2 Definitions and Basic Properties of Extended Riemann Stieltjes Integrals

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1 2 Definitions nd Bsic Properties of Extended Riemnn Stieltjes Integrls 2.1 Regulted nd Intervl Functions Regulted functions Let X be Bnch spce, nd let J be nonempty intervl in R, which my be bounded or unbounded, nd open or closed t either end. Recll tht n intervl is clled nondegenerte if it hs nonempty interior or equivlently contins more thn one point. Let J be the closure of J in the extended rel line [, ]. A function f on J with vlues in X is clled regulted on J, just s for rel-vlued functions in Chpter 1, if the right limit f(t+) := lim s t f(s) exists in X for t J not equl to the right endpoint of J, nd if the left limit f(t ) := lim s t f(s) exists in X for t J not equl to the left endpoint of J. If J = {} = [, ] is singleton these conditions hold vcuously nd we sy tht f is regulted. The clss of ll regulted functions on J with vlues in X will be denoted by R(J; X). Let < b throughout this prgrph (defining quntities f, f +, nd ). For regulted function f on J = [, b], define function f () (t) := f(t ) for t (, b] or f() if t = J. Similrly define f (b) + (t) := f(t+) for t [, b) or f(b) if t = b J. Define + f on J by ( + f)(t) := f(t+) f(t), clled the right jump of f t t, for ll t J except the right endpoint. Similrly, define function f on J by ( f)(t) := f(t) f(t ), clled the left jump of f t t, for ll t J except the left endpoint. Also, let ± f := + f + f, clled the two-sided jump, on the interior of J. For regulted function f on J = [, b], let + J f(t) := + f(t) if t [, b), or 0 if t = / J or t = b; J f(t) := f(t) if t (, b], or 0 if t = b / J or t = ; nd ± J f(t) := J f(t) + + J f(t) (2.1) + f() if t = J, or 0 if t = / J; = ± f(t) if t (, b); f(b) if t = b J, or 0 if t = b / J. R.M. Dudley nd R. Norviš, Concrete Functionl Clculus, Springer Monogrphs in Mthemtics, DOI / _2, Springer Science+Business Medi, LLC

2 18 2 Extended Riemnn Stieltjes Integrls The definition of ± J is mde so tht some lter formuls will not need to hve specil endpoint terms. A step function from n intervl J into Bnch spce X is finite sum m j=1 1 A(j)x j, where x j X nd A(j) re intervls, some of which my be singletons. Clerly step function is regulted. The next fct shows tht function is regulted if nd only if it cn be pproximted uniformly by step functions. Theorem 2.1. Let X be Bnch spce nd < < b < +. The following properties re equivlent for function f : J := [, b] X: () f R([, b]; X); (b) for ech ǫ > 0, there exists Young intervl prtition {(t i 1, t i )} n of [, b] such tht Osc(f; (t i 1, t i )) < ǫ for ech i {1,...,n}; (c) f is uniform limit of step functions. Proof. () (b). Let ǫ > 0. By definition of R(J; X), if J = (, b], f(+) exists, nd if J = [, b), f(b ) exists. Thus there exist 1 > nd b 1 < b such tht Osc(f; (, 1 )) < ǫ nd Osc(f; (b 1, b)) < ǫ. So it suffices to consider the cse tht [, b] is closed, bounded intervl [, b] (specificlly, [ 1, b 1 ]). For ech s (, b), choose δ s > 0 such tht A s := (s δ s, s + δ s ) [, b] nd the oscilltion of f over the open intervls (s δ s, s) nd (s, s + δ s ) is less thn ǫ. For the endpoints, choose δ nd δ b (0, b ) such tht Osc(f; (, +δ )) < ǫ nd Osc(f; (b δ b, b)) < ǫ. Letting A := [, + δ ) nd A b := (b δ b, b], the sets {A s : s [, b]} form cover of the compct intervl [, b] by reltively open sets. Therefore there is finite subcover A s0, A s1,..., A sm of [, b] with = s 0 < s 1 < < s m = b. Tke t 0 := s 0, t 1 A s0 A s1 (s 0, s 1 ), t 2 := s 1,..., t 2m 2 := s m 1, t 2m 1 A sm 1 A sm (s m 1, s m ), nd t 2m := s m. Thus (b) holds with n = 2m + 1. (b) (c). Given ǫ > 0, choose Young intervl prtition {(t i 1, t i )} n of J s in (b). Define step function f ǫ on J by f ǫ (t) := f(s i ) with s i (t i 1, t i ) if t (t i 1, t i ) for some i {1,...,n} nd f ǫ (t i ) := f(t i ) for i {0,..., n} if t i J. Then f ǫ (t) f(t) < ǫ for ech t J. Since ǫ is rbitrry, (c) follows. (c) (). Given ǫ > 0, choose step function f ǫ such tht f ǫ (t) f(t) < ǫ for ech t J. Then for ny s, t J, f(t) f(s) < 2ǫ + f ǫ (t) f ǫ (s). Since f ǫ is regulted, the right side cn be mde rbitrrily smll for ll s, t close enough from the left or right to ny given point of J, proving (). The proof of Theorem 2.1 is complete. The following is n esy consequence of the preceding theorem. Corollry 2.2. If f is regulted on [, b] then f is bounded, Borel mesurble, nd for ech ǫ > 0, { } crd u [, b]: either + [[,b]] f(u) > ǫ or [[,b]] f(u) > ǫ <.

3 2.1 Regulted nd Intervl Functions 19 Proof. By impliction () (b) of Theorem 2.1, there is Young intervl prtition {(t i 1, t i )} n of [, b] such tht Osc(f; (t i 1, t i )) 1 for ech i = 1,...,n. Let t [, b] nd let i be such tht t = t i or t (t i 1, t i ). Then f(t) 1 + mx 1 i n f(t i 1 +) + mx 0 i n f(t i ), proving the first prt of the conclusion. The third prt lso follows from Theorem 2.1(b) since ech jump cn be pproximted rbitrrily closely by increments of f. It then follows tht f is continuous except on countble set, nd so it is Borel mesurble, completing the proof. Intervl functions Let X be Bnch spce with norm, let J be nonempty intervl in R, possibly unbounded, nd let I(J) be the clss of ll subintervls of J. Any function µ: I(J) X will be clled n intervl function on J. The clss of ll X-vlued intervl functions on J is denoted by I(J; X), nd it is denoted by I(J) if X = R. An intervl function µ on J will be clled dditive if µ(a B) = µ(a)+µ(b) whenever A, B I(J) re disjoint nd A B I(J). If µ is n dditive intervl function then clerly µ( ) = 0. An dditive intervl function µ on J is uniquely determined by its restriction to the clss I os (J), i.e. the clss of ll open subintervls nd singletons of J. For intervls s for other sets, A n A will men A 1 A 2 nd n=1 A n = A, while A n A will men A 1 A 2 nd n=1 A n = A, nd A n A will men 1 An (t) 1 A (t) s n for ll t R. Definition An intervl function µ on J will be clled upper continuous if µ(a n ) µ(a) for ny A, A 1, A 2,... I(J) such tht A n A. 2. An intervl function µ on J will be clled upper continuous t if µ(a n ) µ( ) for ny A 1, A 2,... I(J) such tht A n. If J is singleton then the conditions of Definition 2.3 hold vcuously nd ech intervl function on J is upper continuous s well s upper continuous t. In this section we estblish reltions between the clss of dditive upper continuous intervl functions on J nd clsses of regulted functions on J. But first, here is n exmple of n intervl function which is not upper continuous. Exmple 2.4. Let J be nondegenerte intervl. The Bnch spce of ll rel-vlued nd bounded functions on J with the supremum norm is denoted by l (J). Recll tht for Lebesgue mesure λ on J, L (J, λ) is the Bnch spce of λ-equivlence clsses of λ-essentilly bounded functions with the essentil supremum norm. For n intervl A J, let µ(a) := 1 A. Then for 1 A s member of l (J) or of L (J, λ), µ is n dditive intervl function on J, but not upper continuous t. For regulted function h on [, b] with vlues in X there is corresponding dditive intervl function µ h := µ h,[[,b]] on [, b] defined by

4 20 2 Extended Riemnn Stieltjes Integrls µ h,[[,b]] ((u, v)) := h(v ) h(u+) for u < v b, µ h,[[,b]] ({u}) := ± [[,b]] h(u) for u [, b], (2.2) if < b, nd µ h,[,] ({}) := µ h,[,] ( ) := 0 if = b. For < b we hve µ h,(,b] ({b}) = ± (,b] h(b) = h(b). It follows tht if h on (, b] is rightcontinuous on (, b) then for c < d b we hve µ h ((c, d]) = h(d) h(c). Thus the definition (2.2) of µ h just given extends the one given erlier for the definition (1.6) of the Lebesgue Stieltjes integrl. Notice tht µ h does not depend on the vlues of h t its jump points (points where it hs non-zero left or right jump) in (, b). In the converse direction, for ny intervl function µ on [, b], define two functions R µ, nd L µ, on [, b] by { µ( ) if t = [, b] R µ, (t) := µ([, t]) if t (, b], { µ([, t)) if t [, b) L µ, (t) := µ([, b]) if t = b [, b]. (2.3) (Recll tht if µ is dditive, µ( ) = 0.) If µ is upper continuous then R µ, nd L µ, re both regulted point functions on [, b], s will be shown in Proposition 2.6 when < b. The converse is not true s the following shows: Exmple 2.5. Let < b, µ([u, v)) := µ((u, v]) := 0 nd µ([u, v]) := 1 for u v b, nd µ((u, v)) := 1 if lso u < v. Then µ is n dditive intervl function on [, b], R µ, = 1 (,b] nd L µ, = 1 {b} re regulted, but µ is not upper continuous t. For h = R µ, or L µ,, µ µ h. For x J, we will sy tht the singleton {x} is n tom of µ if µ({x}) 0. The following gives chrcteriztion of dditive upper continuous intervl functions on J. In prticulr, such intervl functions cnnot hve more thn countbly mny toms. Proposition 2.6. Let J be nondegenerte intervl nd let µ I(J; X) be dditive. The following five sttements re equivlent: () µ is upper continuous; (b) µ is upper continuous t ; (c) µ(a n ) 0 whenever open intervls A n, nd crd{u J : µ({u}) > ǫ} < for ll ǫ > 0; (2.4) (d) µ(a n ) µ(a) whenever intervls A n A; (e) µ(a n ) µ(a) whenever intervls A n A. If J = [, b] then the bove sttements re equivlent to ech of the following two sttements: (f)r µ, is regulted, R µ, (x ) = µ([, x)) for x (, b], nd R µ, (x+) = µ([, x]) for x [, b);

5 2.1 Regulted nd Intervl Functions 21 (g) L µ, is regulted, L µ, (x ) = µ([, x)) for x (, b], nd L µ, (x+) = µ([, x]) for x [, b). Proof. () (b). Clerly, () implies (b). For (b) (), let intervls A n A. Then A n = B n A C n for intervls B n A C n, with B n nd C n. So µ(a n ) µ(a) by dditivity. (b) (c). The first prt of (c) is cler. For the second prt, suppose there exist ǫ > 0 nd n infinite sequence {u j } of different points of J such tht µ({u j }) > ǫ for ll j. Then there re u J nd subsequence {u j } such tht either u j u or u j u s j. In the first cse, by dditivity, µ({u j }) = µ((u, u j ]) µ((u, u j )) 0 s j, becuse both (u, u j ] nd (u, u j ). This contrdiction proves (2.4) in the cse u j u. The proof in the cse u j u is symmetric. For (c) (b), let intervls A n. Then for some u J, for ll sufficiently lrge n, either A n is left-open t u, or A n is right-open t u. Using dditivity, in ech of the two cses µ(a n ) 0 follows by (c). For (b) (e), let intervls A n A. If A = (u, v) then {u} A n {v} for n lrge enough. For such n, there re intervls C n nd D n with {u} C n A n D n {v} nd C n A n D n = A. Also for such n, we hve either C n = or C n = (u, ], nd either D n = or D n = [, v). Clerly C n. If N := {n: C n } is infinite, there is function j n(j) onto N such tht C n(j). Thus µ(c n ) 0. Similrly µ(d n ) 0. Therefore µ(a n ) = µ(a) µ(c n ) µ(d n ) µ(a). Similr rguments pply to other cses A = (u, v], [u, v), or [u, v]. Clerly, (e) implies (d). For (d) (b), let intervls A n. Then A 1 = B n A n C n for some intervls B n, C n with B n A n C n, nd B n B, C n C for some intervls B, C with A 1 = B C. Since µ is dditive, µ(a n ) 0. (b) (f) (g). The implictions (b) (f) nd (b) (g) re cler. We prove (f) (b) only, becuse the proof of (g) (b) is similr. Let intervls A n. Then for some u nd ll sufficiently lrge n, either A n is left-open t u [, b), or A n is right-open t u (, b]. By ssumption, in the first cse we hve lim n µ(a n ) = R µ, (u+) R µ, (u) = 0 nd in the second cse we hve lim n µ(a n ) = R µ, (u ) R µ, (u ) = 0. The proof of Proposition 2.6 is complete. The clss of ll dditive nd upper continuous functions in I(J; X) will be denoted by AI(J; X). For the next two theorems, recll the definition (2.2) of the intervl function µ h corresponding to regulted function h. Theorem 2.7. For ny regulted function h on nonempty intervl [, b], the intervl function µ h is in AI([, b]; X), nd the mp h µ h is liner. Proof. We cn ssume tht < b. Additivity is immedite from the definition of µ h, s is linerity of h µ h. For upper continuity it is enough to prove sttement (c) of Proposition 2.6. Let open intervls A n. Thus there exist

6 22 2 Extended Riemnn Stieltjes Integrls {u, v n : n 1} [, b] such tht for ll lrge enough n, either A n = (v n, u) with v n u or A n = (u, v n ) with v n u. Then µ h (A n ) = h(u ) h(v n +) 0 s n in the first cse nd µ h (A n ) = h(v n ) h(u+) 0 s n in the second cse. Since µ({t}) = ± [[,b]] h(t) for t [, b], crd{t [, b]: µ({t}) > ǫ} < for ll ǫ > 0 by Corollry 2.2. Thus µ is upper continuous by Proposition 2.6, nd Theorem 2.7 is proved. Theorem 2.8. For n dditive µ I([, b]; X) with < b, the following sttements re equivlent: () µ is upper continuous; (b) µ = µ h on I[, b] for h = R µ, ; (b )µ = µ h on I[, b] for h = L µ, ; (c) µ = µ h on I[, b] for some h R([, b]; X) with h() = 0 if [, b] nd h(+) = 0 otherwise. Proof. () (b): suppose tht n dditive intervl function µ on [, b] is upper continuous. Let h := R µ,. Then by () (f) of Proposition 2.6, nd by the definition of R µ,, h is regulted nd right-continuous on (, b), nd µ h = µ on ll intervls [, t] nd [, t), t [, b]. Since µ nd µ h re dditive, µ = µ h on I[, b]. A proof tht () (b ) is similr. The implictions (b) (c) nd (b ) (c) re immedite. The impliction (c) () follows from Theorem 2.7, proving the theorem. By the preceding theorem, for ech µ AI([, b]; X) with < b there is regulted function h on [, b] such tht µ = µ h. The following will be used to clrify differences between ll such h. Definition 2.9. For ny intervls A J nd Bnch spce (X, ), c 0 (A) := c 0 (A, J) := c 0 (A, J; X) is the set of functions f : J X such tht for some sequence {t n } n=1 A, f(t n) 0 s n nd f(t) = 0 if t t n for ll n. Notice tht if < b nd f c 0 ((, b), [, b]; X)) then f is regulted on [, b] with f () f (b) + 0 on [, b]. Proposition Let g, h R([, b]; X). Then µ g = µ h on I[, b] if nd only if for some constnt c nd ψ c 0 ((, b)) = c 0 ((, b), [, b]; X), g h = c + ψ on [, b]. Moreover, if µ g = µ h on I[, b] nd g() = h() then c = 0. Proof. We cn ssume tht < b. Since ψ(s+) = ψ(t ) = 0 for s < t b, the if prt holds. For the converse impliction suppose tht µ g = µ h. Let D g be the set of ll points t (, b) such tht either g(t) 0 or + g(t) 0. Define D h similrly. Let D := D g D h nd let c := g() h(). Define ψ on [, b] by ψ(t) := g(t) h(t) c for t D nd 0 elsewhere. Since µ g = µ h on I[, b], g(u ) = c + h(u ) nd g(v+) = c + g(v+) for v < u b. Thus

7 2.1 Regulted nd Intervl Functions 23 g h = c on [, b]\d nd g h = c+ψ on D. By Corollry 2.2, ψ c 0 ((, b)), completing the proof of the converse impliction. The regulted function h in Theorem 2.8(c) cn be chosen uniquely if it stisfies dditionl properties. For < b, let D([, b]; X) be the set of ll h R([, b]; X) such tht h is right-continuous on (, b) nd either h() = 0 if [, b] or h(+) = 0 if [, b]. Thus h D([, b]; X) need not be rightcontinuous t, just s R µ, is not if {} is n tom for µ AI([, b]; X). Corollry Let < b nd let X be Bnch spce. The mppings nd D([, b]; X) h µ := µ h AI([, b]; X) (2.5) AI([, b]; X) µ h := R µ, D([, b]; X) (2.6) re one-to-one liner opertors between the vector spces AI([, b]; X) nd D([, b]; X). Moreover, the two mppings re inverses of ech other. Proof. If h D([, b]; X) then µ h AI([, b]; X) by Theorem 2.8(c) (). If µ g = µ h on I[, b] for some g, h D([, b]; X) then h(x) g(x) = µ h ([, x]) µ g ([, x]) for ech x (, b], nd so g h on [, b]. Thus the mpping (2.5) is one-to-one. It clerly is liner. If µ AI([, b]; X) then R µ, D([, b]; X) by Proposition 2.6(f). By the sme proposition, if R µ, = R ν, on [, b] for some µ, ν AI([, b]; X) then µ = ν on I[, b]. Thus the mpping (2.6) lso is one-to-one nd liner, proving the first prt of the conclusion. To prove the second prt of the conclusion, first let h D([, b]; X). By (2.3), (2.2), nd Proposition 2.6(f), if t (, b] then we hve R µh,(t) = h(t). Also, if [, b] then R µh,() = 0 = h(). Thus the composition of (2.5) with (2.6) mps h D([, b]; X) into itself. Since both mps re one-to-one nd (2.6) is onto, the proof of the corollry is complete. Next is property of n upper continuous dditive intervl function similr to the property in Theorem 2.1(b) for regulted functions. For n intervl function µ on [, b] nd n intervl J [, b], let Osc(µ; J) := sup { µ(a) : A I[, b], A J }. (2.7) In the cse J = [, b], we lso write µ sup := Osc(µ; [, b]). Corollry Let X be Bnch spce, µ AI([, b]; X) with < b nd ǫ > 0. There exists Young intervl prtition {(t i 1, t i )} n of [, b] such tht Osc(µ; (t i 1, t i )) ǫ for ech i = 1,...,n. In prticulr, µ is bounded.

8 24 2 Extended Riemnn Stieltjes Integrls Proof. Let h := R µ, be the function from [, b] to X defined by (2.3). By impliction () (f) of Proposition 2.6, h is regulted function on [, b], nd so by impliction () (b) of Theorem 2.1, there exists Young intervl prtition {(t i 1, t i )} n of [, b] such tht Osc(h; (t i 1, t i )) < ǫ for ech i = 1,..., n. By impliction () (b) of Theorem 2.8, for n intervl A (t i 1, t i ), µ(a) = µ h (A), nd so µ(a) ǫ, proving the first prt of the conclusion. The second prt follows similrly becuse h is bounded by Corollry 2.2, completing the proof of the corollry. 2.2 Riemnn Stieltjes Integrls Suppose tht the bsic ssumption (1.14) holds. Let f, h be functions defined on n intervl [, b] with < b < + nd hving vlues in X, Y, respectively. If < b, given tgged prtition τ = ({t i } n i=0, {s i} n ) of [, b], define the Riemnn Stieltjes sum S RS (τ) = S RS (f, dh; τ) bsed on τ by S RS (f, dh; τ) := n f(s i ) [h(t i ) h(t i 1 ) ]. The Riemnn Stieltjes or RS integrl (RS) f dh is defined to be 0 if = b nd otherwise is defined if the limit exists in (Z, ) s (RS) f dh := lim τ 0 S RS (f, dh; τ). (2.8) The refinement Riemnn Stieltjes or RRS integrl (RRS) f dh is defined to be 0 if = b nd otherwise is defined s (RRS) f dh := lim τ S RS (f, dh; τ) (2.9) provided the limit exists in the refinement sense, tht is, (RRS) f dh := A if for every ǫ > 0 there is point prtition λ of [, b] such tht for every tgged prtition τ = (κ, ξ) such tht κ is refinement of λ, S RS (f, dh; τ) A < ǫ. Integrls dh f for both the RS nd RRS integrls re defined symmetriclly vi (1.15). If < b, the Riemnn Stieltjes sum bsed on tgged prtition τ = ({t i } n i=0, {s i} n ) of [, b], with the integrnd nd integrtor interchnged, will be denoted by S RS (dh, f; τ) := n [ h(ti ) h(t i 1 ) ] f(s i ). Then the integrl dh f is defined in the sense of RS or RRS if nd only if the limit (2.8) or (2.9), respectively, exists with S RS (f, dh; τ) replced by S RS (dh, f; τ).

9 2.2 Riemnn Stieltjes Integrls 25 Proposition The refinement Riemnn Stieltjes integrl extends the Riemnn Stieltjes integrl. Proof. For = b, both integrls re 0. For < b, nd prtition λ, refinement κ of λ hs mesh κ λ, so the conclusion holds. Exmple Let f be rel-vlued function on [, b] nd let l c := 1 [c,b] be the indictor function of [c, b] for some < c < b. For ny tgged prtition τ = ({t i } n i=0, {s i} n ) of [, b], we hve S RS(f, dl c ; τ) = f(s i ) for t i 1 < c t i nd s i [t i 1, t i ]. The integrl (RS) f dl c exists nd equls f(c) if nd only if f is continuous t c. Tking c to be prtition point, it follows tht the integrl (RRS) f dl c exists nd equls f(c) if nd only if f is left-continuous t c. Therefore (RRS) 1 [,c] dl c exists nd equls 1, while the sme integrl in the Riemnn Stieltjes sense does not exist. On the other hnd, the integrl (RRS) l c dl c lso does not exist. Two functions f, h on n intervl [, b] will be sid not to hve common discontinuities if for ech t [, b], t lest one of f nd h is continuous t t. Also, f nd h will be sid to hve no common one-sided discontinuities on [, b] if t lest one of f nd h is left-continuous t ech t (, b] nd t lest one is right-continuous t ech t [, b). Proposition Assuming (1.14), for ny functions f nd h from [, b] into X nd Y, respectively, we hve: () If (RS) f dh exists then f nd h hve no common discontinuities; (b) If (RRS) f dh exists then f nd h hve no common one-sided discontinuities. Proof. For () suppose tht the integrl (RS) f dh exists. For ech t [, b] nd prtition κ = {x i } n of [, b], there is n index i such tht t [x i 1, x i ], where we cn tke i to be unique for prtitions with rbitrrily smll mesh. Subtrcting two Riemnn Stieltjes sums bsed on the prtition κ with smll enough mesh κ nd with ll terms equl except for the ith term, one cn conclude tht [f(y i ) f(y i )] [h(x i) h(x i 1 )] is rbitrrily smll for ny y i, y i [x i 1, x i ]. Thus either f or h must be continuous t t. The proof of (b) is similr except tht one needs to restrict to prtitions contining point t [, b]. When f or h is continuous, the RS nd RRS integrls coincide if either exists, s we will show in Theorem Some clssicl sufficient conditions for existence of the two integrls re expressed using the totl vrition or the semivrition of the integrtor. Let X, Y, nd Z be Bnch spces relted by mens of bounded biliner opertor B from X Y into Z s in ssumption

10 26 2 Extended Riemnn Stieltjes Integrls (1.14) nd let h: [, b] Y with < < b < +. The semivrition of h on [, b] is defined by { n w(h; [, b]) :=w B (h; [, b]) := sup x i [h(t i ) h(t i 1 )] : {t i } n i=0 PP[, b], {x i} n } X, mx x i 1, i (2.10) where PP[, b] is the set of ll point prtitions of [, b]. Then h is sid to be of bounded semivrition on [, b] if w B (h; [, b]) < +. As the nottion w B (h; [, b]) indictes, the semivrition depends on the Bnch spces X, Y, nd Z nd the biliner opertor B: (x, y) x y. Recll tht the totl vrition of h on [, b] is the 1-vrition of h on [, b], tht is, v(h; [, b]) := sup { s 1 (h; κ): κ PP[, b] } { n } = sup h(t i ) h(t i 1 ) : {t i } n i=0 PP[, b]. The totl vrition v(h; [, b]) does not depend on X, Z or B. It is cler tht w B (h; [, b]) v(h; [, b]) +. Also, if X = Y, Z = R nd B(x, y) = x(y), then w B (h; [, b]) = v(h; [, b]). However in generl, it is possible tht w B (h; [, b]) < while v(h; [, b]) = +, s the following exmple shows: Exmple For ech t [0, 1], let h(t) := 1 [0,t]. Then h is function from [0, 1] into Y = Z = l [0, 1], the Bnch spce of rel-vlued nd bounded functions on [0, 1] with the supremum norm. For point prtition {t i } n i=0 of [0, 1] nd for ny finite sequence {x i } n X = R of rel numbers in [ 1, 1], we hve n x i [h(t i ) h(t i 1 )] = sup n x i 1 (ti 1,t i] = mx x i 1. sup i Thus h hs bounded semivrition on [0, 1], with w(h; [0, 1]) = 1. Also, h hs unbounded p-vrition for ny p <. Indeed, if 0 < p < nd κ = {t i } n i=0 is point prtition of [0, 1], then s p (h; κ) = n 1 (ti 1,t i] p sup = n. Since n is rbitrry, v p (h; [0, 1]) = +. The function h is not regulted on [0, 1] since it clerly does not stisfy condition (b) of Theorem 2.1. Moreover, for rel-vlued function f on [0, 1], the integrl (RS) 0 1 f dh exists if nd only if f is continuous. Indeed, if f is continuous, then given ǫ > 0, there is δ > 0 such tht f(t) f(s) < ǫ whenever s, t [0, 1] nd s t < δ. Let τ = ({t i } n i=0, {s i} n ) be tgged prtition of [0, 1] with mesh τ < δ.

