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0. [Hint: Find ip'(x) and note ip(x) > 0 as s; > 0 + .] 21. Let fk{t) = (1/fc2 Vt) cos(fc/t). Show that H ° = 1 /o A = lo E£Li A22. Let /fc(<) = efe*  2 e  2 f c t . Show that £ £ 1 /0°° /* ^ /0°° £ £ 1 / fc . 23. Let / : [—1,1]  ^ R b e continuous. Show that lim3._).0+ j _ x sx+tif[t)dt — 0. Let 7 be a gauge such that \S{fk,V) — fj fk\ < £ for every k whenever V « 7. Fix a tagged partition V = {(ti,Ii) : 1 < i < m} which is 7fine. Choose n such that k > n implies /fc(t,) < e/m£(Ii) for i = 1 , . . . ,m. Suppose that J is an arbitrary subinterval of / and k > n. From the Uniform Henstock Lemma (3.8), we have P
= f(ip(P))f(ip(a)) =
f^f.
We now give two examples which like the proof of the FTC illustrate the advantage and generality achieved by allowing the intervals in the gauge to have variable lengths.
Introduction to the Gauge or HenstockKurzweil
Integral
7
Example 7. Suppose f : [a, b] —> R has the constant value c except at a countable number of points E = {zk : k € N}. We show that f is integrable over [a,b] with J f = c(b — a). Let e > 0. If V = {(ti,Ii) : 1 < i < n} is a tagged partition of [a,b], consider (6)
\S(f,V)c{ba)\
£{/&)cK(io i=l
If ti ^ E, the term, (f(ti) — c)l(Ii) in (6) is 0 so if is reasonable to start defining a gauge by setting 7(f) = (t — l,t + l) ift^E. If ti — Zk for some k and ifV « 7 /or some gauge 7, t/ien /(i») — c\t(Ii) < \f(zk) — c\£(j(zk)) So if we choose 5k = e/{\f(zk)  c\)2k+2 and set j(zk) = (zk  h,Zk + fa), then when V « 7 and U = Zk we have \f(U) — c\£(Ii) < s/2k+l. IfV « 7, we have from (6) that fe+i
\S{f,V)c{ba)\<2Yjs/2 fc=i
since each Zk can be the tag for at most 2 subintervals in V. In particular, if f(t) = 0 for t irrational and f(t) = 1 for t rational (the Dirichlet function), then / is integrable over any interval [a, b] with J f = 0. The Dirichlet function is the usual example that is given for a (bounded) function which is not Riemann integrable. The computation above in Example 7 illustrates the advantage of using gauges with variable length. [For a more general result see Example 10.] We next consider an example of an unbounded function which has an improper Riemann (or CauchyRiemann) integral. Example 8. Let f(x) = 1/yfx for x > 0 and /(0) = 0. We claim that f is integrable over [0,1] with L f = 2 as in the calculus. Let e > 0. First consider the function near 0. If0<x
< 2 V / e 2 /16 = e/2
Introduction to Gauge Integrals whenever [0, x\] C 7(0). IfO < u < v < 1, the area under the curve over [u,v] is 2y/v — 2y/u and if u < z
=
<
1
(v
y/z
v—u z
\/z{y/v
V — U
y/z(y/v
y/v +
y/u
+ y/u)

2z

2z
y/v + y/u + y/u)
z V — U \y/v + y/u 
2y/z\
z V —U I
z
V —Z
Z — U
\y/v + y/z
V — U (V — Z
+ y/u + Z —u\
yfz
y/z (v — u)' .2/3
yfz
This suggests that we define 8(z) = ez3/2/4 and set 7(2) = (z — 5{z), z + 5{z)) n (0,2) for 0 < z. Now suppose V = {(<»,£) : 0 < i < n} « 7, with Ii = [xi,xi+i],0 = x0 < x\ < • • • < xn+i = 1. Then to = 0 and from the estimate above, we have \S(f,V)2\
^2{f(U)(xi+1
 Xi)  2 ( ^ + 1  y/x~i)}
i=0
< 2y/x~l + ^T \f(U)(xi+i
 Xi)  2(y/xi+1  yfxl)
i=l
< £
/2 + E
(xi+i
 Xi)2 .3/2
i=i
<£/2 + »=i E = £.
£{xi+i

Xi)
^ __ / 0
t
^
(Xi+i
< ^ + E
 Xi) 25{U) .3/2
Introduction to the Gauge or HenstockKurzweil
Integral
9
Note that near the point 0 where the function / has a singularity the gauge is much "finer" than it is near the point 1 when the function is relatively "flat." That is, by using variable length gauges we are able to take into account the local behavior of the function; the Riemann integral is not able to take this local behavior into account. As noted above the function / has an improper or CauchyRiemann integral. We will show later that any function which has an improper or CauchyRiemann integral is gauge integrable (Theorem 3.4); that is, there are no "improper integrable" functions for the gauge integral (see also Theorem 4.4 for the case of integrals over unbounded intervals). Notes/Remarks The gauge integral was originally introduced by J. Kurzweil and used for studies in ordinary differential equations ([Kl]). The integral was discovered independently by R. Henstock who developed the major convergence theorems (Monotone and Dominated Convergence Theorems) for the gauge integral ([HI]; see [H2] for an exposition of the integral by Henstock). For texts containing expositions of the gauge integral, see [DS], [Gol], [K2], [LPY], [M],
[ML] and [Pfl]. The version of the FTC for both the Riemann and Lebesgue integrals require the assumption that the derivative be either Riemann or Lebesgue integrable, respectively. [See Example 2.12 and [Swl] § 3.3 and 4.3.1.] This led mathematicians to search for an integration theory for which the FTC holds in full generality. In the 1910's Perron and Denjoy developed theories of integration which extended the Lebesgue theory and for which the FTC holds in full generality as in Theorem 5. These integration theories were later proven to be equivalent and are equivalent to the gauge integral. For an exposition of these integrals and proofs of their equivalence, see [Gol]; for a historical discussion of the Perron and Denjoy integrals, see [Pe]. Exercise 10 gives a generalization of the FTC in Theorem 5 (see also [ST]). Other versions of the FTC for the gauge integral are given in [Sw3], and for an interesting version involving parametric derivatives, see [LA2]. Theorem 3 on the existence of 7fine tagged partition is often attributed to Cousin ([Co]). There is more to Theorem 3 than one might suspect at first glance. In fact, Theorem 3 is equivalent to the compactness of a closed bounded interval (see Exercise 4). For suppose that U is an open cover of I = [a, b). For each t there exists Ut S U such that t £ Ut Pick an open interval ^(t) such that t G 7(f) C Ut. Then 7 is a gauge on J so there exists a 7fine tagged partition V = {{U,h) :l
10
Introduction to Gauge Integrals
in particular [M] and [Go5] where Theorem 3 is used to derive many of the fundamental results of real analysis such as the Intermediate Value Theorem, the Extreme Value Theorem and the uniform continuity theorem. See also [Bol], [Bo2], [Shj and [Th] for a result similar to Theorem 3 and applications. In the remarks preceding the definition of the gauge integral, we indicated that the gauge integral can be viewed as being obtained from the Riemann integral by replacing the positive constant 5 in the definition of the Riemann integral by a positive function 5. Indeed, it would be quite simple to restate Definition 2 by replacing the gauge 7 by a positive function 5 and employing condition (4) instead of (5). It is natural to ask how general the function 5 can be in the resulting definition. This question has been addressed by Foran and Meinershagen ([FM]), Pfeffer ([Pf2]), Liu ([L]), and Gordon ([Gol] 9.24, [Go4]). Finally, we give an example generalizing the result in Example 7 which will be used in several starredsections. Definition 9*. A subset E C [a, b] is (Lebesgue) null if for every £ > 0 there is a countable collection of open intervals {Ij} covering E with ^2j£(Ij) < £• See Exercise 9 for several properties of null sets. A property P concerning the points in a subset A is said to hold almost everywhere (abbreviated a.e.) in A if the property P holds for all points in A except those in a null set. Example 10*. Suppose f : [a,b] —>• R is equal to 0 a.e. in I. Then f is integrable with J} f = 0. Let e > 0 and E = {t : f(t) ^ 0}. For each i G N and each E let Ei = {t G E : i  1 <  / ( i )  < i}. Then E = \JZiEi i is null (Exercise 9a). For each i let {/j : j G N} be open intervals covering Ei with Yl%i t(Ij) < z/iZ1 Define a gauge 7 on I by y(t) = R if t <jz E and 7(i) = P if't G Ei and j is the smallest integer such that U G P. Suppose that V = {(tkJk) • 1 < k < m} « 7 and set V{ = {(tk,h) • *fc £ Ei\ for i e N and V0 = {(tk,h) :t<£E}. Then S(f,V0) = 0 and OO
5(/,2?0I<E^) < £ / 2 1 for i > 1 so 00
\S(f,V)\<^e/2* and the result follows.
=e
Introduction to the Gauge or HenstockKurzweil
Integral
11
Exercises 1. Suppose 71 and 72 are gauges on I. Set 7(f) = 71(f) n 72(f) Show that 7 is a gauge on I such that any 7fine tagged partition is also 71 and 72fine. 2. Suppose 71 and 72 are gauges on I such that 71(f) C 72(f) for all t. Show that any 71fine tagged partition is also 72fine. 3. Let a < c < b, I = [a,b\, I\ = [a, c], J2 = [c, b], and let 7 be a gauge on I. If T>i is a 7fine tagged partition of U for i = 1,2, show 2? = X>i U T>2 « 7. 4. Use the following outline to give another proof of Theorem 3. Suppose the theorem is false and bisect the interval I. Use Exercise 3 to construct intervals Io = I D h D h D ... such that £(Ik) < ^(Jfci)/2 and no 7fine tagged partition of Ik exists. Let {x} = DfcLi ^fc a n d obtain a contradiction. 5. Evaluate f* e1'*/t2dt. 6. Suppose g : I —> R is nonnegative and integrable and f : I —> R satisfies l/(f)l 2= d{t)Vt e l . If fj g = 0, show that / is integrable over I and
/ , / = o. 7. Suppose / : I —>• R is such that  /  is integrable over J with J 7  /  = 0. Show that / is integrable over I with fjf = 0. 8. Let a < xo < b. Show that there is a gauge 7 on [a, 6] such that if V « 7 and J is the subinterval in T> containing XQ, then XQ must be the tag for J. Generalize this result to a finite number of points. 9*. Call a subset E C [a, b] (Lebesgue) null if for every e > 0 there is a countable collection of open intervals {Ij} covering E with ^l(Ij) < s. (Definition 9) (a) Show that if E is null and F C E, then F is null. (b) Show that if each {Ej : j G N} is null, then UJEN ^ J ^S n u n (c) Show that any countable set is null. (d) Show that if E C [a, b] is null, then CE is integrable with / CE = 0 where Ce(f) = lift e E and C^(f) = 0 otherwise. 10. (FTC). Let / : [a, b] —>• R be continuous and have a derivative / ' except for countably many points E in [a, b]. Define / : [a, 6] —• R by setting fit)
= f'(t)
when /'(f) exists and f(t)
— 0 otherwise. Show that / is
integrable and J f = f(b) — f(a). Show that the continuity assumption on / cannot be dropped. Given an example where Exercise 10 applies but Theorem 5 does not. 11. Show that in Definition 2 it can always be assumed that the tags are endpoints of the intervals which they tag. Show that if V « 7, then there is a 7fine tagged partition V whose tags are endpoints and 5 ( / , V) = S(f, V).
Chapter 2
Basic Properties of the Gauge Integral In this section we develop the basic properties of the (gauge) integral. Throughout this section let I = [a,b] and f,fi,f2,'I —>• KTheorem 1. Assume / i and fa are integrable over I. (i) (ii) (iii) (iv)
/ l + h is integrable over I with J 7 (/i + fa) = / / / i + / / hFor every t G R i/i is integrable over I with Jt tf\ = t jj fa. If fa > 0 on i", i/ien / 7 / i > 0. / / fa > fa on J, i/ien / , / i > J, / 2 . Proof:
(i): Let e > 0. For i = 1,2 there exist gauges 7J such that 5(/j, V) — Jj / j  < e/2 whenever X> < < 7». Put y(t) = 71 (i) n 7 2 (i). Then 7 is a gauge and if T> << 7, then T> < < 7$ (Exercise 1.2) so
<
S(fa,V) J fa + S(f2,V)Jf:
and the result follows. (ii): is left to Exercise 1. 13
< e
14
Introduction to Gauge Integrals
(iii): Let e > 0. There is a gauge 71 such that 5(/i,23)  / 7 / i  < £ whenever V « 7 i . Since /1 > 0, 0 < 5 ( / i , 2V)< ?)<
Sr/ / ;
so fj /1 > 0. (iv): fi  f2 > 0 on J so (iv) follows from (i), (ii) and (iii). We say that / is absolutely integrable over J if both / and  /  are integrable over / . Corollary 2. If f is absolutely integrable over I, then \ Jj f\ < / 7 \f\. Proof: Since / <  /  and  / <  /  on I, Theorem 1 implies / 7 / < / 7  /  and  / , / = / , (  / ) < / , l/l so I J 7 /  < jj l/lIn contrast to either the Riemann or Lebesgue integrals, we will see later (Example 12) that the absolute integrability assumption is important (see also Exercise 3.9). An application of Theorem 1 and the FTC yields the familiar integration by parts formula. Corollary 3. Let f\ and f2 be differentiable over I. Then f[f2 over I if and only iff\f2 is integrable over I and in this case
is integrable
[ Ah = fi(b)f2{b) ~ fi(a)f2(a)  f hf2 Ja
Ja
Proof: By the product rule we have (fif2)' follows from Theorems 1 and 1.5.
— f[f2 + f\f2
The result now
In the results above we have studied the basic properties of the integral J. f as a function of the integrand / . We now consider the integral as a function of the interval / , i.e., we study the behavior of the integral as a set function. Theorem 4. Let a < c < b. If f is integrable over [a,c] and [c,b], then f is integrable over [a, b] and rb
y Ja
rC
f=
rb
f+ Ja
f J c
Proof: Let e > 0. There exists a gauge 71(72) on [a,c]([c, b}) such that [Cf\<e/2
S(/,2?i)J a
Basic Properties of the Gauge Integral
15
whenever V\ is a 71 fine tagged partition of [a,c]{\S{f,V2)  j c f\ < e/2 whenever T>2 is a 72fine tagged partition of [c, &]). Define a gauge 7 on [a, b] by setting ' (a, c) n 71(f) if t € (a, c), ( c , 6 ) n 7 2 ( t ) i f t e (c,6), 7(*) = < 7i(c) n72(c) iit = c, 71 (a) n (—00, c) if t = a , ^ 72(b) n (c, 00) if i = 6. If 2? is a 7fine tagged partition of [a, b], then T> contains either one subinterval J with c as a tag or V contains two subintervals with c as a tag. (Note that c must be a tag.) In the former case, we can "divide" J at c without changing the Riemann sum S(f, V) and, therefore, obtain the latter case. In the latter case Vi = {{t, J) e V : J C [a,c]} and V2 = {{t,J) € V : J C [c,b]} are tagged partitions of [a,c] and [c,b], respectively, with Vi « 7J. Then
S(f,V)
(0+J»
< S{f,Vl
ff Ja
S(f,V2)
1:f
< £
so the result follows. Remark 5. If J is a subinterval of [a, b] and f is the characteristic function of J it follows from Example 1.7 and Theorem 4 that f is integrable over [a, b] with fj f = ({J) Thus, if f is a step function f = Y^j=i ai^Aj, where a, G R, Aj is a subinterval of I and CA is the characteristic function of Aj, then f is integrable over I with
ff = f^aje(Aj). JI
A 1 J= l
To establish the converse of Theorem 4 we establish the Cauchy criterion for the integral. Similar to the Cauchy criterion for sequences of real numbers, this criterion relieves us of the responsibility of having a value for the integral in hand. Theorem 6. Let f : I —> R. Then f is integrable over I if and only if for every e > 0 there is a gauge 7 on I such that if T>i,T>2 « 7, then \S{f,V1)S{f,V2)\<e. Proof: That the condition is necessary follows from the triangle inequality.
16
Introduction to Gauge Integrals For sufficiency note that for every k there is a gauge ~yk on 7 such that if
VltV2 «lk,
then S(/,Di)S(/,2?2)
We may assume that 71(f) D 72(f) 3 • • • for every t e I (Exercise 1.1). For each k let Vk « fk. If k > j , we have \S(f,Vk)  S{f,Vj)\ < 1/j. Thus, {S(f,T>k)} is a Cauchy sequence in R; let A — l i m 5 ( / , Vk). Then \S(f,Vk)A\
yjy. Then
\S(f, V)  S(f, VN)\ + \S(f, VN)  A\ < 1/N + l/N
<£.
Thus, / is integrable over 7 with Jj f — A. We can now establish the converse of Theorem 4. Theorem 7. Let f : I —>• R be integrable over I. If J is a closed subinterval of 7, then f is integrable over J. Proof: Since we have no candidate for Jj f, we use the Cauchy Criterion of Theorem 6. Let e > 0. There is a gauge 7 on 7 such that if T>\, T>2 are 7fine tagged partitions of 7, then \S(f, T>\) — S(f,T>2)\ < £• Consider the case a < c < d < b and J = [c, d]; the other cases are similar. Let 7' be the restriction of 7 to J and let 71(72) be the restriction of 7 to [a, c]([c, &]). Let T>i(T>2) be a 7i(72)fme tagged partition of [a,c]([c,b}). Now suppose V and £ are 7'fine tagged partitions of J. Then V = V\ U V U V2 « 7 and S' = Vx U £ U T>2 « 7 so \S(f,V)
 S(f,£')\
= \S(f,V)
 S(f,£)\
< £.
It follows from Theorem 6 that / is integrable over J. From Theorems 4 and 7 a function / : 7 —> R is integrable over 7 if and only if/ is integrable over every closed subinterval of 7. Moreover, if {7j : 1 < i < n} is a partition of 7 and / is integrable over I, Jr f = Y^i=i Ii f> ^ s P r o P e r ty of the integral is usually referred to as (finite) additivity for the integral. In Chapter 1 we proved what we referred to as the Fundamental Theorem of Calculus (FTC). Actually, we proved only one half of what is usually referred
17
Basic Properties of the Gauge Integral
to as the FTC, the half concerning the integration of a derivative. The other half concerns the differentiation of the indefinite integral. Using Theorems 1 and 7 we can now address this half of the FTC. Theorem 8 (FTC: Part 2). Letfil^Rbe integrable over I and set F(x) = J f for a < x < b. If f is continuous at x G [a, b], then F is differentiable at x with F'(x) = f(x). Proof: Let e > 0. There exists 6 > 0 such that \f(x)  f(t)\ < e for a; — t\ < S,t e [a, b\. Suppose \y — x\ < 5,y € [a, b] and y > x. Then fix)  e < f{t) < f{x) + e for x < t < y. Using Theorems 1 and 7, integration over [x, y] yields (f{x)  e)iy x)<
f
f(t)dt < ifix) + e)iy  x)
JX
or fix)  e < (F(y)  F(x))/(y  x) < /(x) + e. Since a similar inequality holds for y < x, the result follows. See the Notes/Remarks section for references to more general results. We can employ the Cauchy Criterion of Theorem 6 to show that any continuous function is integrable over a closed, bounded interval. For this we first establish a lemma which is interesting in its own right. Lemma 9. Let / : / —> E and assume that for every e > 0 there exist integrable functions gi,g2 '• I —> E such that g\ < f < gi on I and Jrg2 < Jr 9i + £• Then f is integrable over I. Proof: Let e > 0. By Exercise 1.1 there is a gauge 7 on J such that if V « 7, then \S(guV)  / 7 g» < e for i = 1,2. Suppose V « 7. Then Jgi
e < SiguV)
< Sif,V)
< Sig2,V)
< Jg2
+ e < J9l
+2e.
Thus any Riemann sum for / with respect to a 7fi.ne tagged partition lies within an interval with endpoints [fj gi —e, fj gi + 2e] so any two such Riemann sums differ by at most 3e. It follows from Theorem 6 that / is integrable over Theorem 10. If f : I —¥ M. is continuous, then f is integrable over I.
18
Introduction to Gauge Integrals
Proof: Let s > 0. Since / is uniformly continuous on /, there is a S such that /(x) — f(y)\ < e when x,y £ I, \x — y\ < S. Let V = xo < • • • < xn = b} be a partition of [a,b] such that (XJ — Xj_i) < i = 1 , . . . ,n. For i = 1 , . . . , n put M* = sup{/(i) : aij_i < t < Xi} m, = inf{/(t) : x^_i < t < x{\ and define step functions g\ and 52 by n
>0 {a = S for and
n
gi = m 1 C [ x 0 i X l ] + ^ m i q X i _ 1 , X i ] , g2 = M i C [ x 0 i X l ] + ^ ^ ^ i  i . s i ] • i=2
i—2
Then g\ < f < g2 and 0 < 52 — <7i < £ on / so JI(g2 — 5i) < e(6 —a). Lemma 9 now gives the result. We can employ Theorem 10 to obtain a version of the Mean Value Theorem for integrals. Corollary 11. Let f : I —> R be continuous. There exists t £ I such that rb
I
f =
f(t)(ba).
Ja
Proof: Let M = max{/(t) :t£ I},m = inf{/(i) :te m(ba)
<
f <
I}. Then
M(ba)
m< [fttba) If7(6 a) <M. The result now follows from the Intermediate Value Theorem for continuous functions. We can use the basic properties of the integral developed above along with the FTC to give an example illustrating a property of the gauge integral which it does not share with the Riemann or Lebesgue integrals. Namely, the gauge integral is a "conditional" integral; that is, it admits functions which are integrable but whose absolute values are not integrable. Example 12. ForO
let fit) = t2cos{ix/t2)
and /(0) = 0. Then f is
0
t =0
2icos(7r/i 2 ) + — sin(7r/£2)
0 < t < 1.
Basic Properties of the Gauge Integral
19
By Theorem 1.5 / ' is integrable over [0,1] with J0 f = —1. However, \f'\ is not integrable over [0,1]. To see this let ctk = yj2/(^k + 1) and (3k = 1/V2k. Then the intervals {[ak,/3k]} are pairwise disjoint and \f'\ is integrable over each [ak,Pk] by Theorem 10. By the FTC and Corollary 2, we have rPk
/
r/3k
l/'l> /
/'
=
l/2k.
If l/'l *s integrable over [0,1], we have from the finite additivity of the integral ,1
n
ff}k
/ i/'i>E/ JO
Jak
k=l
n
\f\>Y,li2k fc=l
for every n which is clearly impossible. If / : / — ) • R is integrable over / and  /  is also integrable over /, we say that / is absolutely integrable over J. If / is integrable over / but  /  fails to be integrable over J, we say that / is conditionally integrable over / . Thus, the function in Example 12 is an example of a function which is conditionally integrable. Notes/Remarks There is another version of part 2 of the FTC which implies that the indefinite integral of an integrable function is differentiable at "most" points in [a, b] with derivative equal to / . This result is established in Appendix 3. This result also establishes another important property of an integrable function; namely, that any gauge integrable function is (Lebesgue) measurable. For other proofs of the measurability, see [L], [LPY], [LPY1]. There are also known characterizations of the indefinite gauge integral; e.g., [Gol] 9.17, [LPY] 6.22. The computations in Example 12 show that the FTC does not hold for the Lebesgue integral since the Lebesgue integral is an absolute integral. There are several characterizations of gauge integrability in terms of the Lebesgue integral. First, Liu has given the following necessary condition for gauge integrability. ([Liu]; see also [LCL].) Theorem 13. If f is gauge integrable on [a,b], then there is an increasing sequence of closed subsets {Xk} of [a, b] such that {J^i Xk = [a, b], f is Lebesgue integrable on each Xk and limfc L Jx f = Ja f, where L J denotes the Lebesgue integral.
20
Introduction to Gauge Integrals A necessary and sufficient condition for gauge integrability is given by
Theorem 14 {LCL). f : [a,b] » R is gauge integrable over [a,b] if and only if there is an increasing sequence of closed subsets {Xk} of [a,b] such that Ujt=i %k = [a,b],f is Lebesgue integrable over each Xk and the following condition holds: for every e > 0 there exists N such that if k > N, then there exists a gauge 7fc on [a,b] such that whenever V = {(ij,/,) : 1 < i < m} << 7^, then I 5^ t djf f{U)i{Ii n J)I < £ for every subinterval J. Other characterizations of gauge integrability can be found in [Sc] and
[LAI]. Another interesting connection between the gauge and Lebesgue integrals is given by Theorem 15. Let f : [a, b] —>• R be gauge integrable over [a,b]. If f is gauge integrable over every measurable subset of[a,b] (i.e., ifCsf is gauge integrable over [a,b] for every measurable E C [a, 6]), then f is Lebesgue integrable over [a,b\. See [Gol] p. 146 for a proof.
Exercises 1. Prove Theorem 1 (ii). 2. Evaluate JQ xe~xdx. 3. Let / : J —>• R. Suppose 3A 6 R such that for every e > 0 there are integrable functions g and h with g < f < h and A — e < fTg < fjh < A+e. Show that / is integrable with Jj f = A. 4. Let / : / —» R be integrable over / and suppose that g : I —> R is equal to / except possibly at countably many points in I. Show that g is integrable
with fIg = JI f. 5. Let f(t) = sini if t e [0,1]\Q and f(t) = t if t G Q n [0,1]. Show that / is integrable over [0,1] and calculate JQ f. 6. Can the condition f\(t) > f2(t) for all t e I in Theorem 1 (iv) be relaxed? 7. Suppose J. l/l = 0. Show that / is integrable over I with Jj f = 0. 8. Suppose J.  / — g\ = 0 . Show that / is integrable over I if and only if g is integrable over I with fjf = Jj g. 9. Suppose / : I —> R is nonnegative and continuous on I. It JT f = 0, show that / = 0 on J. Can the nonnegativity assumption be dropped? How about the continuity?
Basic Properties of the Gauge Integral
21
10. Let / , g : I » R be continuous with g > 0. Show that 3 t € I such that
tafa = m
Sta
ll. Let f(t) = 1 for 0 < t < 1 and /(*) =  1 for  1 < t < 0. Is the indefinite integral F of / differentiable at £ = 0? Why doesn't this violate Theorem 8? 12. (Linear Change of Variable) Let / : [a, b] —• R be integrable over [a, 6] and let heR. Define / \ : [a + /i, 6 + h] >• R by / h ( t ) = / ( *  /i). Show that fh is integrable with / o / = / o + / l //,. 13. Let / : R > R be continuous and set fk(x) = Jx f • Show that fk » / pointwise on K and the convergence is uniform if / is uniformly continuous. 14. If / : [a, b] * R is continuous,  /  < 1 and / / = (6 — a), what is / ? 15. Let / : [a, 6] —>• R be continuously differentiable with / ( a ) = /(6) = 0. Show that if / a 6 / 2 = 1, then fitf(t)f'(t)dt =  1 / 2 . 16. Let / : [a, 6] —>• R be continuous on [a, b] and differentiable on (o,6). If / ( a ) = 0, /(6) = —1 and J f = 0, show that 3 c e (a, 6) such that / ' W = 0. 17. Let / : [0,1] —>• R be continuous. Let /o = / and fk+i(t) = J0 fk for fc > 0. Show that if fk = 0 for some k, then / = 0. 18. (Dilation) Let / : [a, b] —> R be integrable over [a, b]. For c 7^ 0 let 3(0 = f{°t), o,/c < t < b/c. Show that g is integrable over [a/c,b/c] vith fif(t)dt = cfyceg{t)dt. 19. Show that continuity cannot be dropped in Corollary 11. 20. Let / : [0,1] —¥ R be continuous and increasing. If g{x) = (1/a;) / 0 / for 0 < x < 1, show that g is increasing. 21. Let / : [a, b] —>• R be positive and continuous with M = max{/(£) : a < t < b}. Show that M = lim(/ a 6 /(i)™^) 1 / n  [ H i n t : / > M  £ on some subinterval.] 22*. Let / : I >• R be integrable over R and equal to g : 7 »• R a.e. Show that 5 is integrable over 7 with fIf = JJ g. [See Definition 1.9.]
Chapter 3
Henstock's Lemma and Improper Integrals One of the most important properties of the gauge integral is the validity of convergence theorems of the form, lim JT fk = / 7 (lim fk), under rather general hypotheses. Such general convergence theorems hold for the Lebesgue integral and are the principal reason that the Lebesgue integral is so much superior to the Riemann integral. We will establish these convergence theorems in Chapter 5. The principal tool used in establishing the convergence theorems is a result usually referred to as Henstock's Lemma. We begin with this result and then give some applications. First we need some terminology. Let I = [a, b]. A partial tagged partition of 7 is a finite collection of pairs J = {(U,Ji):
l
where the {Ji} are nonoverlapping closed subintervals of / and £; 6 Ji (it is not required that UlLi Ji = I) If 7 is a gauge on I and J is a partial tagged partition of I, then J is said to be •yfine, written J « 7, if ti G J, C j(ti) for i = 1 , . . . , n. If J is a partial tagged partition of I and / : I —> R, we write S{f,J) = 5Zi = i f(ti)£(Ji) for the Riemann sum of / with respect to J', and if J = Ur=i *^*> w e w r r t e fj f = Y^i=i Ij f when / is integrable. In case V is a (7fine) tagged partition of I, this agrees with our previous definitions. L e m m a 1 (Henstock). Let f : I —>• K be integrable over I. For e > 0 suppose 7 is a gauge on I such that ifDisa "/fine tagged partition of I, then \S(f, V) — j r f\ < e. If J = {(U, Ji) : 1 < i < n} is any partial tagged partition of I such that J < < 7, then
S(f,J)~
f JJ
23
f\<£, I
24
Introduction to Gauge Integrals
where J = (J"=1 Ji, and
E f&Wi)
If
< 2e.
Jj
i
i=l
Proof: The set J \ IJILi ^» consists of a finite number of disjoint intervals. Let Ki, 1 < i < ra, be the closures of these subintervals. For 77 > 0 the integrability of / over each Ki implies that there exists a 7fine tagged partition Ki of Ki such that
S(f , I C i )  I
<
T]/m.
JKi
Then V = JUtCi U . . . U/Cm is a 7fine tagged partition of I. If J = UiLi ^ > then
s(f,v)£f
= s(f,j) Jj + J2{s(f,iCi) J^ A <
S(f,J) If <£ + E
5
(/>^) / /
£
< e + 771(77/771) = e + 77.
