to f(bn) = 0. Deﬁne f(x) = 0 for all x ∈ In \ Jn. Finally, deﬁne f = 0 on [0, 1] \ G. Since 0 ≤ f(x) ≤ 1 for all x ∈ [0, 1], f is bounded on [0, 1]. Our function can be seen in the following diagram:

Figure 1: HK integrable but not R integrable

21 We claim that f is not Riemann integrable. To see this, ﬁrst let O be an open subset of [0, 1] such that O ∩ K 6= ∅. Since O is open, we can ﬁnd distinct points x and y in O ∩ K. Without loss of generality, suppose that x < y. Then, since K is a ”Cantor-like” set, it contains no isolated points and is totally disconnected.

So, by construction of G, Jn ⊆ In ⊆ (x, y) ⊆ O for some positive integer n.

Let P = {0 = x0 < x1 < . . . < xn = b} be any partition of [0, 1]. Now, let us consider

n [ K \{1} = K \{1} ∩ [xj−1, xj) j=1 n [ = [(K \{1}) ∩ [xj−1, xj)] . j=1

∗ 1 Recall, λ (G) = 2 . So, 1 λ∗(K) = λ∗([0, 1] \ G) = . 2

Thus,

1 = λ∗(K) 2 = λ∗(K \{1}) n X ∗ = λ (K \{1}) ∩ [xj−1, xj)) j=1 X ≤ `((xj−1, xj)).

K∩(xj−1,xj )6=∅

When K ∩ (xj−1, xj) 6= ∅, we can ﬁnd a positive integer n = n(j) such that

Jn ⊆ In ⊆ (xj−1, xj).

Therefore,

Mj = 1 and mj = 0, when K ∩ (xj−1, xj) 6= ∅.

22 Hence,

n X U(P; f) − L(P; f) = (Mj − mj)∆xj j=1 X ≥ (Mj − mj)∆xj

K∩(xj−1,xj )6=∅ X = ∆xj

K∩(xj−1,xj )6=∅ 1 ≥ . 2

Thus, f 6∈ R([0, 1]).

We now show that f is the derivative of the following improper integral. Consider the function

∞ X Z F (x) = f(t)dt. k=1 Jk∩[0,x]

We claim that F 0(x) = f(x) for all x ∈ [0, 1].

Choose an open interval I ⊆ [0, 1] so that I ∩ ([0, 1] \ G) 6= ∅. Then, I ⊇ In ⊇ Jn for some positive integer n. 1 Choose n ∈ N so that I ∩ Jn 6= ∅ and let Sn = `(In). Then, by construction, 0 ≤ Sn ≤ 2 . Hence, from the 1 2 fact that 0 ≤ x ≤ 2 if and only if 2(x − x ) ≥ x, we conclude that

1 1 [S − [S ]2] ≥ S . 2 n n 4 n

2 2 Now, I ∩ ([0, 1] \ G) 6= ∅,I ∩ Jn 6= ∅, and `(Jn) = [`(In)] = [Sn] . So,

1 1 1 1 `(I ∩ I ) ≥ `(I ) − [`(J )] = [S − [S ]2] ≥ S n 2 n 2 n 2 n n 4 n

since In and Jn are both symmetric about the midpoint of In. Therefore,

2 2 16[`(I ∩ In)] ≥ Sn, (2) and by monotonicity,

2 `(I ∩ Jn) ≤ `(Jn) = [Sn] .

23 Hence, with this last inequality together with (2), we conclude

2 2 `(I ∩ Jn) ≤ `(Jn) = [Sn] ≤ 16[`(I ∩ In)] .

Now, consider the set

J = {n ∈ N : I ∩ Jn 6= ∅}.

