Math Real Analysis The Henstock-Kurzweil Integral

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1 Math Real Analysis The Henstock-Kurzweil Integral Steven Kao & Jocelyn Gonzales April 28, Introduction to the Henstock-Kurzweil Integral Although the Rieann integral is the priary integration technique taught to undergraduates, there are several drawbacks to the Rieann integral. I. A lot of functions are not Rieann integrable. Recall that a bounded function is only Rieann integrable if its set of discontinuities has easure zero. For exaple Dirichlet s function: { 1 x Q g(x) = 0 x R\Q is not Rieann integrable since U(f, P) = 1 and L(f, P) = 0. Thus, U(f, P) L(f, P). II. Every derivative is not Rieann integrable. We have seen in class that the function ( ) 1 x 2 sin f(x) = x 2 x 0 0 x = 0 on [ 1, 1] has the derivative f 2 ( ) ( ( )) 1 1 (x) = x cos x 2 + 2x sin x 2 x 0 0 x = 0 (1) (2) however f (x) is unbounded on [ 1, 1], thus it is not Rieann integrable. Using the Rieann integral, the Fundaental Theore of Calculus states if f : [a, b] R is Rieann integrable and F : [a, b] R with F (x) = f(x), x [a, b], then b a f = F (b) F (a). Thus, the Fundaental Theore of Calculus requires that f be Rieann integrable. Since not every derivative is Rieann integrable, the Rieann integral places a constraint on the Fundaental Theore of Calculus. For exaple, f ight have an antiderivative F, but this does not iply that f is Rieann integrable. Thus, the forula fro the Fundaental Theore of Calculus can not be applied. 1

2 The Lebesgue integral was introduced in 1902 by Henry Lebesgue. This integral centers around using the range instead of the doain to integrate functions. Although the Lebesgue integral can integrate a larger class of functions then the Rieann integral, it still has liitations. First, there are still a large nuber of functions which can not be integrated using the Lebesgue integral. Also, the Lebesgue integral does not guarantee that every derivative is integrable. For exaple, equation (1) is not Lebesgue integral. Thus, it also adds additional constraints on the Fundaental Theore of Calculus. The desire to develop an integral that could integrate ore functions and generalize the Fundaental Theore of Calculus otivated atheaticians Ralph Henstock and Jaroslav Kurzweil. In the 1960 s, both Henstock and Kurzweil independently developed the Henstock-Kurzweil (HK) integral. The HK integral can integrate a uch larger class of functions than the Rieann integral and the Lebesgue integral. We will see later that the HK integral can integrate the Dirichlet function. Also, using the HK integral, every derivative is integrable. We will also see that using the generalized Rieann integral. the Fundaental Theore of Calculus can be ade ore general since we can eliinate the assuption that f is Rieann integrable. Figure 1: Classification of Integrable Functions Figure 2: Henstock and Kurzweil 2

3 2 Gauges and δ-fine Partitions We will begin with a few definitions. Tagged Partition: Let P be a partition of the interval [a, b] with P = a = x o < x 1 < x 2 <... < x n = b. A tagged partittion (P, (c k ) n ) is a partition which has selected points c k in each subinterval [x k 1, x k ]. The Rieann su using the tagged partition can be written R(f, P) = f(c k )[x k, x k 1 ]. δ-fine: Let δ > 0. A partition P is δ-fine if every subinterval [x k 1, x k ] satisfies x k x k 1 < δ. The definition for Rieann integrability can be restated as the following: A bounded function f : [a, b] R is Rieann integrable with b f = A if and only if for every ɛ > 0, there exists a δ > 0 such a that, for any tagged partition (P, (c k )) that is δ-fine, it follows that R(f, P) A < ɛ. Gauge: A function δ : [a, b] R is called a gauge on [a, b] if δ(x) > 0 for all x [a, b]. The ajor difference between the HK integral and the Rieann integral is allowing δ to be a function of x rather than a constant. δ(x)-fine: Given a particular gauge δ(x), a tagged partition (P, (c k ) n ) is δ(x)-fine if every subinterval [x k 1, x k ] satisfies x k x k 1 < δ(c k ). If δ(x) is a constant function, then the definition of a δ(x)-fine partition is equivalent to the definition of a δ-fine partition. 3 Exaples of δ(x)-fine tagged partitions Exaple 1 Consider the inverval [0, 1]. Let δ 1 (x) = 1/9. We will find a δ 1 (x)-fine tagged partition on [0, 1]. Since the gauge δ 1 (x) is a constant function. Regardless of the choice of tag, δ 1 (c k ) = 1/9. Thus, any tagged partition (P, (c k ) n ) in which x k x k 1 < 1/9 is a δ 1 (x)-fine tagged partition. Consider the following partition, choosing each tag fro every interval to be any nuber in that inverval: ([ 0, 1 ]) < 1 ([ ([ 1 10, 2 ]) < 1 10, 5 ]) < 1 ([ , 8 ]) < ([ ([ 2 10, 3 ]) < 1 10, 6 ]) < 1 ([ , 9 ]) < ([ ([ 3 10, 4 ]) < 1 10, 7 ]) < 1 ([ ]) , 1 < This is an exaple of a δ 1 (x)-fine tagged partition. 3