Since 77 > 0 is arbitrary, S(/, J) — Jj f\ < e as desired. For the second inequality, let ,7+(j7_) be those (£;, Ji) such that Jj f > 0 (< 0). Then by the first inequality
f(ti)£(Ji)
0 < £ (mew  / / ) = £ /(*0'(*)  / /< £ j+
^
Jji
'
j+
Jji
and
0 <  E (s(uWi)  fj /) = E /eo w  /<7 £ / so the second inequality follows. Henstock's Lemma asserts that if 7 is a gauge on I such that 7fine tagged partitions of I induce Riemann sums which give good approximations to the value of the integral over J, then likewise any 7flne partial tagged partition
Henstock's Lemma and Improper Integrals
25
induces Riemann sums which give good approximations to the value of the integral over the union of the intervals in the partial tagged partition. We now give some immediate applications of Henstock's Lemma. Corollary 2. Let f : I —> R be integrable over I and set F(x) = J f for a < x < b. Then F is continuous on I. Proof: Let e > 0 and x e I. There is a gauge 7 on / such that \S(f, V) — fjf\ < e whenever V « 7. Suppose 7(2;) = (a,/3) and set S — min{/3 x, x — a, e/{\ + /(x))}. Now suppose that \y — x\ < S with y € I. Then from Henstock's Lemma applied to {(x, [y,x])} or {(x, [x,y])} we have
f{x){y
f < e or f(x)(x y)
f
< £ .
Jx
In either case  / J f\ = \F(y)  F(x)\ <£+ \f(x)\\y  a; < £ + £ = 2e so F is continuous at x. We give an application of the second inequality in Henstock's Lemma. Corollary 3. Let f : I —>• R be integrable over I. If J° f = 0 for every c € [a, b], then \f\ is integrable on I with Jr \f\ = 0. Proof: Note that if a < c < d < b, then f* f = f* f  f° f = 0. Let e > 0 and let 7 be a gauge on / such that
S(f,V)J
\S(f,V)\<£
whenever T> = {(ti,Ii)
:1 < i < n } « 7 .
By Henstock's Lemma
YJ\f{U)Wi)<2£. j=l
This implies that  /  is integrable with / 7  /  = 0. Finally, we conclude this section by considering improper integral for the gauge integral. Theorem 4. Let f : [a, b] —> R be integrable over [c, b] for every a < c < b. Then f is integrable over [a, b] if and only if lim c _ >a+ J f = A exists. In this case J f = A.
26
Introduction to Gauge Integrals Proof: For the necessity, let £ > 0 and choose a gauge 7 on [a, b] such that
S(f,V) f
<e
Ja
whenever V «
7. For each c 6 (a, b) there is a gauge 7C on [c, b] such that
S{f,£)
~l
f
< e
whenever £ is a 7fine tagged partition of [c, b]. We may assume jc(t) C 7(i) for all t G [c, b]. Choose c such that c G 7(a) and (c — a)\f(a)\ < e. Suppose s G (a,c). Let £ be a 7sfme tagged partition of [s, b] and set 2? = £ U {(a, [a, s])}. Note V « 7 so /•6
/*6
/ /  / / <
/"6
/*6
/ f~S(f,V)
+ S(f,£)
/ / + /(o)so <3e.
Hence, l i m c ^ a + fcf = fa f. For sufficiency, let {cfc : k = 0 , 1 , . . . } C (a, b] be such that CQ = b, Ck > Ck+i and Cfc —> a. Pick a gauge 71 on [ci,co] such that \S(f, V) — J^° f\ < e/2 whenever V is a 71fine tagged partition of [ci,co]. For k > 2 pick a gauge 7*; on [cfc,Cfc_2J such that
S(f,V)
f
k f <e/2
whenever V is a 7fcfine tagged partition of [cfc,Cfc_2] Choose N such that \A  J^° f\ < e for a < s < CN and /(a)(cjv — a) < e. Now define a gauge 7 by setting
{
(—00, CJV)
t=a
7i(f)n(ci,oo) ci 2. Now suppose P = {(U, Ii) : 1 < i < n} « 7. For each k let X?fc be the subset of V whose tags are in (cfc,Cfc_i] (only a finite number of the Vk are nonempty and no 2 TVs have common elements). Let Jfc be the union of the subintervals belonging to Vk By the definition of 7fc each Vk « 7fc>^i C (ci,c 0 ] and Jk C (cfc,cfc_2) for A; > 2. By
27
Henstock's Lemma and Improper Integrals
Henstock's Lemma \S{f,Vk)  JJk f\ < e/2k. Note that if (x,K) e V is such that the subinterval K = [a, d] is the subinterval in V containing a, then the tag associated with K must be a, i.e., a = x. [Assume that a < x. Then a < Ck < x for some k and x € K c 7(2) C (cfe, Cfc_2). This is impossible since a 6 /ST.] Thus, \AS(f,V)\<\f(a)\£(K)
+
J2([ fs(f,vk) fc=i ^• /J ' fc
00
+
Jd
k=i
It follows that there are no "improper" or "CauchyRiemann" integrals for the gauge integral over bounded intervals. We will show in Chapter 5 that a similar situation occurs for the gauge integral over unbounded intervals. This is in sharp contrast to the situation for both the Riemann and Lebesgue integrals. We can employ Theorem 4 to obtain a "comparison" criterion for the existence of integrals. Corollary 5. Let f,g : [a, b] —> R with \f\ < g on [a, b}. Suppose f is absolutely integrable over [c, b] for a < c < b. (i) If f is nonnegative, then f is integrable over [a, b] if and only if {J f : a < c < b} bounded. (ii) If g is integrable over [a,b], then f is integrable over [a,b]. Proof: (i): Note the function c —> J / i s decreasing on (a, b] so Theorem 4 gives the result immediately. (ii): Put F(s) = J* f, G(s) = fs g for a < s < b. By Theorem 4 it suffices to show that the function F satisfies a Cauchy criterion near a. For a < s < t, \F(t)  F(s)\ < f Js
l/l < f g = G(t) 
G(s).
Js
But, by Theorem 4, the function G satisfies a Cauchy criterion near a so / is integrable over [a, b\. Later, in Corollary 4.11 it will be shown that the absolute integrability assumption in Corollary 5 can be relaxed to integrability. We conclude this section with two examples.
28 Example 6. ForpeR
Introduction to Gauge Integrals let f(t) = tp, 0 < t < 1. For 0 < c < 1 and p ^  1 ,
J f =
(lcp+1)/(p+l).
This expression has a finite limit as c —>• 0 if and only if p+ 1 > 0, and in this case we have JQ tpdt = —jj from Theorem 4. For p = —1, we have f* f =  l n c so t~x is not integrable over [0,1] by Theorem 4. Thus, tp is integrable over [0,1] if and only if p > — 1. We next consider the beta function, often encountered in probability and statistics. Example 7. The beta function is defined to be
B(x,y)=
f ix1(li)y1rfi. Jo
For x > 1 and y > 1 the integrand is continuous so the integral exists. However, for x < l(y < 1) the integrand has a singularity at t = 0(£ = 1). By splitting the integral into two parts, JQ + f^/2, we may consider these cases separately. Now {l—t)v~l is bounded on [0,1/2] andtx~l is integrable on [0,1/2] for x > 0 by Example 6. From Corollary 5 JQ i x _ 1 ( l — t)y~1dt exists for x > 0. The change of variable s = 1 — t can be used to show that similarly / i / 2 ^ x _ 1 ( l ~ t)y~ldt exists for y > 0. Thus, the beta function is defined for x > 0, y > 0. Notes/Remarks Henstock's Lemma (3.1) is sometimes referred to as the SaksHenstock Lemma; Henstock indicates that he adapted Sak's proof of a result for the Burkill integral ([H3] p. 197; [Sa] p. 214). The lemma is used later in the proofs of the Monotone and Dominated Convergence Theorems. Lee TuoYeong, Chew TuanSeng and Lee Peng Yee have established a uniform version of the Henstock Lemma and employed the lemma to give a necessary and sufficient condition for a function to be gauge integrable in terms of the Lebesgue integral ([LCL] Lemma 3). We give their result. It is used later in Chapter 7. Lemma 8 (Uniform Henstock Lemma). Let f : / —> R be integrable over I and £ > 0. Suppose 7 is a gauge on I such that \S(f,T>) — fj f\ < e whenever
29
Henstock's Lemma and Improper Integrals V «
7. If J is any subinterval of I, then
f(U)l(h n J)  / t=l
/}
< 3e
*•
and m
*
V /&)*(/< n J)  /
/
<6e
whenever V = {(ti, !») : 1 < i < m} is a •yfine partial tagged partition of I. Proof: Let V — {(ti, /») : 1 < i < m} be a 7fine partial tagged partition of J and set do = {i : 1 < i < m, l(U n J) > 0}, X>0 = {(ti, /» n J) : i e d 0 } • If
£) {/(tO'(A n J)  / i=1
I
/)
J/iOJ J
then
^ (/((#,n J) /
/}
and we want to show S < 3e. If di = {i e d0 : U e It n J} , then I?! = {(ti, Ii n J ) : i G di} is 7fine so
£ \f(U)l(h n J)  /
/
< £
by Henstock's Lemma (3.1). If ^2 = do\d\, we then need to show that
Y, [fiUWi n J)  I f ied2 L
< 2£.
•'AnJ
[Note {(ti, IiCtJ)} is not 7fine if i € d2 since U £ ^ n J.] For each i £ d2, IiC\J can be written as the difference of 2 intervals whose closures Ai and Bi each
30
Introduction to Gauge Integrals
contain U. [For example, if J = [a, b], It = [di,bi] and at < U < a < 6, < b, then J n U  [a, bi] = [en, bi] \ [ait a), A{ = [«», bi], Bt =
[at,a];
the other cases are similar.] Then {(£j, At) : i e c^} and {(£,,.6^) : i G d 2 } are both 7flne. By Henstock's Lemma, we have
£{/(W
I
/  /
./^nJ
/I
7i3,nJ J
<
+
2e
< £
ied2 ^
^ru
J
as desired. The second inequality in Lemma 8 is obtained as above by using the second inequality in Henstock's Lemma (3.1). The term "uniform" is used in Lemma 10 because the estimates hold uniformly for any subinterval J of I. Although the proof above is phrased in R, the result also holds in R™ (see [LCL]). Lewis and Shisha have related improper Riemann integrals to gauge integrals where the function 6 generating the gauge as described in Chapter 1 satisfies special conditions. For example, it is shown that a function / : [a, b] —> R with a singularity at a is improper Riemann integrable on [a, b] if and only if / is gauge integrable with respect to a function 5 which is increasing on (a, b]. Functions which are integrable with respect to linear functions S are also considered. See [LS] for details.
Exercises 1. Using the notation of Lemma 1 show that
E{i/(*o'(/i)i / / }
<2e.
Henstock's Lemma and Improper Integrals
31
2. Can the condition J / = 0 for all c € [a, b] in Corollary 3 be replaced by 3. 4. 5. 6. 7.
/ „ 6 / = o? Show that 5(1/2,1/2) = n. (Hint: Try t = cos2 u.) Show that B(x,y) = B(y,x). Let / : [a, b] —> R be bounded on [a, 6] and integrable over [c, 6] for every a < c
f3 x2 + 1 (b) / ^5 dx, J2 x2 A io
! z
8. Use Example 6 to show that the product of integrable functions needn't be integrable. 9. Let ak e R for k e N and define / : [0,1] >• R by /(0) = 0, f(t) = 2kak for l/2 fc < t < l/2 f c _ 1 . Show that / is integrable if and only if J2T=i ak converges and in this case L f = J]fcLi ak Give necessary and sufficient conditions for / to be absolutely integrable. [Hint: Use Theorem 4. A direct proof is possible but complicated; see [Gol] p. 320.]
Chapter 4
The Gauge Integral over Unbounded Intervals In this chapter we extend the definition of the gauge integral to unbounded intervals in R. This definition could have been made directly in Chapter 1, but it is best to begin the study of the gauge integral on closed, bounded intervals as was carried out in C h a p t e r s 13. Suppose J is any interval in R, / : I —> R and we wish to define the integral of / over J. We can extend the definition of / to R by setting /(£) = 0 for t E R \ 7 and then the integral of this extension over R would give a definition of the integral of / over I. Thus, we need only consider t h e definition of the integral for functions / : R —>• R. If we wish to extend the definition of the integral for functions / : R —• R, the first thing t h a t we must consider are the meanings of partitions and tagged partitions of R. A partition of R is a finite number of nonoverlapping, closed subintervals of R whose union is R; of course, in any such partition at least one of the subintervals must have infinite length. A tagged partition of R is a finite collection of pairs {(U,Ii) : 1 < z < m } , where {Ii : 1 < i < m } is a partition of R and ti £ Ii for each i. If / : R —> R is a strictly positive function a n d (ti,Ii) is an element of a tagged partition where Ij has infinite length, then the term f(U)£(Ii) would be infinite and the corresponding Riemann sum for this tagged partition would also be infinite. If we associate the integral of a positive function with the area under the graph of this function, this is clearly an undesirable situation. We can easily remedy this situation by considering the extended real numbers. We extend R by adding the two points at infinity, ± o o . We denote this extension by R* and adopt the following order and algebraic properties in R*: 33
34
Introduction to Gauge Integrals —oo < x < oo for every 1 6 R , 00 + 00 = 0 0 ,
—oo — oo = —oo , x ± oo = ±00 for x 6 R, x • (±oo) = ±oo for x > 0, x • (±oo) = =Foo for x < 0. The arithmetic definitions in R* are motivated by the desire to carry forth the properties of limits to functions or sequences with limits in R*. We now also make the somewhat unorthodox definition that 0(±oo) = 0; this definition is motivated by the integration theory which we are about to develop. We refer to intervals of the form [—oo, a], [a, oo], [—oo, oo] as closed subintervals of R* and, similarly, call intervals of the form [—oo, a), (a, oo] open subintervals of R*. For unbounded intervals I of the form above, we set £(I) = oo. Let J be a closed subinterval of R*. A partition of 7 is a finite collection of closed, nonoverlapping subintervals of I whose union is / , and a tagged partition of 7 is a finite collection of ordered pairs V = {(U,Ii) : 1 < i < m} such that {Ii : 1 < i < m} is a partition of I and U € h for each i. As in Chapter 1 we refer to the Ii as the subintervals of V and call U the tag associated with Ii. If I is any subinterval of R and / : I —• E, we always assume that / is extended to R* by setting f(t) = 0 for t e R*\I and that /(±oo) = 0 for any function. If I is a closed subinterval of R*, / : I —>• R and V = {(t,, 1%) : 1 < i < m] is a tagged partition of I, the Riemann sum of / with respect to V is defined to be m
s{f,v) = YJf{uWi). Note that if ±oo is the tag associated with the intervals of infinite length, then S(f, V) is always welldefined and is a real number because of our agreement that 0 • (±oo) = 0. A gauge 7 on I C R* is a function defined on I such that j(t) is an open interval containing t. If V = {(U, Ii) : 1 < i < m} is a tagged partition of I and
35
The Gauge Integral over Unbounded Intervals
7 is a gauge on / , we say that V is 7fine, written V « 7, if U G Jj C 7(ij) for 1 < i < m. Note that if j(t) is a bounded open interval for every t € R n 7, then the tag of any unbounded interval in a 7fine tagged partition V oil must be ±00 so the Riemann sum of any function / : / — » • R with respect to I? is welldefined. We are now in a position to define the integral over an arbitrary closed subinterval J of R*. First we must extend Theorem 1.3 to unbounded intervals. Theorem 1. Let I be a closed interval in R* and 7 a gauge on I. Then there exists a ryfine tagged partition of I. Proof: Consider the case I = [a, 00]. Let 7(00) = (6,00]. If 6 < a, set V = {(00, / ) } . If b > a, there is a 7fine tagged partition £>o of [o, 6 + 1 ] . Then V = VQ U {(00, [b + 1, 00])} is a 7fine tagged partition of I. The other cases of unbounded intervals are similar. We now give the definition of the integral over an arbitrary closed subinterval I of R*. (Exercises 1 and 2 and give other approaches to denning the integral over unbounded intervals.) Definition 2. Let I be a closed interval in R* and f : I —> R. Then f is (gauge) integrable over I if there exists A 6 R such that for every e > 0 there is a gauge 7 on I such that \S(f,V) — A\ < e whenever V is a 'yfine tagged partition of I. As in Theorem 1.4 it is easily established that the value A in Definition 2 is unique (and welldefined by Theorem 1) so we adopt the notation A = Jj f = Jj f{t)dt. If J = [a, oo](I = [—00, a], I = [—00, 00]), we sometimes write JlJ=Ja
Ajoo J' J00 / )•
We give an example illustrating the integral in Definition 2. Example 3. Let f(t) = 1/t2 for t > 1. We claim that f is integrable over [1, 00) with Jx / = 1. Let £ > 0. We will define a gauge 7 on [1, 00] such that 7(i) is bounded for every t £ [l,oo); this will force the tag of an unbounded interval in any tagged partition to be +00. If we define 7(00) = (2/e, 00] and we have [w, 00] C 7(00), then \f(oo)£([w, 00)) — 1/ui] = 1/w < e/2. (Note the area under the graph of f over the interval [w,oo) should be 1/w.) Next, if (z, [u,v]) is a pair in a tagged partition, we want to estimate \f(z)(v — u) — (l/u — l/v)\. (Again the area under the graph of f over [u,v] should be l/u — 1/v.) We have \f(z)(vu)(l/ul/v)\
\l/u
l/v\ <
'(1/vl/u)
36
Introduction to Gauge Integrals
by considering the cases when uv — z2 is positive or negative. Suppose now that V = {(UJi)
:l
is a tagged partition of [l,oo] with Ii = [xii,Xj\,XQ = l,xm From the inequalities above and the fact that
i
i
^ V
i
 i + — = 2^
= oo,tm = oo.
i
+—
we have \S(f,V)
 1 < /(oo)*(J m )  l/a: m _i m— 1
1
+ m—1
2
<^/ +E 1=1
Xi
%i~l
(l/Xii

l/Zj).
/ / we define j(t) = (t — st/4, t + et/4) and assume that V « TJl—l
\S(f,V)  1 < e/2+ £
7, we have
eti
^ ( 1 / * ,  !  1/xO
i=l
e/2 + e/2 ^ ( l / a r i  i  l / i i ) i=l
= e/2 + e / 2 ( l  l / a ; m _ i ) < £ and £/ie claim above is justified. Note that the gauge 7 defined above definitely has intervals of variable length with the length being quite large for t large (consider the graph of f). The basic properties of the integral established in 2.12, 2.47, 2.9 and particularly Henstock's Lemma 3.1 carry forward to the integral over unbounded intervals. The reader is invited to review the proofs of these results. As was the case for bounded intervals, we next show that there are no "improper integrals" over unbounded intervals.
37
The Gauge Integral over Unbounded Intervals
Theorem 4. Let f : [a, oo] = I —> R be integrable over [a, b] for every a < b < oo. Then f is integrable over I if and only i/lim^oo J f = A exists. In the case, A = Jj f. Proof: Let e > 0. Suppose / is integrable over I. There is a gauge 7 on J such that \S(f, V) — Jj f\ < e/2 whenever V « 7. For a < c < 00, there is a gauge 7C on [a, c] such that \S(f, £)  J^ f\ < e/2 whenever £ is a 7fine tagged partition of [a, c]. We may assume that 7 c (t) C ~y(t) for every t € [a, c}. Let 7(00) = (T, 00]. For c > T, let £ be a 7cfine tagged partition of [a, c]. Set V = £u {(00, [c, 00])}. Then V « 7 and
// f / < Jl
Ja
Jl
/fS(f,V) +
S(f,S)
ff Ja
+ /(oo)£([c, 00]) < e/2 + e/2 = e . Hence, lim^oo Jac / = Jt f. Suppose linicxx) J f = A. Choose a sequence {cfc} such that a = CQ < c\ < C2 < . . . and Cfc f 00. Choose a gauge 70 on [CQ, CI] such that \S(f, T>) — J 1 f\ < e/22 whenever T> is a 70fine tagged partition of [c$, C\\. For k > 1, choose a gauge 7fc on [cfc_i,Cfc+i] such that 5(/,2?)  /cCfc+1 /  < e/2k+2 whenever V is a 7fcfine tagged partition of [cfe_i, Ck+i]Choose N such that b> CN implies \ J f — A\ < e/2. Now define a gauge 7 on i" by ' (CAT, 00]
7(*)
i = 00
7fc(t) n (cfe_i,cfe+i)
cfc < i < cfc+i
70(t) n (c0  l,ci)
c0 < t < c\.
k=l,2,.
Suppose that V = {(ti,Ii) : 1 < i < m} is a 7fine tagged partition of J. Assume that Im = [a, 00] is the unbounded interval in T>; note tm = 00 and CN < a. For k > 0 let Vk be the elements in V whose tags are in [cfc,Cfc+i]. By the definition of 7 each Vk is 7/tfine; if Jfc is the union of the subintervals in Vk, then J 0 C [c 0 ,ci) and Jfc C [cki,Ck+i) for k > 1. By Henstock's Lemma (3.1) S(/,27 fc )
/
< e / 2 fc+2
38
Introduction to Gauge Integrals
\As(f,v)\<\A r f\+\ r fJo,
I
OO
S{f,V)
\J a
I

OO
<e/2I JJk n J Jh k=Q
I k=0 A
<e/2 + J2e/2k+2=s. k=0
Of course, an analogous result holds for integrals over intervals of the form [—00,6], [—00,00]. Using Theorem 4, we can obtain an analogue of Corollary 3.5 for unbounded intervals. Corollary 5. Let f : [a, 00] = I —> R be absolutely integrable over [a, b] for every a < b < 00. (i) If f is nonnegative, f is integrable over I if and only if
( fb
sup < /
I /:a<6
(ii) If g : I —• R is nonnegative and integrable over I and \f(t)\ < g(t) for t € I, then f is absolutely integrable over I. Using Corollary 4.11 below it will follow that the absolute integrability assumption in Corollary 5 can be relaxed to integrability. Example 6. Let p e R and set f{t) = l/V> for t ^ 0. Then £ f = (b~P+1 1)/(1— p) forp ^ 1; ifp = 1, j x f = \nb. Thus, from Corollary 5, it follows that f is integrable over [1, 00] if and only if p > 1. [Compare with Example 3.6.] We can employ Corollary 5 to establish the integral test for convergence of series. Proposition 7. Let f : [l,oo] —> R be positive, decreasing and integrable over [1, b] for every 1 < b < 00. Then the integral jx f = A exists if and only if the series J2T=i /CO = & converges. In this case, A < S < A + / ( l ) . Proof: For i < x < i + 1, f{i + 1) < f(x) < f(i). Thus,
f{i + l)
f
The Gauge Integral over Unbounded Intervals
39
and n—1
„n
"1
E/(* + !)< / /<E/wJ i
i=i
i=i
Letting n —>• oo gives the first statement and, in the case of convergence, yields Sf(l)
cost
k ~r~t=
f
r x~]x
cost ~£2~
dt.
The function (cos t)/t2 is integrable over [l,oo] by Corollary 5 (ii) and Example 7 (with p — 2). Since limb_>00(cos6)/6 = 0, it follows from Theorem 4 that the integral of (sin t)/t over [l,oo] exists. However, J1 \^f^\dt does not exist. Indeed, we have
H
smt
nl
(fc+l)7T
dt
sini
dt
Jkv
k=\
(fc+l)ir
nl
E (k + l)ir *:=! for all n. [See also Exercise 8.17.] We next consider absolute integrability for the gauge integral. Recall that the gauge integral admits conditionally integrable functions (Examples 8, 2.12 or Exercise 3.9). We now give necessary and sufficient conditions for absolute integrability. These conditions involve the variation of a function; for those readers unfamiliar with the definitions and properties of functions of bounded variation, they are given in Appendix 2. We begin with the case of functions defined on a closed bounded interval.
40
Introduction to Gauge Integrals
Theorem 9. Let I = [a,b] with —oo R be integrable over I. Let F(x) = J f,a<x
j2\F&)F(xii)\
i=l
l/l
Jxi
Ja
~l
so V < fa  /  < oo. For the converse assume V < oo and let e > 0. There is a partition V — {a = XQ < x\ < • • • <
b}
of / such that V — e < 2™=i I IK f\ — ^ ' w n e r e Ki = lxii,Xi] Note that if V\ = {a = yo < y\ < • • • < ym = b} is a refinement of the partition V(i.e.,Vi D V) and Li = [j/ii,i/i], then < V
(1) i=\
JK
i
JL
i=l
i
(Appendix 2.) Let 71 be a gauge on / such that \S(f, V) — Jj f\ < e whenever V is a 71fine tagged partition of I. By Exercise 3.1 if V = {(ti,Ji) : 1 < i < p} « 71, then (2)
£{i/(*i)i'0>o^/}
< 2s.
Let 7 be a gauge on I such that 7(f) C 71(f) for every t € I,j(t) C Ki when t e K° and 7(2:,) C {xi\,Xi+\) where x_i = —00, xn+i = 00. If £ = {(zi, li) : 1 < i < q} is a 7fine tagged partition of / , then there is a 7fine tagged partition £' = {(^,7,) : 1 < i < r} such that the partition {/,' : 1 < i < r} is a refinement of {K\,... ,Kn} and such that S(\f\,£) = S(\f\,£'). In fact, by the definition of 7, we may take £' = {(ziJiHKj)
: 1 < * < q, 1 < j < nj°
n K° £ 0}
41
The Gauge Integral over Unbounded Intervals (Figure 4.1). Then (1) and (2) give
V£
L
< V and
52{\M)\M)
IA
<2e
so
S(/,£')±£
\S(\f\,£)V\
i=l
L
v
<2e + e
and the result follows.
"i+i Figure 4.1
We next consider the case when the interval J is unbounded. Theorem 10. Let I = [—oo,b], b < oo, and f : J —» R be integrable over I. Let F(x) = J_ f for —oo < x < b. Then \f\ is integrable over I if and only if F is of bounded variation over I. In this case, jj \f\ = Var(F : I). Proof: For —oo < a < b, set Fa(x) = J f for a < x < b and ca = J_ so F(x) = Fa(x) + ca for a < x < b. Suppose  /  is integrable over J. If — oo < a < b, then by Theorem 9
f
/  /  = Var(Fa : [a, b}) = Var(F : [a, b}). Ja By Theorem 4 /
/ =
lim
/
 /  = Var(F : I).
Suppose F is of bounded variation over L Theorem 9 / Ja
If — oo < a < b, then by
 /  = Var(Fa : [a,b]) = Var(F : [a,b}).
42
Introduction to Gauge Integrals
By Theorem 4 Var(F:I)=
lim Var(F : [a, b}) = o  >  oo
lim
f \f\ = /
a^oo^
/.
>/_oo
The cases when 7 = [a, oo] or [—oo, oo] are treated similarly. From Theorems 9 and 10 we have the following comparison theorem for absolute integrability. Corollary 1 1 . Let I be a closed subinterval of R* and let f,g : I —>• R be integrable over I with \f(t)\ < g(t) for every t & I. Then f is absolutely integrable over I with \ Jj f\ < Jj \f\ < / 7 9Proof: Let x0 < xx < ••• < xn with Xi e I. S"=i Ix'
9
=
X
ne Tesu
J " 9 ^
^
now
Then £ " = 1 \ f*'^ f\ <
follows from Theorems 9 and 10.
From Corollary 11 it follows that the absolute integrability assumption in Corollaries 3.5 and 4.5 can be relaxed to integrability. If / , g : I —> R, we write / V g — max(/, g), f A g = min(/, g). In contrast to the situation which obtains for both the Riemann and Lebesgue integrals, if / and g are (gauge) integrable, / V g and / A g may fail to be integrable. Indeed, if / + = / V 0 and / " = (  / ) V 0,  /  = / + + /  , / + = ( / +  /  ) / 2 and / = (—/ +  /  ) / 2 so  /  is integrable if and only if both / + and / are integrable. Thus if f(t) — (sin t)/t for t € I = [1, oo], then by Example 8 both / + and / ~ are not integrable over I. We do have sufficient conditions for the max and min of integrable functions to be integrable. P r o p o s i t i o n 12. Let I be a closed subinterval o/R* and let f,g,h integrable over I.
: I —> R be
(i) If f and g are absolutely integrable over I, then f V g and f A g are integrable over I. (ii) If f < h and g < h, then f V g and f A g are integrable over I. (hi) If h < f and h < g, then / V j and f A g are integrable over I. Proof: (i) follows from the identities fVg=lf
+ g+\fg\}/2<mdfAg=if
the inequality  /  g  <  /  + g and Corollary 11.
+
g\fg\}/2,
The Gauge Integral over Unbounded Intervals
43
To prove (ii) note that f V g
=
2fVgfg<2hfg.
Thus, \f — g\ is integrable by Corollary 11. The result now follows from the identities for / V g and f A g given above. Part (iii) is established in the same way using \f — g\ < f + g — 2h. We conclude this section by considering the integrability of the product of 2 integrable functions. We first establish a lemma which gives a sufficient condition for the integrability of a product; this lemma is sometimes called Dedekind's Test. Lemma 13. Let f,g : [a,b] = I —> R be continuous on (a,b\ with limx_i.0+ g(x) = 0 and g' absolutely integrable over I. Assume that F(x) = J f,a < x < b, is bounded on (a, b]. Then fg is integrable over I. Proof: For a < c < b, we have (Fg)' = —fg + Fg' so Fg' is integrable over [c, b], and since F is bounded on (a,b] and g' is absolutely integrable, Fg' is absolutely integrable over [a, b] by Corollary 11 and 3.5. We have J fg — J Fg' + F(c)g(c), and since F is bounded and limx_>a+ g(x) = 0, fg is integrable over [a, b] with / fg = J Fg' by Theorem 3.4. We now use Dedekind's Test to give an example illustrating a multiplicative property (lackthereof) for the gauge integral. E x a m p l e 14. Let f(t) = t~2 cos(\) = (sin(±))' and g(t) = t1/2 for 0 < t < l(/(0) = 0). By Example 3.6 / and g satisfy the hypotheses of Lemma 13 so h(t) = £~ 3 / 2 cos() is integrable over [0,1]. Let m(t) = cos(j). Then m is integrable over [0,1] (Exercise 16), but mh is not integrable over [0,1] (Exercise 15). This example shows that in general the product of an integrable function and a bounded, integrable function need not be integrable. (See the Notes/Remarks section for further results and references.) Notes/Remarks Two alternative approaches to defining the gauge integral over unbounded intervals are given in Exercises 1 and 2; Exercise 1 is from [LPY] and Exercise 2 from [M].