Then,

X X 2 `(I ∩ Jn) ≤ 16 [`(I ∩ In)] n∈J n∈J " #2 X ≤ 16 `(I ∩ In) n∈J " !#2 ∗ [ ≤ λ I ∩ In n∈J " !#2 ∗ [ = 16 λ I ∩ In n∈J ≤ 16[`(I)]2. (3)

We now show that F 0(x) = f(x) for all x ∈ [0, 1]. Let x ∈ [0, 1] \ G, and let y ∈ [0, 1] with y 6= x. Without loss of generality assume x < y. Then,

X Z X Z F (y) − F (x) = f(t)dt − f(t)dt n∈J n∈J Jn∩[0,y] Jn∩[0,x]   Z Z X   =  f(t)dt − f(t)dt n∈J Jn∩[0,y] Jn∩[0,x]   Z X   =  f(t)dt n∈J Jn∩[x,y] X ≤ `(Jn ∩ [x, y]). n∈J

By assumption, x < y. So F (x) ≤ F (y) and F (y) − F (x) ≥ 0. Therefore, since 0 ≤ f(t) ≤ 1 for all t,

F (y) − F (x) 1 X Z 1 X 1 0 ≤ = f(t)dt ≤ ` ([x, y] ∩ J ) ≤ 16(y−x)2 ≤ 16(y−x), by (3). y − x y − x y − x n y − x n∈J n∈J Jn∩[x,y]

24 Note: By using the closed interval [x, y], rather than the open interval (x, y), we have not changed the outer measure of (x, y), so the result still holds by (3). Thus,

F (y) − F (x) 0 ≤ ≤ 16(y − x). y − x

Then, F (y) − F (x) 0 ≤ F 0(x) = lim ≤ lim 16(y − x) = 0. y→x y − x y→x

Therefore, since f(x) = 0 for all x ∈ [0, 1] \ G,

F 0(x) = f(x) for all x ∈ [0, 1] \ G.

Now, let x0 ∈ G. Consider X Z F (x) = f(t)dt. n∈J Jn∩[0,x]

Since x0 ∈ G, x0 ∈ In for some n ∈ N. Note that if x ∈ In \ Jn for some n ∈ N, then

F 0(x) = f(x).

So, let x ∈ Jn, and suppose x > x0. Then

X Z Z x F (x) − F (x0) = f(t)dt = f(t)dt, n∈J x0 Jn∩[x0,x]

since f ≡ 0 on In \ Jn. So, Z x F (x) = F (x0) + f(t)dt. x0

Since f is continuous on each In, we may apply the Fundamental Theorem of Calculus II, giving us

Z x 0 d d F (x) = (F (x)) = F (x0) + f(t)dt dx dx x0 = 0 + f(x).

0 Thus, F (x) = f(x). If x ≤ x0 we may use the above proof with the identity

Z x F (x0) − F (x) = − f(t)dt. x0

25 Therefore, F 0(x) = f(x) for all x ∈ G, so F 0(x) = f(x) for all x ∈ [0, 1]. Hence, f is the derivative of F on [0, 1]. In other words, F is the primitive of f on [0, 1]. Hence, by The Fundamental Theorem of Calculus for HK integrals, and Hake’s Theorem, f ∈ HK([0, 1]).

The above examples not only show some of the most powerful applications of gauges, but also the deficiencies of the Riemann integral. The previous function f definitely seems like it should be Riemann integrable, we are after all simply adding up the areas of triangles, but we have shown that it is not. The function oscillates much too wildly on a set with positive outer measure. However, the function does belong to HK([0, 1]). Furthermore, in Example (7.2) we showed that the Henstock-Kurzweil integral is able to handle certain pathological functions that the R integral simply cannot. In that example, we constructed a gauge which encloses each point of A in an open interval with arbitrarily small length. By doing that, we obtained a collection of sets whose union has an arbitrarily small outer measure, ultimately rendering the effect that the · set A has on the Riemann sums of a partition P negligible. However, these examples have revealed something deeper. The use of gauges allows us to bridge the very small gap between HK(I) and L(I). In fact, as we will now show, it is only a class of highly oscillatory functions which lie in the space between.

§8 FUNCTIONS WHICH ARE HK INTEGRABLE BUT NOT L INTEGRABLE.

In any basic measure theory class, one learns about the Lebesgue integral in steps. For brevity, we characterize the class of Lebesgue integrable functions in terms of the HK integral. The proof of this assertion can be found in [5].

(8.1) Theorem. Let I = [a, b] ⊆ R be a compact interval. Let f : I → R be a function deﬁned on I. Then, f is Lebesgue integrable if and only if |f| is HK integrable. In either case, the integrals agree.