11 2.2 Riemnn Stieltjes Integrls 27 For ech t [0, 1] there is n i {1,...,n} such tht either t (t i 1, t i ], or t [0, t 1 ]. In either cse S RS (f, dh; τ)(t) f(t) = f(s i ) f(t) < ǫ, nd so (RS) 1 0 f dh exists nd equls f. Now, if f is not continuous then for some t [0, 1], the difference S RS (f, dh; τ)(t) f(t) cnnot be rbitrrily smll for ll tgged prtitions with smll enough mesh, proving the clim. Now we re prepred to prove the following: Theorem Assuming (1.14), if f or h is of bounded semivrition nd the other is regulted, nd they hve no common one-sided discontinuities, then (RRS) f dh exists. Proof. First suppose tht h hs bounded semivrition on [, b] nd f is regulted. Given ǫ > 0, by sttement (b) of Theorem 2.1, there exists prtition λ = {z l } k l=0 of [, b] such tht Osc(f; (z l 1, z l )) < ǫ for l = 1,...,k. (2.11) Since f nd h hve no common one-sided discontinuities, there exists sequence µ = {u l 1, v l : l = 1,...,k} (, b) such tht for ech l = 1,...,k, z l 1 < u l 1 < v l < z l, nd min { Osc(f; [z l 1, u l 1 ]), Osc(h; [z l 1, u l 1 ]) } < ǫ/k (2.12) min { Osc(f; [v l, z l ]), Osc(h; [v l, z l ]) } < ǫ/k. (2.13) Let τ = ({t j } m j=0, {s j} m j=1 ) be tgged refinement of λ µ. For ech j = 1,...,m, let τ j = ({t ji } nj i=0, {s ji} nj ) be ny tgged prtition of [t j 1, t j ]. Then j τ j is n rbitrry tgged refinement of τ. If t j 1 {z 0,..., z k 1 } then f(s j ) [h(t j ) h(t j 1 )] S RS (τ j ) = [ f(sj ) f(t j ) ] [h(t j ) h(t j 1 ) ] n j [ + f(tj ) f(s ji ) ] [h(t ji ) h(t j,i 1 ) ]. The first product on the right side, sy U j, nd the i = 1 term, sy V j, by (2.12), hve the bound U j + V j 4(ǫ/k)mx{ f sup, h sup }. (2.14) At most one of t j 1 nd t j is z i. If t j {z 1,..., z k } then f(s j ) [h(t j ) h(t j 1 )] S RS (τ j ) = [ f(sj ) f(t j 1 ) ] [h(t j ) h(t j 1 ) ] n j [ + f(tj 1 ) f(s ji ) ] [h(t ji ) h(t j,i 1 ) ].

12 28 2 Extended Riemnn Stieltjes Integrls The first product on the right side nd the i = n j term hve the sme bound s in (2.14) by (2.13). We pply the preceding 2k representtions with the bounds (2.14). We bound the sum of the remining terms for ll j by mx l=1,...,k Osc(f; (z l 1, z l ))w B (h; [, b]), nd using (2.11) we get the bound SRS (τ) S RS ( j τ j ) < ǫwb (h; [, b]) + 8ǫ mx{ f sup, h sup }. Since ǫ > 0 is rbitrry, the integrl (RRS) f dh exists. Now suppose tht f hs bounded semivrition nd h is regulted. Given ǫ > 0, choose prtition λ = {z l } k l=0 of [, b] such tht (2.11) with h insted of f holds. Then choose sequence µ = {u l 1, v l : l = 1,..., k} such tht (2.12) nd (2.13) hold. Agin let τ = ({t j } m j=0, {s j} m j=1 ) be tgged refinement of λ µ, nd for ech j = 1,..., m, let τ j = ({t ji } nj i=0, {s ji} nj ) be tgged prtition of [t j 1, t j ]. If t j {z 1,..., z k }, then summing by prts we hve f(s j ) [h(t j ) h(t j 1 )] S RS (τ j ) = [ f(sj ) f(s j1 ) ] [h(t j ) h(t j 1 ) ] + n j 1 [ f(sj,i+1 ) f(s ji ) ] [h(t ji ) h(t j ) ]. If lso t j 1 {z 0,...,z k 1 }, then for the first product on the right side, sy T j, by (2.12), we hve the bound T j 2(ǫ/k)mx{ f sup, h sup }. (2.15) The norm of the sum of the terms T j for which neither t j 1 nor t j is z i hs the bound ǫw B (f; [, b]) by (2.11) with h insted of f. If t j {z 1,..., z k }, then summing by prts we hve f(s j ) [h(t j ) h(t j 1 )] S RS (τ j ) = [ f(sj ) f(s j,nj ) ] [h(t j ) Ah(t j 1 ) ] + n j 1 [ f(sj,i+1 ) f(s ji ) ] [h(t ji ) h(t j 1 ) ]. The first product on the right side hs the sme bound s in (2.15) by (2.13). Applying the preceding representtions for ech j = 1,...,m, using the bounds (2.15) nd (2.11) with h insted of f, we get the bound SRS (τ) S RS ( j τ j ) 4ǫ mx{ f sup, h sup } + 2ǫw B (f; [, b]). As in the first prt of the proof the integrl (RRS) f dh exists, proving the theorem.

13 2.3 The Refinement Young Stieltjes nd Kolmogorov Integrls The Refinement Young Stieltjes nd Kolmogorov Integrls Let f : [, b] X nd h R([, b]; Y ). Recll from Section 1.4 tht for point prtition {t i } n i=0 of [, b] with < b, tgged prtition ({t i} n i=0, {s i} n ) is clled Young tgged point prtition if t i 1 < s i < t i for ech i = 1,...,n. Given Young tgged point prtition τ = ({t i } n i=0, {s i} n ), define the Young Stieltjes sum S YS (f, dh; τ) bsed on τ by, reclling the definition (2.1), S YS (f, dh; τ) n ( n := f ± [,b] h) (t i ) + f(s i ) [h(t i ) h(t i 1 +) ] i=0 (2.16) = n { [f + h](t i 1 ) + f(s i ) [h(t i ) h(t i 1 +) ] } + [f h](t i ). The refinement Young Stieltjes or RYS integrl (RYS) f dh is defined s 0 if = b or s (RYS) f dh := lim τ S YS (f, dh; τ) if < b, provided the limit exists in the refinement sense. The integrl (RYS) df h is defined, if it exists, vi (1.15). Proposition Let f : [, b] X nd h R([, b]; Y ). If < b, given Young tgged point prtition τ = (κ, ξ) of [, b], the Young Stieltjes sum S YS (f, dh; τ) cn be pproximted rbitrrily closely by Riemnn Stieltjes sums S RS (f, dh; τ) bsed on tgged refinements τ of κ such tht ll tgs ξ of τ re tgs of τ. Thus, if (RRS) f dh exists then so does (RYS) b f dh, nd the two re equl. Proof. The lst conclusion holds if = b with both integrls defined s 0, so let < b. For Young tgged point prtition τ = (κ, ξ) = ({x i } n i=0, {y i} n ) of [, b], tke set µ = {u i 1, v i } n (, b) such tht x 0 < u 0 < y 1 < v 1 < x 1 < < x n 1 < u n 1 < y n < v n < x n. Let τ := (κ µ, κ ξ), where ech x i, i = 1,...,n, is the tg for both [v i, x i ] nd [x i, u i ]. For i = 1,...,n, letting u i 1 x i 1 nd v i x i, it follows tht the Riemnn Stieltjes sums S RS (f, dh; τ) converge to the Young Stieltjes sum S YS (f, dh; τ) nd the conclusions follow. Exmple As in Exmple 2.14, for < c < b, let l c be the indictor function of [c, b] nd let f be rel-vlued function, both on [, b]. For ech Young tgged point prtition τ = (κ, ξ) such tht c κ, we hve S YS (f, dl c ; τ) = f(c). Therefore the integrl (RYS) f dl c exists nd hs

14 30 2 Extended Riemnn Stieltjes Integrls vlue f(c). In prticulr, (RYS) l c dl c exists nd hs vlue 1, while the RRS integrl does not exist, s shown in Exmple More generlly, for ny rel number r, let l r c(x) := l c (x) for x c nd l r c(c) := r, so tht l 1 c l c. Then (RYS) f dlr c exists nd hs vlue f(c) for ny r. In fct, the RYS integrl f dh does not depend on the vlues of the integrtor h t its jump points in (, b). This is becuse the vlue of the Young Stieltjes sum (2.16) does not depend on the vlues of h t jump points in the open intervl (, b). The sme is not true for the RRS integrl. Indeed, by chnging the vlue of the integrtor h t jump point t of both f nd h one cn mke h discontinuous from the left or right t t nd so destroy the necessry condition of Proposition The RS integrl, like the RYS integrl, does not depend on vlues of the integrtor t its discontinuity points. This holds becuse the RS integrl is defined only under restrictive conditions. Specificlly, by Proposition 2.15, if (RS) f dh exists nd h hs jump t t (, b) then f must be continuous t t. For u R, let h u := h on [, t) (t, b] nd h u (t) := u. Then for two Riemnn Stieltjes sums we hve { [f(s S RS (f, dh; (κ, ξ)) S RS (f, dh u ; (κ, ξ)) = ) f(s )] [h(t) u] if t κ, 0 if t κ, where s [t δ, t] nd s [t, t + δ] for δ := κ. Due to continuity of f t t the preceding difference cn be mde rbitrrily smll if the mesh κ is smll enough. Theorem Assuming (1.14), if f or h is of bounded semivrition nd the other is regulted, then (RYS) f dh exists. Proof. We cn ssume < b. First suppose tht h hs bounded semivrition on [, b]. Given ǫ > 0, by sttement (b) of Theorem 2.1, there exists prtition λ = {z l } k l=0 of [, b] such tht Osc(f; (z l 1, z l )) < ǫ for l = 1,...,k. (2.17) Let τ = ({t j } m j=0, {s j} m j=1 ) be Young tgged point prtition which is refinement of λ. For ech j = 1,...,m, let τ j = ({t ji } nj i=0, {s ji} nj ) be Young tgged point prtition of [t j 1, t j ]. Then j τ j is Young tgged refinement of τ nd S YS (f, dh; τ) S YS (f, dh; j τ j ) = m j=1 { nj [f(s j ) f(s ji )] [h(t ji ) h(t j,i 1 +)] n j 1 + } [f(s j ) f(t ji )] [h(t ji +) h(t ji )]. Let t j,i 1 < v j,2i 1 < v j,2i < t ji for i = 1,...,n j nd j = 1,...,m. Consider sums

15 2.3 The Refinement Young Stieltjes nd Kolmogorov Integrls 31 2n j x j,r [h(v r ) h(v r 1 )], r=2 where x j,2i = f(s j ) f(s ji ) nd x j,2i+1 = f(s j ) f(t ji ) for i = 1,...,n j except tht for i = n j, x j,2i+1 = 0. Letting v j,2i 1 t j,i 1 nd v j,2i t ji, by (2.17), we get the bound S YS (f, dh; τ) S YS (f, dh; j τ j ) < 2ǫw B (h; [, b]). Given ny two Young tgged point prtitions τ 1 nd τ 2 of [, b], there exists Young tgged refinement τ 3 of both nd S YS (τ 1 ) S YS (τ 2 ) S YS (τ 1 ) S YS (τ 3 ) + S YS (τ 3 ) S YS (τ 2 ). Thus by the Cuchy test under refinement, (RYS) f dh exists. Now suppose tht f hs bounded semivrition nd h is regulted. With the sme nottion for Young tgged point prtitions τ nd τ j, j = 1,...,m, we hve S YS (f, dh; τ) S YS (f, dh; j τ j ) m m = [f(s j ) f(s j1 )] [h(t j ) h(t j 1 +)] + d j, j=1 where for ech j = 1,...,m, d j := (f + h)(t j 1 ) + f(s j1 ) [h(t j ) h(t j 1 +)] + (f h)(t j ) S YS (f, dh; τ j ) = f(s j1 ) [h(t j ) h(t j 1 +)] f(s j,nj ) h(t j ) + f(s j1 ) h(t j 1 +) + n j 1 j=1 { } [f(t ji ) f(s ji )] h(t ji ) + [f(s j,i+1 ) f(t ji )] h(t ji +) = n j 1 { [f(t ji ) f(s ji )] [h(t ji ) h(t j )] } +[f(s j,i+1 ) f(t ji )] [h(t ji +) h(t j )]. By Theorem 2.1(b), given ǫ > 0 one cn choose prtition λ = {z l } k l=1 of [, b] such tht (2.17) with h insted of f holds. If the Young tgged point prtition τ is refinement of λ, then by pproximtions s in the first hlf of the proof, we get the bound S YS (f, dh; τ) S YS (f, dh; j τ j ) < 3ǫw B (f; [, b]). As in the first prt of the proof the integrl (RYS) f dh exists by the Cuchy test under refinement, proving the theorem.

16 32 2 Extended Riemnn Stieltjes Integrls The Kolmogorov integrl Let f be n X-vlued function on nonempty intervl J, which my be singleton, nd let µ be Y -vlued intervl function on J. For tgged intervl prtition T = ({A i } n, {s i} n ) of J, define the Kolmogorov sum S K (J, T ) = S K (f, dµ; J, T ) bsed on T by S K (f, dµ; J, T ) := n f(s i ) µ(a i ). (2.18) For n dditive intervl function µ on J nd n intervl A J, the Kolmogorov integrl = A f dµ is defined s 0 if A =, or if A is nonempty, s the limit = A f dµ := lim T S K (f, dµ; A, T ) if it exists in the refinement sense, tht is, = A f dµ = z Z if for every ǫ > 0 there is n intervl prtition B of A such tht for every refinement A of B nd every tgged intervl prtition T = (A, ξ), S K (f, dµ; A, T ) z < ǫ. Integrls of the form = A dν h re defined, if they exist, vi (1.16). If A = {} = [, ] is singleton then ({}, {}) is the only tgged intervl prtition of A, nd so the Kolmogorov integrl = A f dµ lwys exists nd equls f() µ({}). Note tht the Kolmogorov integrl over singleton need not be 0, wheres for the integrls with respect to point functions considered in this chpter, such integrls re 0. Theorem Let f be n X-vlued function on nonempty intervl J, nd let µ be Y -vlued dditive intervl function on J. For A, A 1, A 2 I(J) such tht A = A 1 A 2 nd A 1 A 2 =, = A f dµ exists if nd only if both = A1 f dµ nd = A2 f dµ exist, nd then = A f dµ = = A 1 f dµ + = A 2 f dµ. (2.19) In prticulr, if the integrl = J f dµ is defined, then I(J) A = A f dµ is Z-vlued dditive intervl function on J. Proof. We cn ssume tht A 1 nd A 2 re nonempty. Let = A f dµ be defined for given A I(J) nd let B be nonempty subintervl of A. To show tht = B f dµ is defined we use the Cuchy test. Given ny two tgged intervl prtitions T 1, T 2 of B, let T 1, T 2 be extensions of T 1, T 2, respectively, to tgged intervl prtitions of A hving the sme subintervls of A \B nd tgs for them. Then S K (B, T 1 ) S K (B, T 2 ) = S K (A, T 1 ) S K(A, T 2 ). By ssumption, the norm of the right side is smll if T 1 nd T 2 re both refinements of suitble prtition B of A. Tking the trce B B of B on B,

17 2.3 The Refinement Young Stieltjes nd Kolmogorov Integrls 33 it follows tht the norm of the left side is smll for tgged intervl prtitions T 1, T 2 which re refinements of B B. Thus the integrl = B f dµ exists by the Cuchy test. Let A, A 1, A 2 I(J) be such tht A = A 1 A 2 nd A 1 A 2 =. Let T 1 nd T 2 be tgged intervl prtitions of A 1 nd A 2, respectively. Then T := T 1 T 2 is tgged intervl prtition of A nd S K (A, T ) = S K (A 1, T 1 ) + S K (A 2, T 2 ). Thus if the two integrls on the right side of (2.19) exist then the integrl on the left side exists, nd the equlity holds. The converse follows from the first prt of the proof, proving the theorem. The following shows tht the intervl function = A f dµ, A I(J), is upper continuous if f is bounded nd µ is upper continuous in ddition to the ssumptions of the preceding theorem. Proposition Let f be bounded X-vlued function on nonempty intervl J, nd let µ be Y -vlued upper continuous dditive intervl function on J. If the integrl = J f dµ is defined, then I(J) A = A f dµ is Z- vlued upper continuous dditive intervl function on J. Proof. The intervl function = A f dµ, A I(J), is dditive by Theorem Let A 1, A 2,... I(J) be such tht A n, nd let s n A n for ech n. Since f is bounded nd µ is upper continuous, it is enough to prove tht D n := = A n f dµ f(s n ) µ(a n ) 0 s n. (2.20) Let J = [, b]. For some u J nd ll sufficiently lrge n, either A n is leftopen t u [, b) or A n is right-open t u (, b]. By symmetry consider only the first cse. Let ǫ > 0. There exists tgged intervl prtition T 0 of (u, b] such tht ny two Kolmogorov sums bsed on tgged refinements of T 0 differ by t most ǫ. Let n 0 be such tht A n0 is subset of the first intervl in T 0, nd let n n 0. Choose Kolmogorov sum bsed on tgged intervl prtition T n of A n within ǫ of = An f dµ. Let T 1 nd T 2 be two tgged refinements of T 0 which coincide with T n nd (A n, {s n }), respectively, when restricted to A n, nd re equl outside of A n. Then the norm of D n does not exceed = f dµ S K (A n, T n) + SK ((u, b], T 1 ) S K ((u, b], T 2 ) 2ǫ. A n Since this is true for ech n n 0, (2.20) holds, proving the proposition. The following is consequence of Proposition 2.6 nd the preceding proposition.

18 34 2 Extended Riemnn Stieltjes Integrls Corollry Let f be bounded X-vlued function on n intervl J := [, b] with < b, nd let µ be Y -vlued upper continuous dditive intervl function on J. If the integrl = J f dµ is defined, then for ech u, v J such tht u < v, lim = f dµ = = t v [[,v) f dµ, lim t u =[[,t]] f dµ = = f dµ, [[,t]] [[,u] lim = f dµ = = t u (u,b]] f dµ, lim t v =[[t,b]] f dµ = = f dµ. [[t,b]] [v,b]] Next we show tht in defining the Kolmogorov integrl = J f dµ with respect to n upper continuous dditive intervl function µ it is enough to tke prtitions consisting of intervls in I os (J) (open intervls nd singletons). We cn ssume tht J is nondegenerte. First consider closed intervl J = [, b] with < b. For point prtition {t i } n i=0 of [, b], the collection { } {t 0 }, (t 0, t 1 ), {t 1 }, (t 1, t 2 ),...,(t n 1, t n ), {t n } (2.21) of subintervls of [, b] is clled the Young intervl prtition of [, b] ssocited to {t i } n i=0. A tgged Young intervl prtition of [, b] is set (2.21) together with the tgs t 0, s 1, t 1, s 2,..., s n, t n, where t i 1 < s i < t i for i = 1,...,n. For nottionl simplicity the tgged Young intervl prtition will be denoted by ({(t i 1, t i )} n, {s i} n ). Now Young intervl prtition of ny nonempty intervl J is ny intervl prtition consisting of singletons nd open intervls. For closed intervl this coincides with the previous definition. A tgged Young intervl prtition of ny intervl J will be Young intervl prtition together with member (tg) of ech intervl in the prtition. A tgged Young intervl prtition of ny nondegenerte intervl is given by specifying the open intervls in it nd their tgs, just s when the intervl is closed. Let {t i } n i=0 be prtition of [u, v] with u < v, so tht u = t 0 < t 1 < < t n = v. Then the ssocited Young intervl prtition of [u, v] is the set (2.21) if [u, v] = [u, v], or is the set (2.21) except for the left nd/or right singletons {u} or {v} respectively if [u, v] is left nd/or right open intervl. Given Young intervl prtition (2.21) of [, b] nd ny subintervl [u, v] of [, b], the trce Young intervl prtition of [u, v] is the one formed by nonempty intervls obtined by intersecting ech subintervl in (2.21) with (u, v) nd djoining the endpoints {u} nd/or {v} if [u, v] is left nd/or right closed intervl. Let J = [, b] with < b, let f be n X-vlued function on J, nd let µ be Y -vlued dditive intervl function on J. For tgged Young intervl prtition T = ({(t i 1, t i )} n, {s i} n ) of J, let S YS (f, dµ; J, T ) := S K (f, dµ; J, T ) (2.22) = n n f(s i ) µ((t i 1, t i )) + f(t i ) µ({t i } J). i=0

19 2.3 The Refinement Young Stieltjes nd Kolmogorov Integrls 35 Assuming tht µ is upper continuous, this sum cn be pproximted by Riemnn Stieltjes sums s follows. Let h := R µ, be the function defined by (2.3) nd let {u i 1, v i } n (, b) be such tht t 0 < u 0 < s 1 < v 1 < t 1 < < t n 1 < u n 1 < s n < v n < t n. Letting κ := J {t i } n i=0, τ := (κ {u i 1, v i } n, κ {s i} n ) is tgged prtition of [c, d] J, where c = if J, or c = u 0 otherwise, nd d = b if b J, or d = v n otherwise, nd ech t i for i = 1,...,n 1 is the tg both for [v i, t i ] nd for [t i, u i ]. Next, s in the proof of Proposition 2.18, letting u i 1 t i 1 nd v i t i for ech i = 1,...,n, the Riemnn Stieltjes sum S RS (f, dh; τ) converges to S YS (f, dµ; J, T ). Thus we hve proved the following: Lemm For f, µ, h, J nd T s bove, the sum S YS (f, dµ; J, T ) cn be pproximted rbitrry closely by Riemnn Stieltjes sums S RS (f, dh; τ) bsed on tgged prtitions τ of subintervls of J such tht ll tgs of T re tgs of τ. Recll tht the clss of ll Y -vlued dditive nd upper continuous intervl functions on nonempty intervl J is denoted by AI(J; Y ). Now we cn show tht in defining the Kolmogorov integrl with respect to n dditive upper continuous intervl function it is enough to tke Young intervl prtitions. Proposition Let J be nondegenerte intervl, let µ AI(J; Y ), nd let f : J X. The integrl = J f dµ is defined if nd only if the limit lim T S YS (f, dµ; J, T ) exists in the refinement sense, nd then the integrl equls the limit. Proof. It is enough to prove the if prt. Let J = [, b] nd let ǫ > 0. Then there re z Z nd Young intervl prtition B of [, b] such tht for ny Young intervl prtition Y which is refinement of B nd ny tgged Young intervl prtition T = (Y, ξ), z S YS (f, dµ; J, T ) < ǫ. Let A be refinement of B consisting of rbitrry subintervls of [, b]. Let {u i } n i=0 nd {v j } m j=0 be the sets of endpoints of intervls in A nd B, respectively. For ech i {0,..., n} define sequence {t ik } k 1 s follows. If {u i } is singleton in A then let t ik := u i for ll k. If n intervl [, u i ] is in A let t ik u i, or if some [u i, ] A let t ik u i, where in either cse t ik re not toms of µ. This cn be done by sttement (c) of Proposition 2.6. For k lrge enough, = t 0k < t 1k < < t nk = b. Then there is unique Young intervl prtition Y k of [, b] ssocited to {t ik } n i=0. Since ech singleton {v j} equls {u i } A for some i, Y k is refinement of B. The intervls in Y k converge to the intervls in A s k, except for singletons in Y k which re not toms of µ, nd hence do not contribute to sums. Let η be set of tgs for A. Then ech tg s of η is eventully in the intervl of Y k corresponding to the intervl of A contining s. For such k, to form set η k of tgs for Y k, tke η nd djoin to it, for ech i (if ny exist) with {u i } / A, t ik with µ({t ik }) = 0. Thus there re Young Stieltjes sums S YS (f, dµ; J, (Y k, η k )) converging to S K (f, dµ; J, (A, η)), proving the theorem in the cse J = [, b]. If J = (, b]

20 36 2 Extended Riemnn Stieltjes Integrls nd/or J = [, b) then the proof is the sme except tht in the first cse, (u 0, c] A for some c nd we define t 0k := u 0 = for ll k, while in the second cse, [d, u n ) A for some d nd we define t nk := u n = b for ll k. The proof of Proposition 2.25 is complete. For the next sttement recll tht R µ, is defined by (2.3). Corollry Let J = [, b] with < b, let µ AI(J; Y ), let h := R µ,, nd let f : J X. Also for ech t [, b], let h(t) if t J, f(t) if t J, h(t) := h(+) if t = J, nd f(t) := 0 if t = J, (2.23) h(b ) if t = b J, 0 if t = b J. Then = J f dµ = (RYS) f d h if either side is defined. Proof. By Proposition 2.6(f), h is regulted, nd so h is well defined. By Theorem 2.8(b), µ = µ h defined by (2.2). Since there is one-to-one correspondence between tgged Young prtitions τ of [, b] nd tgged Young intervl prtitions T of J with S YS ( f, d h; τ) = S YS (f, dµ h ; J, T ), the conclusion follows by Proposition Given regulted function h on [, b], by Theorem 2.7 there is n dditive, upper continuous intervl function µ h := µ h,[,b] corresponding to h defined by (2.2) if < b or s 0 if = b. Let provided the Kolmogorov integrl is defined. b = f dh := = f dµ h,[,b], (2.24) [,b] Proposition Let h be Y -vlued regulted function on [, b], nd let f : [, b] X. Then = f dh = (RYS) b f dh if either side is defined. Proof. We cn ssume tht < b. Since there is one-to-one correspondence between tgged Young prtitions τ nd tgged Young intervl prtitions T with S YS (f, dh; τ) = S YS (f, dµ h ; [, b], T ), the conclusion gin follows by Proposition In the specil cse tht the intervl function µ is Borel mesure on [, b], we will show tht the Kolmogorov integrl = [,b] f dµ grees with the Lebesgue integrl [,b] f dµ whenever both re defined.

