The Do/ciani Mathematical Expositions NUMBER THIRTY-ONE
A Garden of ~ntegra~s
Frank E. Burk California State University at Chico
Published and Distributed by The Mathematical Association of America
© 2007 by The Mathematical Association of America (lncolporated) Library of Congress Catalog Card Number 2007925414 ISBN 978-0-88385-337-5
Printed in the United States of America Current Printing (last digit): 10 9 8 7 6 5 4 3 2 1
To Don Albers
for all his years of dedicated service to the MAA
foreword From quadratures of lunes to quantum mechanics, the development of integration has a long and distinguished history. Chapter 1 begins our exploration by surveying some of the historical highlights-milestones in man's capacity to think rationally. Whether motivated by applied considerations (areas, heat flow, particles in suspension) or aesthetic results such as
..!.. = rr
~ (411)! 1103 + 2639011 9801 L....J (11 !)4 3964n '
.J8
11=0
the common thread has been and will continue to be understanding. Mathematical discoveries are but markers in our quest to understand our place in this universe. In the profession of mathematics, we are all too frequentlynumb led, but we persevere for those rare moments of exhilaration that accompany understanding. That is the nature of mathematics-indeed, the nature of understanding. I hope~that you will personally experience the emotional peaks and valleys that are presented in the pages that follow. If we attain a measure of understanding and an appreciation of OUr mathematical ancestors and their accomplishments, our efforts will have been successful. We will have played a role in our inexorab Ie journey to the stars. -F.B.
ix
Contents Foreword 1
ix
1
An Historical Overview 1.1
Rearrangements
.....
1
1.2
The Lune of Hippocrates .
2
1.3 1.4 1.5
Eudoxus and the Method of Exhaustion . . . . Archimedes I Method . . . . . . . . . . .
4
6
Gottfried Leibniz and Isaac Newton . . .
8
1.6
Augustin-Louis Cauchy ..
1. 7
Bernhard Ri.emann
1.8
]homas Stieltjes . . . . . . . .
12 14
1.9
Henri Lebesgue.
15
11
..... . . . . . . . . . . . .
..
...... .
1.10 The Lebesgue-Stieltjes Integral.
18 19
. . . .
1.11 . Ralph Henstock and Jaroslav Kurzweil . . . . 1.12
Norbert Wiener
1.13 Richard Feynman . .. 1.14 2
References...
22 26 27
... .. . .
..... .
The Cauchy Integral 2.1
Exploring Integration
2.2
Cauchy's Integral.
2.3
Recovering Functions by Integration . .
2.4
Recovering Functions by Differentiation .
2.5
A Convergence Theorem .
2.6
Joseph Fourier. . ..
2.7
. ..
.
..
...
29 29 32
. ...... .
35
. .
..
37
38 . . • •.
40
P. G. Lejeune Dirichlet . . . . . .
. . . . ..
41
2.8
Patrick Billingsley's Example
. . . . ..
43
2.9
Summary...
. . . ..
44
2.10
References..
..
. ....
. . .
44
xi
xii 3
A Garden of Integrals
The Riemann Integral 3.1 Riemann's Integral . . . . . . . . . . . . . . . . . . . .. 3.2 Criteria for Riemann Integrability . . . . . . . . . . .. 3.3 Cauchy and Darboux Criteria for Riemann Integrability 3.4 Weakening Continuity . . . . . . . . . . . . . 3.5 Monotonic Functions Are Riemann Integrable .. 3.6 Lebesgue's Criteria . . . . . . . . . . . . . . 3.7 Evaluating a la Riemann . . . . . . . . . . . . 3.8 Sequences of Riemann Integrable Furtctions . The Cantor Set (1883) . . . . . . . . . . . 3.9 3.10 A Nowhere Dense Set of Positive Measure 3.11 Cantor Functions . . . . . . . . . . . . . . 3.12 Volterra's Example . . . . . . . . . . . . . 3.13 Lengths of Graphs and the Cantor Function . 3.14 Summary . . . . . . . . . 3.15 References.........
45 45 47 49 52 56
58
60 63 65 67
69 70 71
72 73
4
The Riemann-Stieltjes Integral 4.1 Generalizing the Riemann Integral Discontinuities . . . . . . . . . . . . . . . 4.2 Existence of Riemann-Stieltjes Integrals .. 4.3 4.4 Monotonicity of ¢ . . . . . . . . . . . . . . . 4.5 Euler's Summation Formula . . . . . . . . 4.6 Uniform Convergence and R -S Integration References . . 4.7
75 75 77 79 80 82 83 84
5
Lebesgue Measure 5.1 Lebesgue's Idea. 5.2 Measurab Ie Sets . 5.3 Lebesgue Measurable Sets and Caratheodory Sigma Algebras . . . . . . . . . 5.4 5.5 Borel Sets . . . . . . . . . . . . 5.6 Approximating Measurable Sets 5.7 Measurable Functions . . . . . . 5.8 More Measureable Functions . . What Does Monotonicity Tell Us? 5.9 5.10 Lebesgue's Differentiation Theorem 5.11 References..............
85 85 86 89 91 93 94 97 100 104 107 109
Contents
6
7
8
9
The Lebesgue Integral 6.1 Introduction... 6.2 IntegrabiHty: Riemann Ensures Lebesgue I . 6.3 Convergence Theorems . . . . . . . . . . . 6.4 Fundamental Theorems for the Lebesgue Integral. 6.5 Spaces . . . . . . . . . . . . . . . . . . . . . . . . 6.6 L2[-rr. rr] and Fourier Series . . . . . . . . . . . . . 6.7 Lebesgue Measure in the Plane and Fubini's Theorem 6.8 Summary . . . . . . . . . References . . . . . . . . . 6.9
xiii
111 111 115 120 127 136
148 151 152 153
The Lebesgue-Stieltjes Integral 7.1 L-S Measures and Monotone Increasing Functions 7.2 Caratheodory's Measurability Criterion. . . . . 7.3 Avoiding Complacency . . . . . . . . . . 7.4 L-S Measures and Nonnegative Lebesgue Integrable Functions . . . . . . . . . 7.5 L-S Measures and Random Variables . . 7.6 The Lebesgue-Stieltjes Integral . . . . . 7.7 A Fundamental Theorem for L-S Integrals '" 7.8 Reference . . . . . . . . . . . . . . . . . .
155 155 158
The Henstock-Kurzweil Integral 8.1 . The Generalized Riemann Integral . 8.2 Gauges and a-fine Partitions . . . . 8.3 H-K Integrable Functions. . . . . . 8.4 The Cauchy Criterion for H-K Integrability .. 8.5 Henstock's Lemma . . . . . . . . . . . . . . 8.6 Convergence Theorems for the H-K Integral 8.7 Some Properties of the H-K Integral. 8.8 The Second Fundamental Theorem. 8.9 Summary... 8.10 References...
169
The Wiener Integral 9.1 Brownian Motion . . . . . . . . . . . 9.2 Construction of the Wiener Measure . 9.3 Wiener's Theorem . . . 9.4 Measurable Functionals . 9.5 The Wiener Integra] . . .
160 161 164 165
166 167 170
175 176 183 187 189
191 198 203 204
205 205 209
215 220 222
xiv
A Garden of Integrals
9.6 9.7 9.8
Functionals Dependent on a Finite Number of t Values .. 227 Kac's Theorem . . . . . . . . . . . . . . . . . . . . . . . 232 References.......................... 234
10 The Feynman Integral 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10 10.11
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . Summing Probability Amplitudes . . . . . . . . . . . . . A Simple Example . . . . . . . . . . . . . . . . . . . . . The Fourier Transform . . . . . . . . . . . . . . . . . . . The Convolution Product . . . . . . . . . . . . . . . . . . The Schwartz Space . . . . . . . . . , . . . . . . . . . . Solving SchrOdinger Problem A . . . . . . . . . . . . . . An Abstract Cauchy Problem. . . . . . . . . . . . . . . . Solving in the Schwartz Space . . . . . . . . . . . . . . . Solving SchrOdinger Problem B . . . . . . . . . . . . . . References. . . . . . . . . . . . . . . . . . . . . . . . . .
235 235 237 240 244 245 246 250 252 254 263 277
Index
279
About the Author
281
CHAPTER
1
Every one of liS is touched in some way or othel" by the problems of mathematical communication. EvelY one of ZlS can make some contributio12, great or small, within his own proper sphere of activity. And evelY contribution is needed if mathematics is to grow healthily and Ilsefully and beautifully - E. J. McShane
1.1
Rearrangenlents
Figures lea) through Cd) demonstrate the general idea of rearranging a given area to form another shape. In the first example, we have a circle rearranged into a parallelogram by a method that has been known for hundreds of years. Figure 1(e) represents a different manipulation of area called scaling where, despite enlargement or shrinkage, shape and proportion are retained.
21rr
M
21rr
(a)
1
2
A Garden of Integrals
x 2 +1
(b) x 2 +x +1
xl+l
(c)
x 2 +1
x+l
.(d). hla
~
(e)
-b (n:a 2) a
Figure 1. Examples of rearrangements (a-d) and scaling (e) of areas
1.2
The Lune of Hippocrates
Hippocrates (430 B.C.E.), a merchant of Athens, was one of the first to find the area of a plane figure (lune) bounded by curves (circular arcs), The crescent~shaped region whose area is to be determined is shown in Figure 2.
In the figure, ABC and AFC are circular arcs with centers E and D~ respectively. Hippocrates showed that the area of the shaded region bounded by the circular arcs ABC and APC is exactly the area of the shaded square
3
An Historical Overview B
D
Figure 2. Hippocrates' lune whose side is the radius of the circle. The argument depends an the fallowing assumption, illustrated in Figure 3: The areas of the two circles are to each other as the squares of the radii.
Figure 3. Hippocrates' assumption From this assumption we reach twa conclusions: 1. The sectors of twa circles with equal central angles are to each other as the squares of the radii (Figure 4).
Figure 4. Our first conclusion 2. The segments of twa circles with equal central angles are to each ather as the squares of the radii (Figure 5).
Figure 5. Our second conclusion
4
A Garden of Integrals
Hippocrates' argument proceeds as follows (refer to Figure 6). From our second conclusion, Al/A4 r 2 J(..j'2r? ~. Hence Al ~A4 and A2 = tA4 so Al + A2 = A4.
=
Area of the lune
=
=
= Al + A2 + A3 = A4 + A3 = Area of the triangle = ~ ( .J2r)(.J2r) = r2 = Area of the square.
Figure 6. Area of the lune Similar reasoning may be used for the three lunes in Figure 7.
Figure 7. Lune exercises
1.3
Eudoxus and the Method of IElchaustion
Eudoxus (408-355 B.C.E.) was responsible for the notion of approximating curved regions with polygonal regions. In other words, "truth" for polygonal
An Historical Overview
5
regions implies "truth" for curved regions. This notion will be used to show that the areas of circles are to each other as the squares of their diameters, an obvious result for regular polygons. "Truth" is to be based on Eudoxus' Axiom. Axiom 1.3.1 (Axiom of El.ldoxus). Two unequal magnitudes being set out, iffrom the greater there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continuously, there will be left some magnitude that will be less than the lesser magnitude set out.
In modem terminology, let M and E > 0 be given with 0 < E < M. Then form M, M -rM (1- r)M, (l-r)M - r(1-r)M (1-r)2M, ... , where! < r ::: 1. The axiom tells us that for n sufficiently large, say N, (1 - r)N M < t:, a consequence of the set of natural numbers not being bounded above. To get back to what we are trying to show, let c, C be circles with areas a, A and diameters d, D, respectively. We want to show that a/ A = d 2/ D2, given that the result is true for polygons and given the Axiom of Eudoxus. Assume that a/ A > d 2/ D2. Then we have a* < a, so that 0 < a - a* and1a*/ A ..,= d 2/ D2. Let t: < a-a*. Inscribe regular polygons of areas Pn, Pn in circles c, C and consider the areas a - Pn, A - Pn • (See Figure 8.) Now, double the number of sides. What is the relationship between a - pn and a - P2n? See Figure 9. Certainly a - p2n < ~(a - Pn). We subtract more than half each time we double the number of sides. From the Axiom ofEudoXlls, we may determine N so that 0 < a - PN < t: < a - a*; that is, we have a regular inscribed polygon of N sides, where area PN > a*.
