The HenstockKurzweil integral


 Felix Wilkinson
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1 fculteit Wiskunde en Ntuurwetenschppen The HenstockKurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft
2 Abstrct In this thesis we exmine the HenstockKurzweil integrl. First we look t the definition, which only differs slightly from the definition of the common Riemnn integrl; insted of using constnt δ, we use strictly positive function γ. After proving some properties of the HenstockKurzweil integrl, we look t couple exmples. At lst we will see tht the HenstockKurzweil integrl hs some nice benefits over the Riemnn integrl: we note tht ech derivtive is HenstockKurzweil integrble when we look t the Fundmentl theorem nd we cn prove some nice convergence theorems regrding HenstockKurzweil integrls. 2
3 Contents 1 Introduction 3 2 The Riemnn nd the HenstockKurzweil integrl The Riemnn integrl The HenstockKurzweil integrl Properties of the HenstockKurzweil integrl 8 4 Exmples Liner functions Dirichlet s function Modified Dirichlet function Thome s function The Fundmentl Theorem of Clculus 23 6 Convergence theorems 25 7 Conclusion nd discussion 35 8 Acknowledgements 36 1 Introduction Historiclly integrtion ws defined to be the inverse process of differentiting. So function F ws the integrl of function f if F f. Around 1850 new pproch occurred in the work of Cuchy nd soon fter in the work of Riemnn. Their ide ws to not look t integrls s the inverse of derivtives, but to come up with definition of the integrl independent of the derivtive. They used the notion of the re under the curve s strting point for building definition of the integrl. Nowdys this is known s the Riemnn integrl. This integrl hs n intuitive pproch nd is usully discussed in clculus courses. Suppose we hve function f on the intervl [, b] nd we wnt to find its Riemnn integrl. The ide is to divide the intervl into smll subintervls. In ech subintervl [x i 1, x i ] we pick some point t i. For simplicity the point t i is often chosen to be one of the endpoints of the intervl. The vlue f(t i )(x i x i 1 ) is then used to pproximte the re 3
4 under the grph of f on [x i 1, x i ]. The re of ech rectngle tht is obtined this wy is f(t i )(x i x i 1 ). The totl re under the curve on the intervl [, b] is then pproximted by n f(t i)(x i x i 1 ). It is esily seen tht the pproximtion improves s the rectngles get thinner, so if the length of the subintervls [x i 1, x i ] get smller. We tke the limit (if it exists) of this pproximting sums s the length of those subintervls tends to zero. This limit is Riemnn s definition of the integrl f. Riemnn s definition of the integrl turned out to hve some limittions. For exmple, not every derivtive cn be integrted when considering Riemnn s definition. To correct these deficiencies, Lebesgue cme up with new definition of integrtion. Lebesgue s method is complex one nd considerbly mount of mesure theory is required even to define the integrl. Around 1965 Kurzweil cme up with new definition for the integrl which ws further developed by Henstock. In their definition the intuitive pproch of the Riemnn integrl is preserved. They just mde smll djustment to the stndrd ɛ,δdefinition of the Riemnn integrl; insted of constnt δ, Henstock nd Kurzweil used strictly positive function γ. By mking this smll djustment, it turned out tht their integrl lso correct some of the limittions of the Riemnn integrl. The integrl of Kurzweil nd Henstock is known by vrious nmes; the HenstockKurzweil integrl, the generlized Riemnn integrl, nd just the Henstock or just the Kurzweil integrl re exmples. Becuse the strictly positive function γ used in the definition is clled guge, it is lso known s the guge integrl. In this thesis we will look t the integrl of Kurzweil nd Henstock. The integrl will go by the nme the HenstockKurzweil integrl. First we re going to define the Henstock Kurzweil integrl nd compre it with the definition of the Riemnn integrl. In the next section, we will exmine some properties nd we will prove ll those properties. Then we re going to look t some exmples of HenstockKurzweil integrble function. At lst we will investigte the fundmentl theorem nd some convergence theorems regrding HenstockKurzweil integrtion. 2 The Riemnn nd the HenstockKurzweil integrl In this first section we begin with giving the definition of the common Riemnn integrl. Then we expnd this definition to the definition of the HenstockKurzweil integrl in little steps. To get better understnding of wht the vrious definitions ctully men, we will lso give some exmples. 2.1 The Riemnn integrl For Riemnn nd HenstockKurzweil integrtion we first need to divide the intervl [, b] in subintervls. We cll the collection of these subintervls prtition [2, 4]. 4
5 Definition 2.1. A prtition P : {[x i 1, x i ] : 1 i n} of n intervl [, b], is finite collection of closed intervls whose union is [, b] nd where the subintervls [x i 1, x i ] cn only hve the endpoints in common. If we hve for exmple the intervl [0, 3], we could mke prtition by dividing it into three subintervls of equl length: P {[0, 1], [1, 2], [2, 3]}. We see tht this is n finite collection of closed intervls. It is lso cler tht the union of these three subintervls is [0, 3]. Therefore this is indeed prtition. An other prtition could be P {[0, 1 2 ], [ 1 2, 3]}. This is prtition becuse it is gin finite collection of closed subintervls, whose union is [0, 3]. The next step is to tke point t i in ech subintervl. These points t i will be the tgs. These tgs together with our prtition will form collection of ordered pirs which will be clled tgged prtition [2, 4]. Definition 2.2. A tgged prtition P : {(t i, [x i 1, x i ]) : 1 i n} of n intervl [, b], is finite set of ordered pirs where t i [x i 1, x i ] nd the intervls [x i 1, x i ] form prtition of [, b]. The numbers t i re clled the corresponding tgs. For simplicity the left or right endpoints of the subintervls [x i 1, x i ] re often chosen s tgs. We tke our prtition P {[0, 1], [1, 2], [2, 3]} from the previous exmple nd choose the tgs t i to be the left endpoints. The tgged prtition we get by doing this is: P {(0, [0, 1]), (1, [1, 2]), (2, [2, 3])}. We lredy knew tht the subintervls form prtition of [0, 3] nd it is cler tht 0 [0, 1], 1 [1, 2] nd 2 [2, 3], so P is indeed tgged prtition of [0, 3]. We could lso choose the points in the middle of the subintervls to be the tgs. Then we get the tgged prtition P {( 1, [0, 1]), (1 1, [1, 2]), (2 1, [2, 3])} Here it is lso obvious tht ech tg lies in the corresponding subintervl, so tht this P is gin tgged prtition of [0, 3]. The next step tht will led us to the definition of the Riemnn integrl is to define the Riemnn sum [2, 4]. Definition 2.3. If P : {(t i, [x i 1, x i ]) : 1 i n} is tgged prtition of [, b] nd f : [, b] R is function, then the Riemnn sum S(f, P ) of f corresponding to P is S(f, P ) : n f(t i)(x i x i 1 ). For exmple, let s tke the function f(x) x on the intervl [0, 3] together with our tgged prtition P {(0, [0, 1]), (1, [1, 2]), (2, [2, 3])} from the previous exmple. The Riemnn sum will now be given by: S(f, P ) f(0)(1 0) + f(1)(2 1) + f(2)(3 2) The lst step is to tke the limit of this Riemnn sum when the lenght of the intervls tend to zero; so when x i x i 1 0. In the following definition this is given in ɛ,δform [1]. Definition 2.4. For function f : [, b] R, the number A f is clled the Riemnn integrl of f if for ll ɛ > 0, there exists δ ɛ > 0 such tht if P : {(t i, [x i 1, x i ]) : 1 i n} is ny tgged prtition of [, b] stisfying 0 < x i x i 1 δ ɛ for ll i 1, 2,, n, then S(f, P ) A ɛ. 5
6 2.2 The HenstockKurzweil integrl We will now define the HenstockKurzweil integrl. Insted of choosing δ to be constnt, we llow δ to be function which we will cll γ. This smll chnge hs gret benefits s we will see lter on. The only restriction to the function γ is tht is hs to be strictly positive. Such function will be clled guge [2]. Definition 2.5. A function γ on n intervl [, b] is clled guge if γ(x) > 0, for ll x [, b]. We could come up with mny exmples of guges. We could simply tke γ(x) c, where c R >0. We could lso tke the function γ(x) sin x. It is guge on the intervl [1, 3], becuse it is strictly positive on tht intervl. In order to define the HenstockKurzweil integrl we now wnt to use guge γ insted of constnt δ to compre the length of ech subintervl [x i 1, x i ] [2]. Definition 2.6. If γ is guge on the intervl [, b] nd P : {(t i, [x i 1, x i ]) : 1 i n} is tgged prtition of [, b], then P is γfine if x i x i 1 γ(t i ) for ll i 1, 2,, n. Suppose we hve the function γ(x) x + 1 on [0, 3]. This is guge on [0, 3] becuse it is strictly greter thn zero on tht intervl. If we look t our tgged prtition P : {(0, [0, 1]), (1, [1, 2]), (2, [2, 3])} of the intervl [0, 3] from our previous exmples, then we see tht P is γfine, becuse γ(0), γ(1) nd γ(2). So tgged prtition is γfine if the length of ech subintervl is less or equl thn the vlue of γ t the corresponding tg. It turns out tht for every guge γ on n intervl [, b] there exists γfine, tgged prtition. This is the content of Cousin s lemm [8]. Lemm Cousin s lemm. If γ is guge on the intervl [, b], then there exists tgged prtition of [, b] tht is γfine. Before we cn prove Cousin s lemm, we need nother little lemm. Lemm 2.2. Suppose γ is guge nd I 1 nd I 2 re intervls which hve t most one point in common. If I 1 nd I 2 re γfine then the union I : I 1 I 2 is lso γfine. Proof. If I 1 nd I 2 re γfine, then there exist tgged prtitions P 1 : {(t i, [x i 1, x i ] : 1 i n} nd P 2 : {t j, [x j 1, x j ] : 1 j m} of I 1 respectively I 2 which re γfine. So 0 x i x i 1 γ(t i ) nd 0 x j x j 1 γ(t j ) for ll i 1, 2,..., n nd j 1, 2,..., m. If we now tke the union of our tgged prtitions P : P 1 P 2, then the length of ech subintervl is still smller thn the vlue of γ in the corresponding tg, so P is γfine. It is cler tht P : P 1 P 2 is tgged prtition of I : I 1 I 2. So for I : I 1 I 2 there exists tgged prtition tht is γfine nd thus I is γfine. We re now going to use this little lemm to prove Cousin s lemm. 6
7 Proof. Lemm Cousin s Lemm. Let S be the set of points x (, b] such tht there exists γfine tgged prtition of [, x]. We re going to show tht there exists γfine tgged prtition of [, b] by showing tht the supremum of set S exists nd tht it is equl to b. Let x be rel number tht stisfies < x < + γ() nd x < b. Now consider {[, x]} to be prtition of [, x] which consists of one subintervl. If we choose to be tg, then we get tgged prtition of [, x]: {(, [, x])}. Becuse x < + γ() γ(), we know tht the tgged prtition {(, [, x])} is γfine. Thus we know tht x S nd therefore S is nonempty. The xiom of completeness sttes tht every nonempty set of rel numbers tht is bounded bove hs supremum. Since S is bounded bove by b, we know tht S hs supremum. Let β be the supremum of S. Since β is the supremum of S, we hve tht β b nd so the guge γ is defined t β. Since γ(β) > 0 nd since β is the supremum of S, there exists point y S such tht β γ(β) < y < β. Let P 1 be γfine tgged prtition of [, y]; such prtition existst becuse y S. Now we consider {[y, β]} to be prtition of [y, β] which gin consists of one subintervl. We choose β to be tg nd so we get the tgged prtition {(β, [y, β])}. Becuse β y < β (β γ(β)) γ(β) we know tht {(β, [y, β])} is γfine. Now we tke the union: P1 P 1 {(β, [y, β])} nd by lemm 2.1 we know tht this union P 1 is tgged prtition of [, β] tht is γfine. Thus we cn conclude tht β S. Now we need to demonstrte tht β b. To do this, we ssume by wy of contrdiction tht β < b. Since β < b there exists point z (β, b) such tht z < β + γ(β). Let P 2 be γfine tgged prtition of [, β]; such prtition exists becuse β S. We consider {[β, z]} to be prtition of [β, z] nd by tking β s tg we get the tgged prtition {(β, [β, z])}. Becuse z β < β + γ(β) β γ(β) we know tht {(β, [β, z])} is γfine tgged prtition of [β, z]. Now we tke the union P 2 P 2 {(β, [β, z])} nd by pplying lemm 2.1 gin, we know tht this P 2 is γfine tggged prtition of [, z]. Therefore we conclude tht z S. But this is contrdiction, becuse β is the supremum of S. So our ssumption cn not be true nd thus β b. Finlly, since b β is n element of S, we cn conclude tht there exists γfine prtition of [, b]. With the definition of γfine prtition, we cn define the HenstockKurzweil integrl [2]. As sid before, its definition differs only slightly from the definition of the Riemnn integrl. Definition 2.7. For function f : [, b] R the number B f is clled the HenstockKurzweil integrl of f if for ll ɛ > 0 there exists guge γ such tht if P : {(t i, [x i 1, x i ]) : 1 i n} is tgged prtition of [, b] tht is γfine, then S(f, P ) B ɛ. Insted of setting the length of ech subintervl [x i 1, x i ] smller thn constnt δ s in the definition of the Riemnn integrl, we use guge γ to compre the lengths with. This little difference will hve mny benefits. In section 4 we will see some exmples of HenstockKurzweil integrble functions. 7