44
Introduction to Gauge Integrals
Lewis and Shisha have connected the gauge integral over the interval [0, oo] with what they call a "simple integral"; the simple integral essentially uses a constant length gauge and so is an analogue of the classical Riemann integral ([LS]). They also show that a function has an improper Riemann or CauchyRiemann integral if and only if the function is integrable with respect to a monotone function S which generates the gauge as in Chapter 1. Example 14 motivates the following "multiplier problem" for the gauge integral. Characterize those functions g : [a, b] —*• R such that fg is gauge integrable for every gauge integrable function / ; let us refer to each such function g as a multiplier. Example 14 shows that even a bounded, integrable function may not be a multiplier. We show that a function of bounded variation is always a multiplier and then state a partial converse to this result. The proof uses the RiemannStieltjes integral (see [Ru], Chapter 6 for a development); we give a reference to the result employed. Also, employed in the proof is Abel's "partial summation formula": n
n—l
k
^a f c bfc = anBn + ^ S f c ( a f c  ak+i), where Bk = ^ fc=l
fe=l
bj.
s=l
Theorem 15*. Let f : [a, b] — I —» M. be integrable over I and set F(t) = J f. Let g : I —>• K be of bounded variation. Then fg is integrable over I with J fg = F(b)g(b)—J Fdg [this formula is sometimes referred to as "integration by parts"). Proof: Let e > 0. There is a gauge 7 on I such that  5 ( / , 27)  Ja f\ < e whenever T> << 7. Since the Stieltjes integral of F with respect to g exists, there is a 6 > 0 such that m
'VF(ti)(g(xi)g(xi1))
„
Fdg
< e
whenever a = XQ < x\ < • • • < xm = b is a partition of [a, b] with Xj — Xi_i < 8 and Xii
45
The Gauge Integral over Unbounded Intervals A = F(b)g(b) — J Fdg, we have from Henstock's Lemma 3.1 n
fc=i nl
E fe=i
^2f{U)(Xi
Xii) [g(tk)  g{tk+i)}
.1=1
+ ^2f(U)(xi
Xii) g(tn)  A
1
k
nl
<
£
Y2 f(U)(xi  Xii)  F(xk) [g{tk) g(tk+i)}
fc=i nl
„b
+ J2F(xk)[g(tk+1)g(tk)}
/ Fdg
fc=l
nl
+ ^/(UKxiXiJFib)
IffWI
i=l
< £VaT(g : / ) + £ + g(6)£. Hence, the result follows. There is a partial converse to Theorem 15 which gives a characterization of multipliers. We say that a property P holds almost everywhere (abbreviated a.e.) in a subset E C R if the property holds for all the points in E except those in a null subset of E (Definition 1.9). T h e o r e m 16*. If g : [a,b] function
•> R is a multiplier, then g is equal a.e. to a
Si
: [a, 6] ^ R
which is of bounded variation. See [LPY], Theorem 12.9, for a proof which makes extensive use of Lebesgue measure. For a characterization of multipliers in higher dimensions, see [LCL1]. Exercises 6, 8, 10 and 17 give some further results on the integrability of products of integrable functions.
46
Introduction to Gauge Integrals
Exercises 1. Let / : R —> R. Show that / is integrable over R if and only if there exists A G R such that for every e > 0 there exist a < b and a gauge 7 on [a, b] such that \S(f, V) — A\ < e whenever V is a 7fine tagged partition of [a, b}. 2. Let / : R —> R. Show that / is integrable over R if and only if there exists i g l such that for every e > 0 there exist r > 0 and a gauge 7 on R such that if a < — r and b > r, then \S(f,V) — A\ < e whenever V is a 7fine tagged partition of [a, b]. 3. Let ofc G R for k G N. Define / : [1, 00) >• R by f(t) = afc for k < t < k + 1. Show that / is integrable over [l,oo) if and only if £3afc converges. Give necessary and sufficient conditions for / to be absolutely integrable. 4. Let / : [a, 00) —> R be differentiable. Give necessary and sufficient conditions for / ' to be integrable over [a, 00). 5. Show that JQ sin(:r2)d:r exists. (Fresnel Integral). Is the integral absolutely convergent? [Hint: try t = x2.} 6. Let / , g : [a, b] —>• R. Suppose fg and / are integrable over [a, c] for every a < c < b and F(t) = f f for a < t < b. Assume limt_>b F(t) exists and g is differentiable with g' absolutely integrable over [a,b]. Show that fg is integrable over [a, b]. [Hint: Integrate by parts.] 7. (Limit Form of Comparison Test) Let / , g : [a, b] —)• R and assume / and g are integrable over [a,c] for a < c < b with g(t) > 0 for a < t < b. Let l i m ^ b  f(t)/g(t) =L£l*. (i) If L = 0 and g is integrable over [a, b], show / is integrable over [o, b]. (ii) If 0 < L < 00, show g is integrable over [a, b] if and only if / is integrable over [a, b}. (iii) If L — 00 and / is integrable over [a, b], show g is integrable over [a, b]. 8. (Abel) Let / , g : [a, 00) —>• R. Suppose / is continuous and set F(t) = J f. Assume that F is bounded and g is nonnegative, differentiable and decreasing. Show that J"3 fg exists if either (a) lim t _ >00 g(i) = 0 or (b) J"°° / exists. [Hint: Integrate by parts.] 9. Use Abel's Test in Exercise 8 to show J^° ^^dt exists for p > 0. Show that the integral is conditionally convergent for 0 < p < 1. [Recall Example 8.] 10. Let / : [a, 00) —> R be continuous and F{t) = / / bounded on [a, 00). Let g : [a, 00) > 1 be such that g' is nonpositive and continuous on [a, 00) with lim g(t) = t—too
The Gauge Integral over Unbounded Intervals
47
Show that J fg exists. 11. Use Exercise 10 to show f^° f^jdt exists. 12. Suppose / is absolutely integrable over J and let c > 0. Define fc(t) = f(t) if /(*) < c a n d fc{t) = 0 otherwise. Show that fc is absolutely integrable over / . 13*. Extend the definition of null set given in Definition 1.9 and then extend the results in Exercise 1.9 to R. 14. Let / , g be absolutely integrable over J. Show that / + g is absolutely integrable over I. 15. Use the identity cos2 u = (1 — cos2w)/2 and Example 3.6 to show that the function mh in Example 14 is not integrable. 16. Show that cos() is integrable over [0,1]. [Hint: Make the change of variable s = \/t and use Corollary 5 and Example 6.] 17. Let / , g : [a, b] —> R be continuous on (a, b] and g' absolutely integrable over [a, b]. Assume F(t) = Jt / i s bounded. Show that fg is integrable over [a, b] if and only if limc_,.a+ F(c)g(c) exists.
Chapter 5
Convergence Theorems The principal reason that the Lebesgue integral is favored over the Riemann integral is the fact that convergence theorems of the form lim ft fk = J 7 (lim fk) hold for the Lebesgue integral under very general conditions. The major convergence theorems of this type are the Monotone Convergence Theorem (MCT) and the Dominated Convergence Theorem (DCT). In this section we will show that these same major convergence theorems hold for the gauge integral, showing that the gauge integral enjoys the same advantages over the Riemann integral as the Lebesgue integral. We begin by introducing the concept which is at the center of the convergence theorems for the gauge integral, uniform integrability. Throughout this section let / b e a closed interval (bounded or unbounded) in R* and fk, f : I » R for k G N. Further, let I be the family of all closed subintervals oil. Definition 1. {fk} is uniformly integrable over I if each fk is integrable over I and if for every e > 0 there exists a gauge 7 on I such that \S(fk,T>) — fj fk\<s for every k 6 N whenever V « 7. The point of Definition 1 is, of course, that the same gauge works uniformly for all k. For uniformly integrable sequences of integrable functions we have the following convergence theorem. Theorem 2. Let {fk} be uniformly integrable over I and assume that /&—>•/ pointwise. Then f is integrable over I and lim Jj fk = Jt / ( = / 7 (lim fk))Proof: Let e > 0. Let 7 be a gauge on J such that \S(fk,T>) — Jj fk\ < £/3 for every k whenever V « 7, where we may assume that 7(4) is bounded for every f E R. Fix a 7fine tagged partition VQ of / . Pick ./V such that 49
50
Introduction to Gauge Integrals
\S{fi,T>o) — S(fj,V0)\ < e/3 whenever i, j > N; this is possible by the pointwise convergence of {fk} and the fact that /fc(±oo) = 0. If i, j > N, then
jfiffj
^ JfiS(fi,V0 + sUi,v0)
+
j fj
\S(fl,V0)S(fj,V0)\ < £.
Hence, lim JT fk = L exists. Suppose now that I? is a 7fine tagged partition of / . As above pick N such that
\S(fN,V)S(f,V)\<e/3 and also so \L — ff /JV < e/3 Then
\S(f,V)L\<\S(f,V)S(fN,V)\
+ S(fN,V) J fN + j
fNL
< £.
Hence, / is integrable over / with jj f — L = lim \ l fk as desired. We now proceed to establish the major convergence theorems for the gauge integral. Since we are working over arbitrary subintervals of R*, we need a preliminary lemma which will enable us to treat the case of unbounded intervals. L e m m a 3. There exists a strictly positive ip : R —> (0, 00) which is integrable over R and a gauge 7(= •yip) such that 0 < S(ip, V) < 1 for every jfine partial tagged partition V of R. Proof: Pick + t)2)]. whenever V « tagged partition 1/(2TT(1
any strictly positive function p o n l with JRf= l/2[<£>(£) = There is a gauge 7 on R such that \S(ip,V)  1/2 < 1/2 7. Suppose V = {(ij,7j) : 1 < i < m} is a 7fine partial of R and set I = IJl^i ^ By Henstock's Lemma 3.1,
S(ip,V)  j if < 1/2 since
so
0 < S((p,V) < 1/2+ / tp < 1
Convergence Theorems
51
We now prove our first convergence result from which the Monotone and Dominated Convergence Theorems will be derived. Theorem 4. Let fk,f : / —> R be nonnegative with each fk integrable over I and suppose f = YlT=i /fc P°* n * w * se on I w^b, Y^=i Ii fk < °° V s « =
ELi/fc t h e n (i) {sn} is uniformly integrable over I, (ii) / is integrable over I and
(iii) E£Li // A = // / = // EfcLi fkProof: In order to avoid treating special cases we assume in the proof that J = R; this causes no difficulty since we can always extend functions from J to R by setting them equal to 0 on R \ J. Let e > 0. For each n pick a gauge 7„ on R* with 7„(£) bounded for each t e R such that r
S(sn,T>) <e/2 
/ Sn 
whenever V « 7„. Pick no such that Y^kLn JR fk < £ and for every t £ K pick n{t) > no such that k > j > n(t) implies
! > ( * ) < &p(t) where ip and yip are as in Lemma 3. Define a gauge 7 on R* by fn(t)
n
7(*)= [ f l ^ W )
^
for t € R and 7(±oo) = ( p 
T,(±OO)
and set n(±oo) = noSuppose V = {(ti,Ii) : 1 < i < m} « V « 7; for i — 1 , . . . ,n0 implies \S(si,V)
] n 7 ¥ ,(±oo)
7. To establish (i) first note that  / R S j  < e/2l < e. So fix n > n0.
52
Introduction to Gauge Integrals
Set di = {i : 1 < i < m, n(ti) > n} and di = [i : 1 < i < m, n(ti) < n) , and note 2?! = {(£j,/j) : i G d{\ « 7„ by the definition of 7. Set I = \j{U i G di}. We have, using Henstock's Lemma 3.1,
(1)
( snS(sn,V) M
< fsnS{sn,V{) JI
+ V r ( / fj Ui< ied2j=1
< e/2n + E E
fj(U)l(Ii)\ )
fjfj(tiWi)\
ied2 j=i ^Jl* n
>
.
+ E
+
E />(*i)^(/0
»ed 2 j=n(tj) + l
ied.2 j=n(U) + \ '
< e + Ti + T2 + T 3 , with obvious definitions for T\,Ti and T3. We first estimate T3. From Lemma 2,
E E /.*&) *(£) < i£d
ied.2 i = n ( t i ) + l
e
v(*i)*(fc) =
eS(
2
where V2 = {(U,Ii) : i G d 2 }. Next, from the nonnegativity of the fj, Ill
T
It
1,
II
r.
LAJ
n
^EE//^E/^E//,
< e.
Finally, for Ti set s — max{n(£j) : i G cfo} and note {(ti, I{) : n(ti) — k}
«
53
Convergence Theorems 7fc. By Henstock's Lemma 3.1, Ti = ied2 ^
E
E
>
I sn(U) ~ Sn{U){ti)£(Ii.
n{ti)=k
^E E {/'»(*)
sn(ti)(U)i(Ii)
<$>/2fe<£. fc=i
n(*i) = fc
Prom (1) it follows that  / R S n — S^Sm^)! < 4e whenever £> < < 7 and n> TIQ. Hence, (i) is established. Conditions (ii) and (iii) follow from Theorem 2. From Theorem 4 we can immediately obtain one of the major convergence theorems for the gauge integral, the Monotone Convergence Theorem (sometimes called the Beppo Levi Theorem). Theorem 5 {Monotone Convergence Theorem: MCT). Let fk : / —>• R be integrable over R and suppose that fk{t) t f(t) S R /<"" ei/en/ £ € R. If supfc JjfkKoo, then (i) {/fc} is uniformly integrable over I, (ii) / is integrable over I and
(iii) l i m / . / f e ^ / ^ / . O i m / f c ) ) . Proof: Set f0 = 0 and gk = fk ~ fki for A; > 1. Then pfe > 0, £ £ = 1 #fc = /n > / pointwise and J2kLi Si 9k = lim n £ £ = 1 / / ( / * ~ A  i ) = l i m « / / /« = sup„ / 7 / n < 00. Hence, Theorem 4 is applicable and gives the result. Of course, an analogous result holds for decreasing sequences (Exercise 3). The Monotone Convergence Theorem gives a very useful and powerful sufficient condition guaranteeing "passage to the limit under the integral sign", i.e., lim J/fc = /(lim/fc), but the monotone convergence requirement is often not satisfied. We next prove another convergence result, the Dominated Convergence Theorem, which relaxes this requirement. For this we require a preliminary result.
54
Introduction to Gauge Integrals
Definition 6. A sequence /&:!—>• K is uniformly gauge Cauchy over I if for every e > 0 there exist a gauge 7 on I and an N such that i,j > N implies \S(fi,V)  S(fj,V)\ < e whenever V « 7. Proposition 7. Let fk : I > K be integrable over I. Then {fk} is uniformly gauge Cauchy over I if and only if {fk} is uniformly integrable over I and lim fj fk exists. Proof: Let e > 0. Suppose {fk} is uniformly gauge Cauchy over I. Let 7 be a gauge on I and TV be such that \S(fi,T>) — S(fj,V)\ < e/2> whenever i, j > N and V « 7. Let i, j > N. Let 71 be a gauge on / such that
JfiS(fi,V) whenever V «
<e/3and < e/3 and J f3  S(fjtV)
7 ^ Put 72 = 71 H 7. Then if V «
JfiJfi
^
JfiS(fi,V)
+
s{fi,v) Jt Si
<e/3
72,
\S(fi,V)S(fj,V)\ < £
which implies {/7 fi} is Cauchy and, therefore, convergent. Pick M > N such that \ fj fi  fr fj\ < £/3 when i,j > M. Pick a gauge 73 on / such that
S(fM,V)whenever V « S{fi
I
IM < e / 3
73. Put 74 — 73 n 72. If i > M and V «
<\S(fl,D)S(fM,V)\
+
L'"~Lu
+
74, then
S(fM,V)J
/M
< £
so {fk} is uniformly integrable. For the converse, let 7' be a gauge on / such that \S(fi,V) — Jr fo\ < e/3 whenever V « 7' and i € N. Pick N such that \ Jj fi~ Jj fj\ < e/3 whenever
55
Convergence Theorems i,j >N.
UV «
i and i, j > N, then
5(/nP)5(/J;P)<
S(fi,V)Jfi\+ Jfi
fjj
j fiSUj,V) so {fk} is uniformly gauge Cauchy. We now have the necessary machinery for the Dominated Convergence Theorem. Theorem 8 (Dominated Convergence Theorem: DCT). Let fk, f,g : / —>• K and assume fk, g are integrable over I with \fi~fj\ < g for all i,j. If lira fk=f pointwise on I, then (i) {fk} is uniformly integrable over I, (ii) / is integrable over I and
(hi) limfIfk =
fIf(=JI(\\mfk)).
Proof: Set tjtk = V {  / m — fn\ : j < m < n < k}. Then each tjtk is integrable by 4.11 and 4.12 and for each j , {tjtk}1?=i *s a n increasing sequence which converges to the function t, = V {  / m  fn\ : j < m < n} < g. The MCT implies that each t, is integrable with Jt tj < J 7 g. Now tj+\ < tj and tj —)• 0 pointwise so the MCT implies J} tj I 0. Let £ > 0. There exists JV such that Jj t^ < e/2 and there exists a gauge 7 on / such that
S(tN,V)whenever V «
[tN\
Ji
I
<e/2
7. If i,j > N and V << 7, then
\S(fuV)  S(fj,V)\ < S(\fi  fj\,V) < S(tN,V) < JtN + e/2 < e. Hence, {fi} is uniformly gauge Cauchy on J and (i) follows from Proposition 7. Conditions (ii) and (iii) follow from (i) and Theorem 2. Remark 9. The "usual dominating hypothesis" in the DCT for the Lebesgue. integral is that there exists an integrable function g such that  / j  < g for all j . Since \fi — fj\ < \fi\ + \fj\, this hypothesis clearly implies the one in Theorem 8 [see also Exercise 7]. On the other hand, the hypothesis \fi — fj\ < g allows
56
Introduction to Gauge Integrals
the functions fj to be conditionally integrable [just take any conditionally integrable function h and consider the sequence {fj + h}] and is more appropriate for a conditional integral like the gauge integral. Exercise 5 gives another equivalent way of phrasing the dominating hypothesis in Theorem 8. Besides the MCT and DCT another useful convergence theorem for bounded intervals is given in Exercise 1. [The analogue of the result in Exercise 1 for the Riemann integral is the "usual" convergence theorem for the Riemann integral.] See also Exercise 6. We now give some examples illustrating the use of the convergence theorems. For this we derive some results pertaining to integrals which depend upon parameters. Let 7 be a closed subinterval of R* and S a metric space (substitute R or R n for S if necessary). If / : 5 x / > R, we write f(s, •)[/(•, t)} for the function t > f(s,t)[s —• f(s,t)] when s £ S[t G / ] . If f(s, •) is integrable over / for each s 6 S, then F(s) = Jj f(s,t)dt defines a function F : S —• R. Functions which depend upon parameters in this way arise in many areas of mathematics, and it is often important to ask if properties of the function /(•, t) are inherited by the function F. We first consider the property of continuity. Theorem 10. If f(,t) is continuous at so € S for each t € I and f(s, •) is integrable over I for each s £ S and there is an integrable function g : I —> R such that \f(s,t)\ < g(t) for allt € I, then F(s) = fjf(s,t)dt is continuous at soProof: Let Sfc —> s0 in S. Then f(sk,t) >• f{s0,t) and \f{sk,t)\ < g(t) for every t £ I. The DCT implies that F(sk) —• F(SQ) so F is continuous at SQ. Example 11. {Gamma Function) The gamma function is defined by T(x) = L tx~1e~tdt. We show that T is defined and continuous for x > 0. For x > 0 and 0 < t < l,tx~le~t < tx~l so Corollary 3.5 and Example 3.6 imply that the 1 x 1 t integral JJ, t ~ e~ dt exists. For x > 0 and t > \,tx~xe~l = [t x + 1 e _ t ]t~ 2 . For x+1 l x > 0, linit^oo t e~ = 0 so there is a constant M such that tx~1e~t < Mt~2. Now Corollary 4.5 and Example 4.6 imply that J1°° tx~1e~tdt exists. Hence, T(x) exists for x > 0. We next show that T is continuous. Fix xo>O.IfO
57
Convergence Theorems Theorem 12 (Leibniz' Rule). Let S = [a,b]. Suppose (i) gj = Dif exist for all s G S and t £ I, (ii) each / ( s , •) is integrable over I and there is an integrable function g : I R such that
£<••«
=
\D1f{s,t)\
for allt e I,s e S. Then F(s) = Jj f(s,t)dt
is differentiable on S with
F'(s) = J ^(s,t)dt = J Dxf{s,t)dt. Proof: Fix So & S and let {sk} be a sequence in S converging to SQ with Sk ^ so For t e I lim(/(5fc,<)  f{so,t))/{sk
 so) = £>i/(so,*) =
g(so,t)
and the function hk(t)
f(sk,t)

f(s0,t)
Sk ~ SO
is integrable over I. By the Mean Value Theorem, for each pair (k,t) there is a point Zk,t lying between Sk and SQ such that hk(t) = Dif(zk,t,t) so \hk(t)\ < g(t). The DCT implies that limk hk(t) = Jj Dif(s0,t)dt, i.e., F'(s0) = JjDif^o^dt. If / is a bounded interval and D\f is continuous over 5 x 7 , then D\f is bounded and the function g in Theorem 12 can be taken to be a constant. We use this observation in the next example. Example 13. We show A = J0°° e~l dt — \/TT/2. For this we introduce F(x)=
/ e x ( 1 + * 2 ) /(l + i 2 ) ^ Jo
Note that F(0) = arctan 1 = 7r/4, and if x > 0, 0 < Fix) < e~x f —^dt 2 J0 1 + t
= e x 7r/4
58
Introduction to Gauge Integrals
so F(oo) = lim^^oo F(x) = 0. Fix x. We have, by Leibniz' Rule, F'(x) = ex
f e~xt2dt =  ( e " 7 V 5 ) f Jo Jo
e~u'du =
{e~x/y/x)g(Vx),
where g(z) = JQ e~u du. Integrating F' from 0 to oo, gives /oo
roo
TT/4 =  / (ex/y/x)g(y/x')dx Jo
= 2 / Jo
e~z2g(z)dz
/*oo
= 2 / Jo
g'{z)g{z)dz = g(oo)2 = A2 .
Hence, A = \Zn/2. Notes/Remarks Henstock originally established the MCT and DCT for the gauge integral by using what we have referred to as Henstock's Lemma (see the remarks in [H3] § 22). The formal definition of uniform integrability was introduced by Gordon ([Go2]), but it was used implicitly by McLeod in his proofs of the MCT and DCT ([ML] § 3.10). Schwabik used the notion under the name, equiintegrability and also established Theorem 2 ([Sch]). If {fk} is a pointwise convergent sequence of gauge integrable functions, Gordon gives a characterization of uniform integrability of {fk} in terms of their indefinite integrals ([Go] 13.29). It will be shown in Theorem 7.8 that the convergence conclusion of Theorem 3, the MCT and the DCT can be improved to converge in the Alexiewicz norm. Bartle has given an interesting condition (called 7convergence) for the limit / of a sequence of gauge integrable functions {fk} to be integrable and for lim//fc = J f to hold ([Ba2]). See also [Go3] where Gordon discusses 7convergence and alters Bartle's definition slightly. There is a more general convergence theorem for the gauge integral called the Controlled Convergence Theorem; this result implies both the MCT and DCT. However, this theorem involves more technical notions than we are assuming in this text. The interested reader can consult [LPY], Chapter 7, for the statement and several proofs of the Controlled Convergence Theorem; there are also a number of references given in the bibliography. The derivation in Example 13 is from [We]; a generalization of the formula in Example 13 is given in [DS], Example 15.10. Exercise 33 gives an interesting improvement to Theorem 3.4 for "improper integrals."
Convergence Theorems
59
Exercises 1. Suppose 7 is a bounded interval and fk '• I —> R is integrable over 7 with fk—tf uniformly on 7. Show that (i) {fk} is uniformly integrable over I, (ii) / is integrable over 7 and
(in)
Jjh^Jjf.
2. Show that the analogue of the MCT (DCT) is false for the Riemann integral. 3. State and prove a version of the MCT for decreasing sequences {fk}4. Let /fc,/ : I —)• R be nonnegative with each fk integrable over 7 and / = Efcli fk pointwise. Give necessary and sufficient conditions for / to be integrable over 7. 5. Let fk : I —> R be integrable over 7. Show that there exists an integrable function g : 7 —>• R satisfying \fk — fj\ < 9 for all k, j if and only if there exist integrable functions g and h satisfying g < fk < h for all k. 6. (Bounded Convergence Theorem: BCT) Let 7 be a closed, bounded interval and let fk be integrable over I with /it —>• / pointwise. If there exists M > 0 such that /fc(i) < M for all k,t, show that (i) {/fc} is uniformly integrable over 7, (ii) / is integrable over 7 and (hi) / 7 /fc > / 7 / . 7. Let /fc : 7 —>• R be integrable over 7 and fk —> f pointwise. Suppose there exists an integrable function g such that /fc < g for all k. Show that in this case the conclusion of the DCT can be improved to read Jr \fk — f\ —> 0. 8. Let fk,f,g satisfy the hypothesis in Exercise 7 and let 7 = [a, b}. If Fk(t) = / /fc, 77'(i) = / / , show that Fk ^ F uniformly on 7. 9. (Fatou's Lemma) Let fk • I —>• R be nonnegative and integrable and assume that lim/fc is finite on 7. Show that lim/j /fc < /,(lim/fe). Give examples showing strict inequality may hold and the nonnegativity assumption cannot be dropped. [Hint: Consider hk = inf{/j : j > k} and use MCT.] 10. Give an example for which //. —> 0 pointwise, / 7 /fc —> 0 but {/fc} is not dominated by an integrable function. 11. Show that F(x) = J^° J,'"\$ dt is continuous for x £ R. 12. Show that F(x) = /0°° e~xt sini dt defines a continuous function for x > 0. 13. Show that (i) r(a; + 1) = xT(x) for x > 0, (ii) T ( n + 1 ) = n! for n € N,
60
Introduction to Gauge Integrals (iii) limx_>0+ xT(x) = 1,
(iv) T(i) = A Show that T has derivatives of all orders and give a formula for T^k\ 14. Show that j™ x2ne~x2dx = ^ ^ for n = 0 , 1 , 2 , . . . . [Hint: n = 0 is Example 13.] 15. Show that /0°° e~tx"'dx = \sfnJt for t > 0. 16. Show that F(x) = J0°° \r^dt
is differentiable for x > 0.
17. Show that F(x) = /0°° j^tdt
is differentiable for 0 < x < 1.
18. Let / : [0,1] > R be continuous. Show that /„ /(£fc)
1/(0)24. Let /
: [0,00) —>• R be bounded and continuous.
Show that linifc
/o°°T$Ei?* = O f o r p > l . 25. Let / : [a, b] x [c,d] > 1 be continuous. Show that F(x) = J is continuous for a < x < b. 26. Show that limx_>o+ / 0 \r^dt = In 2.
f(x,y)dy
27. Evaluate linx^oo J0°° \^dt. 28. Let / : 1 > R be absolutely integrable and uniformly continuous on R. Show that lim a ._ >00 /(a;) = 0. Can "uniform continuity" be replaced by "continuity" ? 29. (RiemannLebesgue Lemma: junior grade) Let / : [0,1] —> R be continuous. Show that linifc L f(t) s'm(kt)dt = 0. [Hint: First consider the case when / is a step function.] 30. (Laplace Transform) Let / : [0, 00] —> R be integrable over every bounded subinterval. The Laplace transform of / is defined by £{/}(s) = /0°° e~st f(t)dt. The function / is of exponential order if there exist constants a, M such that \f(t)\ < Meat for t > 0. Show that the Laplace transform of such a function exists and defines a continuous function for s > a. 31. Evaluate linifc /Qfc(l + t/k)ke~2tdt. 32. Suppose / : [a, b] >• R is continuous and Ja tkf(t)dt = 0 for fc = 0,1, 2, Show that / = 0. [Hint: Weierstrass Approximation Theorem.]
61
Convergence Theorems
33. (See Theorem 3.4) Let / : [a, b] » R be gauge integrable over [o, b] and let ak I a, a < ak < b. Define fk by fk(t) = f(t) for ak < t < b and fk(t) = 0 for a < t < ak. Show that {fk} is uniformly integrable on [a, b]. Hint: Set F(t) = fa f. Pick 5 > 0 such that F(f)  F(s) < e for t  s\ < 5. Pick 7 such that y(t) C (t  6, t + <5) and \S(f,V)  f* f\ < e whenever V « 7. Suppose V « 7. For each A; let Pfc be the subset of V with tags greater/equal ak. The union of the subintervals in Vk is an interval [y,b] with \y — ak\ < S. Use the Henstock Lemma 3.1 to show
S(fk,D)~
f fk <2e. Ja
34. Use Exercise 33 to give an example of a sequence which is uniformly integrable but does not satisfy the hypothesis in either MCT or DCT. [Hint: Use a conditionally convergent integral.] 35. Let /fc(0) = k and fk(t) = 0 for 0 < t < 1. Show that {/*.} is not uniformly integrable over [0,1]. 36. Let f,g:I—t [0, 00). Assume that / is integrable over J and g is integrable over every bounded subinterval of I. Show that / A g is integrable over I. In particular, / A k is integrable over J for every k € N. [Hint: Use MCT.] 37. Let / : I —> R be absolutely integrable over I. For k e N set fk(t) = k if f(t) < k, fk{t) = k if f{t) > k and / fc (i) = f(t) if  / ( t )  < k [the truncation of / at k]. Show that / fc is absolutely integrable over I. [Hint: consider / A k = g and h — (—k) V g.]