=
=
", 0 ".
,
.
,
Figure 8. Inscribed polygons
a-PI!
a -PlI1
Figure 9. Segments
6
A Garden of Integrals
But PN/PN d 2 /D2, Since a""/A d 2/D2; we have PN/PN = a"" / A, so PN > A. This C8nnot be: PN is the area of a polygon inscribed in the circle C of area A. A similar argUment shows that a/ A cannot be less than d 2 / D2: double reductio ad absurdum.
=
=
1~4 Archimedes' Method The following masterpiece of mathematical reasoning is due to one of the greatest intellects of all time, Archimedes of Syracuse (287-212 B.C.E.). Archi~edes shows that the area of the parabolic segment is ~ that of the inscribed triangle ACB. See Figure 10. A
Figure 10. Archimedes' triangle In our discussion, we will use the symbol 6. to denote "area of," with Xc = (XA + xB)/2). The argument proceeds as follows. In Figure 11, the combined area of triangles ADC and BEC is one-fourth the area of triangle ACB; that is, 1 8ADC + 8BEC = '46.ACB. Repeating,the process, trying to "exhaust" the area between the parabolic curve and the inscribed triangles, we have Area of the parabolic segment
1
1[1
]
= 6ACB + '-(6ACB)
+ -4 -4 (6.ACB) + ...
= 6ACB ( 1 + -1 + -12 4 4
+ '"
4
) = -6.ACB 4 3'
E
Figure 11. Inscribed triangles
7
An Historical Overview
B
D
C
Figure 12. We argue that .6.ADC + .6.BEC = i .6.ACB for the parabola y = ax 2 , a > 0; see Figure 12. S how that the tangent line at C is parallel to AB and that the vertical line through C bisects AB at P. It follows that .6.BEC = i.6.BCP (complete the parallelogram; see Figure 13).
Figure 13.
We note that 1. .6.CEG
= .6.BEG (equal height and base),
2. .6.HGB = i.6.BCP. Thus, we must show that
1 .6.CEG + b.BEG = .6.HGB or .6.BEG = 2".6.HGB. This will be accomplished by showing that FE FE = a
1
Xc (
+2 XB
)
2
2
-
= -a (XB - Xc) • 4
[
2
aXe
= i FH = iQB. Since
+ 2aXe x "21 (X B -
Xc) ]
A Garden of. Integrals
8
B
D
Figure 14. and QB
= aX~ -
[aXE
+ 2aXc (XB - xc)]
= a (XB -
XC)2,
we are done. Show that the area of the parabolic segment is ~ the area of the circumscribed triangle ADB formed by the tangent lines to the parabola at A and B with base AB (Ee = CD). See Figure 14.
1.5
Gottfried Leibniz and Isaac Newton
During the seventeenth and eighteenth centuries the integral was thought of in a descriptive sense, as an ant,iderivative, because of the beautiful Fundamental Theorem of Calculus (FTC)~ as developed by Leibniz (16461716) and Newton (1642-1723). A particular function f on [a b1was integrated by finding an antiderivative F so that F' = f, or by finding a power series expansion and using the FTC to integrate termwise. The Leibniz-Newton integral of J was F(b) - F(a);that is, I
t
where F~ =
1(;C) dx = F(b) - F(a),
J.
We will give an argument of Leibniz and a result of Newton to illustrate their geniuses.
1.5.1
leibniz's Argument
Leibniz argued that n
1
I
1
-=1--+---+··· 4 3 5 7 .
An Historical Overview
9
c
Figure 15. Take the quarter circle (x - 1)2 + y2 = 1, 0 < x ::: I, whose area is 1[14. (See Figure 15.) Leibniz detennined the area of the circular sector in Figure 15 by dividing it into infinitesimal triangles OAB (where A and B are two close points on the circle) and summing. So, how to estimate the area of ,6.0AB? Construct the tangent to the circle at A, with a perpendicular at C passing through the origin. Then ~OAB ~ !AB Oc.. By similar triangles, ABI dx = zloc, so ,6.0AB = dx. Observe that
x
tz
x
That is, x
= 1-
cos e = 2 sin2
= 2x2 1(1 + z2). xz
~2
e
z = tan-.
and
2
Leibniz knew that
=
J
zdx+
J
xdz.
Xl-------~
z
Figure 16.
10
A Garden of Integrals
Hence the area of the circular sector in Figure 15 is
f ~Z = H I! - f dx
xz
x dZ]
=
HI -f
~:2 dZ]
/,1 z2(1 - z2+ z4_ ... ) dz
= -1 -
2
.
(long division)
0
1
1 1 --2-3 + 5-7 + ... and by adding
1
1
(integrating termwise)
t to both sides, NIl
1
"4=1-"3+5-7+· .. · By the way,
1.5.2 Newton's Result Newton tells us that
1
1C
1
1
1
1
-=1+-----+-+--···. 4.J2 3 5 7 9 11 Since Newton routinely integrated series termwise, and since 2
1+ x 1 + x4
= (1 + x 2 )(1 -
x
4
+ x8 -
2
4
... ) = 1 + x - x - x
6
we have 2
1+ x J0 1 + X4 {I
dx = 1 +
~ _ ~ _ ~ + ~ + ~ _ .... 3
5
7
9
11
The argument may be completed by observing that
1++ x ="21(1
1)
2
1
x4
1 - ../2x + x 2 + 1 + .J2x + x 2
and evaluating the appropriate integrals with the substitution
.J2
1
x+ 2: = .J2tane.
+"',
An Historical Overview
1.6
11
Augustin-louis Cauchy
Cauchy (1789-1857) is considered to be the founder of integration theory. In 1823 Cauchy fonnulated a constructive defmition of an integral. Given a general function I on an interval [at b], in contrast to ax2, (1 + x 2)/ (1 + X4), and so on, partition the interval [at b] into subintervals [,,'tk-l. Xk], with a = Xo < Xl < ... < Xn-l < Xn = b. and form the sum
See Figure 17.
a =,;t'o
X"_I
XII
=b
Figure 17. Cauchy's integral
The integral of Cauchy was to be the limit of such sums as the length of the largest subinterval, II6.x II, approaches zero:
1
n
b
C
a
I(x) dx =
lim
IIL\xll-+O
L I(Xk-l){Xk -
Xk-l).
k=l
Cauchy argued that, for continuous functions this limit always exists. As for an evaluative procedure, recovering a function from its derivative (a fundamental result), we have C
lb
F'(x)dx = F(b) - F(a)
for any function F with a continuous derivative. For example, let F(;c) =
1x
3
sin{:rc/x)
o
x
=F 0,
x =0.
Then F'(x) =
l
-rexcos(:rc/X)+3x 2 sin(1f/X) 0
x..J.O, -r x =0
12
A Garden of Integrals
is continuous on the interval [0, 1]. Consequently
c
f
F'(x) dx = F(l) - F(O) = O.
Apparently a finite number of "jump" discontinuities would not cause difficulties. How about a countable number, or even a dense set, of jump discontinuities? Just how discontinuous can a function be and still have an integral?
1.7
Bernhard Riemann
Riemann (1826-1866), having investigated Fourier series, convergence issues, and Dirichlet-type functions (Ion the rationals, 0 on the irrationals, for example), was motivated to develop another constructive definition of an integral (1854). Beginning with a bounded function f on the interval [a, b], we partition (a la Cauchy) into subintervals [Xk-l, Xk], where a = Xo < Xl < ... < Xn = b. Next we "tag" each subinterval with an arbitrary point Ck., where Xk-l < Ck < Xk, and we fonn the sum
See Figure 18.
Figure 18. Riemann's integral Whereas the tag Ck was the left-hand endpoint in the Cauchy definition (Ck = Xk-l), in Riemann's definition we have more variability. Again, as the length of the largest subinterval approaches zero, the limit yields the Riemann integral:
An Historical Overview
13
In 1902 Lebesgue showed that for bounded functions, continuity is both necessary and sufficient for the existence of the Riemann integral (with the possible exception of a set of measure 0). For example, the function f(x} = { l/q
o
x = p (q (PI q relatively prime natural numbers), otherwIse,
is continuous on the irrationals, discontinuous on the rationals, and thus l Riemann integrable. In fact, RJo f(x) dx = O. Riemann also constructed a function with a dense set of discontinuities that was Riemann integrable. (See Exercise 3.4.3.) Every Cauchy integra~le function is Riemann integrable and has the same value. We have a more general Fundamental Theorem of Calculus for recovering a function from its derivative. Theorem 1.7.1 (General FTC for Riemann Integrability). The integral R
t
F'(x)dx = F(b) - F(a)
for any function with a derivative that is bounded and continuous almost evelywhere. For example, F(x} =
1x
2
Sino(rr/x)
x =F 0,
x=o,
has the derivative F'(x) = 1-1CCOS(1C/X)0+ 2xsin(rr/x)
x =F 0, x=O,
bounded and continuous except at x = O. Thus, the Riemann integral of F' exists and
R
f
F'(x) dx = F(l) - F(O) = O.
In 1881 Vito Volterra gave an example of a differentiable function with a bounded derivative that was discontinuous on a set of positive measure and thus not Riemann integrable. (See Section 3.12.) Again, the function x 2 sin(rr / x), modified on a Cantor set of positive measure, sufficed. This example prompted Lebesgue to develop an integral to remedy this defect. It turns out that
L
t
F'(x) dx = F(b) - F(a)
14
A Garden of Integrals
for a differentiable function F with a bounded derivative. More about this later.
1.8 Thomas Stieltjes Stieltjes (1856-1894) was interested in mathematically modelling mass distributions on the real line. Suppose we have point masses distributed as indicated in Figure 19. If ¢(x) denotes the total Plass less than or equal to x, then the graph of ¢ appears as shown in Figure 20. In general, ¢ is a non decreasing function. .."
x /I =6
Figure 19. Masses on the real line
ml+"'+mn o ml + ... + m,,_l o
0
ml
0,---';"--0
I
a=xo
I XII =b
Xl
Figure 20. Mass distribution Now consider the moment of such a mass distribution. The u mass" of [Xk-1. Xk] is t/J(Xk) - t/J(Xk-1) with "arm" Ck, for Xk-1 ~ Ck ~ Xk. This leads to sums of the form
+ C2nl2 + ... + Cnnl" or Cl[t/J(Xl) - t/J(xo)] + ... + cn[t/J(xn) -
q1111
t/J(Xn-l)].
More generally, Stieltjes was led to consider sums - Riemal2l2-Stieitjes slims -like f(cl)ml + ... + f(c,,)I17. n . We have a "weighted" sum. The value of the function at CI, f(cI) is weighted by m1; ... ; the value of the
An Historical Overview
15
function at cn, l(cl1), is weighted by mil' The average would be
L leCk) [rp eXk) -
rp (Xk-l )] [rp(b) - rp(a)}
L I(Ck)rp'(;k)(Xk -
f:
Xk-l)
~ ~----~----------
f:
rp'(x) dx
I(x)rp'(x) dx
~ ~--:b"-------
fa rp'(x) dx for "nice" I and rp. What conditions may we impose on I and
t
f(x)dq,(x)
=R
t
f(x)q,'(x)dx.
The Riema.nn...integral makes sense-.foLfimctions_that are_bounded and_con':.-_ tinuous almost everywhere. So, I continuous and rp' Riemann integrable should work - and it does: see Theorem 4.3.l. Do we have anything new here? Formally, R-S
f.
b
I(x)drp(x)
a
=
n
lim !lfi;tn-t-Q
L I(Ck) [rp (Xk) -
rp(Xk-l)].