An application of this result will help us to show that HK(I) is a space of non-absolutely integrable functions. Consequently, L(I) is a space of absolutely integrable functions. Hence, a function f can be Henstock- Kurzweil integrable without |f| being Henstock-Kurzweil integrable, but this is not true for Lebesgue integrable functions. In this way, we can think of HK(I) as being analogous to the class of conditionally convergent series and L(I) as being analogous to the class of absolutely convergent series. To classify those functions which are absolutely integrable, we employ the following result.

26 (8.2) Theorem. Let I = [a, b] ⊆ R be a compact interval. Let f : I → R be a function deﬁned on I. Let f ∈ HK(I). Then, |f| is HK integrable if and only if the indeﬁnite integral

Z x F (x) = f a has bounded variation on I. In this case,

Z b |f| = V ar(F ; I). a

The proof of this assertion can be found in [1].

Thus, Theorem 8.1 and Theorem 8.2 together show that if F is of bounded variation on I, then F 0 is Lebesgue integrable on I. Using these characterizations, we look at some functions which are HK integrable but not Lebesgue integrable.

(8.3) Example. Consider the function

F : [0, 1] → R given by

  2 π x cos x2 if x ∈ (0, 1] F (x) = 0 if x = 0.

This function has a derivative given by:

  π 2π π 2x cos x2 + x sin x2 if x ∈ (0, 1] f(x) = 0 if x = 0.

We claim that F is not of bounded variation on [0, 1], which will in turn implies that F 0 = f 6∈ L([0, 1]). To

27 see this, consider the sequence x = √1 where k = 1, 2,...,N. Then, k k

N N X X 1 1 |F (xk) − F (xk+1)| = cos kπ − cos(k + 1)π k k + 1 k=1 k=1 N X 1 1 = + k k + 1 k=1 1 1 1 1 1 1 1 = 1 + + + + + + ... + + 2 2 3 3 4 N N + 1 1 1 1 1 = 2(1) + 2 + 2 + 2 − 1 + 2 3 N N + 1 N X 1 1 = 2 − 1 + . k N + 1 k=1

Since the harmonic series diverges, we see that this series goes to +∞ as N → +∞. Thus, F is not of bounded variation on [0, 1]. So, f = F 0 6∈ L([0, 1]). However, according to the Fundamental Theorem of Calculus I for HK integrals, since F is a primitive for f on [0, 1], we see that f = F 0 does belong to HK([0, 1]), and

Z 1 HK f(x)dx = F (1) − F (0) = −1. 0

In the spirit of Theorem (5.2), we will now construct another example of a function which is not Lebesgue integrable but is HK integrable. This example illustrates the strength of non-absolute integrability.

(8.4) Example. Consider ∞ X (−1)n+1 log n . n n=1 We claim that this series is not absolutely convergent. Consider

∞ n ∞ X (−1) log n X log n = . n n n=1 n=1

log n 1 P∞ 1 For all n ≥ 3, n ≥ n . Since the series n=1 n is known to diverge, by the comparison test,

∞ X log n n n=1 diverges as well. Thus, ∞ X (−1)n+1 log n n n=1

28 log n is not absolutely convergent. However, letting an = n , we see that

log n • lim an = lim = 0. n→∞ n→∞ n

log x 0 1−log x • Letting f(x) = x , f (x) = x2 , and

0 • since f (x) < 0 for all x > e, an is a decreasing sequence.

So, by the alternating series test, ∞ X (−1)n+1 log n n n=1 is a convergent series. Thus, by Theorem 5.2, the function κ : [0, 1] → R given by:

 k+1  k (−1) log k 2 k if x ∈ [ck−1, ck) for k ∈ N κ(x) = 0 if x = 1

1 where ck = 1− 2k is HK integrable. However, we claim that κ is not absolutely integrable, hence not Lebesgue integrable. To see this, let Z x K(x) = κ(t)dt 0 be an indeﬁnite integral for κ with base point 0. Note that K(0) = 0 and

n X (−1)k+1 log k K(c ) = . n k k=1

Now, consider

n n−1 X (−1)k+1 log k X (−1)k+1 log k |K(c ) − K(c )| = − n n−1 k k k=1 k=1 log n = n

Using parts of our sequence ck as a partition of [0, 1], we set

P = {0 = y0, y1, . . . , yn = 1} where

y0 = 0, y1 = c1, y2 = c2, . . . , yn−1 = cn−1, yn = 1.

