Generlized Riemnn Integrl Krel Hrbcek The City College of New York, New York khrbcek@sci.ccny.cuny.edu July 27, 2014 These notes present the theory of generlized Riemnn integrl, due to R. Henstock nd J. Kurzweil, from nonstndrd point of view. The key notion we use, tht of -ultrsmll numbers, is due to B. Benninghofen nd M. M. Richter, A generl theory of superinfinitesimls, Fund. Mth. 123: 199 215, 1987, who cll them superinfinitesimls. E. Gordon, Nonstndrd Methods in Commuttive Hrmonic Anlysis, Amer. Mth. Society, 1997, developed n pproch to reltive stndrdness tht is different from tht of Y. Périre; in prticulr, his reltive infinitesimls re the superinfinitesimls. Here we hve combined the two techniques. B. Benninghofen presented n pproch to the generlized Riemnn integrl using superinfinitesimls in Superinfinitesimls nd the clculus of the generlized Riemnn integrl, in Models nd Sets, G. H. Müller nd M. M. Richter, eds., Lecture Notes in Mth. 1103, Springer, Berlin, 1984, pp. 9-52. Our development of the generlized Riemnn integrl follows the excellent exposition in R. Brtle, A Modern Theory of Integrtion, Amer. Mth. Society, 2001, to which the reder is referred for further study of this topic. AUN refers to K. Hrbcek, O. Lessmnn nd R. O Donovn, Anlysis with Ultrsmll Numbers.
1 -Ultrsmll Numbers The fundmentl problem of clculus is to determine the function f from its derivtive f. In AUN, Chpter 4 we solved this problem under the ssumption tht f is continuous; tht is, we lerned to integrte continuous functions. In AUN, Chpter 9 the theory of integrtion is developed for lrger clss of functions, those tht re Riemnn integrble. Among Riemnn integrble functions there re some tht re not continuous, yet it turns out tht the Riemnn integrble functions re precisely those functions tht re continuous lmost everywhere (see Theorem 31). Becuse of this nd other resons, Riemnn theory is not sufficiently generl for mny pplictions in nlysis. In order to see how to go bout formulting generl theory of integrtion, let us revisit the procedure used in AUN, Section 4.1 to recover the originl function f from its derivtive. The ssumption of continuity of f llowed us to use the uniform version of the increment eqution, with dx ultrsmll reltive to f, independent of. This in turn motivted the definition of Riemnn integrl in terms of fine prtitions, tht is, prtitions where ech d is ultrsmll reltive to f, independent of the tg t i. Without the ssumption of continuity of f, the increment eqution requires d to be ultrsmll reltive to f nd. This suggests the following definition: A tgged prtition (P, T ) is superfine (reltive to the level of f) if ech d is ultrsmll reltive to f nd t i. It is n esy exercise to verify tht most of the rguments in AUN, Chpter 9 would go through if one would replce fine prtitions by superfine prtitions in the definition of the Riemnn integrl. The clss of integrble functions would become much lrger nd, in prticulr, the fundmentl theorem of clculus would hold for ll differentible functions. Unfortuntely, in our frmework for reltive nlysis it is not possible to prove tht ny prtitions superfine in this strong sense exist. (See KH, Reltive set theory: Some externl issues, Journ. Logic nd Anlysis 2:8, 2010, 1 37.) The strong version of stbility, s well s of other principles, postulted in our theory requires numbers ultrsmll reltive to given context to be too smll to mke up superfine prtition. There is corser notion of ultrsmll numbers tht does not hve such nice uniform properties, nd therefore is not s suitble for development of nlysis in generl, but is tilor-mde for the specil purpose of generlizing the theory of integrtion. We develop this notion in the rest of the current section nd return to generlized Riemnn integrl in Sections 2 nd 3. Severl rguments in these notes use the Principle of Ideliztion discussed in the Appendix to AUN. The reders should tke look t this mteril before proceeding further, or s needed. Definition 1 Given context nd rel number : (1) A rel number r is -ccessible if r = ϕ() for some observble function ϕ : R R. (2) A rel number s -ultrlrge if x > r for ll -ccessible r > 0. (3) A rel number h 0 is -ultrsmll if h r for ll -ccessible r > 0. 2
Remrk -ccessible, -ultrsmll nd -ultrlrge re externl concepts, nd the conventions from AUN, Section 1.5 pply to them. We sy tht function ϕ is positive if ϕ(x) > 0 for ll n its domin. It follows immeditely from these definitions tht s not -ultrlrge if nd only if x ϕ() for some observble positive ϕ : R R, nd h 0 is -ultrsmll if nd only if h ϕ() for ll observble positive ϕ : R R. We lso note tht the bove sttements remin true if one requires only tht ϕ be defined t, in plce of being defined on ll of R: if ϕ is ny observble positive function defined t, then ϕ : R R defined by { ϕ(x) if defined, ϕ(x) = 1 otherwise, is n observble positive function defined for ll x, nd ϕ() = ϕ(). Theorem 1 (1) If r is -ccessible reltive to p 1,..., p k, then r is observble reltive to, p 1,..., p k. (2) If h is -ultrsmll reltive to p 1,..., p k, then h is ultrsmll reltive to p 1,..., p k. (3) If h is ultrsmll reltive to, p 1,..., p k, then h is -ultrsmll reltive to p 1,..., p k. Proof: (1) If ϕ is observble reltive to p 1,..., p k, then ϕ() is observble reltive to, p 1,..., p k by Closure. (2) For n observble r > 0 let ϕ r be the constnt function with vlue r, ϕ r : x r for ll x R. Then ϕ r is n observble positive function. Hence by definition we hve h ϕ r () = r, which shows tht h is ultrsmll. (3) follows from (1). If h < ϕ() for ll observble positive ϕ, then in prticulr h < n for ny observble n, hence h is of the form δ for some δ 0. It is cler tht if is observble reltive to p 1,..., p k, then h is -ultrsmll if nd only if h is ultrsmll. In prticulr h is not observble reltive to p 1,..., p k. The importnt fct is tht if is not observble reltive to p 1,..., p k, then numbers -ultrsmll reltive to p 1,..., p k cn ctully be observble reltive to, p 1,..., p k, s we prove in the following theorem. Theorem 2 For ny rel number not observble reltive to p 1,..., p k there re some numbers -ultrsmll reltive to p 1,..., p k nd observble reltive to, p 1,..., p k. 3