1
We can show this limit makes sense for I continuous and rp monotone (Theorem 4.4.1). Of course, functions of bounded variation are differences of two monotone functions, so it is true for rp a function of bounded variation. Another question: Does the series L cos(..fii)/n converge? This is an amusing application of the Euler Summation Formula (Section 4.5). The Riemann-Stieltjes integral is very convenient for step functions. By the way, evaluate R f:-l [x - (k - 1) - !]/' (x) dx by parts (for 2 < k < 11), sum the resu1ts, and note that x - [x] - ~ differs from x - (Ie - 1) at a finite number of points. Try I(x) = In x for a n "Stir1ing result.
t
1.9
Henr~
Lebesgue
Where to begin? The Lebesgue integral has affected many areas of mathematics during the past century. Let's begin as Henri Lebesgue (1875-1941)
16
A Garden of Integrals
did, with Volterra's example of a function with a bounded derivative that was not Riemann integrable (see Section 3.12), for it was this example that prompted Lebesgue to develop an integral (1902) that would recover any function from its bounded derivative. That is, L fol F'(x) dx = F(b)-F(a) should be true whenever the derivative F' is bounded. Lebesgue's integral construction was fundamentally different from his predecessors. His simple, but brilliant, idea was to partition the range of the function rather than its domain. Assume a < , < fJ on the interval [a. b]. In place of a = Xo < Xl < ... < Xn b, we have Q! = Yo < Yl < ... < Yn fJ. The sets
=
=
1-1 ([Yk-l. Yk» = {X
E
[a, b] I Yk-l ~ I(x) < Yk}
are disjoint with union [a. b]. Disregarding the empty sets (relabelling if necessary), pick a tag (point Ck) in each nonempty set, and fonn the sum (motivated by areas of rectangles as the height times the length of the base) as follows (see Figure 21): I(cl)-{length of
,-1
([Yo, Yl)}+"'+ f(clI)·{length of
,-1
([Yn-li Yn»)}.
We then have
LYk-l . {length of f- 1 ([Yk-l. Yk»} < L
,-I ,-I
I(ck) . {length of
< LYk . {length of
(fYk-l, Yk»)}
([Yk-l. Yk»)}
Figure 21. Lebesgue's integral construction
17
An Historical Overview
and
What do we mean by the '"length" of 1-1 ([Yk-I, Yk))? Partitioning the range forces us to assign a length, or measure, to possibly unusual sets. For example, suppose we are dealing with the Dirichlet function on the interval [0, 1] (assign a functional value of 1 whenever x is irrational and a value of -1 whenever x is rational). A partitioning of the range, say -l~, 1~, would compel us to assign a length to the sets of rational, I-I([-I~, -t)), and irrational 1- 1 ([!.lt)), numbers in the interval [0, 1]. Both sets, being subsets of [0, 1], should have length less than or equal to 1. Since their union is the interval [0, 1], the sum of their lengths should be 1. Let's see, we could enumerate the rationals, rl, 1'2 • ••• , rll • ••• , and cover each rational with an interval (rll - E/211 , 1"11 + E/2"), which covers all the rationals with an open set of length less than f. Fine; the rationals will have length 0, the irrationals will have length 1. It happens that this Dirichlet function is Lebesgue integrable (not Riemann integrable) and
-!, !,
1.-
L a
I (x) dx
=
[1 . (length of irrationals in [0, 1])]
/.
+ [-
1 . (length of rationals in [0, 1])]
=1. What should the "length" of the numbers in [0, I} without a 5 in their decimal expansion be? How about the Cantor set? Removing intervals would suggest that the Cantor set, even though uncountable, has measure O. A "measure" theory must be developed that is logical and consistent. This integral of Lebesgue, if it is to have any power, suggests that we should be able to measure most sets of real nurribers. Through the efforts of Jordan, Borel, Lebesgue, and Caratheodory, to name a few, a nonnegative countably additive measure was developed - the Lebesgue measure - that measures, in partiCUlar, all Borel sets of real numbers (countable unions of countable intersections of ... open sets). In short, nonmeasurable sets are difficult to construct, so sets without "length" will seldom occur. So, we want 1-1 ([Yk-l, Yk)) to be a measurable set. We want inverse images of intervals to be measurable. We therefore define a measurable
A Garden of Integrals
18
function to be a function for which inverse images of intervals are measurable. Again, "most" functions are measurable, and because the limit of a sequence of measurable functions is measurable, we have some beautiful convergence theorems. For example, enumerate the rationals in the interval [0, 1], and define a sequence of measurable and Riemann integrable functions {lie} by f,,(x) = 1, x = Tl. '2 •... , rle, and lie (x) = 0 otherwise. This sequence is nonnegative, monotone increasing, uniformly bounded by 1; lim lie is a Dirichlet function, and thus it is not Riemann integrable: clearly.limRJ: Ik(X) dx = 0, and R
J:
lim Ik (x) dx is not defined, but
FinallY1 in answer to Volterra, all functions with a bounded derivative 1 are Lebesgue integrable and L J0 F' (x) dx = F(b) - F(a). (See Thea-rem 6.4.2.)
1.10 The lebesgue-Stieltjes Integral The construction of Lebesgue measure begins with the assignment of a measure to an interval, namely its length: the measure of (a, b] is b - a. Just as in the Riemann-Stieitjes integral, where we weighted the interval (a, b} by tfJ(b) - t:/J(a), a particularly fruitful approach to the construction of Lebesgue--Stieltjes measure is to assign a measure of t:/J(b) - tfJ(a) to the interval (a. hJ where tP is a nonnegative, monotone increasing, right~ continuous function on the reais, with lim tfJ(x)
x~-~
= 0,
lim tfJ(x)
X~+OO
= 1.
It turns out, just as with ordinary Lebesgue measure, that a nonnega~ tive, countably additive measure, JL¢, is generated on the Borel sets of real numbers. From this so~cal1ed Lebesgue-Stieltjes measure, we proceed to measurable functions and Lebesgue-Stieltjes integrals. For example, if
x < 0, 0<x:51,
1 < x,
19
An Historical Overview
and if J is the Dirich1et function, what is L-S JR J(x)df.Lr/J? It would be helpful if
L-S
Lf(x)dp.~ la' = L
f(x)(x')' dp.
=L
la' 1.
= L
la' 1. 2xdx = 1.
= L
f(x)2xdx 1 . 2xdx + L
irrationals
1
rationals
o· 2xdx
(See Theorem 7.7.1.)
As it happens, the Lebesgue-Stieltjes integral is crucial in probability.
1.11
Ralph Henstock and
Jaros~av
l(urzweU
Working independently, Henstock (1923-) and Kurzweil (1926-) discovered the generalized Riemann integral in 1961 and 1957, respectively. Their discovery, which is referred to as the H-K integral, is an extension of the Lebesgue integral. All Lebesgue integrable functions are H-K integrable functions, to the same value. What's more, there are HwI{ integrable functions that are not Lebesgue integrable. If a function is Lebesgue integrable then its absolute value must be Lebesgue integrable. Consider the function F (x) = 1
X2
sin(x / x 2 ) 0
x
1= 0,
x = 0,
and its derivative FICx) =
-27T / X cos(1l' / x 2 ) 0
1
+ 2x sin(iT / x 2 )
The Lebesgue integral of [F'[ does not exist.
x
1= 0,
X
=0.
20
A Garden of Integrals Consider the intervals [.J2/(4lc
L
f
+ 3), .j2/(4k + 1)]:
!F'(x)1 dx ?:
t {k 1
IF'(x)1 dx >
t {k 1
Ilk
F'(x)dx
tlk
n
==
L IF(bk) -
F(ak)1
1
[2 . (4k 2+ 1) 4k + 1
==
~
>
~ [ ( 4k ~ 1) (1) -
11
11
SIn
(4k
1C -
2.
4k + 3 sm
(4k
+
2
3) J 1C
~ 3) (-1)]
1
~Lk+l' 1 It turns out that every derivative is H-K integrable. Thus in this example H-K!ol F'(x) dx = F(l) - F(O) = O. (Think about the graph of FJ.) This very powerful integral results from an apparently simple modification of the Riemann integral construction. Rather than partitioning the interval [a, b] into a collection of subintervals of fairly uniform length, and then selecting a tag (point) Ck from each subinterval at which to evaluate the function, we will be guided by the behavior of the function in the assignment of a subinterval. If the function oscillates, or behaves unpleasantly about a point c, we associate a small subinterval with c. If the function is better behaved, we associate a larger subinterval. With the Riemann integral, to obtain accurate approximations by sums of the fonn f(CI)(XI -xo) +... + f(cn)(xn -Xn-I), we required the maximum lengths of the subintervals, lILl-xlI, to be less than some constant O. Witb. the H-K integral, however, the 0 that regulates lengths of subintervals will be a function. A subinterval [u, v] with a tag C must satisfy C -o(c) < u < C :'5 v < C + o(c). A partition of [a, b] will be determined by a positive b, with the requirement function 0(') so that a = Xo < Xl < '., < Xn that Ck - O(Ck) < xk-l ::; c" :!: xk < ck + O(Ck). The H-K sums exhibit the same appearance as the ordinary Riemann sums f(Cl)(Xl -xo) + ... + f(cn)(xn -XII-I), but with the H-K integral 0 is a positive function on [a, b], where Xk-Xk-l < 20(Ck), Xk-l :!: Ck :!: Xk.
=
21
An Historical Overview
Example 1.11.1. For an example, let's begin with the Lebesgue integrable Dirichlet function on the interval [0, 1] that is 1 on the rationals and 0 on the irrationals. Consider any Riemann sum L !(Ck)(Xk - Xk-l)' There will be no contribution to this sum unless the tag Ck is a rational number. We want any interval associated with such a tag to be "small." Enumerate the rationals in [0,1]: 1'1, r2 •... , rn , .... Define a positive function aO on [0,1] by
a(c)
=
I
=
'1:'2, ... ,r
E
C
1
otherwIse.
ll •••••
Then any Riemann sum is nonnegative and 2E. We want convergence here. Redefine
a(c) = {
E/2n+l 1
E
L !(Ck)(Xk -Xk-I) :::; L~ 1· = rll.
otherwise.
Then we may conclude
We have glossed over two difficulties. First, just because we have a positive function 8(·) on the interval [a, b], how do we know there is a partition of [a, b] so that Ck - 8(Ck) < Xk-l < Ck < Xk < Ck + a(Ck)? (This was settled by Cousin in 1885.) Second, to use this integral effectively, we need to be able to construct suitable positive functions 8(·) for a particular function f, from the vague idea that erratic behavior of the function at a point generally requires small subintervals about that point. But in the end we are rewarded handsomely: we have better "Fundamental Theorems," better convergence theorems, and so on, than we have with the Lebesgue integral. Whereas the Lebesgue integral was the integral of the twentieth century, tbe H-K integral may lead to new developments in the twenty-first century. In fact, P. Muldowney (1987) treats two of the integrals we discuss later - the Wiener integral and the Feynman integral as special cases of the H-K integral over function spaces. On the other hand, the Lebesgue integral is particularly suited for LP spaces. Recall L. Carleson's result that the Fourier series of an L2 function converges almost everywhere.
22
1.12
A Garden of Integrals
Norbert \lUiener
The Wiener integral, developed by Norbert Wiener (1894-1964) in the 1920s, was a spectacular advance in the theory of integration. Wiener constructed a measure on a function space - in fact, the Banach space of continuous functions - on the interval [0, 1] beginning at the origin, with IIxll = sUPo::;t:::;ll x (t)l. See Figure 22. position
Figure 22. Wiener's continuous functions This measure arose when Wiener was trying to understand Brownian motion (e.g., that of pollen grains in suspension, moving erratically). Think of a collection of particles at position x (0) = 0 at time 0, moving to position x (1) at time 1 (see Figure 23). position
time
o
Figure 23. Brownian motion Now suppose 0 < tl ::s 1 and al < X(tl) ~ b1- We may think of the set of continuous functions (particles) on [0, 1] that pass through the "window" (aI, b I ] at time t}, a quasi-interval in Wiener's tenus (see Figure 24). We want to measure the fraction of the particles that begin at position 0 at time 0 and pass through the window Cal, bd at time tl' From physical
23
An Historical Overview
position :x:
~~------~----~~------~~~----------~--- timet
Figure 24. Quasi-interval considerations and genius, Wiener assigned a measure, w: w({x(.) E Co I al < XCII) < bllO < 11 < I})
= L
l
bl
.,
C2:rr Il)-I/2e-~i / 2/ 1 d ~l'
al
We have, for example,
l/J = ({x(.) w(",) = L
1°
at 0< tl (21<11)-1/2.-ft/2tld~1 = O. E
Co I 0< xCt]) <
<
i})
and
Certainly all the particles will pass through the large window (-00,00] at 11. So,
L:
w(Co) = w({.x(-) E Co
= L
I -00 < X(tl) ::: 00,
(27rtl)-1/2.-rU2tld~1 =
0 < tl :::
l})
1.