21 2.3 The Refinement Young Stieltjes nd Kolmogorov Integrls 37 Proposition Let µ be finite positive mesure on the Borel sets of [, b], nd let f be rel-vlued µ-mesurble function on [, b]. Then = [,b] f dµ = [,b] f dµ whenever both integrls exist. Proof. We cn ssume tht < b. Suppose tht the two integrls exist nd let ǫ > 0. Since µ is n dditive nd upper continuous intervl function on [, b], by Proposition 2.25, there is Young intervl prtition λ of [, b] such tht if ({(t i 1, t i )} n, {s i} n ) is tgged Young prtition which is refinement of λ, then n f(t i )µ(t i ) + i=0 n f(s i )µ(t n+i ) = f dµ < ǫ, (2.25) [,b] where T i := {t i } for i = 0,...,n nd T n+i := (t i 1, t i ) for i = 1,...,n. Then {T 0,..., T 2n } is decomposition of [, b] into disjoint mesurble sets. For i = 1,...,n, let m i := inf{f(s): s T n+i } nd M i := sup{f(s): s T n+i }. Letting f(s i ) m i nd f(s i ) M i in (2.25) for ech i = 1,...,n, we get tht (2.25) holds with ech f(s i ) replced by m i, or ech f(s i ) replced by M i. Therefore, it follows tht ǫ n f(t i )µ(t i ) + i=0 [,b] n i=0 n m i µ(t n+i ) = f dµ [,b] f dµ = f dµ [,b] n f(t i )µ(t i ) + M i µ(t n+i ) = f dµ ǫ. [,b] Since ǫ is rbitrry, f dµ = = [,b] f dµ. The proof of Proposition 2.28 is complete. Corollry For regulted rel-vlued function f on [, b] nd relvlued function h on [, b], right-continuous on [, b), if (LS) f dh exists then so does = f dh nd the two integrls re equl. Proof. We cn ssume tht < b. Since (LS) f dh exists, h is of bounded vrition, nd so h = h + h, where h + nd h re nondecresing. By Theorem 2.20, (RYS) f dh + nd (RYS) f dh exist. Thus by Proposition 2.27, = f dh+ nd = f dh exist. Since h is right-continuous t, µ h,[,b] ({}) = 0. The conclusion then follows from (2.24), the obvious linerity of h µ h (Theorem 2.7), nd Proposition The following is chnge of vribles theorem for Kolmogorov integrls.

22 38 2 Extended Riemnn Stieltjes Integrls Proposition For n intervl J, let θ be strictly incresing or strictly decresing homeomorphism from n intervl J onto θ(j). Let µ I(θ(J); Y ) be dditive nd upper continuous, µ θ (A) := µ(θ(a)) for A I(J), nd let f : θ(j) X. Then = θ(j) if either side is defined. f dµ = = f θ dµ θ J Proof. The sttement follows from the one-to-one correspondence between tgged intervl prtitions of J nd θ(j), nd equlity of corresponding Kolmogorov sums. The Bochner integrl Next we define the Bochner integrl of Bnch-spce-vlued function with respect to mesure nd compre it to the Kolmogorov integrl. Let S be nonempty set, A n lgebr of subsets of S, nd (X, ) Bnch spce. An A-simple function f : S X will be function f := n x i1 Ai for some x i X, A i A, nd finite n. If µ is subdditive function from A into [0, ], f will be clled µ-simple if µ(a i ) < for ech i. To define the Bochner integrl, for µ-simple function f = n x i1 Ai nd A S, let (Bo) A f dµ := n x iµ(a i A). It cn be shown tht (Bo) f dµ is well defined for µ-simple functions, just s for rel-vlued functions. If (X, d) is ny metric spce, then the function (x, y) d(x, y) is jointly continuous from X X into [0, ). If (X, d) is seprble, then d(, ) is lso jointly mesurble for the Borel σ-lgebr on ech copy of X (e.g. [53, Propositions 2.1.4, 4.1.7]). If X is not seprble, the joint mesurbility my fil (e.g. [53, 4.1, problem 11]). If (S, S, µ) is mesure spce nd (X, ) is Bnch spce, then µ-mesurble function f from S into X will be clled µ-lmost seprbly vlued if there is closed seprble subspce Y of X such tht f 1 (X \ Y ) is µ-null set. Definition Let (S, S, µ) be mesure spce nd let X = (X, ) be Bnch spce. A function f : S X will be clled Bochner µ-integrble if there exist sequence of µ-simple functions {f k } k 1 nd sequence of µ- mesurble rel-vlued functions g k f f k such tht lim k S g k dµ = 0. Theorem Let f : S X be Bochner µ-integrble. Then () f is µ-lmost seprbly vlued nd mesurble for the completion of µ; (b) For ech A S the limit lim f k dµ (2.26) k A exists nd does not depend on the choice of {f k } stisfying the definition.

23 2.3 The Refinement Young Stieltjes nd Kolmogorov Integrls 39 Proof. (): Let {f k } k 1 nd {g k } k 1 be s in Definition Tking subsequence, we cn ssume tht g k dµ < 4 k for ll k. Then there is set B S with µ(b) = 0 such tht for s / B, g k (s) 0, nd so f k (s) f(s). The set of finite rtionl liner combintions of elements in the union of rnges of ll f k is countble, so its closure is seprble subspce of X, in which f(s) tkes vlues for s / B. Thus f is µ-lmost seprbly vlued nd mesurble for the completion of µ, since ech f k is µ-mesurble, e.g. [53, Theorem 4.2.2]. Also, lim k S f f k dµ = 0, (2.27) where the Lebesgue integrls re well defined. (b): For g nd h µ-simple, nd ny mesurble A S, (Bo) A [g h] dµ g h dµ. It follows tht tht if the Bochner integrl of f exists then the S limit (Bo) A f k dµ exists nd does not depend on the sequence of µ-simple functions {f k } stisfying (2.27). So the theorem is proved. If f is Bochner µ-integrble, the limit (Bo) f dµ := lim k A A f k dµ (2.28) is clled the Bochner integrl with respect to µ of f over the set A S. Proposition Let (S, S, µ) be mesure spce nd (X, ) seprble Bnch spce. Let f be µ-mesurble function from S into X. Then f is Bochner µ-integrble if nd only if S f dµ <. Moreover, if f is Bochner µ-integrble, then (Bo) S f dµ S f dµ. (2.29) Proof. Suppose tht f is Bochner µ-integrble, nd let {f k } be sequence of µ-simple functions stisfying Definition By (2.27), for some k lrge enough, we hve f f k dµ < 1, nd then f dµ < 1 + f k dµ <. Moreover, for ech k, we hve (Bo) S fdµ (Bo) S [ f fk ] dµ + S fk dµ. Letting k on the right side, the first term tends to zero by (2.28) nd the second term tends to the right side of (2.29) by (2.27), proving (2.29). For the converse impliction, we cn ssume tht f dµ 1. Let {x j } j=1 be dense sequence in X. For k = 1, 2,..., let A k := {s S : f(s) > 1/k}. Then µ(a k ) k <. If s A k, set g k (s) := 0. If s A k, let g k (s) := x j for the lest j such tht x j f(s) 1/k. Then g k is mesurble nd g k (s) 2 f(s) for ll s S, so g k dµ 2. Also, (g k f)(s) 1/k for ll s, so by dominted convergence, with

24 40 2 Extended Riemnn Stieltjes Integrls (g k f)(s) (1/k)1 Ak (s) + 1 A c k (s) f(s) f(s) for ll s, g k f dµ 0 s k. Let h kr (s) := g k (s) if g k (s) = 0 or x j for some j r; otherwise let h kr (s) := 0. Then h kr is µ-simple function nd h kr (s) g k (s) for ll s, so by dominted convergence, h kr g k dµ 0 s r. Let f k = h kr for r lrge enough so tht h kr g k dµ < 1/k. Then f k f dµ 0 s k, so f is Bochner µ-integrble. We show next tht for regulted function the Bochner nd Kolmogorov integrls both exist nd re equl. Proposition Let X be Bnch spce, f n X-vlued regulted function on [, b], nd µ finite positive mesure on the Borel sets of [, b]. Then (Bo) [,b] f dµ nd = [,b] f dµ both exist nd re equl. Proof. We cn ssume < b. The rnge of f is seprble, so we cn ssume tht X is seprble. Since f is bounded by Corollry 2.2, the Bochner integrl (Bo) f dµ exists by Proposition The Kolmogorov integrl = [,b] f dµ exists by Theorem 2.20, Corollries 2.11 nd 2.26, nd Proposition 2.27 with Y = Z = R nd B(x, y) yx, x X, y R. To show tht the two integrls re equl, let ǫ k 0. By Theorem 2.1 nd the definition of =, there exists nested sequence {λ k } k 1 of Young intervl prtitions λ k = {(t k i 1, tk i )}n(k) of [, b] such tht for ech k 1, mx i Osc(f; (t k i 1, tk i )) < ǫ k nd for ny tgged Young prtition τ k = (λ k, {s i } n(k) ), = f dµ S YS (f, dµ; τ k ) < ǫ k. [,b] For ech k 1, fix tgged prtition τ k nd let n(k) f k := n(k) f(s k i )1 (t k i 1,t k i ) + i=0 f(t k i )1 {t k i }. (2.30) Then for ech k 1, f k is µ-simple function, [,b] f k dµ = S YS (f, dµ; τ k ) nd f f k sup < ǫ k. Thus the two integrls re equl, proving the proposition. We will need the following extension of Lebesgue s differentition theorem to the indefinite Bochner integrl. Theorem Let λ be Lebesgue mesure on [, b] nd let f : [, b] X be Bochner λ-integrble. Then for λ-lmost ll t (, b), 1 [ ] lim (Bo) f dλ (Bo) f dλ = f(t). s t t s [,t] [,s]

25 2.3 The Refinement Young Stieltjes nd Kolmogorov Integrls 41 Proof. By (2.29), it is enough to prove tht for λ-lmost ll t (, b), 1 lim f f(t) dλ = 0, (2.31) u 0 u A t where A t = [t u, t] or A t = [t, t + u]. By Theorem 2.32(), f is λ-lmost seprbly vlued. Let {x n } be countble dense set in f([, b] \ N) for some λ-null set N [, b]. By the Lebesgue differentition theorem (e.g. Theorem in Dudley [53]), we hve 1 lim u 0 u A t f xn dλ = f(t) xn for lmost ll t (, b) nd for ll n. For ny t / N such tht this holds for ll n, it follows tht 1 lim sup f f(t) 1 dλ limsup f x n dλ + x n f(t) u 0 u A t u 0 u A t = 2 x n f(t) for ll n. Given ǫ > 0, one cn choose n such tht x n f(t) < ǫ/2, proving (2.31) for λ-lmost ll t (, b). *The Brtle integrl Next we define the Brtle integrl of vector-vlued function with respect to n dditive vector mesure nd compre it to the Kolmogorov integrl. As in the bsic ssumption (1.14), let X, Y, nd Z be three Bnch spces nd B(, ): X Y Z, x y = B(x, y), bounded biliner opertor with norm 1. Let S be nonempty set, let A be n lgebr of subsets of S, nd let µ be finitely dditive function from A into Y. The semivrition of µ over set A A is defined by { } w(µ; A) := w B (µ; A) := sup x i µ(a i ), where the supremum is tken over ll prtitions {A i } of A into finitely mny disjoint members of A nd ll finite sequences {x i } X stisfying x i 1. As the nottion w B (µ; A) indictes, the semivrition depends on the Bnch spces X nd Z nd the biliner opertor B: (x, y) x y. We sy tht µ hs bounded semivrition if w B (µ) := w B (µ; S) <. The vrition of µ over set A A is defined by { } v(µ; A) := sup µ(a i ), i i

26 42 2 Extended Riemnn Stieltjes Integrls where the supremum is tken over ll prtitions {A i } of A into finitely mny disjoint members of A. (The vrition v(µ; A) does not depend on X, Z, or B.) We sy tht µ hs bounded vrition if v(µ) := v(µ; S) <. The semivrition w B (µ; ) is monotone, subdditive function on A, while the vrition v(µ; ) is monotone, dditive function on A. It is cler tht if A A then 0 w B (µ; A) v(µ; A) +. Also, if X = Y, Z = R, nd B(x, y) = x(y), then w B (µ) = v(µ). However, in generl, it is possible tht w B (µ; ) is not dditive nd tht w B (µ; A) < while v(µ; A) = +, s the following exmples show: Exmple () Let A be the lgebr generted by subintervls of [0, 1], nd let λ be Lebesgue mesure. Let µ be the extension to A of the dditive intervl function of Exmple 2.4 hving vlues in L ([0, 1], λ), tht is, for A A, µ(a) is the equivlence clss contining 1 A. Then µ is n dditive function from A to Z = Y = L. For set A A, finite prtition {A i } of A into finitely mny disjoint sets in A nd for ny finite sequence {x i } X = R of rel numbers in [ 1, 1], we hve i x i µ(a i ) = i x i 1 Ai = mx x i 1. i Thus µ hs bounded semivrition, with w(µ; A) = 1 for λ(a) > 0 nd w(µ; A) = 0 otherwise. Thus w(µ; ) is not dditive. Also, µ hs unbounded vrition. Indeed, if A A nd λ(a) > 0, then one cn tke countble prtition {B i } of A into disjoint sets ech with λ(b i ) > 0. Then for n > 1, the sequence {A i } n defined by A 1 := B 1,..., A n 1 := B n 1 nd A n := k n B k is finite prtition of A nd n n 1 µ(a i ) = 1 Bi + 1 k n B k = n. Since n is rbitrry, v(µ; A) = +. It is esy to see tht the extension to A of the intervl function µ of Exmple 2.4 with vlues in l [0, 1] hs the sme properties. (b) Let H be rel Hilbert spce, X = R, nd Y = Z = L(H, H), the Bnch spce of bounded liner opertors from H into itself, with x y := xy. Let S be set nd A n lgebr of subsets of S. Then finitely dditive projection-vlued mesure will be function from A into Y whose vlues re orthogonl projections onto closed liner subspces (which my be {0} or H) such tht if A nd B in A re disjoint, then µ(a B) = µ(a) + µ(b), where µ(a) nd µ(b) re projections onto orthogonl subspces. It is esily seen tht w(µ; A) = 1 if µ(a) is projection onto non-zero subspce, since otherwise w(µ; A) = 0. If µ(a i ) = 1 for infinitely mny disjoint subsets A i of given set A, then clerly v(µ; A) = +, so µ hs unbounded vrition. Let µ be n dditive Y -vlued function on n lgebr A of subsets of S. A function will be clled µ-simple if it is w(µ)-simple. It is convenient to define n

27 2.3 The Refinement Young Stieltjes nd Kolmogorov Integrls 43 outer form of w(µ; ) on ll subsets of S s follows. If B S, then w (µ; B) is defined to be the infimum of w(µ; A) over ll A A such tht B A. Then w (µ; ) grees with w(µ; ) on A nd w (µ; ) is monotone nd subdditive function on the clss of ll subsets of S. We sy tht subset B of S is µ-null set if w (µ; B) = 0. Notice tht this definition grees with the one given in the cse when µ is mesure on σ-lgebr. The definition of µ-essentilly bounded function (see Section 1.4) extends to the present cse using the new mening of µ-null sets. A sequence of functions {f k } k 1 from S to X will be sid to converge in outer µ-semivrition to function f : S X if for ech ǫ > 0, w (µ; B k ) 0 s k, where B k := {x S: f k (x) f(x) > ǫ}. For µ-simple function f = n x i1 Ai for some n <, x i X, nd A i S, the indefinite integrl I(f, dµ) is defined by I(f, dµ)(a) := A f dµ := n x i µ(a A i ) for ech A S. The indefinite integrl of µ-simple function is independent of the representtion in its definition. The following definition is due to Brtle [11, Definition 1]. Definition Assuming (1.14), let A be n lgebr of subsets of S nd let µ: A Y be n dditive function. A function f : S X is clled Brtle µ-integrble over S if there is sequence {f k } k 1 of µ-simple functions from S to X stisfying the following conditions: () the sequence {f k } converges in outer µ-semivrition to f; (b) the sequence {I(f k, dµ)} of indefinite integrls hs the property tht for ny ǫ > 0 there is δ > 0 such tht if A A nd w(µ; A) < δ, then I(f k, dµ)(a) < ǫ for ll k 1; (c) given ǫ > 0 there is set A ǫ A with w(µ; A ǫ ) < nd such tht if B A nd B S \ A ǫ then I(f k, dµ)(b) < ǫ for ll k 1. We show next tht uniformly for A A, the sequence {I(f k, dµ)(a)} k 1 converges to well-defined limit (B) f dµ := lim I(f k, dµ)(a), k A which will be clled the Brtle integrl of f over the set A. Proposition If f is Brtle µ-integrble then for ech A A the sequence of indefinite integrls {I(f k, dµ)(a)} k 1 in condition (b) converges in the norm of Z uniformly for A A nd the limit does not depend on the sequence of µ-simple functions {f k } k 1 stisfying the conditions of Definition Proof. If w(µ, ) 0, then I(f k, dµ) 0 nd the conclusion holds. Assume tht w(µ, ) 0. Let {f k } be sequence of µ-simple functions stisfying the

28 44 2 Extended Riemnn Stieltjes Integrls conditions of Definition 2.37 nd let ǫ > 0. Tke δ s in condition (b) nd A ǫ A with w(µ; A ǫ ) 0 s in (c). By condition () there is positive integer K ǫ such tht w(µ; U n,m ) < δ for ech n, m K ǫ, where U n,m A nd U n,m {x S : f m (x) f n (x) > (ǫ/w(µ; A ǫ ))}. Let A A nd let n, m K ǫ. Then we hve A f n dµ A f m dµ = I(f n f m, dµ)(a) I(f n f m, dµ)(a U n,m ) + I(f n f m, dµ)(a U c n,m A c ǫ) + I(f n f m, dµ)(a U c n,m A ǫ) 2ǫ + 2ǫ + (ǫ/w(µ; A ǫ ))w(µ; A U c n,m A ǫ ) 5ǫ. This proves the existence nd the uniformity of the limit. To prove the second prt of the conclusion, let {f k }, {g k } be two sequences of µ-simple functions stisfying the conditions of Definition 2.37, nd let ǫ > 0. We cn nd do ssume tht for ech k 1, f k nd g k hve constnt vlues on the sme sets, tht is for some prtition {A k i } of S, f k = i xk i 1 A nd k i g k = i yk i 1 A. For the two sequences {f k}, {g k i k }, tke the minimum of two vlues of δ s in condition (b) nd the union of two sets A ǫ with w(µ; A ǫ ) 0 s in (c). By condition (), there is n integer K ǫ > 0 such tht w(µ; U k ) < δ for ech k K ǫ, where U k A nd U k {x S: f k (x) g k (x) > ǫ/w(µ; A ǫ )}. Let A A. Then for ech k K ǫ, A f k dµ A g k dµ = I(f k g k, dµ)(a) I(f k g k, dµ)(a U k ) + I(f k g k, dµ)(a U c k A c ǫ) + I(f k g k, dµ)(a U c k A ǫ) 2ǫ + 2ǫ + (ǫ/w(µ; A ǫ ))w(µ; A U c k A ǫ ) 5ǫ. This proves the second prt of the conclusion of Proposition For the vector mesure µ of Exmple 2.36() with vlues in L ([0, 1], λ), function f : [0, 1] R is Brtle µ-integrble if nd only if there exists sequence {f k } k 1 of rel-vlued nd µ-simple functions on [0, 1] such tht f k f 0 s k. Indeed, if f k re µ-simple functions stisfying condition () of Definition 2.37, then for A A, A f k dµ = f k 1 A. Thus if f is Brtle µ-integrble then by the preceding proposition, (B) A f dµ = f1 A nd f k f 0 s k, since w(µ; A) = 0 implies λ(a) = 0. The converse is cler by checking the conditions of Definition If µ is the vector mesure A 1 A s in Exmple 2.36() but with vlues in l [0, 1], then function f : [0, 1] R is Brtle µ-integrble if nd only if

29 2.3 The Refinement Young Stieltjes nd Kolmogorov Integrls 45 f is regulted on [0, 1] s follows. The rgument of the preceding prgrph gives tht f : [0, 1] R is Brtle µ-integrble if nd only if there exists sequence {f k } k 1 of rel-vlued nd µ-simple functions on [0, 1] such tht f k f sup 0 s k, since in the present cse w(µ; A) = 0 implies tht A is empty. Moreover, ech µ-simple function is step function, nd so f is regulted on [0, 1] by Theorem 2.1. Proposition Let A be n lgebr of subsets of set S, nd let µ be Y -vlued dditive function on A with bounded semivrition. If f : S X is µ-essentilly bounded nd there is sequence of µ-simple functions converging to f in outer µ-mesure, then f is Brtle µ-integrble nd (B) f dµ f w(µ; S). (2.32) S Proof. Let {f k } k 1 be sequence of µ-simple functions which converge to f in outer µ-mesure. Let ǫ > 0, let M := f + 2ǫ, nd let N be µ-null set such tht {x S : f(x) > f + ǫ} N. Then for ech k 1, {x S : f k (x) > M} {x S : f k (x) f(x) > ǫ} N. (2.33) For ech k 1, let f M k := f k1 Ak, where A k := {x S : f k (x) M}. Then {x S : fk M (x) f(x) > ǫ} {x S : f k (x) f(x) > ǫ} {x S : f k (x) > M}. Since w (µ; ) is monotone nd subdditive, by (2.33) it then follows tht condition () of Definition 2.37 holds for the sequence {fk M} k 1 of µ-simple functions. Since the norm of ech vlue of ech µ-simple function fk M is bounded by M, it follows tht for ech k 1 nd ny A A, I(f k, dµ)(a) Mw(µ; A), (2.34) proving condition (b). Condition (c) holds trivilly becuse w(µ; S) <. Therefore the function f is Brtle µ-integrble. The bound (2.32) follows from (2.34) nd Proposition 2.38 becuse ǫ is rbitrry. The proof of the proposition is complete. Ech dditive intervl function µ: I[, b] Y extends to n dditive function µ on the lgebr A[, b] generted by subintervls of [, b]. We show next tht if µ hs bounded semivrition, then ny regulted function on [, b] is Brtle µ-integrble nd the Brtle integrl grees with the Kolmogorov integrl in this cse. Proposition Assuming (1.14), let µ be Y -vlued dditive intervl function on [, b] such tht µ is of bounded semivrition, nd let f be n X- vlued regulted function on [, b]. Then (B) [,b] f d µ nd = [,b] f dµ both exist nd re equl.

30 46 2 Extended Riemnn Stieltjes Integrls Proof. By Theorem 2.1, f is bounded nd is uniform limit of step functions, nd so it is Brtle µ-integrble by Proposition Let ǫ k 0. By Theorem 2.1 once gin, there exists nested sequence {λ k } k 1 of Young prtitions λ k = {A ki := (t k i 1, tk i )}n(k) of [, b] such tht for ech k 1, mx i Osc(f; A ki ) < ǫ k. As in the proof of Proposition 2.34, for ech k 1, fix tgged prtition τ k = ({A ki } n(k), {sk i }n(k) ) nd let f k be defined by (2.30). Then for ech k 1, f k is step function, [,b] f k d µ = S YS (f, dµ; τ k ), nd f f k sup < ǫ k. By Proposition 2.38, we hve (B) [,b] f d µ = lim k [,b] f k d µ = lim k S YS(f, dµ; τ k ). (2.35) For ny k 1 nd ny tgged refinement τ = ({A j }, {x j }) of τ k, we hve SYS (f, dµ; τ) S YS (f, dµ; τ k ) n(k) [ = f(xj ) f(s k i ) ] µ(a j ) ǫ k w( µ; [, b]). x j A ki Thus the Kolmogorov integrl = [,b] f dµ exists nd equls the Brtle integrl (2.35), completing the proof. 2.4 Reltions between RS, RRS, nd RYS Integrls By Propositions 2.13 nd 2.18, we hve (RS) f dh (RRS) f dh h R (RYS) f dh. (2.36) Here mens tht existence of the integrl to the left of it implies tht of the integrl to the right of it, with the sme vlue, nd h R mens tht the impliction holds for h regulted. In this section we consider the question under wht conditions on f nd h the two rrows in (2.36) cn be inverted. By Proposition 2.15, if the Riemnn Stieltjes integrl exists then the integrnd nd integrtor cnnot hve common discontinuities. This condition is lso sufficient for the first rrow in (2.36) to be inverted (Theorem 2.42). A necessry condition for the existence of the refinement Riemnn Stieltjes integrl is tht the integrnd nd integrtor cnnot hve common one-sided discontinuities. This condition is lso sufficient if the integrnd nd integrtor stisfy suitble p-vrition conditions, s we will show in Corollry Thus it is tempting to guess tht the second rrow in (2.36) is invertible provided f nd h hve no common one-sided discontinuities. However, Exmple 2.44

31 2.4 Reltions between RS, RRS, nd RYS Integrls 47 below shows tht this is not true in generl. We end this section with result giving generl sufficient condition for the second rrow in (2.36) to be invertible (Proposition 2.46). The following definition of full Stieltjes integrl defines it s the (RYS) integrl provided condition (b) holds. Definition Let f nd h be regulted functions on [, b] with vlues in X nd Y, respectively. We sy tht the full Stieltjes integrl (S) f dh exists, or f is full Stieltjes integrble with respect to h over [, b], if () nd (b) hold, where () (RYS) f dh exists, (b) if f nd h hve no common one-sided discontinuities, then (RRS) f dh exists. Then let (S) f dh := (RYS) f dh. By (2.36), whenever the two integrls in () nd (b) exist, they hve the sme vlue. Next we prove reltion stted bove between RS nd RRS integrls. Theorem Let f nd h be bounded functions from [, b] into X nd Y respectively. The integrl (RS) f dh exists if nd only if both (RRS) f dh exists nd f, h hve no common discontinuities. Moreover, if the two integrls exist then they re equl. Proof. Propositions 2.15 nd 2.13 imply the only if prt. For the converse impliction suppose tht the integrl (RRS) f dh exists nd the two functions f, h hve no common discontinuities. Given ǫ > 0, there exists prtition λ = {z j } m j=0 of [, b] such tht S RS(f, dh; (κ, ξ)) (RRS) f dh < ǫ/2 (2.37) for ech tgged prtition (κ, ξ) such tht κ is refinement of λ. Then choose δ > 0 such tht for ech intervl j := [z j δ, z j +δ] [, b], j = 1,...,m 1, Osc(f; j )Osc(h; j ) < ǫ/(2m). Let d := min 1 j m (z j z j 1 ) nd let τ = ({x i } n i=0, {y i} n ) be tgged prtition with mesh τ < d δ. Then ech intervl [x i 1, x i ] contins t most one point of λ. For j = 1,...,m 1, let i(j) be the index in {1,..., n 1} such tht z j (x i(j) 1, x i(j) ]. Let (κ, ξ) be the tgged prtition of [, b] obtined from τ by replcing ech tgged intervl ([x i(j) 1, x i(j) ], y i(j) ), j = 1,...,m 1, by the pir of tgged intervls ([x i(j) 1, z j ], z j ), ([z j, x i(j) ], z j ) if z j < x i(j), or by the tgged intervl ([x i(j) 1, x i(j) ], z j ) otherwise. Then κ is refinement of λ nd

32 48 2 Extended Riemnn Stieltjes Integrls S RS (f, dh; τ) S RS (f, dh; (κ, ξ)) m 1 j=1 f(yi(j) ) f(z j ) h(xi(j) ) h(x i(j) 1 ) < ǫ/2. This in conjunction with (2.37) implies tht (RS) f dh exists nd equls (RRS) f dh. The proof of Theorem 2.42 is complete. In view of the preceding theorem the notion of the full Stieltjes integrl cn be extended s follows: Corollry Let f nd h be regulted functions on [, b] with vlues in X nd Y, respectively. The full Stieltjes integrl (S) f dh exists if nd only if (), (b) of Definition 2.41 nd (c) hold, where (c) if f nd h hve no common discontinuities, then (RS) f dh exists. Moreover, whenever two or more of the integrls in (), (b), nd (c) exist, they hve the sme vlue. The following shows tht there re regulted functions f, h: [0, 1] R hving no common discontinuities such tht the integrl 0 1 f dh exists nd equls 0 s refinement Young Stieltjes integrl but it does not exist s refinement Riemnn Stieltjes integrl. Exmple The functions f nd h will be defined on [0, 1] nd equl to 0 everywhere outside countble disjoint subsets A nd B, respectively. Let B be the union of the sets {i/p: i = 1,...,p 1} over the prime numbers p 5, p {5, 7, 11,... }. Define the set A recursively long prime numbers p 5 by choosing irrtionl points p,i in ech intervl ((i 1)/p, i/p), i = 1,...,p, different from ll such points previously defined. For ech prime p 5, let f( p,2j ) := h(2j/p) := 0 nd f( p,2j 1 ) := h((2j 1)/p) := 1/ p for j = 1, 2,... nd 2j < p. The functions f nd h ech converge to 0 long ny 1 1 enumertion of the sets A, B, respectively. Thus f nd h re regulted nd hve no common discontinuities. The integrl (RYS) 0 1 f dh exists nd equls 0; in fct, ll its pproximting Young Stieltjes sums equl 0. For ny prtition of [0, 1] there re Riemnn Stieltjes sums which re 0. For the prtition κ p := {i/p} p i=0 of [0, 1], for ech prime p 5, consider the Riemnn Stieltjes sum S p := [p/2] 1 j=1 { [ f( p,2j 1 ) h [ +f( p,2j ) h ( 2j 1 ) ( 2j 2 )] h p p ( 2j ) ( 2j 1 )]} h p p = [p/2] 1. p Since S p 1/2 s p, the Riemnn Stieltjes integrl does not exist. The refinement Riemnn Stieltjes integrl lso does not exist by Theorem 2.42.