Proof: We use the Ideliztion Principle. Let {ϕ 1,..., ϕ k } be finite set of positive functions, observble reltive to p 1,..., p k. Then h = min(ϕ 1 (),..., ϕ k ()) > 0 nd h is observble reltive to, p 1,..., p k. In other words, there exists h > 0 observble reltive to, p 1,..., p k such tht h ϕ i () for i = 1,..., k. By Ideliztion, there exists h > 0 observble reltive to, p 1,..., p k such tht h ϕ() holds for ll positive functions observble reltive to p 1,..., p k ; ny such h is -ultrsmll reltive to p 1,..., p k. Exercise 1 (Answer pge 33) Show the following sttements. (1) If r nd s re -ccessible, then r ±s, r s nd r/s (if s 0) re -ccessible. (2) If x nd y re not -ultrlrge, then x ± y nd x y re not -ultrlrge. (3) If h, k re -ultrsmll nd s not -ultrlrge, then h ± k nd x h re -ultrsmll or 0. We need version of the Closure Principle for -ccessible numbers. Theorem 3 (-Closure Principle) Given sttement P(y,, b, p 1,..., p k ) of trditionl mthemtics, nd, b such tht b is -ccessible reltive to given context, where p 1,..., p k re observble: If there exists number y for which the sttement is true, then there exists n -ccessible number y for which the sttement is true. Proof: Let b = ϕ(), for n observble ϕ : R R. We consider the sttement P(y, x, ϕ(x)) with vrible x. Let ψ : R R be function such tht, for ll x R, if there is some y for which P(y, x, ϕ(x)) is true, then ψ(x) is one such y, i.e., P(ψ(x), x, ϕ(x)) holds. We omit the detiled justifiction of the existence of such function (it follows esily from the xioms of Seprtion, Replcement, nd Choice; see the Appendix to AUN for these). By Closure, we cn ssume tht ψ is observble. Then y = ψ() is -ccessible, nd P(ψ(),, ϕ()) is true. It is esy to modify the sttement nd proof of the -closure principle to llow finite list b 1,..., b l of prmeters in plce of the single prmeter b. We next show tht numbers -ultrsmll reltive to f cn replce numbers ultrsmll reltive to f nd in the definition of lim x f(x) (nd hence lso in the definition of continuity of f t nd derivtive of f t ). Theorem 4 The following sttements re equivlent: (1) lim x f(x) = L (2) Reltive to context where f is observble: L is -ccessible nd f( + h) L is -ultrsmll (or 0), for ll -ultrsmll h. 4
Proof: We work in context where f is observble. (1) implies (2): The limit L is observble, hence -ccessible. Let ε > 0 be -ccessible. By the epsilon-delt definition of limit, there exists δ > 0 such tht 0 < x < δ implies f(x) L < ε. Using the -Closure Principle, we cn tke δ to be -ccessible. If now h is - ultrsmll, we set x = + h nd hve x = h < δ, hence f( + h) L < ε. As ε is n rbitrry positive -ccessible number, f( + h) L is -ultrsmll. (2) implies (1): We ssume (1) is flse nd prove tht (2) is flse. So let lim x f(x) L, i.e., there exists ε > 0 such tht ( ) for every δ > 0 there exists x such tht 0 < x < δ nd f(x) L ε. By -Closure Principle we cn ssume tht ε is -ccessible. Let {ϕ 1,..., ϕ k } be n observble finite set of positive functions; we define ϕ by ϕ : x min{ϕ 1 (x),..., ϕ k (x)}. Notice tht ϕ is observble. Let δ = ϕ() in (*). We get tht there exists x such tht 0 < x < ϕ i () nd f(x) L ε is true for i = 1,..., k. Applying Ideliztion, we obtin x such tht 0 < x < ϕ() nd f(x) L ε is true for ll observble positive ϕ. Then h = x is -ultrsmll nd we hve f( + h) L ε, so f( + h) L is not -ultrsmll. Hence (2) fils. We immeditely deduce the following -version of the Increment Eqution. Theorem 5 Reltive context where f is observble: Suppose tht f is differentible t. Let dx be -ultrsmll. Then there is ε which is -ultrsmll or 0, such tht f( + dx) = f() + f () dx + ε dx. We leve the corresponding strddle version of the -Increment Eqution s n exercise. Exercise 2 (Strddle version) (Answer pge 33) Reltive context where f is observble: Suppose tht f is differentible t. Let x 1 x 2 be such tht x 1 nd x 2 re -ultrsmll or 0. Show tht there is ε which is -ultrsmll or 0, such tht f(x 2 ) f(x 1 ) = f ()(x 2 x 1 ) + ε (x 2 x 1 ). Definition 2 A tgged prtition (P, T ) is superfine if ech d is t i -ultrsmll. Existence of superfine prtitions follows from clssicl lemm. 5
Definition 3 Let ϕ be positive function defined on [, b]. We sy tht tgged prtition (P, T ) of [, b] is subordinte to ϕ if d < ϕ(t i ), for ll i = 0,..., n 1. Theorem 6 (Cousin s Lemm) If ϕ is positive function defined on [, b], then there is tgged prtition (P, T ) of [, b] subordinte to ϕ. Proof: We proceed by contrdiction nd ssume tht there is no tgged prtition of I 0 = [, b] subordinte to ϕ. Let c = ( + b)/2 be the midpoint of the intervl I 0. Either the intervl [, c] or the intervl [c, b] hs no prtition subordinte to ϕ; otherwise, we could combine them nd obtin prtition of I 0 subordinte to ϕ. In the first cse we let I 1 = [, c]; otherwise, I 1 = [c, b]; I 1 hs no prtition subordinte to ϕ nd the length of I 1 is (b )/2. Continuing in this mnner, we construct nested sequence of closed intervls I 0, I 1,....I n,..., none of which hs prtition subordinte to ϕ, nd such tht the length of I n is (b )/2 n ; in prticulr, the length of I n converges to 0. By the nested intervl theorem, there is number c tht belongs to every I n. Let n be such tht (b )/2 n < ϕ(c). Then the trivil prtition of I n (tht is, x 0 is the left endpoint of I n, x 1 is the right endpoint of I n ), tgged by t 0 = c, is subordinte to ϕ, contrdiction. Theorem 7 For every < b there exists superfine prtition of [, b]. Proof: We gin use Ideliztion. Let {ϕ 1,..., ϕ k } be set of positive functions defined on [, b] nd observble reltive to nd b; then ϕ(x) = min(ϕ 1 (x),..., ϕ k (x)) is positive function, nd ϕ ϕ i for ll i = 1,..., k. Applying Ideliztion, we obtin positive function ϕ such tht ϕ ϕ for ll observble positive ϕ. By Cousin s Lemm, there exists tgged prtition of [, b] subordinte to ϕ. It is cler tht this prtition is superfine. We conclude with two technicl results bout superfine prtitions. Let x 0 < x 1 < < x n be fine prtition of [, b]; then for ech i there is t most one element in [, +1 ] which is observble reltive to nd b. We now show tht for superfine prtitions we necessrily tke tht element s tg. Theorem 8 Let (P, T ) be superfine prtition of [, b]. c [, b] belongs to T. Every observble rel number Proof: The context is specified by, b. Let c [, b] be observble. Let ϕ be the function defined by { x c if x c; ϕ(x) = 1 if x = c. 6