Now, suppose we have two windows, (al' b I] at tl and (a2. b2] at t2, where a < 11 < t2 < 1. See Figure 25. We have
Wiener assigned a measure of
If the window at time tl is large, (-co, co], no real restriction is imposed on the number of particles, and the measure of
24
A Garden of Integrals
position x
Figure 25. should be the same as the measure of {xC·) that
E
Co I a2 < X(t2) < b2}. Show
Similarly if (a2, b21== (-00,00]. Also,
w({x(.)
E
Co
I-
00 < X(tl) :!:
00, -00
< x (t2) < 00,0 < tl < t2 ~
In
=1. Finally, let K(x t t) == (2nt)-1/2 e-X /2t, with 0 < t1 < t2 < ... < tn ~ 1. See Figure 26. Consider 2
posinonx
,
~----~----~--~---------------+--------~- timet
...
Figure 26. The lHiimensional quasi..interval
1
25
An Historical Overview This quasi-interval will be assigned a measure,
w({x(.) E COlak < X(tk) ::: bkt 1 ::: k < 11,0 < tl < ... < t2 < I})
=Ll
bll
an
d~n"'lbl d~lKC~I.tl)K(~2 -~1It2-tl) a[ " . K(~n - ~n-!. til - tn-I).
Wiener was able to extend tbis measure w on the quasi-intervals (finite number of restrictions) to a measure JJ..w on the sigma-algebra generated by the quasi-intervals. For example, if S = {x(·) E Co I - 1 < xCt) < 1, 0 ::: t :::: I}, let
S2
= {x C·) E Co I - 2 < x (1) < I}, = {x(.) E Co I - ~ < x (~) • x(l)
SII
= xC,)
S1
I
E ICo
<
~11 < X (~l) 2"-
1-
I} ,
::
I, Ie
= 1,2 ... " 2n -
1j.
Then S1 :J S2 :J ... :J Srz :J ... and S = nslZ • The set S is thus measurable as a countable intersection of quasi-intervals. From the.., measure JJ..w we have measurable functionals and, finally, Wiener integrals - "path'" integrals. For example, suppose we have the functional F[x(·)] = x (to); that is, to each element x(·) of the function space Co we assign its value at t = to. x (to). What should fco F[x(·)]dJLw be? Formally,
{ x(to)dJJ..w Jco
=
1 ~(2:n:to)-1/2e-t2/2tod~ = o. 00
-00
The expected value of its position at any t, for 0 < t < 1, should be O. Suppose F[x(·)] = x 2 (to) for 0 < to < 1. Formally,
( x 2(to)dJLw = Jco
1 ~2(2:n:to)-1/2e-~2/2to d~ = 00
to.
-00
The variance is t for 0 < t ::: 1. It would be convenient if F[x(.)] L x2(-r:)d1:, and
Jd
f
fco F[x(·)] dll-w = fco (L x2(~) dT) d/1-w = Such is the case.
f (!co
2 X (T) dll- W )
d~ = L [
.dr
=~.
26
A Garden of Integrals
1.13
Richard Feynman
Consider a quantum mechanical system W that satisfies ScbrOdinger's equation,
aw at
-
in a w 2m at n'
2 i = --V 2
for -
00
<x <
00,
t>0
with '/1(0. x) = lex), -00 < x < 00, and J~coIWI2 dx = 1 for aU t 2: O. To explain the evolution in time of W, Feynman (1918-1988) developed an integral interpretation of W as a limit of Riemann-type sums (1948). It was understood that !Wp is a probability density function, and Feynman concluded that the total probability amplitude W is the sum, over all continuous paths from position Xo at time 0 to position x at time t, of the individual probability amplitudes. That is, K ei ItlA Lx (.)] t 1111 connecllng continuous plllbs
with A. [x (.)] = J~ [imx2(t) - V(x(-r))] d-r; and K a normalizing constant. His idea was to approximate this expression with Riemann sums and take the limit. Thus,
L
Kei/frA[x]
all connecting continuous paths
~
lim K { dXn-l'" ( dxo
n-+oo
JR
JR
!
§" 2 i exp 2n(t/n) ~(Xk - Xk-l) - -:;;
where In
K=
(
21rin(t/n)
)n/2
and
(t)n ~ n
Xu
V(Xk-ll
I
!(xo).
= x.
We have an integral over a function space, ColO, 1], the same function space as the Wiener integral. The integrand is different, however. The exponential term has modulus 1 and becomes highly oscillatory as 11 ~ 00. Furthennore, for the nonnalizing constant K, IKl -7 00 as n -4- 00. Interpretation of this limit, and an explanation of the convergence issues involved, has occupied physicists and mathematicians for over fifty years. In the chapter on the Feynman integral, we will see a brilliant explanation due to Edward Nelson, discovered in the 1960s.
An Historical Overview
'1.14
27
References
1. Carleson, Lennart. On convergence and growth of partial sums of Fourier series. Acta Il'lathematica 116 (1966) 135-157. 2. Muldowney, Panick. A General Theory of Integration in Function Spaces. Essex: Longman Scientific and Technical, and New York: Wiley, 1987. 3. Nelson, Edward. Feynman integrals and the Schrodinger equation Journal of Mathematical Physics 5(1964) 332-43.
CHAPTER
2
The Cauchy
~D1tegra~
The sole aim ofscience ;s the honor of the human mind. andji-om this point of view a question about numbers is as important as a question - C. O. J. Jacobi about the system of the wOI'ld.
2.1
Exploring Integration
Augustin-Louis Cauchy (1789-1857) was the founder of integration theory. Before Cauchy, the emphasis was on calculating integrals of specific functions. For example, in our calculus courses we use the formulae 1+ 2
+ ... + 11 =
2?2
+- +···+n
1
2
IZ (ll
+ 1)
2
and
n(1l + 1)(217 + 1) =---6---
to show, using approximation by interior and exterior rectangles, that the areas of the regions between the .x -axis and the curves y = x and y = x 2 for 0 < x < b are given by x dx = b2 /2 and x 2 dx b 3 /3 results obtained much earlier by Archimedes (287-212 B.C.E). Such beautiful results begged for extension, most accomplished by sheer ingenuity. Here are some examples, formulae owing to Fermat, Wallis, Stirling, and Stieltjes.
J;
2.1.1
J;
=
Fermat's Formula
From Pierre de Fermat (1601-1665) we have
{b
io
q
bq+1
x dx. = q
+ 1'
for q a positive rational number, b > O.
This formula may be justified as follows.
29
30
A Garden of Integrals I
Divide [0, b] into an infinite sequence of subintervals of varying widths withendpointsbrn~O < r < 1,n 0, I,.", and erect a rectangle of height (brn)q over the subinterval [brn+l t brn]. Sum the areas of the "exterior" rectangles, and show that this sum is bq +1 (1-r)/(1- r q +1 ). Evaluate the limit of this sum as r approaches 1.
=
Exercise 2.1.1. Show
/." sin (x) dx = 2, using the familiar limit, lims ......oCsin8)/8 = 1, and Lagrange's identity,
. (1r) + sm. (2H) n + ... + sm. (n1r) n
sm -;;
_ cos (N/2n) - cos ([(211 + 1)1t]/2n) 2 sin (1t/2n)
2.1.2 Wallis's Formula John Wallis (1616-1703) gave us this formula: 2 2 4 4 1 3 3 5
7r
- = - . - . - . - .....
2
2n
2n - 1 2n
Using integration by parts, show that
1 7r
o
Next, since conclude
sinn+2(x) dx = n
2n
+
l1
n +2
J: sin2n+
2
(x) dx :::
J: sin2n+
+1
1C
sinn (x) dx.
0
1
(x) d x :::
J: sin2n
(x) dx, we
J/r
2n 1 <0 sin2n + 1 (x) dx <1. _+ _ 2n + 2 sin2n(x) dx -
I:
Show
'" . 2n+l (x) dx = 2 /.o sm
(2)3' (4)'5' ..
(2n 2+ n1) .
Substituting terms~ we have H
_ .
2
~1r sinZtI + 1 (x) dx 0
IoR' sin2n (x) dx
2 2 4 . _...... 4 =_._._
1 3 3 5
2n 211 - 1 2n + 1 . 2n
11
_ _
31
The Cauchy Integral
Taking the limit as
11 ~ 00,
the result follows.
It is interesting to note that n:2
[1r:/2
8" = 10 =
[1r:/2
x dx =
10
1
[sin- (sin(x))] dx
fl' [sin(x)+ G) sin:(X) + G) (D sin:(X) + ... ] 1
1
= 1 + 32 + 52 + ....
2.1.3 Stirling's Formula James Stirling (1692-1770) provides the formula III --------- -+ 1
as 11 -+
.J2rr 11 (11 le)n
Let f(x) that
j,
= In x, where
1 < x <
11.
00 .
Using the trapezoidal rule, show
1
n
Inx d.x"- -{(In 1 + In2) + (ln2 + In 3) + .. ·+[In(11 - 1) + Inn}}
2
1
= "error terms.
Jt
Because lnx dx = 11 Inll-11+ 1, we have In(11!) where En has a limit as 11 -+ 00. Conclude that
n!
SII
= .fii(n/e)n
. C
--,
,
as
11
-+
= (n+~) Inn-l1 +E
00,
Then by Wallis's formula, we have
S;
2211 (111)2
-=h -+~. S2n .jiZ(211 ) ! Thus Illi [.J2n:n(nle)"] -+ 1 as
11
-+ 00.
2.1.4 Stieltjes' Formula From Thomas Stieltjes (1856-1894) we have that 00
-
.,
r:; vn:
e-x - dx = - , /, o 2
C > O.
n
II ,
32
A Garden of Integrals Using integration by parts, show that 00 /,
_
2
xn+2e x dx =
11
a
+ 1 /,00 xne_x2 dx. a
2
Thus, 00 /,
2
X
2 lie-X
1 . 3· .... (2n - 1)
dx =
a
00 /,
a
x
2n+1 -.x 2
e
dx
=
2n 1 ·2· .... n
2
/,00
2
e- x dx
a
and
.
2 -xx d , we have · 1k = JfOO GlVen a xk e
a 2 h-l
LX> xk-l(a + x),e-x'- dx > 0
+ 2a1k + lk+l =
for all real a. Letting a = -lk/1k+b conclude that ?
lin < 12n - 1 • hn+l l;U+l
< 12n . hn+2
and
= 2n 2+ 1 12n2 ·
Thus 222
- - - 12n + 1 < 12n < hu-l . hn+l. 211 + 1 The argument can be completed by using Stirling's and Wallis's formulae. See Young (1992).
2.2
Cauchy's Integral
What exactly did Cauchy accomplish regarding integration? First, he defined a constructive process (an algorithm, if you will) for calculating integrals, and second, he investigated requirements on the function to be integrated that would make his algorithm meaningful. Here is Cauchy'S 1823 definition of the integral. Given a bounded function f on the interval [a, b], divide [a, b] into a finite number of contiguous subintervals [Xk-l, Xk] with a < Xo < Xl < ... < Xn = b. We shan use the following tenninology. o
The collection of point intervals (xo,[xo,xr]). (XI, [Xl,X2]), ... , (Xn-l, [XII-I. Xu]) is called a Cauchy partition of [at b], denoted by P.
~
We caU the points
Xo, Xl, •••
I
Xn the division points of P.
The Cauchy Integral o
33
The points .."to, Xl •.•. ,Xn-l are called the tags of P.
Form the Cauchy sum, that is,
I
evaluated at the tags,
n
E I(Xk-I)(Xk - Xk-l), E Ih.:x·
k=l
P
The limit (provided it exists) of such sums, as the lengths of the subinter~ vals approach 0, is said to be the Cauchy integral 01 lover [a, b], written C I(x) dx. That is, a bounded function I on the interval [a, bJ is said to be Cauchy integrable 017 [a, b] iff there is a number A with the property that for each E > 0 there exists a positive constant 8 such that, for any Cauchy partition P of [a, b] whose subintervals have length less than 8,
J:
E I(Xk-l)(Xk -
Xk-l) - A <
E.
P
We write C
2.2.1
J:
I(x) dx = A.