4 Exaple 2 Again considering the interval [0, 1], let δ 2 (x) = { 1/4 x = 0 x/3 0 < x 1. Consider the following partition, choosing the first tag fro the first interval to be 0 and each tag fro every other interval to be the right hand end-point of that interval: ([ 0, 1 ]) = 1 ( ) 5 5 < δ 2 0 = 1 ([ 2 4 5, 1 ]) = 1 ( ) < δ 2 = ([ 1 5, 1 ]) = 1 ( ) < δ 2 = 1 ([ , 3 ]) = 1 ( ) < δ 2 = ([ 1 4, 1 ]) = 1 ( ) < δ 2 = 1 ([ , 3 ]) = 3 ( ) < δ 2 = ([ 1 3, 2 ]) = 1 ( ) < δ 2 = 2 ([ ]) , 1 = 1 ( ) 4 < δ 2 1 = 1 3 This is an exaple of a δ 2 (x)-fine tagged partition. 4 Theore 1 Before we state and prove Theore 1, we will first prove the Nested Interval Property, which will be used in the proof of Theore 1. Nested Interval Property For each n N, assue we are given a closed interval I n = [a n, b n ] = {x R : a n x b n }. Assue also that each I n contains I n+1. Then, the resulting nested sequence of closed intervals I 1 I 2 I 3... has a nonepty intersection; that is, n=1 I n. Proof. Let Let A = {a n : n N} be the left-hand endpoints of each interval. Since every interval is contained inside of the previous interval, the b n ters are upper bounds for A. By the existence of a least upper bound, since A has an upper bound, it ust have a least upper bound. Let x = sup A. Because x is an upper bound for A, then a n x, n. Also, since x is a least upper bound of A, x b n, n. Then we have a n x b n, n. Thus x I n for every choice of n N. Thus x n=1 I n and the intersection is not epty. Theore 1: Given a gauge δ(x) on an interval [a, b], there exists a tagged partition (P, (c k ) n ) that is δ(x)-fine. Proof. Let δ(x) be a gauge on [a, b]. An algorith to find a δ(x)-fine tagged partition is as follows. First consider the trivial partition P o = {a = x o < x 1 = b}. Then check to see if c o [a, b] \ I such that b a < δ(c o ). If such c o exists, then choose (P o, (c o )). Thus there exists a tagged partition which is δ(x)- fine. If no c o exists such that b a < δ(c o ), then bisect the interval into two equal halves. Consider the partition P 1 = {a = x o < x 1 < x 2 = b}. Then apply the algorith to each new half. 4

5 We will now prove the algorith ust terinate in a finite nuber of steps. Assue that the algorith does not terinate in a finite nuber of steps. Then we have infinite nested intervals (I n ) where (I n ) 0. Since the algorith has not terinated, this iplies that δ(x) (I n ), x I n. By the Nested Interval Property, we know x o n=1 I n. Then δ(x o ) (I n ), n N. But since (I n ) 0, this iplies that δ(x o ) = 0. We draw a contradiction since by definition δ(x) > 0. Thus this algorith ust terinate after a finite nuber of steps. Thus we can always create a tagged partition of [a, b] that is δ(x)-fine for a given gauge. 5 Introducing the Henstock-Kurzweil (HK) integral We now have the achinery to define the Henstock-Kurzweil integral: Definition: A function f : [a, b] R is Henstock-Kurzweil (HK) integrable if A R st. ε > 0, a gauge δ : [a, b] R st. for each tagged partition (P, (c k ) n ) that is δ(x)-fine, R(f, P) A < ε In this case, we write b a f = A, or HK b f = A (if we need to distinguish between different versions of a integrals) and say that f has an HK integral value of A. Our first order of business is to ake sure that if a function has an HK integral value, it can have only one such value: Theore 2: If a function is HK integrable, its value is unique. Proof. Suppose f : [a, b] R is HK integrable and has values A 1 and A 2. Let ε > 0 ε/2 > 0. By definition of HK integrable, a gauge δ 1 : [a, b] R st. tagged partitions (P 1, (a k ) n1 ) which are δ 1 (x)-fine, R(f, P 1 ) A 1 < ε/2 (3) Siilarly, a gauge δ 2 : [a, b] R st. tagged partitions (P 2, (b k ) n2 ) which are δ 2(x)-fine, R(f, P 2 ) A 2 < ε/2 (4) Define δ : [a, b] R to be δ(x) := in(δ 1 (x), δ 2 (x)). Clearly δ is a gauge since δ(x) = δ 1 (x) or = δ 2 (x), both of which > 0, x [a, b]. Now let (P, (c k ) n ) be an arbitrary tagged partition which is δ(x)-fine. Notice, (P, (c k) n ) is necessarily δ 1 (x)-fine and δ 2 (x)-fine. This is because for any subinterval [x k 1, x k ] with endpoints taken fro P, we have x k x k 1 < δ(c k ) δ 1 (c k ) x k x k 1 < δ(c k ) δ 2 (c k ) and Therefore, 0 A 1 A 2 = ( A 1 R(f, P) ) + ( ) R(f, P) A 2 iddle-an trick A 1 R(f, P) + R(f, P) A 2 triangle ineq. < ε/2 + ε/2 fro (3), (4) since R(f, P) is δ 1 (x) and δ 2 (x)-fine = ε And since since ε is arbitrary A 1 A 2 = 0 A 1 = A 2. 5

