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The Henstock-Kurzweil

integral

Bachelorthesis Mathematics

June 2014

Student: E. van Dijk

First supervisor: Dr. A.E. Sterk

Second supervisor: Prof. dr. A. van der Schaft

Abstract

In this thesis we examine the Henstock-Kurzweil integral. First we look at thedefinition, which only differs slightly from the definition of the common Riemannintegral; instead of using a constant δ, we use a strictly positive function γ. Af-ter proving some properties of the Henstock-Kurzweil integral, we look at a coupleexamples. At last we will see that the Henstock-Kurzweil integral has some nice ben-efits over the Riemann integral: we note that each derivative is Henstock-Kurzweilintegrable when we look at the Fundamental theorem and we can prove some niceconvergence theorems regarding Henstock-Kurzweil integrals.

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Contents

1 Introduction 3

2 The Riemann and the Henstock-Kurzweil integral 4

2.1 The Riemann integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.2 The Henstock-Kurzweil integral . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Properties of the Henstock-Kurzweil integral 8

4 Examples 15

4.1 Linear functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4.2 Dirichlet’s function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4.3 Modified Dirichlet function . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4.4 Thomae’s function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5 The Fundamental Theorem of Calculus 23

6 Convergence theorems 25

7 Conclusion and discussion 35

8 Acknowledgements 36

1 Introduction

Historically integration was defined to be the inverse process of differentiating. So a func-tion F was the integral of a function f if F ′ = f . Around 1850 a new approach occurredin the work of Cauchy and soon after in the work of Riemann. Their idea was to not lookat integrals as the inverse of derivatives, but to come up with a definition of the integralindependent of the derivative. They used the notion of the ’area under the curve’ as astarting point for building a definition of the integral. Nowadays this is known as the Rie-mann integral. This integral has an intuitive approach and is usually discussed in calculuscourses.Suppose we have a function f on the interval [a, b] and we want to find its Riemann inte-gral. The idea is to divide the interval into small subintervals. In each subinterval [xi−1, xi]we pick some point ti. For simplicity the point ti is often chosen to be one of the end-points of the interval. The value f(ti)(xi − xi−1) is then used to approximate the area

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under the graph of f on [xi−1, xi]. The area of each rectangle that is obtained this way isf(ti)(xi− xi−1). The total area under the curve on the interval [a, b] is then approximatedby∑n

i=1 f(ti)(xi−xi−1). It is easily seen that the approximation improves as the rectanglesget thinner, so if the length of the subintervals [xi−1, xi] get smaller. We take the limit(if it exists) of this approximating sums as the length of those subintervals tends to zero.

This limit is Riemann’s definition of the integral∫ baf .

Riemann’s definition of the integral turned out to have some limitations. For example, notevery derivative can be integrated when considering Riemann’s definition. To correct thesedeficiencies, Lebesgue came up with a new definition of integration. Lebesgue’s method isa complex one and a considerably amount of measure theory is required even to define theintegral. Around 1965 Kurzweil came up with a new definition for the integral which wasfurther developed by Henstock. In their definition the intuitive approach of the Riemannintegral is preserved. They just made a small adjustment to the standard ε,δ-definitionof the Riemann integral; instead of a constant δ, Henstock and Kurzweil used a strictlypositive function γ. By making this small adjustment, it turned out that their integralalso correct some of the limitations of the Riemann integral.The integral of Kurzweil and Henstock is known by various names; the Henstock-Kurzweilintegral, the generalized Riemann integral, and just the Henstock or just the Kurzweil in-tegral are examples. Because the strictly positive function γ used in the definition is calleda gauge, it is also known as the gauge integral.In this thesis we will look at the integral of Kurzweil and Henstock. The integral will goby the name the Henstock-Kurzweil integral. First we are going to define the Henstock-Kurzweil integral and compare it with the definition of the Riemann integral. In the nextsection, we will examine some properties and we will prove all those properties. Thenwe are going to look at some examples of Henstock-Kurzweil integrable function. At lastwe will investigate the fundamental theorem and some convergence theorems regardingHenstock-Kurzweil integration.

2 The Riemann and the Henstock-Kurzweil integral

In this first section we begin with giving the definition of the common Riemann integral.Then we expand this definition to the definition of the Henstock-Kurzweil integral in littlesteps. To get a better understanding of what the various definitions actually mean, we willalso give some examples.

2.1 The Riemann integral

For Riemann and Henstock-Kurzweil integration we first need to divide the interval [a, b]in subintervals. We call the collection of these subintervals a partition [2, 4].

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Definition 2.1. A partition P := {[xi−1, xi] : 1 ≤ i ≤ n} of an interval [a, b], is a finitecollection of closed intervals whose union is [a, b] and where the subintervals [xi−1, xi] canonly have the endpoints in common.

If we have for example the interval [0, 3], we could make a partition by dividing it intothree subintervals of equal length: P = {[0, 1], [1, 2], [2, 3]}. We see that this is an finitecollection of closed intervals. It is also clear that the union of these three subintervals is[0, 3]. Therefore this is indeed a partition. An other partition could be P = {[0, 1

2], [1

2, 3]}.

This is a partition because it is again a finite collection of closed subintervals, whose unionis [0, 3].The next step is to take a point ti in each subinterval. These points ti will be the tags.These tags together with our partition will form a collection of ordered pairs which will becalled a tagged partition [2, 4].

Definition 2.2. A tagged partition P := {(ti, [xi−1, xi]) : 1 ≤ i ≤ n} of an interval[a, b], is a finite set of ordered pairs where ti ∈ [xi−1, xi] and the intervals [xi−1, xi] form apartition of [a, b]. The numbers ti are called the corresponding tags.

For simplicity the left or right endpoints of the subintervals [xi−1, xi] are often chosenas tags. We take our partition P = {[0, 1], [1, 2], [2, 3]} from the previous example andchoose the tags ti to be the left endpoints. The tagged partition we get by doing thisis: P = {(0, [0, 1]), (1, [1, 2]), (2, [2, 3])}. We already knew that the subintervals form apartition of [0, 3] and it is clear that 0 ∈ [0, 1], 1 ∈ [1, 2] and 2 ∈ [2, 3], so P is indeed atagged partition of [0, 3]. We could also choose the points in the middle of the subintervalsto be the tags. Then we get the tagged partition P = {(1

2, [0, 1]), (11

2, [1, 2]), (21

2, [2, 3])}.

Here it is also obvious that each tag lies in the corresponding subinterval, so that this Pis again a tagged partition of [0, 3].The next step that will lead us to the definition of the Riemann integral is to define theRiemann sum [2, 4].

Definition 2.3. If P := {(ti, [xi−1, xi]) : 1 ≤ i ≤ n} is a tagged partition of [a, b] andf : [a, b]→ R is a function, then the Riemann sum S(f, P ) of f corresponding to P isS(f, P ) :=

∑ni=1 f(ti)(xi − xi−1).

For example, let’s take the function f(x) = x on the interval [0, 3] together with ourtagged partition P = {(0, [0, 1]), (1, [1, 2]), (2, [2, 3])} from the previous example. The Rie-mann sum will now be given by: S(f, P ) = f(0)(1 − 0) + f(1)(2 − 1) + f(2)(3 − 2) =0 · 1 + 1 · 1 + 2 · 1 = 3.The last step is to take the limit of this Riemann sum when the lenght of the intervals tendto zero; so when xi − xi−1 → 0. In the following definition this is given in ε,δ-form [1].

Definition 2.4. For a function f : [a, b]→ R, the number A =∫ baf is called the Riemann

integral of f if for all ε > 0, there exists a δε > 0 such that if P := {(ti, [xi−1, xi]) : 1 ≤i ≤ n} is any tagged partition of [a, b] satisfying 0 < xi − xi−1 ≤ δε for all i = 1, 2, · · · , n,then |S(f, P )− A| ≤ ε.

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2.2 The Henstock-Kurzweil integral

We will now define the Henstock-Kurzweil integral. Instead of choosing δ to be a constant,we allow δ to be a function which we will call γ. This small change has great benefits aswe will see later on. The only restriction to the function γ is that is has to be strictlypositive. Such a function will be called a gauge [2].

Definition 2.5. A function γ on an interval [a, b] is called a gauge if γ(x) > 0, for allx ∈ [a, b].

We could come up with many examples of gauges. We could simply take γ(x) = c, wherec ∈ R>0. We could also take the function γ(x) = sin x. It is a gauge on the interval [1, 3],because it is strictly positive on that interval.In order to define the Henstock-Kurzweil integral we now want to use a gauge γ instead ofa constant δ to compare the length of each subinterval [xi−1, xi] [2].

Definition 2.6. If γ is a gauge on the interval [a, b] and P := {(ti, [xi−1, xi]) : 1 ≤ i ≤ n}is a tagged partition of [a, b], then P is γ-fine if xi − xi−1 ≤ γ(ti) for all i = 1, 2, · · · , n.

Suppose we have the function γ(x) = x + 1 on [0, 3]. This is a gauge on [0, 3] becauseit is strictly greater than zero on that interval. If we look at our tagged partition P :={(0, [0, 1]), (1, [1, 2]), (2, [2, 3])} of the interval [0, 3] from our previous examples, then wesee that P is γ-fine, because 1− 0 = 1 ≤ 1 = 0 + 1 = γ(0), 2− 1 = 1 ≤ 2 = 1 + 1 = γ(1)and 3 − 2 = 1 ≤ 3 = 2 + 1 = γ(2). So a tagged partition is γ-fine if the length of eachsubinterval is less or equal than the value of γ at the corresponding tag.It turns out that for every gauge γ on an interval [a, b] there exists a γ-fine, tagged partition.This is the content of Cousin’s lemma [8].

Lemma 2.1 - Cousin’s lemma. If γ is a gauge on the interval [a, b], then there exists atagged partition of [a, b] that is γ-fine.

Before we can prove Cousin’s lemma, we need another little lemma.

Lemma 2.2. Suppose γ is a gauge and I1 and I2 are intervals which have at most onepoint in common. If I1 and I2 are γ-fine then the union I := I1 ∪ I2 is also γ-fine.