Cauchy's Theorem (1823)
This definition of the Cauchy integral raises an obvious question: Is boundedness of I sufficient to guarantee the existence of A? In a first-of-its-kind theorem, Cauchy argued that continuity of the function would guarantee a successful outcome to this complicated limiting process, a number A. Theorem 2.2.1 (Cauchy, 1823). If I is continuous on the interval [a, b], then I is Cauchy integrable on [a, b]. Proof. The argument hinges on two ideas: 1. Continuous functions on closed bounded intervals are uniformly continuous. 2. Cauchy sequences are useful when we do not have a limit in hand, in this case, the number A in the definition for Cauchy integrability.
Lp
From these ideas, we will show that the Cauchy sums, I h.x, converge to a limit A, and that this limit does not depend on the particular choice of partitions, except for the requirement that the lengths of the subintervals of the partitions approach O. We will compare Cauchy sums of any two partitions.
34
A Garden of Integrals
Begin with an easier problem: Compare the Cauchy sums of a partition P and another partition P obtained by adding a finite number of additional division points to P, a so-called refinement of P. Suppose a Xo < Xl < ... < X" b are the division points of P with
=
=
= Xit
Xo
=
Xl
= Y20, Y21. Y22, ... , Y2ia = X2.
Xn-i
=
Y10, Y11. YI2,.·., Ylil
Yno. Ynl. Yn2 •. ··, Yni n
the division points of
= Xn.
P.
Then,
n
- L !(Xk-l)(Xk -
n
Xk-l) -
k=l n
ik
LL
l(Ykj-l)(Ykj - Ykj-l)
k=l J=l ik
- LL
[f(Xk-l) - l(Ykl-l)](Ykj - Ykj-l)
k=l 1=:1
Because the function I is uniformly continuous on the interval [a, b], given an E > 0 we have a positive number 0 so that If(x) - 1(Y)1 < E when x and yare any points of the interval [a, b] within 0 of each other. By requiring that the subintervals of P have length less than this 0, we may now conclude that
LI6.x - Lf6.y < e(b -a). p
p
Cauchy sums for refinements are within e(b -a) of each other whenever the subintervals of the partition all have length less than the 0 determined by uniform continuity. Back to the original problem. Suppose that PI and P2 are any partitions of the interval [a, b] whose subintervals have length less than O. Together, PI U P21 We have a refinement of each, whose subintervals have length less than Thus the associated sums, LPI 16.x and LPa !6.x, are within e(b - a) of LPI UP2 16.x and hence within 2E(b - a) of each other.
o.
The Cauchy Integral
35
Any two Cauchy sums differ arbitrarily little from each other iff the lengths of the associated partitions' subintervals are sufficiently small. So take any sequence of partitions PI1 of the interval [at b], with the length of the longest subintervals approaching 0 as n -7 O. The sequence of Cauchy sums, {LPn f8. nx}t are all within 2tE(b - a) of each other, for n sufficient1y large: LPIl f 8. nx -7 A. Show that this limit does not depend on the particular choice of partitions as long as the requirement of the length of the largest subinterval approaching 0 is met. We have shown that the Cauchy process for continuous functions is well defined. 0
2.2.2 Cauchy Criteria for Cauchy Integrability In the preceding argument, we actually have the following result. Theorem 2.2.2. A continuous function f on the interval [a, b] is Cauchy integrable iff/or each E > 0 there exists a positive constant 8 so that, if PI and Pa are any Cauchy pal1itions of [a, b] whose subintervals have length less than 8, the associated Cauchy sums are within E of each other:
If f is continuous on [at b], then If I is continuous on [a, b], so Cauchy integrability of f implies Cauchy integrability of If I· The converse is false: If f (x:) = 1 x ~atio.nal, -1 x uratIonal,
1
then If I is continuous and C follf(x)1 dx
2.3
= 1.
RecoverAng functions by Integrration
Now that we have an integration process for continuous functions, is there some means of ca1culating the Cauchy integral that does not require the genius of Femlat, Stieltjes, et aL? In many cases the answer is yes: The first part of the so-called Fundamental Theorem of Calculus (FTC) recovers a function from its derivative by integration. Theorem 2.3.1 (FTC for the Cauchy Integral). If F is a differentiable junction. on the interval [at b]. an.d F' is contin.uolls on [at b]. then
36
A Garden of Integrals
1. F' is Cauchy integrable on [a b], and I
2. C
J: F'(t)dt = F(x) - F(a) for each x in the inte1''Val [a, b].
Proof The first conclusion follows from Theorem 2.2.1. To show the second conclusion involves the uniform continuity of F and the mean value theorem for derivatives. Suppose F(Xk) - F(Xk-l) = F'(Ck)(Xk - Xk-l), for Xk-l < Ck < Xk. Let E > 0 be given. From the Cauchy integrability'of F', we have a positive. number 01 so that if P is any Cauchy partition of the interval [a, xl with subintervals of length less than Ol:r then
L F' ~X
-
C
P
1.x F'(t)dt
<
E.
a
Because the derivative F' is continuous by assumption, and thus uniformly continuous on the interval [a. b], we have a positive number 02 so that IF'(e) - F'(d)1 < E whenever c and d are points of the interval [a, b] satisfying Ic - dl < 02. Letting &be the smaller of the two numbers 81 and 82, we may conclude that
\F(X) - F(a) - C
f
F'(t)dt
n
-
n
I)F(Xk) - F(Xk-l)]-
L
k=l
k=1
+ <
t
t
F'(XIc-l)(XIc - Xk-l)
F'(Xk-l)(Xk - Xk-l) - C
k=1
!F'(Ck) - F'(Xk-l)! (Xk - Xk-l)
1.% F'(t)dt a
+
< e(b -a)
L F' ~X - c f F'IJ)dt P
k=l
a
+ E,
for all partitions P with subintervals of length less than O. The proof is complete. 0 Compare Theorems 3.7.1, 6.4.2, and 8.7.3. Exercise 2.3.1. Redo Exercise 2.1.1 in light of this result, and conclude
The Cauchy Integral
37
that and
c
i"
sinxdx
Exercise 2.3.2. Let F(x) =
!
,,,3
= COS O-cosJr = 2.
sino(rc/ x)
0 < x < 1, x =0.
a. Calculate F'. b. Show that F' is continuous on [0, 1]. c. Calculate C Jot F' (x) dx. Cauchy not only gave us the existence of the integral for a large class of functions (continuous), but also gave us a straightforward means of calculating many integrals.
2.4
Recovering Functions by Differentiation .
In addition to the idea of recovering a function from its derivative by integration, we have the notion of recovering a function from its integral by differentiation, the second part of the Fundamental Theorem of Calculus. Let's examine some properties of the Cauchy integral. Is it continuous? Is it differentiable? ..
Theorem 2.4.1 (Another FTC for the Cauchy Integral). If f is a continuous function on the interval [at b], and we define a function F Oli [a, b] by F(x) = C J~'C f(t) dt, then
1. F is differentiable on [a. b],
2. F' =
f
on [a, b], and
3. F is absolutely continuous on [a, b]. Proof. We have shown that the Cauchy integral of a continuous function is well defined, so F makes sense. To show that F is differentiable on [a, b], and that the derivative of F, F', is in fact equal to f, entails estimating the familiar expression
F(x
+ '~: -
F(x) _ f(x)
38
A Garden of Integrals
where h. is small enough so that x + h belongs to the interval (a, b). Because f is continuous at x by assumption, we have a 0 > 0 so that If(t) - f(x)1 < e whenever It -xl < 8 and t belongs to the interval [a, b]. Then, F(x
+ h) -
F(x) _ f(X)\
h
= .!... C h
r + [f(t) Jx x h
f(x)1 dt < e.
J:
In other words, (C f(t) dt)' = f(x), and ~his is what we wanted to show. As for the absolute continuity, see Definition 5.8.2 and note that F' is continuous and thus bounded on the interval [a, b]:
Compare Theorems 3.7.2, 6.4.1, and 8.8.1.
Exercise 2.4.1. Given f(x) = {
a. Determine F(x) = C
I;/2x f(t) dt.
I-x 2x-2
cIt f(t)dt,
O~x:::Sl.
1 ::s x
~
2.
G(x) = CI~x f(t)dt, H(x)
b. Using the definition of the derivative, show that F is differentiable at x = 1 and F'(l) = f(l).
c. Determine G' and H'. Integration is a smoothing process: Continuous functions become differentiab Ie functions under the integration process.
2.5
A Convergence Theorem
Another means of calculating integrals is using convergence theorems, comparing the integrals of a sequence of functions with the integral of the limit of the sequence of functions. This is valid with some restrictions.
Theorem 2.5.1 (Convergence for Cauchy Integrable Functions).
If {fk}
is a sequence of continuoZls functions converging uniformly to the function f on [a, b]. then f is Cauchy integrable on [at b] and C f(x) dx =
lim C
I:
J:
fk (x) dx.
39
The Cauchy Integral
Proof That the function
I
is continuous follows from Weierstrass's Theorem: The unifonn limit of a sequence of continuous functions is continuous:
II (x) - f(y)1 ::: I/ex) -
Ik(X)1
+ IlkeX) -
/k(Y)1
+ I/k(Y) - 1(Y)I.
the first and third tenns are "small" by uniform convergence, and the second term is "smalln by continuity of Ik. Thus I is Cauchy integrable on [a. b]. As for the second conclusion,
and the right-hand side of this inequality can be made arbitrarily small by unifonn convergence: Given E > 0, we have a natural number K so that Ilk - II < E whenever k > K, throughout the interval [a, b]. 0 Exercise 2.5.1. Consider a sequence of continuous functions {/k} given by
0::: x
<
i,
!<x
II (x)
o <x< ~,
8 -16x
!<x<.2. 2 - 8' .2.<x
16x -12
o
12 (x) 16x - 12 14 -16x
o
b. Show that the sequence {Ik} converges uniformly on [0, 1], where Ilk - Ik+ll :::: 22-k. c. Calculate C f01 lim Ik (x) dx. Exercise 2.5.2. Let the function Ik(X)
= xl/k, where Ie = 1,2, ..
O<x
= I. Is the convergence unifonn?
J.i Ik (x) dx.
and
40
2.6
A Garden of Integrals
Joseph Fourier
We now go back to the year 1822, in which year the book Theone analytique de la chaleur, by Joseph Fourier (1768-1830), appeared. Fourier was looking for a steady-state temperature function T (x , y) in the strip, o < x < x, 0 < y, satisfying the partial differential equation,
a2 T
a2 T
for 0 < x <
-8 =0 x 2 +-8 y2
1r I
°< y.
with temperature 0 when x = 0 or x = x, and an initial temperature distribution ¢ on the base where y = O.
Example 2.6.1. Assuming that T(x, y)
= X(x)Y(y)
(separation of variky ables), we argue that Tk(x) = Xk(X)Yk(y) = e- sin(kx), for k = 1,2, ... , satisfies the partial differential equation and that Tk(O) = Tk(Jr) = O. Heuristically, T(x y) = E bke-ky sin(kx), the constants bk to be determined.Tosatisfythebaseconditionaty = Oweneed¢(x) = Ebksin(kx) for 0 < x < TC. Multiply both sides of the equality by sin(nx). Assuming that the integrals are meaningful and that we can integrate termwise, show that the coefficients bn are given by bll = 2/x ¢(x) sin(nx) dx, for O<x<x. I
J;
What kind of integral are we discussing here? Certainly if ¢ is continuous, the Cauchy integral would be sufficient. But periodicity and sine suggest an odd extension of ¢. If ¢(O) # 0, we have problems at x = 0. Because Fourier series successfully model a wide variety of physical phenomena, we want to be able to deal with discontinuous behavior.
Exercise 2.6.1. a. Suppose f(x) = {
o < x < 1 or 1 < x < 2, ~ x=l.
Because! is discontinuous, ! is not Cauchy integrable. On the other hand, show that for any Cauchy partition P of [0, 2],
L !(Xk-l)i::l.x -
2
<8.
p
b. Suppose
f(x)
=
I
°
1 <x < I, 2 1 <x < 2.
Show that for any Cauchy partition of [0,21, IE! i::l.x -
31
<
o.
41
The Cauchy Integral
Argue that for a bounded function with a fmite number of jump discontinuities, obvious modifications to the Cauchy integral could be made.