6 6 Dirichlet s function, revisited Now that we ve shown the value of an HK integral for an HK integrable function is well-defined, we are ready to calculate HK 1 g, where g : [0, 1] R is the Dirichlet function: 0 { 1, if x Q g(x) := 0, if x R \ Q Clai: HK 1 0 g = 0. Proof. Let ε > 0. In order to prove the clai, we ust construct a gauge δ : [0, 1] R st. given an arbitrary tagged partition (P, (c k ) n ) which is δ(x)-fine, R(f, P) 0 = g(c k )(x k x k 1 ) < ε. (5) Since the rationals are countable, we can list the as {r 1, r 2, r 3,... }. Now define, { ε/2 i+1, if x = r i Q δ(x) := 1, if x R \ Q. Notice the following three observations: (i) The function δ is a gauge because δ(x) > 0, x [0, 1] since ε, 2, and 1 are all positive. (ii) Because g(c k ) = 0 when c k is irrational, we can ignore these irrational tags and pass to the subsequence (c kj ) j=1 consisting of all the rational eleents of (c k) n when evaluating the finite su in (5). (iii) Because a tag can be an endpoint of a subinterval, it s possible for two different tags to have the sae value, corresponding to the right endpoint of one subinterval and the left endpoint of the successive subinterval. It is not possible to have equal tags fro non-successive subintervals, nor is it possible to have three or ore tags with the sae value. This is true for both the original sequence of tags (c k ) n and for the rational subsequence of tags (c kj ) j=1. For exaple, suppose we have the sequence (c 1, c 2,..., c 10 ). Then it s possible for c 3 = c 4, but not for c 3 = c 5 nor c 3 = c 4 = c 5. Siilarly, if we have the subsequence (c k1, c k2,..., c k6 ), it s possible for c k3 = c k4 but not for c k3 = c k5 nor c k3 = c k4 = c k5. By observation (ii), we can rewrite the lefthand side of inequality (5) as: R(f, P) 0 = g(c kj )(x kj x kj 1) = < = j=1 (x kj x kj 1) g(c kj ) = 1 since c kj is rational j=1 δ(c kj ) j=1 j=1 < 2 = 2ε since (P, (c k ) n ) is δ(x)-fine ε 2 ij+1 def. of δ(x) where the i j s are finite subseq. of N + i=1 ε 2 i+1 by observation (iii) 1/4 = ε geoetric series, 1 1/2 6