Proof. If I1 and I2 are γ-fine, then there exist tagged partitions P1 := {(ti, [xi−1, xi] : 1 ≤i ≤ n} and P2 := {tj, [xj−1, xj] : 1 ≤ j ≤ m} of I1 respectively I2 which are γ-fine. So0 ≤ xi − xi−1 ≤ γ(ti) and 0 ≤ xj − xj−1 ≤ γ(tj) for all i = 1, 2, ..., n and j = 1, 2, ...,m.If we now take the union of our tagged partitions P := P1 ∪ P2, then the length of eachsubinterval is still smaller than the value of γ in the corresponding tag, so P is γ-fine. Itis clear that P := P1 ∪ P2 is a tagged partition of I := I1 ∪ I2. So for I := I1 ∪ I2 thereexists a tagged partition that is γ-fine and thus I is γ-fine.

We are now going to use this little lemma to prove Cousin’s lemma.

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Proof. Lemma 2.1 - Cousin’s Lemma. Let S be the set of points x ∈ (a, b] such that thereexists a γ-fine tagged partition of [a, x]. We are going to show that there exists a γ-finetagged partition of [a, b] by showing that the supremum of set S exists and that it is equalto b.Let x be a real number that satisfies a < x < a + γ(a) and x < b. Now consider {[a, x]}to be a partition of [a, x] which consists of one subinterval. If we choose a to be a tag,then we get a tagged partition of [a, x]: {(a, [a, x])}. Because x− a < a+ γ(a)− a = γ(a),we know that the tagged partition {(a, [a, x])} is γ-fine. Thus we know that x ∈ S andtherefore S is nonempty. The axioma of completeness states that every nonempty set ofreal numbers that is bounded above has a supremum. Since S is bounded above by b, weknow that S has a supremum. Let β be the supremum of S.Since β is the supremum of S, we have that β ≤ b and so the gauge γ is defined at β.Since γ(β) > 0 and since β is the supremum of S, there exists a point y ∈ S such thatβ − γ(β) < y < β. Let P1 be a γ-fine tagged partition of [a, y]; such a partition existstbecause y ∈ S. Now we consider {[y, β]} to be a partition of [y, β] which again consists ofone subinterval. We choose β to be a tag and so we get the tagged partition {(β, [y, β])}.Because β − y < β − (β − γ(β)) = γ(β) we know that {(β, [y, β])} is γ-fine. Now we takethe union: P̃1 = P1∪{(β, [y, β])} and by lemma 2.1 we know that this union P̃1 is a taggedpartition of [a, β] that is γ-fine. Thus we can conclude that β ∈ S.Now we need to demonstrate that β = b. To do this, we assume by way of contradictionthat β < b. Since β < b there exists a point z ∈ (β, b) such that z < β + γ(β). Let P2

be a γ-fine tagged partition of [a, β]; such a partition exists because β ∈ S. We consider{[β, z]} to be a partition of [β, z] and by taking β as a tag we get the tagged partition{(β, [β, z])}. Because z − β < β + γ(β) − β = γ(β) we know that {(β, [β, z])} is a γ-finetagged partition of [β, z]. Now we take the union P̃2 = P2 ∪ {(β, [β, z])} and by applyinglemma 2.1 again, we know that this P̃2 is a γ-fine taggged partition of [a, z]. Therefore weconclude that z ∈ S. But this is a contradiction, because β is the supremum of S. So ourassumption can not be true and thus β = b.Finally, since b = β is an element of S, we can conclude that there exists a γ-fine partitionof [a, b].

With the definition of a γ-fine partition, we can define the Henstock-Kurzweil integral[2]. As said before, its definition differs only slightly from the definition of the Riemannintegral.

Definition 2.7. For a function f : [a, b] → R the number B =∫ baf is called the

Henstock-Kurzweil integral of f if for all ε > 0 there exists a gauge γ such thatif P := {(ti, [xi−1, xi]) : 1 ≤ i ≤ n} is a tagged partition of [a, b] that is γ-fine, then|S(f, P )−B| ≤ ε.

Instead of setting the length of each subinterval [xi−1, xi] smaller than a constant δ asin the definition of the Riemann integral, we use a gauge γ to compare the lengths with.This little difference will have many benefits. In section 4 we will see some examples ofHenstock-Kurzweil integrable functions.

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3 Properties of the Henstock-Kurzweil integral

In this section we will look at some properties of the Henstock-Kurzweil integral. We willbegin with comparing it with the Riemann integral. Next thing we will see is that theHenstock-Kurzweil integral is unique. Then some linearity properties will follow. Afterthat we are going to look at the Cauchy criterion for Henstock-Kurzweil integrability andwe are going to use this criterion to prove a theorem considering the integrability on subin-tervals of [a, b]. At last we are looking at the Henstock-Kurzweil integrals of functions thatare zero, of two functions that are equal to each other and finally of a function that is lessor equal than another Henstock-Kurzweil integrable function.We saw in the previous section that the definitions of the Riemann integral and theHenstock-Kurzweil integral are very similar. Because the definitions are so similar wecould wonder if a Riemann integrable function is also Henstock-Kurzweil integrable andthe other way around.It turns out that indeed every Riemann integrable function is Henstock-Kurzweil inte-grable. The idea behind proving this statement is to choose our gauge γ to be the constantδ from the definition of the Riemann integral. This statement is formally presented in thenext theorem.

Theorem 3.1. If f : [a, b] → R is Riemann integrable on [a, b] with∫ baf = A, then f is

Henstock-Kurzweil integrable on [a, b] with∫ baf = A.

Proof. Let ε > 0. Since f is Riemann integrable, there exists a constant δε > 0 such thatif P := {(ti, [xi−1, xi]) : 1 ≤ i ≤ n} is a tagged partition of [a, b] such that xi − xi−1 ≤ δεfor all i = 1, 2, · · · , n then |S(f, P ) − A| ≤ ε. Now we choose our gauge to be γ(x) = δε.Then we know that if P is a tagged partition of [a, b] that is γ-fine then

|S(f, P )− A| ≤ ε.

Because our ε was arbitrary, we know this holds for all ε > 0 and thus we can concludethat f is Henstock-Kurzweil integrable with

∫ baf = A.

The converse of this statement is false. Not every Henstock-Kurzweil integrable functionis Riemann integrable. An example of a function that is not Riemann integrable, butis Henstock-Kurzweil integrable is Dirichlet’s function. We will discuss this function insection 4.2.The following theorem states that if a function f is Henstock-Kurzweil integrable, then itsintegral is unique [6].

Theorem 3.2 - Uniqueness of Henstock-Kurzweil integral. If f : [a, b] → R isHenstock-Kurzweil integrable on the interval [a, b], then the Henstock-Kurzweil integral off on [a, b] is unique.

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Proof. We assume that there are two numbers B1, B2 ∈ R that are both the Henstock-Kurzweil integral of f on [a, b]. Let ε > 0. Since B1 is Henstock-Kurzweil integrable, thereexists a gauge γ1 on [a, b] such that if P1 is a γ1-fine partition of [a, b] then

|S(f, P1)−B1| ≤ε

2.

Similarly, there exists a gauge γ2 on [a, b] such that if P2 is a γ2-fine partition of [a, b] then

|(S(f, P2)−B2| ≤ε

2.

We now define a new gauge: γ := min{γ1, γ2}. Suppose now that P is a tagged partitionof [a, b] that is γ-fine. Because of the construction of γ, we know that this P is also γ1-and γ2-fine. Thus we have:

|B1 −B2| = |B1 − S(f, P ) + S(f, P )−B2|≤ |S(f, P )−B1|+ |S(f, P )−B2|

≤ ε

2+ε

2= ε.

Because our ε was arbitrary, we conclude that B1 = B2 and thus the Henstock-Kurzweilintegral of f on [a, b] is unique.

Now we will look at some linearity properties of the Henstock-Kurzweil integral.

Theorem 3.3 - Linearity properties. Let f and g be Henstock-Kurzweil integrable on[a, b], then

1. kf is Henstock-Kurzweil integrable on [a, b] for each k ∈ R with∫ bakf = k

∫ baf ;

2. f + g is Henstock-Kurzweil integrable on [a, b] with∫ ba

(f + g) =∫ baf +

∫ bag.

Proof. 1. Let ε > 0. By assumption f is Henstock-Kurzweil integrable. So for ε|k| there

exists a gauge γ such that if P is a tagged partition of [a, b] that is γ-fine, then∣∣∣∣S(f, P )−∫ b

a

f

∣∣∣∣ ≤ ε

|k|.

Now we have that∣∣∣∣S(kf, P )− k∫ b

a

f

∣∣∣∣ =

∣∣∣∣∣n∑i=1

kf(ti)(xi − xi−1)− k∫ b

a

f

∣∣∣∣∣= |k|

∣∣∣∣S(f, P )−∫ b

a

f

∣∣∣∣≤ |k| ε

|k|= ε.

Because our ε was arbitrary, this holds for all ε > 0 and therefore kf is Henstock-Kurzweil integrable on [a, b] with

∫ bakf = k

∫ baf .

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2. Let ε > 0. By assumption f and g are Henstock-Kurzweil integrable. So for ε2

thereexist gauges γf and γg such that whenever Pf and Pg are tagged partitions of [a, b]that are γf -fine respectively γg-fine, then∣∣∣∣S(f, Pf )−

∫ b

a

f

∣∣∣∣ ≤ ε

2

and ∣∣∣∣S(g, Pg)−∫ b

a

g

∣∣∣∣ ≤ ε

2.

Now we define γ := min{γf , γg}. Furthermore we have:

S(f + g, P ) =n∑i=1

[(f + g)(ti)(xi − xi−1)]

=n∑i=1

[f(ti)(xi − xi−1) + g(ti)(xi − xi−1)]

=n∑i=1

[f(ti)(xi − xi−1)] +n∑i=1

[g(ti)(xi − xi−1)]

= S(f, P ) + S(g, P ).

If we now have a partition P that is γ-fine, then because of the construction of γ thisP is also γf - and γg-fine. Therefore we know that∣∣∣∣S(f + g, P )− (

∫ b

a

f +

∫ b

a

g)

∣∣∣∣ ≤ ∣∣∣∣S(f, P )−∫ b

a

f

∣∣∣∣+

∣∣∣∣S(g, P )−∫ b

a

g

∣∣∣∣≤ ε

2+ε

2= ε.