2.7 P. G. Lejeune Dirichlet Several mathematicians tried to discover conditions that would guarantee convergence of the Fourier series. A successful argument was given by P.O. Lejeune Dirichlet (1805-1859).
Theorem 2.7.1 (Dirichlet's Convergence Theorem for Fourier Series, 1829). Suppose that f is a bounded function, piecewise continuous (finite nu.mber ofjump discontinuities) and piecewise monotonic an the interval [-1I' •1I']. with period 2il'. The Fourier series representation of f at x.
1
co
2
1
-an + I)a" cosnx + b" sinllx) with
111: f(x) cos(nx) dx. 1 1'1t f(x) sin(nx) dx,
an = -1
...
bn =
1I'
-'It
11:
-'It
converges to f(x
II
= 0,1,2, .. ,' alld
n =
+ 0) + f(x -
1. 2, ... ,
0)
2 where, as usual, lim}'~;t+ fey) = f(x+O) andIim}'~.~- fey) = f(x-O) For a discussion and proof, see Bressoud (1994).
Exercise 2.7.1. Assuming Dirichlet's result, with f(x) = X < 1r, extended periodically, show the following: 1I'
4[
()
Ix I ="2 -
1I'
il'2
1
1
1[2
1
1
cos
X
+
cos (3x) 32
+
cos(5x) 52
+ ...
]
Ixl
for -
for
11:
-1[
<
:5 x :5
1[
8"' = 1 + 32 + 52 + ... · "6 = 1 + 22 + 32 + ... .
(Euler)
Exercise 2.7.2. Suppose f(x) = cos(ax), for -11: :5 x < il' and integer, extended periodically (From Courant and John (1965).)
O!
not an
42
A Garden of Integrals
a. Show that
cos(ax) = 2a sin(ax) (~_ cos(x) 1! 2a 2 a 2 - 12 b. Let x =
+ cos(2x)
_ ... ) .
(X2 -
22
1
+....
and conclude
11:
cot(1I:x) - -
1 = --2x (1 12 -
1C X
c. For 0 < x
~
11:
x2
+?
x2
2- -
)
{3 < 1, show that the series converges uniformly, Then
integrate term by term to show
In (Sin(1!X)) = lim ~ In 1! X n L.." k=l
(1 _lcX2) . 2
So sin(3t'x) =
(1- X2) (1 _ X2) (1 _X2) ... 12 22
for 0 < x < 1.
I
32
1!X
d. Show that (Wallis) While studying convergence problems for Fourier series, Dirichlet began to consider functions that assumed one value on tbe rationals and a different value on the irrationals. For example, suppose
I(x) = {I
o
x ~atio~al,
x matlOnal,
and
1
g(x) = {
~
x rational, x irrational,
Certainly 1 = C Jo (I + g)(x) dX, and linearity of the integral would require that 1 C I(x) dx + C g(x) dx. But these two Cauchy integrals are not defined. Dirichlet had discussions with Riemann to try to find an integration process that would overcome that difficulty. Riemann did not fmd such an integration process (that would be discovered by Lebesgue), but did develop another integration process more powerful than Cauchy's, Riemann's process is the subject of the next chapter.
=
J;
J;
43
The Cauchy Integral
2.8 Patrid{ Billingsley's Example Lest we become blase about results concerning continuous functions, here is Patrick Billingsley's (1982) continuous nowhere oiifferentiablefimction. Suppose ¢ (x) is the distance from x to the nearest integer. Define a function B by B(x) = E~o ¢(2tl x)/2". That B is continuous follows from the Weierstrass M -test.
Exercise 2.8.1. Calculate C fol B(x) dx.
In Exercise 5.9.2 we shall see that if a function then for every
f
f
has a derivative at Xo, > 0 we may determine a positive number ~ so that
If(8) - f(a) - f'(xo)(fJ - a)1 <
f(fJ -
a),
if a ¥= fJ and Xo - ~ < a < Xo < fJ < Xo + 8. To show B is not differentiable, we will pick a point Xo in the interval [0,1) and approximate it with binary expansions. Because Xo is between 0 and or and 1, we have
! !
al
al
1
-<xo<-+2 2 2'
for a 1 = 0 or 1.
Now divide [0, 1] into four subintervals of equal length 1/22. Then
Continuing, al
aN
2" + ... + 2N
< Xo <
al
aN
1
2" + ... + 2N + 2N
for
aj
= 0 or 1.
Let
and fom] the difference quotient
We have 2/1 fJN = (m+ 1)/2N - 1I and 211 aN = m/2N - ll for In an integer. Also, ¢ is linear with slope ± 1 on the interval [m/2 N - n , (pi + 1)/2N - 1l ] J
44
A Garden of Integrals
as long as N a sum of ±1s:
11 ~
1, that is, when 0 :5 n ::: N - 1. Thus the first sum is
The second sum vanishes because 2n {JN = (m + 1)21J - N and 2n a.N = m2n - N are integers for n > N. Letting N -)0 00, (J N - aN -)0 0, differentiability of B at Xo would yield B'(xo) = L ±l.
2.9
Summary
Two Fundamental Theorems of Calculus for the CaZichy Integral: If F is differentiable on [a, b] and if F' is continuous on [a, b], then
1. F' is Cauchy integrable on [a. b] and 2. C
J: F'(t)dt = F(x) - F(a), for a
< x < b.
If f is continuous on [a. b] and F(x) = C J~"( f(t)dt, then 1. F is differentiable on [a, b],
2. F' =
f
on [a, b], and
3. F is absolutely continuous on [at b].
2.10
References
1. Billingsley, Patrick. Van der Waerden's contInuous nowhere differentiable function. Ame1'ican Mathematical Monthly 89 (1982) 691.
2. Bressoud, David. A Radical Approach to Real Analysis. Washington: Mathematical Association of America, 1994. 3. Courant, Richard, and Fritz John. Introduction to Calculus and Analysis. Vol. 1. New York: Wiley Interscience, 1965.
4. Young, Robert M. Excursions ill Calculus' An Interplay of the Continuous and the Discrete. Dolciani Mathematical Expositions. No. 13. Washington: MathematIcal Association of America, 1992.
CHAPTER
3
The Ruemawon
~ntegra~
Reason with a capital R = Sweet Reason, the newest and rarest thing in human life, the most delicate child of human history. -
Edward Abbey
We are nature's unique experiment to make the rational intelligence prove itself sounder than the rejle.:'C. Knowledge is our destiny. -
Jacob Bronowsld
The Riemann integral (1854) - the familiar integral of calculus developed by Bernhard Riemann (1826-1866) - was a response to various questions raised by Dirichlet about just how discontinuous a function could be and sti II have a well-defined integral.
3.1
Riemann's Integral
Given a bounded function f on the interval [a, b], divide [a, b] into a finite number of contiguous subintervals [Xk-l, Xk], with a = Xo < Xl < ... < Xn = b, and pick a point Ck in [Xk-l t Xk]. As with Cauchy integration, we begin by defining some tenninology. The collection of point intervals
is called a Riemann partition of [a, b], to be denoted by P, the points xo. Xl, ••• ,XII are called the divisionpoints of P, and the points Cl, C2, ••• I en are called the tags of P. Fonn the Riemann sum LZ=l f(Ck)(Xk - Xk-l), which we will write as Lp f llx. The limit (provided it exists) of such sums, as the lengths
45
46
A Garden of Integrals
of the subintervals approach zero, is said to be the Riemann integral of I over [a. b], written I(x) dx. That is, a bounded function I on the interval [a, b] is said to be Riemann that for each integrable on [a. b] if there is a number A with the properor \ e > 0 there exists a positive constant 8 such that for any Riemann partition P of [a, b] whose subintervals have length less than 8,
RJ:
L/AX-A <e. p
We write R!: I(x) dx = A. Dirichlet-type functions - say, 1 on the irrationals and 0 on the rationals - show again that some restrictions must be imposed on I besides boundedness. In comparing Riemann's and Cauchy's integration processes, we notice that Riemann's I may be evaluated at any point Ck in the interval [Xk-l, Xk], whereas Cauchy's I is evaluated at the left-hand endpoint of that interval, Xk-l. For a given partition P, how much variability can we have in the associated Riemann sums as Ck varies throughout the interval [Xk-ll Xk]? Certainly
L p
inf [Xk-l,Xk]
I AX:S
L I(c) AX::S L p
p
sup
I Ax.
[Xk-l,xk]
Thus the absolute value of the difference of any two Riemann sums for the same partition is bounded above by L p (sup f - inf I) Ax, where the quantity sup I - in! f denotes SUP[Xk_t,Xk] I - infrxk-l,x.d I· On the other hand, we have points C and d in the interval [Xk-l, xkl so that sup I - inf I - 2e < I(ek) - I(dk) :S sup I - inf I for e an arbitrary positive number. Then L(sup I - inf I) Ax - 2e(b - a) < p
L I(c) Ax - L I(d) Ax p
< L(sup I
p
- wi) Ax.
p
Thus L p (sup I - inf f) Ax is the smallest upper bound of the absolute value of the difference of any two Riemann sums associated with the partition P . (The reader may want to review the Cauchy integral discussion in Section 2.2.)
47
The Riemann Integral
Proceeding in the same manner, but omitting most of the details, we again have that for any refinement P of P, n
n
k
I: I(ck) fj.x - L L
I(ckj) fj.y
k=l j=l
k=l
11
=
L
ilL
L[f(Ck) - f(Ckj)] fj.y
k=l j=l
<
tt(
k=l j=l
f -
sup
[;tJ.::-I.·"tk}
inf [·"tk-hXJ.::]
f)
n
=
L (sup f
- inf f) fj.x
k=l
= L(sup f - inf f) fj.x. p
That is, for a partition P of [a, b] and any of its refinements, the absolute value of the difference of any two associated Riemann sums has a smallest upper bound of p (sup f - inf f) fj.x. Apparently we have an integrability condition...
L
3.2 Criteria foU'
R~emann
Integrability
Let us examine the criteria for what makes a function Riemann integrable.
Theorem 3.2.1 (Riemann Integrability Criteria). A bounded fllnction f on the interval [at b] is Riemann integrable iff given an E > 0 we may
Lp
determine a positive number 8 so that (sup f - inr f) fj.x < e for all partitions P whose subintervals have length less than 8. Proof If we assume that f is Riemann integrable on the interval [at b], then given an e ;.. 0 we have a 8 > 0 and a number A so that for any partition of [a, b] whose subintervals have length less than 8, every Riemann sum is between A -e/4 and A + E/4. But then the abso1ute value of the difference of any two Riemann sums is less than e/2, and (sup -inf f) fj.x < E. On the other hand, suppose that given an e > 0 we have a 8 > 0 so that (sup f - inf f) fj.x < e/2 for all partitions P whose subintervals have lengths less than 8. Let PI and P2 be two such partitions:
Lp
I
Lp
L(sup f - inf f) fj.x <:. PI
2
and
L(sup f - inf f)fj.y < :.. P'l
2
48
A Garden of Integrals
Since PI U P2 is a refinement of PI and P2, all of its subintervals will have lengths less than 8. Fwthermore,
<
Lf
~x -
L
f ~z +
L
f ~z -
Pl~Pl
p\UPl
PI
< L(supJ -infJ) ~x
L J ~y P2
+ L(supJ -infJ)~y < E. P2
Pi
Associated Riemann sums for partitions whose subintervals have lengths less than ~ are "close." Consider any sequence of partitions Pn of the interval [at b], with 8n approaching 0 as n -+ 00, and form a sequence of associated Riemann sums: {LPn J ~nX}. For 11 sufficiently large, ~n <~, and consequently
LJ~nX-LJ~mX <E. Pn
The sequence
{L Pn J
~nx}
Pm
is a Cauchy sequence. We have that
Show that another sequence fin of partitions with 8n approaching 0 as n -+ 00 leads to the same result. This completes the argument. 0 Eltercise 3.2.1. Suppose that J(x) = X2, 0 :5 x ::: 1. Show that J is Riemann integrable on [0, 1.}. Hint: Lp (x~ - X~_l) ~X < LP 2Xlr:~XAX < 28. Exercise 3.2.2. If f is continuous on [at b], then f is Riemann integrable on [a, b]. Also, if f is Cauchy integrable on [at b], then f is Riemann integrable on [a, b1. Hint: Let e > 0 be given. Because J is uniformly continuous on [a, bL we have a 8 > 0, so that IJ(c) - J(d)1 < e whenever Ic-dl < 8. Let P be any partition of [a, b] whose subintervals have length less than 8.