29 Then, n X log 2 log 3 log(n − 2) log(n − 1) |K(y ) − K(y )| ≥ 0 + + + ... + + . i i−1 2 3 n − 2 n − 1 i=1

P∞ log n Taking n → ∞, we see that K 6∈ BV([0, 1]), since n=1 n diverges. Therefore, |κ| is not HK integrable on P∞ (−1)n+1 log n [0, 1]. Hence, κ is not Lebesgue integrable on [0, 1]. However, by Theorem (5.2), since n=1 n does converge, κ ∈ HK([0, 1]) and

∞ Z 1 X (−1)n+1 log n 1 HK κ = = γ log 2 − [log 2]2 n 2 0 k=1 where γ is the Euler-Mascheroni constant given by

" n # X 1 γ = lim − log n ≈ 0.5772 (see[8].) n→∞ k k=1

Here, κ fails to be Lebesgue integrable because |κ| is not HK integrable. However, upon deeper inspection, we R x realize that |κ| fails to be HK integrable because 0 κ 6∈ BV([0, 1]). Indeed, by simply looking at the ﬁrst few partial sums we can see that the function oscillates wildly. The following diagram shows the graph of K(c20):

Figure 2: K(c20)

We will now look at one ﬁnal function which belongs to HK(I) but not L(I). In the process we show the power of Hake’s Theorem.

(8.5) Example. Consider the following function.

30   1 1  x sin x3 if x 6= 0 f(x) = 0 if x = 0.

We cannot express the integral Z x f 0 with elementary functions. However, consider the function

Z x sin t g(x) = dt. 0 t

It is a commonly known fact that lim g(x) = π [9]. Armed with this fact, we can now evaluate the integral x→∞ 2 1 of f as an improper Riemann integral in the following manner. Let t = x3 , then

Z 1 1 1 1 Z ∞ sin t sin 3 dx = dt 0 x x 3 1 t 1 = lim g(x) − g(1) 3 x→∞ 1 = (π − 2g(1)) < ∞. 6

1 1 Now, let u = x and dv = sin x dx. Then, du = − x2 and v = − cos x, and we have

Z A A Z A sin x cos x cos x dx = − − 2 dx 1 x x 1 1 x cos 1 cos A Z A cos x = − − 2 dx. 1 A 1 x

Now, Z A cos x Z ∞ cos x lim 2 dx = 2 dx A→∞ 1 x 1 x exists since cos x 1 ≤ on [1, ∞), x2 x2 and Z ∞ 1 2 dx < ∞. 1 x

31 So, Z A cos x lim 2 dx A→∞ 1 x converges. Hence,

! Z 1 1 1 1 Z ∞ sin t 1 cos 1 cos A Z A cos x sin 3 dx = dt = lim − − 2 dx 0 x x 3 1 t 3 A→∞ 1 A 1 x

1 Z ∞ cos x = cos 1 − 2 dx . 3 1 x

So, f is improper Riemann integrable, and Hake’s Theorem guarantees that it is a “proper” HK integral. However, f is not Lebesgue integrable. To see this, we use the fact that a Lebesgue integrable function must be absolutely integrable. For k ∈ N consider

Z (k+1)π k Z (j+1)π sin t X sin t dt = dt. Then, letting u = t − jπ, du = dt. So t t π j=1 jπ k X Z π | sin(u + jπ)| = du u + jπ j=1 0 k X Z π sin u = du, since 0 ≤ u ≤ π. u + jπ j=1 0

1 1 So, since u+jπ ≥ (j+1)π for 0 ≤ u ≤ π, we have

k k X Z π sin u X 1 Z π du ≥ sin u du u + jπ (j + 1)π j=1 0 j=1 0 k 2 X 1 ≥ . π j + 1 j=1

So, k Z (j+1)π k X sin t 2 X 1 dt = . t π j + 1 j=1 jπ j=1

32 Therefore,

Z 1 Z ∞ 1 1 1 sin t sin 3 dx = dt 0 x x 3 1 t Z π n Z (j+1)π 1 sin t 1 X sin t = dt + dt 3 t 3 t 1 j=1 jπ Z π n 1 sin t 2 X 1 = dt + 3 t 3π j + 1 1 j=1 n 2 X 1 ≥ . 3π j + 1 j=1

Taking n → ∞, we see that Z 1 1 1 sin 3 dx 0 x x diverges since the harmonic series is divergent. So, f is not absolutely integrable hence not Lebesgue integrable. R x Again the function is not L integrable because 0 f 6∈ BV([0, 1]).