8 3 Properties of the HenstockKurzweil integrl In this section we will look t some properties of the HenstockKurzweil integrl. We will begin with compring it with the Riemnn integrl. Next thing we will see is tht the HenstockKurzweil integrl is unique. Then some linerity properties will follow. After tht we re going to look t the Cuchy criterion for HenstockKurzweil integrbility nd we re going to use this criterion to prove theorem considering the integrbility on subintervls of [, b]. At lst we re looking t the HenstockKurzweil integrls of functions tht re zero, of two functions tht re equl to ech other nd finlly of function tht is less or equl thn nother HenstockKurzweil integrble function. We sw in the previous section tht the definitions of the Riemnn integrl nd the HenstockKurzweil integrl re very similr. Becuse the definitions re so similr we could wonder if Riemnn integrble function is lso HenstockKurzweil integrble nd the other wy round. It turns out tht indeed every Riemnn integrble function is HenstockKurzweil integrble. The ide behind proving this sttement is to choose our guge γ to be the constnt δ from the definition of the Riemnn integrl. This sttement is formlly presented in the next theorem. Theorem 3.1. If f : [, b] R is Riemnn integrble on [, b] with f A, then f is HenstockKurzweil integrble on [, b] with f A. Proof. Let ɛ > 0. Since f is Riemnn integrble, there exists constnt δ ɛ > 0 such tht if P : {(t i, [x i 1, x i ]) : 1 i n} is tgged prtition of [, b] such tht x i x i 1 δ ɛ for ll i 1, 2,, n then S(f, P ) A ɛ. Now we choose our guge to be γ(x) δ ɛ. Then we know tht if P is tgged prtition of [, b] tht is γfine then S(f, P ) A ɛ. Becuse our ɛ ws rbitrry, we know this holds for ll ɛ > 0 nd thus we cn conclude tht f is HenstockKurzweil integrble with f A. The converse of this sttement is flse. Not every HenstockKurzweil integrble function is Riemnn integrble. An exmple of function tht is not Riemnn integrble, but is HenstockKurzweil integrble is Dirichlet s function. We will discuss this function in section 4.2. The following theorem sttes tht if function f is HenstockKurzweil integrble, then its integrl is unique [6]. Theorem Uniqueness of HenstockKurzweil integrl. If f : [, b] R is HenstockKurzweil integrble on the intervl [, b], then the HenstockKurzweil integrl of f on [, b] is unique. 8
9 Proof. We ssume tht there re two numbers B 1, B 2 R tht re both the Henstock Kurzweil integrl of f on [, b]. Let ɛ > 0. Since B 1 is HenstockKurzweil integrble, there exists guge γ 1 on [, b] such tht if P 1 is γ 1 fine prtition of [, b] then S(f, P 1 ) B 1 ɛ 2. Similrly, there exists guge γ 2 on [, b] such tht if P 2 is γ 2 fine prtition of [, b] then (S(f, P 2 ) B 2 ɛ 2. We now define new guge: γ : min{γ 1, γ 2 }. Suppose now tht P is tgged prtition of [, b] tht is γfine. Becuse of the construction of γ, we know tht this P is lso γ 1  nd γ 2 fine. Thus we hve: B 1 B 2 B 1 S(f, P ) + S(f, P ) B 2 S(f, P ) B 1 + S(f, P ) B 2 ɛ 2 + ɛ 2 ɛ. Becuse our ɛ ws rbitrry, we conclude tht B 1 B 2 nd thus the HenstockKurzweil integrl of f on [, b] is unique. Now we will look t some linerity properties of the HenstockKurzweil integrl. Theorem Linerity properties. Let f nd g be HenstockKurzweil integrble on [, b], then 1. kf is HenstockKurzweil integrble on [, b] for ech k R with kf k f; 2. f + g is HenstockKurzweil integrble on [, b] with (f + g) f + g. Proof. 1. Let ɛ > 0. By ssumption f is HenstockKurzweil integrble. So for ɛ there k exists guge γ such tht if P is tgged prtition of [, b] tht is γfine, then S(f, P ) f ɛ k. Now we hve tht S(kf, P ) k f kf(t i )(x i x i 1 ) k k S(f, P ) f ɛ k k ɛ. Becuse our ɛ ws rbitrry, this holds for ll ɛ > 0 nd therefore kf is Henstock Kurzweil integrble on [, b] with kf k f. 9 f
10 2. Let ɛ > 0. By ssumption f nd g re HenstockKurzweil integrble. So for ɛ 2 there exist guges γ f nd γ g such tht whenever P f nd P g re tgged prtitions of [, b] tht re γ f fine respectively γ g fine, then nd S(f, P f) S(g, P g ) f ɛ 2 g ɛ 2. Now we define γ : min{γ f, γ g }. Furthermore we hve: S(f + g, P ) [(f + g)(t i )(x i x i 1 )] [f(t i )(x i x i 1 ) + g(t i )(x i x i 1 )] [f(t i )(x i x i 1 )] + S(f, P ) + S(g, P ). [g(t i )(x i x i 1 )] If we now hve prtition P tht is γfine, then becuse of the construction of γ this P is lso γ f  nd γ g fine. Therefore we know tht S(f + g, P ) ( f + g) S(f, P ) ɛ 2 + ɛ 2 ɛ. f + S(g, P ) Since our ɛ ws rbitrry, this holds for ll ɛ > 0 nd therefore f + g is Henstock Kurzweil integrble on [, b] with (f + g) f + g. g The next thing we re going to look t is the Cuchy criterion for HenstockKurzweil integrls [4]. This theorem is nice one, becuse with this theorem one cn prove tht certin function is HenstockKurzweil integrble without knowing the vlue of the integrl. It will be used in some proofs lter on. Theorem Cuchy criterion. A function f : [, b] R is HenstockKurzweil integrble on [, b] if nd only if for every ɛ > 0 there exists guge γ on [, b] such tht if P 1 nd P 2 re tgged prtitions of [, b] tht re γfine, then S(f, P 1 ) S(f, P 2 ) ɛ. 10
11 Proof. First we ssume tht f is HenstockKurzweil integrble on [, b]. Let ɛ > 0. We know tht for ɛ, there exists guge γ on [, b] such tht if P 2 1 nd P 2 re tgged prtitions of [, b] tht re γfine then, S(f, P 1) f ɛ 2 nd S(f, P 2 ) Now it follows tht S(f, P 1 ) S(f, P 2 ) S(f, P 1) ɛ 2 + ɛ 2 ɛ. f ɛ 2. b f + f S(f, P 2 ) Becuse our ɛ ws rbitrry, we know tht this holds for ll ɛ > 0. Conversely, we ssume tht for every ɛ > 0 there exists guge γ on [, b] such tht if P 1 nd P 2 re tgged prtitions of [, b] tht re γfine, then S(f, P 1 ) S(f, P 2 ) ɛ. For ech n N, choose guge γ n such tht S(f, P 1 ) S(f, P 2 ) 1 n whenever P 1 nd P 2 re γ n fine. We my ssume tht the sequence {γ n } is nonincresing. For ech n, let P n be tgged prtition of [, b] tht is γ n fine. If m > n N, then γ N γ n γ m. Now P n is γ n fine nd thus lso γ N fine. The sme holds for P m : P m is γ m fine nd thus γ N fine. Now we hve tht S(f, P n ) S(f, P m ) 1 N for m > n N. So the sequence {S(f, P n )} is Cuchy sequence. Now let A be the limit of this sequence nd let ɛ > 0. Choose n integer N N such tht 1 N ɛ 2 nd S(f, P n ) A ɛ 2 for ll n N. Now let P be prtition of [, b] tht is γ N fine, then S(f, P ) A S(f, P ) S(f, P N ) + S(f, P N ) A 1 N + ɛ 2 ɛ 2 + ɛ 2 ɛ. Becuse our ɛ ws rbitrry, we conclude tht f is HenstockKurzweil integrble on [, b]. 11