7 which is what we needed to show. Reark: Having a gauge allows us to control the length of each subinterval, enabling us to encapsulate sets of easure zero. So essentially, the characteristics of the Lebesgue integral is built into the HK integral. Also notice, if ε < 1, it is not possible to for a tagged partition which is δ(x)-fine and which consists only of rational tags, because when you add up the lengths of the subintervals in such a partition, the su will be less than 1. 7 The Fundaental Theore of Calculus Now we coe to the ain feature of the Henstock-Kurzweil integral, that which sets it apart fro both the Rieann and Lebesgue integrals and allows for a larger class of functions to be integrable. Theore 3: (The Fundaental Theore of Calculus). Let F : [a, b] R and f : [a, b] R be functions. Suppose F is differentiable on [a, b] st. F (x) = f(x), x [a, b]. Then, HK b a f = F (b) F (a). Reark: Notice that the Fundaental Theore of Calculus for HK integrals does NOT require any additional assuptions on f, unlike the Rieann integral version which requires f to be Rieann integrable and even the Lebesgue version, which requires f to be absolutely continuous. ε 2(b a) Proof. Assue conditions and let ε > 0 > 0 since > 0. As before, we ust construct a gauge δ : [0, 1] R st. given an arbitrary tagged partition (P, (c k ) n ) which is δ(c)-fine, (F (b) F (a)) R(f, P) < ε (6) (The reason why we switched to using c as our variable instead of x will soon be clear.) There are three key observations. First, since F (c) is differentiable c [a, b] with derivative equal to f(c), by Newton s approxiation: δ(c) > 0 st. x c δ(c), F (x) F (c) f(c)(x c) ε x c. (7) In other words, the value of δ depends on the choice of c, since we only have regular continuity in our assuption, not unifor continuity. Since δ(c) > 0 c [a, b], it is exactly this δ(c) that will be our gauge! Second, for an arbitrary subinterval [x k 1, x k ] and its tag c k [x k 1, x k ], we have the following: Fro (7) and (8) we get: x k x k 1 = x k x k 1 since x k > x k 1 = (x k c k ) + (c k x k 1 ) adding 0 and associativity = x k c k + c k x k 1 since x k c k and c k x k 1 x k c k x k x k 1 < δ(c k ), by transitivity and because (8) c k x k 1 x k x k 1 < δ(c k ) (P, (c k ) n ) is δ(c)-fine (9) F (x k ) F (c k ) f(c k )(x k c k ) ε x k c k = ε(x k c k ), (10) 7

8 since x k > c k. And fro (7) and (9) we get: F (x k 1 ) F (c k ) f(c k )(x k 1 c k ) ε x k 1 c k = ε(c k x k 1 ) 1 F (x k 1 ) F (c k ) f(c k )(x k 1 c k ) ε(c k x k 1 ) F (x k 1 ) + F (c k ) f(c k )( x k 1 + c k ) ε(c k x k 1 ) (11) Putting all this together, F (x k ) F (x k 1 ) f(c k )(x k x k 1 ) = F (x k ) + ( F (c k ) + F (c k )) F (x k 1 ) f(c k )(x k + ( c k + c k ) x k 1 ), by the iddle-an trick. F (x k ) F (c k ) f(c k )(x k c k ) + F (x k 1 ) + F (c k ) f(c k )( x k 1 + c k ), by triangle inequality. So, F (x k ) F (x k 1 ) f(c k )(x k x k 1 ) ε(x k c k ) = ε(x k x k 1 ) + ε(c k x k 1 ) (12) by (10) and (11). Suing (12) over the tagged partition (P, (c k ) n ): F (x k ) F (x k 1 ) f(c k )(x k x k 1 ) ε(x k x k 1 ) = ε(x n x 0 ) ε(b a) = = ε 2 (13) because the su is telescoping and x n = b and x 0 = a by construction of our partition P. Third, notice that: (F (x k ) F (x k 1 )) = (F (x 1 ) F (x 0 )) + (F (x 2 ) F (x 1 )) + + (F (x n ) F (x n 1 )) = F (x n ) F (x 0 ) telescoping finite su = F (b) F (a) by construction of our partition P (14) 8

9 Finally, we are ready to show (6) holds: (F (b) F (a)) R(f, P) = (F (x k ) F (x k 1 )) f(c k )(x k x k 1 ) = F (x k ) F (x k 1 ) f(c k )(x k x k 1 ) F (x k ) F (x k 1 ) f(c k )(x k x k 1 ) which is what we needed to show. 8 Closing Thoughts ε < ε by (13), 2 by (14) & def. of Rie. su by finite triangle inequality Obviously, we have just barely scratched the surface of the power of the Henstock-Kurzweil integral. But just to give a glipse of its versatility, we leave the reader with two facts (which we will not prove) of the HK integral: (i) The HK integral integrates: (a) all functions which have anti-derivatives (b) all Rieann integrable functions (c) all Lebesgue integrable functions (d) all functions that can be obtained as iproper integrals (ii) The HK integral has theores which generalize (a) the Monotone Convergence Theore (b) the Doinated Convergence Theores of easure theory 9

10 9 References Abbott, Stephen. Understanding Analysis. New York: Springer, Print. Bartle, Robert Gardner. A Modern Theory of Integration. Providence, RI: Aerican Matheatical Society, Print. Herschlag, Greg. A Brief Introduction to Gauge Integration. ay/vigre/vigre2006/papers/herschlag.pdf.[23 Apr Web.] McKinnis, Erik O. Gauge Integration. (2002): Web. Oberwolfach Photo Collection. Details: Ralph Henstock, Jaroslav Kurzweil. Web. 28 Apr Prove the Derivative Is Not Lebesgue Integrable. Math Stack Exchange. Web. 28 Apr Tao, Terence. Analysis I. New Delhi: Hindustan, Print. 10