Appendix B. Criterion of Riemann-Stieltjes Integrability

Transcription

1 Appendx B. Crteron of Remann-Steltes Integrablty Ths note s complementary to [R, Ch. 6] and [T, Sec. 3.5]. The man result of ths note s Theorem B.3, whch provdes the necessary and suffcent condtons for Remann-Steltes ntegrablty of f wth respect to α n terms of sets of pont of dscontnuty of these functons. In an equvalent form, ths result s contaned n [H, Theorem C]. Here we gve a more drect proof, whch does not use explctly the Lebesgue measure. Let α = α(x) be a monotoncally non-decreasng functon on a fnte nterval [a, b], and let f = f(x) be a bounded real functon on [a, b]. For an arbtrary partton := {a = x 0 x 1... x n 1 x n = b} of [a, b], we defne the upper and lower sums as follows: (1) U(, f, α) := M α, L(, f, α) := where m α, (2) M := sup f m := nf f, α := α(x ) α(x 1 ) for = 1, 2,..., n. [x 1,x ] [x 1,x ] For any two parttons 1 and 2, ther common refnement := 1 2 satsfes (see [R, Theorem 6.4]) (3) U( 1, f, α) U(, f, α) L(, f, α) L( 2, f, α). Therefore, we always have (4) nf U(, f, α) sup L(, f, α). Defnton B.1. The functon f s Remann-Steltes ntegrable wth respect to α on [a, b] f both sdes of (4) are equal. In ths case, we wrte f R(α) on [a, b] and defne the Remann- Steltes ntegral (5) b a f dα := nf U(, f, α) = sup L(, f, α). Theorem B.2 ([R], Theorem 6.6). f R(α) on [a, b] f and only f for every ε > 0 there exsts a partton such that (6) U(, f, α) L(, f, α) < ε. roof. For every ε > 0 there are parttons 1 and 2 such that U( 1, f, α) < nf U(, f, α) + ε 2, L( 2, f, α) > sup U(, f, α) ε 2. If f R(α) on [a, b], then we have equalty n (4), whch mples 0 U( 1, f, α) L( 2, f, α) < ε 2 + ε 2 = ε, and (6) follows from (3) wth = := 1 2. On the other hand, f we have (6), then the dfference between nf and sup n (4) s less than ε. Snce ε > 0 s arbtrary, we must have the equalty,.e. f R(α) on [a, b]. B 1

2 Further, snce α(x) s non-decreasng on [a, b], there are one-sded lmts α(p ) := lm α(y), a < p b; α(p+) := lm α(y), a p < b, y p y p+ and α(p ) α(p) α(p+). Theorem B.3. Let f be a bounded real functon on [a, b]. Then f R(α) on [a, b] f and only f f and α satsfy both propertes (I) and (II) below. (I) () If α(p ) < α(p), a < p b, then f(p ) = f(p). () If α(p) < α(p+), a p < b, then f(p+) = f(p). (II) Let S f and S α denote the sets of ponts of dscontnuty of f and α correspondngly. Then for every ε > 0 there exsts a (fnte or countable) sequence of ntervals (a, b ), 1, such that (7) S := (S f \ S α ) ( (a, b ), and α(b ) α(a ) ) < ε. Here the ntervals (a, b ) are not necessarly contaned n [a, b]. We extend f f(a), α α(a) on (, a) and f f(b), α α(b) on (b, + ), so that the last expresson, and also the expresson n (9) below, are well defned n any case. Remark B.4. The property (I) smply says that f f R(α) on [a, b], then f and α cannot be both left-dscontnuous, or both rght-dscontnuous at same pont. Of course, ths property s redundant f α s contnuous on [a, b]. By change of varable ([R, Theorem 6.19]), ths case can be reduced to α(x) x. In ths partcular case, our theorem s contaned n [T, Theorem 3.5.6]. Defnton B.5. The llaton of f on a set A, (8) A f := sup A f nf f = sup f(x) f(y). A x,y A If f s defned on [a, b], then the llaton of f at a pont p [a, b], (9) ω f (p) := lm h 0+ [p h,p+h] Lemma B.6. () f s contnuous at p f and only f ω f (p) = 0; () f(p ) = f(p) f and only f f 0 as h 0+; () f(p+) = f(p) f and only f [p,p+h] f 0 as h 0+. We skp the proof, because t s very elementary (see [T, Theorem 3.5.2]). Lemma B.7. If f R(α) on [a, b], then f and α satsfy the propertes (I) n Theorem B.3. roof. Let p be a pont such that α(p ) < α(p), a < p b. By Theorem B.2, for every ε > 0 there s a partton := {a = x 0 x 1... x n = b} (dependng on α) such that (10) U(, f, α) L(, f, α) = (M m ) α < ε. Next, for small h (0, p a), the nterval (p h, p) does not contan pont x. From (3) (wth 1 = 2 = ) t follows that the refned partton := {p h, p} satsfes U(, f, α) L(, f, α) U(, f, α) L(, f, α) < ε B 2 f.