12 If we hve function f tht is HenstockKurzweil integrble on n intervl [, b], then intuitively we suppose tht it is lso HenstockKurzweil integrble on subintervls of [, b]. We lso suppose by intuition tht if function is HenstockKurzweil integrble on [, c] nd on [c, b], then it is lso HenstockKurzweil integrble on [, b]. It turns out tht these two sttements re indeed true. This is the content of the following theorem. We re going to use the previous theorem, the Cuchy criterion, to prove this. [4]. Theorem 3.5. Let f : [, b] R nd c (, b). 1. If f is HenstockKurzweil integrble on [, c] nd on [c, b], then f is HenstockKurzweil integrble on [, b]. 2. If f is HenstockKurzweil integrble on [, b], then f is HenstockKurzweil integrble on every subintervl [α, β] [, b]. Proof. 1. Let ɛ > 0. Becuse f is HenstockKurzweil integrble on [, c] we know tht for ɛ there exists guge γ 2 such tht if P 1 nd P 2 re tgged prtitions of [, c] tht re γ fine, then S(f, P 1 ) S(f, P 2 ) ɛ 2. We lso pply the Cuchy criterion to the intervl [c, b]. We know tht for ɛ there 2 exists guge γ b such tht S(f, P 1 ) S(f, P 2 ) ɛ 2 whenever P 1 nd P 2 re tgged prtitions of [c, b] tht re γ b fine. Now we define new guge: γ : mx{γ, γ b }. If we now hve prtition tht is γ fine or γ b fine, then it is lso γfine. Now we tke P : P 1 P 1 nd P b : P 2 P 2 ; these P nd P b re tgged prtitions of [, b] tht re γfine. Now we hve: S(f, P ) S(f, P b ) S(f, P 1 ) + S(f, P 1 ) [S(f, P 2 ) + S(f, P 2 )] S(f, P 1 ) S(f, P 2 ) + S(f, P 1 ) + S(f, P 2 ) ɛ 2 + ɛ 2 ɛ. Becuse our ɛ ws rbitrry, this holds for ll ɛ > 0 nd therefore we conclude by the Cuchy criterion tht f is HenstockKurzweil integrble on [, b] 2. Let ɛ > 0. Choose guge γ such tht S(f, P 1 ) S(f, P 2 ) ɛ whenever P 1 nd P 2 re γfine. We cn do this becuse of the Cuchy criterion. Now fix prtitions P α of [, α] nd P β of [β, b] tht re γfine. Let P 1 nd P 2 be tgged 12
13 prtitions of [α, β] tht re γfine. Such prtitions exist becuse of Cousin s lemm (lemm 2.1). Now define P 1 : P α P 1 P β nd P 2 : P α P 2 P β. P 1 nd P 2 re now tgged prtitions of [, b] tht re γfine. So we get: S(f, P 1 ) S(f, P 2 ) S(f, P α ) + S(f, P 1 ) + S(f, P β ) [S(f, P α ) + S(f, P 2 ) + S(f, P β )] S(f, P 1 ) S(f, P 2 ) ɛ. Becuse our ɛ ws rbitrry, we hve tht for every ɛ > 0 there exists guge γ on [α, β] such tht S(f, P 1 ) S(f, P 2 ) ɛ when P 1 nd P 2 re γfine tgged prtitions of [α, β]. By using the Cuchy criterion gin, we conclude tht f is Henstock Kurzweil integrble on [α, β]. We re now going to look t functions tht re zero lmost everywhere on n intervl [, b]. It turns out tht such functions re HenstockKurzweil integrble nd tht their integrls re, s expected, equl to zero [4]. Theorem 3.6. Let f : [, b] R. If f 0 except on countble number of points on [, b], then f is HenstockKurzweil integrble on [, b] with f 0. Proof. Let ɛ > 0. First we define the set where f(x) 0: A : { n : n Z >0 } {x [, b] : f(x) 0}. Now we re going to define n pproprite guge on [, b]: { 1 if x [, b] nd x A γ(x) :. if x A ɛ2 n f( n) Suppose tht P {(t i, [x i 1, x i ]) : 1 i n} is tgged prtition of [, b] tht is γfine. So x i x i 1 γ(t i ) for ll 1 i n. Now let π be the set of ll indices i such tht the tgs t i A. Choose n i such tht t i ni. Let σ be the set of indices i such tht t i A. We know tht if we hve such tg, then f(t i ) 0 nd thus f(t i )(x i x i 1 ) 0. So tgs 13
14 tht re not in A do not contribute to S(f, P ). We hve S(f, P ) f(t i )(x i x i 1 ) f(t i )(x i x i 1 ) + f(t i )(x i x i 1 ) i π i π f( ni ) ɛ2 n i f( ni ) i σ i π ɛ2 n i ɛ 2 n i ɛ nd becuse our ɛ ws rbitrry, this holds for ll ɛ > 0. Thus we cn conclude tht f is HenstockKurzweil integrble on [, b] with f 0. With the previous theorem we cn prove tht if we hve two functions tht re the sme lmost everywhere on n intervl [, b] nd if one of them is HenstockKurzweil integrble on tht intervl, then the other is lso HenstockKurzweil integrble nd their integrls re the sme. This is the content of the next theorem [4]. Corollry 3.1. Let f : [, b] R be HenstockKurzweil integrble on [, b] nd let g : [, b] R. If f g except on countble number of points on [, b], then g is Henstock Kurzweil integrble on [, b] with g f. Proof. We define new function: h : g f on [, b]. We hve h 0 except on countble number of points on [, b]. So from theorem 3.6 we conclude tht h is HenstockKurzweil integrble on [, b] with h 0. Since f nd h re both HenstockKurzweil integrble on [, b], from theorem 3.3 we know tht g f + h is HenstockKurzweil integrble on [, b] with g f + h f. The lst property we re going to look t considers two HenstockKurzweil integrble functions on n intervl [, b]. If one of them is less or equl thn the other lmost everywhere on the intervl [, b], then the sttement is tht the integrl of the smllest function is less or equl thn the biggest [8]. We will lso use this theorem when we re proving convergence theorems in section 6. Corollry 3.2. If f nd g re HenstockKurzweil integrble on [, b] nd f(x) g(x) except on countble number of points on [, b], then f g. 14