Notice, in the above example, we have inadvertently found a function which does not belong to HK([0, 1]). Namely, 1 1 h(x) = sin . x x3

For if h ∈ HK([0, 1]), then both 1 1 h(x) = sin x x3 and   1 1  x sin x3 if x 6= 0 f(x) = 0 if x = 0 belong to HK([0, 1]), which would imply that f ∈ L([0, 1]). However, as we just showed, this is not the case. Therefore, h 6∈ HK([0, 1]). This illustrates an important point. Although the HK integral is a very powerful tool, it cannot integrate every function.

Since it is a well known fact that R(I) $ L(I), we have shown that

R(I) $ L(I) $ HK(I).

We conclude the paper by discussing various strengths of the HK integral over the R and L integrals in applied mathematics.

33 §9 APPLICATIONS OF THE HK INTEGRAL.

Until now, we have mainly examined the differences between the HK integral and the classic Riemann and Lebesgue integrals by using pathological functions. However, Kurzweil originally developed the HK integral as a way to solve complicated differential equations. Indeed, due to the improved Fundamental Theorem of Calculus, the HK integral is a powerful tool for solving differential equations especially those involving highly oscillatory functions.

(9.1) Example. Recall from Example (8.3), the function F : [0, 1] → R be given by:

  2 π x cos x2 if x ∈ (0, 1] F (x) = 0 if x = 0 with derivative   π 2π π 2x cos x2 − x sin( x2 ) if x ∈ (0, 1] F 0(x) = 0 if x = 0.

Now, consider the initial value ODE:

y0(t) = t2y(t) + F 0(t), y(0) = 0, and 0 ≤ t ≤ 1.

Recall that F 0 is not Lebesgue integrable. So, F˜(x, t) = t2y(t) + F 0(t) cannot be solved by Lebesgue or Riemann integration. However, consider the following function:

3 Z t 3 t − s 0 y(t) = e 3 HK e 3 F (s)ds . 0

Then, y(0) = 0 and

3 Z t 3 3 Z t 3 0 2 t − s 0 t d − s 0 y (t) = t e 3 HK e 3 F (s)ds + e 3 HK e 3 F (s)ds . 0 dt 0

Now, by the Fundamental Theorem of Calculus II for HK integrals,

Z t 3 3 d − s 0 − t 0 HK e 3 F (s)ds = e 3 F (t), dt 0

34 and

3 Z t 3 3 3 3 Z t 3 0 2 t − s 0 t − t 0 2 t − s 0 0 y (t) = t e 3 HK e 3 F (s)ds + e 3 e 3 F (t) = t e 3 HK e 3 F (s)ds + F (t), 0 0 or equivalently, y0(t) = t2y(t) + F 0(t), 0 ≤ t ≤ 1, and y(0) = 0. Thus, 3 Z t 3 t − s 0 y(t) = e 3 HK e 3 F (s)ds 0 is indeed a solution to the aforementioned ODE.

Although diﬀerential equations play an important part in the applicability of the HK integral, working mathematicians who regularly study the HK integral and its applications are also making headway in areas such as probability, statistics, physics and ﬁnance. For example, a Brownian motion X = (Xt)(0 < t, X0 = 0) with drift rate (rate at which the average changes) µt and variance σt can be constructed from a standard

Brownian motion W = (Wt) by the equation:

Z t Z t Xt = σsdWs + µsds. 0 0

Although the ﬁrst integral is L integrable, the second, which is called a “Stochastic integral,” is not. Almost all of the sample paths given by x(s)(0 < s ≤ t, x(0) = 0) - which are continuous functions between topological spaces that produce a set of values according to the “rules” of the process - are of unbounded variation. So, for any path x if

µx((u, v]) is even as simple as x(v) − x(u), then the integral Z t dµx 0 does not exist. This is due to the fact that the Lebesgue integral is an absolutely convergent integral. When considered seperately, the sums of all paths x so that

x(v) − x(u) ≥ 0

35 and x(v) − x(u) < 0 diverge to ∞ and −∞ respectively. However, when considered as a HK integral, the integral exists. Indeed, given a partition

P = {0 = u0, u1, . . . , un = t} of [0, t],

n X µx((uj, uj−1]) = x(u1) − x(u0) + x(u2) − x(u1) + ... + x(un) − x(un−1) j=1

= x(un) − x(u0) = x(t).