33 2.4 Reltions between RS, RRS, nd RYS Integrls 49 Lemm For regulted functions f nd h on [, b] the following two sttements hold: (1) If t ech point of [, b) t lest one of f nd h is right-continuous, then the integrl (RRS) f dh exists whenever in its definition the Riemnn Stieltjes sums converge for tgged prtitions ({x i } n i=0, {y i} n ) with tgs y i (x i 1, x i ], i = 1,...,n. (2) If t ech point of (, b], t lest one of f nd h is left-continuous, then the integrl (RRS) f dh exists whenever in its definition the Riemnn Stieltjes sums converge for tgged prtitions ({x i } n i=0, {y i} n ) with tgs y i [x i 1, x i ), i = 1,...,n. Proof. We prove only (1) becuse proof of (2) is symmetric. Thus we hve tht there is n A Z with the following property: given ǫ > 0 there exists prtition λ = {z j } m j=0 of [, b] such tht SRS (f, dh; (κ, ξ )) A < ǫ (2.38) for ech tgged prtition (κ, ξ ) = ({x i } n i=0, {y i} n ) such tht κ is refinement of λ nd y i (x i 1, x i ] for i = 1,...,n. Choose set ζ = {u j } m j=1 (, b) such tht z j 1 < u j < z j for j = 1,...,m nd mx Osc(f; [z j 1, u j ])Osc(h; [z j 1, u j ]) < ǫ/m. (2.39) 1 j m Suppose tht (κ, ξ) = ({x i } n i=0, {y i} n ) is tgged prtition of [, b] such tht κ is refinement of λ ζ. Define ξ e := {t i } n by t i := y i for i odd, t i := x i for i even. Define ξ o := {s i } n by s i := y i for i even, s i := x i for i odd. Also, let ξ r := {x i } n. Then letting S(ξ) := S RS(f, dh; (κ, ξ)), we hve S(ξ) + S(ξ r ) = S(ξ e ) + S(ξ o ). Thus S(ξ) A S(ξr ) A + S(ξe ) A + S(ξo ) A. (2.40) By (2.38) with ξ = ξ r, the first term on the right does not exceed ǫ. If ech t i in ξ e is not x i 1 then the sme bound holds for the second term on the right. Otherwise, let I e be the set of indices i {1,...,n} such tht x i 1 = y i = t i ξ e. Thus for ech i I e, i must be odd, so t i 1 = x i 1 = t i. Let J {0,..., n} be the set of indices i such tht x i λ. For i I e, if i 1 J then we cn nd do replce the two tgged intervls ([x i 2, x i 1 ], t i 1 ) nd ([x i 1, x i ], t i ) by the tgged intervl ([x i 2, x i ], x i 1 ) without chnging the Riemnn Stieltjes sum nd where now the tg is in the interior of the intervl nd the prtition is still refinement of κ. For i I e, if i 1 J then if we replce t i (= x i 1 ) by x i, we chnge the sum by t most ǫ/m for ech i by (2.39), nd since there re t most m such vlues of i, we chnge the totl sum by t most ǫ. Therefore it follows by (2.38) tht S(ξ e ) A 2ǫ. Similrly one cn show tht the lst term in (2.40) does not exceed 2ǫ, so S(ξ) A < 5ǫ. Since ǫ is rbitrry, the integrl (RRS) f dh exists nd equls A, proving the lemm.

34 50 2 Extended Riemnn Stieltjes Integrls Proposition Let regulted functions f, h on [, b] be such tht f is left-continuous on (, b] nd h is right-continuous on [, b), or vice vers. If (RYS) f dh exists then so does (RRS) f dh, nd the two integrls hve the sme vlue. Proof. We cn ssume tht neither f nor h is identiclly 0, since if one is, the conclusion holds. We prove the proposition when f is left-continuous on (, b] nd h is right-continuous on [, b). A proof of the other cse is symmetric. In tht cse by (1) of Lemm 2.45 nd since f is left-continuous on (, b], it is enough to prove the convergence of Riemnn Stieltjes sums bsed on Young tgged prtitions of [, b]. To this im we show tht differences between Riemnn Stieltjes nd Young Stieltjes sums bsed on the sme Young tgged prtition cn be mde rbitrrily smll by refinement of prtitions. Let ǫ > 0. By definition of the refinement Young Stieltjes integrl there exists prtition λ = {z j } m j=0 of [, b] such tht S YS(f, dh; τ) (RYS) f dh < ǫ (2.41) for ny Young tgged prtition τ = (κ, ξ) such tht κ is refinement of λ. Choose set ζ = {v j 1, u j : j = 1,...,m} (, b) such tht for ech j = 1,...,m, z j 1 < v j 1 < u j < z j, mx( h sup Osc(f; [u j, z j ]), f sup Osc(h; [z j 1, v j 1 ])) < ǫ/(2m). (2.42) Let (κ, ξ) = ({x i } n i=0, {y i} n ) be Young tgged prtition of [, b] such tht κ is refinement of λ ζ. Then S RS (f, dh; (κ, ξ)) S YS (f, dh; (κ, ξ)) = S 1 +S 2, where S 1 := S 2 := n [ f(yi ) f(x i ) ] [h(x i ) h(x i 1 ) ], n [ f(xi ) f(y i ) ] [h(x i ) h(x i 1 ) ]. (2.43) Let ξ = {y i }n be nother set of tgs for κ such tht x i 1 < y i < x i for i = 1,...,n. By (2.41) with τ = (κ, ξ) nd τ = (κ, ξ ), we get the bound n [ f(y i ) f(y i ) ] [h(x i ) h(x i 1 ) ] < 2ǫ. Letting y i x i for i = 1,...,n, it follows tht in (2.43), S 2 2ǫ. Let J {0,...,n} be the set of indices i such tht x i λ, nd let I := {0,..., n} \ J. To bound S 1 define prtition κ = {x i }n i=0 of [, b] by x i := x i if i J nd tke ny x i (x i, y i+1 ) if i I. Then κ is refinement of λ nd (κ, ξ) is Young tgged prtition. Let ξ = {y i }n be nother set of tgs for κ

35 2.5 The Centrl Young Integrl 51 defined by y i := y i if i J nd y i := x i if i I. By (2.41) with τ = (κ, ξ) nd τ = (κ, ξ ), we get the bound [ f(yi ) f(x i ) ] [h(x i ) h(x i 1 )] < 2ǫ. i I Letting x i x i for i I, it follows tht the sum of ll terms in S 1 with i I does not exceed 2ǫ. By (2.42), the norm of ech term of S 1 with i J does not exceed ǫ/m, nd the norm of the totl sum of such terms does not exceed ǫ. Therefore in (2.43), S 1 3ǫ. The proof of Proposition 2.46 is complete. 2.5 The Centrl Young Integrl L. C. Young [244] defined, up to endpoint terms given by Young [247], n integrl which we will cll the centrl Young integrl, or the CY integrl. The ide of the CY integrl is to use the RRS integrl, voiding its lck of definition when f nd h hve common one-sided discontinuities by tking right-continuous version of f nd left-continuous version of h, or vice vers, nd dding sums of jump terms to restore the desired vlue of f dh. For regulted functions the CY integrl extends the RYS integrl, s will be shown in Theorem 2.51, nd thus in turn the RRS integrl by Proposition Define functions f (b) + nd f() on [, b] with < b by nd f (b) + (x) := { f+ (x) := f(x+) := lim z x f(z) if x < b, f(b) if x = b, f () (x) := { f (x) := f(x ) := lim z x f(z) if < x b, f() if x =. Reclling the definitions of +,, nd ± in nd before (2.1), the CY integrl hs two equivlent forms, defined next: Definition Assuming (1.14), let f nd h be regulted functions on [, b] with vlues in the Bnch spces X, Y respectively. If < b define the Y 1 integrl (Y 1 ) f dh := (RRS) = (RRS) f (b) + dh () [,b] + [,b] f ± [,b] h f (b) + dh() [ + f + h]() (2.44) + [f h](b) (,b) + f ± h

36 52 2 Extended Riemnn Stieltjes Integrls if the RRS integrl exists nd the sum converges unconditionlly in Z s defined in Section 1.4. If = b define the Y 1 integrl s 0. Similrly, if < b define the Y 2 integrl (Y 2 ) f dh := (RRS) = (RRS) f () dh (b) + + [,b] [,b] f ± [,b] h f () dh(b) + + [f + h]() (2.45) + [ f h](b) + (,b) f ± h if the RRS integrl exists nd the sum converges unconditionlly in Z. If = b define the Y 2 integrl s 0. Integrls (Y j ) df h, j = 1, 2, re defined symmetriclly, if they exist, s in (1.15). It will be shown tht the Y 1 integrl exists if nd only if the Y 2 integrl does, nd if so, the two integrls hve the sme vlue. This will be done by giving n lterntive representtion of the two integrls in terms of the refinement Young Stieltjes integrl. To this im, for < b we define the functions f (,b) f (,b) (x) := nd f (,b) + on [, b] respectively by f() if x =, f(x ) if x (, b), f(b) if x = b, f() if x =, nd f (,b) + (x) := f(x+) if x (, b), f(b) if x = b. (2.46) Proposition Assuming (1.14) nd < b, for regulted functions f nd h on [, b] with vlues in X nd Y, respectively, (Y 1 ) f dh = (RYS) f (,b) + dh (,b) + f ± h (2.47) nd (Y 2 ) f dh = (RYS) f (,b) dh + (,b) f ± h (2.48) if either side is defined, where the right side is defined if the refinement Young Stieltjes integrl exists nd the sum converges unconditionlly in Z. Proof. We prove only (2.47) becuse proof of (2.48) is symmetric. By Propositions 2.18 nd 2.46, the refinement Riemnn Stieltjes integrl in (2.44) exists if nd only if it exists s refinement Young Stieltjes integrl, nd then both hve the sme vlue. Let τ = ({x i } n i=0, {y i} n ) be Young tgged prtition of [, b]. Then

37 2.5 The Centrl Young Integrl 53 n S YS (f (b) +, dh() ; τ) = f(y i +) [h(x i ) h(x i 1 +) ] + [ f + + h ] () n 1 [ + f+ ± h ] (x i ) = S YS (f (,b) +, dh; τ) + [ + f + h ] () [ f h ] (b). Therefore (RYS) f (b) + dh () exists if nd only if (RYS) f (,b) + dh does. Also, (2.47) holds, proving the proposition. The following sttement reltes the two refinement Young Stieltjes integrls in (2.47) nd (2.48) with the integrl (RYS) f dh. Lemm For regulted functions f nd h on [, b] with < b, the following three sttements re equivlent: () (RYS) f dh exists; (b) (RYS) f (,b) + dh nd (RYS) [f b (,b) + f] dh both exist; (c) (RYS) f(,b) dh nd (RYS) b [f f(,b) ] dh both exist. If ny one of the three sttements holds then (RYS) f dh = (RYS) = (RYS) f (,b) + dh (,b) + f ± h f (,b) dh + (,b) f ± h, (2.49) where the two sums converge unconditionlly. Proof. We prove only the impliction () (b) nd (2.49) becuse the converse impliction follows by linerity of the RYS integrl nd proof of the rest is symmetric. Thus suppose tht () holds. It is enough to show tht (RYS) [f(,b) + f] dh exists nd equls the sum in (2.49). First we show tht the sum in (2.49) converges unconditionlly. Let ǫ > 0. By definition of the refinement Young Stieltjes integrl there exists prtition λ of [, b] such tht S YS(f, dh; τ) (RYS) f dh < ǫ (2.50) for ny Young tgged prtition τ = (κ, ξ) of [, b] such tht κ is refinement of λ. Let ζ = {u j } m j=1 (, b) be set disjoint from λ. Choose refinement κ = {t i } n i=0 of λ such tht ech u j is in (t i(j) 1, t i(j) ), j = 1,...,m, for some t i(j) 1, t i(j) κ \ λ, with i(j) < i(j + 1) 1 for j = 1,..., m 1. Define

38 54 2 Extended Riemnn Stieltjes Integrls ξ = {s i }n nd ξ = {s i }n by u j = s i(j) < s i(j) < t i(j) for j = 1,...,m nd tke ny s i = s i (t i 1, t i ) for the other indices i. By (2.50) with τ = (κ, ξ ) nd τ = (κ, ξ ), we get tht m j=1 [ f(uj ) f(s i(j) )] [h(t i(j) ) h(t i(j) 1 +) ] = SYS (f, dh; (κ, ξ )) S YS (f, dh, (κ, ξ )) < 2ǫ. Letting s i(j) u j, t i(j) u j nd t i(j) 1 u j for j = 1,..., m, s is possible, it follows tht ζ + f ± h 2ǫ for ny finite set ζ (, b) disjoint from λ. Thus the sum in (2.49) converges unconditionlly in Z. Now to prove the existence of (RYS) [f b (,b) + f] dh notice tht for ny Young tgged prtition τ = ({t i } n i=0, {s i} n ) of [, b], S YS (f (,b) + f, dh; τ) + f ± h (,b) n + f(s i ) [h(t i ) h(t i 1 +) ] + (,b) n 1 + f ± h [ + f ± h](t i ). Using (2.50) nd the unconditionl convergence just proved, one cn mke the right side rbitrrily smll by wy of refinements of prtitions, proving the lemm. Now we re redy to prove the equlity of Y 1 nd Y 2 integrls. Theorem Let f R([, b]; X) nd h R([, b]; Y ). Then (Y 1 ) f dh = (Y 2 ) f dh if either side is defined. Proof. We cn ssume tht < b. Suppose tht the Y 2 integrl exists. Hence by Proposition 2.48, the refinement Young Stieltjes integrl (RYS) f (,b) dh exists nd (2.48) holds. Then by Lemm 2.49 pplied to f = f (,b), the integrl (RYS) f (,b) + dh exists nd (RYS) f (,b) dh = (RYS) f (,b) + dh (,b) ± f ± h, where the sum converges unconditionlly. Also, by linerity of unconditionl convergence, the sum (,b) + f ± h = ± f ± h f ± h (,b) (,b)

39 2.5 The Centrl Young Integrl 55 converges unconditionlly. Therefore the right side of (2.47) is defined, proving the existence of the Y 1 integrl nd the reltion, by (2.48), (Y 2 ) f dh = (RYS) = (RYS) f (,b) dh + (,b) f ± h f (,b) + dh (,b) + f ± h = (Y 1 ) f dh. A proof of the converse impliction is symmetric nd we omit it. Since the Y 1 nd Y 2 integrls hve been shown to coincide for regulted functions f nd h, if either is defined, we now define the centrl Young integrl, or CY integrl, by (CY ) f dh := (Y 1 ) f dh = (Y 2 ) The CY integrl extends the RYS integrl, s the following shows. f dh. (2.51) Theorem Assuming (1.14), for f R([, b]; X) nd h R([, b]; Y ) the following hold: () if (RYS) f dh exists then so does (CY ) b f dh, nd the two re equl; (b) letting ˆf := [f (,b) + f (,b) + ]/2 (cf. (2.46)), if < b nd (CY ) b f dh exists then so does (RYS) ˆf dh, the sum (,b) [f ˆf] ± h converges unconditionlly in Z, nd (CY ) f dh = (RYS) ˆf dh + f ˆf] (,b)[ ± h. (2.52) Proof. For (), suppose tht (RYS) f dh exists. By the impliction () (b) nd reltion (2.49) of Lemm 2.49, the right side of (2.47) is defined. Hence the CY integrl exists nd equls the RYS integrl, proving (). For (b), suppose tht (CY ) f dh exists. By Proposition 2.48, the right sides of (2.47) nd (2.48) re defined. By linerity, (RYS) ˆf dh exists nd the sum (,b) [f ˆf] ± h converges unconditionlly in Z. Then (2.52) follows by dding (2.47) nd (2.48). If function f hs vlues t its jump points in (, b) which re normlized, tht is, f = ˆf = [f (,b) +f(,b) + ]/2, then by the preceding theorem (CY ) b f dh nd (RYS) f dh both exist or both do not. In generl the equivlence is not true, s the following shows:

40 56 2 Extended Riemnn Stieltjes Integrls Proposition There exist functions f nd h on [0, 1], where f c 0 ((0, 1)) (cf. Definition 2.9) nd h is continuous, for which (Y 2 ) 0 1 f dh exists, while (RYS) 0 1 [f f (0,1) ] dh nd hence (RYS) 0 1 f dh do not exist. Proof. Let f(t) := k 1/2 if t = 1/(3k) for k = 1, 2,..., nd f(t) := 0 otherwise. Let h(1/(3k + 1)) := 0 nd h(1/(3k 1)) := k 1/2 for k = 1, 2,.... Let h(0) = h(1) := 0 nd let h be liner in between, i.e., liner on ech closed intervl where h is so fr defined only t the endpoints. Since h is continuous nd f (0) 0, (Y 2) 0 1 f dh exists nd is 0. Since f(0,1) show the nonexistence of the integrl (RYS) 0 1 f dh. Let λ = {u j } m j=0 0, it is enough to be ny prtition of [0, 1]. Tke the smllest m such tht t m := 1/(3m + 1) u 1. If κ m := λ {t m } then the contribution to ny Young Stieltjes sum bsed on κ m coming from [0, t m ] is 0. For n > m, consider prtitions κ n := κ m {1/(3k + 1), 1/(3k 1): k = m + 1,...,n}. We form Young Stieltjes sum bsed on κ n by letting it be the sme s one for κ m on [t m, 1], nd by evluting f t 1/(3k) for k = m + 1,...,n. Then the prt of our Young Stieltjes sum for κ n coming from [0, t m ] is n k=m+1 f = ( ) [ ( 1 1 h 3k + 0 n k=m+1 [ h 3k 1 ( 1 3k 2 ) ( 1 h ) h k 1/2 k 1/2 )] 3k + 1 )] ( 1 3k + 1 s n. Thus two Young Stieltjes sums for (RYS) 0 1 f dh, both bsed on refinements of λ, differ by n rbitrrily lrge mount. So (RYS) 0 1 f dh does not exist. 2.6 The Henstock Kurzweil Integrl A guge function on [, b] is ny function defined on [, b] with strictly positive vlues. Given guge function δ( ) on [, b] with < b, tgged prtition ({x i } n i=0, {y i} n ) of [, b] is δ-fine if y i δ(y i ) x i 1 y i x i y i + δ(y i ) for i = 1,...,n. Let TP(δ, [, b]) be the set of ll δ-fine tgged prtitions of [, b]. Lemm Let δ( ) be guge function defined on n intervl [, b] with < b. Then:

41 2.6 The Henstock Kurzweil Integrl 57 () TP(δ, [, b]) is nonempty, i.e. there exists tgged prtition ζ := ({x i } n i=0, {y i } n ) TP(δ, [, b]). (b) For ny finite set F [, b], ζ cn be chosen so tht F {x i } n i=0. (c) In (b), ζ cn insted be chosen so tht F {y i } n. Proof. (): The system of open intervls {(y δ(y), y + δ(y)): y [, b]} is n open cover of [, b]. Since [, b] is compct, there is finite subcover {J i := (y i δ(y i ), y i + δ(y i )): i = 1,...,n} of [, b] such tht ny n 1 of the J i do not cover [, b]. Since y i for i = 1,...,n re distinct, we cn ssume tht y 1 < < y n. Also, J i J i+1 for i = 1,...,n 1, J i J j = for i j > 1, i, j = 1,..., n, J 1 nd b J n. Hence one cn find numbers = x 0 < x 1 < < x n = b such tht x i J i J i+1 for i = 1,...,n 1 nd y i [x i 1, x i ] for i = 1,...,n. Thus ({x i } n i=0, {y i} n ) TP(δ, [, b]), proving (). (b): We cn ssume tht F (, b). Let F = {ξ j } k j=1 where < ξ 1 < < ξ k < b. Let ξ 0 := nd ξ k+1 := b. By prt (), ech intervl [ξ j, ξ j+1 ], j = 0, 1,..., k, hs δ-fine tgged prtition. Combining these, we get one for [, b], proving (b). (c): Given finite set F [, b], [, b] \ y F (y δ(y)/2, y + δ(y)/2) is finite union of closed subintervls [ j, b j ] of [, b]. Ech of these subintervls hs δ-fine tgged prtition. Also, [, b] \ j [ j, b j ] is finite union of disjoint intervls (which my be open or closed t either end). Decomposing such intervls into smller ones, we obtin set of such intervls ech contining exctly one y F nd included in (y δ(y), y + δ(y)). Putting together the tgged prtitions of ech [ j, b j ] nd the tgged intervls indexed by y F, we get δ-fine tgged prtition of [, b] in which ech y F is tg, proving (c). Definition Assuming (1.14), let f : [, b] X nd h : [, b] Y. If < b, the Henstock Kurzweil integrl, or HK integrl, (HK) f dh is defined s n I Z, if it exists, with the following property: given ǫ > 0 there is guge function δ( ) on [, b] such tht S RS (f, dh; τ) I < ǫ for ech δ-fine tgged prtition τ of [, b]. If = b, let (HK) f dh := 0. The integrl (HK) df h is defined, if it exists, vi (1.15). Proposition 2.55 (Cuchy test). Let < b, f : [, b] X, nd h : [, b] Y. The Henstock Kurzweil integrl (HK) f dh is defined if nd only if given ǫ > 0 there is guge function δ( ) on [, b] such tht for ech τ 1, τ 2 TP(δ, [, b]). S RS (f, dh; τ 1 ) S RS (f, dh; τ 2 ) < ǫ (2.53)

42 58 2 Extended Riemnn Stieltjes Integrls Proof. To prove the if prt, for ech n 1, choose guge function δ n ( ) on [, b] such tht (2.53) holds with ǫ = 1/n for ech τ 1, τ 2 TP(δ n, [, b]). Replcing δ n by min{δ 1,..., δ n }, we cn ssume tht δ 1 δ 2. For n = 1, 2,..., let S n := S RS (f, dh; τ n ) for some τ n TP(δ n, [, b]). Then {S n : n 1} is Cuchy sequence in Z, nd hence convergent to limit I. Given ǫ > 0 choose n n such tht n > 2/ǫ nd I S n < ǫ/2. Let δ := δ n. Then for ech τ TP(δ, [, b]), I SRS (f, dh; τ) I Sn + S n S RS (f, dh; τ) < ǫ/2 + 1/n < ǫ. Thus (HK) f dh is defined. Since the converse impliction is cler, the proof is complete. Proposition Let f : [, b] X nd h : [, b] Y. Let < c < b. Then (HK) f dh exists if nd only if (HK) c f dh nd (HK) b c f dh both exist, nd then we hve (HK) f dh = (HK) c f dh + (HK) c f dh. (2.54) The proof of this property is esy once we observe the following: Lemm For ny guge function δ( ) on [, b] nd c [, b], there is guge function δ ( ) on [, b] such tht δ δ nd ech δ -fine tgged prtition hs c s tg, in the interior of its intervl in the reltive topology of [, b]. Proof. Suppose tht c b nd δ( ) is guge function on [, b]. Let δ (c) := δ(c), nd let δ (x) := min{δ(x), c x /2} for x [, b] \ {c}. For δ -fine tgged prtition ({x i } n i=0, {y i} n ) of [, b], c = y i (x i 1, x i ) if c [x i 1, x i ] (, b), c = y 1 < x 1 if c =, or c = y n > x n 1 if c = b. Proof of Proposition Suppose tht I 1 := (HK) c f dh nd I 2 := (HK) c b f dh both exist. Given ǫ > 0, tke guge functions δ 1( ) on [, c] for I 1 nd ǫ/2, nd δ 2 ( ) on [c, b] for I 2 nd ǫ/2. By the preceding lemm, there is guge function δ( ) on [, b] such tht δ δ 1 on [, c), δ δ 2 on (c, b], δ(c) min{δ 1 (c), δ 2 (c)}, nd c is tg for ech δ-fine tgged prtition of [, b]. Let τ = ({x i } n i=0, {y i} n ) be δ-fine tgged prtition of [, b] nd let c = y i for some i {1,..., n}. Writing f(y i ) [h(x i ) h(x i 1 )] = f(y i ) [h(y i ) h(x i 1 )] + f(y i ) [h(x i ) h(y i )], we hve S RS (f, dh; τ) = S RS (f, dh; τ 1 ) + S RS (f, dh; τ 2 ) for suitble τ 1 TP(δ 1, [, c]) nd τ 2 TP(δ 2, [c, b]). Thus I 1 + I 2 S RS (f, dh; τ) < ǫ, proving the existence of (HK) f dh nd reltion (2.54). Now suppose tht I := (HK) f dh exists. Let δ( ) be guge function on [, b] for I nd ǫ/2. By the preceding lemm we cn nd do ssume tht

43 2.6 The Henstock Kurzweil Integrl 59 c is tg for ech δ-fine tgged prtition of [, b]. Let δ 1 ( ) nd δ 2 ( ) be the restrictions of δ( ) to [, c] nd [c, b], respectively. Let τ 1, τ 1 TP(δ 1, [, c]) nd let τ 2 TP(δ 2, [c, b]). Then τ := τ 1 τ 2 nd τ := τ 2 τ 2 re two δ-fine tgged prtitions of [, b], nd S RS (f, dh; τ 1 ) S RS(f, dh; τ 1 ) = S RS (f, dh; τ ) S RS (f, dh; τ ) < ǫ. Thus (HK) c f dh exists by the Cuchy test. Similrly it follows tht (HK) c b f dh exists. Since τ 1 τ 2 TP(δ, [, b]) for ech τ 1 TP(δ 1, [, c]) nd τ 2 TP(δ 2, [c, b]), (2.54) holds. The Henstock Kurzweil integrl shres some elementry properties with other Stieltjes-type integrls considered bove, s will be shown in Theorems 2.72 nd Essentilly the next sttement sys tht the HK integrl exists if nd only if corresponding improper integrls exist. Proposition 2.76 will show tht this property is not shred by the RRS, RYS, or CY integrl. Proposition Assuming (1.14), let f : [, b] X nd h: [, b] Y with < b. The integrl (HK) f dh is defined if nd only if the integrl (HK) x b f dh is defined for ech x (, b) nd the limit { } lim x (HK) x f dh + f() [h(x) h()] (2.55) exists. Also, the integrl (HK) f dh is defined if nd only if (HK) x f dh is defined for ech x (, b) nd the limit { x } (HK) f dh + f(b) [h(b) h(x)] lim x b exists. In either cse, the limit nd the integrl over [, b] re equl. Proof. We prove only the first prt of Proposition 2.58 becuse proof of the second prt is symmetric. Suppose tht I := (HK) f dh is defined. Then I(x) := (HK) x b f dh is defined for ech x (, b] by Proposition Given ǫ > 0, let δ( ) be guge function on [, b] such tht ny two Riemnn Stieltjes sums bsed on δ-fine tgged prtitions differ by t most ǫ. For x (, +δ()], by Proposition 2.56 nd Lemm 2.53(), there is δ-fine tgged prtition τ x of [, x] such tht the Riemnn Stieltjes sum bsed on τ x is within ǫ of (HK) x f dh. Also, let τ x be ny δ-fine tgged prtition of [x, b] nd σ x := ({, x}, ). Let τ 1 nd τ 2 be two tgged prtitions of [, b] which coincide with τ x nd σ x, respectively, when restricted to [, x], nd which both equl τ x when restricted to [x, b]. Then τ 1 nd τ 2 re δ-fine nd I I(x) f() [h(x) h()] x (HK) f dh S RS (f, dh; τ x ) + SRS (f, dh; τ 1 ) S RS (f, dh; τ 2 ) < 2ǫ.