Since our ε was arbitrary, this holds for all ε > 0 and therefore f + g is Henstock-Kurzweil integrable on [a, b] with

∫ ba(f + g) =

∫ baf +

∫ bag.

The next thing we are going to look at is the Cauchy criterion for Henstock-Kurzweilintegrals [4]. This theorem is a nice one, because with this theorem one can prove that acertain function is Henstock-Kurzweil integrable without knowing the value of the integral.It will be used in some proofs later on.

Theorem 3.4 - Cauchy criterion. A function f : [a, b] → R is Henstock-Kurzweilintegrable on [a, b] if and only if for every ε > 0 there exists a gauge γ on [a, b] such that ifP1 and P2 are tagged partitions of [a, b] that are γ-fine, then |S(f, P1)− S(f, P2)| ≤ ε.

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Proof. First we assume that f is Henstock-Kurzweil integrable on [a, b]. Let ε > 0. Weknow that for ε

2, there exists a gauge γ on [a, b] such that if P1 and P2 are tagged partitions

of [a, b] that are γ-fine then, ∣∣∣∣S(f, P1)−∫ b

a

f

∣∣∣∣ ≤ ε

2

and ∣∣∣∣S(f, P2)−∫ b

a

f

∣∣∣∣ ≤ ε

2.

Now it follows that

|S(f, P1)− S(f, P2)| ≤∣∣∣∣S(f, P1)−

∫ b

a

f

∣∣∣∣+

∣∣∣∣∫ b

a

f − S(f, P2)

∣∣∣∣≤ ε

2+ε

2= ε.

Because our ε was arbitrary, we know that this holds for all ε > 0.Conversely, we assume that for every ε > 0 there exists a gauge γ on [a, b] such that if P1

and P2 are tagged partitions of [a, b] that are γ-fine, then |S(f, P1)− S(f, P2)| ≤ ε. Foreach n ∈ N, choose a gauge γn such that

|S(f, P1)− S(f, P2)| ≤1

n

whenever P1 and P2 are γn-fine. We may assume that the sequence {γn} is non-increasing.For each n, let Pn be a tagged partition of [a, b] that is γn-fine. If m > n ≥ N , thenγN ≥ γn ≥ γm. Now Pn is γn-fine and thus also γN -fine. The same holds for Pm: Pm isγm-fine and thus γN -fine. Now we have that

|S(f, Pn)− S(f, Pm)| ≤ 1

N

for m > n ≥ N . So the sequence {S(f, Pn)} is a Cauchy sequence. Now let A be the limitof this sequence and let ε > 0. Choose an integer N ∈ N such that 1

N≤ ε

2and

|S(f, Pn)− A| ≤ ε

2

for all n ≥ N . Now let P be a partition of [a, b] that is γN -fine, then

|S(f, P )− A| ≤ |S(f, P )− S(f, PN)|+ |S(f, PN)− A|

≤ 1

N+ε

2

≤ ε

2+ε

2= ε.

Because our ε was arbitrary, we conclude that f is Henstock-Kurzweil integrable on [a, b].

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If we have a function f that is Henstock-Kurzweil integrable on an interval [a, b], thenintuitively we suppose that it is also Henstock-Kurzweil integrable on subintervals of [a, b].We also suppose by intuition that if a function is Henstock-Kurzweil integrable on [a, c]and on [c, b], then it is also Henstock-Kurzweil integrable on [a, b]. It turns out that thesetwo statements are indeed true. This is the content of the following theorem. We are goingto use the previous theorem, the Cauchy criterion, to prove this. [4].

Theorem 3.5. Let f : [a, b]→ R and c ∈ (a, b).

1. If f is Henstock-Kurzweil integrable on [a, c] and on [c, b], then f is Henstock-Kurzweilintegrable on [a, b].

2. If f is Henstock-Kurzweil integrable on [a, b], then f is Henstock-Kurzweil integrableon every subinterval [α, β] ⊆ [a, b].

Proof. 1. Let ε > 0. Because f is Henstock-Kurzweil integrable on [a, c] we know thatfor ε

2there exists a gauge γa such that if P1 and P2 are tagged partitions of [a, c] that

are γa-fine, then

|S(f, P1)− S(f, P2)| ≤ε

2.

We also apply the Cauchy criterion to the interval [c, b]. We know that for ε2

thereexists a gauge γb such that ∣∣∣S(f, P̃1)− S(f, P̃2)

∣∣∣ ≤ ε

2

whenever P̃1 and P̃2 are tagged partitions of [c, b] that are γb-fine. Now we define anew gauge: γ := max{γa, γb}. If we now have a partition that is γa-fine or γb-fine,then it is also γ-fine. Now we take Pa := P1 ∪ P̃1 and Pb := P2 ∪ P̃2; these Pa and Pbare tagged partitions of [a, b] that are γ-fine. Now we have:

|S(f, Pa)− S(f, Pb)| =∣∣∣S(f, P1) + S(f, P̃1)− [S(f, P2) + S(f, P̃2)]

∣∣∣≤∣∣∣S(f, P1)− S(f, P2)

∣∣∣+∣∣∣S(f, P̃1) + S(f, P̃2)

∣∣∣≤ ε

2+ε

2= ε.

Because our ε was arbitrary, this holds for all ε > 0 and therefore we conclude by theCauchy criterion that f is Henstock-Kurzweil integrable on [a, b]

2. Let ε > 0. Choose a gauge γ such that

|S(f, P1)− S(f, P2)| ≤ ε

whenever P1 and P2 are γ-fine. We can do this because of the Cauchy criterion. Nowfix partitions Pα of [a, α] and Pβ of [β, b] that are γ-fine. Let P̃1 and P̃2 be tagged

12

partitions of [α, β] that are γ-fine. Such partitions exist because of Cousin’s lemma(lemma 2.1). Now define P1 := Pα ∪ P̃1 ∪ Pβ and P2 := Pα ∪ P̃2 ∪ Pβ. P1 and P2 arenow tagged partitions of [a, b] that are γ-fine. So we get:∣∣∣S(f, P̃1)− S(f, P̃2)

∣∣∣ =∣∣∣S(f, Pα) + S(f, P̃1) + S(f, Pβ)− [S(f, Pα) + S(f, P̃2) + S(f, Pβ)]

∣∣∣= |S(f, P1)− S(f, P2)|≤ ε.

Because our ε was arbitrary, we have that for every ε > 0 there exists a gauge γ on

[α, β] such that∣∣∣S(f, P̃1)− S(f, P̃2)

∣∣∣ ≤ ε when P̃1 and P̃2 are γ-fine tagged partitions

of [α, β]. By using the Cauchy criterion again, we conclude that f is Henstock-Kurzweil integrable on [α, β].

We are now going to look at functions that are zero almost everywhere on an interval[a, b]. It turns out that such functions are Henstock-Kurzweil integrable and that theirintegrals are, as expected, equal to zero [4].

Theorem 3.6. Let f : [a, b] → R. If f = 0 except on a countable number of points on

[a, b], then f is Henstock-Kurzweil integrable on [a, b] with∫ baf = 0.

Proof. Let ε > 0. First we define the set where f(x) 6= 0: A := {an : n ∈ Z>0} = {x ∈[a, b] : f(x) 6= 0}. Now we are going to define an appropriate gauge on [a, b]:

γ(x) :=

{1 if x ∈ [a, b] and x 6∈ Aε2−n

|f(an)| if x ∈ A.

Suppose that P = {(ti, [xi−1, xi]) : 1 ≤ i ≤ n} is a tagged partition of [a, b] that is γ-fine.So xi − xi−1 ≤ γ(ti) for all 1 ≤ i ≤ n. Now let π be the set of all indices i such that thetags ti ∈ A. Choose ni such that ti = ani

. Let σ be the set of indices i such that ti 6∈ A.We know that if we have such a tag, then f(ti) = 0 and thus f(ti)(xi − xi−1) = 0. So tags

13

that are not in A do not contribute to S(f, P ). We have

|S(f, P )| =

∣∣∣∣∣n∑i=1

f(ti)(xi − xi−1)

∣∣∣∣∣≤

∣∣∣∣∣∑i∈π

f(ti)(xi − xi−1)

∣∣∣∣∣+

∣∣∣∣∣∑i∈σ

f(ti)(xi − xi−1)

∣∣∣∣∣≤∑i∈π

|f(ani)| ε2

−ni

|f(ani)|

=∑i∈π

ε2−ni

≤ ε∞∑i=1

2−ni = ε

and because our ε was arbitrary, this holds for all ε > 0. Thus we can conclude that f isHenstock-Kurzweil integrable on [a, b] with

∫ baf = 0.

With the previous theorem we can prove that if we have two functions that are the samealmost everywhere on an interval [a, b] and if one of them is Henstock-Kurzweil integrableon that interval, then the other is also Henstock-Kurzweil integrable and their integralsare the same. This is the content of the next theorem [4].

Corollary 3.1. Let f : [a, b] → R be Henstock-Kurzweil integrable on [a, b] and let g :[a, b] → R. If f = g except on a countable number of points on [a, b], then g is Henstock-

Kurzweil integrable on [a, b] with∫ bag =

∫ baf .

Proof. We define a new function: h := g−f on [a, b]. We have h = 0 except on a countablenumber of points on [a, b]. So from theorem 3.6 we conclude that h is Henstock-Kurzweil

integrable on [a, b] with∫ bah = 0. Since f and h are both Henstock-Kurzweil integrable on

[a, b], from theorem 3.3 we know that g = f + h is Henstock-Kurzweil integrable on [a, b]

with∫ bag =

∫ baf +

∫ bah =

∫ baf .

The last property we are going to look at considers two Henstock-Kurzweil integrablefunctions on an interval [a, b]. If one of them is less or equal than the other almost every-where on the interval [a, b], then the statement is that the integral of the smallest functionis less or equal than the biggest [8]. We will also use this theorem when we are provingconvergence theorems in section 6.

Corollary 3.2. If f and g are Henstock-Kurzweil integrable on [a, b] and f(x) ≤ g(x)

except on a countable number of points on [a, b], then∫ baf ≤

∫ bag.