Then ϕ is positive nd observble. Let i be such tht c [, +1 ]. We show tht t i = c. If not, then since d is t i -ultrsmll, we must hve d < ϕ(t i ) = t i c, i.e., +1 < t i c, so c / [, +1 ], contrdiction. This shows tht c T. Let (P, T ) be tgged prtition. We define new tgged prtition (P, T ) s follows: whenever < t i < +1, we split the intervl [, +1 ] into [, t i ] nd [t i, +1 ], nd let t i be the tg for both. We note tht f(t i )(+1 ) = f(t i )(+1 t i ) + f(t i )(t i ), so (f; P, T ) = (f; P, T ). The prtition (P, T ) hs the property tht the tg for ech subintervl is either the left or the right endpoint. If (P, T ) is superfine, then (P, T ) is superfine nd, by the previous proposition, every c [, b] from the context is specified by one of the points x 0,..., x n of the prtition P. Theorem 9 Let, b R nd system of open intervls {I k } k=1 pper t the observtion level. If (P, T ) is superfine prtition of [, b], then for ech t i k=1 I k there is some k such tht [, +1 ] I k. Proof: Let us write I k = ( k, b k ), for some k < b k. We define ϕ : [, b] R by { min(x k, b k x) where k is lest such tht t i ( k, b k ); ϕ(x) = 1 if no such k exists. Notice tht ϕ is well-defined. It is positive function t the observtion level. Let t i k=1 ( k, b k ) nd let k be the lest index such tht t i ( k, b k ). The prtition is superfine, so using the definition of ϕ we hve +1 = d < ϕ(t i ) min(t i k, b k t i ). It follows tht, +1 ( k, b k ), i.e., [, +1 ] I k. 2 The generlized Riemnn integrl The generliztion of Riemnn integrl tht we present here ws developed independently by Rlph Henstock nd Jroslv Kurzweil; it is sometimes clled Henstock-Kurzweil integrl. Definition 4 A function f defined on [, b] is generlized Riemnn integrble on [, b] (or simply integrble on [, b]) if there is n observble number R such tht (f; P, T ) R, for ll superfine tgged prtitions (P, T ) of [, b]. In this is the cse, we write f(x) dx = R. 7
Superfine prtitions re fine, so it follows immeditely tht ll Riemnn integrble functions re integrble in the new, generlized sense. In prticulr, ll continuous functions nd ll monotone functions defined on [, b] re integrble. The first two theorems re nlogs of AUN Theorems 132 nd 133 for the Riemnn integrl. They re proved by replcing the word fine with superfine in the proofs from AUN, Chpter 9. Theorem 10 (Linerity) Let f nd g be integrble on [, b] nd let λ, µ be rel numbers. Then λ f +µ g is integrble on [, b] nd (λ f + µ g)(x) dx = λ f(x) dx + µ g(x) dx. Theorem 11 (Monotonicity) Let f nd g be integrble on [, b]. Assume tht f(x) g(x), for ll x b. Then f(x) dx g(x) dx. Theorem 12 (Cuchy Test) Let f be defined on [, b]. Then f is integrble on [, b] if nd only if (f; P, T ) (f; P, T ), for ll superfine tgged prtitions (P, T ), (P, T ) of [, b]. Proof: If f is integrble, then (f; P, T ) f(x) dx (f; P, T ), so f hs the Cuchy property. For the converse, ssume tht f hs the Cuchy property. The context is specified by f, nd b. It suffices to show tht the numbers (f; P, T ) re not ultrlrge; we cn then let R be the observble neighbor of (f; P, T ), nd the Cuchy property implies tht f is integrble nd f(x) dx = R. We fix one superfine prtition (P 0, T 0 ) nd let m = (f; P 0, T 0 ) 1 nd M = (f; P0, T 0 ) + 1. For every superfine prtition (P, T ) we hve (f; P, T ) (f; P0, T 0 ) nd hence m < (f; P, T ) < M. The following sttement is therefore true: There exist m, M such tht m < (f; P, T ) < M, for ll superfine prtitions. The sttement is internl nd its prmeters re f,, b; therefore, by Closure, There exist observble m, M such tht m < (f; P, T ) < M, for ll superfine prtitions. 8
This is precisely the ssertion tht no (f; P, T ) is ultrlrge. Theorem 13 (Additivity) Let < c < b; then f is integrble on [, b] if nd only if f is integrble on [, c] nd on [c, b]. If this is the cse, then f(x) dx = c f(x) dx + c f(x) dx. Proof: Assume tht f is integrble on [, b]. Let (P 1, T 1 ) nd (P 2, T 2 ) be two superfine prtitions of [, c]. We extend them to superfine prtitions (P 1, T 1 ) nd (P 2, T 2 ) of [, b] in such wy tht they coincide on [c, b]. We then hve (f; P1, T 1 ) (f; P 2, T 2 ) = (f; P 1, T 1 ) (f; P 2, T 2 ) 0, becuse f is integrble on [, b]. Hence f hs the Cuchy property on [, c] nd is integrble on [, c]. Assume tht f is integrble on [, c] nd [c, b]. Let (P, T ) be superfine prtition of [, b]; by the comment following Theorem 8 we cn ssume without loss of generlity tht c is prtition point of P. Let (P 1, T 1 ) nd (P 2, T 2 ) be the restrictions of (P, T ) to [, c] nd [c, b], respectively. Then (f; P, T ) = (f; P1, T 1 ) + c (f; P 2, T 2 ) f(x) dx + so f is integrble on [, b] nd c f(x) dx, f(x) dx = c f(x) dx + c f(x) dx. Theorem 14 (Fundmentl Theorem of Clculus) If f is differentible on [, b], then f is integrble on [, b] nd f (x) dx = f(b) f(). Proof: Let (P, T ) be superfine prtition of [, b]. By the strddle version of the -Increment Eqution (Exercise 2), f(+1 ) f( ) = f (t i ) d + ε i d with ε i which is t i -ultrsmll or 0. Hence f (t i ) d = (f(+1 ) f( )) ε i d. But (f(+1) f( )) = f(x n ) f(x 0 ) = f(b) f(). Moreover, ll ε i re ultrsmll (or 0) nd d re ultrsmll with d = b, so ε i d 0, s usul. This shows tht f is integrble nd f (x) dx = f(b) f(). 9
Theorem 14 fulfills our gol of solving the fundmentl problem of clculus for ll differentible functions. The reder will notice tht, up to this point, the proofs in this section closely resemble those of the nlogous results in AUN, Chpter 9. However, there re powerful theorems bout the generlized Riemnn integrl tht go fr beyond Chpter 9. Their proofs tend to be more subtle nd, in some cses, involve combintion of epsilon delt nd ultrsmll rguments. We derive severl such results in the rest of this section. Theorem 15 Let f be function nd R rel number. The following sttements re equivlent: (1) f is integrble on [, b] nd f(x) dx = R. (2) For every ε > 0 there is positive function δ such tht (f; P, T ) R < ε, for ll tgged prtitions (P, T ) subordinte to δ. Proof: The context is specified by f, nd b. (1) implies (2): Let ε > 0 be observble. In the proof of Theorem 7 we showed tht there is positive function ϕ with the property tht every prtition (P, T ) subordinte to ϕ is superfine, nd hence (f; P, T ) R < ε. Letting δ = ϕ proves (2) for observble ε > 0. By Closure, (2) is true for ll ε > 0. (2) implies (1): Let (P, T ) be superfine, nd let ε > 0 be observble. By Closure there is n observble positive function δ with the property in (2). As (P, T ) is subordinte to δ, we hve (f; P, T ) R < ε. This is true for ll observble ε > 0, so (f; P, T ) R. Definition 5 A prtilly tgged prtition of [, b] is prtition P = {x 0, x 1,..., x n } nd prtil tgging T = {t j : j J}, where J {0, 1,..., n 1} nd t j [, +1 ] for ll j J. As for tgged prtitions, we use the nottion (f; P, T ) for f(t j) d. Definition 6 We sy tht prtilly tgged prtition is subordinte to ϕ if We cll it superfine if d < ϕ(t j ), for ll j J. d is t j -ultrsmll, for ll j J. 10