49
The Riemann Integral
Exercise 3.2.3. Cauchy integrable functions are Riemann integrable ftmctions. Do the integrals have the same value? Hint:
ct
f{x) dx - R
t
f{x) dx
=
[c t
:E f{xk-1) ax] + [L:: f(Xk-l) fj.x - L f(Ck) fj.x ] f{x) dx -
+ [:Ef{Ck) ax -R and use the uniform continuity of
3.3
f
t
f(X)dX] '
on [a, b].
Cauchy and Darboux Criteria for Riemann Integrability
Just how discontinuous can a bounded function be and maintain Riemann integrability? We have more work to do before we can answer this question. First, we will somewhat improve the Riemann Integrability Criteria.
Theorem
~.3.1
(Cauchy Criteria for Riemann Integrability). A bounded function f on the interval [a, b] is Riemann integrable ifffor each E > 0 there e;asts a Riemann partition P o/[a, b1so that Lp (sup f -inf f) fj.x < E.
Proof Let E > 0 be given, with P'" a Riemann partition of [a, bl, where a = x~ < < ... < x = b, so that Lp"(SUP f - inff) fj.x < E. We will construct a positive constant 0 so that any partition P of [a, b] whose subintervals have length less than 0 satisfies Lp (sup f - inf f) fj.x < E. Suppose P has division points a .:to < Xl < ... < Xn b. Form the refinement PUP'" of P. Initially, suppose this refinement has one more division point than P, say Xk-l < xj < Xk for some k. In this case,
xi
M
=
+ ( [xjinf ,Xk]
=
f-
:::: 2B (Xk - xk-d , where
If I <
B on the interval [a. b].
inf
[Xk-I.Xk]
f)
(Xk - X
j)
50
A Garden of Integrals
Now, since P'" consists of M + 1 division points 2 we can add at most M - 1 such points to P different from a and b. Thus for the partition P,
L
inf I t:,.z -
pUP"
L inf I
L inf I t:,.x
P < (M - 1)2B· (maximum length of the subintervals of P).
t:,.x'" -
L infI
P'"
t:,.x
P
< 2B(M - 1)· (maximum length of the subintervals of
Pl.
The reader may show
L sup I P
t:,.x -
L sup I t:,.x
11
p./l
< 2B(M - 1). (maximum length of the subintervals of P). Adding, we have L(sup 1- inf I) t:,.x < L(sup 1- inf/) t:,.x ~ p
+ 4B(M -
1)
p~
. (maximum length of the subintervals of P)
<
E +E,
provided we choose 8 so that 4B(M -1)0 < E. This is not a problem, since we began with III < B and a partition P* with M + 1 division points. The proof in the other direction follows from Theorem 3.2.1. 0
Exercise 3.3.1. Use the preceding result to show that R fol x 2 dx = Hint: Consider 0, lin, 2/11, ... ,11/11.
t·
Theorem 3.3.1 suggests another approach to the Riemann integral (DarboliX, 1875). For a function I bounded on [as bl and any partition P of
[a, b], inf I(b - a) ::: [a.b]
L infj t:,.x ~ LSup I t:,.x :s [a,b] sup I(b - a).
The lower (upper) sum of I with respect to the partition P is defined as
L(/, P)
== L
(U(j; P) ==
inf I t:,.x,
L sup I
t:,.x).
The Riemann Integral
51
The lower sums are bounded above by sUP[a,b] I(b - a), and the upper sums are bounded below by inf[a,bJ I(b - a). Define the lower Darboux integral of I on [a, b] by
J
b I(x)dx
= sup L(/, P) P
:.........a.
and the upper Darboux integral of f on [a, b] as -b
J a
The bounded function
I
-dJ
I
I(x)dx = infU(/, Pl· P
is said to be Darboux integrable on [a, b] if the
numbers Jb I (x )dx, J a I (x )dx are equal. In this case we write D J: f(x )dx. ~
Exercise 3.3.2. Show the following. 3.
For any partition P of [a, b], L(f, P) < U(f, Pl.
b. Refinements do not decrease lower sums or increase upper sums: If p;. is a refinement of P, L(/, P) < L(/. P U P*) and U(J, P U P~) < U(J, Pl· Hint: P'" = P U {z}, XJ-I < Z < xJ. c. Any 10JVer sum does not exceed any upper sum: L(J, PI) < U(J, P2 ). Hint: Part b, PI U P2 . b
-b
d. J I(x)dx < Jal(x)dx. Hint: L(j. PI) < U(J, P2 ) for any parti:.-a
-dJ
tions of [a,b]. "Fix" PI and vary P2 • Thus L(f,P1 ) < Jal(x)dx. Vary Ph and so on. Theorem 3.3.2 (Darboux. Integrability Criteria). Aboundedlunction I on the interval [a, b] is Darbollx integrable iff for each E > 0 there e.xists a Riemannpartition Pfi o/[a, b] so that Lpt;(suP 1- inff)6.x < E.
Proof Let E > 0 be given. Assume we have a partition Pfi of [a, b] so that Lpe(suP 1- inff)6.x < E; that is, U(f, Pe ) - L(/, Pfi) < E. However, b
L(f, PE) < L -dJ
-b
f(x)dx < J
b
Thus Jal(x)dx - J I(x)dx < =-a
b
E.
Q
f(x)dx < U(j. PE)'
We may conclude -b
Lf(X)dX = J/(x)dX.
52
A Garden of Integrals
We have Darboux integrability. Now assume we have Darboux integrability: b
Lf(X)dX = Let
E
f
-b
=D 1.
b
/(x)dx
f(x)dx.
> 0 be given. Because b
sup L(/. P) = L f(x)dx = D
J.
~
b
f(x)dx =
f /(x)dx = infU(/.
P),
we have partitions PI. P2 of [a, b] so that D
t
f(;x)dx -
i< LU,
< U(/. Pz) < D
f
PI U Pz) ::: U(/. PI U Pz)
f(x)dx
+
We have a partition PI U P2 = PE of [a, bl with U(j, that is L(sup I - inf/)~x < E. 0
i·
p,J -
L(f, P E) <
f,
PI!;
Note: Theorems 3.3.1 and 3.3.2 tell us that Riemann integrability is equivalent to Darboux integrability.
3.4
Weakening Continuity
Lp
To achieve smallness of the sum (sup f - mf I) ~x we have required smallness of sup I - inf I on every subinterval of a partition. Is this necessary? For example, divide a given partition into two collections of subintervals. In the first collection, put the subintervals with "small" sup I - inf I. This, we hope, will be most of the subintervals in terms of total length close to b-a. In the other collection, put the subintervals with "large" sup I -infl (although bounded by sUP[a.b] I -infra.b] f). Their total length is, we hope, "small" in comparison with b - a. This sorting was Riemann's idea. Theorem 3.4.1 (Riemann's Theorem). Let I be abounded fimction on the interval [a, b], and let OJ and 1 be any positive numbers. 112en I is Riemann integrable on [a, b] iff we have a positive constant 8 so that, lor
53
The Riemann Integral
any partition of [a, b] into subintervals of length less than 0, the sum of the lengths 01 the subintervals [Xk-I. Xk] with SUP[Xk_l,x,d f - inf[xk_t.xkl I greater than w is less than l.
Proof. We begin by assuming that I is Riemann integrable OD. [a, b], with w and I positive numbers. Apply the integrability criteria from Theorem 3.2.1. Then, given an E > 0, we have a 0 > 0 so that if P is any partition of [a, b] whose subintervals have leDgths less than 0, then Lp(SUP I -inf/) 6.x < E.
=
Let e- wI. We have a 8 for this E so that if P is any partition of [a, b] whose subintervals have lengths less than 8,
L (sup I - inf I) 6.x p
(sup I
-
inf I) 6.x
+
sup / -inf / ~w
(sup I
-
inf f) 6.x
sup / -inf />w
< wI. But then (sup 1- inf I) 6.x < wl.
6.x < sup f -inf />aJ
sup f -inf / >w
That is, the sum of the lengths of the subintervals with sup I is less than [. For the other direction, let E
w=---
E
-
inf I > w
> 0 be given, and let
1=
2(b - a)'
E
2 (SUP[a,b]
f -- inf[a,b] f) .
The theorem is trivial if I is constant, so we may assume SUP[a,bl f inf[a,b] I > O. By assumption we have a 0 > 0 so that for any partition of [a, b] into subintervals whose lengths are less than 0 we can describe the sum of the lengths of the subintervals with sup f - inf f greater than e-/[2(b - a)]: " L.J sup f -inF /> 2ll-a}
6.x <
(
2 sUP[a,b]
E
I -
inf[a,b]
I ).
54
A Garden of Integrals
But then L(sup f - inf f) 6.x p
sup f -inf'
':5
(sup f - inf f) 6.x 2\6!..0.)
(sup f
+
-
inff) 6.x
supf-inrf>~
L
< 2 (b €- a ) 6.x + (sup rb] La, €
f - f) inf
[a,b}
L
sup f -inff
6.x >---!i..2lb.;;;;a)
€
<:2+ :2' This concludes the argument.
D
Exercise 3.4.1. Given
Ilx) 9..
= {
~
1 1 2n < x < 2n _ I' n = I, 2, ... , otherwise.
Use Riemann's Theorem to show that f is Riemann integrable on [0, 1]. Hint~ Because sup f -inff < 1 on any subinterval of [0, 11 and L 6.x = 1, we assume 0 < w, I < 1. Divide [0,1] into subintervals [0, 1/2], [1/2, 1]; choose N > 3 so that 1/ N < 1/2; and define 8 = I/N 2 :
o
1
Because f has N - 1 "jumps" on [1/ N, 1], we have' fewer than N subintervals on [1/2, 1] where f jumps by 1. Since each subinterval has length less than 0 = 1/ N2, the sum of the lengths of the subin" tervals with sup f - inf f > W is less than N(l/ N2) = 1/ N < 1/2. Of course we have an infinite number of jumps on [0,1/2], but the total length of such subintervals cannot exceed 1/2. Thus, the total length of subintervals, for any partition of [0, 1] with 8 = 1/ N2 having sup f - inf f > W, is less than I. b. Use Theorem 3.3.1 to show lIN < E',
f
is Riemann integrable on [0,
11, Hint:
55
The Riemann Integral
J
P
E
= lO,
1111 N - 2N2' N' N
1 1111111 + 2N2' '" , "3 + 2N2' 2N2' + 2N2 ' 1
2-
2.' 2
Exercise 3.4.2. Given
J
I (x) = 1 with I > points.
0 1/q
x = 0, I, or irrational for 0 < x < 1, x = p / q, for p, q relatively prime numbers,
t at one point, I
>
i
at three points, and
I
> l/n at most?
o. Show that I is continuous at the irrational numbers and discontinuous at the rational numbers. Hint: J = {x I I(x) > E} is a finite set. For Q! irrational, choose an interval about Q! that contains no points of J. b. Using Riemann's Theorem, show that I is Riemann integrable on [0, IJ. Hint: Choose N so that 1/ N < wand 1/ N < l. Define = 1/ N3. The jumps of I e.."{ceed 1/ N on fewer than N2/2 intervals, 1 each having length less than 1/ N 3 . What is the value ofR J0 I(x) dx? Despite a dense set of discontinuities, I is Riemann integrable.
a
Exercise 3.4.3. In this exercise we construct Riemann's example of a function with a dense set of discontinuities that is Riemann integrable. Let
I(x)
=! o -! ±2>~ x
x =
be extended periodically by I(x) = I(x Riemann's function
R(x)
<
<
~,
+ 1).
=EI(I~). n n=l
R is bounded by L~l 1/2n2.
Because I(x) is discontinuous iff x is an odd multiple of -;}, 1{IiX) is discontinuous iff I1X is an odd multiple of that is, iff x = (2i - 1)/212 is a rational number with an odd numerator and even denominator. Restricting our attention to [0, 1], we have discontinuities at
t.