14

Proof. Define two new functions:

f̃(x) =

{f(x) ∀x such that f(x) ≤ g(x)

0 ∀x such that f(x) > g(x)

and

g̃(x) =

{g(x) ∀x such that f(x) ≤ g(x)

0 ∀x such that f(x) > g(x).

Because f(x) = f̃(x) and g(x) = g̃(x) almost everywhere on [a, b], we know from corollary

3.1 that∫ baf =

∫ baf̃ and

∫ bag =

∫ bag̃. Because g̃ ≥ f̃ we know g̃− f̃ ≥ 0 and thus it follows

that∫ ba(g̃ − f̃) ≥ 0. Now we have

∫ bag −

∫ baf =

∫ bag̃ −

∫ baf̃ =

∫ ba(g̃ − f̃) ≥ 0, so it follows

that∫ bag ≥

∫ baf .

4 Examples

Now we have seen al those nice properties of the Henstock-Kurzweil integral in the previoussection, we are going to look at some explicit examples of Henstock-Kurzweil integrablefunctions. For all the examples that we treat, we will give a proof that they are Henstock-Kurzweil integrable mostly by defining an appropriate gauge.We will start easy: first we are going to look at linear functions. Secondly we are going toprove that Dirichlet’s function is not Riemann integrable, but that it is Henstock-Kurzweilintegrable as said before. Then we will modify Dirichlet’s function and see what happenswith the integrability. At last we are going to investigate Thomae’s function.

4.1 Linear functions

We begin here with investigating the easiest linear function. We want to find out if thefunction f(x) = c with c ∈ R is Henstock-Kurzweil integrable on the interval [a, b]. We

know that it is Riemann integrable with∫ baf = c(b− a). So we know by theorem 3.1 that

it is also Henstock-Kurzweil integrable with∫ baf = c(b − a). In the following proposition

we are going to prove this by looking at the definition.

Proposition 4.1. The function f(x) = c with c ∈ R is Henstock-Kurzweil integrable on

the interval [a, b] with∫ baf = c(b− a).

Proof. Let ε > 0. We know that f(x) = c, ∀x ∈ [a, b]. So for any tagged partition P of[a, b], we have

S(f, P ) =n∑i=1

f(ti)(xi − xi−1) =n∑i=1

c(xi − xi−1).

15

This is a telescoping sum, so∑n

i=1 c(xi − xi−1) = c(b− a). Therefore we have that

|S(f, P )− c(b− a)| = |c(b− a)− c(b− a)|= 0 < ε

for any tagged partition P . So this certainly holds if P is a tagged partition that is γ-finefor any gauge γ. Since our ε is arbitrary, it holds for all ε > 0 and thus f(x) = c is

Henstock-Kurzweil integrable on [a, b] with∫ baf = c(b− a).

We will now look at the function f(x) = x. We know that this function is Riemannintegrable and therefore Henstock-Kurzweil integrable. Here we are going to prove it byfinding an appropriate gauge.

Proposition 4.2. The function f(x) = x is Henstock-Kurzweil integrable on the interval

[a, b] with∫ baf = 1

2(b2 − a2).

Proof. Let ε > 0. We take γ(x) :=√

2εn

to be our gauge, where n is the number of tags and

thus the number of subintervals in the tagged partition. Suppose P is a tagged partition of

[a, b] that is γ-fine. So xi − xi−1 ≤ γ(ti) =√

2εn

for all 1 ≤ i ≤ n. We can write 12(b2 − a2)

as a telescoping sum:1

2(b2 − a2) =

1

2

n∑i=1

[x2i − x2i−1].

If we look at the distance from a tag ti to the middle of a subinterval [xi−1, xi], we knowit is always less or equal to half the length of the subinterval. So:∣∣∣∣ti − xi + xi−1

2

∣∣∣∣ ≤ 1

2|xi − xi−1| .

16

Now we have:∣∣∣∣S(f, P )− 1

2(b2 − a2)

∣∣∣∣ =

∣∣∣∣∣n∑i=1

[ti(xi − xi−1)]−1

2

n∑i=1

[x2i − x2i−1]

∣∣∣∣∣=

∣∣∣∣∣n∑i=1

[ti(xi − xi−1)−1

2(xi − xi−1)(xi + xi−1)]

∣∣∣∣∣=

∣∣∣∣∣n∑i=1

[(xi − xi−1)(ti −xi + xi−1

2)]

∣∣∣∣∣≤

n∑i=1

|xi − xi−1|∣∣∣∣ti − xi + xi−1

2

∣∣∣∣≤

n∑i=1

1

2|xi − xi−1|2

≤n∑i=1

1

2

√2ε

n

2

=n∑i=1

ε

n= ε.

Because our ε was arbitrary, we know that this holds for all ε > 0 and thus f(x) = x is

Henstock-Kurzweil integrable on [a, b] with∫ baf = 1

2(b2 − a2).

Because we know now that the functions f(x) = c with c ∈ R and f(x) = x are Henstock-Kurzweil integrable on an arbitrary interval [a, b], we know that all linear functions areHenstock-Kurzweil integrable on an interval [a, b]. This follows directly from theorem 3.3and propositions 4.1 and 4.2 and the fact that a linear function is always of the formf(x) = dx+ c with c, d ∈ R.

4.2 Dirichlet’s function

Dirichlet’s function h : [a, b]→ R is the characteristic function of the rational numbers. Itis a function which is discontinuous everywhere on the interval [a, b]. Dirichlet’s functionis given by:

h(x) =

{1 if x ∈ Q0 if x 6∈ Q

This function is bounded, but not Riemann integrable. We will prove that now.

Proposition 4.3. Dirchlet’s function h(x) is not Riemann integrable on [a, b].

17

Proof. Recall from section 2 that the number A is the Riemann integral of h(x) on [a, b]if for all ε > 0 there exists a δε > 0 such that if P is any tagged partition of [a, b]satisfying xi − xi−1 ≤ δε for all i = 1, 2, · · · , n then |S(h, P )− A| ≤ ε. Suppose thatP = {(ti, [xi−1, xi]) : 1 ≤ i ≤ n} is a tagged partition of [a, b] that satisfies xi − xi−1 ≤ δεfor all i = 1, 2, · · · , n. Because the rationals and irrationals are both dense in R, eachsubinterval contains a point that is rational and a point that is irrational. We distinguishbetween two cases: the first case is when A 6= 0 and the second is when A = 0.First, assume A 6= 0. We choose all our tags to be irrational, so ti 6∈ Q. If we do this weget:

|S(h, P )− A| =

∣∣∣∣∣n∑i=1

h(ti)(xi − xi−1)− A

∣∣∣∣∣=

∣∣∣∣∣n∑i=1

0(xi − xi−1)− A

∣∣∣∣∣= |A| .

So we can not always make the expression |S(h, P )− A| smaller than ε for any taggedpartition P that satisfies xi − xi−1 ≤ δε. Therefore we can conclude that h(x) is notRiemann integrable on [a, b].Now we assume that A = 0. Now we choose all our tags to be rational, so ti ∈ Q. Now weget:

|S(h, P )| =

∣∣∣∣∣n∑i=1

h(ti)(xi − xi−1)

∣∣∣∣∣=

∣∣∣∣∣n∑i=1

1(xi − xi−1)

∣∣∣∣∣= |b− a| .

So again, we can not always make the expression |S(h, P )− A| smaller than ε for anytagged partition P that satisfies xi − xi−1 ≤ δε. Therefore we conclude that h(x) is notRiemann integrable on [a, b].

We have now shown that Dirichlet’s function is not Riemann integrable on an interval[a, b]. Dirichlet’s function is actually Henstock-Kurzweil integrable. We will prove thatstatement now [2].

Proposition 4.4. Dirichlet’s function h : [a, b]→ R is Henstock-Kurzweil integrable with∫ bah = 0.

Proof. Let ε > 0. First we enumerate the rational numbers in [a, b] as {r1, r2, · · · }. Nowwe define our gauge γε:

γε(x) :=

{ε2i

if x = ri

1 if x 6∈ {r1, r2, · · · }.

18

Note that our γε is not a constant value. Suppose P = {(ti, [xi−1, xi]) : 1 ≤ i ≤ n} is atagged partition of [a, b] that is γε-fine. If ti ∈ {r1, r2, · · · } is a tag, then h(ti) = 1 andxi − xi−1 ≤ γε(ti) = ε

2iwhere [xi−1, xi] is the subinterval corresponding to the tag ti. Thus

we have: h(ti)(xi − xi−1) ≤ ε2i

for tags ti ∈ {r1, r2, · · · }. If ti 6∈ {r1, r2, · · · } is a tag, thenh(ti) = 0 and xi− xi−1 ≤ γε(ti) = 1 where [xi−1, xi] is the subinterval corresponding to thetag ti. Therefore: h(ti)(xi − xi−1) = 0. So the subintervals with tags ti 6∈ {r1, r2, · · · } donot contribute to S(h, P ). Let π be the set of indices i such that ti ∈ {r1, r2, · · · } and letσ be the set of indices i such that ti ∈ {r1, r2, · · · }. We conclude that:

|S(h, P )| =

∣∣∣∣∣n∑i=1

h(ti)(xi − xi−1)

∣∣∣∣∣≤

∣∣∣∣∣∑i∈π

h(ti)(xi − xi−1)

∣∣∣∣∣+

∣∣∣∣∣∑i∈σ

h(ti)(xi − xi−1)

∣∣∣∣∣≤

∞∑i=1

ε

2i

= ε∞∑i=1

1

2i= ε.

Since our ε was arbitrary, we know that this holds for all ε and thus Dirichlet’s function his Henstock-Kurzweil integrable with

∫ bah = 0.

In section 3 we saw that every Riemann integrable function is Henstock-Kurzweil inte-grable. Here we have shown that the converse statement is false by giving an example ofa function that is Henstock-Kurzweil integrable, but not Riemann integrable.

4.3 Modified Dirichlet function

We are now going to look what happens if we adjust Dirichlet’s function a little bit. Wedefine a new function g by

g(x) :=

{x if x ∈ Q0 if x 6∈ Q.

We are now going to investigate the Riemann and the Henstock-Kurzweil integrability ofthis function. It turns out that this little modification does not affect the integrability: thefunction g is still not Riemann integrable but it is Henstock-Kurzweil integrable. To provethese two statements, we can follow the proofs of Dirichlet’s function and we only need tomake a few adjustments.