15 Proof. Define two new functions: { f(x) x such tht f(x) g(x) f(x) 0 x such tht f(x) > g(x) nd g(x) { g(x) x such tht f(x) g(x) 0 x such tht f(x) > g(x). Becuse f(x) f(x) nd g(x) g(x) lmost everywhere on [, b], we know from corollry 3.1 tht f f nd g g. Becuse g f we know g f 0 nd thus it follows tht ( g f) 0. Now we hve g f g tht g f. f ( g f) 0, so it follows 4 Exmples Now we hve seen l those nice properties of the HenstockKurzweil integrl in the previous section, we re going to look t some explicit exmples of HenstockKurzweil integrble functions. For ll the exmples tht we tret, we will give proof tht they re Henstock Kurzweil integrble mostly by defining n pproprite guge. We will strt esy: first we re going to look t liner functions. Secondly we re going to prove tht Dirichlet s function is not Riemnn integrble, but tht it is HenstockKurzweil integrble s sid before. Then we will modify Dirichlet s function nd see wht hppens with the integrbility. At lst we re going to investigte Thome s function. 4.1 Liner functions We begin here with investigting the esiest liner function. We wnt to find out if the function f(x) c with c R is HenstockKurzweil integrble on the intervl [, b]. We know tht it is Riemnn integrble with f c(b ). So we know by theorem 3.1 tht it is lso HenstockKurzweil integrble with f c(b ). In the following proposition we re going to prove this by looking t the definition. Proposition 4.1. The function f(x) c with c R is HenstockKurzweil integrble on the intervl [, b] with f c(b ). Proof. Let ɛ > 0. We know tht f(x) c, x [, b]. So for ny tgged prtition P of [, b], we hve S(f, P ) f(t i )(x i x i 1 ) c(x i x i 1 ). 15
16 This is telescoping sum, so n c(x i x i 1 ) c(b ). Therefore we hve tht S(f, P ) c(b ) c(b ) c(b ) 0 < ɛ for ny tgged prtition P. So this certinly holds if P is tgged prtition tht is γfine for ny guge γ. Since our ɛ is rbitrry, it holds for ll ɛ > 0 nd thus f(x) c is HenstockKurzweil integrble on [, b] with f c(b ). We will now look t the function f(x) x. We know tht this function is Riemnn integrble nd therefore HenstockKurzweil integrble. Here we re going to prove it by finding n pproprite guge. Proposition 4.2. The function f(x) x is HenstockKurzweil integrble on the intervl [, b] with f 1 2 (b2 2 ). 2ɛ Proof. Let ɛ > 0. We tke γ(x) : to be our guge, where n is the number of tgs nd n thus the number of subintervls in the tgged prtition. Suppose P is tgged prtition of 2ɛ [, b] tht is γfine. So x i x i 1 γ(t i ) for ll 1 i n. We cn write 1 n 2 (b2 2 ) s telescoping sum: 1 2 (b2 2 ) 1 2 [x 2 i x 2 i 1]. If we look t the distnce from tg t i to the middle of subintervl [x i 1, x i ], we know it is lwys less or equl to hlf the length of the subintervl. So: t i x i + x i x i x i 1. 16
17 Now we hve: S(f, P ) 1 2 (b2 2 ) [t i (x i x i 1 )] 1 [x 2 i x 2 2 i 1] [t i (x i x i 1 ) 1 2 (x i x i 1 )(x i + x i 1 )] [(x i x i 1 )(t i x i + x i 1 )] 2 x i x i 1 t i x i + x i x i x i ɛ 2 n ɛ n ɛ. Becuse our ɛ ws rbitrry, we know tht this holds for ll ɛ > 0 nd thus f(x) x is HenstockKurzweil integrble on [, b] with f 1 2 (b2 2 ). Becuse we know now tht the functions f(x) c with c R nd f(x) x re Henstock Kurzweil integrble on n rbitrry intervl [, b], we know tht ll liner functions re HenstockKurzweil integrble on n intervl [, b]. This follows directly from theorem 3.3 nd propositions 4.1 nd 4.2 nd the fct tht liner function is lwys of the form f(x) dx + c with c, d R. 4.2 Dirichlet s function Dirichlet s function h : [, b] R is the chrcteristic function of the rtionl numbers. It is function which is discontinuous everywhere on the intervl [, b]. Dirichlet s function is given by: { 1 if x Q h(x) 0 if x Q This function is bounded, but not Riemnn integrble. We will prove tht now. Proposition 4.3. Dirchlet s function h(x) is not Riemnn integrble on [, b]. 17
18 Proof. Recll from section 2 tht the number A is the Riemnn integrl of h(x) on [, b] if for ll ɛ > 0 there exists δ ɛ > 0 such tht if P is ny tgged prtition of [, b] stisfying x i x i 1 δ ɛ for ll i 1, 2,, n then S(h, P ) A ɛ. Suppose tht P {(t i, [x i 1, x i ]) : 1 i n} is tgged prtition of [, b] tht stisfies x i x i 1 δ ɛ for ll i 1, 2,, n. Becuse the rtionls nd irrtionls re both dense in R, ech subintervl contins point tht is rtionl nd point tht is irrtionl. We distinguish between two cses: the first cse is when A 0 nd the second is when A 0. First, ssume A 0. We choose ll our tgs to be irrtionl, so t i Q. If we do this we get: S(h, P ) A h(t i )(x i x i 1 ) A 0(x i x i 1 ) A A. So we cn not lwys mke the expression S(h, P ) A smller thn ɛ for ny tgged prtition P tht stisfies x i x i 1 δ ɛ. Therefore we cn conclude tht h(x) is not Riemnn integrble on [, b]. Now we ssume tht A 0. Now we choose ll our tgs to be rtionl, so t i Q. Now we get: S(h, P ) h(t i )(x i x i 1 ) 1(x i x i 1 ) b. So gin, we cn not lwys mke the expression S(h, P ) A smller thn ɛ for ny tgged prtition P tht stisfies x i x i 1 δ ɛ. Therefore we conclude tht h(x) is not Riemnn integrble on [, b]. We hve now shown tht Dirichlet s function is not Riemnn integrble on n intervl [, b]. Dirichlet s function is ctully HenstockKurzweil integrble. We will prove tht sttement now [2]. Proposition 4.4. Dirichlet s function h : [, b] R is HenstockKurzweil integrble with h 0. Proof. Let ɛ > 0. First we enumerte the rtionl numbers in [, b] s {r 1, r 2, }. Now we define our guge γ ɛ : { ɛ if x r 2 γ ɛ (x) : i i 1 if x {r 1, r 2, }. 18
19 Note tht our γ ɛ is not constnt vlue. Suppose P {(t i, [x i 1, x i ]) : 1 i n} is tgged prtition of [, b] tht is γ ɛ fine. If t i {r 1, r 2, } is tg, then h(t i ) 1 nd x i x i 1 γ ɛ (t i ) ɛ where [x 2 i i 1, x i ] is the subintervl corresponding to the tg t i. Thus we hve: h(t i )(x i x i 1 ) ɛ for tgs t 2 i i {r 1, r 2, }. If t i {r 1, r 2, } is tg, then h(t i ) 0 nd x i x i 1 γ ɛ (t i ) 1 where [x i 1, x i ] is the subintervl corresponding to the tg t i. Therefore: h(t i )(x i x i 1 ) 0. So the subintervls with tgs t i {r 1, r 2, } do not contribute to S(h, P ). Let π be the set of indices i such tht t i {r 1, r 2, } nd let σ be the set of indices i such tht t i {r 1, r 2, }. We conclude tht: S(h, P ) h(t i )(x i x i 1 ) h(t i )(x i x i 1 ) + h(t i )(x i x i 1 ) i π i σ ɛ 2 i 1 ɛ 2 ɛ. i Since our ɛ ws rbitrry, we know tht this holds for ll ɛ nd thus Dirichlet s function h is HenstockKurzweil integrble with h 0. In section 3 we sw tht every Riemnn integrble function is HenstockKurzweil integrble. Here we hve shown tht the converse sttement is flse by giving n exmple of function tht is HenstockKurzweil integrble, but not Riemnn integrble. 4.3 Modified Dirichlet function We re now going to look wht hppens if we djust Dirichlet s function little bit. We define new function g by { x if x Q g(x) : 0 if x Q. We re now going to investigte the Riemnn nd the HenstockKurzweil integrbility of this function. It turns out tht this little modifiction does not ffect the integrbility: the function g is still not Riemnn integrble but it is HenstockKurzweil integrble. To prove these two sttements, we cn follow the proofs of Dirichlet s function nd we only need to mke few djustments. Proposition 4.5. The modified Dirichlet function g(x) is not Riemnn integrble on [, b]. 19