The next key lemm shows tht, for integrble functions, the Riemnn sums give good pproximtion of the integrl not only over the whole intervl [, b], but over ny collection of subintervls from the prtition s well. Theorem 16 (Sks-Henstock Lemm) Let f be integrble on [, b] nd let ε > 0. There exists positive function δ such tht f(t j ) d xj+1 f(x) dx < ε, for every prtilly tgged prtition (P, T ) subordinte to δ. In fct, for ny such prtilly tgged prtition we lso hve f(t j ) d xj+1 f(x) dx < 2ε, nd even f(t j ) d xj+1 f(x) dx < 2ε. Proof: Since f is integrble, we cn find positive function δ in V(f,, b, ε) such tht (f; P, T ) f(x) dx < ε, for every tgged prtition (P, T ) subordinte to δ. Consider now prtilly tgged prtition (P, T ) subordinte to δ. The context contins f,, b, ε, nd this prtilly tgged prtition (P, T ). In prticulr, J is observble. Let i J. Since f is integrble over [, +1 ], we hve f(x) dx (f; P i, T i ) for ny superfine prtition (P i, T i ) of [, +1 ] (reltive to the context ugmented by, +1 ). Select one such prtition for ech i / J (this is justified by the Principle of Finite Choice, see the Appendin AUN.) The union of the prtitions (P i, T i ) together with (P, T ) is prtition of [, b] subordinte to δ. Therefore, f(x) dx f(t j ) d (f; Pi, T i ) < ε. i / J On the other hnd, by the dditivity property of the integrl we hve f(x) dx = xj+1 f(x) dx + i / J Substituting this into the previous inequlity, we get xj+1 f(x) dx. f(x) dx f(t j ) d + [ f(x) dx ] (f; P i, T i ) < ε. i / J 11
As {i : i / J} is finite nd observble, nd for ech i we hve +1 f(x) dx (f; Pi, T i ), the quntity between the squre brckets is ultrclose to 0. It follows tht xj+1 f(x) dx f(t j ) d < ε. For the second clim, we consider the two prtilly tgged prtitions determined by nd + J + = {j J : f(t j ) d xj+1 J = {j J : j / J + }. + f(x) dx} Applying the first result seprtely to J + nd J, we get xj+1 f(t j ) d f(x) dx = ( f(t j ) d nd xj+1 f(t j ) d xj+1 ) f(x) dx < ε f(x) dx = ( xj+1 ) f(x) dx f(t j ) d < ε. The second clim follows by dding the two lines. For the finl clim, we use the tringle inequlity f(t j ) d f(t j ) d xj+1 f(x) dx + xj+1 so f(t j ) d xj+1 f(t j ) d f(x) dx + < xj+1 f(x) dx + 2ε, nd similrly, from xj+1 one deduces tht f(x) dx, xj+1 xj+1 f(x) dx f(t j ) d + f(x) dx f(t j ) d xj+1 f(x) dx < f(t j ) d + 2ε. f(x) dx 12
Exercise 3 (Answer pge 33) Show tht if f is integrble on [, b] nd (P, T ) is superfine prtilly tgged prtition of [, b], then f(t j ) d xj+1 f(x) dx. Moreover f(t j ) d xj+1 f(x) dx. We re redy to prove the centrl result in the theory of generlized Riemnn integrl. Theorem 17 (Monotone Convergence Theorem) Let (f n ) n=1 be monotone sequence of integrble functions on [, b] such tht f(x) = lim n f n(x) exists for ech x [, b]. If the sequence ( f n(x) dx) n=1 is bounded, then f is integrble on [, b] nd f(x) dx = lim n f n (x) dx. Proof: We give proof for the cse when (f n ) n=1 is incresing. The context is specified by the sequence (f n ) n=1,, nd b. Hence the function f is observble. Let k N. Since f k is integrble, there is positive function δ k such tht f k (t i ) d i J f k (x) dx 1 2 k, ( ) for every prtilly tgged prtition subordinte to δ k. Since this is trditionl mthemticl sttement, we cn choose one such δ k for ech k, nd by Closure, we my ssume tht the sequence (δ k ) k=1 is observble. By the monotonicity property of the integrl, the sequence ( f n(x) dx) n=1 is incresing. By ssumption, it is bounded, so by the Monotone Convergence Theorem for sequences, this sequence converges. Let n observble R be such tht R = lim n f n (x) dx. Let (P, T ) be superfine prtition. By the Locl Stbility Principle (see the Appendix to these notes), there is n ultrlrge N such tht (P, T ) remins superfine reltive to the context ugmented by N. Since N is ultrlrge, we hve R f N (x) dx. 13
For ech x [, b], let k(x) N be the lest such tht f k(x) (x) f(x) 1 N. In prticulr, f k(ti)(t i ) f(t i ), for i = 0,..., n. Notice tht the function x k(x) is observble reltive to the context ugmented by N. We define δ(x) = δ k(x) (x), so δ is positive function observble reltive to the context ugmented by N. Since (P, T ) is superfine reltive to the context ugmented by N, it is subordinte to δ. We need to show tht f(t i ) d R. First, notice tht f(t i ) d f k(ti)(t i ) d, (*) since f(t i ) = f k(ti)(t i ) + ε i, with ε i 0, nd ε i d 0. Second, using the Sks-Henstock lemm, we must hve To see this, define f k(ti)(t i ) d J p = {i : k(t i ) = p}, for ech p N. f k(ti)(x) dx. (**) (Of course, J p = for ll but finitely mny vlues of p.) We consider the prtilly tgged prtition (P, T p ) obtined from (P, T ) by restricting T to J p. Now since (P, T ) is subordinte to δ, if i J p we hve d < δ(t i ) = δ ki(t)(t i ) = δ p (t i ), by definition of δ. This shows tht (P, T p ) is subordinte to δ p. By ( ) we deduce tht f k(ti)(t i ) d f k(ti)(x) dx 2 2 p = 1 2 p 1. i J p As ech k(t i ) is equl to some p N, dding these inequlities gives 14
f k(ti)(t i ) d since N is ultrlrge. Finlly, we show tht f k(ti)(x) dx f k(ti)(t i ) d = f k(ti)(t i ) d p N i J p 1 0, 2p 1 p N Let K = mx{k(t i ) : 0 i < n}. Then K N nd f k(ti)(x) dx f k(ti)(x) dx f k(ti)(x) dx R. (***) f N (x) f k(ti)(x) f K (x), for ll x [, b]. Hence nd R f N (x) dx f N (x) dx f k(ti)(x) dx f k(ti)(x) dx f K (x) dx, f K (x) dx R. (The lst step holds becuse K is ultrlrge.) This estblishes (***). Now putting together (*), (**), nd (***) yields the conclusion. We derive few consequences of Monotone Convergence Theorem here, nd severl more in the next section. Given n intervl I = (, b), we denote by l(i) the length of I; i.e., l(i) = b. Definition 7 Let A R. We sy tht A is null set if there re open intervls I n such tht A n=1 I n nd l(i n ) 0. n=1 Exmple Let C = {c n } n=1 be countble set. Fix ε 0 nd, for ech n, let ( I n = c n ε 2 n, c n + ε ) 2 n. 15