3 Therefore, replacng by f necessary, we can assume that p h, p,.e. Then from (10) t follows p h = x 0 1 < p = x 0 for some 0 {1, 2,..., n}. f α 0 = ( ) M 0 m 0 α0 < ε. Snce α 0 = α(p) α(p h) α(p) α(p ) > 0, and ε > 0 can be chosen arbtrarly small, we conclude that f 0 as h 0+. By Lemma B.6(), we have f(p ) = f(p). The proof of part () n (I) s complete. art () can be proved qute smlarly. Lemma B.8. If f R(α) on [a, b], then f and α satsfy the property (II) n Theorem B.3. roof. By Lemma B.6(), the set of ponts of dscontnuty of f, (11) S f = { p [a, b] : ω f (p) > 0 } = F k, where F k := { p [a, b] : ω f (p) 2 k}. Fx ε > 0. By Theorem B.2, for every k = 1, 2,..., there exsts a partton := {a = x 0 x 1... x n = b} (dependng on k) such that (12) U(, f, α) L(, f, α) = (M m ) α < ε k := 4 k ε. Note that f p F k \, then for some {1, 2,..., n} we have p (x 1, x ), and M m ω f (p) 2 k. Let A k denote the set of all such ndces. Then (13) (F k \ ) (x x 1 ), and α 2 k (M m ) α < 2 k ε. A k A k A k Further, F k \ S α s contaned n (F k \ ) ( \ S α ). Snce α(x) s contnuous at every pont p \ S α, one can cover such pont by ntervals (p h, p + h) wth arbtrarly small α(p + h) α(p h). Together wth (x 1, x ), A k, these ntervals compose a fnte famly of ntervals (a k,, b k, ) such that (F k \ S α ) ( (a k,, b k, ), and α(bk, ) α(a k, ) ) < 2 k ε. Fnally, by vrtue of (11), (S f \ S α ) = (F k \ S α ) (a k,, b k, ), and ( α(bk, ) α(a k, ) ) < 2 k ε = ε. Snce the countable set of ntervals {(a k,, b k, )} can be renumbered as {(a, b )}, we get the desred property (7). The followng lemma, together wth the prevous Lemmas B.7 and B.8, completes the proof of Theorem B.3. Lemma B.9. Let f be a bounded functon on [a, b] satsfyng the propertes (I) and (II) n Theorem B.3. Then f R(α) on [a, b]. B 3