Proposition 4.5. The modified Dirichlet function g(x) is not Riemann integrable on [a, b].

19

Proof. Recall again that the number A is the Riemann integral of g(x) on [a, b] if forall ε > 0 there exists a δε > 0 such that if P is any tagged partition of [a, b] satisfying0 ≤ xi − xi−1 ≤ δε for all i = 1, 2, · · · , n then |S(g, P )− A| ≤ ε. Suppose that P ={(ti, [xi−1, xi]) : 1 ≤ i ≤ n} is a tagged partition of [a, b] that satisfies 0 ≤ xi − xi−1 ≤ δεfor all i = 1, 2, · · · , n. Because the rationals and irrationals are both dense in R, eachsubinterval contains a point that is rational and a point that is irrational. Again, wedistinguish between two cases: A 6= 0 and A = 0.The proof of the first case, when A 6= 0, is identical to the proof of that of Dirichlet’sfunction, so we will skip that part here.If A = 0, we choose our xi ∈ Q. Next we choose our tags to be the midpoints of thesubintervals: ti = xi−1+xi

2∈ Q. It could be that a = x0 and b = xn are no rational

numbers. Now we get

|S(g, P )| =

∣∣∣∣∣n∑i=1

g(ti)(xi − xi−1)

∣∣∣∣∣=

∣∣∣∣∣g(t1)(x1 − x0) +n−1∑i=2

g(ti)(xi − xi−1) + g(tn)(xn − xn−1)

∣∣∣∣∣=

∣∣∣∣∣g(t1)(x1 − a) + g(tn)(b− xn−1) +n−1∑i=2

xi−1 + xi2

(xi − xi−1)

∣∣∣∣∣=

∣∣∣∣∣g(t1)(x1 − a) + g(tn)(b− xn−1) +n−1∑i=2

1

2(x2i − x2i−1)

∣∣∣∣∣=

∣∣∣∣g(t1)(x1 − a) + g(tn)(b− xn−1) +1

2(x2n−1 − x21)

∣∣∣∣We can not make the term 1

2(x2n−1 − x21) arbitrarily small by refining our partition. So we

can not make the whole expression |S(g, P )− A| smaller than ε for any tagged partition Pthat satisfies xi−xi−1 ≤ δε. Therefore we can conclude that g(x) is not Riemann integrableon [a, b].

Proposition 4.6. The modified Dirichlet function g : [a, b] → R is Henstock-Kurzweil

integrable with∫ bag = 0.

Proof. Let ε > 0. We start again with enumerating the rationals in [a, b] as {r1, r2, · · · }.Now we define our gauge γε:

γε(x) :=

{∣∣ εx2i

∣∣ if x = ri

1 if x 6∈ {r1, r2, · · · }.

Suppose P = {(ti, [xi−1, xi]) : 1 ≤ i ≤ n} is a tagged partition of [a, b] that is γε-fine. If

ti ∈ {r1, r2, · · · } is a tag, then g(ti) = ti and xi − xi−1 ≤ γε(ti) =∣∣∣ εti2i

∣∣∣ where [xi−1, xi]

20

is the subinterval corresponding to the tag ti. If ti 6∈ {r1, r2, · · · } is a tag, then g(ti) = 0and xi − xi−1 ≤ γε(ti) = 1 where [xi−1, xi] is the subinterval corresponding to the tag ti.Therefore: g(ti)(xi − xi−1) = 0. So the subintervals with tags ti 6∈ {r1, r2, · · · } do notcontribute to S(h, P ). Let π be the set of integers i such that ti ∈ {r1, r2, · · · } and let σbe the set of integers i such that ti ∈ {r1, r2, · · · }. We can conclude that:

|S(g, P )| =

∣∣∣∣∣∑i∈π

g(ti)(xi − xi−1)

∣∣∣∣∣+

∣∣∣∣∣∑i∈σ

g(ti)(xi − xi−1)

∣∣∣∣∣≤∑i∈π

|ti|∣∣∣∣ εti2i

∣∣∣∣≤

∞∑i=1

ε

2i= ε.

Since our ε was arbitrary, we know this holds for all ε > 0 and thus is the modified Dirichletfunction g Henstock-Kurzweil integrable on [a, b] with

∫ bag = 0.

4.4 Thomae’s function

The next function we will look at is Thomae’s function. It is given by:

T (x) =

1 if x = 01n

if x = mn∈ Q \ {0} is in lowest terms with n > 0

0 if x 6∈ Q.

This function is a special one, because it has the property that it is continuous at every ir-rational point and discontinuous at every rational point. To show this we look at sequences[1].

Proposition 4.7. Thomae’s function T is continuous at every irrational point and dis-continuous at every rational point.

Proof. We will first look at the rational points; so we look at c ∈ Q. We know that T (c) > 0.We can find a sequence (yn) 6∈ Q which converges to c, because the irrationals are dense inR. Because (yn) 6∈ Q, we have T (yn) = 0. So while (yn)→ c, T (yn)→ 0 6= T (c). Becausec was an arbitrary rational number, we conclude that T (x) is not continuous at all rationalpoints.Now we are going to look at the irrational points; so we look at d 6∈ Q. Because therationals are also dense in R we can find a sequence (zn) ∈ Q which converges to d. Thecloser a rational number is to a fixed irrational number, the larger its denominator mustnecessarily be. If the denominator is large, then its reciprocal is close to zero. So if (zn)→ dit follows that T (zn) → 0 = T (d). Because d was an arbitrary irrational number, we canconclude that T (x) is continuous at all irrational points.

21

Now we want to investigate if Thomae’s function is Henstock-Kurzweil integrable. Itturns out that it is. We will prove that statement now. We are doing this again by findingan appropriate gauge. This gauge will look like the gauges we used for Dirichlet’s functionand the modified version of it: we distinguish for the different cases just like in Thomae’sfunction itself.

Proposition 4.8. Thomae’s function T is Henstock-Kurzweil integrable on the interval[a, b] with

∫ baT = 0.

Proof. Let ε > 0. Suppose that P = {(ti, [xi−1, xi]) : 1 ≤ i ≤ n} is a tagged partition of[a, b]. Let π be the set of indices i such that the tags ti are rational; so π := {i : ti = mi

ni∈

Q \ {0} is in lowest terms, with ni > 0}. Let ρ be the set of indices i such that the tags tiare irrational; so ρ := {i : ti 6∈ Q}. Let σ be the set of indices i such that the tags ti arezero; so σ := {i : ti = 0}. The set σ contains at most two elements, because there can beat most be two intervals that have the same tag. This tag should be the right endpoint ofone of the subintervals and the left endpoint of the other subinterval. Now we define ourgauge γε as follows:

γε(x) =

ε4

if x = 0nε2q

if x = mn∈ Q \ {0} is in lowest terms with n > 0

1 if x 6∈ Q

where q is the number of elements of the set π. Now suppose our tagged partition P isγε-fine. Then:

|S(T, P )| =

∣∣∣∣∣n∑i=1

T (ti)(xi − xi−1)

∣∣∣∣∣≤

∣∣∣∣∣∑i∈π

T (ti)(xi − xi−1)

∣∣∣∣∣+

∣∣∣∣∣∑i∈ρ

T (ti)(xi − xi−1)

∣∣∣∣∣+

∣∣∣∣∣∑i∈σ

T (ti)(xi − xi−1)

∣∣∣∣∣=

∣∣∣∣∣∑i∈π

1

ni(xi − xi−1)

∣∣∣∣∣+

∣∣∣∣∣∑i∈σ

(xi − xi−1)

∣∣∣∣∣≤

∣∣∣∣∣∑i∈π

1

ni

niε

2q

∣∣∣∣∣+

∣∣∣∣∣∑i∈σ

ε

4

∣∣∣∣∣≤

∣∣∣∣∣∑i∈π

ε

2q

∣∣∣∣∣+∣∣∣ ε2

∣∣∣=∣∣∣ ε2

∣∣∣+∣∣∣ ε2

∣∣∣ = ε

Because our ε was arbitrary, this holds for all ε > 0 and therefore we conclude thatThomae’s function T (x) is Henstock-Kurzweil integrable on [a, b] with

∫ baT = 0.

22

5 The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if F is differentiable on [a, b] and F ′ = f ,

then∫ baf = F (b) − F (a). In the case of Riemann integration it is necessary to add the

additional condition that f must be integrable. This condition is necessary, because notevery derivative turns out to be Riemann integrable. This is one of the shortcomings ofthe Riemann integral. We are going to investigate if this problem also occurs when we areworking with the Henstock-Kurzweil integral.To show that not every derivative is Riemann integrable, we will now present an example.We consider the following function:

F (x) :=

{x2 sin( 1

x2) if x 6= 0

0 if x = 0.

We claim that this function is continuous and differentiable everywhere. For x 6= 0 this iseasy to see: F is a product of a composition of continuous, differentiable functions. Forx = 0 we have that limx→0 F (x) = 0 and F (0) = 0, so F (x) is continuous at x = 0 and

because limx→0F (x)−F (0)

x−0 = limx→0F (x)x

= limx→0 x sin( 1x2

) = 0, F (x) is also differentiableat x = 0. By using the basic rules of differentiation we get:

F ′(x) :=

{2x sin( 1

x2)− 2

xcos( 1

x2) if x 6= 0

0 if x = 0.

Lebesgue’s theorem states that a function is Riemann integrable if and only if it is boundedand continuous almost everywhere [8]. We are going to show that F ′(x) is not Riemannintegrable on an interval [0, b] (where b ∈ R+) by showing that it is unbounded [8].

Proposition 5.1. The function F ′(x) as given above is unbounded on the interval [0, b].

Proof. Let n be a positive integer and let x = n2

2π. The Archimedean property states that

for every real number x, there exists a natural number N such that N > x. Our x = n2

2πis a

real number, so there exists N1 ∈ N such that N1 > x. This is equivalent to the statementthat n <

√2πN1. Now take y = 1

2πb2. This y is again a real number, so there exists N2 ∈ N

such that N2 > y and this is equivalent to 1√2πN2

< b. Now we take N = max{N1, N2}.Then n <

√2πN and 0 < 1√

2πN< b. So we have that 1√

2πN∈ [0, b] and we are going to

evaluate F ′(x) at this value:∣∣∣∣F ′( 1√2πN

)

∣∣∣∣ =

∣∣∣∣ 2√2πN

sin(2πN)− 2√

2πN cos(2πN)

∣∣∣∣= 2√

2πN > n.