44 60 2 Extended Riemnn Stieltjes Integrls Since x (, + δ()] nd ǫ > 0 re rbitrry, the limit (2.55) exists nd equls I. Thus the only if prt of the proposition holds. To prove the converse impliction we need the following uxiliry sttement, sometimes clled Henstock s lemm. Lemm Suppose < b nd I := (HK) f dh is defined. Given ǫ > 0, let δ( ) be guge function on [, b] for I nd ǫ, nd let ({x i } n i=0, {y i} n ) be δ-fine tgged prtition of [, b]. Then for ny subset J {1,...,n}, { f(y i ) [h(x i ) h(x i 1 )] (HK) i J xi } f dh ǫ. (2.56) x i 1 Proof. Let k :=crdj for J := {1,..., n} \ J, nd let ǫ be n rbitrry positive number. By Proposition 2.56, for ech i J, there is δ-fine tgged prtition τ i of [x i 1, x i ] such tht xi S RS(f, dh; τ i ) (HK) f dh x i 1 < ǫ /k. Let τ := i J τ i i J ({x i 1, x i }, y i ). Then τ is δ-fine tgged prtition of [, b], nd the left side of (2.56) is equl to S RS (f, dh; τ) (HK) f dh xi }, {S RS (f, dh; τ i ) (HK) f dh i J x i 1 which is less thn ǫ + ǫ. Since ǫ is rbitrry, (2.56) holds. Continution of the proof of Proposition Suppose tht the limit (2.55) exists. Let I be its vlue. Then I(x) := (HK) x b f dh is defined for ech x (, b). Given ǫ > 0, we will construct guge function δ( ) on (, b] such tht for ech x (, b), + δ(x) < x nd I(x) S RS (f, dh; τ x ) ǫ for ny δ-fine tgged prtition τ x of [x, b]. Suppose we hve such guge function δ( ) on (, b]. Define δ() so tht I I(x) f() [h(x) h()] < ǫ for ny x (, +δ()]. Let τ = ({x i } n i=0, {y i} n ) be δ-fine tgged prtition of [, b]. Then by construction of δ( ), we hve tht y 1 =, τ x1 := ({x i } n, {y i} n i=2 ) is δ-fine tgged prtition of [x 1, b], nd I SRS (f, dh; τ) I(x1 ) S RS (f, dh; τ x1 ) + I I(x1 ) f() [h(x 1 ) h()] < 2ǫ. Thus the integrl (HK) f dh is defined nd equls I.

45 2.6 The Henstock Kurzweil Integrl 61 To construct the desired guge function δ( ) on (, b], let {u n : n 0} be decresing sequence of numbers such tht u 0 = b nd lim n u n =. Given ǫ > 0, let δ 1 ( ) be guge function on [u 1, u 0 ] for (HK) u u0 1 f dh nd ǫ/2. For n 2, let δ n ( ) be guge function on [u n, u n 2 ] for (HK) u un 2 n f dh nd ǫ/2 n. For z (u 1, u 0 ] define δ(z) so tht 0 < δ(z) δ 1 (z) nd u 1 < z δ(z). For z (u n, u n 1 ] with n > 1, define δ(z) so tht 0 < δ(z) δ n (z) nd u n < z δ(z) < z + δ(z) < u n 2. Thus δ( ) is defined on (, b]. Let x (, b) nd let τ x = ({x i } m i=0, {y i} m ) be δ-fine tgged prtition of [x, b]. For ech n 1, let I n := {i: y i (u n, u n 1 ]}, nd let J n := i In [x i 1, x i ]. By construction, τ x restricted to J n is δ n -fine tgged prtition of J n. Since J 1 (u 1, u 0 ] nd J n (u n, u n 2 ) for ech n 2, by Lemm 2.59, for ech n 1, i I n { xi } ǫ f(y i ) [h(x i ) h(x i 1 )] (HK) f dh x i 1 2 n. Thus (HK) x f dh S RS (f, dh; τ x ) ǫ/2 n = ǫ. n=1 As noted erlier, this completes the proof of Proposition Proposition Let h be rel-vlued function on [, b]. If (HK) f dh exists whenever f = 1 J for n intervl J, then h is regulted on [, b]. Proof. Suppose not. Then we cn ssume tht for some u k c [, b), the h(u k ) do not converge. If h(u k ) re unbounded let ǫ = 1. If they re bounded let ǫ := 3 1 [limsup k h(u k ) liminf k h(u k )]. Let f c := 1 (c,b]. Tke guge function δ( ) on [, b] for ǫ nd (HK) f c dh. By Lemm 2.57, there exist guge function δ δ nd δ -fine tgged prtition τ of [, b] with c s the tg for n intervl [x i 1, x i ] nd c < x i. For ll k lrge enough we hve u k < x i. Then tke δ -fine tgged refinement τ k of τ where c is the tg for [x i 1, u k ], by Lemm 2.53() for [u k, x i ]. The corresponding Riemnn Stieltjes sums S RS (f c, dh; τ k ) = h(b) h(u k ) vry by more thn 2ǫ s k, contrdiction. Suppose continuous function f : [, b] R is differentible everywhere on (, b). It my be tht f is not Lebesgue integrble, e.g. if [, b] = [0, 1], f(x) = x 2 sin(π/x 2 ), 0 < x < 1, f(x) = 0 elsewhere. Between 1910 nd 1920, Denjoy nd Perron (see the Notes) defined integrls such tht for every such f, x f (t)dt = f(x) f(), x b. The next fct shows tht the lter-invented Henstock Kurzweil integrl hs this property:

46 62 2 Extended Riemnn Stieltjes Integrls Theorem Let f be continuous function from [, b) to R hving derivtive f (x) for < x < b except for t most countbly mny x. Then (HK) x f (t)dt exists nd equls f(x) f() for x < b. Proof. The sttement follows from Theorems 7.2 nd 11.1 of Gordon [84]. For nother exmple, f(x) := x 2 sin(e 1/x ), x 0, f(0) := 0, stisfies the conditions of Theorem 2.61 lthough f hs very wild oscilltions. *2.7 Wrd Perron Stieltjes nd Henstock Kurzweil Integrls Before defining the Wrd Perron Stieltjes integrl we need to define wht re clled mjor nd minor functions. Given two rel-vlued functions f, h on [, b], we cll M mjor function of f with respect to h if M() = 0, M hs finite vlues on [, b], nd for ech x [, b] there exists δ(x) > 0 such tht M(z) M(x) + f(x)[h(z) h(x)] if x z min{b, x + δ(x)}, M(z) M(x) + f(x)[h(z) h(x)] if mx{, x δ(x)} z x. (2.57) Thus for mjor function M there exists positive function δ M ( ) = δ( ) on [, b] which stisfies (2.57). Let U(f, h) be the clss of ll mjor functions of f with respect to h, nd let { inf{m(b): M U(f, h)} if U(f, h), U(f, h) := + if U(f, h) =. A function m is minor function of f with respect to h if m U( f, h). Let L(f, h) be the clss of ll minor functions of f with respect to h. Thus m L(f, h) provided m() = 0, m hs finite vlues on [, b], nd there exists positive function δ( ) = δ m ( ) on [, b] such tht m(z) m(x) + f(x)[h(z) h(x)] if x z min{b, x + δ(x)}, m(z) m(x) + f(x)[h(z) h(x)] if mx{, x δ(x)} z x. (2.58) Let L(f, h) := Then the following is true: Lemm L(f, h) U(f, h). { sup{m(b): m L(f, h)} if L(f, h), if L(f, h) =.

47 *2.7 Wrd Perron Stieltjes nd Henstock Kurzweil Integrls 63 Proof. The sttement is true if either of the sets L(f, h) or U(f, h) is empty. Suppose tht L(f, h) nd U(f, h) re nonempty. Let m L(f, h) nd M U(f, h). Also, let δ := δ m δ M nd w(x) := M(x) m(x) for x [, b]. By (2.57) nd (2.58), for ech x (, b], it follows tht w(z) w(x) if x z min{b, x + δ(x)}, w(z) w(x) if mx{, x δ(x)} z x. Therefore inf { M(x): M U(f, h) } sup { m(x): m L(f, h) } is nondecresing function of x. Then the sttement of the lemm holds becuse m() = M() = 0. Now we re redy to define the Wrd Perron Stieltjes integrl. Definition If U(f, h) = L(f, h) is finite then the common vlue will be denoted by (WPS) f dh nd clled the Wrd Perron Stieltjes integrl, or the WPS integrl. Note tht if = b then U(f, h) = L(f, h) = 0, nd so the integrl (WPS) f dh equls 0. First we show tht the Wrd Perron Stieltjes integrl extends the refinement Riemnn Stieltjes integrl. Theorem If (RRS) f dh exists then so does (WPS) f dh, nd the two re equl. Proof. We cn ssume tht < b. For u < v b, let M(u, v) be the supremum of ll Riemnn Stieltjes sums S RS (f, dh; τ) bsed on tgged prtitions τ of [u, v], with M(u, u) 0. Then M(u, y) M(u, x) + f(x)[h(y) h(x)] if u x y, M(u, y) M(u, x) + f(x)[h(y) h(x)] if u y x. (2.59) Suppose (RRS) f dh exists nd ǫ > 0. Then there exists λ = {z j : j = 0,...,m} PP[, b] such tht (RRS) f dh ǫ < S RS (f, dh; τ) < (RRS) f dh + ǫ (2.60) for ech tgged refinement τ of λ. For ech x (, b], let j(x) be the lrgest integer such tht z j(x) x. Let M() := 0, nd for ech x (, b] let j(x) M(x) := M(z j 1, z j ) + M(z j(x), x), j=1 where the sum over the empty set is 0. Define positive function δ( ) on [, b] by δ(x) := min{x z j(x), z j(x)+1 x} if x (z j(x), z j(x)+1 ), δ() := z 1,

48 64 2 Extended Riemnn Stieltjes Integrls δ(b) := b z m 1, nd δ(z j ) := min{z j z j 1, z j+1 z j } for j = 1,...,m 1. By (2.59), the functions M nd δ so defined stisfy (2.57). Hence M is mjor function of f with respect to h. By (2.60), it follows tht M(b) (RRS) f dh + ǫ. Similrly one cn define minor function m such tht m(b) (RRS) f dh ǫ. Since ǫ is rbitrrily smll, this proves the theorem. Next is the min result of this section, which sys tht the Henstock Kurzweil nd Wrd Perron Stieltjes integrls coincide. Theorem The integrl (WPS) f dh exists if nd only if (HK) b f dh exists, nd then their vlues re equl. Proof. We cn ssume tht < b. Suppose (WPS) f dh is defined. Then given ǫ > 0, there exist mjor function M U(f, h) nd minor function m L(f, h) such tht M(b) ǫ < (WPS) f dh < m(b) + ǫ. (2.61) Let δ := δ M δ m nd let τ = ({t i } n i=0, {s i} n ) be ny δ-fine tgged prtition of [, b], which exists by Lemm 2.53(). By (2.57), we hve M(t i ) M(s i ) f(s i )[h(t i ) h(s i )], M(s i ) M(t i 1 ) f(s i )[h(s i ) h(t i 1 )] (2.62) for ech i = 1,...,n. Adding up the 2n inequlities (2.62), we get the upper bound M(b) S δ (τ) for the Riemnn Stieltjes sum S δ (τ) := S RS (f, dh; τ). Similrly, using (2.58), we get the lower bound S δ (τ) m(b). By (2.61), it then follows tht (WPS) f dh ǫ < S δ (τ) < (WPS) f dh + ǫ. Thus (HK) f dh exists nd hs the sme vlue s the (WPS) integrl. Now suppose (HK) f dh is defined. Given ǫ > 0 there exists guge function δ( ) such tht for ech δ-fine tgged prtition τ, the Riemnn Stieltjes sum S δ (τ) := S RS (f, dh; τ) hs bounds (HK) f dh ǫ < S δ (τ) < (HK) f dh + ǫ. (2.63) For ech x (, b], let δ x ( ) be the guge function δ( ) restricted to the intervl [, x], nd m(x) := inf { S δx (τ): τ TP(δ x, [, x]) }, M(x) := sup { S δx (τ): τ TP(δ x, [, x]) }. Let m() = M() := 0. Then m(x) nd M(x) hve finite vlues for ech x (, b]. Let x (, b) nd z (x, x + δ(x)

49 *2.7 Wrd Perron Stieltjes nd Henstock Kurzweil Integrls 65 b]. Then for ech τ = ({ = t 0,..., t m = x}, {s i } m i=0 ) TP(δ x, [, x]), ({t 0,..., t m, z}, {s 1,...,s m, x}) TP(δ z, [, z]). Thus S δx (τ) + f(x)[h(z) h(x)] M(z) for ny τ TP(δ x, [, x]). Thus the first inequlity in (2.57) holds. Similrly one cn show tht the second one nd (2.58) lso hold. Therefore m nd M re minor nd mjor functions respectively. By Lemm 2.62, the definitions of L(f, h) nd U(f, h), nd (2.63), it then follows tht (HK) f dh ǫ m(b) L(f, h) U(f, h) M(b) (HK) f dh + ǫ. Since ǫ > 0 is rbitrry, L(f, h) = U(f, h) nd equls the HK integrl. The proof of Theorem 2.65 is complete. Next we show tht for integrtors in the clss c 0 ((, b)) = c 0 ((, b), [, b]; R) with < b (cf. Definition 2.9), the Henstock Kurzweil nd Wrd Perron Stieltjes integrls differ from the refinement Young Stieltjes, centrl Young, nd Kolmogorov integrls. Proposition For f R[, b] nd ny h c 0 ((, b)) if < b or h R[, ], i.e. h is ny function {} R if = b, (RYS) f dh = (CY ) f dh = = f dµ h,[,b] = 0. [,b] Proof. This follows from the definitions of the integrls becuse h () ± [,b] h µ h,[,b] 0 if < b nd µ h,[,] 0. h (b) + We will show in Proposition 2.68 tht the nlogous sttement for the Henstock Kurzweil integrl does not hold. Lemm For ech h c 0 ((, b)) with < b there exists rightcontinuous f R[, b] such tht f h on (, b). Proof. Let h = i c i1 {ξi}. For ech i 1, let f i (x) := 0 if x [, ξ i ) [ξ i +δ i, b] for some δ i > 0, let f i (ξ i ) := c i, nd let f i be liner on [ξ i, ξ i + δ i ]. If δ i 0 fst enough s i so tht ech (ξ i, ξ i +δ i ) contins no ξ j with c j > c i /2, then f := i f i converges uniformly on [, b] nd f hs the stted properties. Proposition There exist h c 0 ((0, 1)) nd f R[0, 1] such tht (HK) 0 1 f dh is undefined.

50 66 2 Extended Riemnn Stieltjes Integrls Proof. For k odd, k = 1, 3,...,2 n 1, n = 1, 2,..., let h(k/2 n ) := 1/2 n/3 nd h = 0 elsewhere. Then h c 0 ((0, 1)). By Lemm 2.67, tke rightcontinuous f R[0, 1] such tht f(k/2 n ) = 1/2 n/3 for k = 1, 3,...,2 n 1 nd n = 1, 2,.... Let δ( ) be ny strictly positive function on [0, 1]. By the ctegory theorem (e.g. Theorem in Dudley [53]), there is n intervl J = [c, d] [0, 1] of length ǫ for some ǫ > 0 such tht δ(y) > ǫ for dense set S of y in J, nd the endpoints of J re not dydic rtionls. So, ll tgged prtitions of J with tgs in S re δ-fine. For r = 1, 2,..., let M(r) be the set of ll binry rtionls k/2 n in J for integers n, 1 n r, nd odd integers k 1. As n +, the number of such k/2 n for fixed n is symptotic to 2 n 1 ǫ. For given r write M(r) = {x 2j 1 } m j=1 where x 1 < x 3 < < x 2m 1 for some m = m(r). For ech j {1,...,m 1}, choose x 2j (x 2j 1, x 2j+1 ) which is not dydic rtionl, nd let x 0 := c, x 2m := d, reclling tht J = [c, d]. For ech j {1,..., m}, choose y 2j 1 S (x 2j 2, x 2j 1 ) nd y 2j S (x 2j 1, x 2j ) close enough to x 2j 1 so tht [f(y 2j ) f(y 2j 1 ) f(x 2j 1 )]h(x 2j 1 ) < 1/ M(r), where M(r) = crd (M(r)). Then ζ r := ({x j } 2m j=0, {y j} 2m j=1 ) is δ-fine tgged prtition of J nd S RS (f, dh; ζ r ) m f(x 2j 1 )h(x 2j 1 ) < 1. j=1 Since the ltter sums re unbounded s r increses, so re the former Riemnn Stieltjes sums. Then combining ζ r with δ-fine tgged prtition of [0, c] (if 0 < c) nd δ-fine tgged prtition of [d, 1] (if d < 1), which exist by Lemm 2.53(), we get unbounded Riemnn Stieltjes sums for δ-fine tgged prtitions of [0, 1]. Since the guge function δ( ) on [0, 1] is rbitrry, (HK) 0 1 f dh does not exist. Under some restrictions on the integrtor, the Henstock Kurzweil integrl will be shown to extend the refinement Young Stieltjes integrl. Note tht in the following, h my be function R µ, s in Proposition 2.6(f), rightcontinuous on (, b) but not necessrily t, so the RYS integrl cn be replced by the Kolmogorov integrl ccording to Corollry Theorem Suppose f nd h re regulted on [, b], nd if < b, h is right-continuous or left-continuous on (, b). If (RYS) f dh exists then (HK) f dh exists nd hs the sme vlue. Proof. We cn ssume tht < b nd h is right-continuous on (, b) by symmetry. Given ǫ > 0, let λ = {z j } m j=0 be prtition of [, b] such tht for ny Young tgged refinement τ of λ, S YS (f, dh; τ) (RYS) f dh < ǫ. (2.64)

51 *2.7 Wrd Perron Stieltjes nd Henstock Kurzweil Integrls 67 Recll + (,b) f defined before (2.1) nd f(,b) + defined in (2.46). Then + (,b) f f (,b) + f on [, b]. By Lemm 2.49 we cn ssume tht (2.64) holds lso for f replced by f (,b) + or by + (,b) f, tking λ s common refinement λ = λ λ of prtitions λ for f (,b) + nd ǫ/2, nd λ for + (,b) f nd ǫ/2. So it suffices to prove the theorem for f (,b) + nd for + (,b) f. Define guge function δ( ) on [, b] s follows: if t λ, let δ(t) := min j t z j /2. Then ny δ( )-fine tgged prtition must contin ech z j s tg. For j = 0,...,m, define δ(z j ) such tht δ(z j ) min i j z i z j /3 nd f m j=1 { Osc ( h; [z j δ(z j ), z j ) ) + Osc ( h; (z j 1, z j 1 + 2δ(z j 1 )] )} < ǫ. (2.65) Let τ = ( {t i } n i=0, {s i} n ) be ny δ( )-fine tgged prtition of [, b] nd let S δ (τ) := S RS (f, dh; τ) = n f(s i ) [ h(t i ) h(t i 1 ) ] (2.66) be the Riemnn Stieltjes sum bsed on τ. We cn ssume tht s i = t i = s i+1 never occurs, since if it did, the ith nd (i+1)st term in S δ (τ) could be replced by f(s i )[h(t i+1 ) h(t i 1 )], nd we would still hve Riemnn Stieltjes sum bsed on δ( )-fine tgged prtition equl to S δ (τ). Since the tgs {s i } n must contin the points of λ, for ech index j {0,...,m} there is n index i(j) {1,...,n} such tht z j = s i(j). Let µ := {i(j): j = 0,...,m} {1,...,n}. For j = 1,...,m 1, by definition of δ( ), we must hve t i(j) 1 < s i(j) < t i(j). (2.67) We lso hve i(0) = 1, s 1 = z 0 = t 0 = nd i(m) = n, s n = z m = t n = b. Now consider the cse tht f is right-continuous on (, b), which holds when f = f (,b) +. We will show tht one cn find Young Stieltjes sum, bsed on refinement of λ, which is rbitrrily close to the Riemnn Stieltjes sum S δ (τ). To this im we will replce the vlues h(t i ) in S δ (τ), i = 1,...,n 1, by vlues h(x i ) t continuity points x i of h, close to t i. Then the Young Stieltjes sum terms f(x i ) ± h(x i ) will be 0. Define Riemnn Stieltjes sum T by, for ech i = 1,...,n 1 such tht t i < s i+1, replcing t i in S δ (τ) by slightly lrger x i > t i which is continuity point of h with x i < s i+1. Then T S δ (τ) < ǫ. Define nother Riemnn Stieltjes sum U equl to T except tht for ech i such tht s i+1 = t i nd 1 i n 1, we replce s i+1 by y i+1 nd t i by x i with t i < x i < y i+1 < t i+1, where x i is continuity point of h nd y i+1 t i is s smll s desired. Since f is right-continuous on (, b), we cn mke U T < ǫ. In either cse we cn ssume tht s i t i < x i s i + 2δ(s i ) for i = 1,...,n 1. Let x 0 := nd x n := b. Let y i := s i for ech i = 2,...,n 1 for which y i ws not previously defined. Thus y i(j) = s i(j) for

52 68 2 Extended Riemnn Stieltjes Integrls j = 1,..., m 1 by (2.67). Choose ny y 1 (, t 1 ) nd y n (x n 1, b). Let V := S YS (f, dh; ζ) be the Young Stieltjes sum bsed on the Young tgged point prtition ζ := (ξ, η) := ({x i } n i=0, {y i} n ). Then V U = f() + h() + f(y 1 )[h(x 1 ) h(+)] f()[h(x 1 ) h()] + f(b) h(b) + f(y n )[h(b ) h(x n 1 )] f(b)[h(b) h(x n 1 )] = [f(y 1 ) f()][h(x 1 ) h(+)] + [f(y n ) f(b)][h(b ) h(x n 1 )] < 2ǫ by (2.65). Now let κ := λ ξ = {u l } k l=0, prtition of [, b] with k = m+n 1. Ech intervl (x i 1, x i ) for i = 2,...,n 1 contins z j if i = i(j) µ, nd otherwise contins no z j nd becomes n intervl (u l 1, u l ) for some l. In the ltter cse let v l := y i. In the former cse we hve for some l, u l 2 = x i 1 < z j = u l 1 < x i = u l. Choose ny v l 1 := w j with u l 2 < v l 1 < u l 1 nd v l := r j with u l 1 < v l < u l. Let W := S YS (f, dh; (κ, σ)) be the Young Stieltjes sum bsed on the Young tgged point prtition (κ, σ), where σ := {v l } k l=1. Then W V = m 1 j=1 { f(w j )[h(z j ) h(x i(j) 1 )] + f(z j ) h(z j ) + f(r j )[h(x i(j) ) h(z j )] f(z j )[h(x i(j) ) h(x i(j) 1 )]} m 1 j=1 + [f(wj ) f(z j )][h(z j ) h(x i(j) 1 )] m 1 j=1 [f(r j ) f(z j )][h(x i(j) ) h(z j )] < 2ǫ by (2.65). Here W is Young Stieltjes sum bsed on the Young tgged refinement (κ, σ) of λ, so W (RYS) f dh < ǫ. Thus S δ(τ) (RYS) f dh < 7ǫ. The conclusion follows when f is right-continuous on (, b). Next consider the cse tht f c 0 ((, b)) (cf. Definition 2.9), which holds when f = + (,b) f. There is set of tgs {w j} m j=1 for the prtition λ = {z j} m j=0 such tht (λ, {w j } m j=1 ) is Young tgged point prtition of [, b] nd f(w j) = 0 for ll j. Thus by (2.64), m 1 j=1 f(z j ) h(z j ) (RYS) f dh < ǫ. (2.68) By (2.67), t i λ only for i = 0 nd n. For ny set ν {1,...,n 1}, consider prtitions κ = {u l } k l=0, k = m + n 1, of [, b] consisting of λ, t i for i ν, nd for ech i {1,...,n 1} \ ν, continuity point of h close to t i. Let l(1) < < l(n 1) be such tht {u l(i) } n 1 re the u l not in λ. Let

53 *2.7 Wrd Perron Stieltjes nd Henstock Kurzweil Integrls 69 σ = {v l } k l=1 be set of tgs for κ such tht (κ, σ) is Young point prtition of [, b] nd f(v l ) = 0 for ll l. Then from (2.68) nd (2.64) we obtin f(t i ) h(t i ) < 2ǫ. (2.69) i ν Consider lso prtitions (κ, σ) defined in the sme wy except tht for ξ {1,..., n 1}, if i ξ \ µ nd i is even, i / µ implies l(i 1) = l(i) 1. We tke u l = u l(i) to be continuity point of h little lrger thn t i, while u l(i) 1 is continuity point of h little smller thn t i 1 nd v l(i) = t i 1. For ech i = 1,...,n 1 such tht u l(i) is not yet defined, let it be continuity point of h ner t i, which is then true for ll i = 1,...,n 1. Let f(v l ) = 0 for other l s before. For even i ξ \ µ, letting u l(i) t i nd u l(i) 1 t i 1, it follows tht { f(ti 1 )[h(t i ) h(t i 1 )]: i ξ \ µ, i even } 2ǫ. The sme holds likewise for i odd. Thus f(t i 1 )[h(t i ) h(t i 1 )] 4ǫ. i ξ\µ With ν = {i 1: i ξ \ µ} in (2.69) this gives f(t i 1 )[h(t i ) h(t i 1 )] 6ǫ. i ξ\µ If s i = t i 1 then i µ by (2.67). Thus { f(ti 1 )[h(t i ) h(t i 1 )]: t i 1 = s i, i = 2,...,n 1 } 6ǫ. Here f(t i 1 ) could be replced by f(t i 1 ) f(y i ), where t i 1 < y i < t i nd f(y i ) = 0. So the Riemnn Stieltjes sum (2.66) differs by t most 6ǫ from the Riemnn Stieltjes sum S := n f(w i )[h(t i ) h(t i 1 )], where w i := y i if t i 1 = s i nd i = 2,...,n 1, w i := s i otherwise. Then t i 1 < w i t i for i = 2,...,n 1. The rest of the proof follows s in the cse when f is right-continuous nd t i < s i+1 for ll i = 1,..., n 1, so tht U = T nd y i = s i for ll i = 2,..., n 1. The proof of Theorem 2.69 is complete.