Chapter 6
Integration over More General Sets: Lebesgue Measure In previous chapters we considered integrating functions which are gauge integrable over subintervals (or finite unions of subintervals). In this chapter we consider integrating functions which are gauge integrable over more general subsets. The sets which we consider are the Lebesgue measurable subsets. This chapter uses only results from previous chapters but is not necessary for reading the unstarred chapters which follow. There is a natural notion of length for subintervals of R, and we now seek to extend the length function in a natural way to certain subsets of K. We say that a subset E of R is (Lebesgue) integrable if CE is (gauge) integrable over R and define the (Lebesgue) measure of E to be \{E) = J^CE Let £ be the family of all integrable subsets of R. We have the following properties of £ and the set function A. Proposition 1. (i) All bounded intervals I belong to £ and \(I) = £(I). In particular, <j> € £ and \((f>) = 0. (ii) If E,F e £, then E U F, E n F and E\F belong to £. (iii) If Eke C for every k N , then f  ^ i Ek e C. (iv) E £ £ if and only if E + a G £ for every a € R and, in this case X(E) = \(E + a) (translation invariance). (v) / / {Ei} C £ is pairwise disjoint and E = U S i ^*' ^ e n E & C if and only J/^Zi^i ^(Ei) < °° and, in this case, \(E) = Yl^Li ^(Ei) (countable additivity). (vi) / / E,F e £ and E C F, then \(E) < \(F). (vii) / / E is null, then E £ C and X(E) = 0. 63
64
Introduction to Gauge Integrals
Proof: (i) is clear, (ii) follows from the identities CEUF = CE VCV, CEHF = CE A CF and CE\F = CE — CEHF (Proposition 4.12). For (iii), let £ = f£Li #fe a n d Fk = fljLi Ei T h e n Fk e C by (ii) and CFk I CE so CE is integrable by MCT (Theorem 5.5), i.e., E £ C. (iv) follows from Exercise 2.12. For (v), CE = X)fcli ^Ek so this part is immediate from MCT. (vi) follows from (ii) and (v). (vii) follows from Example 1.10 or Exercise 1.9. We say that a subset E c R i s (Lebesgue) measurable if E n / is integrable for every bounded interval / . We denote the family of all measurable subsets of R by M. and note that C C M by Proposition 1. We extend A to M by setting \{E) = oo if E £ M\C and call A Lebesgue measure on R. We have the following properties for M. and A. Proposition 2. (i) (p £ M. and \{<j>) = 0. (ii) E £ M implies Ec = R\E £ M. (iii) EiEM forieN implies  J ~ i Ei, f)Zi Ei e M(iv) If I is an interval (bounded or unbounded), X(I) = £(I). (v) If {Ei} C M. is pairwise disjoint, then AflJ^i Ei) = Y^Hi M^*) {countable additivity). (vi) E £ M. if and only if E + a € M /or every a £ R and, in i/iis case, A(iJ) = A(i? + a) (translation invariance). Proof: (i) is clear. For (ii) let / be a bounded interval. Then I = (I H E) U (/ n Ec) and I £ C, I n E £ C imply V" n Ec £ C by Proposition 1 (ii). For (iii), if J is a bounded interval and A,B £ M., then I (1 (A U B) =
(lnA)u(lnB)
and/n( J 4nB) = ( / n i ) n ( / n B ) s o i u B a n d J 4 n B
belong to M by Proposition 1. If {Ei} C A4, we construct a pairwise disjoint sequence {Fk} from M. whose union is E = \J°°.± Ei. For this, set F\ = E\ and Ffc+i = E^+i \ Ui=i Ei. Then the {.Ffc} are pairwise disjoint, Fk £ M by (ii) and the observation above and E = Ufcli Fk From Proposition 1 Efe = 1 A(Ffc n I) < X(I) for each n so ^ T ^ A(Ffc n I) < \(I) and E n J £ £, i.e., £ e X . Since f l ^ ^ = ( U S i ^iT = F, F £Mby (ii) and what has just been established. (iv) is clear, (v) follows from Proposition 1 (v) and (vi) follows from Proposition 1 (iv). At this point it is not clear how large the families C and M are. Since every open subset of R is a countable union of pairwise disjoint open intervals ([DS] 16.6), it follows from Proposition 2 that every open set is measurable and its
Integration over More General Sets: Lebesgue Measure
65
measure is the sum of the lengths of the open intervals whose union is the open set. Also, from Exercise 3 every bounded open set is integrable. It now follows from Proposition 2 (ii) that every closed subset of E is also measurable. We now introduce some useful terminology from measure theory. A family of subsets ^) of a set S is called a aalgebra if J^ satisfies conditions (i)(iii) of Proposition 2. If T is any family of subsets of a set S, there is a smallest calgebra containing T called the cralgebra generated by T (Exercise 1). The cralgebra B = B(M) generated by the open subsets of E is called the family of Borel sets of E. From Proposition 2 and the observations above, we see that M is a cralgebra containing the Borel sets B. It is the case that B ^ M and assuming the Axiom of Choice that M ^ "P(R), where 'P(R) is the power set of E (see Notes/Remarks for references to these results). We will show later (Theorem 5 and Corollaries 7,8) that every measurable set is approximated in some sense by a Borel set. A set function v defined on a cralgebra ^ is said to be countably additive if v : ^ —> E* takes on at most one of the values {±oo}, v(
66
Introduction to Gauge Integrals
t must be contained in 7(2). Then either Jfc^i) € £fci or J^it) € Ufcl~i ^kHence, E C U{J : J 6 £}. Arrange the elements of {Jk(t) : t e E} into a (possibly finite) sequence {Jfc : k e M } . Then for each k £ M there exists ifc G E n Jfc such that Jfc C 7(*k). Then {(ifc, Jk) : k e M} is the desired family. We can now establish an approximation theorem for bounded measurable sets. Theorem 4. Let E be a bounded, measurable set contained in the bounded, closed interval I and let e > 0. There exists an at most countable family {Jk : k 6 M} of closed, nonoverlapping subintervals of I such that E C U{Jfc : k € M} and EfcgM *( Jfc) < A (£) + £Proof: There exists a gauge 7 on 7 such that  5 ( C B , 2 3 ) — A(£) < £ whenever V « 7 (Exercise 3). Let {(ifc, Jfc) '• k 6 M } be the family in Lemma 3 relative to e, E and the gauge 7, where we assume M C N. By Henstock's Lemma for every q G M,
^2{\(EnJk)cE(tk)e(Jk)}
^{A(£nJfe)W}
fc=i
fc=i
< £ .
Therefore, for every q e M, J2l=i £(Jk) < E L i X(E n Jfc) + £ < A(£) + £, and, hence, ^2keM ^(^k) < A(J5) + £ so the result follows. From Theorem 4 it is now easy to obtain an approximation theorem for general measurable subsets. Theorem 5. Let E be measurable. For every e > 0 there exists an open set G D E such that X(G\E) < e. Proof: First assume that E bounded interval I. We can extend interval, still denoted by Jfc, to be £ by e/2 and extend each Jk by a so G is open and E C G. Also, by
is bounded and is contained in a closed, each interval Jfc in Theorem 4 to be an open such that YlkeM ^k) < A(.E) + e [replace factor of e/2k+1]. Set G = U{ Jk : k € M } Exercise 2,
X(E) < X(G) < J2 A(Jfc) ^ A ( £ ) + £ > fceM
and by Exercise 2, A(G \ E) = \{G)  X(E) < £ .
67
Integration over More General Sets: Lebesgue Measure
If E is measurable, let Ek = En [—k, k]. By the first part, for every k there exists an open set Gu D Ek such that X(Gk \ Ek) < e/2k. Set G = Ufcli Gk so G is an open set containing E = Ufcli ^k Since G \ E c Ufc=i(^fc \ ^fc)> Exercise 2 gives oo
X(G\E) < ^
oo
A(Gfc \ Efc) < £
£ /2
fc
= £.
fc=i fc=i
We now derive several corollaries from Theorem 5. The first gives a useful characterization of null sets. Corollary 6. E is null if and only if E is integrable and X(E) — 0. Proof: =>: follows from Example 1.10 or Exercise 1.9. •4=: Let £ > 0. By the argument in Theorem 5 for every k there exists a sequence of open intervals {Jj : j 6 N} such that Ek C U j l i Jj a n ^ Ef=i l(Jj) < £/2fcso E is null.
Then
{J, ^ J ' e N} covers E and Y^=i Y!?=i *(Jj) < £
A subset H C R is called a 5(5 if ff is a countable intersection of open sets. Corollary 7. Let E be measurable. Then there exists a Qs set H D E such that \{H \E)=0. Proof: By Theorem 5 for every fe € N there exists an open set Gk D E such that A(Gfc \ E) < 1/k. Put H = HfcLi ^fc Then H is a Qs, and since H\EcGk\E for every jfe, A(H \ B) < 1/ife so X(H \E) = 0. Corollary 8. E C M. is measurable if and only if E = B U Z, where B is a Borel set and Z is null. Proof: Since B C M. and any null set is measurable, the condition is sufficient. Conversely, if E is measurable, let H be as in Corollary 7 and let Z = Since H G B, the result follows from Corollary 7.
H\E.
We now establish a result which illustrates the utility of the class of measurable sets. We say that a function / : J —»• R is integrable over a subset E C / if fCs is integrable over I and set
Lf'LfC°T h e o r e m 9. Let I be a closed interval in R* and f : I —> R. / / / is absolutely integrable over I, then f is (absolutely) integrable over every measurable subset
68
Introduction to Gauge Integrals
of I. Moreover, if Aii is the aalgebra of measurable subsets of I and u(E) = JE f for E E Aii, then v is countably additive. Proof: Since f = f+ — f~ and both / + and / ~ are integrable over I, we may assume that / is nonnegative. Let E E Aii. Now fh = fA (kCtk,k]nE) is integrable over / for every k (Proposition 4.12), fk t fCE and 0 < fk < f so fCE is integrable by DCT (Theorem 5.8). Suppose {Ek} is a pairwise disjoint sequence from Mi with E = IJfcLi EkThen fCE = J X X fCEk so u(E) = ZT=i u(Ek) by the MCT (Theorem 5.5). The absolute integrability assumption in Theorem 9 is important; see Exercise 8. See also Theorem 13 in Notes/Remarks. See Exercise 12 for a partial converse to Theorem 9. We now give improvements in the MCT and DCT which are sometimes useful in applications. For the MCT we need a preliminary lemma. Lemma 10. Let fk : / —>• R be integrable over I for every k with 0 < fk < fk+i If M = sup Jj fk < oo and if f(t) = \im fk(t) [limit in R*], then f is finitevalued a.e. in I. Proof: Let E = {t E I : f(t) = oo}. For each j and k, fk/j so (fk/j) A 1 is integrable [Exercise 5.36] with
j{fk/j)M<
is integrable
J(fk/j)<M/j.
For each Jifk/j)
A 1 f (//j) A 1
as k —> oo so the MCT implies
(*) \im J (fk/j) A 1 = J(f/j) A 1 < Ml 3 • Now / > 0 so (f/j) A 1  and lira, Jj(f/j) A 1 = 0 by (*). If t E I \ E, (/W/J')A1 = f(t)/j for large j so (f(t)/j)M ^ 0; if tEE, (f(t)/j) Al = 1 so (f(t)/J)A1 >• 1 Therefore, (f/j)M » CE and by the MCT, CE is integrable with Jj CE = 0. Corollary 6.6 now gives the conclusion. We can now establish a generalization of the MCT. Theorem 11 (General MCT). Let fk : I >• R be integrable over I with fk < fk+i and supfc JT fk < oo. Then there exists an integrable function f : I —> R such that fktf a.e. in I and Jj fkt Jj f
69
Integration over More General Sets: Lebesgue Measure
Proof: By replacing fk by fk~fi, if necessary, we may assume /fc > 0. Let E = {t £ I : lim fk(t) < oo}. Put gk = CEfk so fk = gk a.e. and 0 < gk < gk+i by Lemma 10. Let / = lim#fc and note f(t) < oo for every t € I. Exercise 2.22 implies that lim JT gk = lim J 7 fk = fj f.
The MCT and
We can also give a more general form of the DCT. Theorem 12 (General DCT). Let fk,f,9'I> K Assume that fk, k e N and g are integrable over I with /fc —> / a.e. and \fk\ < g a.e. Then f is integrable over I with lim Jj fk — Jj fWe leave the proof for Exercise 13. Notes/Remarks The construction of Lebesgue measure can be legitimately viewed as an attempt to extend the natural length function defined on subintervals of R to a more general class of subsets of R. Of course, such an extension should possess some of the properties of the natural length function; for example, it is natural to require that such an extension satisfy the translation invariance property of Proposition 2 and also the countable additivity property of Proposition 2. We have obtained such an extension by using our construction of the gauge integral, but it would be much more desirable to have a purely geometric construction of the extension. There are several such constructions which are "wellknown"; see for example, [Swl] 2.5, [Nl] III, [Ro] § 3. In these extensions, Theorem 5, Corollary 7 and their converses hold and give geometric characterizations of the class of (Lebesgue) measurable sets (the class of subsets of R to which the length function is extended). It is known that such an extension is essentially unique ([Swl] 2.5.6), and if M is again the class of measurable subsets, then B C M C V(R) ([Swl] 2.5.9, 1.3.1). See also [Nl], [Swl] 2.5 for other interesting properties of Lebesgue measure. Exercise 8 shows that the absolute integrability of / in Theorem 9 is important. In fact, the converse of Theorem 9 holds; however, its proof seems to require the measurability of a gauge integrable function. Theorem 13*. Let I be a closed interval in R* and f : I —> R. If f is integrable over every measurable subset of I, then f is absolutely integrable over I. Proof: Since / is measurable (Appendix 3), P={tel
: f(t) > 0} and N = {t e I: f(t) < 0}
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Introduction to Gauge Integrals
are measurable. Therefore, / + = fCp and / = /Cjv are integrable over J by hypothesis and  /  = / + + / ~ is integrable over I. It follows from Theorem 13 that a function which is integrable over an interval I is not, in general, integrable over every measurable subset of I. In fact, a function which is integrable over an interval J needn't be integrable over a countable disjoint union of subintervals of J ([LPY] p. 115). Theorem 3 of [YL] gives a necessary and sufficient condition for a function which is integrable over an interval I to be integrable over a pairwise disjoint union of subintervals of I. A more general result guaranteeing the integrability over a pairwise disjoint union of measurable subsets is given in Theorem 7 of [LPY1].
Exercises 1. If T is any family of subsets of a set S, show that there is a smallest cralgebra containing T• 2. Let Yl be a cralgebra of subsets of a set S and assume fi : Y, —• [0, oo] is a measure. Show: (i) if B , F e ^ , £ c F , then //(F) < //(F) [monotonicity], and if //(F) < oo, then / / ( F \ F ) = //(F)  //(F), and (ii) if { F J C £ , then //(Ufci Ei) < "Y^L\ M^i) [countable subadditivity]. [Hint: For (ii), disjointify the {Ei\ as in the proof of Proposition 2 (Hi).] 3. Show that every bounded measurable set in R is integrable. 4. Let / : R >• R and A, B be measurable with A C B. Show that if / is absolutely integrable over B, then / is absolutely integrable over A. Show the absolute integrability hypothesis cannot be replaced by integrability. [Hint: For / > 0, consider fk = f AfcCjin[fc,fc]; Exercise 3.9 or Exercise 4.3.] 5. Give an example of a Borel set which is neither open or closed. 6. Let F be measurable and e > 0. Show that there exists a closed (compact) subset F C E such that A(F \ F) < e. 7. A subset H of M is called an T„ set if H is a countable union of closed sets. If E is measurable, show that there exists an Ja set H C F such that X(E \H) = 0. 8. Show that the absolute integrability assumption in Theorem 9 cannot be replaced by integrability. [Hint: Exercise 3.9 or Exercise 4.3.] 9. (Chebychev Inequality) Let 7 be a bounded interval and / : I —¥ [0, oo) be integrable over / . Let r > 0 and set A = {t G I: /(£) > r } . Show that A is integrable and
\(A)
f.
r)+ and note fk > CA, 0 < fk < 1.]
Integration over More General Sets: Lebesgue Measure
71
10. Let 7 be a bounded interval and / : 7 —• R absolutely integrable over 7. Show that ft  /  = 0 if and only if f(t) = 0 for all t € 7 except those in a null subset. [Hint: {t :  / ( t )  ^ 0} = \J™=1{t : /(i) > 1/fc}; use Chebychev's Inequality.] 11. Let 7 be a closed interval in R* and suppose / : 7 —> R is absolutely integrable over every bounded subinterval of 7. Set E = {t 6 7 : f(t) ^ 0}. Show that E is measurable. [Hint: Let J be a bounded subinterval of 7 and set fk = (k\f\) A Cj and note fk t C ^ r v ] 12. In Exercise 11 show that when JF \f\ exists, then E fl F is measurable. 13. Prove Theorem 12. [Hint: Amalgamate all of the null sets in the statement.] 14. State and prove a generalization of Theorem 5.4 in the spirit of Theorem 11.
Chapter 7
The Space of Gauge Integrable Functions The space of gauge integrable functions is a vector space under the operations of pointwise addition and scalar multiplication, and the space also has a natural seminorm, called the Alexiewicz norm. In this chapter we describe a few of the basic properties of the space and give references to some of the more technical properties. This chapter requires a basic knowledge of normed linear spaces and employs more advanced techniques (including several starred results from the text) than are required for the reading of the remainder of this text. We treat the case when / = [a, b], —oo < a < b < oo. Let HK{I) be the space of all gauge integrable functions defined on / . Then HK(I) is a vector space under the usual operations of pointwise addition and scalar multiplication. There is a natural seminorm on 'HK(I) originally defined by Alexiewicz ([A]) via ll/H = sup{ / o */ : a < t < b} (Exercise 1). If / = 0 a.e. in I, then ll/H = 0, and, conversely, if / = 0, it follows from Theorem 2 of Appendix 3 that / = 0 a.e. in / . Thus, if we identify functions which are a.e. in J, then ~HK(I) is a normed space under the Alexiewicz norm. In contrast to the case of the space of Lebesgue integrable functions with the L 1 norm, the space 7iK(I) is not complete under the Alexiewicz norm. E x a m p l e 1. Let p : [0,1] —• R be continuous and nowhere differentiable with p(0) = 0 ([D51] 11.20). Pick a sequence of polynomials {pk} such that pk —> p uniformly and Pfc(0) = 0 (Weierstrass Approximation Theorem ([DS] 18.8)). By the FTC Pk(t) = f0 p'k for every t g [0,1] so {p'k} is a Cauchy sequence in HK([0,1]) with respect to the Alexiewicz norm. If there exists f £ /HK([0,1]) such that \\p'k — f\\ —> 0, then Pk{t) = J0 p'k —> J0 f uniformly for t G [0,1] 73
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Introduction to Gauge Integrals
so p{t) = J0 f. Theorem 2 of Appendix 3 then implies that p is a.e. which is a contradiction. Hence, UK (I) is not complete.
differentiate
We give a description of the completion of UK(I) as a subspace of distributions. For a distribution T € V(R), we write DT for its distributional derivative ([Sw2], § 26.4.12, [Tr] § 23). Let T(I) = {T e V'(R) : DT = f, where / : R >• R is continuous, / = 0 on (co, a] and f(t) = f(b) for t > b}. Note that if / G UK (I), then / induces an element F of T{I) by F(t) = J* f, a < t < b, F(t) = 0 for t < a and F(t) = f* f for t > b (Theorem 2 of Appendix 3 and Corollary 2.3); i.e., the map f —> F imbeds UK (I) into J{I). We define a norm on T(I) by setting T = sup{/(i) : a < t < b}. Now T{I) is complete under this norm and the norm restricted to UK(I) is just the Alexiewicz norm so J(I) is the completion oiUK(I). For details, see [MO]. Since UK(I) is not complete, it is natural to ask if UK(I) is second category (or a Baire space; [DS] § 24). This is, however, not the case; UK (I) is first category in itself. To see this let CQ(I) be the set of all continuous functions on / which vanish at a equipped with the supnorm, / = sup{/(t) : a < t < b}, f € C0{I). We can identify UK (I) with the subspace {F : F(t) = f*f,a < t assume that the boundary of X,dX, is countable with dX = {rj : i € N}. Assume that H(J) = J2T=i JjCjkfk exists for every subinterval J of I and lim^(j)_>o H(J) = 0. If f = E f c = i ^ 4 / * : [pointwise], then f G UK (I) and fjf = H(J) for every subinterval J. Proof: Let e > 0. For each j let 7j be a gauge on I such that \S(fj,T>) — j t fj\ < e/23 whenever V « 7j and let 5j > 0 be such that £(J) < Sj implies \H(J)\ < e/2j. Define a gauge 7 on I by y(t) = IjitfnJj for t € J,, y{t) = X°
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The Space of Gauge Integrable Functions
for t G X° and if t = rj choose 7(f) to be a symmetric neighborhood of t with £(7(£)) < 5j. Note the following properties relative to 7: (i) if t € Jj and t G J C 7(f), then /(*) = fj(t) and # ( J ) = /_, /,; (ii) if * G X° and i € J C X°, then / ( / ) = 0 and H(J) = 0; (iii) if t = rj and r 3 e J c 7(7^), then /(<) = 0 and \H(J)\ < e/2J. Suppose V = {(U,Ii) 2 = 1,2,... ,
: 1 < i < m} «
7. Set di = {k : tk G J{\ for
do = {k : tk = rj for some j} , 0
dx = {k:tke
X } and Vj = {(U, h) : i G d,} for j =  1 , 0 , 1 , . . . . Note
(iv) Vj « 7 , f o r j = 1,2,.... Then (1)
S(/,2>)tf(/) =
E EtffoW1*)1^7*)} E {/(**) W  H(Jfc)}
<
fcgd_!
+ ^{m)e(ik)H(ik)} k£d0
+ — i?_i + i?o + Rl , with obvious definitions for i?,. First, i?_i = 0 by (ii). Next, if tk = rj for some j , then by (iii) f(tk) = 0 and t{Ik) < Sj so \H(Ik)\ < e/2j, and since rj is the tag for at most 2 subintervals in V, R0 < 2 YlJLi £/^j = 2e. Finally, if tk G dj for i > 1, then by (i) and (iv) and Henstock's Lemma,
^imwk) kec
 H{ik)}
^{MtkWk)
 [ ft) Jlk
kedi
S(fi,Vi)
f kedi
fi
<e/T
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Introduction to Gauge Integrals
Thus, Ri<e and (1) implies that \S(f,V)  H(I)\ < 3s. Hence, / is integrable over I with / 7 / = H(I). The same computation works for every subinterval J; just begin with V being a 7fine tagged partition of J. For our next result we need to the Alexiewicz norm. Let X 11/11' = sup{ Jjf\:J el} for / 2/ for / G UK(I) so   and
another norm on UK(I) which is equivalent be the family of all subintervals of I and set G UK (I). It is easily seen that / < /' <  ' are equivalent (Exercise 2).
Proposition 3. Let fk G UK(I) with J^kLi HAH' < °° an^ ^ { A } be a pairwise disjoint sequence of open subintervals of I. Set f = Y2k=i Cjkfk [pointwise]. (i) Then H(J) = Yl'kLi Ij Cjkfk converges uniformly for J e l . (ii) \ime{J)^0 H{J) = 0. (iii) Iff€UK(I), then \\f52nk=1CJkfk\\i>0 [i.e., Jj f = T,Zi uniformly for J e l ) .
JjCjJk
Proof: Since  J j Cjfc All < AII' f° r every J G X, (i) and (iii) follow immediately. For (ii), let e > 0. Choose n such that Yl'kLn+i HAH' < £ / 2 and choose 6 > 0 such that £(J) < S implies  JjCjkfk\ < £/(2n) for k = 1 , . . . ,n [this follows from the uniform continuity of the definite integral F(t) = Ja f of a gauge integrable function, Corollary 3.2]. Then £(J) < 5 implies i/(J) < e. To establish the barrelledness of UK{I) we require the following result called the AntosikMikusinski Matrix Theorem. T h e o r e m 4. Let aij G R for i, j G N and A = [a,ij}. Suppose (i) lim; aij = 0 for every j and (ii) for every increasing sequence of positive integers {nij} there is a subsequence {rij} of {m,j} such that lim^ ^,=1 ainj exists. Then limi a^ = 0 uniformly for j G N. In particular, lira; an = 0. For a proof (of a more general result) see [Swl] 2.8.2 or [Sw2] 9.2. A normed space Y is barrelled if every weak* bounded subset of the dual space Y' is norm bounded ([Sw2] 24.5, [Tr] 33.1). We use this dual space characterization of barrelled normed spaces, but we do not require a concrete description of the dual of UK (I). T h e o r e m 5. Let B C UK{I)' [i.e., UK(I) is barrelled].
be weak* bounded. Then B is norm bounded
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The Space of Gauge Integrable Functions
Proof: Let A C UK {I) be bounded with a = sup{/ : / £ A}. It suffices to show that (3 = sup{(f,/) : u £ B, f £ A} < oo. Suppose /3 = oo. Then there exist z^ € B, f\ £ A such that  ( ^ i , / i )  > 2. By Exercise 4, £ —» (^i, C[ a t j/i) is continuous so there exists ai such that (^i,C[ a > a i ]/i) > 1 and(^i,C( a i i 6 ]/i> > 1 Either [a, ai] or (au b] satisfies sup{(i/,C[ 0 ) a i ]/) : v £ B, f £ A} = oo or sup{(i/, C( a i i b]/) : v £ B,f £ A} = oo. Pick one of these intervals which satisfies this condition and label it I\ and put J\ = I\I\. Note \(vi,CjJi)\ > 1. Now there exist v2 £ B,f2 £ A such that \{v2,Chf2)\ > 24. By the argument above there is a partition of I\ into 2 disjoint subintervals A2,B2 such that \{u2,CAJ2)\ > 23,\(u2,CBj2)\ > 2 3 and sup{\(v, CAJ)\ : v £ B,f £ A} — oo. Put J2 — B2,I2 — A2 and continue this construction. This produces a pairwise disjoint sequence of intervals {Jj}, {fj} C A, {VJ} C B such that
(2)
\(uj,Cjifj)\>f.
Let Kj be the interior of JJ; then (2) holds with Jj replaced by Kj. Set hj = £fj. Then ll^ll < ±\\fj\\ < a/j2. Hence £ ° l i   ^   < oo. Consider the matrix M = [m^] = \\{v%,Cxjhj}. We claim that M satisfies (I) and (II) in Theorem 4. First, (I) holds by the weak* boundedness of B. Next, Theorem 2 and Proposition 3 are applicable to the sequences {Kj} and {hj} since if X = I \ {J"L1 Kj, the boundary of X consists of the endpoints of the Kj and possibly 2 other limit points. Therefore, if h = Y^jLi Cvijhj [pointwise], then h £ HK(I) and the series converges in the norm of UK (I). By the continuity of each Vi, Y^7Limij = {\ui^) a n d {\vi,h) —>• 0 by the weak* boundedness of B. Since the same argument can be applied to any subsequence of {hj}, condition (II) is satisfied. Theorem 4 implies ma —> 0 contradicting (2). Thus, HK(I) supplies another example of a first category, barrelled normed space (see [Sw2], Exer. 24.6, for another, more familiar, example). We give a description of the dual of UK (I). Theorem 6. F £ TLK{I)' if and only if there is a right continuous function f of bounded variation with f(b) = 0 such that (F, (f) = JT ip(t)f(t)dt for every if £ UK {I). Moreover, \\F\\ = Var(f : I). See [A], Theorem 1 for a proof. For references to other description of the dual, see the Notes/Remarks section. We show that the step functions are dense in 'HK(I) (a step function is a linear combination of characteristic functions of intervals). Theorem 7. The step functions are dense in
%K(I).
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Introduction to Gauge Integrals
Proof: Let / e HK(I) and e > 0. Pick a gauge 7 on / such that \S(f, V) Jjf\ < e whenever V « 7. Fix V = {{UJi) : 1 < i < m} « 7. Set V = E™ 1 f{U)Cu. By the Uniform Henstock Lemma (3.8) if J £ I ,
/ /  / * = £ { / ff A
£
/  /(*i)€(/i n J)
<3e.
Jn/i
Hence,   / 
JJ
m
fk <J2 i=1
„
Jjnu
m
fkfk(ti)£(iinj)
+Yl\fk(ti)\l(iinj) i=1
<6e + s = 7e. Hence, if k > n, then /fc < Is. It follows from Theorem 8 that the convergence conclusions in the MCT and DCT can be improved to norm convergence, i.e., limj^/fc = Jj f uniformly for all subintervals J. Notes/Remarks For another example showing that 'HK(I) is not complete, see [LPY] 11.1. Sargent first showed that the space HK(I) is barrelled. Actually Sargent showed that the space of Denjoy integrable functions on / is barrelled under the Alexiewicz norm ([Sari]); since the Denjoy integral is equivalent to the gauge integral ([Gol]), it follows that HK(I) is barrelled. Sargent's proof
The Space of Gauge Integrable Functions
79
used a characterization of the dual space which is not required in the proof of Theorem 5. Thomson has given another proof of the barrelledness of TLK(I) which does not use the dual characterization of barrelledness ([Th2]). See also [Gi] for another proof of barrelledness (and actually a stronger topological property). The space %K{I) has a stronger property than being barrelled which is also of interest. A topological vector space X is called a Sargent space (/3space by Sargent) if there exists no sequence of subsets {Ek} satisfying (SI) 0 e Ek, EkEkC
Ek+1
(S2) x = u r = i ^ (S3) Every Ek is nowhere dense. A second category, metrizable space is clearly a Sargent space, but TCK(I) furnishes an example of a first category Sargent space ([Sar2]; see also [LPY] 11.3). Sargent spaces are important because the Uniform Boundedness Principle and the BanachSteinhaus Theorem hold for such spaces ([Sar2], [Sw2] 9.1415, [Tr] § 33). For other descriptions of the dual of 'HK(I) other than that given in Theorem 6, see [Sari], Theorem 4, [LPY], Theorems 12.3 and 12.7 and [MO] 2.1. In particular, Sargent describes the dual of 7iK(I) as the space EBV{I) of functions which are equivalent to a function of bounded variation; i.e., / 6 EBV(I) if and only if there is a function g : I —• IR of bounded variation on / such that f — g a.e. in I. A function / on / is a multiplier for WK(I) if and only if fg is integrable over / for every g e TLK(I). (See Notes/Remarks in Chapter 4.) Sargent shows that / is a multiplier if and only if / e EBV(I) so the dual space and the space of multipliers coincide ([Sar3]; [LPY] 12.9). For the higher dimensional case see [LCL1]. When / is an unbounded interval let TLK(I) be the space of all gaugeintegrable functions denned on / . If / = [—00,00], the (Alexiewicz) norm on UK {I) is defined to be / = s u p {  / ! 0 O /  :  0 0 < t < 00}. If X is the family of all subintervals of I, then a norm equivalent to the Alexiewicz norm is given by /' = sup{ fjf\:J€ I } . [Similar definitions are made in case / = [—00, a] or J = [a, 00].] The completion of HK(I) can be described as above. The barrelledness of 'HK(I) in the case when / is unbounded can be established as in Theorem 5. However, there does not seem to be a concrete description of the dual space of HK(I) in the case of an unbounded interval. Finally, Theorem 7 should be compared with Theorem 5.10 of [LPY] where it is shown that every gauge integrable function is the a.e. pointwise limit of step functions.
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Introduction to Gauge Integrals
Exercises 1. Show that the Alexiewicz norm satisfies the axioms of a seminorm, i.e., 11/ + 9\\< 11/11 + ff and t/ = t/. 2. Show that the expression for  ' following Theorem 2 defines a seminorm on UK(I) with H/ll < 11/11' < 2   /   . 3. Show that UK{I) is separable. 4. Let / € UK {I). Define / : /  > • UK (I) by f(t) = C [a ,t]/ Show / is continuous. 5. Let Y be a dense subspace of the normed linear space X. Show that the category of Y in X is the same as the category of Y in itself. ([AM] p. 201). 6. Let g e BV(I). Use Theorem 4.15 to show that (G,f) = Jjfg defines a continuous linear functional G on UK {I) with G < Var(g : I) + \g(b)\.