Because our n was an arbitrary integer, we conclude that F ′(x) is unbounded on theinterval [0, b].

23

So we have found a function that is a derivative and that is unbounded. Therefore thisderivative is not Riemann integrable. So the condition that f = F ′ is Riemann integrableis necessary in the Fundamental Theorem.Now we are going to show that we do not need this extra condition when we are consideringHenstock-Kurzweil integration, because it turns out that every derivative is Henstock-Kurzweil integrable. We will see this while we are proving the Fundamental Theoremconsidering Henstock-Kurzweil integration [2, 7].

Theorem 5.1 - The Fundamental Theorem. If F : [a, b]→ R is differentiable at every

point of [a, b] then f = F ′ is Henstock-Kurzweil integrable on [a, b] with∫ baf = F (b)−F (a).

Proof. Let ε > 0. Suppose t ∈ [a, b]. Because f(t) = F ′(t) exists, we know from thedefinition of differentiability that for ε

b−a there exists a δt such that if 0 < |z − t| ≤ δt forz ∈ [a, b] then ∣∣∣∣F (z)− F (t)

z − t− f(t)

∣∣∣∣ ≤ ε

b− a.

Now we let these δt’s form our gauge:

γ(t) := δt.

Note that this gauge γ is not a constant, because δt is not the same for all t; δt dependson t. Now if |z − t| ≤ γ(t) for z, t ∈ [a, b] then

|F (z)− F (t)− f(t)(z − t)| ≤ ε

b− a|z − t| .

If a ≤ α ≤ t ≤ β ≤ b and 0 < |β − α| ≤ γ(t) then by the triangle inequality, we get

|F (β)− F (α)− f(t)(β − α)| ≤ |F (β)− F (t)− f(t)(β − t)|+ |F (t)− F (α)− f(t)(t− α)|

≤ ε

b− a|β − t|+ ε

b− a|t− α|

=ε

b− a|β − α| .

Let P := {(ti, [xi−1, xi]) : 1 ≤ i ≤ n} be a tagged partition of [a, b] that is γ-fine. We canwrite F (b) − F (a) as a telescoping sum: F (b) − F (a) =

∑ni=1(F (xi) − F (xi−1)). Now we

24

have:

|S(f, P )− (F (b)− F (a))| = |F (b)− F (a)− S(f, P )|

=

∣∣∣∣∣n∑i=1

(F (xi)− F (xi−1))−n∑i=1

f(ti)(xi − xi−1)

∣∣∣∣∣=

∣∣∣∣∣n∑i=1

(F (xi)− F (xi−1)− f(ti)(xi − xi−1))

∣∣∣∣∣≤

n∑i=1

|F (xi)− F (xi−1)− f(ti)(xi − xi−1)|

≤n∑i=1

ε

b− a|xi − xi−1|

=ε

b− a(b− a) = ε.

Because our ε was arbitrary, this holds for all ε > 0 and thus we conclude that f = F ′ isHenstock-Kurzweil integrable on [a, b] with

∫ baf = F (b)− F (a).

We see that every derivative is Henstock-Kurzweil integrable, because we could takeγ(t) := δt as gauge, where δt stems from the definition of differentiability. So the short-coming that not every derivative is Riemann integrable does not occur when we are con-sidering Henstock-Kurzweil integrability and the Fundamental theorem does not need anextra condition.

6 Convergence theorems

Another limitation of the Riemann integral arises when we consider limits of sequences offunctions. To illustrate this we look at Dirichlet’s function again. Recall from subsection4.2 that Dirichlet’s function is the characteristic function of the rationals and is given by:

h(x) =

{1 if x ∈ Q0 if x 6∈ Q.

Now we focus on the interval [0, 1]. Let {r1, r2, ...} be an enumeration of the rational pointsin this interval. Now define h1(x) = 1 if x = r1 and h1(x) = 0 otherwise. Next, defineh2(x) = 1 if x = r1 or x = r2 and h2(x) = 0 otherwise. By continuing this, we get:

hn(x) =

{1 if x ∈ {r1, r2, · · · , rn}0 if x 6∈ {r1, r2, · · · , rn}

.

25

Each hn has a finite number of discontinuities and is therefore Riemann integrable with∫ bahn = 0. We also have that hn → h pointwise. We saw in subsection 4.2 that Dirichlet’s

function is not Riemann integrable. Thus we have that

limn→∞

∫ 1

0

hn =

∫ 1

0

h

does not hold, because the value on the right-hand side does not exist [1].In this section we are going to look at convergence theorems for the Henstock-Kurzweil in-tegral. We want to investigate if we can overcome shortcomings as above while consideringHenstock-Kurzweil integration instead of Riemann integration.Before we begin our investigation regarding convergence and Henstock-Kurzweil integra-tion, we define uniformly Henstock-Kurzweil integrability [4]. Then we continue with ourfirst convergence theorem which will involve uniformly Henstock-Kurzweil integrals. Thenext theorem will be about uniform convergence and Henstock-Kurzweil integrability andat last we are going to look at pointwise convergence.

Definition 6.1. Let {fn} be a sequence of Henstock-Kurzweil integrable functions on [a, b].The sequence {fn} is uniformly Henstock-Kurzweil integrable on [a, b] if for all ε > 0there exist a gauge γ on [a, b] such that if P is a tagged partition of [a, b] that is γ-fine then∣∣∣S(fn, P )−

∫ bafn

∣∣∣ ≤ ε for all n.

What this definition actually means is that if we can find one gauge γ that works for allfn, then we call the sequence {fn} uniformly Henstock-Kurzweil integrable.Now we are going to prove our first convergence-theorem [4].

Theorem 6.1. Suppose {fn} is a sequence of Henstock-Kurzweil integrable functions on[a, b] and that {fn} converges pointwise to f . If {fn} is uniformly Henstock-Kurzweil inte-

grable on [a, b], then f is Henstock-Kurzweil integrable on [a, b] with∫ baf = limn→∞

∫ bafn.

Proof. Let ε > 0. Since {fn} is uniformly Henstock-Kurzweil integrable there exists agauge γ on [a, b] such that if P̃ is a tagged partition of [a, b] that is γ-fine, then∣∣∣∣S(fn, P̃ )−

∫ b

a

fn

∣∣∣∣ ≤ ε

3

for all n. Since {fn} converges pointwise on [a, b], there exists an integer N such that∣∣∣S(fn, P̃ )− S(fm, P̃ )∣∣∣ ≤ ε

3

for all m,n ≥ N . Now it follows that:∣∣∣∣∫ b

a

fn −∫ b

a

fm

∣∣∣∣ ≤ ∣∣∣∣∫ b

a

fn − S(fn, P̃ )

∣∣∣∣+∣∣∣S(fn, P̃ )− S(fm, P̃ )

∣∣∣+

∣∣∣∣S(fm, P̃ )−∫ b

a

fm

∣∣∣∣≤ ε

3+ε

3+ε

3= ε

26

for all m,n ≥ N . This means that {∫ bafn} is a Cauchy sequence. Let A be the limit of

this sequence. Our claim is now that∫ baf = A.

Let ε > 0. Since {∫ bafn} is a Cauchy sequence, we can choose an integer N such that∣∣∣∣∫ b

a

fn − A∣∣∣∣ ≤ ε

3

for all n ≥ N . Since {fn} is uniformly Henstock-Kurzweil integrable, there exists a gaugeγ such that if P is a tagged partition of [a, b] that is γ-fine, then∣∣∣∣S(fn, P )−

∫ b

a

fn

∣∣∣∣ ≤ ε

3

for all n. Now suppose P is a tagged partition of [a, b] that is γ-fine. Since {fn} convergespointwise to f there exist k ≥ N such that

|S(f, P )− S(fk, P )| ≤ ε

3.

Thus, it follows that:

|S(f, P )− A| ≤ |S(f, P )− S(fk, P )|+∣∣∣∣S(fk, P )−

∫ b

a

fk

∣∣∣∣+

∣∣∣∣∫ b

a

fk − A∣∣∣∣

≤ ε

3+ε

3+ε

3= ε.

Since our ε was arbitrary, we conclude that f is Henstock-Kurzweil integrable on [a, b] with∫ baf = A = limn→∞ fn.

The next convergence theorem we are going to look at is about uniform convergence incombination with Henstock-Kurzweil integrability.

Theorem 6.2. If {fk} is a sequence of Henstock-Kurzweil integrable functions on [a, b] thatconverges uniformly to f on [a, b], then it follows that f is Henstock-Kurzweil integrable on

[a, b] with∫ baf = limk→∞

∫ bafk.

Before we can prove this theorem, we need to prove a lemma.

Lemma 6.1. If f : [a, b]→ R is Henstock-Kurzweil integrable and |f(x)| ≤M with M ∈ Rfor all x ∈ [a, b] then

∣∣∣∫ ba f ∣∣∣ ≤M(b− a).

Proof. |f(x)| ≤ M means that −M ≤ f(x) ≤ M . In section 4.1 we saw that the function

f(x) = c with c ∈ R is Henstock-Kurzweil integrable on [a, b] with∫ baf = c(b− a). So we

know that M and −M are Henstock-Kurzweil integrable. Applying theorem 3.3, we get∫ b

a

M = M(b− a)

27

∫ b

a

−M = −∫ b

a

M = −M(b− a).

Corollary 3.2 states that if f and g are Henstock-Kurzweil integrable on [a, b] and f(x) ≤g(x) on [a, b], then

∫ baf ≤

∫ bag. We apply this corollary to −M ≤ f(x) and to f(x) ≤ M

and thus we get ∫ b

a

−M = −M(b− a) ≤∫ b

a

f

and ∫ b

a

f ≤∫ b

a

M = M(b− a).

We conclude that ∣∣∣∣∫ b

a

f

∣∣∣∣ ≤M(b− a).