20 Proof. Recll gin tht the number A is the Riemnn integrl of g(x) on [, b] if for ll ɛ > 0 there exists δ ɛ > 0 such tht if P is ny tgged prtition of [, b] stisfying 0 x i x i 1 δ ɛ for ll i 1, 2,, n then S(g, P ) A ɛ. Suppose tht P {(t i, [x i 1, x i ]) : 1 i n} is tgged prtition of [, b] tht stisfies 0 x i x i 1 δ ɛ for ll i 1, 2,, n. Becuse the rtionls nd irrtionls re both dense in R, ech subintervl contins point tht is rtionl nd point tht is irrtionl. Agin, we distinguish between two cses: A 0 nd A 0. The proof of the first cse, when A 0, is identicl to the proof of tht of Dirichlet s function, so we will skip tht prt here. If A 0, we choose our x i Q. Next we choose our tgs to be the midpoints of the subintervls: t i x i 1+x i Q. It could be tht x 2 0 nd b x n re no rtionl numbers. Now we get S(g, P ) g(t i )(x i x i 1 ) g(t n 1 1)(x 1 x 0 ) + g(t i )(x i x i 1 ) + g(t n )(x n x n 1 ) i2 g(t n 1 x i 1 + x i 1)(x 1 ) + g(t n )(b x n 1 ) + (x i x i 1 ) 2 i2 g(t n 1 1 1)(x 1 ) + g(t n )(b x n 1 ) + 2 (x2 i x 2 i 1) i2 g(t 1)(x 1 ) + g(t n )(b x n 1 ) (x2 n 1 x 2 1) We cn not mke the term 1 2 (x2 n 1 x 2 1) rbitrrily smll by refining our prtition. So we cn not mke the whole expression S(g, P ) A smller thn ɛ for ny tgged prtition P tht stisfies x i x i 1 δ ɛ. Therefore we cn conclude tht g(x) is not Riemnn integrble on [, b]. Proposition 4.6. The modified Dirichlet function g : [, b] R is HenstockKurzweil integrble with g 0. Proof. Let ɛ > 0. We strt gin with enumerting the rtionls in [, b] s {r 1, r 2, }. Now we define our guge γ ɛ : { ɛ if x r x2 γ ɛ (x) : i i 1 if x {r 1, r 2, }. Suppose P {(t i, [x i 1, x i ]) : 1 i n} is tgged prtition of [, b] tht is γ ɛ fine. If t i {r 1, r 2, } is tg, then g(t i ) t i nd x i x i 1 γ ɛ (t i ) ɛ t i 2 where i [xi 1, x i ] 20
21 is the subintervl corresponding to the tg t i. If t i {r 1, r 2, } is tg, then g(t i ) 0 nd x i x i 1 γ ɛ (t i ) 1 where [x i 1, x i ] is the subintervl corresponding to the tg t i. Therefore: g(t i )(x i x i 1 ) 0. So the subintervls with tgs t i {r 1, r 2, } do not contribute to S(h, P ). Let π be the set of integers i such tht t i {r 1, r 2, } nd let σ be the set of integers i such tht t i {r 1, r 2, }. We cn conclude tht: S(g, P ) g(t i )(x i x i 1 ) + g(t i )(x i x i 1 ) i π i σ t i ɛ t i π i 2 i ɛ 2 i ɛ. Since our ɛ ws rbitrry, we know this holds for ll ɛ > 0 nd thus is the modified Dirichlet function g HenstockKurzweil integrble on [, b] with g Thome s function The next function we will look t is Thome s function. It is given by: 1 if x 0 1 T (x) if x m Q \ {0} is in lowest terms with n > 0. n n 0 if x Q This function is specil one, becuse it hs the property tht it is continuous t every irrtionl point nd discontinuous t every rtionl point. To show this we look t sequences [1]. Proposition 4.7. Thome s function T is continuous t every irrtionl point nd discontinuous t every rtionl point. Proof. We will first look t the rtionl points; so we look t c Q. We know tht T (c) > 0. We cn find sequence (y n ) Q which converges to c, becuse the irrtionls re dense in R. Becuse (y n ) Q, we hve T (y n ) 0. So while (y n ) c, T (y n ) 0 T (c). Becuse c ws n rbitrry rtionl number, we conclude tht T (x) is not continuous t ll rtionl points. Now we re going to look t the irrtionl points; so we look t d Q. Becuse the rtionls re lso dense in R we cn find sequence (z n ) Q which converges to d. The closer rtionl number is to fixed irrtionl number, the lrger its denomintor must necessrily be. If the denomintor is lrge, then its reciprocl is close to zero. So if (z n ) d it follows tht T (z n ) 0 T (d). Becuse d ws n rbitrry irrtionl number, we cn conclude tht T (x) is continuous t ll irrtionl points. 21
22 Now we wnt to investigte if Thome s function is HenstockKurzweil integrble. It turns out tht it is. We will prove tht sttement now. We re doing this gin by finding n pproprite guge. This guge will look like the guges we used for Dirichlet s function nd the modified version of it: we distinguish for the different cses just like in Thome s function itself. Proposition 4.8. Thome s function T is HenstockKurzweil integrble on the intervl [, b] with T 0. Proof. Let ɛ > 0. Suppose tht P {(t i, [x i 1, x i ]) : 1 i n} is tgged prtition of [, b]. Let π be the set of indices i such tht the tgs t i re rtionl; so π : {i : t i m i n i Q \ {0} is in lowest terms, with n i > 0}. Let ρ be the set of indices i such tht the tgs t i re irrtionl; so ρ : {i : t i Q}. Let σ be the set of indices i such tht the tgs t i re zero; so σ : {i : t i 0}. The set σ contins t most two elements, becuse there cn be t most be two intervls tht hve the sme tg. This tg should be the right endpoint of one of the subintervls nd the left endpoint of the other subintervl. Now we define our guge γ ɛ s follows: ɛ if x 0 4 nɛ γ ɛ (x) if x m Q \ {0} is in lowest terms with n > 0 2q n 1 if x Q where q is the number of elements of the set π. Now suppose our tgged prtition P is γ ɛ fine. Then: S(T, P ) T (t i )(x i x i 1 ) T (t i )(x i x i 1 ) + T (t i )(x i x i 1 ) + T (t i )(x i x i 1 ) i π i ρ i σ 1 (x i x i 1 ) n i π i + (x i x i 1 ) i σ 1 n i ɛ n i π i 2q + ɛ 4 i σ ɛ 2q + ɛ 2 i π ɛ + ɛ ɛ 2 2 Becuse our ɛ ws rbitrry, this holds for ll ɛ > 0 nd therefore we conclude tht Thome s function T (x) is HenstockKurzweil integrble on [, b] with T 0. 22