Chapter 8
Multiple Integrals and Fubini's Theorem Except for the usual cumbersome notation it is straightforward to define the gauge integral for functions defined on intervals in ndimensional Euclidean space, R n . In this chapter we make such a definition and study some of the basic properties of this multiple integral in R™. Most of the proofs in 1 dimension carry forward to higher dimensions except for the usual notational complications; we do not repeat the proofs which are obvious modifications of earlier proofs in 1 dimension. The computation of integrals over subsets of R" (multiple integrals) is usually carried out by doing iterated 1dimensional integrals. Theorems which relate multiple integrals to iterated integrals are usually referred to as Fubini Theorems. We give 2 versions of Fubini's Theorem. The first version (junior grade) uses only basic properties of the gauge integral and should suffice for an introductory real analysis course. The second version is more general but uses results relating null sets and the gauge integral from Chapter 6 on Lebesgue measure; it is included to show the generality of the gauge integral in R" for those readers who have gone through Chapter 6 [at least the part up to Corollary 6.6]. Let R*™ = R* x • • • x R* where there are n factors in the product. An interval I in R*n is a product Ji x • • • x J n where each Ik is an interval in R*; we say that I is a closed (open) interval if each Ik is closed (open). The interior of 7, written 1°, is the product of the interiors of the Ik • The volume of an interval I, denoted v(I), is the product v(I) = rifc=i^(^fc)> where we continue to use the convention that 0 • oo = 0. [In R 2 , the term area would be more appropriate.] 81
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Introduction to Gauge Integrals
If J is a closed interval, a partition of J is a finite collection of closed subintervals of I, {Ik : k = 1,... ,m}, such that 1% n 1° = 0 if k ^ j [i.e., the {Ik} do not overlap] and I = Ufcli^fc A tagged partition V of 7 is a finite collection of pairs {(:TJ,/J) : 1 < i < m} such that {Ii : 1 < i < m} is a partition of I and Xi G 7^ for 1 < i < m; we call the /j subintervals of V and call #» the tag associated with Ii. A gauge 7 on J is a function defined on I such that 7(2;) is an open interval containing x for every x £ I. A tagged partition T> = {{xi, Ii) : 1 < i < m} is 7fine if i j £ /; C 'j(xi) for 1 < i < m; we write V << 7 as before. We next proceed to show that every gauge 7 defined on a closed interval in R*n has a 7fine tagged partition. For simplicity and ease of understanding, we write out most of the proofs for n = 2; the proofs can be adapted to Wn with the usual notational complications. We give a proof of the existence of 7fine tagged partitions based on their existence in R; the construction is used later in the proof of Fubini's Theorem 13 [a more straightforward proof for compact intervals is outlined in Exercise 1]. Let J and K be closed intervals in R* and I = J x K. Let 7 be a gauge on / . If Pi and P2 are the projections from R*2 onto the first and second coordinates, respectively, then li{x,y)
= Pi7{x,y)(^2{x,y)
=
P2j{x,y))
has the property that 7i(, 2/)(72(^, •)) defines a gauge on J(K) for each y G K{x G J ) [Draw a picture!]. Lemma 1. For y E K let Vy be a 7i(, y)fine tagged partition of J. (i) 7/72(2/) = n{72(a;,j/) : (x, A) £ Vy}, then y2 *s a gauge on K. (ii) If K, = {(yj,Kj) : 1 < j < m} is a Y2~fine tagged partition of K and if Vy =
{(puj,P?):l<j
then V = {{{pf, yj),P?J
x^):l<j<m,l<«<
pVj}
is a 7fine tagged partition of I. V is called the compound tagged partition generated by {Vy : y 6 K} and /C. [See Figure 1.] Proof: (i) is clear since the intersection in (i) is finite. For (ii), if {p\\yj) G Pij x Kh then y, G Kj C i2{y3) C y2(p?,Vj) and
P\' eP? C7i(pf,%)
Multiple Integrals and Fubini's Theorem
83
Hence, P f x Kj C 7 i ( p r . V i ) x 72(pf,%) = 7 ( p f , % )
Figure 8.1
Corollary 2. Every gauge j on I has a jfine
tagged partition.
If V = {(xi,Ii) : 1 < i < m} is a tagged partition of a closed interval I and / : I —> R, the Riemann sum of / with respect to V is defined to be S(f, T>) = YlTLi f(xi)v(Ii)Here we are assuming, as was the case in R*, that all functions vanish at infinite points and 0 • oo — 0. Definition 3. Let I be a closed interval in R*n and f : I —> K. Then f is (gauge) integrable over I if there exists A G K such that for every e > 0 there is a gauge 7 on I such that \S(f, V) — A\ < e whenever V « 7. From Corollary 2, Definition 3 is meaningful and the uniqueness of A follows as in Theorem 1.4. A is called the integral of / over I and is denoted by /
/
/
=
/ / / ( * ) < * * •
The basic properties of the integral such as linearity, additivity, positivity, etc. carry forward to R*™ as before; we do not repeat their statements or proofs. There is one difference between the integral in R*n and R* that should be noted. A tag in a tagged partition can be the tag for 2™ subintervals for intervals in R*n; this can lead to some slightly different estimates for Riemann sums. We give a simple example illustrating the integral in R 2 . If / is an interval in R*™, its closure, written / , is the product of the closures of the component intervals defining / .
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Introduction to Gauge Integrals
Example 4. Let I be a bounded interval in W1. Then Ci is integrable over I with JRn Ci = v(I). We sketch the proof for n = 2. Let e > 0. Pick a closed interval F C I and an open interval G D I such that v(G) — v(F) < e. Define a gauge on R2 by setting 7(2;) = G when x £ I and choosing 7(2:) such that 7(2;) D F — 0 when x £ I. Suppose V = {(xi,Ii) : 1 < i < m} is a 7fine tagged partition of R*2. We may assume Xj G I for i = 1 , . . . ,p and x\ <£ I for i = p + 1 , . . . , m. Then S(d,V) = £ L i v(It). For i = 1 , . . . ,p,h C G so £ ? = 1 v(Ii) < «(G) and, therefore, S(Ci,T>) < v(I) + e. Suppose x e F. Then x e It for some i and /i c 7(a;*) implies 2; e 7(2^) D F . But, if i > p, 7(2;*) D F = 0 so 1 < i < P Therefore, U L i ^ D F a n d Ef=i «(7») > KF) a n d > therefore, S(Ci,T>) > v(I) — e. Hence, \S(Ci,V) — v(I)\ < e as desired. [Here we have tacitly used the "intuitively obvious" fact that the volume function is finitely additive over intervals. This is surprisingly cumbersome to prove in dimensions larger than 1. See for example [MS2] IV.2.1.] It follows from Example 4 and linearity of the integral that if (p is a step function, a linear combination of characteristic functions of bounded intervals, then tp is integrable over M*ra and its integral is given in a natural way. We next give a further example which shows that the gauge (double) integral in higher dimensions also admits conditionally convergent integrals (Exercise 6). Example 5. Let J2'kLiak Set Kj = \\j2\
be a convergent real series and I = [0,1] x [0,1]. l/2j1},
Lj = Kj x Kj and Jj = L2j .
Then {Jj} is a pairwise disjoint sequence of closed subintervals of I with v{Jj) = l/2 4 ; '. Let 0 < e < 1. Pick a pairwise disjoint sequence {Oj} of open subintervals of I such that Oj D Jj and v(Jj) +
e/2V>v(Oj)>v{Jj)
[draw a picture}]. Put f — Y^i o,j2^Cj.. We show that f is integrable over I with fjf = Yl'jLi aj Pick N such that \ Ei^fc ai\ < £ an^ \ak\ < e for k > N and let M — sup{afc : k € N}. We now define a gauge 7 on I. If x € Jj, set 7(x) = Oj\ if x £ Jj for any j and x ^ 0, pick 7(2;) to be an open interval containing x such that 0 ^ 7(2;) and 7(2;) n J, = 0 for every j ; for x = 0, let + l and 7(0) n JN = 0. 7 (0) = (5,5) x (8, d) where 7(0) DJkfork>N Now suppose that V = {(xi,Ii) : 0 < i < m} is •yfine with 0 € IQ; note Xo = 0. Let q be the largest integer such that Jq is not contained in the interior
Multiple Integrals and Fubini's Theorem
85
of IQ; note that q > N since IQ C 7(0) and 7(0) contains all Jk for k > N + 1. Note also that m
s(/,p) = £/(*>(/,) = £ E f{*iWi) i=l
k=l XjEJk
= E^ 24fc E v&) fc=l
Xj^Jk
since f(x) ^ 0 if and only if x G J^ and afc ^ 0 for some k. If k < q, then v(Jk) + e/25k = l/2 4fc + e/2 5fc > v(Ok) > v(Jk) = l/2 4fc and
v(Ok) > £
(1) e/2fc > 24fc J2 v(Jj)  1 > 0 /orfc< g. For k = q, J2
v(Ij) < v(Oq) < 1/2 4 ' + £ /2 5 « < 2/2 4 "
so
(2)
a g 2 4 « ^
E /(*>&•)
«(/,) < 2a, < 2e
since q > N. From (1) and (2), we /iaue 91
5(/,2?)^afc fc=i
< fc=l
+
ag24« ^
„(/,) + aQ +
E afc
k=q+l
91
< ^ fc=i
a fc  £ /2 fc + 2s + s + e < Ms + As = ( M + 4)e.
86 Hence, / 7 / =
Introduction to Gauge Integrals YlkLiak
We leave it to Exercise 6 to show that the function / is absolutely integrable if and only if the series ^Zafc i s absolutely convergent. We next establish the ndimensional analogue of Theorem 2.10 on the integrability of continuous functions. We first require a preliminary result. Lemma 6. Let K C R™ be compact and I a compact interval containing K. If f : K —» R is continuous and nonnegative on K and if f is extended to I by setting / = 0 on I\K, then there is a sequence of nonnegative step functions {sfc} vanishing outside I such that Sk(x) 4 / ( I ) V I G J, where the convergence is uniform for x € K. Proof: For each k G N divide I into a finite collection Tk of pairwise disjoint intervals such that the diameter of each subinterval in Tk is smaller than 1/k. Make the choice such that Tk+i is a refinement of Tk, that is, each subinterval in Tk+i is contained in some interval of TkConstruct Sk from Tk as follows: for x G / there is a unique J G Tk\ set Sk(x) = sup{/(a;) : x € J } and Sk{x) = 0 for x ^ J. Note Sk{x) I since Tk+i is a refinement of Tk • We claim that Sk{x) I f(x): First, suppose that x € I\K. Then, there is an open interval Ix containing x and such that Ix C\ K = 0. Any interval of sufficiently small diameter containing x will be contained in Ix. Since / is zero outside K, this implies that Sk(x) = 0 for large k so Sk(x) J. f(x). For x G K, let £ > 0. There is an open interval Ix containing x such that \f{x) — f(y)\ < e for y G K C\ Ix. Thus, 0 < f(x) and f(y) < f(x) + e for all y £ Ix since / is zero outside K. For large k, the interval of Tk which contains x will be a subset of Ix so Sk(x) < f(x) + e for large k. Since f{x) < Sk(x) for all x, it follows that Sk(x) I f(x). The uniform convergence on K follows from the uniform continuity of / on K, that is, the estimate Sk(x) < f(x) + e holds uniformly for x G K and large k. Note that when the function / is extended from K to / , the extended function will not, in general, be continuous on J or even K. If / is not nonnegative, we may decompose f = f+ — f~ and apply Lemma 6 to both / + and / ~ . We then obtain a sequence of step functions vanishing outside J, {sk}, which converges pointwise to / on / and Sk t f uniformly on K; however, the convergence is now not monotone.
87
Multiple Integrals and Fubini's Theorem
Theorem 7. Let K C Rn be compact and f : K ¥ R continuous on K. If f is extended to Wn by setting / = 0 outside of K, then f is integrable over R*n. Proof: By considering / = / + — / ~ , we may assume that / is nonnegative. Choose a compact interval I containing K and let {s^} be the sequence of step functions converging to / given in Lemma 6. Each Sk vanishes outside / so each is integrable. Since s/t 4 / and fjSk > 0 for every k, the MCT implies that / is integrable over R*n. Theorem 7 gives a reasonably broad class of integrable functions which are sufficient for many of the applications encountered in introductory real analysis courses. FUBINI'S THEOREM: J U N I O R G R A D E In this section we establish a version of Fubini's Theorem which is adequate for most applications encountered in an introductory real analysis course. Again we establish the results in R*2, but the methods work equally well in R*n. Let J,K be closed intervals in R*, I = J x K and / : /  > • R. The usual method for evaluating (double) integrals over I, Jt / , is to evaluate an iterated integral, fK fj f(x,y)dxdy. A theorem which asserts the equality of a double integral and an iterated integral is often referred to as a "Fubini Theorem". For a Fubini Theorem to hold we must establish the existence of the integral Jj / ( x , y)dx for each y and then show the function y —»• Jj f(x, y)dx is integrable over K with integral equal to Jt f. Our next result gives conditions sufficient for this to hold. Theorem 8. Suppose J and K are bounded closed intervals and f : I —> M. is bounded. Suppose there exists a sequence of step functions {sfc} such that Sk —> f pointwise on I. Then (i) for each y 6 K,f(,y) is integrable over J, (ii) the function F(y) = fjf(x,y)dx is integrable over K, and (hi) the function f is integrable over I with
f= JI
/
f(x,y)dxdy.
JK J J
Proof: First, suppose that / is a characteristic function of an interval, / = CAXB,A and B intervals in R. For (i) and (ii) we have fjf(,y) = fj CACB{y) = CB(y)e(A n J ) and /
[ f(;y)dy=
JK JJ
[ CB(y)£(A n J)dy = i(B n K)£( A n J ) . JK
88
Introduction to Gauge Integrals
Since / J / = u ( 7 r i i 4 x B ) = £(B n K)£(A n J ) , (hi) follows. The validity of the theorem for step functions now follows immediately from the linearity of the integral. Now consider the general case. Let M > 0 be such that \f(z)\ < M for z e I. For z € I, set M sk(z) = < sk(z) M
sk(z) > M M < sk(z) < M sk(z) <
M.
Then {sk} is a sequence of step functions which converges pointwise to / on I and is uniformly bounded by M. By the first part of the proof, sk(,y) is integrable over J for every y G K and {sk(,?/)} converges pointwise to f(,y) on J . By the Bounded Convergence Theorem (Exercise 5.6), f(,y) is integrable over J with lim JjSk(,y) — Jjf(^y)Also, by the first part, each function y —>• fjSk(,y) is integrable over K, and we have just shown that this sequence of functions converges pointwise to the function y —> fj f(,y). Since \Jjf(,y)\ < M£(J), the Bounded Convergence Theorem implies that V *• fj /(•> V) is integrable over K with lim JK fj sk(, y)dy = JK fj /(•, y)dy. Since {sk} converges to / pointwise on I and the sequence is uniformly bounded by M, another application of the Bounded Convergence Theorem and the first part of the proof give lim / sk = lim / Ji
/ sk{x,y)dxdy
JKJJ
=
f = JI
/
f(,y)dy,
JKJJ
and the result is established. Of course, by symmetry, Theorem 8 is also applicable to the other iterated integral, and, in particular, implies that the 2 iterated integrals JK J, f(x,y)dxdy and fjfKf(x,y)dydx are equal. This equality does not hold in general [see Exercise 2]. It is also the case that both iterated integrals may exist and are equal but the function may fail to be integrable [Exercise 3]. Note also that Lemma 6 gives a sufficient condition for the hypothesis in Theorem 8 to be satisfied. We next establish a result which removes the boundedness assumptions from Theorem 8. The result also establishes a sufficient condition for integrability in terms of an iterated integral; such results are sometimes referred to as a "Tonelli Theorem".
89
Multiple Integrals and Fubini's Theorem
Theorem 9. Let g : / »• R be such that \f(z)\ < g(z) for all z £ I and the iterated integral JK Jj g(x,y)dxdy exists. Assume there exists a sequence of step functions, {sk}, such that Sk —>• / pointwise on I. Then (i) for each y e K, Jj f(x, y)dx exists, (ii) the function F(y) = J, f(x,y)dx is integrable over K, and (iii) / is integrable over I with
f= JI
/ f(x> V)dxdy. JK J J
Proof: If/ satisfies the hypothesis of the theorem, then so do the functions / + and / ~ so we may assume that / is nonnegative. For each fc set Ik = {(x,y) : \x\ < fc, \y\ < fc} C\ I and define fk by fk = f A (kCik); that is, fk is altered by setting fk = 0 outside Ik and truncating / at k [draw a picture]. Each fk satisfies the hypothesis of Theorem 8 over the interval Ik If Jfc = J n [—k,k] and Kk = K n [—k,k], then Theorem 8 gives Jr fk = JK Jj fk(x,y)dxdy. Since fk vanishes outside Ik, we have / / A = fj IK /fc(x> v)dxdVThe sequence {/fc} increases pointwise to / on / so both sequences I / A I and
/ fk{x,y)dx
\ ,y e K,
are increasing with J{ fk = Jj JK fk(x,y)dxdy < Jj JK g(x,y)dxdy and Jj fk(x,y)dx < Jj g(x,y)dx for y G K. The MCT implies that / is integrable over / with lim Jj fk = Jj f a n d that f(,y) is integrable over J with \imjjfk(,y) = Jjf(,y). Also, the sequence {Jj fk(,y)} is increasing with JK Jj fk(x,y)dxdy < JK Jj g(x,y)dxdy so another application of MCT gives lim /fir Ijfk(x,y)dxdy = JK Jj f(x,y)dxdy. Hence, Jj f = JK Jj f(x,y)dxdy as desired. Remark 10. If f is a continuous function on a closed unbounded interval I, then the hypothesis in Theorem 9 is satisfied. For let {Kj} be a sequence of compact subintervals of I such that Kj C Kj+i and I = \JCXL1 Kj. First assume that f is nonnegative. By Lemma 6 for every j there is a step function Sj vanishing outside Kj such that \SJ(Z) — f(z)\ < 1/j for z € Kj. Then Sj —> f pointwise on I. For general f we can decompose f = f+ — f~ and apply the observation above to both f+ and f~.
90
Introduction to Gauge Integrals
In applying Theorem 9 a good candidate for the function g is /. In particular, if fKfj \f(x,y)\dxdy exists, then Theorem 9 implies that both / and  /  are integrable over I and the conclusion of Theorem 9 holds for both / and /. Exercise 2 shows that it is important that the function g in Theorem 9 be nonnegative. As an application of Theorem 9 we show the existence of the convolution product of 2 continuous, absolutely integrable functions. Let / , g : R —»• R be continuous and absolutely integrable over R. The convolution product of / and g is defined to be / * 9{x) =
f(x
y)g(y)dy
JR
when the integral exists. We show f * g exists for all x € R. Indeed, we have /
/ \f(x  y)\\g(y)\dxdy = f \g(y)\ [
JR JR
JR
\f(x)\dxdy
JR
= f\g\ [\f\, JR
JR
and by Theorem 9 the function (x,y) —> f(x — y)g(y) is absolutely integrable over R 2 , x
f(x~
~*
y)g{y)dy = / * g{x)
JR.
exists for all x and f*g is integrable over R. Moreover, the computation above shows
Im\f * 9\ < Ju\f\ JR\g\
F U B I N I THEOREM*: G E N E R A L CASE In this section we establish a very general form of the Fubini Theorem for the gauge integral which does not require the restrictive assumptions in Theorems 8 and 9. The proof does, however, use the property of null sets and integrable functions established in Corollary 6.6. We first establish a lemma required for the existence of an iterated integral. Lemma 1 1 . Let f : I —> R be integrable over I and let N = {y e K : f(,y) Then N is null.
is not integrable over J} .
Multiple Integrals and Fubini's Theorem
91
Proof: y G N if and only if there exists s > 0 such that for every gauge 7 J on J there exist 7jfine tagged partitions V, Q of J such that
(*)
S(f(;y),P)S(f(.,y),Q)>e.
For each z G N let Ni be the set of all y e N which satisfy condition (*) with e = 1/i. Then JV* C N+i and AT = U^=i N so it suffices to show that each TV; is null (Exercise 1.9). Fix i and let e > 0. To show Ni is null it suffices to show there exists a gauge 7x on K such that S(CjVi,72.) < £ whenever 72. is a 7i<fine tagged partition of J, i.e., JK CMI — 0 [6.6]. There exists a gauge 7 on / such that (**)
\S(f,V)S(f,S)\<e/i
whenever V and £ are 7fine tagged partitions of I. If y £ Ni, "/y(x) = Pil{xi y) defines a gauge 7 y on J. By definition of N there exist 2 7j,fine tagged partitions Vy and Qy of J such that
(***)
S(f(;y),Vv)S(f(;y),Qy)>l/i.
It y £ K \ Ni, "fy(x) = Piry(x,y) still defines a gauge on J so pick a 7yfine tagged partition Vy of J and set Vy = Qy [note S(f(,y), Vy)S(f(,y), Qy) = 0 for y e K\Ni}. If Vy = {(pi P?) :l<j
7 2 (y) = [ f 
/ Qy
\
P*I(PIV)
Then 72 is a gauge on K. Suppose that 72 = {(yk,Kk) K. By construction T> = {((Pf, !/0. Pf
n
f ^7(9^,y)
: 1 < k < m} is a 72fine tagged partition of
xKk):l<j
92
Introduction to Gauge Integrals
and £ = {((#* ,J/fe),Qf x Jffe) : 1 < j < qyk,l<
k<m}
are both 7fine. Hence, by (**) and (***), e/i>\S(f,V)S(f,£)\ 1yk
fc=l
£ ; £(Kk){S(f(; yk), Vyk)  S(f(, yk), Qyk)} fc=l VkZNi
>l/i
^2 Z(Kk) = l/iY,CNMZ{Kk) k=i
=
lliS(CNt,K)
fe=l
Vk^Ni
so S(CjVi, 7?.) < £ as desired. Exercise 22 shows that the conclusion of Lemma 11 is best possible in some sense. Let the notation be as in Lemma 11 and set F(y) = Jj f(x, y)dx for y $_ N'. If we are to obtain a Fubini Theorem in the form of Theorem 8 or 9 we must extend the function F to all of K in such a way that the extended function is integrable over K with integral equal to Jr f. Since N is a null set, if such an integrable extension exists it is independent of the way in which the extension is defined on K (Exercise 2.22). In the interest of simplicity, one usually agrees to define F(y) = 0 for y € N. The resulting function, F, is then shown to be integrable over K with integral equal to / 7 / , and one writes
h= I
I F{y)dy= ( f f(x,y)dxdy. JK
JK J J
The following technical lemma is useful in the proof of the general Fubini Theorem. Lemma 12. If N CM. is null, then N x l is null (in M.2).
93
Multiple Integrals and Fubini's Theorem
Proof: It suffices to show that Nk = N x (k, k) is null for every k G N since N x R = U{7Vfc :fc€ iV} (Exercise 1.9). Fix A; and let e > 0. There exists a sequence of open intervals
{J,} in K such that N C U°li ^ and E ^ i ^ ( ^ ) < £•
Then
{*»} = ijj
(/c,fc)} is a sequence of open intervals in R 2 such that Nk C \JJL\ h S^=i u (^j) < e ^ Hence, iVfc is null.
x anc
^
We now state and prove a general form of Fubini's Theorem for the gauge integral; the proof uses the construction of compound tagged partitions given in Lemma 1. Theorem 13 (Fubini). Let f : I —>• R be integrable over I and F(y) = Jj f(x, y)dx, for y e K, be defined as above. Then F is integrable over K with
f f= [ F(y)dy= [ f f(x,y)dxdy. i
JK
JK JJ
Proof: From Lemmas 11 and 12 there is a function g : I —> R such that f = g a.e. in J, f(,y) = g(,y) whenever f(,y) is integrable over J, g(, y) is integrable over J for every y G K and F{y) = Jj g(x, y)dx for y G K (Exercise 2.22). Thus, we may assume that /(•, y) is integrable for every y G K. Let £ > 0 and choose a gauge 7 on / such that \S(f, V) — Jj f\ < e whenever V is a 7fine tagged partition of / . By Lemma 5.3 pick a function ip : K —> (0, 00) which is integrable on K and a gauge 7^ on K such that 0 < S(ip, V) < 1 whenever V is a 7^fine tagged partition of K. For each y G K pick a 7i(, y)fine tagged partition Vy of J such that S(f(,y),Vy)[integrability of f(,y)\.
/ f(x,y)dx
< e
Put
h(y) = J
f(x,y)dxS{f(;y),Vy).
Define 7 2 (y) = <^{l(x, y) : (x,P) G Vy} as in Lemma 1, and set 72 = 7 V f]j'2. We show that \S(F,K.) — / 7 /  < e whenever K is a 72fine tagged partition of K, and this will establish the result. Let V be the 7fine compound tagged
94
Introduction to Gauge Integrals
partition generated by K. and {Vy : y £ K} as in Lemma 1. Then 5(F,/C).
<\S(F,fC)S(f,V)\
<
J2
+
S(f,V)J.
{F(y)£(K)  S(f(,y),Vy)£(K)}
+ £
(y,K)€ic
(y,K)eK
eS(ip,K.)+e
(y,K)eic
< 2e.
Theorem 13 has an interesting application to the computation of Lebesgue measure in E 2 . [It is straightforward to extend the definition and terminology of (Lebesgue) integrable and measurable subsets of 2 as was done in Chapter 6.] As in Chapter 6 let A be Lebesgue measure on , and let A2 be Lebesgue measure on K2. If E C R 2 and t/€M, the ysection of E at y is defined to be E» = Note CE{x,y) obtain:
CEV(X)
{x:(x,y)eE}.
for i , j 6 1 . From Lemma 11 and Theorem 13 we
Corollary 14. Let B c R 2 be integrable. Then Ey is integrable over R for all y, except those in a null set. With the same notation as in Theorem 13 the function y —» X(EV) is integrable over E with \2{E) = JRX(Ey)dy. Notes/Remarks The Fubini Theorem along with the convergence theorems (MCT, DCT) are the great successes of the Lebesgue integration theory. The versions of the Fubini/Tonelli Theorems given in Theorems 8 and 9 although applicable to a much smaller class of functions than in the Lebesgue case, are sufficient for an introductory real analysis course since most functions encountered are continuous (or "nearly continuous") (see [Swl] 3.9.3, 3.9.4 for the Lebesgue case). The general form of the Fubini Theorem for the gauge integral given in Theorem 13 illustrates the generality of the gauge integral since the theorem applies to conditionally convergent integrals (Exercises 6 and 25) unlike the absolutely convergent Lebesgue integral.
Multiple Integrals and Fubini's Theorem
95
The Fubini Theorem gives a method for computing a "double integral" as an iterated integral assuming the function is integrable. A Tonelli Theorem is a theorem which establishes the integrability of a function from assumptions about the existence of iterated integrals; Theorem 9 is such a "Tonelli result." We establish a more general form of the Tonelli Theorem for measurable functions. Theorem 15* (Tonelli). Let f : I > R be measurable. If JK Jj \f(x,y)\dxdy exists and is finite, then f is integrable over I with Jj f = JK Jj f(xi y)dxdy. Proof: By decomposing / = f+—f~, we may assume that / is nonnegative. For each k set h = {(x, y) : \x\ < k, \y\ < k} (11. There exists a sequence {tfk} of nonnegative, simple functions such that fk t / • ([Swl] 3.1.1.2.) Set fk = Cikifk Then fk is nonnegative, simple, integrable and fk t /• By Corollary 14 and the linearity of the integral, the conclusion of Fubini's Theorem 8 holds for fk The proof of the theorem now follows as the proof of Theorem 9. For further examples and counterexamples relative to Fubini's Theorem, see [Swl] §3.9. As noted earlier it is straightforward to extend the development of Lebesgue measure in R that was carried out in Chapter 6 to R n . For more details on Lebesgue measure in R n , see [Swl], [N2]. The convolution product of 2 functions as defined following Remark 10 is applicable to a much wider class of functions than considered there. For details on the existence of convolution products, see [Swl] 3.11.1, 6.1.23, [HS]; for applications of the convolution product, see [Swl] §3.11, [DS] §18, [HS].
Exercises 1. Let J, K be compact intervals in R and I = J x K. Let 7 be a gauge on / . Use the following argument to show that there exists a 7fine tagged partition of / . [Suppose the result false and divide / into 4 "equal" subintervals. The result must be false for one of the subintervals. Now continue to divide. In R n the same construction works by dividing into 2 n "equal" subintervals.] 2. Let f(x,y) = e" x «  2 e  2 ^ . Show that J* J™ f(x,y)dxdy > 0 and /i°°/o
f{x,y)dydx<0.
96
Introduction to Gauge Integrals
3. Let f(x, y) = xy/(x2 Show that rl
/
+ y2)2 when (x, y) ^ (0,0) and / ( 0 , 0) = 0. /•!
/
rl
f(x,y)dxdy
=
rl
/
f(x,y)dydx
=0
but / is not integrable over [—1,1] x [—1,1]. [Hint: Show that / is not integrable over [0,1] x [0,1] by showing /
/
f(x,y)dxdy
=  1 / 4 I n 2 + l/41n (1 + 1/n 2 )  1/2 In 1/n
J 1/n JO
JEn
where En = {(x, y) : 0 < x < 1,1/n < y < 1}, and {JB / } is unbounded.] 4. Let J, if be closed intervals in R* and I = J xK. Let J7\ /C be tagged partitions of J, K, respectively. Show I = {((x, y), A x B) : (x, A) € J, (y, B) € /C} is a tagged partition of I. Give an example of a tagged partition of I which is not of this form. Let 7 J ( 7 X ) be a gauge on J{K). Show that l{x,y) = ){x) x 7(3/) is a gauge on 7, and if J « jj, K. « JK, then X < < 7. Give an example of a gauge on I which is not of this form. 5. Use only the definition of the integral to show that C^i] is integrable over R2. 6. Show that the function / in Example 5 is absolutely integrable if and only if the series Ylak ls absolutely convergent. 7. Extend the definition of a null set to M2(R'1). If N is a null subset of R 2 , use only the definition of the integral to show that CN is integrable over R2. 8. Suppose / satisfies the hypothesis of Lemma 6. Show that there exists a sequence of nonnegative step functions {sfe} such that Sk{x) f f(x) for x £ I, where the convergence is uniform on K. 9. Let a, (3 : [a,b]—tR be continuous with a(t) < (3{t) for a < t < b. Let E = {{x, y) :a<x
a{x)
(3(x)} .
Suppose / : R 2 —>• R is continuous. Show that / is integrable over E and b
kf = I aGP^v)dydx.
10. Find the integral of the function f(x, y) = x3y2 over the region in the first quadrant inside the ellipse 9x 2 + 25y2 = 2 5  9 and outside the circle a;2 + y 2 = 1.
