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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( 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 / 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 «
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) R be continuous and lim^oo /(£) = L. What can you say about linifc JQ f(kt)dt? 20. Evaluate 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(• 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 + 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(k be the closed subintervals of J obtained by dividing / into 2k equal subintervals (i.e., bisect I and keep bisecting). Let E\ be the family of J € P i such that there exists t G E n J with J C 7(t). Let £2 be the family of all J e V2 such that J is not contained in an element of £\ and there exists t £ E D J with J C j(t). Continuing this procedure produces a sequence (£k) of families of closed subintervals of / (some £k may be empty). Then £ = (JfcLi £fc i s a n a * most countable family of nonoverlapping, closed intervals of J. We claim that E C U{ J : J G £}. Let t € E. There exists an integer fci such that if k > k\, then the closed interval Jk it) of T>k containing
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
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}
70
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
\(A) CA, 0 < fk < 1.]
Integration over More General Sets: Lebesgue Measure
71
10. Let 7 be a bounded interval and / : 7 —• R absolutely integrable over 7. Show that ft  /  = 0 if and only if f(t) = 0 for all t € 7 except those in a null subset. [Hint: {t :  / ( t )  ^ 0} = \J™=1{t : /(i) > 1/fc}; use Chebychev's Inequality.] 11. Let 7 be a closed interval in R* and suppose / : 7 —> R is absolutely integrable over every bounded subinterval of 7. Set E = {t 6 7 : f(t) ^ 0}. Show that E is measurable. [Hint: Let J be a bounded subinterval of 7 and set fk = (k\f\) 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
74
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°
75
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 /()tf(/) =
E EtffoW1*)1^7*)} E {/(**) W  H(Jfc)}
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
76
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
77
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).
78
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)
• 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
82
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 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) > £ vW = V24fc XjEJk
(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)
• 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 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 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 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 : • 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 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,£)\ R be McShane integrable over I and c > 0. Then c A / and (—c) V / are McShane integrable over I. Proof: Let g(t) = c A t for t G R. Then g satisfies a Lipschitz condition on R (Exercise 8). Thus, Theorem 7 is applicable and gives that c A / is integrable over I. The other part is similar using the function t > (—c) V t. We next consider the Fundamental Theorem of Calculus (FTC) for the McShane integral. First, we need a result which is just a rephrasing of the FTC for the gauge integral (Theorem 1.5). Lemma 9. Let f : [a, b] —>• R be differentiable on [a, b] and e > 0. There exists a gauge 7 on [a,b] such that \S{f ,V) — (f(b) — / ( a ) )  < £ whenever V is a "ffine (gauge) tagged partition of\a,b]. Theorem 10 (FTC; Part 1). Let f : [a, b] > R be differentiable on [a:b] and assume that / ' is McShane integrable over \a,b]. Then Ja f = f(b) — f(a). Proof: Let e > 0. There exists a gauge 71 on [a, b] such that
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))\ 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 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• 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 9k t • M be differentiable in [a,b] and assume that f and g' are McShane integrable over [a,b]. Then fg and fg' are McShane integrable over [a,b] with J fg' = fg\ba — f fg. Proof: / and g are bounded and continuous and are, therefore, McShane integrable. By Theorem 29 fg' and fg are McShane integrable. Since (fg)' = fg + fg': (fg)' 1S McShane integrable and Theorem 10 gives / (fg)' = fg\ha =
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
/ 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