54 70 2 Extended Riemnn Stieltjes Integrls The following gives prtil converse to the preceding theorem. We sy tht the Young Stieltjes sums for f nd h re unbounded on ny subintervl of [, b] with < b if for ll c < d b, sup{ S YS (f, dh; τ) } = +, where the supremum is tken over ll Young tgged prtitions τ of [c, d]. Then the sums bsed on prtitions which re refinements of ny given prtition of [c, d] re lso unbounded. Proposition Let < b. Given f left- or right-continuous on (, b) nd h R[, b], suppose tht the Young Stieltjes sums for f nd h re unbounded on ny subintervl of [, b]. Then (HK) f dh does not exist. Proof. Suppose tht (HK) f dh exists. Let δ( ) be guge function nd I number such tht for ny δ-fine tgged prtition ({x i } n i=0, {y i} n ) of [, b], n f(y i )[h(x i ) h(x i 1 )] I < 1. (2.70) By the ctegory theorem (e.g. Theorem in Dudley [53]), there re n intervl [c, d] nd positive integer m such tht δ(x) > 1/m for ll x in dense set S in [c, d]. Replcing [c, d] by subintervl of itself, we cn ssume tht 0 < d c < 1/m nd tht c, d S. Thus every tgged prtition ({ξ i } k i=0, {η j} k j=1 ) of [c, d] with tgs η j S for j = 1,...,k is δ-fine. Tke δ-fine tgged prtition s in (2.70), where we cn tke some y i = c < y l = d by Lemm 2.53(c). We cn lso tke x i = c nd x l = d, s follows. If initilly y i = c < x i, then let x r := x r for r = 0, 1,...,i 1 nd y r := y r for r = 1,...,i 1. Let x i := y i := c nd for r = i,..., n, let y r+1 := y r nd x r+1 := x r. Then ({x i }n+1 i=0, {y i }n+1 ) is δ-fine tgged prtition of [, b] giving the sme sum s in (2.70). We cn tke d s some x l likewise. Let κ be the resulting δ-fine tgged prtition of [, b]. We show next tht the prt of κ corresponding to [c, d] cn be replced by tgged prtition of [c, d] such tht the corresponding Riemnn Stieltjes sum is rbitrrily lrge nd the resulting tgged prtition of [, b] is δ-fine. For ny M <, by ssumption, there is Young tgged prtition τ = ({t j } k j=0, {s j} k j=1 ) of [c, d] such tht S YS(f, dh; τ) > M. Since h is regulted, there exist v j < t j, j = 1,...,k, nd u j > t j, j = 0,...,k 1, such tht t j 1 < u j 1 < s j < v j < t j for j = 1,...,k nd k 1 f(c)[h(u 0 ) h(c)] + f(t j )[h(u j ) h(v j )] + j=1 k f(s j )[h(v j ) h(u j 1 )] + f(d)[h(d) h(v k )] > M. j=1 Now since f is right- or left-continuous, nd S is dense, we cn replce s j by some s j S s close to s j s desired, mking smll chnge in the sum. Likewise we cn replce t j by some t j S, where u j 1 < s j < v j < t j < u j for

55 2.8 Properties of Integrls 71 ech j = 1,...,k 1. Recll tht c, d S; the endpoints t 0, t k re not replced. We thus obtin Riemnn Stieltjes sum Σ for δ-fine tgged prtition ζ of [c, d] with Σ > M. Then joining ζ with the δ-fine tgged prtitions of [, c] (if < c) nd [d, b] (if d < b) given by the fixed tgged prtition κ, we get unbounded Riemnn Stieltjes sums for δ-fine tgged prtitions of [, b], contrdicting (2.70). Wrd [238] stted nd Sks [201, Theorem VI.8.1] gve proof of the fct tht the integrl (WPS) f dh is defined provided the corresponding Lebesgue Stieltjes integrl (LS) f dh is defined, nd then they re equl. By Theorem 2.65, the Henstock Kurzweil integrl is in the sme reltion with the LS integrl. The following gives conditions under which the converse holds. Theorem Let f be nonnegtive on [, b] nd let h be nondecresing on [, b] nd right-continuous on [, b). Then (HK) f dh exists if nd only if (LS) f dh does, nd then the two re equl. Proof. If = b both integrls re 0, so we cn ssume < b. The Mc- Shne integrl is defined s the HK integrl except tht in tgged prtition ({t i } n i=0, {s i} n ) the tgs s i need not be in the corresponding intervls [t i 1, t i ]. By Corollry of Pfeffer [185, p. 113], under the present hypotheses (HK) f dh exists if nd only if the corresponding McShne integrl exists, nd then the two re equl. The equivlence between the McShne nd Lebesgue Stieltjes integrls when f 0 nd h is nondecresing ws proved by McShne [167, pp. 552, 553]. Also, it follows from Theorem 4.4.7, Proposition , nd Theorem of Pfeffer [185]. The proof of the theorem is complete. 2.8 Properties of Integrls In this section, by = integrl we will men n integrl = f dh s opposed to n integrl =f dµ for n intervl function µ. Suppose tht the bsic ssumption (1.14) holds. For functions f : [, b] X nd h: [, b] Y, consider the following properties of n integrl f dh: I. For u 1, u 2 K nd f 1, f 2 : [, b] X, (u 1 f 1 + u 2 f 2 ) dh = u 1 f 1 dh + u 2 f 2 dh, where the left side exists provided the right side does.

56 72 2 Extended Riemnn Stieltjes Integrls II. For v 1, v 2 K nd h 1, h 2 : [, b] Y, f d(v 1 h 1 + v 2 h 2 ) = v 1 f dh 1 + v 2 f dh 2, where gin the left side exists provided the right side does. III. For < c < b, f dh exists if nd only if both c f dh nd b c f dh exist, nd then f dh = c f dh + c f dh. (2.71) IV. If < b nd f dh exists then for ech t [, b], tking limit for s [, b] if t = or b, I(t) = I(f, dh)(t) := t f dh exists nd { } I(t) I(s) f(t) [h(t) h(s)] = 0. (2.72) lim s t Properties I nd II imply tht the opertor V W (f, h) f dh Z is biliner for ny function spces V, W on which it is defined. If property IV holds, the indefinite integrl I(f, dh)(t), t [, b], is regulted function whenever the integrtor h is regulted, nd then I(t) = f(t) h(t) nd + I(s) = f(s) + h(s) for s < t b. Theorem The RS, RRS, RYS, S, CY, HK nd = integrls stisfy properties I nd II. Proof. We prove only property I becuse the proof of property II is symmetric. We cn ssume tht < b. Let u 1, u 2 K, f 1, f 2 : [, b] X, nd h: [, b] Y. For the RS, RRS, nd HK integrls, property I follows from bilinerity of the mpping (f, h) f h from X Y to Z, nd from the equlity S RS (u 1 f 1 + u 2 f 2, dh; τ) = u 1 S RS (f 1, dh; τ) + u 2 S RS (f 2, dh; τ), vlid for ny tgged prtition τ of [, b]. The sme rgument with Riemnn Stieltjes sums replced by Young Stieltjes sums (2.16) implies property I for the YS integrl, nd so it holds for the full Stieltjes S integrl. Also, property I holds for the = integrl by Proposition For the CY integrl given by the Y 1 integrl (2.44), let f 1, f 2 be regulted functions. On the intervl [, b), we hve (u 1 f 1 +u 2 f 2 ) + = u 1 (f 1 ) + +u 2 (f 2 ) + nd + (u 1 f 1 + u 2 f 2 ) = u 1 + f 1 + u 2 + f 2. Property I for the CY integrl follows, since it holds for the RRS integrl nd by the linerity of unconditionlly convergent sums. The proof of Theorem 2.72 is complete. Theorem The RRS, RYS, S, CY, HK, nd = integrls stisfy property III.

57 2.8 Properties of Integrls 73 Proof. Let < c < b nd let f : [, b] X nd h: [, b] Y. Suppose tht (RRS) c f dh nd (RRS) c b f dh exist. For tgged prtition τ 1 = ({x i } n i=0, {t i} n ) of [, c] nd tgged prtition τ 2 = ({y j } m j=0, {s i} m j=1 ) of [c, b], let τ := ({x i } n i=0 {y j} m j=1, {t i} n {s j} m j=1 ), tgged prtition of [, b]. Then the equlity S RS (f, dh; τ) = S RS (f, dh; τ 1 ) + S RS (f, dh; τ 2 ) (2.73) yields tht (RRS) f dh exists nd (2.71) holds for the RRS integrl. For the converse suppose tht (RRS) f dh exists. Applying the Cuchy test for pirs of tgged prtitions τ nd τ of [, b] which re refinements of {, c, b} nd induce the sme tgged prtition of [c, b], it follows tht (RRS) c f dh exists. Likewise (RRS) c b f dh exists. The proof of property III for the RRS integrl is complete. The proof of property III for the RYS integrl is the sme except tht the equlity S YS (f, dh; τ) = S YS (f, dh; τ 1 ) + S YS (f, dh; τ 2 ) for Young tgged prtitions s in (2.16) is used insted of (2.73). Thus property III lso holds for the = integrl by Proposition Since property III holds for the RRS nd RYS integrls it lso holds for the full Stieltjes integrl (S) f dh by definition 2.41 of the (S) integrl nd since f nd h hve common one-sided discontinuity on [, b] if nd only if they lso hve one on t lest one of [, c] or [c, b]. Since property III for the HK integrl is proved by Proposition 2.56, it is left to prove property III for the CY integrl. To this im suppose tht functions f, h re regulted nd tht the Y 1 integrl (Y 1 ) f dh exists. By the sme property for the RRS integrl lredy proved, it follows tht the following three integrls exist nd (RRS) f (b) + dh() = (RRS) c f + dh () + (RRS) c f (b) + dh. On the right side the integrnd f + over [, c] depends on f(c+) nd the integrtor h over [c, b] depends on h(c ). For the two integrls (RRS) c f + dh () nd (RRS) c f (c) + dh (), the difference between Riemnn Stieltjes sums for ech bsed on the sme tgged prtition τ = ({t i } n i=0, {s i} n ) of [, c] is f(s n +) [h(t n ) h(t n 1 )] f (c) + (s n) [h(t n ) h(t n 1 )] = + f(t n ) [h(t n ) h(t n 1 )] if s n = t n or is 0 if s n [t n 1, t n ). Tking limit under refinement of tgged prtitions of [, c], we cn get t n 1 t n = c, nd so h(t n ) h(t n 1 ) 0, while f is bounded. Therefore the following two integrls both exist nd re equl:

58 74 2 Extended Riemnn Stieltjes Integrls c c (RRS) f + dh () = (RRS) f (c) + dh (). Next, for the two integrls (RRS) c b f (b) + dh nd (RRS) c b f (b) + dh (c), the difference between Riemnn Stieltjes sums for ech bsed on the sme tgged prtition τ = ({t i } n i=0, {s i} n ) of [c, b] is f(s 1 +) [h(t 1 ) h(c )] f(s 1 +) [h(t 1 ) h(c)] = f(s 1 +) h(c) f(c+) h(c) s t 1 t 0 = c, which one cn obtin by tking limit under refinements of tgged prtitions of [c, b]. Therefore the following two integrls both exist nd the eqution (RRS) c f (b) + dh = (RRS) c f (b) + dh(c) + f(c+) h(c) holds. By the definition of the Y 1 integrl (2.44), it then follows tht the Y 1 integrl exists over the subintervls [, c] nd [c, b], nd we hve (Y 1 ) f dh = (RRS) c f + dh () + (RRS) c f (b) + dh [ + f + h]() (,c) + f ± h [ + f ± h](c) = (RRS) +(RRS) = (Y 1 ) c c c (c,b) + f ± h + [f h](b) f (c) + dh() [ + f + h]() (,c) + f ± h + [f h](c) f (b) + dh(c) [ + f + h](c) (c,b) + f ± h + [f h](b) f dh + (Y 1 ) c f dh. The converse impliction follows by pplying the sme rguments, nd so property III holds for the CY integrl. The proof of Theorem 2.73 is complete. The Riemnn Stieltjes integrl does not stisfy property III. Indeed, for the indictor functions f := 1 [1,2] nd h := 1 (1,2] defined on [0, 2], the RS integrl exists over the intervls [0, 1] nd [1, 2], but not over the intervl [0, 2]. Notice tht f nd h for this exmple hve common discontinuity t 1 [0, 2]. This feture of the Riemnn Stieltjes integrl ws observed by Pollrd [187], who lso showed tht the refinement Riemnn Stieltjes integrl does stisfy property III. A weker form of this property cn be stted s follows:

59 2.8 Properties of Integrls 75 III.For < c < b, if f dh exists then both c f dh nd b c f dh exist, nd (2.71) holds. Then we hve the following result: Proposition The RS integrl stisfies property III. Proof. Suppose tht (RS) f dh exists nd let < c < b. Adjoining equl tgged subintervls of [c, b] to ny two given tgged prtitions τ 1, τ 2 of [, c], one cn form two tgged prtitions τ 1, τ 2 of [, b], without incresing τ 1 τ 2, nd such tht S RS (τ 1 ) S RS(τ 2 ) = S RS(τ 1 ) S RS (τ 2 ). Thus one cn pply the Cuchy test to prove the existence of (RS) c f dh. Similrly one cn show tht (RS) c b f dh lso exists. The dditivity reltion (2.71) then follows from the equlity S RS ((κ, ξ)) = S RS ((κ, ξ )) + S RS ((κ, ξ )), vlid for ny tgged prtitions such tht κ = κ κ, ξ = ξ ξ, nd κ, κ re prtitions of [, c] nd [c, b] respectively. Theorem The RS, RRS, RYS, nd HK integrls stisfy property IV. Proof. Suppose tht (RS) f dh exists. The indefinite integrl I RS(u) := (RS) u f dh is defined for u (, b] by Proposition 2.74 nd is 0 when u = by the definition of the RS integrl. Also by Proposition 2.74, we hve I RS (u) I RS (v) = sgn(u v)(rs) u v u v f dh. In light of symmetry of proofs, we prove only tht { lim (RS) v u u v } f dh S RS (f, dh; σ v ) = 0 (2.74) for ny < u b, where for < v < u, σ v is the tgged prtition ({v, u}, {u}) of [v, u]. Let u (, b]. Given ǫ > 0 there is δ (0, u ] such tht ny two Riemnn Stieltjes sums bsed on tgged prtitions of [, u] with mesh less thn δ differ by t most ǫ. For u δ v < u, choose Riemnn Stieltjes sum S RS (τ v ) = S RS (f, dh; τ v ), bsed on tgged prtition τ v of [v, u], within ǫ of (RS) u v f dh. Let τ nd τ be two tgged prtitions of [, u] with mesh less thn δ which re equl when restricted to [, v] nd which coincide with τ v nd σ v, respectively, when restricted to [v, u]. Then (RS) u (RS) < 2ǫ f dh S RS (f, dh; σ v ) v u v f dh S RS (τ v ) + SRS (τ ) S RS (τ ) (2.75) for ech v with u δ v < u. Since ǫ > 0 is rbitrry, (2.74) holds, proving property IV for the RS integrl.

60 76 2 Extended Riemnn Stieltjes Integrls If (RRS) f dh exists, we rgue similrly. To prove (2.74) with RRS insted of RS, let ǫ > 0. Then there is prtition λ = {t j } m j=0 of [, u] such tht ny two Riemnn Stieltjes sums bsed on tgged refinements of λ differ by t most ǫ. For v [t m 1, u), choose Riemnn Stieltjes sum, bsed on tgged prtition τ v of [v, u], within ǫ of (RRS) v u f dh. Let τ nd τ be two tgged refinements of λ which coincide with τ v nd σ v, respectively, when restricted to [v, u]. Then (2.75) with RRS insted of RS holds for ech v [t m 1, u), proving property IV for the RRS integrl. Suppose now tht (RYS) f dh exists. Let u (, b], nd for < v < u, let σ v be Young tgged prtition of the form ({v, u}, {s}) of [v, u]. Since h is regulted by definition of the RYS integrl, we hve { } lim S YS (f, dh; σ v ) f(u) [h(u) h(v)] v u { } = lim [f(s) f(u)] [h(u ) h(v+)] + [f(v) f(u)] + h(v) = 0. v u Therefore nd by property III gin, we hve to show tht { lim (RYS) v u u v } f dh S YS (f, dh; σ v ) = 0. The proof is the sme s for the RRS integrl except tht Riemnn Stieltjes sums re replced by Young Stieltjes sums. Since property IV for the HK integrl follows from Proposition 2.58, the proof of Theorem 2.75 is complete. Property IV for the HK integrl grees with the only if prt of Proposition 2.58, which sys tht existence of the HK integrl implies existence of improper versions. The following shows tht the if prt of Proposition 2.58 does not hold for the RRS integrl, for the RYS integrl by Propositions 2.18 nd 2.46, nor for the CY integrl by its definition. Proposition There re continuous rel-vlued functions f, h on [0, 1] such tht lim t 1 (RRS) t 0 f dh exists, but (RRS) 1 0 f dh does not. Proof. Let s m := 1 1/m for m = 1, 2,.... For ech m, let f(s 4m 2 ) = f(s 4m ) := 0, f(s 4m 1 ) := m 1/2, f(0) = f(1) := 0, nd let f be liner nd continuous on intervls between djcent points where it hs been defined. Let h(s 4m 3 ) := 0, so h(0) = 0, h(s 4m 2 ) = h(s 4m ) := m 1/2, h(1) := 0, nd let h lso be liner between djcent points where it hs been defined. Then f nd h re both continuous. Since h is constnt on ech intervl where f is non-zero, we hve for 0 < t < 1 tht (RRS) 0 t f dh = 0, tking prtition contining t nd ll points s m < t. But for ny prtition κ = {t i } n i=0 of [0, 1], there exist rbitrrily lrge Riemnn Stieltjes sums for f, h bsed on refinements of κ, s follows. Let m 0

61 2.8 Properties of Integrls 77 be the smllest m such tht s 4m 3 > t n 1. Form prtition κ N by djoining to κ ll points s 4m 3, s 4m for m = m 0, m 0 +1,...,m 0 +N, nd form Riemnn Stieltjes sums contining terms f(s 4m 1 )[h(s 4m ) h(s 4m 3 )] = 1/m for ll such m, nd 0 terms f(s 4m )[h(s 4m+1 ) h(s 4m )]. As N these sums become rbitrrily lrge, s climed. The following fct is chnge of vribles formul for the full Stieltjes integrl. Proposition Let φ: [, b] R be strictly monotone continuous function with rnge [c, d] := rn(φ). Assuming (1.14), let f : [c, d] X nd h: [c, d] Y. Then (S) (f φ) d(h φ) exists if nd only if (S) d c f dh does, nd { b (S) d (S) (f φ) d(h φ) = c f dh if φ is incresing, (S) d f dh if φ is decresing. c Proof. We cn ssume tht < b. First suppose tht φ is incresing. The inverse function φ 1 of φ is lso continuous nd incresing from [c, d] onto [, b]. Therefore for the two integrls the limits under refinement of prtitions both exist or not simultneously, nd if they do exist they re equl, proving the first prt of the proposition. Now suppose tht φ is decresing on [, b]. Then letting θ(x) := x for x [ b, ], it follows tht φ := φ θ is n incresing continuous function on [ b, ]. Let τ = ({t i } n i=0, {s i} n ) be tgged prtition of [, b]. Then τ := ({ t n i } n i=0, { s n i} n 1 i=0 ) is tgged prtition of [ b, ] nd S RS (f φ, dh φ; τ) = S RS (f φ, dh φ; τ). Also, if σ is tgged refinement of τ then σ is tgged refinement of τ. Thus (RRS) f φ dh φ = (RRS) b f φ dh φ, where the two integrls exist or not simultneously. Now pplying the first prt of the proposition to the integrl on the right side, it follows tht (RRS) b f φ dh φ = (RRS) φ() φ(b) f dh, provided either integrl is defined. The sme is true for refinement Young Stieltjes integrls, proving the proposition. Next is fct which llows interchnge of integrtion with liner mp for sclr-vlued integrtor.