Multiple Integrals and Fubini's Theorem
97
11. Define / : [0,1] x [0,1] > R by f{x,y) = 1 if x,y € Q and f{x,y) = 0 otherwise. Verify Theorem 13 for / . Do any of the integrals exist as Riemann integrals? What can you conclude? 12. Show that the function f(x, y) = e~x siny is integrable over [0, oo] x [0, 2it] and calculate the integral. 13. Is the function f(x, y) = l/(x + y) integrable over [0,1] x [0,1]? What about f(x,y) = l/(x2 + y2)? 14. Give an example of a function / such that JQ JQ f(x,y)dxdy exists but Jo1 lo f(x,y)dydx does not. 15. Let / : J > R,g : K > R and define h : I = J x K ^ R by h{x,y) = f(x)g(y). Assume both / and g are pointwise limits of step functions and both are absolutely integrable. Show that h is absolutely integrable with
fih=fjfSi(9
16. Evaluate JQ L ex dxdy. [Hint: Change the order of integration.] 17. Let En = [0,n] x [0,n]. Show that /0°° s^dx = n/2 by evaluating l i m / ^ e~xy sin x in 2 different ways. 18. Let / ( * , y) = ye^+^v2. Show /0°° /0°° f(x, y)dxdy = /0°° /0°° f(x, y)dydx and use this to show that JQ e~x dx = \pHj1. 19. Let f(x, y) = e~xy/(l + x2). Show that / is integrable over [0, oo] x [0,1]. 20. Let f(x,y) = e~xy sinxy. Show that / is integrable over [0,oo] x [1,2]. 21. Extend the first part of Corollary 14 to measurable subsets. 22. Assume the existence of a subset P C [0,1] which is not Lebesgue measurable ([Swl] 1.3.1). Let Z C [0,1] be null and set E = Z x P. Show that CE is integrable over [0,1] x [0,1] but CE(,V) is not integrable over [0,1] for all y. [See Lemma 11.] 23. Let P be as in Exercise 22. Show that [0,1] x P is not integrable (measurable) in R 2 . 24. Analogous to Exercise 1.12 show that when n = 2 it can be assumed in Definition 8.3 that all tags are vertices of the subintervals which they tag. What about general n? 25. Verify the conclusion of the Fubini Theorem 13 for the function in Example 5. 26. (Improper Integral) Let Ik = [0,l/2 fe ], k = 0 , 1 , . . . , and Jfc = Ik x Ik. Suppose that / : Jo —• R is integrable over Jo \ Jfc for k > 1 and lim Jj , j f = L exists. Show that / is integrable over Jo with fjf = L. (Compare with Theorem 3.4.) 27. Apply Exercises 26 to Example 5. 28. (Linear Change of Variable) Let / : R n —> R be absolutely integrable and
Introduction to Gauge Integrals
98
A : R™ —»• R n be linear. Show that / o A is absolutely integrable with JK„ f o A\ det A = JK„ / . [Hint: A is a product of transformations of the form (xi,... ,xn) »• ( t e i , . . . ,xn),(xi,... ,xn) > (xi + a ^ , ^ , . . . ^ n ) , ^ ^ 1 j . . . , X{ j . . . , iCj j . . . 5 *^n/
^ V*^l 5 • • • j ^ j i • • • j *^i) • • • j •En)•
U s e ITUDini S
Theorem.] 29. Show that absolute integrabiHty is important in Exercise 27. [Hint: Let h = [js^T.pJforfc £ Nand Jk = Ikxlk. LetA fc = {(*i,t 2 ) e Jk : t2 >U}, Bk = {(^1,^2) £ Jfe : <2 < *i} Construct a function / which is 0 outside (JfcLo ^k a n d i s integrable over each Jk with fA f = — JB f = 1/k. Use Exercise 26 to show / is integrable over R 2 . Define A : R 2 >• R 2 by A(ti,t2) = i ^ ^ , h^z) [rotation through TT/4]. Show that / o A is not integrable.]
Chapter 9 T h e M c S h a n e Integral 9.1
Definition and Basic Properties
In this chapter we describe another gaugetype integral due to E.J. McShane, which we will call the McShane integral. McShane alters the definition of the gauge or HenstockKurzweil integral by not requiring that the tag associated with a subinterval in a tagged partition belong to the associated subinterval, that is, the tag is free to belong anywhere in the domain of the function being considered. This requirement is somewhat nonintuitive; it does not seem to be known what led McShane to this definition, but it does have a profound effect on the subsequent integration theory which is developed. In fact, the McShane integral is equivalent to the Lebesgue integral in Euclidean spaces [Appendix 4]. We will begin by defining the McShane integral over closed subintervals in R*. Let 7 be a closed interval (bounded or unbounded) in M*. Again if / is a realvalued function, it is always assumed that / is extended to all of R* with /(±oo) = 0. A partition of 7 is a finite collection of closed, nonoverlapping subintervals of 7 whose union is 7. A free tagged partition of 7 is a finite collection of pairs V = {(U,Ii) : 1 < i < m} such that {7, : 1 < i < m} is a partition of 7 and U € 7; we again refer to the point i, as being the tag associated with the subinterval 7;. We refer to such tagged partitions as free tagged partitions to distinguish them from the tagged partitions used in the gauge or HenstockKurzweil integral. If / : 7 —• R and V = {(ti,Ii) : 1 < i < m} is a free tagged partition of 7, we define the Riemann sum of / with respect to V as before to be S(f, V) = YT=i f(U)£(Ii)If 7 is a gauge on 7, we say that a free tagged partition V = {(ti,Ii) : 1 < i < m} is 'yfine if 7^ C "fiU) for 1< i < m, and as before we write T> < < 7 when V is 7fine. 99
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Introduction to Gauge Integrals
Definition 1. A function f : I —> R is McShane integrable over I if there exists A € R such that for every e > 0 there is a gauge 7 on I such that \S(f, V) — A\ < e whenever T> is a jfine free tagged partition of I. From Theorem 1.3 or Theorem 4.1 it follows that the definition of the McShane integral makes sense [i.e., any gauge has a 7fine (free) tagged partition], and as was the case for the gauge integral, the number A is unique and is called the McShane integral of / over / . In this chapter we will only consider the McShane integral so we will write A = Jt f, etc., as before. In situations where other integrals are considered we will adopt notation which will make it clear which integral is being considered [e.g., Chapter 10 and Appendix 4]. Henceforth, we will refer to the HenstockKurzweil integral as the gauge integral to distinguish it from the McShane integral. Since there are more free tagged partitions than (gauge) tagged partitions, it is clear that any McShane integrable function is gauge integrable and the two integrals are equal. We will see below there are functions which are gauge integrable but not McShane integrable so the use of free tagged partitions does lead to a less general integration theory; in particular, we will see that the use of free tagged partitions implies that the McShane integral is an "absolute integral" [Theorem 5] in contrast to the "conditional" gauge integral (recall Examples 2.12 and 4.8). There are a couple of observations concerning the use of free tagged partitions which should be made. First, one can no longer assume that each tag only appears at most twice or that the tag appears only as an endpoint (Exercise 1.12) as was the case for (gauge) tagged partitions (see Exercise 1). Although the tag is not required to belong to its associated subinterval in a free tagged partition, a gauge 7 does somehow require the tag to be "near" its associated subinterval. The proofs of many of the basic properties of the McShane integral are essentially identical to the proofs of the corresponding properties of the gauge integral. We will list these basic properties in Theorem 2 below; the reader is invited to check the corresponding proofs for the gauge integral. Theorem 2. Let f,g:I—>M.
be McShane integrable over I.
(i) / + 9 *s McShane integrable over I with fI(f + g) = fIf + Jj 9(ii) For every a 6 R a / is McShane integrable over I with J, af — a J, f. (iii) If f> 9 on I, then Jj f > Jj 9For the proof, see Theorem 2.1.
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The McShane Integral Theorem 3. Let f : I > R.
(i) (Cauchy Criterion) f is McShane integrable over I if and only if for every e > 0 there is a gauge 7 on I such that \S(f, V\) — S(f, X>2) < e whenever T)\ and T>2 are jfine free tagged partitions of I. (ii) If f is McShane integrable over I and J is a closed subinterval of I, then f is McShane integrable over J. (iii) If I = [a, b] and  0 0 0 there exist McShane integrable functions g,h : I » R with g < f < h and L(h — g) < e, then f is McShane integrable over I. (v) If I is a closed, bounded interval and f is continuous on I, then f is McShane integrable over I. (vi) (Henstock's Lemma). Suppose f is McShane integrable over I and for s > O7 is a gauge on I such that \S(f,V) — Jj f\ < £ whenever V is a jfine free tagged partition of I. If J = {(U, Ji) : 1 < i < m} is any jfine free partial tagged partition of I and if J — U»=i ^ii then S(f,J)~
f ^sandY* JJ
f{U)e(Ji)~1
/
/ < 2e.
JJi
[Free partial tagged partitions, Riemann sums for free partial tagged partitions and jfine free partial tagged partitions are defined as in Chapter 3.] For the proofs of (i)(vi), see, respectively, Theorems 2.6, 2.7, 2.4, Lemma 2.9, Theorem 2.10 and Lemma 3.1. We now come to the basic difference in the McShane and gauge integrals; the McShane integral is an "absolute" integral. For the proof of this result we need a preliminary lemma; the reader should take note of the use of free tagged partitions in the proof. Lemma 4. Let f : I —> R be McShane integrable over I and let e > 0. Suppose 7 is a gauge on I such that for every 7/me free tagged partition T> of I, we have \S(f,V)Jj f\ < e. IfV = {(ti,Ii) :l
n
Proof: Put T = {It n J, = Ki3; : 1 < i < m, 1 < j < n,I° D J° ^ 0}. Define 2 7fine free tagged partitions V and £' of I as follows: if /(£,) > f(sj),
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Introduction to Gauge Integrals
put Uj = U and Sij = SJ; if f(ti) < f(sj), put Uj = Sj and Sij = tj. Note f(Uj)  f(sij) = \f{U)  f{Sj)\. Now set V = {(U^Kij) : Ktj € T} and £' = {(sij,Kij) : Kij 6 .T7}. Note V and £' are 7fine so by hypothesis
\S(f,V')S(f,£')
E {/(*o)  /(*;)}«*) KiidT
i=l i = l
Theorem 5. If f : I —t R is McShane integrable over I, then \f\ is McShane integrable over I with  J 7 /  < / 7 /. Proof: We show that the Cauchy criterion is satisfied for /. Let e > 0. Let 7 be a gauge on I such that  5 ( / , V) — fjf\ < e whenever V is a 7fine free tagged partition of I. Let V = {(£j, Jj) : 1 < i < m} and £ = {(SJ, Jj) : 1 < j < n} be 7fine free tagged partitions of J. Then by Exercise 1 and Lemma 4. (*) \S(\f\,V)S(\f\,£)\
^ E E i/(**)  M O W * n J i) ^2£ so the Cauchy criterion is satisfied. Since / <  /  and — / < /, the last inequality follows directly from Theorem 2 (iii). Theorem 5 should be contrasted with Theorem 4.10 for the gauge integral; recall Examples 2.12, 4.8 and Exercises 3.9, 4.3. Corollary 6. Let f,g:I^R be McShane integrable over I. Then f A g and f' V g are McShane integrable over I. Proof: fVg=(f
+ g+\f
g\)/2 and / A g =  ( (  / ) V (3)).
This result should also be compared with Proposition 4.12; recall Examples 2.12 and 4.8.
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The McShane Integral
The proof of Theorem 5 actually allows us to establish a slightly more general result which has some interesting corollaries. A function g defined on an interval J satisfies a Lipschitz condition if there exists L > 0 such that \g(t)  g(s)\ < L\s — t\ for all s,t G J; the parameter L is called a Lipschitz constant for g. [See Exercise 6 for examples of functions satisfying Lipschitz conditions.] Theorem 7. Let f : I —> R be McShane integrable over I and let g : J —>• R satisfy a Lipschitz condition over the interval J. Assume J D {/(*) '• t £ I}. Then g o f is integrable over I. Proof: If we replace / with g o f in the inequality (*), we obtain \S(gof,V)S(gof,£)\
S(f',V)
f f
<e
Ja whenever V is a 71fine free tagged partition of [a,b]. Let 7 be the gauge in Lemma 9. Put 72 = 71 n 7 and let V be a 7fine (gauge) tagged partition of
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Introduction to Gauge Integrals
[a,b]. From Lemma 9, we have
If
(/(&) " /(a)) < ! f'~ S(f,V)
Ja
Ja
\S(f,V)(f(b)f(a))\<2e and the result follows. This version of the FTC should be contrasted with the FTC for the gauge integral (Theorem 1.5) where it is not necessary to assume the integrabihty of the derivative. Example 2.12 along with Theorem 5 shows that the assumption of the McShane integrabihty of the derivative in Theorem 10 cannot be dropped. For part 2 of the FTC we have the analogue of Theorem 2.8. Theorem 11 (FTC; Part 2). Let f : [a, b] > R be McShane integrable over [a,b] and set F(i) = f f. If f is continuous at to £ [a,b], then F is differentiable at t0 with F'(to) = f(t0). See the proof of Theorem 2.8. We next consider an interesting approximation result which is also used later in a convergence theorem. Theorem 12. Let f : I —> R be McShane integrable over I and e > 0. There exists a McShane integrable step function g : I —> R such that fj \f — g\ < e. Proof: There exists a gauge 71 on I with 71 (t) bounded for every t e R such that \S(f,T>) — Jj f\ < e/3 whenever V is a 71fine free tagged partition of / . Fix such a partition V — {(£*, /,) : 1 < i < m). Define a step function g : I —> R by setting g — J^lLi f(U)Cii [note ±00 must be a tag for any unbounded interval and /(±oo) = 0]. Now \f — g\ is McShane integrable over / (Theorems 2 and 5) so there exists a gauge 72 on / such that \S(\f — g\,£) — ft \f — g\\ < e/3 whenever S is a 72fine free tagged partition of I. Set 7 = 71 n 72. In each subinterval Ik, let £k be a 7fine (gauge) tagged partition of h (Theorems 1.3 and 4.1). Set £ = Ufcli £fc a n d n ° t e that £ is a 7fine (gauge) tagged partition of/. Suppose that £ = {(sfc, Jk) '• 1 < k < n}. For each k, 1 < k < n, there is a unique jk such that Jfc C Ijk • Consider the pair (tjk, Jk)' we have s^ 6 4 C Ijk C 7i(i Jfc ) s o P = {(tjk,Jk) : I < k
n
fc=ii=i
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The McShane Integral But, £(Ji n Jfc) = 0 if i ^ k so
k=i
However, g(sk) = /(*jfc) since Sfc € Ijk so we have n
£ I/O*)  ff(sk)IW = S(\f g\,£)< 2£/3. fc=i
Since £ is also 72fine, we have
J\fg\<S(\fg\,£)+s/3<e.
9.2
Convergence Theorems
In this section we establish versions of the major convergence theorems (MCT, DCT) for the McShane integral. Due to the fact that the McShane integral is an absolute integral (Theorem 5), the conclusions of the convergence theorems for the McShane integral take on a stronger form than those for the gauge integral. Throughout this section let 7 be a closed subinterval of R*, and let fk, f '• I —¥ R. Some of the proofs or parts of the proofs for the convergence theorems are identical to the corresponding results for the gauge integral and will not be repeated, but references to the corresponding proofs will be given. We begin with an analogue of Theorem 4.10. T h e o r e m 13. Let f : (a, 00) = / —> M be McShane integrable over I and set F(t) = fa f,a < t < 00. Then Var(F : / ) = / 7 /. Proof: Let a = x0 < X\ < ... < xn = b be a partition. Then
j= l
j = lJ*il
J
I
soVar{F:I)<Sj\f\. For the reverse inequality, assume first that / is a step function / = XL = i ^CUj i where Aj = [a.j,bj] and the {A,} are nonoverlapping, closed
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Introduction to Gauge Integrals
subintervals with union J C I. Then
I
52\F(bj)F(aj)\ = 52 3= 1
3= 1
f
J a*
\tj\i(AJ) = Jj\f\<J\f\. 3=1
Therefore, Var(F : J) > fj \f\ and Var(F : I) = fx \f\ when / is a step function. If / is McShane integrable over 7, then for every k there is a step function gk such that J 7 \f — gk\ < 1/A; (Theorem 12). Put Gk(t) = J gk Then, by the first part of the proof, \Var(Gk : I)  Var(F : 7) < Var(Gk  F : I) < J \gk  f\ < \/k so Var(Gk : I) »• Var(F : 7). Since / , \gk\ >• / , \f\ and Var(Gk : I) = J 7 \gk\, we have Var(F : I) = Jt \f\. This theorem and its proof should be contrasted to Theorem 4.10. The result for arbitrary subintervals of R* can be similarly formulated and established. Analogous to Definition 5.1, we have Definition 14. Let fk:I^B.be McShane integrable over I for each k € N. The sequence {fk} is uniformly McShane integrable over I if for every e > 0 there is a gauge 7 on 7 such that \S(fk, V) — Jj fk\ < £ for all k and all jfine free tagged partitions T> of I. We next establish the analogue of Theorem 5.2 (7.8) for the McShane integral. Theorem 15. Let fk : I —>• R Suppose {fk} is uniformly McShane integrable over I and {fk} converges pointwise to the function f. Then f is McShane integrable over I and Jj \fk — f\ —> 0. Proof: As in the proof of Theorem 5.2, \imjrfk = L exists and / is McShane integrable with fj f = L. We do the remainder of the proof for the case when 7 = [a, 00]; the other cases use the analogue of Theorem 13 for other subintervals. Put gk = fk — f so gk —> 0 pointwise, and since / is McShane integrable, {gk} is uniformly McShane integrable. Let e > 0. There exists a gauge 7
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The McShane Integral
on J with 7(4) bounded for every f 6 R such that \S{gk,T>) — Jjgk\ < £ for all k whenever P is a 7fine free tagged partition of / . Fix such a partition V = {(ti, Ii) : 1 < i < m} and assume that £1 = 00, I\ = [b, 00]. Let b < a and let {Jj : 1 < j < n} be a partition of [a,a]. Then £ = {(U,Ii n Jj) : 1 < i < m, 1 < j < n) is 7fine so Henstock's Lemma (3(vi)) implies
f! / , EE lit,
lb
I
i=l 7=1
n
9k 
9k{U)t{Ii^Jj)
<£.
J
\ hnJj
Therefore, (#) n
9k
< £
££^(*<)^ J < nJ i) i = l .7 = 1
£ + $3<7fc(*i)^n[a,a])
< £ + max
\gk(ti)\l([a,b]).
2
The last term on the right hand side of (#) can be made < e for large k since <7fc —>• 0 pointwise. Hence, ( # ) implies that Var(Gfc : [a, a]) < 2£ for large k. Since a > b is arbitrary, Var(Gfc : J) < 2e for large k. By Theorem 13, / / Iflfcl < 2e for large fc as desired. Theorem 15 should be compared to Theorem 5.2 (7.8) for the gauge integral. In particular, the much stronger conclusion should be noted; indeed, fj\fk — f\ —> 0 implies f,(fk — f) —> 0 uniformly over all subintervals J ot I (Theorem 7.8). We now give an analogue of Theorem 5.4 for the McShane integral. Again, due to the fact that the McShane integral is an absolute integral and Theorem 15 allows a stronger conclusion than its counterpart for the gauge integral [Theorem 5.2], the statement and conclusion are somewhat different. Theorem 16. Let fk : I —*• K be McShane integrable over I for every k and suppose that f — Y^kLi fk pointwise on I with Y^k=i Si \M ^ °° V s n = Y!k=ifk, then (i) {sn} is uniformly McShane integrable over J, (ii) / is McShane integrable over I with J 7 / = Sfcli J/ fk and
(hi) JI\sn~f\ =
JI\EZn+lfk\^0.
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Introduction to Gauge Integrals
The proof of this result is essentially the same as the proof of Theorem 5.4 so we do not repeat the argument. However, a few remarks are in order. Since the McShane integral is an absolute integral, we do not need to assume that the {fk} are nonnegative as in Theorem 5.4, and we then replace the assumption that Y^kLi Ii fk < °° w * t n Efcli / / l/fcl < °° The proof of Theorem 16 (i) then proceeds as the proof of Theorem 5.4 (i) except that in the estimation of the term Ti the absolute integrability of fk is used. Conclusions (ii) and (iii) of Theorem 16 now follow from part (i) and Theorem 15. Since
E / / *  / / = E //* ^ E /iM.
fc=l '
fc=n+l
fc = 71 + l "
the conclusion in Theorem 16 (iii) is stronger than the conclusion in Theorem 5.4 (iii). As in Theorem 5.5 the MCT now follows readily from Theorem 16. Theorem 17 {MCT). Let fk : I >• K be McShane integrable over I and suppose that fk(t) T f(t) € K /or eiien/1 € I. If supk Si fk < °°> then (i) {/fc} is uniformly McShane integrable over 7, (ii) / is McShane integrable over I and
(iii) nmfIfk =
JIf(=fI(limfk)).
Proof: As before set /o = 0 and gk = fk — /fei for A: > 1. Then #fc > 0, E L i 5 f c = / «  > / pointwise and £feLi / 7 ^fe = lim„£fc=i //(/fc ~ /fci) = lim„ Jj fn = sup n fjfn < oo. Hence, Theorem 16 is applicable and gives the result. The DCT for the McShane integral can be derived exactly as in Chapter 5 for the gauge integral. We can, however, obtain a stronger conclusion in the DCT for the McShane integral. We repeat the essential steps. Definition 18. Let fk'.I*M.. The sequence {fk} is uniformly McShane Cauchy over I if for every e > 0 there exist a gauge 7 on I and an N such that if i,3 > N, then \S(fi,V) — S(fj,V)\ < e whenever V is a •yfine free tagged partition of I. Then exactly as in Proposition 5.7, we have Proposition 19. Let fk : I >• M. be McShane integrable over I. Then {fk} is uniformly McShane Cauchy over I if and only if {fk} is uniformly McShane integrable over I and lim L fk exists.
The McShane Integral
109
Theorem 20 (DCT). Let /&, R 6e McShane integrable over I with \fk\ < 9 on I. If fk —* f pointwise on I, then (i) {/fc} is uniformly McShane integrable over I, (ii) / is McShane integrable over I and
(iii) / 7 \fk  f\ )• 0. The proof of Theorem 20 (i) proceeds as the proof of Theorem 5.8 (i). Parts (ii) and (iii) then follow immediately from Theorem 15. A few remarks are in order. First, the domination assumption \fk\ < g implies \fk — fj\ < 2g as in Theorem 5.8; however, the domination assumption \fk — fj\ < 9 m Theorem 5.8 allows for conditionally convergent integrals in the case of the gauge integral, whereas the assumption /fc < g does not. Next, since \ Jj fk — / / / I < Jj l/fc — /> the conclusion (iii) in Theorem 20 implies conclusion (iii) in Theorem 5.8. The continuity and differentiation results for integrals containing parameters given in Theorems 5.10 and 5.12 can be obtained in exactly the same way for the McShane integral. We do not repeat the statements. We next establish a result for "improper" McShane integrals. Theorem 21. Let f : [a, b] —)• R and let a < bk < 6, 6fe f ° Suppose that f is McShane integrable over [a, bk] for every k. Then f is McShane integrable over [a, b] if and only ifsup{f k \f\ : k} < oo. In this case, l i m / k f — L exists and
Proof: If / is integrable over [a,b],J \f\ > Jak \f\ for every k. Conversely, assume sup{J k \f\ : k} < oo. First, consider the case when / > 0 and set fk = C[aMf. Then fk is integrable over [a,b},f*k f = f* fk and fk t /• The MCT implies that / is integrable over [a, b] with f*k f t / a 6 / . If / is integrable over [a, b], both / + and / ~ are integrable over [a,b] so the result just established can be applied to both / + and / ~ . Theorem 21 should be compared to Theorem 3.4 for the gauge integral. A comparison theorem for the McShane integral is given in Exercise 15. Finally, we conclude this section by establishing a result for the indefinite McShane integral. Definition 22. Let g : [a, b] —• R. Then g is absolutely continuous if for every e > 0 there exists a 5 > 0 such that if {[ak,bk] • 1 < k < n} is any finite family of nonoverlapping subintervals of [a,b] with X)fe=i(&fc — ak) < 5, then
ELib( 6 fc)  s K ) l < £ 
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Introduction to Gauge Integrals
For example, any function satisfying a Lipschitz condition on [a, b] is obviously absolutely continuous. [See Exercise 18 for the converse.] Proposition 23. Let f : [a,b] —» R be absolutely continuous. Then f is uniformly continuous and of bounded variation on [a, b]. Proof: / is obviously uniformly continuous. Let £ = 1 and S be as in Definition 22. Partition [a, b] by V = {a = x0 < xi < • • • < xn = b} with Xk — Xki < 6,1 < k < n. Then Var(f : [xki,Xk]) < 1 so Var(f : [a, b}) < n. The converse of Proposition 23 is false; the standard counterexample is the socalled Cantor function (see [Swl] 4.4.9). Theorem 24. Let f : [a,b] —>• R be McShane integrable over [a, b] and set F(t) = J f,a
Y,\FQ>i)n*i)\
i= l
Ja
<
i=l
/
Ja
i
i=\
Ja
'
"i
b
(\f\fk) +
kJ2(bica)<e.
%=\ Theorem 24 should be compared to Theorem 4.9 for the gauge integral.
9.3
Integrability of Products and Integration by Parts
In this section we consider the integrability of the product of two McShane integrable functions (see Exercise 24). As an application we derive an integration by parts formula for the McShane integral. The results of this section are not used later and may be skipped. Throughout this section let J be a closed subinterval of R* and let / , g : I —>• R. We begin with a preliminary result on products. Proposition 25. Let f be bounded and McShane integrable over I. Then fn is McShane integrable over I for every n £ N. Proof: The function h(t) — tn satisfies a Lipschitz condition on every bounded interval (Exercise 6). Hence, h o f — fn is McShane integrable over J by Theorem 7.
The McShane Integral
111
Corollary 26. Let f,g be bounded and McShane integrable over I. Then fg is McShane integrable over I. Proof: Since 2fg = (f+g)2—f2—g2,
the result follows from Proposition 25.
To remove the boundedness hypothesis in Corollary 26, we establish 2 lemmas. Lemma 27. Let f be nonnegative and McShane integrable over I. Then there exists an increasing sequence of nonnegative, bounded, McShane integrable functions {fk} such that fkt f on IProof: Set fk = fAk
and apply Corollary 8.
Lemma 28. Let f,g be nonnegative and McShane integrable over I. If there exists a McShane integrable function h : I —> R such that fg
Ibaf'9 + Jbaf9'.
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Introduction to Gauge Integrals
Theorem 30 should be compared with Corollary 2.3. Similarly, we can use Theorem 30 to derive a result on integration by substitution or change of variable. Theorem 31 (Integration by Substitution). Let f : [c, d] —> R be continuous and let g : [a, b] —>• [c, d] be differentiate on [a, b] with g' McShane integrable over [a, b]. Then (fog)g' is McShane integrable over [a, b] with J y ? / = ja(f° g)g'. Proof: Set F(u) = Qa) f for c < u < d. Then F' = f on [c, d] by Theorem 11. By Theorem 29 (/ o g)g' is integrable since / o g is continuous on [a, b]. But, (Fog)' = (fog)g' so Theorem 10 gives fa(fog)g' = Fog\ba = /»£> f. Theorem 31 should be compared to Theorem 1.6 for the gauge integral.
9.4
More General Convergence Theorems
In this section we give a more general form of the MCT which is very useful in many applications. We also use the generalization to establish a version of Fatou's Lemma and a more general form of the DCT. Recall that a subset E C R is null or is a null set if for every e > 0 there is a cover of E by open intervals {Gj} such that Y^7Li^(Gj) < £ (Definition 1.9). Theorem 32. E is null if and only if CE is McShane integrable over R with /RCE=0. Proof: Suppose E is null and let e > 0. Let {Gj} be a sequence of open intervals covering E with J27Li ^(Gj) < £ P u t s « = CGI V • • • V Con. Since sn ^,h = lim sn exists. Note that 0 < h < 1 and CE < h. Also, s n < E L i CGk implies that / K sn < Y2=i l(Gk) ^ Efcli t(Gk) < £• Hence, the MCT implies that h is integrable with fRh < s. Since 0 < CE < h, CE is integrable and JR CE = 0 by Theorem 3 (iv). For the converse let J be an arbitrary closed, bounded interval and put A = I n E. It suffices to show that A is null. Note that CA is integrable with JRCA = JJCA = 0. There exists a gauge 7 on R such that \S(CA,T>)\ < e whenever V is a 7fine free tagged partition of R. Let {(tk,Ik) • k € M} be the cover from Lemma 6.3 relative to A, I, e, 7. For n e N, let Mn = {k e M : 1 < k < n}. Since {(tk,h) '• k € Mn} « 7, by Henstock's Lemma we have
2 keM„
cA{tkWk) fceM„
The McShane Integral
113
for every n. Hence, ^keM £(Ik) < £• Each Ik can be expanded to an open interval Gk containing 7fc with Y^,k£M Z(Gk) < 2e. Since A C (JfceM h C UfcgM ^fc> this m e a n s that A is null. Theorem 32 and particularly its proof should be compared with Example 1.10 and Corollary 6.6 for the gauge integral. Recall that property P concerning the points of a subset A C R is said to hold almost everywhere (a.e.) in A if the property P holds for all points in A except those in a null set. For the more general version of MCT we first require some preliminary results. Lemma 33. Let fk '• I —> R be nonnegative and McShane integrable over I with fk < fk+i on I Let f{t) be the limit of {fk(t)} in the extended reals, R*. 7/supfe Jj fk = M < oo, then f is finite a.e. in I. Proof: Let E = {t € 7 : f(t) = oo}. For each j and k, fk/j is integrable so (fk/j) A 1 is integrable (Corollary 8) with Jjifk/j) A 1 < Jj(fk/j) < M/j. For each j(fk/j) A l t (f/j) A 1 as k >• oo so the MCT implies
(*) limfc J (fk/j) A 1 = J (f/j) A 1 < M/j. Now / > 0 so (f/j) A 1  and lim, JT(f/j) A 1 = 0 by (*). If t G I\E, (f(t)/j) A 1 = f(t)/j for large j so (f(t)/j) Al ^ 0; U teE, (f(t)/j) A 1 = 1 so (f(t)/j) A 1 > 1. That is, (f/j) A 1 >• CE pointwise so by the MCT CE is integrable with Jr CE = 0. Theorem 32 now gives the result. Lemma 34. Let f : I —> R be McShane integrable over I and g : I —>• R be such that f = g a.e. in I. Then g is McShane integrable over I with Jr f = J, g. Proof: Put h = f — g so h = 0 a.e. in 7. It suffices to show that h is integrable with fjh = 0. Let E = {t : h(t) ^ 0}. Put hn = \h\ A n. Then hn < UCE implies hn is integrable with Jj hn = 0 (Exercise 4). Since hn f \h\, the MCT implies that \h\ is integrable with Jj\h\ = 0. Thus, h is integrable with Jj h = 0 (Exercise 4). This lemma and its proof should be compared with Example 1.10 (Exercise 2.22) for the gauge integral. We now have a more general form for Theorem 16. Theorem 35. Let each fk : 7 —• R be McShane integrable over I with Ylk=i Ii\fk\ < °° There exists a McShane integrable function f : I —> R such that if sn = X)fe=i A> ^ e n sn —• / a  e  a n ^ / / \sn — f\ —>• 0.
114
Introduction to Gauge Integrals Proof: Set gn = YTk=i \fk\ Then gn t T,T=i \fk\ and since .
n
oo

/ 9n = £ / \h\ < £ /fk\\h
Jl
k=lJl
k=lJl
Lemma 33 implies that ]Cfct=i l/fcl converges in R a.e. in / . Let E = {t : E r = i IA(*)I < °°} a n d s e t hk = CEfk, /(*) = EZi fk(t) for t € E and /(f) = 0 for t i E. Then fk = hk a.e., J 7 fk = Jj hk (Lemma 34), 5Zfc=i hk > / pointwise on / and ^2^=1 Ii I'1* I < oo. Theorem 16 now applies so / is McShane integrable with
/
X>/
l X>/
*:=i
>0.
fc=i
Our general form of the MCT now follows easily from Theorem 35. Theorem 36 (MCT). Suppose each fk : / —» K is McShane integrable over I with fk < fk+i and supfc J, fk < oo. TTien i/iere exists a McShane integrable function f : I —> R such that /fc —>• / a.e. and Jj fkt Jj fProof: Set /o = 0 and gk = fk — /fci for fc > 1. Then gk > 0 and oo
V
«
n
/ 3fc = lim V
~
/ (fk  fki)
„
= lim / /„
< oo.