23 5 The Fundmentl Theorem of Clculus The Fundmentl Theorem of Clculus sttes tht if F is differentible on [, b] nd F f, then f F (b) F (). In the cse of Riemnn integrtion it is necessry to dd the dditionl condition tht f must be integrble. This condition is necessry, becuse not every derivtive turns out to be Riemnn integrble. This is one of the shortcomings of the Riemnn integrl. We re going to investigte if this problem lso occurs when we re working with the HenstockKurzweil integrl. To show tht not every derivtive is Riemnn integrble, we will now present n exmple. We consider the following function: { x 2 sin( 1 ) if x 0 x F (x) : 2 0 if x 0. We clim tht this function is continuous nd differentible everywhere. For x 0 this is esy to see: F is product of composition of continuous, differentible functions. For x 0 we hve tht lim x 0 F (x) 0 nd F (0) 0, so F (x) is continuous t x 0 nd F (x) F (0) F (x) becuse lim x 0 lim x 0 lim x 0 x sin( 1 ) 0, F (x) is lso differentible x 2 x 0 t x 0. By using the bsic rules of differentition we get: F (x) : x { 2x sin( 1 x 2 ) 2 x cos( 1 x 2 ) if x 0 0 if x 0. Lebesgue s theorem sttes tht function is Riemnn integrble if nd only if it is bounded nd continuous lmost everywhere [8]. We re going to show tht F (x) is not Riemnn integrble on n intervl [0, b] (where b R + ) by showing tht it is unbounded [8]. Proposition 5.1. The function F (x) s given bove is unbounded on the intervl [0, b]. Proof. Let n be positive integer nd let x n2. The Archimeden property sttes tht 2π for every rel number x, there exists nturl number N such tht N > x. Our x n2 is 2π rel number, so there exists N 1 N such tht N 1 > x. This is equivlent to the sttement tht n < 2πN 1. Now tke y 1. This y is gin rel number, so there exists N 2πb 2 2 N 1 such tht N 2 > y nd this is equivlent to 2πN2 < b. Now we tke N mx{n 1, N 2 }. Then n < 2πN nd 0 < 1 1 2πN < b. So we hve tht 2πN [0, b] nd we re going to evlute F (x) t this vlue: F 1 ( ) 2πN 2 sin(2πn) 2 2πN cos(2πn) 2πN 2 2πN > n. Becuse our n ws n rbitrry integer, we conclude tht F (x) is unbounded on the intervl [0, b]. 23
24 So we hve found function tht is derivtive nd tht is unbounded. Therefore this derivtive is not Riemnn integrble. So the condition tht f F is Riemnn integrble is necessry in the Fundmentl Theorem. Now we re going to show tht we do not need this extr condition when we re considering HenstockKurzweil integrtion, becuse it turns out tht every derivtive is Henstock Kurzweil integrble. We will see this while we re proving the Fundmentl Theorem considering HenstockKurzweil integrtion [2, 7]. Theorem The Fundmentl Theorem. If F : [, b] R is differentible t every point of [, b] then f F is HenstockKurzweil integrble on [, b] with f F (b) F (). Proof. Let ɛ > 0. Suppose t [, b]. Becuse f(t) F (t) exists, we know from the ɛ definition of differentibility tht for there exists δ b t such tht if 0 < z t δ t for z [, b] then F (z) F (t) f(t) z t ɛ b. Now we let these δ t s form our guge: γ(t) : δ t. Note tht this guge γ is not constnt, becuse δ t is not the sme for ll t; δ t depends on t. Now if z t γ(t) for z, t [, b] then F (z) F (t) f(t)(z t) ɛ z t. b If α t β b nd 0 < β α γ(t) then by the tringle inequlity, 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) ɛ β t + ɛ t α b b ɛ β α. b Let P : {(t i, [x i 1, x i ]) : 1 i n} be tgged prtition of [, b] tht is γfine. We cn write F (b) F () s telescoping sum: F (b) F () n (F (x i) F (x i 1 )). Now we 24
25 hve: S(f, P ) (F (b) F ()) F (b) F () S(f, P ) (F (x i ) F (x i 1 )) f(t i )(x i x i 1 ) (F (x i ) F (x i 1 ) f(t i )(x i x i 1 )) F (x i ) F (x i 1 ) f(t i )(x i x i 1 ) ɛ b x i x i 1 ɛ (b ) ɛ. b Becuse our ɛ ws rbitrry, this holds for ll ɛ > 0 nd thus we conclude tht f F is HenstockKurzweil integrble on [, b] with f F (b) F (). We see tht every derivtive is HenstockKurzweil integrble, becuse we could tke γ(t) : δ t s guge, where δ t stems from the definition of differentibility. So the shortcoming tht not every derivtive is Riemnn integrble does not occur when we re considering HenstockKurzweil integrbility nd the Fundmentl theorem does not need n extr condition. 6 Convergence theorems Another limittion of the Riemnn integrl rises when we consider limits of sequences of functions. To illustrte this we look t Dirichlet s function gin. Recll from subsection 4.2 tht Dirichlet s function is the chrcteristic function of the rtionls nd is given by: { 1 if x Q h(x) 0 if x Q. Now we focus on the intervl [0, 1]. Let {r 1, r 2,...} be n enumertion of the rtionl points in this intervl. Now define h 1 (x) 1 if x r 1 nd h 1 (x) 0 otherwise. Next, define h 2 (x) 1 if x r 1 or x r 2 nd h 2 (x) 0 otherwise. By continuing this, we get: { 1 if x {r 1, r 2,, r n } h n (x) 0 if x {r 1, r 2,, r n }. 25
26 Ech h n hs finite number of discontinuities nd is therefore Riemnn integrble with b h n 0. We lso hve tht h n h pointwise. We sw in subsection 4.2 tht Dirichlet s function is not Riemnn integrble. Thus we hve tht lim n 1 0 h n does not hold, becuse the vlue on the righthnd side does not exist [1]. In this section we re going to look t convergence theorems for the HenstockKurzweil integrl. We wnt to investigte if we cn overcome shortcomings s bove while considering HenstockKurzweil integrtion insted of Riemnn integrtion. Before we begin our investigtion regrding convergence nd HenstockKurzweil integrtion, we define uniformly HenstockKurzweil integrbility [4]. Then we continue with our first convergence theorem which will involve uniformly HenstockKurzweil integrls. The next theorem will be bout uniform convergence nd HenstockKurzweil integrbility nd t lst we re going to look t pointwise convergence. Definition 6.1. Let {f n } be sequence of HenstockKurzweil integrble functions on [, b]. The sequence {f n } is uniformly HenstockKurzweil integrble on [, b] if for ll ɛ > 0 there exist guge γ on [, b] such tht if P is tgged prtition of [, b] tht is γfine then S(f n, P ) ɛ for ll n. f n Wht this definition ctully mens is tht if we cn find one guge γ tht works for ll f n, then we cll the sequence {f n } uniformly HenstockKurzweil integrble. Now we re going to prove our first convergencetheorem [4]. Theorem 6.1. Suppose {f n } is sequence of HenstockKurzweil integrble functions on [, b] nd tht {f n } converges pointwise to f. If {f n } is uniformly HenstockKurzweil integrble on [, b], then f is HenstockKurzweil integrble on [, b] with f lim n f n. Proof. Let ɛ > 0. Since {f n } is uniformly HenstockKurzweil integrble there exists guge γ on [, b] such tht if P is tgged prtition of [, b] tht is γfine, then S(f n, P ) f n ɛ 3 for ll n. Since {f n } converges pointwise on [, b], there exists n integer N such tht S(f n, P ) S(f m, P ) ɛ 3 for ll m, n N. Now it follows tht: b f n f m f n S(f n, P ) + S(fn, P ) S(f m, P ) + S(f m, P ) ɛ 3 + ɛ 3 + ɛ 3 ɛ h f m