62 78 2 Extended Riemnn Stieltjes Integrls Proposition Let K be either the field R or the field C, nd let L be bounded liner mpping from Bnch spce X to nother Bnch spce Z. For f : [, b] X nd h: [, b] K, if (RS) f dh exists then so does (RS) (L f) dh nd ( L (RS) ) f dh = (RS) (L f) dh. (2.76) Proof. We cn ssume tht < b. For ny tgged prtition τ of [, b], we hve L ( S RS (f, dh; τ) ) = S RS (L f, dh; τ). Since L is continuous, if (RS) f dh exists then the right side hs limit s the mesh of τ tends to zero nd (2.76) holds, proving the proposition. Next it will be seen tht the previous proposition does not extend to cses where both the integrnd nd integrtor hve multidimensionl vlues, for ny integrl we consider. Exmple Let X = Y = R 2, let Z = R, nd let the biliner form from X Y into Z be the usul inner product. Let g nd h be ny relvlued functions on [0, 1] nd ν ny rel-vlued intervl function on [0, 1]. Let f(t) := (g(t), 0) R 2 nd H(t) := (0, h(t)) R 2 for ny t [0, 1]. For ny intervl J [0, 1] let µ(j) := (0, ν(j)) R 2. Let L(x, y) := (y, x) for ech (x, y) R 2. Then for ech form of integrtion we hve defined, the integrls f dh nd f dµ exist nd re 0, where = 1 0 or [0,1], but the integrls (L f) dh nd (L f) dµ my not exist, or my exist with ny vlue. For ll the integrls f dg defined so fr nd properties proved for them, we hve the symmetricl properties of integrls df g by (1.15). Integrtion by prts We strt with the clssicl integrtion by prts formul for the Riemnn Stieltjes nd refinement Riemnn Stieltjes integrls. Theorem Assuming (1.14), let f : [, b] X nd g: [, b] Y. For # = RS or # = RRS, if (#) f dg exists then so does (#) df g, nd (#) f dg + (#) df g = f(b) g(b) f() g() =: f g b. (2.77) Proof. We cn ssume tht < b. Let τ = ({t i } n i=0, {s i} n ) be tgged prtition of [, b], nd let s 0 := t 0 =, s n+1 := t n = b. Summing by prts, we hve

63 n [ f(ti ) f(t i 1 ) ] g(s i ) = f g tn t Properties of Integrls 79 n f(t i ) [g(s i+1 ) g(s i ) ]. (2.78) For # = RS, notice tht if mx i (t i t i 1 ) < δ for some δ > 0, then mx i (s i+1 s i ) < 2δ. Therefore if (RS) f dg exists then by (2.78) nd (1.15), (RS) df g exists nd (2.77) holds for # = RS. For # = RRS, notice tht the sum on the right side of (2.78) cn be written s the sum n i=0 i=0 { f(t i ) [g(s i+1 ) g(t i ) ] + f(t i ) [g(t i ) g(s i ) ]} =: S. Thus if {t i } n i=0 is refinement of prtition λ of [, b] then S is Riemnn Stieltjes sum bsed on tgged refinement of λ. Hence if (RRS) f dg exists then by (2.78) nd (1.15), (RRS) df g exists nd (2.77) holds for # = RRS. The proof is complete. Next is n integrtion by prts formul for the centrl Young integrl s defined in Section 2.5. Theorem Assuming (1.14), let f R([, b]; X) nd g R([, b]; Y ). If < b, suppose tht t lest one of the conditions () or (b) holds: () the sums (,b) + f + g nd (,b) f g converge unconditionlly in Z; (b) on (, b), f = [f + f + ]/2 nd g = [g + g + ]/2. If (CY ) f dg exists then so does (CY ) df g, nd (CY ) f dg + (CY ) df g = f g b + A, (2.79) where A = 0 if = b nd otherwise A = [ + f + g ] ()+ [ f g ] (b)+b with { (,b) B := + f + g + (,b) f g if () holds 0 if (b) holds. Proof. We cn ssume tht < b. By the definitions of the CY integrl (2.51) nd the Y 1 integrl (2.44), (CY ) f dg = (RRS) f (b) + dg() [ + f + g ] () + [ f g ] (b) (,b) + f ± g,

64 80 2 Extended Riemnn Stieltjes Integrls where the RRS integrl exists nd the sum converges unconditionlly in Z. If () holds, by linerity of unconditionl convergence, the sum (,b) ± f g converges unconditionlly in Z, nd + f + g. (2.80) (,b) ± f g (,b) + f ± g = (,b) f g (,b) If (b) holds, since is biliner, + f ± g = + f (2 g) = (2 + f) g = ± f g. Thus gin the sum (,b) ± f g converges unconditionlly in Z, nd + f ± g. (2.81) (,b) ± f g = (,b) Also, in cse () or (b) the integrtion by prts theorem for the RRS integrl (Theorem 2.80 with # = RRS) yields tht (RRS) df (b) + g () exists nd (RRS) f (b) + dg() + (RRS) df (b) + g() = f(b) + g() b (2.82) = f(b) g(b ) f(+) g(). Hence by definition of the Y 2 integrl (2.45), (CY ) df g exists nd (CY ) df g = (RRS) df (b) + g() + [ + f g ] () + [ f g ] (b) + (,b) ± f g. Then by (2.80) in cse () or (2.81) in cse (b), it follows tht (CY ) f dg + (CY ) df g = (RRS) f (b) + dg() + (RRS) df (b) + g() +f g b f(b) g(b ) + f(+) g() [ + f + g ] () + [ f g ] (b) + A. This together with (2.82) implies formul (2.79), proving the theorem. For the refinement Young Stieltjes integrl, n integrtion by prts theorem nlogous to the preceding one does not hold. Recll tht by Proposition 2.52 there re rel-vlued functions f nd h on [0, 1], where f is in c 0 ((0, 1)) nd h is continuous, for which (CY ) 0 1 f dh exists, while (RYS) 0 1 f dh does not exist. For this pir of functions, (RYS) 0 1 h df exists nd equls 0 becuse f (0) f(1) + 0, nd + f + h = f h = 0 since h is continuous. However, by Theorem 2.51() nd Theorem 2.81, we hve the following corollry:

65 2.8 Properties of Integrls 81 Corollry Under the conditions of Theorem 2.81, if (RYS) f dg nd (RYS) df g both exist then (2.79) with CY replced by RYS holds. For regulted function f on [, b] with < b nd ˆf := [f (,b) we hve [ f ˆf] ± g + (,b) (,b) ± f [g ĝ] = (,b) + f + g + (,b) + f (,b) + ]/2, f g, provided tht the sums converge unconditionlly. Therefore by Theorem 2.51(b) nd Theorem 2.81, the following form of integrtion by prts for the refinement Young Stieltjes integrl holds: Corollry Assuming (1.14) nd < b, let f R([, b]; X) nd g R([, b]; Y ) be such tht t lest one of the two conditions () nd (b) of Theorem 2.81 holds. If either (CY ) f dg or (CY ) df g exists then (RYS) ˆf dg nd (RYS) df ĝ both exist, nd (RYS) ˆf dg + (RYS) df ĝ = f g b [ + f + g ] () + [ f g ] (b), where ˆf = f nd ĝ = g if the condition (b) of Theorem 2.81 holds. (2.83) Bounds for integrls Here we give bounds for integrls ssuming tht either the integrnd or integrtor hs bounded vrition. Theorem Assuming (1.14), f : [, b] X, nd h: [, b] Y, we hve () if f is regulted nd h is of bounded vrition then the full Stieltjes integrl (S) f dh is defined nd for ech t [, b], (S) f dh f(t) [h(b) h()] Osc(f; [, b])v 1 (h; [, b]); (2.84) (b) if h is regulted nd f is of bounded vrition then the full Stieltjes integrl (S) f dh is defined nd for ech t [, b], (S) f dh f(t) [h(b) h()] Osc(h; [, b])v 1 (f; [, b]). (2.85) Proof. We cn ssume tht < b. In both cses if f nd h hve no common discontinuities then the integrl f dh exists in the Riemnn Stieltjes sense by Theorems 2.42 nd If f nd h hve no common one-sided discontinuities then the integrl f dh exists in the refinement Riemnn Stieltjes sense by Theorem Under the ssumptions of the present theorem the integrl

66 82 2 Extended Riemnn Stieltjes Integrls f dh lwys exists in the refinement Young Stieltjes sense by Theorem 2.20, nd so the full Stieltjes integrl (S) f dh is defined. To prove () it then suffices to prove it for the refinement Young Stieltjes integrl. Letting τ = ({t i } n i=0, {s i} n ) be Young tgged prtition of [, b] nd t [, b], we hve the bound S YS (f, dh; τ) f(t) [h(b) h()] n [f(t i ) f(t)] ( ± [,b] h)(t n i) + [f(s i ) f(t)] [h(t i ) h(t i 1 +)]. i=0 For ech i = 1,...,n, letting u i 1 t i 1 nd v i t i, one cn pproximte + h() by h(u 0 ) h(), h(b) by h(b) h(v n ), ± h(t i ) by h(u i ) h(v i ) for i = 1,...,n 1 nd h(t i ) h(t i 1 +) by h(v i ) h(u i 1 ) for i = 1,...,n. For n rbitrry ǫ > 0, this gives the further bound ( ǫ+ mx f(t i) f(t) h(u 0 ) h() + h(b) h(v n ) + 0 i n + mx 1 i n f(s i) f(t) n 1 ) h(u i ) h(v i ) n h(v i ) h(u i 1 ) ǫ + Osc(f; [, b])v 1 (h; [, b]). Since ǫ > 0 is rbitrry, the limit under refinements of Young tgged prtitions gives the bound (2.84) for the refinement Young Stieltjes integrl, proving (). As in (), it will suffice to prove (b) only for the refinement Young Stieltjes integrl. Let τ = ({t i } n i=0, {s i} n ) be Young tgged prtition of [, b]. For j = 1,...,2n, let j f := f(u j ) f(u j 1 ) with {u j } 2n j=0 = {t 0, s 1, t 1,..., s n, t n } nd let {v j } 2n j=1 := {t 0+, t 1, t 1 +,...,t n }. Summtion by prts gives S YS (f, dh; τ) = f(b) h(b) f() h() 2n j=1 j f h(v j 1 ). Adding to nd subtrcting from the right side first f(b) h() nd then f() h(b), we get two representtions S YS (f, dh; τ) = f(b) [h(b) h()] = f() [h(b) h()] 2n j=1 2n j=1 j f [h(v j 1 ) h()] j f [h(v j 1 ) h(b)]. (2.86) Thus for t {, b}, we hve the bound SYS (f, dh; τ) f(t) [h(b) h()] Osc(h; [, b])v1 (f; [, b]). (2.87)

67 2.8 Properties of Integrls 83 Since τ is n rbitrry Young tgged prtition of [, b], (2.85) holds if t {, b}. Suppose tht t (, b) nd let t = t ν for some ν {1,...,n 1}. Then τ 1 := ({t i } ν i=0, {s i} ν ) nd τ 2 := ({t i } n ν, {s i } n ν+1) re Young tgged prtitions of [, t] nd [t, b] respectively. Applying the first representtion in (2.86) to τ 1 nd the second representtion to τ 2, it follows tht S YS (f, dh; τ) = S YS (f, dh; τ 1 ) + S YS (f, dh; τ 2 ) ν = f(t) [h(b) h()] j f [h(v j 1 ) h()] n i=ν+1 j f [h(v j 1 ) h(b)], nd so (2.87) holds for ny t (, b) nd ny τ contining t s prtition point. Thus (2.85) holds for ny t [, b]. The proof of Theorem 2.84 is complete. Corollry Under the ssumptions of the preceding theorem, the indefinite full Stieltjes integrl I S (f, dh)(t) := (S) t f dh, t [, b], is defined nd hs bounded vrition. Proof. We cn ssume < b. The indefinite Riemnn Stieltjes integrl exists by the preceding theorem nd property III (see Proposition 2.74). The indefinite refinement Riemnn Stieltjes nd Young Stieltjes integrls exist by the preceding theorem nd property III (see Theorem 2.73). Let κ = {t i } n i=0 be prtition of [, b]. Then using dditivity of the three integrls proved s the lredy mentioned properties III nd III, we get the bound ti n s 1 (I S (f, dh); κ) (S) f dh f(t i 1 ) [h(t i ) h(t i 1 )] t i 1 n + f(t i 1 ) h(t i ) h(t i 1 ) (Osc(f; [, b]) + f sup ) v 1 (h; [, b]), completing the proof of the corollry. A substitution rule Here is substitution rule for the Riemnn Stieltjes integrls with integrtors hving bounded vrition.

68 84 2 Extended Riemnn Stieltjes Integrls Proposition For Bnch spce X, let h W 1 [, b], g R([, b]; X), nd f R[, b] be such tht the two pirs (h, g) nd (h, f) hve no common discontinuities, nd gf : [, b] X is the function defined by pointwise multipliction. Then the following Riemnn Stieltjes integrls (including those in integrnds) re defined nd (RS) di RS (g, dh) f = (RS) gf dh = (RS) g di RS (dh, f), (2.88) where I RS (g, dh)(t) := (RS) t g dh X nd I RS(dh, f)(t) := (RS) t f dh for t [, b], nd denotes the nturl biliner mpping X R X. Proof. We cn ssume tht < b. Since the pirs (g, h) nd (f, h) hve no common discontinuities, the indefinite integrls I RS (g, dh) nd I RS (dh, f) exist nd re in W 1 ([, b]; X) nd W 1 [, b], respectively, by Corollry Since the discontinuities of I RS (g, dh) nd I RS (dh, f) re subsets of those of h, nd so the pirs (I RS (g, dh), f) nd (g, I RS (dh, f) hve no common discontinuities, the leftmost nd the rightmost integrls in (2.88) exist by the sme corollry. Since gf is regulted function on [, b] nd the pir (gf, h) hve no common discontinuities, the middle integrl in (2.88) exists by the sme corollry. We prove only the first equlity in (2.88), since the proof of the second one is symmetric. Let τ = ({t i } n i=0, {s i} n ) be tgged prtition of [, b]. Then SRS (di RS (g, dh), f; τ) S RS (gf, dh; τ) f sup R(τ), where R(τ) := n (RS) It is enough to prove tht ti t i 1 g dh g(s i ) [h(t i ) h(t i 1 ) ]. lim R(τ) = 0. (2.89) τ 0 Let ǫ > 0. By Theorem 2.1, there exists point prtition {z j } m j=0 of [, b] such tht Osc(g; (z j 1, z j )) < ǫ for j {1,...,m}. (2.90) Let u z j v b for some j. Splitting the intervl [u, v] into the prts [u, z j ] nd [z j, v] if z j is not n endpoint of [u, v], nd then pplying properties III nd IV for the Riemnn Stieltjes integrl (see respectively Proposition 2.74 nd Theorem 2.75), we hve tht (RS) v u g dh g(z j ) [h(v) h(u)] is smll provided v u is smll enough. Therefore nd since g nd h hve no common discontinuities, there exists δ > 0 such tht if z j [t i 1, t i ] for some j nd mesh τ < δ, then

69 ti (RS) (RS) g dh g(s i ) [h(t i ) h(t i 1 )] t i 1 ti g dh g(z j ) [h(t i ) h(t i 1 )] t i Properties of Integrls 85 + [g(z j ) g(s i )] [h(t i ) h(t i 1 )] < ǫ/(m + 1). This together with (2.90) nd the bound of Theorem 2.84 yields R(τ) ǫv 1 (h; [, b]) + (m + 1)ǫ/(m + 1) = ǫ [ v 1 (h; [, b]) + 1 ] if the mesh of the tgged prtition τ is less thn δ. The proof of (2.89), nd hence of Theorem 2.86, is complete. A chin rule formul Here we give representtion of the full Stieltjes integrl s defined in Definition 2.41, with respect to composition F f, where f hs bounded vrition nd F is smooth. Theorem Let f : [, b] R d be of bounded vrition, let F be relvlued C 1 function on R d nd let h R[, b]. Then the composition F f is of bounded vrition nd (S) h d(f f) = (S) h ( F f ) df + (,b]h { (F f) ( F f ) } f + [,b)h { + (F f) ( F f ) } + f, (2.91) where the two sums converge bsolutely if < b nd equl 0 if = b. Proof. We cn ssume tht < b. Since F is Lipschitz function on ny bounded set (the rnge of f), F f is of bounded vrition. Thus the two full Stieltjes integrls in (2.91) exist by Theorems 2.20, 2.17, nd Agin since F is Lipschitz nd F is bounded on the rnge of f, the two sums in (2.91) converge bsolutely. By Propositions 2.13 nd 2.18, it is enough to prove the equlity (2.91) for the RYS integrls. Let τ = ({t i } n i=0, {s i} n ) be Young tgged prtition of [, b]. Also let S (τ) := { h ( F f ) ( F f ) } f {t 1,...,t n} nd S + (τ) := {t 0,...,t n 1} { h +( F f ) ( F f ) } + f.

70 86 2 Extended Riemnn Stieltjes Integrls Then we hve the identity S YS (h, df f; τ) = S YS (h ( F f ), df; τ) + S (τ) + S + (τ) + R(τ), (2.92) where R(τ) is the sum n [F(f(ti h(s i ){ )) F(f(t i 1 +)) ] ( F(f(s i )) ) [f(t i ) f(t i 1 +) ]}. Let ǫ > 0 nd let B be bll in R d contining the rnge of f. Since F is uniformly continuous on B, there is δ > 0 such tht F(u) F(v) < ǫ whenever u v < δ nd u, v B. Moreover, since f is regulted, by Theorem 2.1(b), there exists prtition λ = {z j } m j=0 of [, b] such tht Osc(f; (z j 1, z j )) < δ for j = 1,...,m. Let τ be Young tgged refinement of λ. Then for ech i = 1,...,n, by the men vlue theorem for λ F(λf(t i ) + (1 λ)f(t i 1 +)), 0 λ 1, nd by the chin rule of differentition there is λ i [0, 1] such tht F(f(t i )) F(f(t i 1 +)) = F(θ i ) [f(t i ) f(t i 1 +) ], where θ i := λ i f(t i ) + (1 λ i )f(t i 1 +). Thus R(τ) n h(s i ) F(θi ) F(f(s i )) f(ti ) f(t i 1 +) n ǫ h sup f(ti ) f(t i 1 +) ǫ h sup v 1 (f; [, b]). Since ech of the four other sums in (2.92) converges under refinements of prtitions to the respective four terms in (2.91), the stted equlity (2.91) holds, proving the theorem. Tking the C 1 function F defined by F(u, v) := uv, u, v R, we obtin from the preceding theorem: Corollry Let f, g be two rel-vlued functions on [, b] hving bounded vrition, nd let h R[, b]. Then the product function fg hs bounded vrition, nd (S) h d(fg) = (S) hf dg + (S) hg df h f g + (,b] [,b)h + f + g, (2.93) where the two sums converge bsolutely if < b nd equl 0 if = b.

71 2.9 Reltions between Integrls 87 Proof. We cn ssume tht < b. The two full Stieltjes integrls on the right side of (2.93) exist by Theorems 2.20, 2.17, nd By linerity, their sum gives the integrl (S) h( F (g, f)) d(g, f) with F(u, v) := uv for u, v R. Thus (2.93) follows from (2.91), proving the corollry. Tking h 1 in the preceding corollry, we recover (for f, g of bounded vrition) the integrtion by prts formul for the RYS integrl obtined in Corollry Reltions between Integrls f W p, h W q, 1 p + 1 q > 1 = (RS) f dh (LS) f dh f R, h D h R (RRS) b f dh f 0, h D, h =b f dh := [,b] f dµ h,[,b] (HK) f dh f R, h D (RYS) f dh Fig Implictions for integrls f R 1 p + 1 q = 1 (CY ) f dh In this section we explore whether, under some hypotheses, if one integrl exists, then so does nother, with the sme vlue. Figure 2.1 summrizes some implictions nd Tble 2.1 gives references for proofs. In Figure 2.1, mens tht existence of the integrl to the left of it implies tht of the integrl to the right of it, with the sme vlue. For left nd right re interchnged. The mrking f R or h R mens tht the impliction holds for regulted f or h, respectively. The mrking h D mens tht the impliction holds for regulted nd right-continuous h. Finlly, = mens tht the condition to the left of it implies existence of the integrl to the right of it. The condition 1 p + 1 q = 1 on the right hs no rrows from (or to) it,

72 88 2 Extended Riemnn Stieltjes Integrls Tble 2.1. References to proofs of the implictions shown in Figure 2.1: (RS) f dh (RRS) f dh Proposition 2.13 f W q, h W p, > 1 = p q =b f dh Corollry 3.95 (RRS) f dh =b f dh, h R Propositions 2.18 nd 2.27 (LS) f dh =b f dh, f R, h D Corollry 2.29 (LS) f dh (HK) f dh, f 0, h D, h Theorem 2.71 = b f dh (HK) f dh, f R, h D Theorem 2.69 = b f dh (RYS) b f dh Proposition 2.27 (RYS) f dh (CY ) b f dh Theorem 2.51() f W p, g W q, = 1, 1 < p < does p q not imply tht ny of the bove integrls exists Proposition signifying tht there exist f W p nd h W q with 1 p + 1 q = 1, 1 < p, q <, such tht f dh does not exist for ny of the definitions given. The figure is for rel-vlued functions nd for < b, lthough the implictions not bout LS or HK integrls hold for f nd h X- nd Y -vlued respectively. Proposition Let µ be n upper continuous dditive X-vlued intervl function on [, b], nd let ν be Y -vlued intervl function on [, b] such tht ν( ) = 0. Let h be the point function on [, b] defined by h(t) := µ([, t]) for t b, nd let g be Y -vlued regulted function on [, b] such tht ν([, t)) = g(t ) for < t b. The Kolmogorov integrl = [,b] ν([, )) dµ exists if nd only if (RRS) g() dh does, nd if they exist then for ech t b, = ν([, )) dµ = (RRS) [,t] t g () dh. Also for c < b, = (c,b] ν([, )) dµ exists if nd only if (RRS) c b g(c) dh does, nd if they exist then for ech t (c, b], = ν([, )) dµ = (RRS) (c,t] t c g (c) dh. Proof. We cn ssume tht < b. By Proposition 2.6(f), since µ is upper continuous nd dditive, h is regulted nd right-continuous on [, b]. By Propositions 2.18 nd 2.46, since for ny c t b, g (c) is left-continuous on (c, t], the integrl (RRS) c t g (c) dh exists if nd only if (RYS) c t g (c) dh does, nd the two integrls hve the sme vlue if they exist. Since ν( ) = 0 nd + h() = 0, we hve tht S YS (ν([, )), dµ; [, t], τ) = S YS (g (), dh; [, t], τ) for ny tgged Young intervl prtition τ of [, t]. Therefore by Proposition 2.25 nd the definition of the (RYS) integrl, the Kolmogorov integrl

73 2.9 Reltions between Integrls 89 = [,t] ν([, )) dµ exists if nd only if (RRS) t g () dh does, which follows from existence of either integrl for t = b, nd the two integrls hve the sme vlue for ech t if they exist, proving the first prt of the lemm. To prove the second prt, gin let c < t b. Since + h(c) = 0, we hve tht S YS (ν([, )), dµ; (c, t], τ) = S YS (g (c), dh; [c, t], τ) for ny tgged Young intervl prtition τ of [c, t]. Therefore by Proposition 2.25 gin, the Kolmogorov integrl = (c,t] ν([, )) dµ exists if nd only if (RYS) c t g(c) dh does, which follows from existence of either integrl for t = b, nd the two integrls hve the sme vlues for ech t. The proof of the proposition is complete. The following reltion between integrls of prticulr forms is used in Chpter 9. It follows from the preceding proposition nd from definitions (1.15), (1.16) of the integrls with the reversed order. Proposition Let µ be n upper continuous dditive X-vlued intervl function on [, b], nd let ν be Y -vlued intervl function on [, b] such tht ν( ) = 0. Let h be the point function on [, b] defined by h(t) := µ([, t]) for t b, nd let g be Y -vlued regulted function on [, b] such tht ν([, t)) = g(t ) for < t b. The Kolmogorov integrl = [,b] dµ ν([, )) exists if nd only if (RRS) dh g() t b, = dµ ν([, )) = (RRS) [,t] does, nd if they exist then for ech t dh g (). Also for c < b, = (c,b] dµ ν([, )) exists if nd only if (RRS) c dh g(c) does, nd if they exist then for ech t (c, b], = dµ ν([, )) = (RRS) (c,t] t c dh g (c). The dditivity property of the Kolmogorov integrl in conjunction with the preceding proposition implies the following dditivity property. Corollry Let h R([, b]; X) be right-continuous nd g R([, b]; Y ). For < c < b, the integrl (RRS) dh g() exists if nd only if both (RRS) c dh g () (RRS) nd (RRS) c dh g (c) dh g () = (RRS) c exist, nd then dh g () + (RRS) c dh g (c).

74 90 2 Extended Riemnn Stieltjes Integrls Proof. Let µ := µ h,[,b] nd ν := µ g,[,b] be the intervl functions defined by (2.2), nd let h(t) := µ([, t]) for t b. Using Proposition 2.6 nd the definition (2.3) of R µ,, it then follows tht h = R µ, = h h(), µ is n upper continuous dditive intervl function, ν( ) = 0, nd g(t ) = R ν, (t ) = ν([, t)) for < t b. Therefore by dditivity of the Kolmogorov integrl (Theorem 2.21 extended to the integrl (1.16) with the reversed order) nd by Proposition 2.90, it follows tht (RRS) dh g () exists if nd only if both (RRS) c dh g() nd (RRS) b c dh g(c) exist, nd then (RRS) dh g () The proof is complete. = (RRS) d h g () = = dµ ν([, )) [,b] = = dµ ν([, )) + = dµ ν([, )) [,c] (c,b] = (RRS) c dh g () + (RRS) c dh g (c) Bnch-Vlued Contour Integrls nd Cuchy Formuls Throughout this section we ssume tht < b. For suitble curves ζ( ) in the complex plne C nd functions f, which my be Bnch-vlued, integrls ζ( ) f(ζ)dζ will be defined nd treted. The curves will be just s in the clssicl theory of holomorphic functions into C, s follows. Definition A curve is continuous function ζ( ) from nondegenerte intervl [, b] into C. A C 1 rc is curve ζ( ) such tht for ζ ξ+iη with ξ nd η rel-vlued, the derivtive ζ (t) = ξ (t) + iη (t) exists nd is continuous nd non-zero for < t < b, nd hs limits ζ (+) = lim t ζ (t) 0 ζ (b ) = lim t b ζ (t). A piecewise C 1 curve is curve ζ on [, b] such tht for some prtition = t 0 < t 1 < < t n = b, ζ( ) is C 1 rc on ech [t i 1, t i ], nd such tht for ech i = 1,...,n 1, the rtio ζ (t i +)/ζ (t i ) is not rel nd negtive, nor is ζ (+)/ζ (b ). A closed curve is curve ζ( ) from [, b] into C such tht ζ() = ζ(b). The closed curve ζ( ) is simple if ζ(s) ζ(t) for s < t < b. If z 0 is point not in the rnge of closed curve ζ( ), tht is, z 0 rn(ζ), then we cn write ζ(t) z 0 = r(t)e iθ(t) for some rel r(t) > 0 nd θ(t) which re continuous functions of t, with θ(b) θ() = 2nπ for some integer n, clled the winding number w(ζ( ), z 0 ) of ζ round z 0. For exmple, if [, b] = [0, 1] nd ζ(t) = e 2πit, then w(ζ( ), z) = 1 if z < 1 nd 0 if z > 1.