Theorem 35 implies there is a McShane integrable function f : I —¥ that YlT=i 9k = Hmn / „ = / a.e. and YlT=i Si 9k = Hm„ / 7 / „ = / 7 / .
such
We can now employ Theorem 36 to obtain a general form of Fatou's Lemma. Lemma 37 (Fatou). Let fk • I —> [0, oo) be McShane integrable over I. Suppose that lim J, fk < oo. Then lim ft = f is finite a.e. and if g : I —>• R is such that f = g a.e. in I, then g is McShane integrable over I with frg < lim/j/fcProof: Put gk = / i A • • • A fk. Then gk > 0, gk t Um/fc, 0 < gk < fk and each gfc is integrable (Corollary 6). Therefore, limj^tjfc < lim/ 7 /fc < oo. By Lemma 33 lim ffc is finite a.e. and by Theorem 36 / 7 g = lim J 7
The McShane Integral
115
Finally, we obtain a general form of the DCT involving convergence a.e. Theorem 39 (DCT). Let fk,f, g '• I —> K with fk,g McShane integrable over I. Assume \fk\ < g a.e. for each k and /fc —> / a.e. in I. Then f is McShane integrable over I with Jj \fk — /  —> 0. Proof: g — fk > 0 a.e. and l i m / ^ — fk) = Jj g — lim Jj fk < oo since I / / /fc < Ji\fk\ < Jj g Fatou's Lemma implies that / is integrable with
[M(gfk)= Jif(9f)= Ji[g Ji[ f <M Jif(gfk) = Ji fg^[f Ji k
Ji
or lim Jj fk > Jj f. Similarly, fk + g > 0 a.e. and lim Jj (fk + g) = lim Jj fk + Jj g < oo implies by Fatou's Lemma JT(f + g) < Hm Jj fk + Jj g or lim Jj fk > Jjf. Hence lim Jj fk exists and equals Jj f. Note that the conclusion of Theorem 39 is not as sharp as the conclusion of the version of the DCT given by Theorem 20 (i).
9.5
The Space of McShane Integrable Functions
In this section we consider the space of McShane integrable functions. We show that there is a natural seminorm defined on the space and in contrast to the Alexiewicz seminorm defined on the space of gauge integrable functions in Chapter 7, the space of McShane integrable functions is complete under this seminorm. This section requires only a very basic knowledge of (semi) normed spaces and the general MCT and DCT given in Sec. 9.4. Let J be a closed interval in R* and let M. (I) be the space of all functions which are McShane integrable over J. Then M(I) is a (real) vector space and has a very natural seminorm defined on it via /i = Jj /;  i is often called the L1norm. Note that  l^ = 0 if and only if / = 0 a.e. in I (Lemma 34 and Exercise 19); it is often the case that functions in M.(I) are identified if they are equal a.e. in / so  i can then be considered to be a norm on M.(I). In contrast to the situation with the space of gauge integrable functions under the Alexiewicz norm (7.1), we show that (Ai(I), \\ i) is complete. Theorem 40. (A^(/), i) is complete. Proof: Let {fk} C M(I) be Cauchy with respect to  i. It suffices to show that there exist a subsequence {fnk} of {fk} and a function / € M.(I) such that \\fnk  /   i )• 0. Pick a subsequence nk such that /„ fc+1  /n fc i < l/2 fc
116
Introduction to Gauge Integrals
and put gk = /„ fc+1  fnk. Then YUkLi Ii \9k\ < °o so by Theorem 35 there exists g G M{I) such that Yl'kLi 9k = 9 a  e  a n d
/
/ \fnj+1
fnig\*o.
Thus, iif = fni+g,f€ M(I) and \\fnj  /   i > 0. Note from Theorem 12 that the subspace of step functions is dense in M(I) with respect to  i.
9.6
Multiple Integrals and Fubini's Theorem
As was the case for the gauge integral, except for the usual notational difficulties, there is no problem in denning the McShane integral for functions defined on intervals in R n . In this section we give the definition. Most of the basic results are analogous to those for the McShane integral in R 1 and the versions of the Fubini Theorems for the gauge integral carry forward to the McShane integral. Let R*n = R* x • • • x R* where there are n factors in the product. An interval I is a product I\ x • • • x In where each Ik is an interval in R*; we say that I is a closed (open) interval if each Ik is closed (open). The interior of I, written 1°, is the product of the interiors of the Ik • The volume of an interval J, denoted v(I), is the product v(I) = Hfe=i ^(Ik), where we continue to use the convention that 0 • oo = 0. [In R 2 , the term area would be more appropriate.] If I is a closed interval, a partition of I is a finite collection of closed subintervals of I, {Ik : 1 < k < m}, such that 1% fl 1° = 0 if k ^ j [i.e., the {Ik} do not overlap and 7 = UfcLi^fc] A free tagged partition V of J is a finite collection of pairs {(a;,, Ii) : 1 < i < m} such that {Ii : 1 < i < m} is a partition of I and Xi G I for 1 < i < m; we call the 7, the subintervals of V and call Xi the tag associated with Ii. Note again that in a free tagged partition the tag Xi is not required to belong to its associated subinterval Ii. A gauge 7 on / is a function defined on I such that 7(3:) is an open interval containing x for every x G I. A free tagged partition V = {(xi, Ii) : 1 < i < m} is jfine if Ii C 7(3;;) for 1 < i < m; we write V « 7 as before. If / is a function denned on an interval 7, we assume that / vanishes at all infinite points. If / : 7 > R and V = {(XJ,7J) : 1 < % < m} is a free tagged partition of 7, the Riemann sum of / with respect to V is defined to be S(f, V) = YHLI f(xi)v{Ii) [again we assume 0 • 00 = 0].
The McShane Integral
117
Definition 41. Let I be a closed subinterval of R*n and / : J » R. Then f is McShane integrable over I if there exists A 6 R such that for every e > 0 there exists a gauge 7 on I such that \S(f, V) — A\ < e whenever T> is a yfine free tagged partition of I. From Corollary 8.2 it follows that Definition 41 makes sense and as before the number A is unique, is called the McShane integral of / over I and is denoted by A = Jr f, etc. The basic properties of the integral such as linearity, additivity, positivity, etc., carry forward to the McShane integral over R n . We do not repeat the statements or proofs (see Theorems 2 and 3). In particular, the analogue of Theorem 5 on absolute integrability holds in R*". The proof in Example 8.4 is valid for free tagged partitions so it follows that any step function in R™ is McShane integrable. Also, Theorem 8.7 is valid for the McShane integral and gives a large class of functions which are McShane integrable. The junior grade versions of Fubini's Theorem (Tonelli's Theorem) given in Chapter 8 (Theorems 8 and 9) for the gauge integral hold with no change for the McShane integral. The proofs given in Chapter 8 utilize the integrability of step functions, the Bounded Convergence Theorem and MCT, all of which are valid for the McShane integral. Again, we do not repeat the statements or proofs but refer the reader to Chapter 8. Similarly, the general form of Fubini's Theorem given in Chapter 8 (Theorem 8.13) for the gauge integral holds for the McShane integral. This version requires the analogue of Lemma 8.1 for free tagged partitions and the use of null sets and the analogue of Corollary 6.6 given in Theorem 32. It should be pointed out that Theorem 8.13 for the gauge integral admits conditionally integrable functions whereas its analogue for the McShane integral does not. Notes/Remarks As was the case for the gauge integral, there is a stronger form of Part 2 of the FTC than that given in Theorem 11. If / : [a, b] » R is McShane integrable over [a, b] and F(t) = Ja f, a < t < b, is the indefinite integral of / , then F is differentiable a.e. in [a, b] with F' = f a.e. This follows from the result in Appendix 3 for the gauge integral since any McShane integrable function is gauge integrable. There is also a converse to Theorem 24 which gives a characterization of indefinite McShane integrals. If F : [a,b] —> R is absolutely continuous on [a, b], then F is differentiable a.e. in [a, b] with F(t)  F(a) = J* F' [h ere r is defined to be 0 where F is not differentiable]. This is a wellknown result
118
Introduction to Gauge Integrals
for the Lebesgue integral [see [Swl] 4.4.7] and is sometimes referred to as the RadonNikodym Theorem for the Lebesgue integral. [See Appendix 4.] Theorem 40 on the completeness of the space of McShane integrable functions is usually referred to as the RieszFischer Theorem for the Lebesgue integral ([Swl] 3.5.1). There is in some sense a converse to the result in Theorem 29. If g : I —> R is measurable (Lebesgue) and fg is McShane integrable over I for every / 6 •M(I), then g is equal a.e. to a bounded measurable function (such functions are said to be essentially bounded). This solves what might be called the "multiplier problem" for the McShane integral (see the Notes/Remarks section of Chapter 4). This is a wellknown result for the Lebesgue integral ([Swl] Exer. 3.2.21), and since the McShane and Lebesgue integrals are equivalent (Appendix 4), this means the same result holds for the McShane integral.
Exercises 1. Let T> = {(U, Ii) : 1 < i < m} be a free tagged partition of I and let J = {Jj : 1 < j < n} be a partition of J. Show V = {(£;, Ii D Jj) : 1 < i < m, l<j
The McShane Integral
119
11. Let / : [a, oo) > R be McShane integrable. Show that lim^oo /6°°  /  = 0. [Hint: Pick7 such that 7(i) is bounded for t € K and S(/,X>)/ a °°  /   < e whenever V « 7. Fix such a V = {{U, J») : 1 < i < m} with ii = 00, h = [b, 00]. Consider J^°  /  for c > b.} 12. Can you formulate and prove a version of Exercise 11 for a sequence of uniformly McShane integrable functions?
13. Let / : I 4 K be McShane integrable. Show that lim^jj^o Jj l/l = 0[Hint: Pick 7 such that 5(/,X>)  / f  /   < £ when V « 7. Fix such a V = {(ti,Ii) : 1 < i < m} and put M  max{/(ij) : 1 < i < m}. Let J be a subinterval of J. Consider £ — {(U, h n J ) : 1 < i < m} and use Henstock's Lemma to see how to choose S so that £(J) < S implies fj\f\<*]
14. For what values of p is tp McShane integrable over [0,1]? 15. (Comparison Theorem) Let / , g : [a,b] > R, a < bk < b, bk t b. Suppose / is McShane integrable over [a, bk] for every k and g is McShane integrable over [a, b] with  /  < g on [a, 6]. Show that / is McShane integrable over [0,6]. 16. Establish analogues of Theorem 21 and Exercise 15 for unbounded intervals (improper integrals). 17. Use Exercise 16 to determine the values of p for which tp is McShane integrable over [1, 00). 18. Show that g(t) = i 1 / 3 ,!) < t < 1, is absolutely continuous but does not satisfy a Lipschitz condition. 19*. Let /:/—>• [0, 00) and suppose / is integrable over / with fr f — 0. Show that / = 0 a.e. in I. [Hint: Let E = {t: f(t) > 0} and put fk = (kf) A CE (Exercise 6.10).] 20*. Show that the nonnegativity assumption in Fatou's Lemma is important. 21*. Show that strict inequality can hold in Fatou's Lemma. [Hint: Consider fk = C(o,2) f° r & °dd and fk = C(i,3) for k even.] 22*. Let / : R >• R be McShane integrable over R. Let M be the class of Lebesgue measurable subsets of R and define v : M. —> R by v{E) — JE f (Chapter 6). Show that v is countably additive (compare with 6.9 and Exer. 6.8 for the gauge integral). 23*. Let / = [a, b}. Show that the inclusion map M.(I) —>• 'HK(I) is continuous with respect to the norm  i on M(I) and the Alexiewicz norm on UK (I). 24. Show that the product of McShane integrable functions need not be McShane integrable. Hint: tp. 25. Show that the assumption in Theorem 29 that g is bounded and McShane integrable over I can be replaced by the assumption that g is bounded and
120
Introduction to Gauge Integrals
McShane integrable over every bounded sub interval of I. [Hint: Consider gCIk where Ik = [k, k] n /.] 26. Give an example where Exercise 25 applies but Theorem 29 does not. 27. Show that M(I) is separable under  i. 28. Let / : R >• E be McShane integrals. For k e N let
(m
m\
fkit) = { k k
f(t) > k f(t) < k
be the truncation of / at k. Show that each fk is McShane integrable and JR fk —> / K /• Show that an analogous result is false for the gauge integral. [See [B] for a discussion of this truncation for the gauge integral; another truncation for the gauge integral is considered in [LPY] § 18.]
Chapter 10
McShane Integrability is Equivalent to Absolute HenstockKurzweil Integrability It is clear from the definitions (1.2, 4.2, 9.1) that if a function is McShane integrable, then it is HenstockKurzweil (gauge) integrable and both integrals have the same value. Example 2.12 furnishes an example of a function which is HenstockKurzweil integrable but is not McShane integrable (see also Exercises 3.9, 4.3). In this chapter we establish the basic connection between the two integrals. In particular, we give an elementary proof of the fact that a function is McShane integrable over a closed, bounded interval if and only if it is absolutely HenstockKurzweil integrable. We begin by fixing the notation which will be employed. Let / = [a, b], — oo < a < b < oo, and /:/—>• R. It is convenient in this chapter to consider a gauge to be a positive function 5 : I —»• (0, oo) as in Chapter 1 (recall that such a function generates a gauge 7 by setting j(t) = (t — S(t),t + 6(t)) so a tagged partition V = {(U,Ii) : 1 < i < m} is <5fine if and only if U £ Ii C (U — S(ti), ti + S(U)) for 1 < i
:Vens[a,x]},
ms(x) = inf{5(/, V) : V € irs[a, x}} , 121
122
Introduction to Gauge Integrals
for a < x < b; the function Ms{ms) is called a major [minor) function for / [see the Notes/Remarks section, Lemma 1 and Exercise 2 which give a motivation for these definitions and also the method of proof]. If / is HenstockKurzweil (gauge) integrable over / , we write H J / f o r its integral, and if / is McShane integrable over / , we write M J f for its integral. Lemma 1. (i) If x  5(x) f(x)(vu). (ii) If x — S(x) < u < x < v < x + 6(x), then ms{v) — ms(u) < f(x)(v — u). (iii) Ms — ms is nonnegative and increasing on I. (iv) If f >0, then both Ms and ms are nonnegative and increasing. (v) If f is HenstockKurzweil integrable over I and if e > 0 and 6 is a gauge on I such that
S(f,V)H
f f
< £
Ja
whenever V g ixs[a, b], then 0 < Ms(b) — ms(b) < 2e. Proof: (i): Let V £ Trs[a,u]. Then V U {(x, [u, v])} € ns[a,v] so Ms(v)>S(f,V)
+
f(x)(vu).
Hence, M5{v) > Ms(u) + f(x)(v  u) and (i) follows. (ii) is similar to (i). (iii) Let e > 0 and a < u < v < b. There exists T>, V s 7T{[a, u] such that Ms(u)  m5(u) < S(f, V)  S{f, V) + e. Let T e irs{u, v]. Then £ = V U T, £' = V U T e n5[a, v] so M6(u)  ms(u) < S(f, V)  S(f, V) + e = S(f, £)  S(f, £') + e < Ms(v)
ms(v)+e,
and therefore, Ms(u) — ms(u) < Ms(v) — ms(v) so Ms  ms is increasing. Clearly, Ms  ms is nonnegative.
McShane
123
Integrability
(iv) is similar to (iii). For (v), if £>,£ e ws[a,b], then \S(f,V)S(f,S)\
< 2e so Ms(b) ms(b)
<
2s. The proof below makes repeated use of Fatou's Lemma. We use the version of Fatou's Lemma given in Remark 9.38 so it needs to be realized that there are functions involved which may take on infinite values on a null set. Theorem 2. f is McShane integrable over I if and only if f is absolutely HenstockKurzweil integrable over I. Proof: As noted earlier since the McShane integral is an absolute integral, if / is McShane integrable, then / is absolutely HenstockKurzweil integrable. For the converse, it suffices to show that if / is nonnegative and HenstockKurzweil integrable, then / is McShane integrable. For convenience, extend / to [a, oo) by setting f(t) = 0 for b < t < oo. Let e > 0 and pick a gauge 5 on J such that if V is a 5fine tagged partition of J, then \S(f, V)  H Ja f\ < e. Extend Ms and ms by setting Ms(t) = Ms(b), ms(t) — ms(b) for t > b. Define Hn(t) = n(Ms(t + 1/n)  Ms(t)) and hn(t) = n(ms(t + l/n) 
ms(t)).
Since Ms and ms are increasing (Lemma 1 (iv)), Hn and hn are McShane integrable (Exercise 9.2). Set H — ]\mHn, h — lim/i n . We claim that (1) 0 < M f
Hn < Ms(b).
Ja
Indeed, we have rb
/
rb+l/n
Hn=n[M Ja
pa+l/n
MsM \
Jb
Ja
Ms
>
M5 ,
= Ms(b)
(Exercise 9.3 and Lemma 1. (iv)). Fatou's Lemma (Remark 9.38) implies that H is McShane integrable and MjbaH<M5(b). We next claim that h is McShane integrable. For this suppose t is fixed and n > l/6(t). Then, by Lemma 1 (i) and (ii), for large n, f(t) < n{Ms{t + l/n)  Ms(t)) and f(t) > n(ms{t + l / n ) 
ms(t))
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Introduction to Gauge Integrals
so f(t) < H(t), f(t) > hn(t). Therefore, \jm(H  hn)+ = H  h. As in (1), 0 < M Ja hn < ms(b) so Fatou's Lemma implies that H  h is McShane integrable and, hence, h is McShane integrable. Note that from the computations above we have (2) h(t) < f(t) < H{t). Finally, we claim that (3) 0 < M / [Hh)<
2e.
Ja
As in (1), it follows that 0 < M f (Hnhn)<
Ms(b) 
ms{b).
Ja
Since H — h < lim(ij" — hn), Fatou's Lemma and Lemma 1 (v) give pb
(4) M / (H h) Ja
pb
< M
rb
lmi(Hn  hn) < IjmM / (Hn  hn) < M5(b) Ja
Ja
mS(b) < 2e. It follows from (2) and (3) that / is McShane integrable (9.3 (iv)). Of course, it is also the case that H J f — M Ja f. Notes/Remarks The proof of Theorem 2 is due to Vyborny ([V]). There were several earlier proofs of the result, but all of these proofs used Lebesgue measure/integration — a concept not present in the definitions of either integral (see, for example, [ML] § 8.3, [LPY] 16.10, [LW], [Pfl] 6.3.4, [C]). Another elementary proof is given in [CWT]. The definitions of the major and minor functions are taken from [BV] but essentially appeared in [KlJ. The appearance of major/minor functions in the proof of the equivalence of the McShane and absolute HenstockKurzweil integrals is not surprising. It is known that the HenstockKurzweil integral is equivalent to the Perron integral ([Gol], [LPY] § 8) and major/minor functions are employed in the definition of the Perron integral [the definition of major/minor functions for the Perron integral is different from the definitions used above, but the functions M$,m$
McShane
Integrability
125
are major/minor functions in the Perron sense; see the definition of the Perron integral below and Exercise 2j. In order to appreciate the proof of Theorem 2 above we give a brief description of the Perron integral. A function H : I —>• R is a major function of / if H(a) = 0 and D_H(t) > f(t) for t e I; h : / > K is a minor function for / if h(a) = 0 and £)/(£) < f(t) for t € / [here ~Df(D_f) is the upper (lower) derivative of / ] . Definition 3. / is Perron integrable over I if f has both major and minor functions and —oo < ini{H(b) : Ha major function = sup{/i(6) : ha minor function
for / } for / } < oo.
The common value A in Definition 3 is called the Perron integral of / over / and is denoted by P J f = A. We have the following criterion for Perron integrability. Theorem 4. / is Perron integrable over I if and only if for every e > 0 there exist major and minor functions, H and h, for f such that 0 < H(b) — h(b) < e. Note that Eq. (4) is just the sufficient condition of Theorem 4. For the basic properties of the Perron integral, including Theorem 4, see [N2] Chapter XVI or [Gol].
Exercises 1. Extend Theorem 2 to unbounded intervals. 2. Show that the functions M$,m$ are major, minor functions for / in the sense of Definition 3 for the Perron integral.
Appendix 1: The Riemann Integral In this appendix we repeat Riemann's definition of the Riemann integral and give references where the interested reader can find expositions of the integral. Let J = [a, b\ and / : J >• K. Definition 1. The function f is Riemann integrable over I if there exists A € R such that for every e > 0 there is a S > 0 such that if V = {a = x 0 < X\ < • • • < xn = b} is a partition of I with x\ — Xi\ < 5 for 1 < i < n and Xii
< ti < Xi, then
^
f{U){xi  Xii)v  A < e.
The term 5Z™=i f(ti)(xi — Xii) is called a Riemann sum of / with respect to V and the choice of points i,. The number A is the Riemann integral of / over / and is usually denoted by Jj f, J f, Ja f{t)dt or Jj f(t)dt. For an exposition of the Riemann integral using Definition 1, see Hoffman, [Ho] Chapter 4. There is another approach to the Riemann integral due to Darboux that is more popular in the current introductory texts on real analysis; this approach, based on upper and lower integrals, is also discussed in Hoffman and the 2 approaches are shown to be equivalent (see also [Ru]).
Exercises 1. If / is Riemann integrable over [a, b], show that / is bounded on [a, b]. 2. Define / : [a, b] —> R by f{i) = 0 if t is irrational and fit) = 1 if t is rational [/ is often called the Dirichlet function]. Show that / is not Riemann integrable.
127
Appendix 2: Functions of Bounded Variations For the reader who may not be familiar with the basic properties of functions of bounded variation, we record them in this appendix. Let / : [a, b] > R. If IT = {a = x0 < x\ < • • • < xn = b} is a partition of [a, b], the variation of / over IT is nl
var(f :ir) = J2 \f(xi+i)
~ f(xi)\ >
i=0
and the variation of / over [a, b] is Var(f
: [a,b]) = sup var(f : TT) ,
where the supremum is taken over all possible partitions, IT, of [a, b}. If Var{f : [a, 6]) < oo , / is said to have bounded variation; the class of all such functions is denoted by BV[a, b]. The variation measures the amount the function oscillates in [a, b]. As the example below illustrates, even a continuous function can fail to belong to BV[a, b]. Example 1. Let f(t) = tsin(l/i) forO < t < 1 and /(0) = 0. Set xn = l / ( n + l/2)7r. Then f(xn) = l / ( n + l/2)7r if n is even, and f(xn) = — l / ( n + l/2)7r if n is odd. If nn is the partition {0 < xn < xn\ < • • • < x\ < 1}, then
soVar{f
: [0,1]) = oo. 129
130
Introduction
P r o p o s i t i o n 2. If f e BV[a,b], Proof: Let x e (a,b). \f(x)
to Gauge
Integrals
then f is bounded on [a,b].
Then
 f(a)\ + \f(b)  f(x)\
< Var(f
: [a, b])
so
2\f(x)\<\f(a)\
+ \f(b)\ +
Var(f:[a,b})
and / is bounded. We consider properties of Var(f : J) as a function of the interval J. First, as a consequence of the triangle inequality, we have L e m m a 3 . Let f : [a, b] —>• M.. If TT and TT' are partitions then var(f : TT) < var(f : TT').
of [a, b] with TT C IT',
P r o p o s i t i o n 4 . Let f : [a, b] —> R and a < c < b. Then Var(f
: [a, b]) = Var(f
: [a, c]) + V o r ( / : [c, 6]).
Proof: Let 7r be a partition of [a, b] a n d TT' the partition obtained by adding the point c to 7r. Let TTI and 7T2 be the partitions of [a, c] and [c, 6], respectively, induced by TT'. Then by Lemma 3, var(f
: TT) < var(f
: TT') = var(f
: TT\) + var(f
: 7^) < Var(f
: [a, c\)
+ Var(f:[c,b}) so
Var(f
: [a, b}) < Var(f
: [a, c}) + Var(f
: [c, b}).
If TT\ and TT2 are partitions of [a, c] and [c,fo],respectively, then TT = TT\ U TT2 is a partition of [a, 6] so var(f
: TT) = var(f
: TTI) + var(f
: TT^) < Var(f
: [a, b\).
Hence, Var(f
: [a, c]) + V a r ( / : [c, 6]) < V o r ( / : [a, 6]).
Next, we consider how the variation, Var(f
: I), depends on the function
131
Appendix 2: Functions of Bounded Variations Proposition 5. Let f,g G BV[a,b].
Then
(i) f + ge BV[a, b] with Var(f + g:[a,b})< Var{f : [a, b}) + Var(g : [a, b}), (ii) for teR,tf€ BV[a,b] with Var(tf : [a, b}) = \t\Var(f : [a,b]). Proof: (i) follows from the triangle inequality and (ii) is clear. Thus, 5V[a, 6] is a vector space under the usual pointwise addition and scalar multiplication of functions. Let / € BV[a,b], We define the total variation of / by Vf(t) = Var(f : [a,t]) if a < t < b and V)(a) = 0. Proposition 6. Vf and Vf — f are increasing on [a, b}. Proof: Vf is increasing by Proposition 4. Let a < x < y
: [x,y]) 
f(y)
implies 9{y)  9(x) = Vf{x)  f(y) + Var(f =
: [x, y\)  Vf(x) + f(x)
Var(f:[x,y])(f(y)f(x))>0.
Proposition 7. If f £ BV[a,b] is (right, left) continuous at x, then Vf is (right, left) continuous at x. Proof: Let e > 0. Suppose / is right continuous at x < b. There is a partition TT of [x, b] such that var(f : re) > Var(f : [x, b}) — e. Since / is right continuous at x, we may add a point x\ to 7r to obtain a partition 7r' = {x < xi < • • • < x n } of [x, b] such that /(x)  / ( x i )  < e. Then nl
e + var(f,TT') = e + /(x 0 ) + / ( x ^ l + £
\f(xl+l)

f(x%)\
i=\
<2e +
Var(f:[xub})
so Var(f
: [x, b]) < var(f : IT) + e < var(f : w') + e < 2e + Var(f
: [xi, b}).
132
Introduction to Gauge Integrals
Thus, 0 < Vf(xi)  Vf(x) < 2e, and since Vf t,lim y _,. x + Vf(y) = Vf(x), i.e., Vf is right continuous. The statement about left continuity is similar. Since f = Vf — (Vf — / ) , Propositions 6 and 7 along with Exercise 1 and Proposition 5 give the following characterization of functions of bounded variation. Theorem 8. Let f : [a, b] >• R. Then f e BV[a, b] if and only if f = g  h, where g, h t . / / / is (right, left) continuous, then g and h can be chosen to be (right, left) continuous. We now extend the notion of bounded variation to functions defined on unbounded intervals. Let / : R —>• R. Definition 9. Var(f, R) = sup{Var(f Var(f
: [a, b}) :  o o < a < b < oo} ,
: [a, oo)) = sup{Var(f
Var(f : (oo, b]) = sup{Var(f
: [a, b}) : a < b < oo} , : [a, b]) :  o o < a < b} .
We say that / has bounded variation over the interval / if Var(f we denote the space of all such functions by BV(I). The following properties are easily derived.
: I) < oo;
Proposition 10. (i)
Var(f
: R) = lim Var(f
:
[a,a}),
a—»oo
Var(f
: [a, oo)) = lim Var(f
: [a, b]),
b—Kxi
Var(f
: (oo, 61) =
lim Var(f
: [a, b}).
a—> — oo
(ii) For a
Var(f
: R) = Var(f
: (oo, a}) + Var(f
: [a, oo)),
Var(f : (oo, b}) = Var(f
: (oo, a}) + Var(f
: [a, b}),
Var(f
: [a, oo)) = Var(f
: [a, b]) + Var(f
: [b, oo)).
We have the following further properties of the variation function.
133
Appendix 2: Functions of Bounded Variations Proposition 1 1 . Assume Var(f,R)
< oo. Then
(i) / is bounded and (ii) limxyoo Var(f : (oo, x}) = 0, l i m ^ ^ Var(f
: [x, oo)) = 0.
Proof: (i): Fix a e E. For x > a, \f(x)  f(a)\ < Var(f
: {a,x}) < Var(f
: E);
: [x, a}) < Var(f
: E).
similarly, for x < a, \f(a)  f(x)\ < Var(f Hence,
\f(x)\<\f(a)\+Var(f:R). (ii): Let e > 0. There exists o such that Var(f Since Var(f
: E) = Var(f
so lim x ^oo Var(f
: (oo, a]) + Var(f
: E) < Var(f
: [a, oo)), Var(f
: (oo, a})+e.
: [a, oo)) < e
: [a;, ex))) = 0. The other statement is similar.
It is easily seen that the analogue of Proposition 5 holds for unbounded intervals. If / G BV(R), we define the total variation of / by Vf(t) = Var(f : (—oo,i]). The analogues of Propositions 6 and 7 then hold for the total variation function and, thus, Theorem 8 likewise holds for functions of bounded variation on E.
Exercises 1. If / is increasing on [a, b], show Var(f : [a,b]) = f(b) — f(a). If / is increasing on E, give necessary and sufficient conditions for / to be of bounded variation on E. 2. Give necessary and sufficient conditions for / to satisfy Var(f : \a,b}) = 0. Repeat for Var(f : R) = 0. 3. Let J be a closed interval in E and let / , g G BV(I). Show that fg,  /  , / V 5 ) / A <; belong to BV{I). 4. Prove Proposition 10. 5. Let / G BV(R). Show that limx^ioo f(x) exists.
Appendix 3: Differentiating Indefinite Integrals In this appendix we consider the other half of the Fundamental Theorem of Calculus for the gauge integral, differentiating indefinite integrals. Let / = [a, &],—oo < a < fr < oo. If / € HK(I), the space of gauge integrable functions on / , and F(t) = Ja f, a < t < b, we show that F is differentiable a.e. with F' = f a.e. For the proof we require a covering theorem of Vitali. If E C / , a family J of closed subintervals is a Vitali covering of E if for every t e E and 5 > 0 there exists J € J such that t € J and £( J) < 5. Let A (A*) denote Lebesgue (outer) measure on R. T h e o r e m 1 (Vitali Covering Theorem). Let E
 f{tk)(xk
 x fc _!)}
< £.
fc=i
We may assume that j(t) = (t  S(t),t + 5(t)) with S(t) > 0 for t E I. For every t € E there exists rj(t) > 0 such that for every 5 < S(t) there is a point 135
Introduction to Gauge Integrals
136 s £ I with 0 < s — t < S and \F(t)F{a)f(t){8t)\>V(t)(st).
For n e N, set En = {t <E E : r](t) > 1/n}. To show X(E) = 0 it suffices to show that X(En) = 0 for every n. Fix n and let J be the family of all closed intervals [t, s] with t £ En and s as above. Then J is a a Vitali cover of £?„. By the Vitali covering theorem, there exists {[ifc,Sfc] : 1 < k < m} C J pairwise disjoint with X*(En) < XX=i(sfc _ *fc) + E N o w ((**;> [*fc' sk\) : 1 < A: < m} is a 7fine partial tagged division of / so by Lemma 9.3 we obtain m
X*(En) < ^2 \F(tk) ~ F(sk)  f(tk)(sk
 tk)\/v(tk)
+e<2ne
+ e.
fe=i
Hence, X*(En) = 0 as desired. We can treat the points in / where F has a lefthand derivative in a similar manner thus establishing the result. A consequence of Theorem 2 is that any gauge integrable function is (Lebesgue) measurable. Theorem 3. If f € UK (I), then f is (Lebesgue) measurable. Proof: F is continuous by Corollary 3.2 and therefore measurable. By Theorem 2
n« + ./»)FM fc>oo
1/K
for all t except those in a null set so / is measurable. Notes/Remarks For other proofs of the measurability see [LPY] 5.10, [LPY1] or [Pfl] 6.3.3.