27 for ll m, n N. This mens tht { f n} is Cuchy sequence. Let A be the limit of this sequence. Our clim is now tht f A. Let ɛ > 0. Since { f n} is Cuchy sequence, we cn choose n integer N such tht f n A ɛ 3 for ll n N. Since {f n } is uniformly HenstockKurzweil integrble, there exists guge γ such tht if P is tgged prtition of [, b] tht is γfine, then S(f n, P ) f n ɛ 3 for ll n. Now suppose P is tgged prtition of [, b] tht is γfine. Since {f n } converges pointwise to f there exist k N such tht S(f, P ) S(f k, P ) ɛ 3. Thus, it follows tht: S(f, P ) A S(f, P ) S(f k, P ) + S(f k, P ) ɛ 3 + ɛ 3 + ɛ 3 ɛ. f k + f k A Since our ɛ ws rbitrry, we conclude tht f is HenstockKurzweil integrble on [, b] with f A lim n f n. The next convergence theorem we re going to look t is bout uniform convergence in combintion with HenstockKurzweil integrbility. Theorem 6.2. If {f k } is sequence of HenstockKurzweil integrble functions on [, b] tht converges uniformly to f on [, b], then it follows tht f is HenstockKurzweil integrble on [, b] with f lim k f k. Before we cn prove this theorem, we need to prove lemm. Lemm 6.1. If f : [, b] R is HenstockKurzweil integrble nd f(x) M with M R for ll x [, b] then b f M(b ). Proof. f(x) M mens tht M f(x) M. In section 4.1 we sw tht the function f(x) c with c R is HenstockKurzweil integrble on [, b] with f c(b ). So we know tht M nd M re HenstockKurzweil integrble. Applying theorem 3.3, we get M M(b ) 27
28 M M M(b ). Corollry 3.2 sttes tht if f nd g re HenstockKurzweil integrble on [, b] nd f(x) g(x) on [, b], then f g. We pply this corollry to M f(x) nd to f(x) M nd thus we get nd We conclude tht M M(b ) f M M(b ). b f M(b ). f Now we hve proven this lemm, we cn prove theorem 6.2 [3]. Proof. Theorem 6.2. Let ɛ > 0. Becuse {f k } converges uniformly to f, there exists K N such tht ɛ f k f 2(b ) for ll x [, b] whenever k K. Consequently, if h, k K then we hve f k f h f k f + f f h ɛ 2(b ) + ɛ 2(b ) ɛ (b ). By ssumption f k nd f h re HenstockKurzweil integrble nd thus follows from theorem 3.3 tht f k f h is HenstockKurzweil integrble. Now we cn pply lemm 6.1 to f k f h : (f k f h ) ɛ (b ) ɛ. (b ) Becuse of the linerity properties of the HenstockKurzweil integrl, we know tht (f k f h ) f k f h ɛ. Becuse our ɛ ws rbitrry, this mens tht the sequence { f k} is Cuchy sequence. Let A be the limit of this sequence. Our clim is gin tht f A. Let ɛ > 0 nd let P {(t i, [x i 1, x i ]) : 1 i n} be tgged prtition of [, b]. Since { f k} is Cuchy sequence with limit A, there exist n integer N N such tht for k N, we hve b f k A ɛ 3. 28
29 Since {f k } converges uniformly to f, there exists n K N such tht if k K, then f k f ɛ 3(b ) for ll x [, b]. Becuse of this, it follows tht S(f k, P ) S(f, P ) f k (t i )(x i x i 1 ) f(t i )(x i x i 1 ) [(f k (t i ) f(t i ))(x i x i 1 )] (f k (t i ) f(t i ))(x i x i 1 ) ɛ 3(b ) (x i x i 1 ) ɛ 3(b ) (b ) ɛ 3. All f k re HenstockKurzweil integrble. So for ech f k there exists guge γ k on [, b] such tht whenever P is γ k fine we hve S(f k, P ) f k ɛ 3. Now we choose k such tht k K nd k N. Now we combine ll the previous nd we get S(f, P ) A S(f, P ) S(f k, P ) + S(f b k, P ) f k + f k A ɛ 3 + ɛ 3 + ɛ 3 ɛ. Becuse our ɛ ws rbitrry, this holds for ll ɛ > 0 nd we conclude tht f is Henstock Kurzweil integrble with f lim k f k. We sw here tht uniform convergence gurntees the limit function to be Henstock Kurzweil integrble s well. Now we re going to investigte if this lso holds for pointwise convergence. As it turns out, supposing tht ech function in sequence tht converges pointwise is Riemnn integrble is not enough to gurntee tht the limit function is lso Riemnn integrble. We sw counterexmple of this t the beginning of this section. The monotone convergence theorem provides set of conditions tht gurntees tht the limit of pointwise convergent sequence of Henstock integrble functions is lso Henstock integrble. This theorem does not hold for the Riemnn integrl; the sequence of functions 29
30 h n (x) t the beginning of this section turns out to be counterexmple gin. Before we cn prove the monotone convergence theorem, we need to tke look t the SksHenstock lemm [3]. This lemm is bout subprtitions. A subprtition of [, b] is prtition tht does not need to cover the whole intervl [, b]. So the union of the subintervls of subprtition does not hve to be [, b]. Definition 6.2. A subprtition P : {[x j 1, x j ] : 1 j m} of n intervl [, b] is finite collection of closed intervls such tht [x j 1, x j ] [, b] nd where the subintervls [x j 1, x j ] cn only hve the endpoints in common. Definition 6.3. A tgged subprtition P : {(t j, [x j 1, x j ]) : 1 j m} of n intervl [, b] is finite set of order pirs where t j [x j 1, x j ] nd the intervls [x j 1, x j ] form subprtition of [, b]. The numbers t j re gin clled the tgs. The SksHenstock lemm sttes tht if we hve subprtition of the intervl [, b], then we could mke the difference between the Riemnn sum nd the integrl over the subintervls of the subprtition lso smller thn ɛ. It lso sys tht if we first tke bsolute vlues nd then summrize, this term will be smller thn 2ɛ. Lemm SksHenstock lemm. Let f be function tht is HenstockKurzweil integrble on [, b] nd let for ɛ > 0 γ ɛ be guge on [, b] such tht if P {(t i, [x i 1, x i ]) : 1 i n} is tgged prtition of [, b] tht is γ ɛ fine, then S(f, P ) b f ɛ. If P {(t j, [x j 1, x j ]) : 1 j s} {(t j, J j ) : 1 j s} is ny γ ɛ fine subprtition of [, b], then 1. S(f, P ) [ J 1 f + + J s f] s j1 [f(t j)(x j x j 1 ) x j x j 1 f] ɛ nd 2. s f(t j )(x j x j 1 ) x j 2ɛ. j1 x j 1 f Proof. 1. Let K 1,, K m be closed subintervls in [, b] such tht {J j } s j1 {K k } m k1 forms prtition of [, b]. Now let α > 0 be rbitrry. Becuse ech K k [, b], we know tht f is HenstockKurzweil integrble on ech K k by theorem 3.5. So for ech K k, there exists guge γ α,k such tht if Q k is γ α,k fine tgged prtition of K k then S(f, Q k) α m. If we hve tht γ α,k (x) γ ɛ (x) for some k, then we could dd positive constnt to γ ɛ (x) such tht we get new guge tht is greter thn γ α,k (x): γ ɛ γ ɛ + c with c R >0 such tht γ α,k (x) γ ɛ (x) for ll x K k. A prtition tht is γ ɛ fine is then lso γ ɛ fine. So this γ ɛ is lso guge tht stisfies the ssumption in the sttement of this theorem. So we my ssume tht γ α,k (x) γ ɛ (x) for ll x K k. Then we K k f 30
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