75 2.10 Bnch-Vlued Contour Integrls nd Cuchy Formuls 91 Proposition For piecewise C 1 curve ζ( ) on [, b] (not necessrily simple) nd point z rn(ζ), we hve w(ζ( ), z) = 1 2πi (RS) dζ(t) ζ(t) z. Proof. In the representtion ζ(t) = z+r(t)e iθ(t), r( ) > 0 nd θ( ) re piecewise C 1 functions on [, b]. Using elementry clcultions, we get 1 b dζ(t) 2πi ζ(t) z = 1 2πi = 1 2πi [ dr(t) b r(t) + de iθ(t) ] e iθ(t) [ log r(b) r() + i(θ(b) θ()) ] = w(ζ( ), z) since r(b) = r(), proving the proposition. A topologicl spce (S, T ) is connected if it is not the union of two disjoint nonempty open sets. A subset C S is connected iff it is connected in its reltive topology. Holomorphic functions with vlues in Bnch spce re defined s follows: Definition Let X be complex Bnch spce nd let U be n open connected set in C. A function f : U X hs Tylor expnsion round point u U if there re n r > 0 nd sequence {h k } k 0 X such tht for ll z with z u < r, we hve z U nd f(z) = (z u) k h k, (2.94) k=0 where the series converges in X nd is clled the Tylor series of f round u. A function f : U X is holomorphic on U if it hs Tylor expnsion round ech point of U. We ssume tht holomorphic functions re defined on connected open sets, even if this ssumption is not used until we del with nlytic continution, s in Theorems 5.33 nd 5.34, for the following resons. Let U nd V be disjoint nonempty open subsets of C. Let g nd h be ny two holomorphic functions from U nd V respectively into C. Let f(z) := g(z) for z U nd f(z) := h(z) for z V. Then f would be holomorphic on U V. But suppose g nd h cn ech be extended to be entire functions, holomorphic on ll of C, nd g h on C. When function g hs n entire extension, such n extension is unique nd it seems unnturl to define holomorphic function in wy tht could contrdict n entire extension. More specificlly, consider sums f(z) = k 1 k/(z z k ) with k k <, clled Borel series. Suppose tht z k 1 for ll k nd tht the set of ll

76 92 2 Extended Riemnn Stieltjes Integrls limits of subsequences z kj with k j s j is the unit circle T 1 := {z : z = 1}. Then the series converges on U V where U := {z : z < 1} nd V := {z : z > 1} \ {z j } j 1. The sum f(z) is holomorphic on U nd on V. Ech of U nd V is open nd connected, nd they re disjoint. It cn hppen tht f 0 on U but f 0 on V, e.g. [196, Theorem 4.2.5]. So gin, nlytic continution from U would conflict with the vlues on V. Other holomorphic functions on open sets U C hve no entire extension nd non-unique mximl holomorphic extensions in C. Let C = {z = x + iy: x, y R} nd U := {z C: x > 0}. Consider the function g(z) = z for z U (which hs brnch point t 0). It hs holomorphic extension to the complement {z C: y = 0, x 0} c nd nother to {z C: x = 0, y 0} c. Neither extension cn be extended holomorphiclly to ny further point. To obtin nturl domins for holomorphic functions with brnch points one tkes Riemnn surfces (cf. [1, 2nd ed., ]). We show next tht Tylor expnsion if it exists is unique nd the series converges bsolutely nd uniformly. Lemm Let X be Bnch spce over K nd let {h k } k 0 X. If the power series k 0 tk h k converges bsolutely nd its sum is equl to zero for ech t K such tht t δ for some δ > 0, then h k 0 for k = 0, 1,.... Proof. Tking t = 0 we get h 0 0. Suppose there is lest integer k 0 1 such tht h k0 0. For ech t with 0 < t < δ, we hve k k 0 t k k0 h k = 0, nd so h k0 h k t k k0 = 1 δ k0 k>k 0 k>k 0 h k δ k ( t ) k k0 C δ δ k0 t (δ t ), where C := sup{ k k δ k : k 0} <. Letting t 0 it follows tht h k0 = 0, contrdiction, proving tht h k = 0 for ll k = 1, 2,.... Proposition Let X be complex Bnch spce, nd let U be n open connected set in C. If f : U X hs Tylor expnsion (2.94) round u U nd the Tylor series converges on B(u, r) for some r > 0, then it converges bsolutely on B(u, r) nd uniformly on B(u, s) for ny s < r. In prticulr, f is continuous on B(u, r) nd its Tylor expnsion round u is unique. Proof. If 0 < s < r then s k h k 0 s k, nd so limsup k h k 1/k 1/s. Since s (0, r) is rbitrry, limsup k h k 1/k 1/r. Thus the Tylor series (2.94) converges bsolutely by the root test for series of positive numbers. By Lemm 2.95, the Tylor expnsion (2.94) is unique. It lso follows tht the Tylor series converges uniformly on B(u, s) for ny s < r, nd so f is continuous on B(u, r), proving the proposition. Let ζ( ) be piecewise C 1 curve from [, b] into C. Let X be Bnch spce over C nd let f be continuous function into X from set in C including the rnge of ζ( ). Then the contour integrl over ζ( ) is defined by

77 2.10 Bnch-Vlued Contour Integrls nd Cuchy Formuls 93 ζ( ) f(ζ)dζ := (RS) f(ζ(t)) dζ(t), where denotes the nturl biliner mpping X C X. The Cuchy integrl formul will first be extended to Bnch-vlued functions for curves which re circles. Proposition Let f be holomorphic function from n open disk U := {z: z w < r} C into complex Bnch spce X. Let 0 < t < r nd ζ(θ) := w + te iθ for 0 θ 2π. Then for ny z C with z w < t, f(z) = 1 f(ζ)dζ 2πi ζ( ) ζ z. (2.95) Proof. Let L be ny continuous C-vlued liner functionl on X (L X ), so tht for some K <, L(x) K x for ll x X. For ech convergent series f(v) = k=0 (v w)k h k, we hve convergent series L(f(v)) = k=0 L(h k)(v w) k, nd so L f is holomorphic from U into C. Thus (2.95) holds for L f in plce of f by the clssicl Cuchy integrl formul (e.g. Ahlfors [1, 2nd ed., , Theorem 6]). Now L cn be interchnged with the integrl by Proposition If x, y X nd L(x) = L(y) for ll L X, then x = y by the Hhn Bnch theorem. The conclusion follows. For function f from n open set U C into complex Bnch spce X nd z U, the derivtive f (z) is defined, if it exists, s f(w) f(z) lim X. w z w z Let f (1) := f nd n 1. If the nth derivtive f (n) is defined in neighborhood of z, then f (n+1) (z) is defined s (f (n) ) (z), if it exists. The clssicl Cuchy integrl formul for derivtives extends directly to Bnch spces: Theorem Under the hypotheses of Proposition 2.97, f (n) (z) exists for ll n = 1, 2,... nd f (n) (z) = n! f(ζ)dζ 2πi ζ( ) (ζ z) n+1. The conclusion follows from Proposition 2.97 nd the next lemm since holomorphic function is bounded on C by Proposition Lemm Suppose tht g is continuous on the rnge C of piecewise C 1 curve ζ( ). Then for ech n 1, the function

78 94 2 Extended Riemnn Stieltjes Integrls g(ζ) dζ I n (z) = I n (g; z) := ζ( ) (ζ z) n hs derivtive on the complement C c of C, nd I n(z) = ni n+1 (z). Proof. We prove first tht I 1 is continuous on C c. Let z 0 C c nd let δ > 0 be such tht the open bll z z 0 < δ does not intersect C. By restricting z to the smller bll z z 0 < δ/2, we hve tht ζ(t) z > δ/2 for ech t [, b]. By linerity of the RS integrl, we hve I 1 (z) I 1 (z 0 ) = (z z 0 ) Using the bounds (2.84), it follows tht ζ( ) g(ζ) dζ (ζ z)(ζ z 0 ). I 1 (z) I 1 (z 0 ) z z 0 (12/δ 2 ) g ζ [,b],sup v 1 (ζ; [, b]), nd so I 1 = I 1 (g) is continuous t z 0, nd this is true for ll continuous g on C. Applying the first prt of the proof to the function h(ζ) := g(ζ)/(ζ z 0 ), we conclude tht I 1 (z) I 1 (z 0 ) z z 0 = ζ( ) g(ζ) dζ (ζ z)(ζ z 0 ) = I 1(h; z) I 1 (h; z 0 ) = I 2 (g; z 0 ) s z z 0. This proves the conclusion in the cse n = 1. The conclusion for n rbitrry n is proved by induction. Suppose tht for some n 2, I n 1(g; z) = (n 1)I n (g; z) for z C c, for ll continuous g on C. Agin let z 0 C c. Using the identity I n (z) I n (z 0 ) [ g(ζ) dζ = ζ( ) (ζ z) n 1 (ζ z 0 ) g(ζ) dζ ] ζ( ) (ζ z 0 ) n g(ζ) dζ +(z z 0 ) (ζ z) n (ζ z 0 ), ζ( ) one cn conclude tht I n is continuous t z 0. Indeed, defining h(ζ) := g(ζ)/(ζ z 0 ), the first term is equl to I n 1 (h; z) I n 1 (h; z 0 ) nd tends to zero s z z 0 by the induction hypothesis, while in the second term the integrl is bounded for z in neighborhood of z 0 s shown in the first prt of the proof. Now dividing the identity by z z 0 nd letting z z 0, the quotient in the first term tends to derivtive I n 1(h; z 0 ), which by the induction hypothesis equls (n 1)I n (h; z 0 ) = (n 1)I n+1 (g; z 0 ). The remining fctor I n (h; z) is continuous s before, nd so hs the limit I n (h; z 0 ) = I n+1 (g; z 0 ). Thus I n (z 0) exists nd equls ni n+1 (z 0 ). The proof of the lemm is complete.

79 2.10 Bnch-Vlued Contour Integrls nd Cuchy Formuls 95 Proposition Let X be complex Bnch spce, nd let f be holomorphic function on disk B(u, r) C with vlues in X for some u C nd r > 0. Then for ech z B(u, r), f(z) = f(u) + (z u) k f (k) (u) k! k 1 is the Tylor expnsion of f round u. Proof. Let 0 < s < t < r, let ζ(θ) := u + te iθ for 0 θ 2π, nd let z B(u, t). Then by the Cuchy integrl formul (Proposition 2.97), f(z) = 1 f(ζ)dζ 2πi ζ z. For ech ζ rn(ζ), the series 1 ζ z ζ( ) = (ζ u) 1 k=0 ( z u ) k ζ u converges bsolutely, nd uniformly if z u s. Since f is continuous on the rnge rn(ζ) by Proposition 2.96, it is bounded. Then term-by-term integrtion yields f(z) = (z u) k k=0 2πi ζ( ) f(ζ)dζ (ζ u) k+1. Using the bound (2.84) for the RS integrl nd the root test, it follows tht the series converges bsolutely. Now the conclusion follows by the Cuchy integrl formul for derivtives (Theorem 2.98) nd by the uniqueness of Tylor expnsion (Proposition 2.96). Proposition If f is holomorphic from n open connected set U C into Bnch spce X nd w U, then for { f(z) f(w) g(z) := z w, if z w, f (w), if z = w, g is lso holomorphic on U. Proof. It is esily seen tht g is holomorphic on U \ {w}, since z 1/(z w) is, so it suffices to find power series expnsion for g round w. Let f(z) = k=0 (z w)k x k for z w < r, where r > 0 nd x k := f (k) (w)/k! X by Proposition Then g(z) = k=0 (z w)k x k+1, lso for z w < r, with g(w) = x 1 = f (w), finishing the proof.

80 96 2 Extended Riemnn Stieltjes Integrls A topologicl spce (S, T ) is loclly connected iff T hs bse consisting of connected open sets. Clerly, ny normed vector spce with its usul topology is connected nd loclly connected. In prticulr, R nd C re loclly connected. Theorem For ny topologicl spce (S, T ) nd ny nonempty connected set A S there is unique connected set B A such tht B is mximl connected set for inclusion. Such set B is lwys closed. Proof. The collection of connected sets including A is prtilly ordered by inclusion. The union of ny inclusion-chin of connected sets is esily seen to be connected. Thus by Zorn s lemm there is connected set B 1 A which is mximl connected set for inclusion. Suppose B 2 is nother such set. The union of two non-disjoint connected sets is esily seen to be connected. This gives contrdiction unless B 1 = B 2, so B is unique. Moreover, the closure of ny connected set is esily shown to be closed. So by mximlity, B is closed nd the theorem is proved. A mximl connected subset B of S is clled component. If A is singleton {x}, it is lwys connected, nd the B from the preceding theorem is clled the component of x. Proposition Let (S, T ) be loclly connected topologicl spce. Then ny component F of S is open s well s closed. Thus S hs unique decomposition s union of disjoint open nd closed components. Proof. Let F be component of S nd x F. Let V be connected open neighborhood of x by locl connectedness. Then F V is connected since F nd V re nd F V {x}. So V F nd F is open. The rest follows. Clerly, n open set in loclly connected spce is loclly connected, so it is in unique wy union of disjoint nonempty connected open sets. An open set U C will here, following Ahlfors [1, 2nd ed., 4.4.2], be clled simply connected if nd only if U c hs no bounded component. A more generl definition is bsed on the following. Definition A closed curve ζ( ) from [, b] into set U is nullhomotopic in U if there exists jointly continuous function H from [, b] [0, 1] into U such tht H(, 0) ζ( ), for some w U, H(, 1) w, nd for 0 u 1, H(, u) = H(b, u), so tht ech H(, u) is closed curve. Theorem Let U C be connected nonempty open set. Then the following re equivlent: () ny closed curve ζ( ) with rnge in U is null-homotopic in U;

81 2.10 Bnch-Vlued Contour Integrls nd Cuchy Formuls 97 (b) for every closed curve ζ( ) into U nd z U c, w(ζ( ), z) = 0; (c) U c hs no bounded component; (d) U is homeomorphic to the open unit disk {z C: z < 1}. Proof. The implictions (d) () (b) nd (c) (b) re esy. The other implictions re not s esy nd re given in Newmn [178, 6.6 nd 7.9]. Complex nlysis texts lso give proofs, of the stronger fct tht in (d), unless U = C, the homeomorphisms cn be tken to be holomorphic with holomorphic inverses (Riemnn mpping theorem, see e.g. Berdon [12, Theorem nd 11.2]). In Theorem 2.105, () is the definition of simply connected for generl topologicl spces. It is not equivlent to (c) in R 3, for exmple. Here is Cuchy integrl theorem for Bnch-vlued functions. Theorem Let f be holomorphic from n open connected set U C into complex Bnch spce X nd let ζ( ) be piecewise C 1 closed curve whose rnge is included in U. If (i) the winding number w(ζ( ); z) = 0 for ech z U, e.g. if (ii) U is simply connected, then f(ζ)dζ = 0. (2.96) ζ( ) Proof. Condition (ii) implies (i) by Theorem 2.105, (c) (b), or () (b) for the generl definition of simply connected. So we cn ssume (i). By Proposition 2.96, f is continuous on U, nd so the integrl (2.96) is defined. For ny continuous liner functionl L X, z L(f(z)) is holomorphic from U into C. Thus by Proposition 2.78, we hve ( ) L f(ζ)dζ = L(f(ζ))dζ. ζ( ) ζ( ) The integrl on the right is zero by the Cuchy theorem for multiply connected sets (e.g., Ahlfors [1, 2nd ed., Theorem 18]). Thus by the Hhn Bnch theorem (e.g. [53, Corollry]), it follows tht (2.96) holds, proving the theorem. Here re Cuchy integrl formuls with winding numbers. Theorem Let f be holomorphic from connected nd simply connected open set U C into complex Bnch spce X. Let ζ( ) be piecewise C 1 closed curve whose rnge C is included in U nd let z U \ C. Then w(ζ( ), z)f(z) = 1 f(ζ)dζ 2πi ζ( ) ζ z, (2.97)

82 98 2 Extended Riemnn Stieltjes Integrls nd for n = 1, 2, 3,..., w(ζ( ), z)f (n) (z) = n! 2πi ζ( ) f(ζ)dζ. (2.98) (ζ z) n+1 Proof. On the right side of (2.97) write f(ζ) = [f(ζ) f(z)] + f(z). For the f(ζ) f(z) term, the integrl is 0 by Proposition nd Theorem For the f(z) term, we get the left side of (2.97) by Proposition 2.93, proving (2.97). We thus hve (2.98) for n = 0. For given ζ( ) nd z C, w(ζ( ), u) is constnt for u in neighborhood of z, so the multipliction by w(ζ( ), u) cn be interchnged with d n /du n u=z. On the right, we cn use induction on n nd Lemm 2.99, so (2.98) is proved. If w(ζ( ), z) = 1, we get the fmilir form of the formul, now for rther generl closed curves ζ( ) Notes Notes on Section 2.1. More properties of rel-vlued function equivlent to being regulted re given by Theorem 2.1 in [54, Prt III]. The correspondence between right-continuous regulted point functions nd dditive upper continuous intervl functions given by Corollry 2.11 is similr to the well-known correspondence between right-continuous point functions of bounded vrition nd σ-dditive set functions on the Borel σ- lgebr. Notes on Section 2.2. In T. J. Stieltjes s tretment of his integrl f dh in [227], the integrnd f is continuous, while the integrtor h is monotone incresing function. The Riemnn Stieltjes integrl (RS) f dh, for ny f, h such tht it exists, ppered in G. König [122], who pprently hd been using the RS integrl in his lectures for some time, s stted in [228, p. 247] nd in [168, p. 314]. Szénássy [228] hs chpter on König, where on p. 247 of the English trnsltion, König s generl definition of the RS integrl is quoted. Interest in Stieltjes-type integrls flourished fter F. Riesz [193] showed tht the Stieltjes integrl provides representtion of n rbitrry bounded liner functionl on the spce C[0, 1] of ll continuous functions on [0, 1]. Pollrd [187] introduced the refinement Riemnn Stieltjes integrl, showed tht it extends the Riemnn Stieltjes integrl, nd proved tht for h nondecresing nd f bounded, the integrl (RRS) f dh exists if nd only if the Drboux Stieltjes integrl does. The ltter integrl is defined s sup κ L(f, h; κ) = inf κ U(f, h; κ) if the equlity holds, where for prtition κ = {t i } n i=0 M(f; A) := sup{f(t): t A}, { L(f, h; κ) := n m(f; [t i 1, t i ])[h(t i ) h(t i 1 )], of [, b], m(f; A) := inf{f(t): t A} nd U(f, h; κ) := n M(f; [t i 1, t i ])[h(t i ) h(t i 1 )]. (2.99)

83 2.11 Notes 99 F. A. Medvedev [168] gives n extensive ccount of the erly history of Stieltjes-type integrls, nd in prticulr gives some detils bout König s work on the RS integrl. S. Pollrd [187] gve detiled tretment of the refinement Riemnn Stieltjes integrl. Erlier, theory of limits bsed on refinements ws treted by E. H. Moore [175]. Thus the RRS integrl sometimes is clled the Pollrd Moore Stieltjes integrl (see e.g. [94, p. 269]). The books of Gochmn [82] nd Hildebrndt [97, 9 18] lso contin detiled expositions of the RRS integrl. For rel-vlued functions, Propositions 2.13 nd 2.15 were proved in Pollrd [187, p. 90] nd Smith [223], respectively. Notes on Section 2.3. The refinement Young Stieltjes integrl developed in stges over long period. This integrl ws rediscovered by severl uthors, nd hence it is known by different nmes. According to T. H. Hildebrndt [94], the RYS integrl originted in the work of W. H. Young [254]. In this work W. H. Young extended his pproch to defining the Lebesgue integrl, nd the RYS integrl ppered there s n uxiliry tool. At tht time there ws considerble interest in enlrging the clss of functions integrble with respect to monotone function h. Lebesgue [134], for exmple, showed tht such clss could be the clss of ll Lebesgue summble functions with respect to dh when Stieltjes-type integrl is defined by mens of the Lebesgue integrl using chnge of vribles formul. Lebesgue suggested tht it would be difficult to extend the Stieltjes integrl to such generl integrnds by ny other mens. Recll tht Rdon s work which led to the Lebesgue Stieltjes integrl ppered little lter [190]. However, W. H. Young [254] showed tht his method of monotonic sequences used in connection with the Lebesgue integrl extends to n integrtion with respect to ny function h of bounded vrition lmost without chnges. The min chnge concerned the definition of the integrl f dh for step function f. In this cse, W. H. Young set { [f + h](x i 1 ) + c i [h(x i ) h(x i 1 +)] + [f h](x i )] }, f dh := i (2.100) provided f(x) = c i for x (x i 1, x i ). The method of monotonic sequences of W. H. Young ws lter extended by Dniell [37] to integrls of functions defined on bstrct sets. A concise theory bsed on n integrl of Stieltjes type ws provided by W. H. Young s son L. C. Young in the form of textbook [243], first published in 1927 (see lso the presenttion in Hildebrndt [97]). The integrl, defined s limit of Young Stieltjes sums if it exists when the mesh of tgged prtitions tends to zero, ws mentioned by R. C. Young [251]. Full use of the refinement Young Stieltjes integrl, in the context of Fourier series, is due to L. C. Young [247]. Motivted by the wekness of the Riemnn Stieltjes integrl, Ross [195] suggested n extension of the Drboux Stieltjes integrl, replcing (2.99) by sums reminiscent of Young Stieltjes sums. Nmely, let f be rel-vlued func-

84 100 2 Extended Riemnn Stieltjes Integrls tion on [, b], nd let h be nondecresing function on [, b]. For prtition κ = {t i } n i=0 of [, b], let { n L(f, dh; κ) := J(f, dh; κ) + m(f; (t i 1, t i ))[h(t i ) h(t i 1 +)] U(f, dh; κ) := J(f, dh; κ) + n M(f; (t i 1, t i ))[h(t i ) h(t i 1 +)], where J(f, dh; κ) := n f(t i) ± [,b] h(t i), nd m(f, A), M(f, A) re defined s in (2.99). We sy tht f is Ross Drboux Stieltjes, or RDS, integrble on [, b] with respect to h, if U(f, dh) := inf κ U(f, dh; κ) = sup κ L(f, dh; κ) =: L(f, dh), nd then let (RDS) f dh := U(f, dh) = L(f, dh). By Theorem in [195], bounded function f on [, b] is RDS integrble with respect to h if nd only if there is C R such tht given ǫ > 0 there exists δ > 0 such tht S YS (f, dh; τ) C < ǫ for ech tgged Young intervl prtition τ = ({t i } n i=0, {s i} n ) of [, b] such tht mx{h(t i ) h(t i 1 +): i = 1,..., n} < δ. Then using Theorem 2.1(b), it is esy to see tht if (RDS) f dh exists then so does (RYS) f dh nd the two re equl. The Brtle integrl given by Definition 2.37 is generl biliner integrl of vector function with respect to n dditive vector mesure. According to Diestel nd Uhl [41, p. 58], Brtle [11] lunched theory of integrtion tht includes most of the known integrtion procedures tht hve ny clim to qulity. His integrl specilizes to include the Bochner integrl but does not include the generl Pettis integrl. Possibly workers in the theory of vector mesures would be better off if they ttempted to use the Brtle integrl rther thn inventing their own. A survey of the history of vector integrtion cn be found in Hildebrndt [95] nd Brtle [11]. There re severl different constructions of n integrl of Bnch-spcevlued function with respect to nonnegtive finite sclr-vlued mesure µ. The Bochner integrl given by Definition 2.31 is one of the best known mong such integrls. S. Bochner [21] defined his integrl. The extension of Lebesgue s differentition theorem to Bnch-vlued functions, Theorem 2.35, is given e.g. by Diestel nd Uhl [41, Theorem II.2.9, p. 49]. A function f tking vlues in Bnch spce X is integrble with respect to µ in the sense of Dunford [56] if there exists sequence of µ-simple functions {f k } k 1 such tht it converges to f µ-lmost everywhere, nd the sequence { A f k dµ} k 1 converges in X for ech A S. The integrl (D) A f dµ is then defined to be the limit lim k A f k dµ, nd is clled the second Dunford integrl. The integrl does not depend on {f k } since it is specil cse of the Brtle integrl by Theorem 9 in [11], which is well defined by Proposition The second Dunford integrl clerly extends the Bochner integrl. Hildebrndt [95, p. 123] showed tht for µ-mesurble functions f, the second Dunford integrl coincides with severl other integrls. Exmple 7 of Birkhoff [16, p. 377] gives µ-mesurble function f such tht f( ) is not summble but is integrble in the sense of Dunford. For sclr-vlued mesurble function f nd vector-vlued mesure µ, n integrl nlogous to the second Dunford integrl is treted in Dunford nd Schwrtz [57, Definition IV.10.7].

85 2.11 Notes 101 Notes on Section 2.4. The if prt of Theorem 2.42 ws proved in Smith [223, p. 495]. Notes on Section 2.5. L. C. Young [244, p. 263] defined n extension of the Riemnn Stieltjes integrl for regulted complex-vlued functions f nd h by (Y 0 ) f dh := (RRS) f + dh + t b [ f(t) f(t+) ][ h(t+) h(t ) ] if the RRS integrl exists nd the sum converges bsolutely. However, the vlue of the Y 0 integrl depends on h( ), f(b+), nd h(b+), which need not be defined. If in the Y 0 integrl we replce f(t+) by f (b) + (t), h(t+) by h(b) + (t), nd h(t ) by h () (t), then we get the Y 1 integrl defined by (2.44). L. C. Young in 1938 gve such convention for vlues t the endpoints in footnote 6 on p. 583 of [247], so one my suppose tht he hd these definitions in mind lredy in his 1936 pper. The Y 1 nd Y 2 integrls hve been defined nd proved equl when both exist in Dudley [49]. An extension of the Y 1 nd Y 2 integrls to integrls of the form f dh g with two integrnds f nd g ws given in [54, Prt II]. In Section 9.8 below such n extension is given for the RYS nd Kolmogorov integrls. Theorem 2.50 for functions with vlues in Bnch lgebr is specil cse of Theorem 3.7 in [54, Prt II]. Notes on Sections 2.6 nd *2.7. Sks [201] defines integrls due to Perron nd Denjoy nd gives references to their work, published beginning in The Perron integrl nd one form of the Denjoy integrl turned out to be equivlent to the Henstock Kurzweil integrl for integrls g(x)dx: Gordon [84, Chpter 11]. Wrd [238] in 1936 defined n integrl clled Perron Stieltjes integrl which includes both the Lebesgue Stieltjes nd refinement Riemnn Stieltjes integrls. Theorem 2.64 is Theorem 5 of Wrd [238]. Kurzweil [129, Section 1.2] in 1957 defined the Henstock Kurzweil integrl nd proved Theorem 2.65 on the equivlence of the W P S nd HK integrls [129, Theorem 1.2.1]. Kurzweil ([129], [130], [131]), Henstock ([90], [91], [92]), nd McShne [166] consider extended integrls for functions U(, ) of two vribles, the second nd third uthors lso on generl spces X, where (HK) f dh is the specil cse U(x, y) f(y)h(x) nd X = [, b]. A discussion of Proposition 2.58 nd its proof when h(x) = x cn be found in McLeod [165, Sections 1.5, 2.8, nd 7.3]. While there hs been reltively little literture bout the refinement Young Stieltjes nd centrl Young integrls, there hs been much more bout Henstock Kurzweil (or guge) integrls, e.g. Lee Peng-Yee [184]. A 1991 book by Henstock [92] hs reference list of more thn 1200 ppers nd books, minly on the theme of non-bsolute integrtion, if not necessrily bout

86 102 2 Extended Riemnn Stieltjes Integrls the HK integrl itself. In fct, Henstock [92] trets integrtion over generl spces ( division spces ). Schwbik [211, Theorem 3.1] proved Theorem 2.69 when h is of bounded vrition nd f is not necessrily regulted but either () f is bounded, or (b) f is rbitrry nd for ny t if h(t ) = h(t+) then h(t) = h(t ). Notes on Section 2.8. According to Hildebrndt [94, p. 276], the integrtion by prts formul (2.79) for the refinement Young Stieltjes integrl ws proved by de Finetti nd Jcob [66] ssuming tht f nd g re of bounded vrition. Hewitt [93] proved n integrtion by prts formul similr to (2.83) for the Lebesgue Stieltjes integrl. Corollry 2.83 contins the result of Love [145], who clls the refinement Young Stieltjes integrl the refinement Ross- Riemnn Stieltjes integrl, or the R 3 S-integrl. Notes on Section Ahlfors [1, 2nd ed., 4.4.4] wrote: It ws E. Artin who discovered tht the chrcteriztion of homology by vnishing winding numbers ties in precisely with wht is needed for Cuchy s theorem. This ide hs led to remrkble simplifiction of erlier proofs. Proposition 2.93 is given e.g. in Ahlfors [1, 2nd ed., 4.2.1].

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