Now we have proven this lemma, we can prove theorem 6.2 [3].

Proof. Theorem 6.2. Let ε > 0. Because {fk} converges uniformly to f , there exists aK ∈ N such that

|fk − f | ≤ε

2(b− a)

for all x ∈ [a, b] whenever k ≥ K. Consequently, if h, k ≥ K then we have

|fk − fh| ≤ |fk − f |+ |f − fh|

≤ ε

2(b− a)+

ε

2(b− a)=

ε

(b− a).

By assumption fk and fh are Henstock-Kurzweil integrable and thus follows from theorem3.3 that fk − fh is Henstock-Kurzweil integrable. Now we can apply lemma 6.1 to fk − fh:∣∣∣∣∫ b

a

(fk − fh)∣∣∣∣ ≤ ε

(b− a)(b− a) = ε.

Because of the linearity properties of the Henstock-Kurzweil integral, we know that∣∣∣∫ ba (fk − fh)

∣∣∣ =∣∣∣∫ ba fk − ∫ ba fh∣∣∣ ≤ ε. Because our ε was arbitrary, this means that the sequence {∫ bafk} is a

Cauchy sequence. Let A be the limit of this sequence. Our claim is again that∫ baf = A.

Let ε > 0 and let P = {(ti, [xi−1, xi]) : 1 ≤ i ≤ n} be a tagged partition of [a, b].

Since {∫ bafk} is a Cauchy sequence with limit A, there exist an integer N ∈ N such that

for k ≥ N , we have ∣∣∣∣∫ b

a

fk − A∣∣∣∣ ≤ ε

3.

28

Since {fk} converges uniformly to f , there exists an K ∈ N such that if k ≥ K, then

|fk − f | ≤ε

3(b− a)

for all x ∈ [a, b]. Because of this, it follows that

|S(fk, P )− S(f, P )| =

∣∣∣∣∣n∑i=1

fk(ti)(xi − xi−1)−n∑i=1

f(ti)(xi − xi−1)

∣∣∣∣∣=

∣∣∣∣∣n∑i=1

[(fk(ti)− f(ti))(xi − xi−1)]

∣∣∣∣∣≤

n∑i=1

|(fk(ti)− f(ti))(xi − xi−1)|

≤n∑i=1

ε

3(b− a)(xi − xi−1)

=ε

3(b− a)(b− a) =

ε

3.

All fk are Henstock-Kurzweil integrable. So for each fk there exists a gauge γk on [a, b]such that whenever P is γk-fine we have∣∣∣∣S(fk, P )−

∫ b

a

fk

∣∣∣∣ ≤ ε

3.

Now we choose a k such that k ≥ K and k ≥ N . Now we combine all the previous and weget

|S(f, P )− A| ≤ |S(f, P )− S(fk, P )|+∣∣∣∣S(fk, P )−

∫ b

a

fk

∣∣∣∣+

∣∣∣∣∫ b

a

fk − A∣∣∣∣

≤ ε

3+ε

3+ε

3= ε.

Because our ε was arbitrary, this holds for all ε > 0 and we conclude that f is Henstock-Kurzweil integrable with

∫ baf = limk→∞

∫ bafk.

We saw here that uniform convergence guarantees the limit function to be Henstock-Kurzweil integrable as well. Now we are going to investigate if this also holds for pointwiseconvergence. As it turns out, supposing that each function in a sequence that convergespointwise is Riemann integrable is not enough to guarantee that the limit function is alsoRiemann integrable. We saw a counterexample of this at the beginning of this section.The monotone convergence theorem provides a set of conditions that guarantees that thelimit of a pointwise convergent sequence of Henstock integrable functions is also Henstockintegrable. This theorem does not hold for the Riemann integral; the sequence of functions

29

hn(x) at the beginning of this section turns out to be a counterexample again.Before we can prove the monotone convergence theorem, we need to take a look at theSaks-Henstock lemma [3]. This lemma is about subpartitions. A subpartition of [a, b]is a partition that does not need to cover the whole interval [a, b]. So the union of thesubintervals of a subpartition does not have to be [a, b].

Definition 6.2. A subpartition P̃ := {[xj−1, xj] : 1 ≤ j ≤ m} of an interval [a, b] is afinite collection of closed intervals such that [xj−1, xj] ⊂ [a, b] and where the subintervals[xj−1, xj] can only have the endpoints in common.

Definition 6.3. A tagged subpartition P̃ := {(tj, [xj−1, xj]) : 1 ≤ j ≤ m} of an interval[a, b] is a finite set of order pairs where tj ∈ [xj−1, xj] and the intervals [xj−1, xj] form asubpartition of [a, b]. The numbers tj are again called the tags.

The Saks-Henstock lemma states that if we have a subpartition of the interval [a, b],then we could make the difference between the Riemann sum and the integral over thesubintervals of the subpartition also smaller than ε. It also says that if we first takeabsolute values and then summarize, this term will be smaller than 2ε.

Lemma 6.2 - Saks-Henstock lemma. Let f be a function that is Henstock-Kurzweilintegrable on [a, b] and let for ε > 0 γε be a gauge on [a, b] such that if P = {(ti, [xi−1, xi]) :

1 ≤ i ≤ n} is a tagged partition of [a, b] that is γε-fine, then∣∣∣S(f, P )−

∫ baf∣∣∣ ≤ ε. If

P̃ = {(tj, [xj−1, xj]) : 1 ≤ j ≤ s} = {(tj, Jj) : 1 ≤ j ≤ s} is any γε-fine subpartition of[a, b], then

1.∣∣∣S(f, P̃ )− [

∫J1f + · · ·+

∫Jsf ]∣∣∣ =

∣∣∣∑sj=1[f(tj)(xj − xj−1)−

∫ xjxj−1

f ]∣∣∣ ≤ ε and

2.∑s

j=1

∣∣∣f(tj)(xj − xj−1)−∫ xjxj−1

f∣∣∣ ≤ 2ε.

Proof. 1. Let K1, · · · , Km be closed subintervals in [a, b] such that {Jj}sj=1 ∪ {Kk}mk=1

forms a partition of [a, b]. Now let α > 0 be arbitrary. Because each Kk ⊆ [a, b], weknow that f is Henstock-Kurzweil integrable on each Kk by theorem 3.5. So for eachKk, there exists a gauge γα,k such that if Qk is a γα,k-fine tagged partition of Kk then∣∣∣∣S(f,Qk)−

∫Kk

f

∣∣∣∣ ≤ α

m.

If we have that γα,k(x) ≥ γε(x) for some k, then we could add a positive constantto γε(x) such that we get a new gauge that is greater than γα,k(x): γ̃ε = γε + c withc ∈ R>0 such that γα,k(x) ≤ γ̃ε(x) for all x ∈ Kk. A partition that is γε-fine is thenalso γ̃ε-fine. So this γ̃ε is also a gauge that satisfies the assumption in the statementof this theorem. So we may assume that γα,k(x) ≤ γε(x) for all x ∈ Kk. Then we

30

know that if Qk is γα,k-fine, then it is also γε-fine. Now let P ∗ := P̃ ∪Q1 ∪ · · · ∪Qm.P ∗ is a tagged partition of [a, b]. If P ∗ is γε-fine, then∣∣∣∣S(f, P ∗)−

∫ b

a

f

∣∣∣∣ ≤ ε.

Furthermore we have

S(f, P ∗) = S(f, P̃ ) + S(f,Q1) + · · ·+ S(f,Qm)∫ b

a

f =

∫J1

f + · · ·+∫Js

f +

∫K1

f + · · ·+∫Km

f.

From this, it follows that∣∣∣∣S(f, P̃ )− [

∫J1

f + · · ·+∫Js

f ]

∣∣∣∣ =

∣∣∣∣∣[S(f, P ∗)−m∑k=1

S(f,Qk)]− [

∫ b

a

f −m∑k=1

∫Kk

f ]

∣∣∣∣∣≤

∣∣∣∣∣S(f, P ∗)−∫ b

a

f −m∑k=1

[S(f,Qk)−∫Kk

f ]

∣∣∣∣∣≤∣∣∣∣S(f, P ∗)−

∫ b

a

f

∣∣∣∣+m∑k=1

∣∣∣∣S(f,Qk)−∫Kk

f

∣∣∣∣≤ ε+

m∑k=1

α

m

= ε+mα

m= ε+ α.

Because α was arbitrary, we have∣∣∣S(f, P ∗)− [

∫J1f + · · ·+

∫Jsf ]∣∣∣ ≤ ε as desired.

2. Let P̃+ be the pairs (tj, [xj−1, xj]) of P̃ such that f(tj)(xj − xj−1)−∫ xjxj−1

f ≥ 0 and

let P̃− be the pairs (tj, [xj−1, xj]) of P̃ such that f(tj)(xj − xj−1)−∫ xjxj−1

f < 0. P̃+

and P̃− are again subpartitions of [a, b] which are γε-fine so we are going to apply

31

the first part of this lemma to both P̃+ and P̃−:

∑P̃+

∣∣∣∣∣f(tj)(xj − xj−1)−∫ xj

xj−1

f

∣∣∣∣∣ =∑P̃+

[f(tj)(xj − xj−1)−∫ xj

xj−1

f ]

=

∣∣∣∣∣∣∑P̃+

[f(tj)(xj − xj−1)−∫ xj

xj−1

f ]

∣∣∣∣∣∣≤ ε∑

P̃−

∣∣∣∣∣f(tj)(xj − xj−1)−∫ xj

xj−1

f

∣∣∣∣∣ = −∑P̃−

[f(tj)(xj − xj−1)−∫ xj

xj−1

f ]

=

∣∣∣∣∣∣∑P̃−

[f(tj)(xj − xj−1)−∫ xj

xj−1

f ]

∣∣∣∣∣∣≤ ε.

Now we combine this and we get

s∑j=1

∣∣∣∣∣f(tj)(xj − xj−1)−∫ xj

xj−1

f

∣∣∣∣∣ =∑P̃+

∣∣∣∣∣f(tj)(xj − xj−1)−∫ xj

xj−1

f

∣∣∣∣∣+∑P̃−

∣∣∣∣∣f(tj)(xj − xj−1)−∫ xj

xj−1

f

∣∣∣∣∣≤ ε+ ε = 2ε

as desired.