So, it turns out that Z t dµx 0 is ﬁnite and is equal to x(t). In terms of Brownian motion, the stochastic integral on (0, t] is given by

Z t dWs = Wt. 0

Furthermore, much research has been done on the uses of the HK integral in quantum mechanics. Particularly when considering transforms and Feynman paths, which are often highly oscillatory by nature.

§10 FINAL COMMENTS.

The HK integral is one of the most powerful methods of integration currently being researched by mathematicians. By using gauges, one can evaluate functions on more of a local level than one can with the traditional Riemann integral. This seemingly small change to the traditional deﬁnition of the Riemann integral has proven to have far reaching consequences. For example, the HK integral makes integration and diﬀerentiation truly inverse processes. The fact that the HK integral is a non-absolutely convergent integral makes it ideal for integrating functions which oscillate wildly, a feature not always available with the L integral. This allows one to look at the integration process as a whole, rather than being forced to consider the negative and nonnegative cases separately as is often the case in Lebesgue integration theory. However, this advantage does have its drawbacks. To date no one has developed a suitable norm for the space HK(I). However, all of the above results can be generalized to Rn along with HK versions of “Fubini’s Theorem” for iterated

36 integrals and the “Divergence Theorem” similar to the ones learned in second or third year calculus courses. In addition, Hake’s theorem tells us that there are no improper HK integrals, another result unavailable in R and L integration theory. Furthermore, the deﬁnition of the Henstock-Kurzweil integral is relatively simple and requires no knowledge of measure theory making it a viable alternative to the traditional Riemann integral. In fact, there is a petition to replace the R integral in undergraduate analysis courses with the HK integral. The use of gauges may help students develop a deeper understanding of epsilon-delta style proofs, and give them some rudimentary intuition about outer measure. The HK integral’s ability to integrate functions of a highly oscillatory nature has proven to be a valuable tool in many areas of applied mathematics including diﬀerential equations, stochastic probability and quantum mechanics. Although, no one has found a suitable norm for the space of Henstock-Kurzweil integrable functions, there are semi-norms available. Even so, one cannot deny the Henstock integral’s value as an important supplement to the theory of integration.

Although the HK integral may not be the apotheosis of integration theory, it is an important mathematical tool with far reaching consequences. Its sheer simplicity and ability to generalize all previous integrals make it an important addition to integration theory, and although the HK integral cannot totally replace the Lebesgue integral, it is a tool that should be considered by any serious student of integration theory.

37 References

[1] Bartle, R. G., A Modern Theory of Integration, American Mathematical Society, Rhode Island 2001.

[2] Goﬀman, C., A Bounded Derivative Which is Not Riemann Integrable, The American Mathematical Monthly, Vol. 84 No. 3, March 1977, pps. 205-206.

[3] Rudin, W., Principles of Mathematical Analysis, McGraw-Hill Inc., New York, 1976.

[4] Royden, H.L., Fitzpatrick, P.M, Real Analysis, New Jersey, 2010.

[5] Kurtz, D.S., Swartz, C.W., Theories of Integration: The Integrals of Riemann, Lebesgue, Henstock- Kurzweil, and McShane, World Scientiﬁc, New Mexico, 2004.

[6] Myers, T., The Gauge Integral and its Relationship to the Lebesgue Integral, Google Books, California, 2007

[7] Dummit, D.S., Foote, R.M.,Abstract Algebra, John Wiley and Sons Inc., New Jersey, 2004.

[8] Gourdon,X., Sebah,P., (April 2004), Numbers, Constants and Computation, numbers.computation.free.fr, http://www.numbers.computation.free.fr/Constants/Gamma/gamma.pdf

[9] LOYA, P., (Feb. 2005), Dirichlet and Fresnel Integrals via Iterated Integration, Mathematics Magazine, VOL. 78, NO. 1,http://www.math.binghamton.edu/loya/papers/LoyaMathMag.pdf.