4 roof. Step 1. We have f M = const < on [a, b]. By Theorem B.2, t suffces to show that for an arbtrary ε > 0, there exsts a partton := {a = x 0 < x 1 <... < x n 1 < x n = b} of [a, b] satsfyng the nequalty (6) for gven f and α. Ths nequalty can be wrtten n the form (14) U(, f, α) L(, f, α) = f α < ε, where I := [x 1, x ]. Step 2. Fx a constant ε 1 > 0. Note that snce α(x) s a monotone functon, ts set of ponts of dscontnuty S α s at most countable: S α := {c 1, c 2,...}. From the assumpton (I) n Theorem B.3 t follows that for each = 1, 2,..., one can choose a small constant h > 0 such that (15) I 1, f α < 2 ε 1, I 1, I + 1, f α < 2 ε 1, for = 1, 2,..., I + 1, where I 1, := [c h, c ], I + 1, := [c, c + h ]. Obvously, we also have (16) S α := {c 1, c 2,...} V 1 := 1 I 1,, where I 1, := (a 1,, b 1, ) := (c h, c + h ). Step 3. Based on the constant ε 1 > 0, defne the set (17) F := {p [a, b] : ω f (p) ε 1 > 0}. We clam (as n [T, Lemma 3.5.4]) that F s compact. Indeed, f p F and p p 0 [a, b] as, then for an arbtrary h > 0 there s such that p p 0 < h/2. For such, we have (p h/2, p + h/2) (p 0 h, p 0 + h), hence by (8) and (9), the llaton of f, and f f ω f(p ) ε 1, [p 0 h,p 0 +h] [p h/2, p +h/2] ω f (p 0 ) := lm h 0+ Ths argument proves the compactness of F. f ε 1 > 0,.e. p 0 F. [p 0 h,p 0 +h] Step 4. Further, note that F S f the set of ponts of dscontnuty of f. Therefore, by our assumpton (II), for the gven constant ε 1 > 0, there exsts a sequence of ntervals I 2, := (a 2,, b 2, ) such that (18) (F \ S α ) (S f \ S α ) V 2 := ( I 2,, and α(b2, ) α(a 2, ) ) < ε 1. Step 5. From (16) and (18) t follows F (V 1 V 2 ), so that the compact set F s covered by the unon of two famles of open ntervals {I 1, } and {I 2, }. Therefore, one can choose fnte subfamles {I 1,} {I 1, } and {I 2,} {I 2, } such that (19) F (V 1 V 2), where V 1 := I 1,, V 2 := I 2,. Consder another compact set F := [a, b] \ (V 1 V 2). Snce F does not ntersect F, we have ω f (p) < ε 1 for every p F. By defnton of ω f (p) n (9), (20) [p h,p+h] f < ε 1 for every p F wth some h = h(p) > 0. B 4

5 The famly of the correspondng open ntervals {(p h, p + h), p F } covers the compact F. Therefore, ths famly contans a fnte subfamly {I 3, := (a 3,, b 3, )} such that (21) F V 3 := I 3,, and [a 3,,b 3, ] f < ε 1 for each. Step 6. It s easy to see that (19) and (21) mply [a, b] (V 1 V 2 V 3), so that [a, b] s covered by the unon of three fnte famles of open ntervals {I 1,}, {I 2,}, and {I 3,}. Let := {a = x 0 < x 1 <... < x n 1 < x n = b} be a partton of [a, b], whch ncludes the pont a, b, all the endponts of ntervals I 1,, I 2,, I 3,, and also the centers c of the ntervals I 1, := (c h, c + h ), whch belong to (a, b). Denote I := [x 1, x ] for = 1, 2,..., n. Note that I are closed ntervals, whereas I 1,, I 2,, I 3, are open. However, all the estmates (15), (18), and (21), hold true for closed ntervals. Let A 1 denote the set of all ndces {1, 2,..., } such that I V 1, A 2 the set of all / A 1 such that I 2 V 2, and A 3 the set of all the remanng, for whch we automatcally have I V 3, because [a, b] (V 1 V 2 V 3). For each A 1, we have ether I I 1, or I I + 1, for some, hence by vrtue of (15), (22) f α < 2 A 1 2 ε 1 = 2ε 1. Smlarly, snce f M, we have f 2M, and the last nequalty n (18) mples (23) f α 2M α < 2M ε 1. I A 2 A 2 Fnally, from (21) and monotoncty of α t follows (24) f α ε 1 α ( α(b) α(a) ) ε 1. I A 3 A 3 Snce A 1 A 2 A 3 = {1, 2,..., n}, the estmates (22) (24) yeld f α ( 2 + 2M + α(b) α(a) ) ε 1 < ε, provded 0 < ε 1 < ( 2 + 2M + α(b) α(a) ) 1 ε. Thus we have the desred estmate (14) and lemma s proved. =1 References [H] H. J. Ter Horst, Remann-Steltes and Lebesgue-Steltes Integrablty, Amer. Math. Monthly, vol. 91, 1984, pp [R] W. Rudn, rncples of Mathematcal Analyss, 3rd edton. [T] W. F. Trench, Introducton to real analyss. B 5