Now we can use the Saks-Henstock lemma to prove the monotone convergence theorem[8].

Theorem 6.3 - Monotone convergence theorem. Let {fn} be a monotone sequence ofHenstock-Kurzweil integrable functions on [a, b] that converges pointwise to a limit function

f . If limn→∞∫ bafn exists, then f is Henstock-Kurzweil integrable and

∫ baf = limn→∞

∫ bafn.

Proof. We suppose {fn} is increasing. Let ε > 0 and let A = limn→∞∫ bafn. Then there

exists an N ∈ N such that for all n ≥ N∣∣∣∣∫ b

a

fn − A∣∣∣∣ ≤ ε

3.

32

Since fn is Henstock-Kurzweil integrable for each n, there exists a gauge γn on [a, b] foreach n such that ∣∣∣∣S(fn, P )−

∫ b

a

fn

∣∣∣∣ ≤ ε

3 · 2n

if P is γn-fine. Without loss of generality, we suppose that γn ≥ γn+1. Since {fn} convergespointwise to f on [a, b], we choose for each x ∈ [a, b] an integer Mx ≥ N such that

|fn(x)− f(x)| ≤ ε

3(b− a)

for n ≥Mx. Now we define our gauge: γ(x) := γMx(x) and we let P = {(ti, [xi−1, xi]) : 1 ≤i ≤ m} be a γ-fine tagged partition of [a, b]. Note that our γ is not a constant. Now wedivide the set of tagged subintervals of P into classes based on the length of subintervals.Pm is the collection of γMtm

-fine tagged partitions and let for each m− 1 ≥ i ≥ 1 Pi be thecollection of γMti

-fine partitions of P −{⋃mj=i+1 Pj}. The set of Pi for 1 ≤ i ≤ m is disjoint

and the union of elements in this set is P . Let for each 1 ≤ i ≤ m, Ki be the collection ofintervals used in the collection of tagged intervals of Pi.By the triangle inequality we get:

|S(f, P )− A| ≤

∣∣∣∣∣S(f, P )−m∑i=1

S(fMti, Pi)

∣∣∣∣∣+

∣∣∣∣∣m∑i=1

S(fMti, Pi)−

m∑i=1

∫Ki

fMti

∣∣∣∣∣+

∣∣∣∣∣m∑i=1

∫Ki

fMti− A

∣∣∣∣∣ . (1)

We are now going to look at those three terms seperately.We can use the fact that {fn} converges pointwise to f to estimate the first term. Mti ≥ Nby definition, so

∣∣fMti− f

∣∣ ≤ ε3(b−a) . Using this, we get:∣∣∣∣∣S(f, P )−

m∑i=i

S(fMti, Pi)

∣∣∣∣∣ =

∣∣∣∣∣m∑i=1

(f(ti)− fMti(ti))(xi − xi−1)

∣∣∣∣∣≤

m∑i=1

∣∣f(ti)− fMti(ti)∣∣ (xi − xi−1)

≤m∑i=1

ε

3(b− a)(xi − xi−1)

=ε

3(b− a)(b− a) =

ε

3.

Now we are going to look at the second term of (1). Here we are going to use Henstock’slemma. The collection of intervals PMti

is by definition γMti-fine for each 1 ≤ i ≤ m.

Because for each n fn is Henstock-Kurzweil integrable, fMtiis Henstock Kurzweil integrable

and thus ∣∣∣∣S(fMti, P )−

∫ b

a

fMti

∣∣∣∣ ≤ ε

3 · 2Mti

33

if P is a γMti-fine partition of [a, b]. Using Henstock’s lemma, we get:∣∣∣∣S(fMti

, Pi)−∫Ki

fMti

∣∣∣∣ ≤ ε

3 · 2Mti

for each 1 ≤ i ≤ m. Now we use this for the second term:∣∣∣∣∣m∑i=1

S(fMti, Pi)−

m∑i=1

∫Ki

fMti

∣∣∣∣∣ ≤m∑i=1

∣∣∣∣S(fMti, Pi)−

∫Ki

fMti

∣∣∣∣≤

m∑i=1

ε

3 · 2Mti≤ ε

3.

Now we are going to look at the last term of (1). We know that {fn} is an increasingsequence of functions. So fMti

≥ fN because Mti ≥ N . Because {fn} converges pointwise

to f from below, we have that {∫ bafn} converges to A from below and we have∣∣∣∣∫

Ki

fMti− A

∣∣∣∣ ≤ ∣∣∣∣∫Ki

fN − A∣∣∣∣

for each 1 ≤ i ≤ m. Therefore we have∣∣∣∣∣m∑i=1

∫Ki

fMti− A

∣∣∣∣∣ ≤∣∣∣∣∣m∑i=1

∫Ki

fN − A

∣∣∣∣∣=

∣∣∣∣∫ b

a

fN − A∣∣∣∣

≤ ε

3.

Now we are combining the three terms again and we get:

|S(f, P )− A| ≤

∣∣∣∣∣S(f, P )−m∑i=1

S(fMti, Pi)

∣∣∣∣∣+

∣∣∣∣∣m∑i=1

S(fMti, Pi)−

m∑i=1

∫Ki

fMti

∣∣∣∣∣+

∣∣∣∣∣m∑i=1

∫Ki

fMti− A

∣∣∣∣∣≤ ε

3+ε

3+ε

3= ε.

Because our ε was arbitrary, we conclude that f is Henstock-Kurzweil integrable with∫ baf = A = limn→∞

∫ bafn.

Of course, a similar result holds when the sequence {fn} is decreasing. The proof issimilar, and hence omitted.

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The example in the beginning of this section shows that the monotone convergence theoremdoes not hold for Riemann integrability. Recall the sequence of functions defined by

hn(x) =

{1 if x ∈ {r1, r2, · · · , rn}0 if x 6∈ {r1, r2, · · · , rn}

on the interval [0, 1]. For this sequence of functions it is clear that hn ≤ hn+1, but as wesaw the limit function is Dirichlet’s function, which is not Riemann integrable.The dominated convergence theorem provides an other set of conditions that also guaran-tees the pointwise limit f of a sequence {fn} to be Henstock-Kurzweil integrable. Againa similar theorem for the Riemann integral does not exist. We do not prove this theoremhere. For those who would like to see a proof we refer to Gordon’s book The Integrals ofLebesgue, Denjoy, Perron, and Henstock [4].

Theorem 6.4 - Dominated convergence theorem. Suppose {fk} is a sequence ofHenstock-Kurzweil integrable functions on [a, b] that converges pointwise to f(x). If g(x)and h(x) are Henstock-Kurzweil integrable functions on [a, b] such that g(x) ≤ fk(x) ≤ h(x)

for all k, then f(x) is Henstock-Kurzweil integrable with∫ baf = limk→∞

∫ bafk.

To show that the dominated convergence theorem does not hold for Riemann integrablefunctions, we could again take a look at the sequence of functions hn(x). These functionsdo only take the values 0 and 1. So lets take r(x) = c with c ∈ R and c ≤ 0 and s(x) = dwith d ∈ R and d ≥ 1. Then we have that r(x) ≤ hn(x) ≤ s(x) for all n and we know thatr(x) and s(x) are Riemann integrable. But still Dirichlet’s function, the limit function ofhn, is not Riemann integrable.

7 Conclusion and discussion

We have seen that making a small change in the definition of the Riemann integral re-sulted in a more general integral: the Henstock-Kurzweil integral. We saw that using thegauge γ instead of a constant δ led to certain benefits over the Riemann integral. Firstwe saw examples of functions that are Henstock-Kurzweil integrable, but not Riemannintegrable. So by using the Henstock-Kurzweil definition of integrals, we enlarged therange of integrable functions. Next thing we saw is that not every derivative is Riemannintegrable. So in the fundamental theorem we have to include the extra condition thatf = F ′ is integrable. When we examined the fundamental theorem regarding Henstock-Kurzweil integrals, we concluded that we do not need this extra condition; it turned outthat every derivative is Henstock-Kurzweil integrable. In the last section we saw that forthe Henstock-Kurzweil integral there exist some nice convergence theorems. For examplewe examined the monotone convergence theorem and the dominated convergence theorem.These theorems turned out to work for Henstock-Kurzweil integrals, but not for Riemannintegrals.

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When we do take measure theory in account, we could also compare the Henstock-Kurzweilintegral with Lebesgue’s integral. Furthermore we could use the monotone convergencetheorem and the dominated convergence theorem for Lebesgue’s integral to prove themonotone convergence theorem and the dominated convergence theorem for the Henstock-Kurzweil integral. This is the way R.A. Gordon proved the convergence theorems in hisbook ”The integrals of Lebesgue, Denjoy, Perron, and Henstock”. He also shows equiva-lence between various integrals. For simplicity’s sake we did not take the measure theory inaccount. We focused on the easy and intuitive approach of the Henstock-Kurzweil integraland its similarity with the Riemann integral. We wanted to focus on the ε,γ-definition of theHenstock-Kurzweil integral and investigate the advantages it has above the ε,δ-definitionof the Riemann integral.

8 Acknowledgements

I would like to thank my supervisor dr. A.E. Sterk for helping me during writing thisthesis. I appreciate his comprehensive comments and his good guidance throughout thewhole process very much.

References

[1] Abbott, S. (2001). Understanding Analysis. Springer.

[2] Bartle, R. G. (1996). Return to the Riemann Integral. American MathematicalMonthly, October.

[3] Bartle, R.G. (2000). Modern theory of Integration. Graduate studies in Mathematics,vol 32, American Mathematical Society.

[4] Gordon, R. A. (1994). The Integrals of Lebesgue, Denjoy, Perron, and Henstock. Amer-ican Mathematical Society.

[5] Herschlag, G. (2006). A Brief Introduction to Gauge Integration. Duke University.

[6] Lee, T. Y. (2011). Henstock-Kurzweil Integration on Euclidean Spaces. World ScientificPublishing Co. Pte. Ltd.

[7] McInnis, E. O. (2002). Gauge Integration. Thesis, Naval Postgraduate School, MotereyCalifornia.

[8] Wells, J. (2011). Generalizations of the Riemann Integral: An Investigation of theHenstock Integral. Senior Project University of Oregon.

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