ε Then l(i n ) = 2 2 nd n n=1 l(i n) = 2 ε 0. Thus, ll countble sets re null sets. The Cntor set is n exmple of n uncountble null set. We sy tht sttement bout rel numbers is true lmost everywhere if the set of those x for which the sttement is flse is null set. Theorem 18 If f is integrble on [, b] nd f(x) = g(x) lmost everywhere, then g is integrble on [, b] nd g(x) dx = f(x) dx. Proof: Let h(x) = g(x) f(x). Then there is null set E such tht h(x) = 0 for ll x / E. It suffices to prove tht h is integrble nd h(x) dx = 0. The context contins h,, b, nd E. By Closure, we cn find n observble {I k } k=1 such tht E k I k nd l(i k ) 0. We first show the result in the cse when h is bounded. By Closure, there is then n observble M such tht h(x) M, for ll x [, b]. If (P, T ) is superfine reltive to this level, then by Theorem 9, for every t i E, we hve [, +1 ] I ki for some k i. If t i / E, then h(t i ) = 0. These observtions give (h; P, T ) = h(t i ) d + h(t i ) d t i E M d t i E M l(i k ) 0. k=1 (In the lst step we used the fct tht the intervls [, +1 ] re non-overlpping, so k d=k i l(i k ).) Hence h is integrble nd h(x) dx = 0. Now ssume tht h 0, but possibly unbounded. For ech n N we define t i / E h n (x) = min(h(x), n), for x [, b]. Clerly h n (x) = 0 for ll x / E nd h n is bounded. By the first prgrph, h n is integrble nd h n(x) dx = 0. As {h n } n=1 is incresing nd lim n h n (x) = h(x), the Monotone Convergence Theorem shows tht h is integrble nd h(x) dx = lim n h n (x) dx = 0. Now let h be rbitrry. Write h = h + h ; clerly h + (x), h (x) equl 0 except for x E, nd h + 0, h 0. Thus both h + nd h re integrble 16
nd h+ (x) dx = h (x) dx = 0 by the previous two prgrphs. It follows by dditivity tht h is integrble nd h(x) dx = h + (x) dx h (x) dx = 0. Exmple The Dirichlet function f : [0, 1] R is defined by { 1 if x Q; f(x) = 0 otherwise. As ll countble sets re null sets, it follows tht the Dirichlet function is generlized Riemnn integrble nd f(x) dx = 0. We sw in AUN, Chpter 9 tht the Dirichlet function is not Riemnn integrble. We now give clssicl exmple showing tht the monotone convergence theorem fils for Riemnn integrble functions. Fix n enumertion of Q = {q n : n N}. Define f n : [0, 1] Q by { 1 if x = q k, for some k n; f n (x) = 0 otherwise. Then the sequence (f n ) is incresing nd its limit is the Dirichlet functionf. Moreover, f n is continuous everywhere except on finite set, so f n is Riemnn integrble nd f n(x) dx = 0. Thus, lim n f n(x) dx = 0, but the limit function f is not Riemnn integrble. The finl result of this section is substntilly strengthened version of the Fundmentl Theorem of Clculus, which llowed us to recover the function from its derivtive. It sttes essentilly tht if continuous function is differentible everywhere except on countble set, then one cn still recover the function from its derivtive. (Recll tht if function is differentible everywhere on [, b], then it is continuous.) Theorem 19 (Fundmentl Theorem of Clculus, Strong Version) Let f be continuous on [, b]. Let C [, b] be countble set. Let g be function defined on [, b] such tht g(x) = f (x) for ll x C. Then g is integrble nd g(x) dx = f(b) f(). Proof: The context is specified by f, g,, b nd n enumertion {c k } k=1 of C. The set C is null set, nd hence, by Theorem 18, we my ssume without loss of generlity tht g(x) = 0 for ll x C. Let (P, T ) be superfine prtition of [, b]. If t i / C, then g(t i ) d = f (t i ) d = f(+1 ) f( ) ε i d where ε i 0, by the strddle version of the Increment Eqution. If t i C, then g(t i ) d = 0. Hence g(t i ) d = (f(+1 ) f( ) ε i d ) (f(+1 ) f( )). t i / C 17 t i / C
It remins to prove tht t i C (f(+1) f( )) 0. Let ε > 0 be observble. From the ssumption tht f is continuous t c k it follows tht there is δ k > 0 such tht x c k < δ k implies f(x) f(c k ) < ε 2 k. We define the function ϕ s follows { δ k if x = c k ; ϕ(x) = 1 otherwise. Then ϕ is n observble positive function. Suppose tht t i = c k. Then, since the prtition is superfine we hve Hence d < ϕ(c k ) = δ k. f(+1 ) f( ) f(+1 ) f(c k ) + f(c k ) f( ) ε 2 k + ε 2 k = ε 2 k 1. Ech c k cn be equl to t most two tgs (t i nd t i+1, if it so hppens tht t i = c k = t i+1 ), so we hve ε f(+1 ) f( ) 2 = 4ε. 2k 1 t i C k=1 As ε ws n rbitrry observble positive number, we hve f(+1 ) f( ) 0. t i C 3 The Lebesgue integrl The most populr dvnced theory of integrtion is due to Henri Lebesgue. We show tht the min theorems bout Lebesgue integrl follow from the results of the preceding section. Definition 8 A function f : [, b] R is Lebesgue integrble on [, b] if both f nd f re generlized Riemnn integrble on [, b]. Exercise 4 (Answer pge 33) Show tht Lebesgue integrbility is equivlent to the generlized Riemnn integrbility of f + nd f. We will show lter in this section tht the function f defined by { 1 f(x) = x sin( 1 x ) if x 0; 0 if x = 0. 18
is integrble on [0, 1], but not Lebesgue integrble on [0, 1]. Theorem 20 (Comprison Test) Let f nd g be integrble on [, b]. If f g, then f is Lebesgue integrble on [, b]. Proof: The context is specified by f, g,, nd b. Let (P, T ) be superfine prtition of [, b]. From g f g we derive tht g(x) dx f(x) dx g(x) dx, i.e., f(x) dx g(x) dx. Hence, +1 f(x) dx g(x) ds not ultrlrge. Let R be the observble neighbor of f(x) dx. Now ( f ; P, T ) R f(t i ) d + f(x) dx R. f(x) dx The first term is ultrclose to 0 by the Sks-Henstock lemm nd the second lso, by definition of R. We conclude tht f is integrble. Theorem 21 (Linerity) Let f nd g be Lebesgue integrble on [, b] nd let λ, µ be rel numbers. Then λ f + µ g is Lebesgue integrble on [, b]. Proof: The function λ f + µ g is integrble by Theorem 10. Furthermore, λ f(x) + µ g(x) λ f(x) + µ g(x) nd the function on the right is integrble. The Comprison Test implies tht λ f(x) + µ g(x) is integrble. Exercise 5 (Additivity) (Answer pge 33) Let < c < b; show tht f is Lebesgue integrble on [, b] if nd only if f is Lebesgue integrble on [, c] nd on [c, b]. Theorem 22 Let f, g be integrble nd f g. Then f is Lebesgue integrble if nd only if g is Lebesgue integrble. 19