Appendix 4: Equivalence of Lebesgue and McShane Integrals In this section we show the equivalence of the Lebesgue and McShane integrals over closed, bounded intervals. Again A denotes Lebesgue measure. Theorem 1. Let f : [a, b] —> R be Lebesgue integrable over [a,b\. Then f is McShane integrable over [a, b] and the values of the two integrals are equal. Proof: We may assume that / is nonnegative ([Swl] 3.2.11). All integrals in the proof below will be Lebesgue integrals. Let e > 0. Choose S > 0 such that X(A) < S implies JAf <e ([Swl] 3.2.17). Let a = min{l,e/((J + b  a)}. For n e N, let En = {t:(n
l)a < f(t) < no.}
so {En : n G N} are pairwise disjoint with [a, b] = U^Li ^n For each n pick Gn open such that Gn D En and X(Gn\En) < S/2n(n + 1). Define a gauge 7 on [a, b] by choosing S(t) to be an open interval containing t and such that S(t) C Gn whenever t £ En. Now suppose that T> = {{U, Ii) : 1 < i < m} is a 7fine free tagged partition of [a,b]. We show
S(f,V) [ f
< 3e,
Ja
completing the proof. For each i = 1 , . . . ,m, let n; be such that U G Eni. Then 137
138
Introduction to Gauge Integrals b
(1)
.
TO
/
/ < E / i/(*o/(*)i

\nu)  m\dt
<E/ +E /
/(*<)<** + E /
/(*)*
R\ + i?2 + R3 i with obvious definitions of Ri. To estimate i?i, note that if £,, £ G S n o then both /(tj) and f(t) belong to the interval [(n* — l ) a , n j a ) so aA
Ri < E
( ^ n £•„;) < a(6  a) < £ .
Since <» G .Eni and Ii C Gni(T> is 7fine), 00
.
00
f dt
#2 = E E / n=l
i
^ E E (n+i)«*(Wn<
^
Jh\Eni
n = 1
i
< j > + I ) « A ( G „ \ K ) < $ > + 1 ) « w ^  i ) ^ E £ =£ • n=l
V
n=l
'
n=l
For i? 3 put A = U £ i ( M £ n J  Since J4 C G n i , m
00
\(A) = E ( AKJ = E E A(MK.) i=l
A /
n=l
i
rii=n
< V A ( G n \ # n ) < V —7
Thus,
r < <5
R3<JAf<e.
From (1), we have \S(f,V)
— J f\ < 3£ as desired.
Theorem 2. If f : [a, b] —> R is McShane integrable, then f is Lebesgue integrable over [a, b] and the two integrals are equal.
Appendix 4: Equivalence ofLebesgue and McShane Integrals
139
Proof: All integrals in the proof below will be McShane integrals. Let F(t) — J f be the indefinite integral of / . Since / is McShane integrable, / is HenstockKurzweil integrable so F is differentiable a.e. in [a, b] with F' = f a.e. in [a,b] (Appendix 3.2). If we show that F € BV[a,b], then / is Lebesgue integrable ([Swl] 4.2.5 and A.1.8).
Let 7 be a gauge on [a, b] such that \S(f, V)  ja f\ < 1 whenever V is a 7fine free tagged partition of [a,b]. Fix such aV — {(U,Ii) : 1 < i < m}. It sufficies to show that F is of bounded variation on each /;. Fix i, 1 < i < m. Let xo < x\ < • • • < xn be a partition of 7j. Then E = {(U, [XJ\, Xj}) : 1 < j < n) is a 7fine free tagged partition of I{. By Lemma 9.3,
E
/ ( * i ) ( ^j Xji)~
/
/
< 2
JXji
which implies £ " = 1 F(a; : ,)F(a; j _i) < 2+\f(ti)\e(Ii). variation on i j . The 2 integrals are equal by Theorem 1.
Hence, F is of bounded
Remark 3. The construction of the free tagged partition £ in the proof above shows the utility of free tags. Notes/Remarks The proof of Theorem 1 is that of Davies and Schuss ([DSc]). The proof of Theorem 2 is taken from Mendoza ([Me]). Other proofs of the equivalence of the Lebesgue and McShane integrals can be found in [Lin], [Gol] 10.11, [Ku].
Exercise 1. Extend Theorems 1 and 2 to unbounded intervals.
Appendix 5: Change of Variable in Multiple Integrals In this appendix we establish several versions of the change of variable theorem for multiple integrals. The first version is for continuous functions and requires only basic results for the gauge integral. A general version is then established which requires fairly extensive use of results for the Lebesgue integral. We begin by considering the case of linear transformations on R n . First, consider the following special linear transformations on R". (i) For t ^ 0, define dt : K n > R n by dt{x\,... ,xn) = {tx\,... ,xn). (ii) Define a : R" —> R71 by a(x\,... , xn) = (xi + X2, £2, • • • , xn). (iii) For 1 < i < j < n, define Sij : W1 )• Mn by Sij \Xi,...
, Xiz...
, Xj,...
, xn j — yX\,...
, Xj,...
, xi,...
, xnj
.
If A is an n x n matrix, its determinant is denoted by det A. Lemma 1. If A is any of the linear transformations in (i),(ii) or (iii) and f : R" —• R is absolutely integrable, then f o A is absolutely integrable with / R n f(y)dy = / R „ /(Ac) I det A\dx. Proof: Each of the cases in (i), (ii) or (iii) follows easily from Fubini's Theorem 8.13 and the linear change of variable formula in R 1 . (Exercise 1.) For example, if t > 0, then / JR"
f(dtx)dx=
/
••• /
J00
/"OO
OO
/
f{txi,..
••• / 00
.xn) dxi • • dxn
J00 1
T / O E I , . .  ,xn)dxi
J—00
= (1/ldetdtl) / 141
t
f(x)dx.
•••dxn
142
Introduction to Gauge Integrals
Every nonsingular linear transformation A : R n —> R™ is a product of linear transformations of the form in (i), (ii) and (hi) ([HK] 1.6.12) so we obtain Theorem 2. / / A : R™ >• R™ is linear and f : R n > R is absolutely integrable, then f o A is absolutely integrable with
(1) /
f(y)dy=
[
f(Ax)\detA\dx.
We denote Lebesgue measure on R™ by A. From Theorem 2 we have Corollary 3. If B C R™ is a bounded Borel set and A : R n >• R n is linear and nonsingular, then X(A1B)^(detA1)X{B).
(2)
Proof: Put / = CB in (1) to obtain f f = X(B) = f
f (Ax)\ det A\dx = /
CAiB(x)\
det A\dx
= A(A 1 B)detA. We now consider the case of nonlinear transformations on R n . We assume the reader is familiar with the basic properties of differentiable functions from R n > R n (for example, see [DS] §29, [Ru] §9). We begin by setting down the basic assumptions and notations. Let D\ and D!—>• R n a continuously differentiable function from D\ onto Di with a continuously differentiable inverse g : D^ —> D\. If / : D\ 4 R, we denote the j t h partial derivative of/ by djf, and we denote the differential of h by Dh. Thus, Dh is the linear transformation on R" whose (Jacobian) matrix is given by Dh = [djhi] and by the Chain Rule Dg is the inverse of Dh. For the case of a general transformation h we consider dividing the region D\ into cubes. If C is any cube inside D\ and x € C, then h(x) is approximated by Dh(x), and if h were linear, then by Corollary 3 X(h(C)) =  det Dh(x)\\(C). Thus, it is reasonable to assume that in the nonlinear case X(h(C)) is approximated by  det Dh(x)\X(C). We establish an upper bound for the approximation in Lemma 4 and use a device of J. Schwartz ([Sz]) to establish the transformation formula (3) / JD2
f(y)dy=
f
f (h(x))\ det Dh(x)\dx
J Di
for continuous functions.
Appendix 5: Change of Variable in Multiple Integrals
143
In what follows it is convenient to use the supnorm, a; =   ( x i , . . . ,xn)\\ =max{a; i  : 1 < i < n} , on R™. The closed (ndimensional) cube with center x and side length 2c is then given by C(x, c) = {y : \\x  y\\ < c}. If A = [dy] is an n x n matrix, we define the norm of A to be (
n
\\A\\  max < ^2 ay  : 1 < i < If x e R", we then have the inequality Ar < i4a;. Let x e C and let C = C{x,c) be a cube inside D\. By the Mean Value Theorem n
hi(y)  hi(x) = ^djhiix
+ 6j(y)(y  x)){yj  XJ)
3=1
where 0 < 6j(y) < 1. Thus, \\h(y)  h{x)\\
(4) \{h{C))< ^maxUDM^IIJ A(C). Lemma 4. / / A : W1 —> R n is linear and C is any cube inside D\, then (5) X(h(C)) < \detA\ (maxA 1 D/i(z) > )
A(C).
Proof: Apply (4) to the map A_1h to obtain A(A _1 /i(C)) < ( m a x p  1 ^ / ! ^ ) ! ! )
X(C).
But, h{C) is a bounded Borel set so (2) gives {det A"1) X(h{C))<
(max\\ A'1 Dh(z)\\\
X(C),
and the result follows. Lemma 5. Given x 6 D\ and e > 0 there exists S > 0 such that X(h{Q)) <  det Dh{x)\(l + e)A(Q), where Q is any cube inside C(x,5) C D\.
144
Introduction to Gauge Integrals Proof: Put A = Dh{x) in (5) to obtain
(6) \{h{C))<\detDh{x)\(m&yi\\Dh{x)1Dh{z)\\\
\{C)
where C is any cube inside D\. Now z —• \\Dh(x)~l Dh(z)\\ is continuous on D\ and approaches 1 = 7 as z —> x so the inequality in (6) gives the result. We next require a variant of the Covering Lemma 6.3 for open sets in R n . We say that a cube in R n is leftclosed if it has the form [a, b) x • • • x [a, b) (n factors). The proof of Lemma 6.3 is easily adapted in R" to give the following covering lemma (see also [M] 13.4, [MS2] 5.3, [ML] § 5.5). Lemma 6. (Covering Lemma) Let G e l " be open and contained in a rectangle I. Let 7 be a gauge on I. There exists an at most countable, pairwise disjoint family {Qk : k G M} of leftclosed cubes and {xk : k € M} C G such that xk e Qk C y(xk) and G = U{Qk • k e M}. Lemma 7. Let D\ and Di be bounded and f : D2 —> [0, 00) be continuous and bounded along with Dh and Dg. Then (7) / JD2
f(y)dy=
f
f(h(x))\detDh(x)\dx.
JDI
Proof: Both integrals in (7) exist by Exercise 2. Let e > 0. For each x € Di let S be as in Lemma 5 and define a gauge 71 on D\ by 7 i(a;)
= { . z € D i : \\x  z\\ < 6} .
For each x £ D± there exists an open cube 72(2) centered at x and contained in D\ such that \f(h(y) — f(h(x))\ < e and \f(h(y))\detDh(y)\
 f(h(x))\ det Dh(x)\\ < e
foryG72(x). Define a g a u g e 7 on Dx by 7(2;) = 71(^)072(3;). By the Covering Lemma there exist pairwise disjoint halfclosed cubes {Qk} and {xk} C D\ such that xk € Qk C ~f{xk) C £>i and £>i = UQk, D2 = Uh(Qk). If y e h(Qk), then y — h(x) for some x £ Qk so f(y) < f(h(xk)) + e, and if x € Qk, then £ + f(h(x))\ det Dh(x)\ > f(h{xk))\
det
Dh{xk)\.
Appendix 5: Change of Variable in Multiple Integrals
145
Thus, by Lemma 5, / f(y)dy < I if(h(xk)) Jh(Qk) Jh(Qk) < f(h(xk))(l
+ e]dy = [f(h(xk) + e)]X{h(Qk))
+ s)\ det Dh(xk)\X(Qk)
= (1 + e) / f(h(xk))\detDh(xk)\dx JQk
+ +
< ( l + e ) / \f(h(x))\detDh{x)\+e}dx JQk = (1 + e) f
e\(h(Qk)) e\(h(Qk)) + sX{h{Qk))
f(h{x))\ det Dh{x)\dx + e(l + e)X(Qk)
JQk
+
s\(h(Qk)).
Using the Monotone Convergence Theorem and the countable additivity of A, adding the inequality above gives /
f(y)dy
JD2
< (1 + e) [
f(h(x))\ det Dh(x)\dx
JDi
+ £(1 + £)A(D 1 )+ £ A( J D 2 ). Letting e —> 0 gives (8) /
f(y)dy<
[
f(h(x))\ det Dh(x)\dx.
In (8) we may interchange D\ with Di and h with g and replace / with / o h\ det Dh\ in the left hand side of (8). Then the integral on the right hand side of (8) becomes / and gives /
f (h(x))\ det Dh(x)\dx<
JDi
[
f(y)dy
JDI
so the result follows. Theorem 8. Let f : Di 4 R be continuous. Then / is integrable over D2 if and only if f o h\ det Dh\ is integrable over D\ and (7) holds. Proof: First assume / > 0. For each k put Gk = { i £ Dj : a; < k, f(x) < k, \ det Dh(x)\ < k, \det Dg(x)\ < k} .
146
Introduction to Gauge Integrals
Then each Gk is open, bounded and increases to D\. Lemma 7 applies to each Gfc and gives /
f(y)dy=
Jh(Gk)
f
f(h(x))\detDh(x)\dx.
JGk
The Monotone Convergence Theorem now gives the conclusion. If / : Z?2 —> R, the part above applies to each / + and / ~ and gives the result. We now establish a more general form of the transformation formula (7) for Lebesgue integrable functions. This will require fairly extensive use of results from Lebesgue measure/integration for which we will supply appropriate references. We establish a preliminary lemma. Lemma 9. If Z C D\ is {Lebesgue) null, then h(Z) is null. Proof: Suppose that Z is bounded and is contained in the closed cube / . Given e > 0 there exist cubes {Qk} inside / such that Z C UQfc and SfcLi ^(Qk) < £ \\Dh()\\ is bounded on / by, say, M. By Lemma 5, h(Z) C Uh(Qk) and \{h(Qk)) < Mn\(Qk). This implies h(Z) is null. If Z is unbounded, then Z = LlZk, where Zk = {x G Z : \\x\\ < k}. Each Zk is null by the part above so Z is null. Theorem 10. Let f : Dz —> IR he absolutely integrable over D^ Then f o h\ det Dh\ is absolutely integrable over D\ and (7) holds. Proof: Let fk : £>2 —> R be continuous, integrable over D?, such that /fc  /111 ^ 0 with fk > / a.e. ([Swl] 3.5.6, 3.7.1, 3.6.3). By Theorem 8, / JD2
fk(y)dy=
f JDI
fk(h(x))\
det Dh(x)\dx+
f
f(y)dy
JD2
and the sequence {fkoh\ det Dh\} is Cauchy in L1(D1). The sequence therefore converges in  i to some ip G LX{D\) ([Swl] 3.5.1). But, fk ° h\detDh\ —> f o h\ det Dh\ a.e. by Lemma 9 so (p = f o h\ det Dh\ a.e. and the result follows. For the complete analogy of Theorem 8, see Exercise 3. For other proofs of the transformation formula for the gauge integral see [M], [MS]. For the Lebesgue integral see [Sz] and the references therein.
Appendix 5: Change of Variable in Multiple Integrals
147
Exercises 1. Complete the proof of Lemma 1. 2. Let G C K" be bounded and open and / : G —> R be bounded and continuous. Show that / is integrable over G. 3. Suppose / : £>2 —>• K is measurable and such that /ofc det Dh\ is absolutely integrable over D\. Show that / is absolutely integrable over D2 and (7) holds. [Hint: Repeat the proof of the last part of the Lemma 7.] 4. If E C IRn is measurable, show that h(E) is measurable. 5. Let I = JQ e _ t dt. Show that / = ^/n/2 by using polar coordinates (x, y) = (r cos#, y sin 6) to compute I2 as a double integral.
Bibliography [A]
A. Alexiewicz, Linear Functional on Denjoy integrable functions, Colloq. Math. 1 (1948), 289293. [AM] C. Adams and A. Morse, On the Space (BV), Trans. Amer. Math. Soc. 42 (1937), 194205. [Bal] R. Bartle, Return to the Riemann Integral, Amer. Math. Monthly 103 (1996), 625632. [Ba2] R. Bartle, A convergence theorem for generalized Riemann integrals, Real Anal. Exchange 20 (1994/95), 119124. [B] R. Bianconi, On the Convergence of Integrals of a Truncated HenstockKurzweil Integrable Function, Real Anal. Exchange 23 (1997/98), 247250. [Boas] R. P. Boas, A Primer of Real Functions, Carus Math. Monograph 13, Math Assoc. Amer., Providence, 1960. [Bol] M. Botsko, Unified Treatment of Various Theorems in Elementary Analysis, Amer. Math. Monthly 94 (1987), 450452. [Bo2] M. Botsko, The Use of Full Covers in Analysis, Amer. Math. Monthly 96 (1989), 328333. [BBT] A. Bruckner, J. Bruckner and B. Thomson, Real Analysis, Prentice Hall, Upper Saddle River, N.J., 1997. [B V] P. Bullen and R. Vyborny, Some Applications of a Theorem of Marcinkiewicz, Canadian Math. Bull. 34 (1991), 165174. [C] Chew TuanSeng, Another proof that every absolutely Henstock integrable function is McShane integrable, Functional Analysis and Global Analysis, Proc. Conference, Manila, 1996, 5861. [CWT] Chew TuanSeng, Wong WoonKwong and Tan GeokChoo, On Absolutely Henstock Integrable Functions, Real Anal. Exchange 23 (1997/98), 799804. [Co] P. Cousin, Sur les fonctions de n variables complexes, Acta Math. 19 (1894). 149
150 [DSc] [DS] [FM] [GLZ] [Gi] [Gol] [Go2] [Go3] [Go4] [Go5] [Go6] [HI] [H2] [H3] [HS] [Ho] [HK] [Ku]
Bibliography R. Davies and Z. Schuss, A Proof that Henstock's Integral includes Lebesgue, J. London Math. Soc. 2 (1970), 561562. J. DePree and C. Swartz, Introduction to Real Analysis, Wiley, N.Y., 1987. J. Foran and S. Meinershagen, Some Answers to a Question of Bullen, Real Anal. Exchange 13 (1987/88), 265277. I. J. L. Garces, P. Y. Lee and D. Zhao, MooreSmith Limits and the Henstock Integral, Real Anal. Exchange 24 (1998/99), 447456. A. Gilioli, Natural ultrabornological, noncomplete, normed function spaces, Arch. Math. 61 (1993), 465477. R. Gordon, The Integrals of Lebesgue, Denjoy, Perron and Henstock, Amer. Math. Soc, Providence, 1994. R. Gordon, Another Look at a Convergence Theorem for the Henstock Integral, Real Anal. Exchange 15 (1989/90), 724728. R. Gordon, An Iterated Limits Theorem Applied to the Henstock Integral, Real Anal. Exchange 21 (1995/96), 774781. R. Gordon, Another Proof of Measurability (of S) for the Generalized Riemann Integral, Real Anal. Exchange 15 (1989/90), 386389. R. Gordon, The Use of Tagged Partitions in Elementary Real Analysis, Amer. Math. Monthly 105 (1998), 107117, 886. R. Gordon, Some Comments on the McShane and Henstock Integrals, Real Anal. Exchange 23 (1997/98), 329341. R. Henstock., Definitions of Riemann Type of Variational Integral, Proc. London Math. Soc. 11 (1961), 402418. R. Henstock, A Riemann Type Integral of Lebesgue Power, Canadian J. Math. 20 (1968), 7987. R. Henstock, Lectures on the Theory of Integration, World Scientific Publ., Singapore, 1988. E. Hewitt and K. Stromberg, Real and Abstract Analysis, SpringerVerlag, N.Y., 1965. K. Hoffman, Analysis in Euclidean Space, PrenticeHall, Englewood Cliffs, 1975. K. Hoffman and R. Kunze, Linear Algebra, PrenticeHall, Englewood Cliffs, N.J., 1971. Y. Kubota, A Note on McShane's Integral, Math. Japonica 1 (1985), 5762.
Bibliography [Kl]
151
J. Kurzweil, Generalized Ordinary Differential Equations and Continuous Dependence on a Parameter, Czech. Math. J. 82 (1957), 418449. [K2] J. Kurzweil, Nichtabsolut Konvergente Integrale, Tuebner, Leipzig, 1980. [LAI] J. Lamoreaux and G. Armstrong, A Necessary and Sufficient Condition for Gauge Integrability, Real Anal. Exchange 19 (199394), 254255. [LA2] J. Lamoreaux and G. Armstrong, The Fundamental Theorem of Calculus for Gauge Integrals, Math. Mag. 71 (1998), 208212. [LPY] Lee Peng Yee, Lenzhou Lectures on Henstock Integration, World Scientific Publ, Singapore, 1989. [LPY1] Lee Peng Yee, Measurability and the Henstock Integral, Proc. International Math. Conference 1994, Ed: Yuen Fong et al, World Scientific Publ., Singapore, 1996. [LW] Lee Peng Yee and Wang Pujie, Every Absolutely Henstock Integrable Function is McShane Integrable, J. Math. Study 27 (1994), 4751. [LCL] T. Lee, T. Chew and P. Y. Lee, On Henstock integrability in Euclidean space, Real Anal. Exchange 22 (199697), 382389. [LCL1] T. Lee, T. Chew and P. Y. Lee, Characterization of Multipliers for the Double Henstock Integrals, Bull. Austral. Math. 54 (1996), 441449. [LS] J. Lewis and O. Shisha, The Generalized Riemann, Simple, Dominated and Improper Integrals, J. Approx. Theory 38 (1983), 192199. [Lin] Lin YingJian, On the Equivalence of McShane and Lebesgue Integrals, Real Anal. Exchange 21 (1995/96), 767770. [L] Liu Genqian, The Measurability of S in Henstock Integration, Real Anal. Exchange 13 (1987/88), 446450. [Liu] Liu Genqian, On Necessary Conditions for Henstock Integrability, Real Anal. Exchange 18 (1992/93), 522531. [M] J. Mawhin, Analyse, 2nd Edition, DeBoeck Universite, 1997. [ML] R. McLeod, The Generalized Riemann Integral, Math. Assoc. Amer., Providence, 1980. [MSI] E. J. McShane, A Unified Theory of Integration, Amer. Math. Monthly 80 (1973), 349359. [MS2] E. J. McShane, Unified Integration, Acad. Press, N.Y., 1983. [Me] J. Mendoza, On Lebesgue Integrability of McShane Integrable Functions, Real Anal. Exchange 18 (1992/93), 456458.
152 [MO] [Nl] [N2] [Pe] [Pfl] [Pf2] [RN] [Ro] [Ru] [Sa] [Sari] [Sar2] [Sar3] [Sc] [Sch] [Sz] [Sh] [Shi] [Si] [Swl]
Bibliography P. Mikusinski and K. Ostaszewski, The space of Henstock integrable functions II, Lecture Notes #1419, SpringerVerlag, Heidelberg, 1990. I. Natanson, Theory of Functions of a Real Variable I, Ungar, N.Y., 1955. I. Natanson, Theory of Functions of a Real Variable II, Ungar, N.Y., 1960. I. Pesin, Classical and Modern Integration Theories, Acad. Press, N.Y., 1970. W. Pfeffer, The Riemann Approach to Integration, Cambridge University Press, Cambridge, 1993. W. Pfeffer, A Note on the Generalized Riemann Integral, Proc. Amer. Math. Soc. 103 (1988), 11611166. F. Riesz and B. Nagy, Functional Analysis, Ungar, N.Y., 1955. H. Royden, Real Analysis, MacMillan, N.Y., 1988. W. Rudin, Principles of Mathematical Analysis, McGrawHill, N.Y., 1964. S. Saks, Sur les fonctions d'intervalle, Fund. Math. 10 (1927), 211224. W. L. C. Sargent, On Linear Functionals in Spaces of Conditionally Integrable Functions, Quart. J. Math. 1 (1950), 288298. W. L. C. Sargent, On Some Theorems of Hahn, Banach and Steinhaus, J. London Math. Soc. 28 (1953), 438451. W. L. C. Sargent, On the Integrability of a Product, J. London Math. Soc. 23 (1948), 2834. A. Schurle, A Function is Perron Integrable if it has Locally Small Riemann Sums, J. Austral. Math. Soc. 41 (1986), 224232. S. Schwabik, Henstock's Condition for Convergence Theorems and Equiintegrability, Real Anal. Exchange 18 (1992/93), 190205. J. Schwartz, The Formula for Change of Variables in a Multiple Integral, Amer. Math. Monthly 61 (1954), 8185. P. Shanahan, Unified Proof of Several Basic Theorems of Real Analysis, Amer. Math. Monthly 79 (1972), 895898. O. Shisha, Tests of Existence of Generalized Riemann Integrals, Probability and Related Fields, Plenum Press, N.Y., 1994, 439441. M. Sion, Introduction to the Methods of Real Analysis, Holt, N.Y., 1968. C. Swartz, Measure, Integration and Function Spaces, World Scientific Publ., Singapore, 1994.
Bibliography [Sw2]
153
C. Swartz, An Introduction to Functional Analysis, Marcel Dekker, N.Y., 1992. [Sw23] C. Swartz, Even More on the Fundamental Theorem of Calculus, Proy. Revista Mat. 12 (1993), 129135. [ST] C. Swartz and B. Thomson, More on the Fundamental Theorem of Calculus, Amer. Math. Monthly 95 (1988) 644648. [Thl] B. Thomson, On Full Covering Properties, Real Anal. Exchange 6 (1980/81), 7793. [Th2] B. Thomson, Spaces of Conditionally Integrable Functions, J. London Math. Soc. 2 (1970), 358360. [Tr] F. Treves, Topological Vector Spaces, Distributions and Kernels, Acad. Press, N.Y., 1967. [V] R. Vyborny, KurzweilHenstock Absolute Integrable Means McShane Integrable, Real Anal. Exchange 20 (199495), 363366. [We] R. Winstock, Elementary Evaluations of f e~x dx, fo cosx2dx and /o°° sin x2dx, Amer. Math. Monthly 97 (1990), 3942. [YL] Ye Guoju and Lee Peng Yee, A Version of Harnack Extension for the Henstock Integral, J. Math. Study 28 (1995), 106108. [Za] A.C. Zaanen, Linear Analysis, NorthHolland, Amsterdam, 1956.
Index convolution, 90 countable additivity, 63 countably additive, 65 Covering Lemma, 65
a.e., 10, 45, 113 A = ///,100 Abel, 46 absolutely continuous, 109 absolutely integrable, 14, 19 additivity, 16 Alexiewicz norm, 73 almost everywhere, 10, 45, 113 AntosikMikusinski Matrix Theorem, 76 area, 81, 116
£ > « 7 , 5, 35, 82, 99, 116 DCT, 55, 109, 115 Dilation, 21 Dirichlet function, 7, 127 essentially bounded, 118 exponential order, 60 Ey, 94
B, 65 B(x,y), 28 barrelled, 74 Beppo Levi Theorem, 53 beta function, 28 Borel sets, 65 Bounded Convergence Theorem: BCT, 59 bounded variation, 129, 132 BV(I), 132 BV[a,b], 129
n 0 , 74 f*g{x), 90 /+,42 .T,42 Fatou, 114 Fatou's Lemma, 59 / A 5 , 42 ^,70 free tagged partition, 99, 116 Fresnel Integral, 46 FTC, 11 FTC; Part 1, 103 FTC; Part 2, 17, 104 Fubini, 93 Fubini Theorem, 87 / V 5 , 42
Cantor function, 110 Cauchy criterion, 15, 101 characteristic function, 15 Chebychev Inequality, 70 compound tagged partition, 82 conditionally integrable, 19 155
156 Fundamental Theorem of Calculus, 1, 103 Gamma Function, 56 7fine, 5, 35, 82, 99, 116 gauge, 5, 34, 82, 116 gauge integral, 4 General DCT, 69 General MCT, 68 generalized Riemann integral, 4 Gs, 67 Henstock, 23 Henstock's Lemma, 23, 101 HenstockKurzweil integral, 4 UK{T), 73 HK{iy, 77 improper integrals, 25, 36 integrable, 5, 35, 63, 83 integral test, 38 Integration by Parts, 111 integration by parts formula, 14 Integration by Substitution, 6, 112 interior, 81, 116 interval, 81, 116 7°, 81, 116 J « 1 , 23 C, 63 L 1 norm, 115 A, 135 A*, 135 Laplace Transform, 60 Lebesgue measure, 64, 65 Leibniz' Rule, 57 Linear Change of Variable, 21, 118 Lipschitz condition, 103 Lipschitz constant, 103
Index M, 64 M(I), 115 major, 122 major function, 125 McShane integrable, 100, 117 MCT, 53, 108, 114 measurable, 64 measure, 63, 65 M5, ms, 121 minor, 122 minor function, 125 Monotone Convergence Theorem, 53 multiplier, 44 multiplier problem, 118 null, 10, 112 Pi, 82 A , 82 partial tagged partition, 23 partition, 33, 34, 82, 99, 116 Perron integrable, 125 Perron integral, 124 7T4[a,x], 121 Riemann complete integral, 4 Riemann integrable, 127 Riemann sum, 4, 83, 99, 116, 127 RiemannLebesgue Lemma, 60 /a6/, 5 s\f,V), 4, 83, 116 Sargent space, 79 aalgebra, 65
fKfjf(x'y}dxdy'87 step function, 15, 84 Straddle Lemma, 2 tag, 4, 34, 82, 116 tagged partition, 4, 33, 34, 82
Index Tonelli, 95 Tonelli Theorem, 88 total variation, 131 translation invariance, 63 Uniform Henstock Lemma, 28 uniformly gauge Cauchy, 54 uniformly integrable, 49 uniformly McShane Cauchy, 108 uniformly McShane integrable, 106 v(I), 81, 116 Var(f: [a, 6]), 129 var(f : TT), 129 variation, 129
Vf(t), 131 Vitali covering, 135 Vitali Covering Theorem, 135 volume, 81, 116 Vyborny, 124 ysection, 94
INTRODUCTION
TO
Gauge Integrals by C Swartz This book presents the Henstock/Kurzweil integral and the McShane integral. These two integrals are obtained by changing slightly the definition of the Riemann integral. These variations lead to integrals which are much more powerful than the Riemann integral. The Henstock/Kurzweil integral is an unconditional integral f o r which the fundamental theorem of calculus holds in full generality, while the McShane integral is equivalent to the Lebesgue integral in Euclidean spaces.
A basic knowledge of introductory real analysis is required of the reader, who should be familiar with the fundamental properties of the real numbers, convergence, series, differentiation, continuity, etc.
ISBN 981024239^51
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