Proof: Suppose f nd g re integrble nd f is Lebesgue integrble. Then g = (g f) + f where g f is integrble nd nonnegtive; so g f is Lebesgue integrble. Hence g is Lebesgue integrble by Theorem 21. The proof when g is supposed to be Lebesgue integrble is similr. Exercise 6 (Answer pge 33) In the Monotone Convergence Theorem, show tht if one of the functions f n is Lebesgue integrble, then f = lim f n is Lebesgue integrble. Exercise 7 (Answer pge 34) Show tht if f is Lebesgue integrble nd f(x) = g(x) lmost everywhere, then g is Lebesgue integrble. Theorem 23 If f, g, h re integrble nd h f, h g, then the functions min{f(x), g(x)} nd mx{f(x), g(x)} re integrble. Proof: We first observe tht min{f(x), g(x)} = 1 (f(x) + g(x) f(x) g(x) ) 2 nd mx{f(x), g(x)} = 1 (f(x) + g(x) + f(x) g(x) ) 2 (check seprtely the cses f(x) g(x) nd f(x) g(x)). Next, from the first eqution, f(x) g(x) = f(x)+g(x) 2 min(f(x), g(x)) f +g 2h, so f(x) g(x) is integrble by the Comprison Test. It follows tht min{f(x), g(x)} nd mx{f(x), g(x)} re integrble. The next theorem is cornerstone of Lebesgue integrtion. Suppose tht (f k ) is sequence of functions bounded below, i.e., there is function h such tht f k h for ll k. Then h k (x) = inf l k f l (x) exists, nd the sequence (h k (x)) is incresing, for ll x [, b]. We define the function lim inf f k by (lim inf f k )(x) = lim inf f k (x) = lim k h k(x), for ech x where the limit exists (i.e., where the sequence (h k (x)) is bounded). Theorem 24 (Ftou s Lemm) Let f k be integrble functions, for k = 1, 2,.... Let h be integrble nd such tht h f k, for ll k. Suppose tht lim inf f k is defined for ll x [, b]. If lim inf f k(x)dx <, then lim inf f k is integrble nd lim inf f k (x) dx lim inf f k (x) dx. If h is Lebesgue integrble, then lim inf f k is Lebesgue integrble. 20
Proof: For fixed k nd n k, let g k n = min(f k,..., f n ). Then g k n h. An esy induction using the previous theorem shows tht g k n is integrble. Moreover, g k n g k n+1, so the sequence (g k n) n 1 is decresing. But h(x) dx g k n(x) dx g k k(x) dx, so ( gk n(x) dx) n k is bounded. By the Monotone Convergence Theorem, we hve tht h k = inf f l = lim l k n gk n is integrble. The sequence (h k ) k 1 is incresing nd h h k f k, so h(x) dx h k (x) dx f k (x) dx, for ech k. Hence h(x) dx lim inf h k(x) dx lim inf f k(x) dx <. But the sequence ( h k(x) dx) k 1 is incresing, so lim inf h k (x) dx = lim k h k (x) dx < nd ( h k(x) dx) k 1 is bounded. Applying the Monotone Convergence Theorem once gin, we deduce lim inf f k (x) dx = The lst clim follows from Exercise 6. lim h k(x) dx = lim k k lim inf h k (x) dx f k (x) dx. Theorem 25 (Dominted Convergence Theorem) Let f n be integrble functions for n = 1, 2,... nd let f(x) = lim n f n (x) exist for ll x [, b]. Suppose tht there re integrble functions h 1, h 2 such tht h 1 (x) f n (x) h 2 (x), for ll x [, b]. Then the function f is integrble nd f(x) dx = lim n f n (x) dx. If t lest one of h 1, h 2 is Lebesgue integrble, then f is Lebesgue integrble. Proof: The ssumptions of Ftou s lemm re stisfied, so we deduce tht f is integrble nd lim f n (x) dx lim inf 21 f n (x) dx.
Similrly, i.e., lim f n (x) dx = lim sup lim inf lim( f n (x)) dx ( f n (x)) dx = lim sup f n (x) dx lim f n (x) dx. f n (x) dx, It follows tht lim inf f n(x) dx = lim sup f n(x) dx, i.e., lim f n(x) dx exists nd equls lim f n(x) dx. Exercise 8 (Men Convergence Theorem) (Answer pge 34) Under the ssumptions of the Dominted Convergence Theorem, show tht the functions f f n re Lebesgue integrble nd lim n f(x) f n (x) dx = 0. For the next exmple we need corollry of the Dominted Convergence Theorem. Theorem 26 Let f : [, b] R be such tht f(x) h(x), for ll x [, b], for some h integrble on [, b]. Suppose tht f is integrble on [r, b], for every < r < b. Then f is integrble on [, b] nd f(x) dx = lim r + r f(x) dx. Moreover, if b = r 0 > r 1 > r 2 >... nd lim k r k =, then f(x) dx = k=0 rk r k+1 f(x) dx. Proof: We define f k by f k (x) = { f(x) for r k x b, 0 otherwise. Then ech f k is integrble on [, b], lso f k (x) h(x), nd lim k f k (x) = f(x), for ll x [, b]. By the dominted convergence theorem, f is integrble 22
on [, b] nd f(x)dx = lim k = lim k k 1 = lim k = ri f k (x)dx r k f(x) dx ri r i+1 f(x) dx r i+1 f(x) dx. Exmple We show in this exmple tht the function f defined by { 1 f(x) = x sin( 1 x ) if x 0; 0 if x = 0 is integrble on [0, 1], but not Lebesgue integrble on [0, 1]. We first show tht f is not Lebesgue integrble on [0, 1]. Assume, for contrdiction, tht f is integrble on [0, 1]. Let k = 1 kπ nd note tht [0, 1] = {0} ( k+1, k ] ( 1, 1]. k=1 Let n 1. By monotonicity nd dditivity, we hve 1 0 1 x ( 1 x) sin dx 1 n 1 x ( 1 x) sin dx = k=1 k k+1 1 x ( 1 x) sin dx. We use the substitutions u = 1 x nd v = u kπ to deduce k ( 1 1 kπ ( k+1 x x) sin dx = u sin(u) 1u ) (k+1)π 2 du π 1 = sin(v) dv 0 v + kπ 1 π 2 sin(v) dv = (k + 1)π (k + 1)π. Hence But k=1 1 0 1 x ( 1 sin x 1 k+1 =, so 1 1 0 x ) dx 2 π k=1 0 1, for ech n N. k + 1 sin( 1 x ) dx =, contrdiction. 23
We now show tht f is integrble: The function t 1 t sin( 1 t ) is continuous on (0, 1], nd integrtion by prts gives 1 x ( ) 1 1 t sin dt = t 1 x ( ( t) 1 ( 1 t 2 sin t )) dt = t cos ( 1 t ) 1 x 1 x cos ( ) 1 dt, t for ech 0 < x 1. As t cos( 1 t ) is bounded by the constnt function with vlue 1, which is integrble on [0, 1], it follows from the corollry to the dominted convergence theorem tht t cos( 1 t ) is integrble on [0, 1] nd 1 0 cos( 1 t ) dt = 1 lim x 0+ x cos( 1 t ) dt. The function G(x) = 1 x cos( 1 t ) dt is thus continuous t 0. As x cos( 1 x ) is continuous on (0, 1], we hve G (x) = cos 1 x, for ll x (0, 1], by the Fundmentl Theorem of Clculus for continuous functions. We now let F be defined by { x cos( 1 F (x) = x ) + G(x) for 0 < x 1, 0 if x = 0. Note tht F is continuous on [0, 1] nd F (x) = 1 ( ) 1 x sin = f(x), for 0 < x 1. x So f is integrble by the strong version of the Fundmentl Theorem of Clculus (Theorem 19). With this exmple we end the systemtic development of the theory of integrtion. Our gol in the next few pges is to define some concepts tht ply key role in the trditionl expositions of Lebesgue integrl. Definition 9 A collection Σ of subsets of set S is clled n lgebr if (1) S Σ; (2) If X, Y Σ, then X Y Σ; (3) If X Σ, then S \ X Σ. Exercise 9 (Answer pge 34) Let Σ be n lgebr of subsets of S nd X 1,..., X n Σ. Show tht (1) X 1... X n Σ, (2) X 1... X n Σ, (3) If X, Y Σ, then X \ Y Σ. 24
Definition 10 An lgebr Σ is clled σ-lgebr if X n Σ, n 1 for every sequence (X n ) n 1 of sets from Σ. Exercise 10 (Answer pge 34) Let Σ be σ-lgebr. Show tht if X n Σ (n = 1, 2,... ), then n 1 X n Σ. Definition 11 The chrcteristic function χ A of set A is the function defined by χ A (x) = { 1 if x A, 0 otherwise. Definition 12 A set A [, b] is Lebesgue mesurble if χ A is integrble on [, b]. We denote by M[, b] the collection of ll Lebesgue mesurble subsets of [, b]. With these definitions we hve: Theorem 27 Let [, b] be closed intervl. Then (1) M[, b] is σ-lgebr. (2) For every c d b, the closed intervl [c, d] is Lebesgue mesurble. (3) All null subsets of [, b] re Lebesgue mesurble. Proof: If X, Y M, then X Y M: χ X Y = mx(χ X, χ Y ), so it is integrble by Theorem 23. If X k M for k = 1, 2,..., we let Y n = X 1... X n. By the bove nd induction, Y n M. We note tht (χ Yn ) n 1 is incresing nd χ X = lim n χ Yn. So χ X is integrble, by monotone convergence theorem. The remining clims re left s n exercise. 25
Definition 13 Let Σ be n lgebr of subsets of S. A function µ : Σ [0, ] defined for ll sets in Σ is clled finitely dditive mesure on Σ if (1) µ(x) 0 for ll X Σ; (2) µ( ) = 0, µ(s) > 0; (3) µ(x Y ) = µ(x) + µ(y ) whenever X nd Y re disjoint. If, in ddition, Σ is σ-lgebr nd (4) µ( k 1 X k) = k 1 µ(x k) holds for every sequence (X k ) of pirwise disjoint sets from Σ, then µ is clled σ-dditive mesure on Σ. Observe tht n dditive mesure is monotone: If X Y, with X, Y Σ then µ(x) µ(y ). To see this, write Y = X (Y \ X) (recll tht Y \ X Σ), so µ(y ) = µ(x) + µ(y \ X) µ(x). Definition 14 The Lebesgue mesure is the function m : M[, b] R defined by m(a) = χ A (x)dx. The next theorem justifies the nme. The proof is left s n esy exercise. Theorem 28 Let [, b] be closed intervl. Then (1) The Lebesgue mesure m is σ-dditive mesure on M[, b]. (2) For c < d b, we hve m([c, d]) = d c. (3) If E is null set, then m(e) = 0. Theorem 29 For every set A M[, b] nd every ε > 0 there is n open set O such tht A O nd m(o) m(a) ε. Proof: Let ϕ be positive function. We show first tht there is (finite or infinite) sequence (I k ) of non-overlpping closed intervls, nd sequence of tgs (t k ), such tht ech t k I k A, l(i k ) < ϕ(t k ), nd A I k. We strt with I = [, b] nd split it into two subintervls, I 0 = [, m] nd I 1 = [m, b], where m = ( + b)/2 is the midpoint of I. Similrly, we split I 0 into I 00 nd I 01 using the midpoint of I 0, nd I 1 into I 10 nd I 11. Continuing 26
in this mnner, we construct countbly infinite system of intervls I η (η is finite sequence of 0 nd 1) such tht [, b] = l(η)=n I η nd l(i η ) = b, if η hs length n. 2n For ech x A, we cn find n such tht (b )/2 n < ϕ(x), nd therefore we cn find sequence η = η(x) of length n such tht I η (x ϕ(x), x + ϕ(x)). We let t η = x. Let (η k ) be n enumertion of the sequences η we hve used, mking sure tht no sequence η k extends η n, for k > n. Then the intervls I k = I ηk re non-overlpping (by removing the extensions, we mde sure tht no intervl I k in the list is contined in n intervl I n for n < k) nd their union contins A. Letting t k = t ηk, ll the requirements re stisfied. The context is specified by A. Let ϕ be positive function s in the proof of Theorem 7, i.e., every prtition (P, T ) subordinte to ϕ is superfine, nd hence (χa ; P, T ) χ A (x) dx = m(a). Let (I k ) nd (t k ) correspond to ϕ s in the first prgrph. It is cler tht, for ech n N, (I k ) n k=1 nd (t k) n k=1 determine prtilly tgged subprtition (P n, T n ) of [, b] subordinte to ϕ (hence superfine). We write I k = [x k, x k+1 ]. Then n l(i k ) = (χ A ; P n, T n ) (since ech t k A) k=1 = n k=1 xk+1 χ A (x) dx (Exercise 3) x k n m(a I k ) m(a). k=1 Now A I k, nd therefore l(ik ) m(a). Let ε > 0 be observble. To complete the proof, it remins only to replce ech closed intervl I k = [x k, x k+1 ] by slightly lrger open intervl J k = (x k ε, x 2 k+2 k+1 + ε ), nd let O = J 2 k+2 k. Then A O nd m(o) m(j k ) = l(i k ) + ε m(a) + ε. 2k+1 This is true for ech observble ε > 0, so for ech ε > 0 by closure. Theorem 30 If m(n) = 0, then N is null set. If {N k } k=1 re null sets, then k=1 N k is null set. Proof: This is corollry of the preceding theorem. 27
This result estblishes tht our mesurble sets coincide with the usul ones, nd tht null sets re precisely the sets of mesure 0. We give one more definition. Definition 15 A function f : [, b] R is (Lebesgue) mesurble if, for ll c < d, the set is Lebesgue mesurble. {x [, b] : c f(x) d} Every nonnegtive mesurble function is limit of n incresing sequence of simple mesurble functions, i.e., functions of the form n i χ Ai, i=1 where A i [, b] re mesurble, nd i re rel numbers (for the proof, see e.g. H. R. Royden, Rel Anlysis, Third Edition, Prentice Hll, 1988.) It is n esy exercise to show tht simple functions re Lebesgue integrble nd ( b n ) n i χ Ai (x) dx = i m(a i ). i=1 By the Monotone Convergence Theorem it follows tht nonnegtive mesurble function is integrble if nd only if there is n incresing sequence (f k ) k 1 of simple mesurble functions such tht lim f k = f k nd ( i=1 ) f k (x)ds bounded. k 1 Reders fmilir with the usul pproches to Lebesgue integrl should be ble to conclude from the bove observtions tht functions Lebesgue integrble ccording to the trditionl definitions re precisely the functions Lebesgue integrble ccording to our definition, nd the vlues of the integrls re the sme. In this book we restricted ourselves to the theory of integrtion for functions defined on bounded intervl [, b]. It is importnt to be ble to integrte functions defined on R (or subset of R), but this is now esy. We give only the key definitions. Definition 16 A function f : R R is integrble on R if there is n observble number R such tht s A set A R is mesurble if r f(x) dx R, for ll r, s. A [r, s] is mesurble, for ll r, s. 28
We leve it s chllenging exercise for the reder to formulte nd prove nlogs of the results of sections 2 nd 3 for R in plce of [, b]. The min difference is tht bounded mesurble functions on R re not necessrily integrble. There re even mesurble sets A for which the chrcteristic function is not integrble (for exmple, A = R). 4 Lebesgue Theorem We finish with clssicl result. Theorem 31 (Lebesgue Theorem) A bounded function f : [, b] R is Riemnn integrble if nd only if it is continuous lmost everywhere. We first introduce chrcteriztion of continuous functions tht is used in the proof. Definition 17 Let f : A R be function. Let be limit point of A nd let L be rel number. We sy tht L is limit vlue of f t if L f(x) for some x, x, x A. The context of the previous definition is, A, f, L. Theorem 32 Let f : A R be function, be limit point of A, nd L be rel number. The following conditions re equivlent: (1) lim x f(x) = L. (2) L is the unique limit vlue of f t nd f(x) is limited for ll x, x, x A. Proof: We show (1) implies (2). The context is given by f,, A, nd L. Assume lim x f(x) = L. Then L f(x) for ll x, x, x A. But such x exist since is limit point of A. Hence, L is limit vlue of f t nd L is limited. Let L be nother limit vlue. Consider n extended context f,, A, L, L, nd write + when working reltive to it. By definition, there is x +, x, such tht f(x ) + L. But f(x) + L lso, by Closure, so L + L. This shows tht L = L since L, L re observble reltive to f,, A, L, L. To see tht (2) implies (1), notice tht the ssumed unique limit vlue L of f t is observble reltive to f,, A, by Closure. If x, x, x A, then f(x) is limited, nd so the observble neighbor of f(x) is defined nd is lso limit vlue of f t, i.e., it is equl to L. Hence L f(x), which shows tht lim x f(x) = L. Definition 18 Let f : A R be function nd limit point of A. We let L denote the set of ll limit vlues of f t. 29
The set L is observble reltive to f,, A. Theorem 33 Let f : A R be function nd be limit point of A. The set L is closed. Proof: The context is given by f,, A. Let M be observble nd such tht M L for L L. Consider the extended context f,, A, L nd write + when working reltive to it. By definition L + f(x) for some x +, x, x A. But then M f(x) where x, x, x A, so M L. Definition 19 Let I be n intervl nd let f : I R be bounded. (1) We sy tht the function f : I R defined by f (x) = mx(l x {f(x)}), for ech x I, is the upper envelope of f. (2) We sy tht the function f : I R defined by is the lower envelope of f. f (x) = min(l x {f(x)}), for ech x I, (3) We sy tht the function ω : I R defined by is the vrition of f. ω(x) = f (x) f (x), for ech x I, We note tht the functions f, f nd ω re defined on I becuse the sets L x re closed nd bounded. Further, ll these functions nd observble reltive to f, I. Theorem 34 A bounded function f : I R is continuous t I if nd only if ω() = 0. Proof: If f is continuous t, then f() is the only limit vlue of f t, nd f () = f () = f(), so ω() = 0. Conversely, if f () = f (), then f() is the only limit vlue of f t. (Note tht f hs t lest one limit vlue t, becuse f(x) is limited, for ny x.) As f is bounded, lim x f(x) = f(). Proof of Lebesgue Theorem: Let D = {x [, b] : f is not continuous t x}. This set is observble. Assume first tht f is Riemnn integrble. Let D k = {x [, b] : ω(x) 1 }, for k N. k 30
By Theorem 34, D = k=1 D k, so it suffices to show tht ech D k is null set. The context contins f,, b nd k. Recll (AUN, Chpter 9) tht, if f is Riemnn integrble on [, b], then there exists fine prtition P such tht S(P) s(p) = (F i f i ) d 0, where F i (f i, resp.) is the supremum (infimum, resp.) of f on the intervl [, +1 ]. Let K = {i : F i f i 1 k }; s S(P) s(p) i K (F i f i ) d 1 k d, nd 1 k is observble, it follows tht i K d 0. Finlly, we note tht D k i K (, +1 ) {x 0,..., x n }. It follows tht D k is null set. For the converse, ssume tht D is null set. Let {J k } k=1 be sequence of open intervls such tht i K D k=1 J k nd l(jk ) 0. We now ugment the context by the sequence {J k } k=1, nd let (P, T ) be ny prtition of [, b] superfine reltive to this ugmented context. In the nottion of the first prt of this proof, we will show tht f i d F i d (*) This is enough (see the direction Drboux Riemnn of AUN, Theorem 139). Let I 1 = {i : t i D} nd I 2 = {i : t i / D}. By Theorem 9, for ech i I 1 there is k i such tht [, +1 ] J ki. As the intervls of the prtition re non-overlpping, we conclude tht i I 1 d k=1 l(j k) 0 (see the rgument in the proof of Theorem 19), nd hence i I 1 (F i f i ) d 0, i.e., i I 1 F i d i I 1 f i d. Now for i I 2 the function f is continuous t t i, hence f(x) f(t i ) is t i -ultrsmll (or 0) for every x [, +1 ]. This implies tht F i f i for i I 2 nd thus i I 2 F i d i I 2 f i d s usul. Together we hve (*). 31
Appendix The xioms of RBST do not give complete description of the universe of reltive set theory. (See KH, Reltive set theory: Internl view, Journ. Logic nd Anlysis 1:8, 2009, 1-108.) Other xioms of some prcticl usefulness cn be dded to RBST. We give n exmple tht is used in the proof of Theorem 17. Locl Trnsfer Principle. Let P(x 1,..., x k ; S α ) be ny sttement in the S-lnguge. If P(x 1,..., x k ; S α ) holds, then there exists γ α such tht P(x 1,..., x k ; S β ) holds for ll β with α β γ. The point is tht x 1,..., x k re rbitrry; they do not hve to be observble reltive to α! Also, for every α β there exists N N which is ultrlrge reltive to α nd observble reltive to β. 32
Answers to Exercises. Answer to Exercise 1, pge 4 (1) Let r nd s be -ccessible. Let f, g be observble nd such tht r = f() nd s = g(). Then r ± s = (f + g)(), r s = (f g)() nd r/s = (f/g)() (if s = g() 0). But f ± g, f g, nd f/g re observble by Closure, so r ± s, r s, nd r/s re -ccessible. (2) Let x nd y be -limited. Let r, s be positive -ccessible numbers such tht x r nd y s. Then x ± y x + y r + s nd r + s is -ccessible so x ± y is -limited. Also x y = x y r s, nd r s is -ccessible so x y is -limited. (3) Let h, k be -ultrsmll. Let r > 0 be -ccessible. Then r/2 is - ccessible (since 2 is -ccessible). Thus h ± k h + k r/2 + r/2 = r. So h ± k is -ultrsmll or 0. Let x be -limited. Let r > 0 be -ccessible such tht x r. Let s > 0 be -ccessible. Then r/s > 0 is -ccessible by (1). Now x h = x h r r/s = r. This shows tht x h is -ultrsmll or 0. Answer to Exercise 2, pge 5 The proof is identicl to the proof of the strddle version from the Increment Eqution, but one uses the opertions on -ultrsmlls insted. Answer to Exercise 3, pge 13 The context is given by f,, b. Let (P, T ) be superfine prtilly tgged prtition of [, b]. Let ε > 0 be observble. By Closure, we cn find positive observble δ like in the Sks-Henstock lemm. Since (P, T ) is superfine, it is subordinte to δ so f(t j ) d xj+1 f(x) dx < ɛ. Since ɛ > 0 ws rbitrry, the quntity between the bsolute vlues is ultrsmll or 0 nd so f(t j) d xj+1 f(x) dx. The second clim is proved similrly. Answer to Exercise 4, pge 18 Assume tht f + nd f re generlized Riemnn integrble. Then by linerity, f = f + f nd f = f + f re generlized Riemnn integrble. For the converse, notice tht f + = f +f 2 nd f = f f 2. Hence, if f nd f re generlized Riemnn integrble, then so re f + nd f, by linerity. Answer to Exercise 5, pge 19 This is cler since f nd f re integrble on [, b] if nd only if they re integrble on [, c] nd [c, b]. Answer to Exercise 6, pge 20 If one of the functions f n is Lebesgue integrble then they ll re by Theorem 22, so in prticulr f + n nd f n re integrble, so lso f n. But f + (x) = lim n f + n (x) nd f (x) = lim x f n (x). By the Monotone Convergence Theorem pplied to f + nd f we hve tht f + nd f re integrble, so f + nd f re integrble. Hence f is Lebesgue integrble. 33