1. 1. 3 1 3 5. . 1 3 2' 4' 4' 6' 6' 6' ... , 2n • 217
1" • ,
211 - 1 . 217 I
and so on. We have odd multiples of ~, odd multiples of ~, odd multiples of I odd mUltiples of 1/21Z, ... I a dense subset of [0, 1]. The set {x I x = (2i - 1)/21Z; i ,n = ±1, ±2, ... } is a dense subset of the reals. Riemann's function has a dense set of discontinuities. Again, consider [0, 1]. (The periodic nature of I extends the discussion to the real line.)
i, ...
56
A Garden of Integrals
t.
a. Suppose x is an odd multiple of Then (lex) is an odd multiple of iff k = 1,3,5, .... So f(kx+) - f(kx-) = = -1 iff k = 1,3,5, .... Because we have uniform convergence (Weierstrass M-test),
!
-! - !
i.
b. Suppose x is an odd mUltiple of Then (kx) is an odd multiple of t iff k = 2· 1,2· 3,2 . 5, .... So f(kx+) - f(kx-) = -1 iff k - 2,6, 10, .. .. .
+
_
1
R(x ) - R(x ) = - 22 -
1 1 1 :n: 2 62 - 102 - ... = - 22 . S·
t.
c. Suppose x is an odd mUltiple of Then (kx) is an odd multiple of iff k = 3 . 1,3 . 3,3 . 5, .... So f(kx+) - f(kx-) = -1 iff k = 3,9,15, ....
t
+
_
1 R(x ) - R(."C ) = - 3 2
-
1 92
-
1 152
-
...
= -
1 32
Te 2 • S.
d. Suppose x is an odd multiple of I/2n. Then (kx) is an odd multiple of ~ iff k = n· I,n· 3,n· 5, .... So f(kx+) - f(kx-) = -1 iff Ie = 11, 311, 5n, ....
+
_
R(x ) - R(x )
1
= - JZ2
-
1
32n2 -
1
52112 - •••
1
= -11 2
(Te8 2 ) .
In summary, R is a bounded function with a dense set of discontinuities. e. Show R is Riemann integrable on [0, 1]. Hint: The interval [0, 1] has a finite number of points (rational numbers of the fonn (2i - 1)/2n) where R "jumps" by rr/8n 2 > w: 0 < 1'1 < r2 < ... < rK < 1. We have shown that continuous functions are Riemann integrable. These last exercises show that, in some cases, a countable number of discontinuities may not prevent Riemann integrability.
3.5
MonotoDlOc Functions Are Riemann Integrable
We begin by reviewing some characteristics of monotonic functions pertinent to Riemann integrability.
The Riemann Integral
57
Exercise 3.5.1. Show that monotonic functions have limits at each point of their domain.
right~
and left-hand
Exercise 3.5.2. Show that monotonic functions have a countable number of discontinuities. Hint: J(x-) < T < J(x+), for r a rational number. Noting that Lp (sup J -inf J) 6.x < IJ(b)-J(a)18, conclude the Riemann integrability of monotone functions.
In spite of this relatively straightforward argument for integrability, monotonic functions may be very interesting. For the following exercises, let TIl T2. T3,'" be any enumeration of the rational numbers in (0,1). Define J on [0, IJ by J(x) = L{nlrn<x} 1/2"', where 0 < x < 1, J(O) = 0, and J(1) = 1. Clearly J is monotone increasing and thus Riemann integrable on [0, 1J. Exercise 3.5.3. Show that lim,x-+fK J(x) = J(ri) = J(rK). Hint: Fix a rational number rK in (0. 1). Since monotonic functions have right- and lefthand limits, certainly J(ri) < J(rK). If J(Ti) < J(rK), choose E > 0 so that JeTi) + E" < J(rK), and then choose a natural number N so that 1/2N < E/2. Now select 8 > 0 so that (rK -8, rK) n {rII r2 •... , rN} = cp. For r K - 8 < x < r K,
J(rK) - Jex)
=
" ~
1
lIE + ... = 2N < 2"
2'" < 2N+1
or
{nlx:srn
J(rK) -
2'
_
< J(x) :::.: J(rK) < JerK) -
E,
and we have a contradiction. Exercise 3.5.4. Show that J(rk) + 1/21c = J(rt) for all k. Hint: For rk < x, J(x) - Jerk) = L{"'lrk.=:;rll<.'~} 1/211 and
That is, J has a jump of 1/2k at each rational I'k. Exercise 3.5.5. Show that J is continuous on the irrationals.Hint: Suppose (X is an irrational number in (0, 1), and let E > 0 be given. Choose N so that 1/2N < E. Construct an open interval about (X that does not contain 1'1, 1'2, .•• rN. Every rational in this subinterval has a subscript larger than N. If x is a point of this interval, IJ(x) - J(a)1 < LN+l 1/2" < E, and J is a strictly increasing function that is continuous on the irrationals and has a jump of 1/21c at 1"k. I
58
A Garden of Integrals
Again, a countable number of discontinuities does not prevent Riemann integrability.
3.6
Lebesgue's Criteria
Finally, in 1902, the French mathematician Henri Lesbesgue determined necessary and sufficient conditions for a bounded function to be Riemann integrable. (See Section 6.1.2.) Theorem 3.6.1 (Lebesgue, 1902). Suppose I -is a bounded function 01 fa, b]. Then I is Riemann integrable on [a I b] iff I is continuous almost evelywhe1"e.
Proof. To say
I
is continuous almost everywhere on [a. b] means we can "cover" the discontinuities of I with a countable collection of open intervals h, 12,"" whose length :Ll(1k) can be made arbitrarily small- that is, a set of measure zero. The idea here is that where I is continuous ("most" of the interval [a, b] by length), sup I - inf I will be small. Where f is discontinuous such intervals comprise a "small" subset of [a, b]. Integrability is determined by the behavior of f on "most" of the interval. We begin with the assumption that the bounded function I is Riemann integrable on [a, b}. We will show that the set of discontinuities of f is a set of measure zero. Show the set of discontinuities of f is the union of the sets
1~ x
E
[a, b] I lim ( 8-+0
sup (x-8,x+8)
I(z) -
inf (x-o,x+l1)
fez)) > .!.l, n~
for n = 1, 2, .... It will be sufficient to show that such sets have measure zero, Let E > 0 be given and fix n. From Cauchy's Criteria for Riemann integrability (Theorem 3.3.1) we have a partition P of [a, b] so that L(sup f
-
inf I) 8x <
P
.!...-. 411
If the subinterval [Xk-lt xkl contains a point d with
lim (
8~
sup
~~~~)
I(x) -
inf
~~~~)
I(X))
>
.!.n
59
The Riemann Integral
in its interior, then (sup f
and
-
~:
inf f) Ax :::
on this subinterval. But
L
(sup f - inf f) Ax
L
+
"jale:nor"
(sup f - inf f)
E
< 4
/).x
otherwise:
• II
We may conclude that L:"inle:nor" Ax < E/2. Otherwise, such points are division points - a finite set - and can be covered by a finite number of intervals whose total length is less than E/2. We have covered the set dE [a b] I
I
I
lim (
8-+0
sup
(d-8,d+6)
f -
inf
(d-8.d+6)
f)
>
.!.l n
with a finite set of intervals of total length less than E. By the arbitrary nature of E, we conclude that this set and the points of discontinuity of the bounded function f have measure zero. For the other direction, we assume the set of discontinuities of the bounded fUnction f has measure zero (/ is continuous almost everywhere). We will show that / is Riemann integrable On [a, bJ. We have a cover of the discontinuities of f by open intervals 11 .12 • ••• , with E l (Ik) < E. At each point Jl: of continuity of f, we have an open interval J:x: containing x so that sup f - inf f over the closure of this open interval is less than E. The collection of open intervals I}, 12 •... Jx is an open cover of the compact set [a, b1. By the Heine-Borel Theorem, we have a finite subcollection that covers [a, b]. The endpoints of these subintervals (that are in [a, b]) and {a, b} are division points for a partition P of [a, b]. Thus, I
L(sUp f p
-
inf f)
/).x
< (sup
f -
< (sup
f - f)
[a,b]
[a,b]
inf
[a,b]
inf
[a,b]
f)
L l(lk) + L l{i E
E
x)
+ E(b -a).
Appealing to Cauchy's Criteria for Riemann Integrability (Theorem 3.3.1), we may conclude that f is Riemann integrable on [a. b}. The proof is complete. 0
60
A Garden of Integrals
The reader should compare this result with the corresponding result for Cauchy integrals (Theorem 2.2.1).
Exercise 3.6.1. a. Settle the Riemann integrability of the functions of Ex.ercises 3.4.1, 3.4.2, and 3.4.3 using this result of Lebesgue. b. If I is Riemann integrable on [a, b], then III is Riemann integrable on [a, b]. Show that the converse is false. Hint: Let
lex) = {
~1
x rational, x irrational,
and consider the interval [0, 1].
3.7
Evaluating it la Riemann
N ow that we have existence of the Riemann integral for bounded, continuous almost everywhere functions, may we recover a function from its derivative using the Riemann integration process, as with the Cauchy integral? A Fundamental Theorem of Calculus offers some answers.
Theorem 3.7.1 (FTC for the Riemann Integral). If F is a differentiable function on the interval [a, b], and F' is bounded and continuous almost evelywhere on [a, b], then 1. FI is Riemann integrable on [a, b], and 2. RJ: F'(t) dt
= F(x) -
F(a) for each x in the interval [a, b].
Proof. The frrst conclusion follows from Lebesgue's result (Theorem 3.6.1). For the second part, existence of the Riemann integral of F' means that, given an € > 0, we have a partition P of [a, x] so that L p sup FI 6.x and Lp inf F' 6.x are within € of each other. By the mean value theorem for derivatives, F(x) - F(a) =
L F(Xk) -
F(Xk-l) =
L F'(Ck) 6.x. p
Since LP FI(Ck) 6.x is between Lp infF' 6.x and Lp sup F' 6.x, we have that F(x)- F(a) is between Lp inf F' 6.x and LP sup F' 6.x, along with RJ: FI(t) dt. Thus IF(x) - F(a) - RJ: F'(t) dtl < 2c:. 0
The Riemann Integral
61
Compare this with Theorem 2.3.1. The conditions on the derivative have been weakened from continuous to bounded and continuous almost everywhere. Look also at Theorems 6.4.2 and 8.7.3.
Exercise 3.7.1. a. Given F(x)
=
I
show that RIol F'(x) dx b. Given F(x) =
2
x sino(rc/x)
0 < x < 1,
x =0.
= O.
I
2
x Sino(rc/x'l)
0 < x < 1, .'"C
= 0,
show that F' exists but F' is not Riemann integrable. What are some of the properties of the Riemann integral? Can we recover a function from its integral using differentiation? Let's look at another Fundamental Theorem of Calculus for the Riemann integral. Theorem ~ 7.2 (Another FTC for the Riemann Integral). Suppose J is a bounded and continuolls almost everywhere JUJlction on the interval [a, b]. Define F on [a, bl by F(x) = J(t) dt.
RI:
1. Then F is continuoZls (il1 fact, absolutely continuous)
011
[a, b].
2. If J is cOlItinuoZls at a point Xo in [a, b]. then F is differentiable at Xo and F'(xo) = J(xo). 3 F' =
J
almost evelywhere.
Proof The function F is well defined by Lebesgue's result (Theorem 3.6.1). To show continuity of F, select B so that If I < B on the interval [a, b]. Then
IF(y) - F(x)1
=
Llet) Y
R
dl < ElY -
xl.
Absolute continuity follows easily (see Definition 5.8.2). Now we assume that J is continuous at Xo. Then given an € > 0, we have a 0 > 0 so that f(xo) - E < J(t) < J(xo) + €, for f E (xo - 8, Xo + 8) n [a, b]. Constants are Riemann integrable, and J is Riemann integrable by assumption.
62
A Garden of Integrals
From integration, then,
R1
X
xo
[f(xo) - €] dt <
Rl
x
xo
f(t)dt
x
< Rl [f(xo) xo for x E [XO, Xo
+ e] dt,
+ 8) n [a, h}_ Thus -E
:5 F(x) - F(xo) - f(xo) < e. X-Xo
Argue x E (xo - 8, xo]
n [a, b], and the conclusion follows.
0
Compare with Theorems 2.4.1, 6.4.1, and 8.8.1. Exercise 3.7.2. Given lnx
-= J; (1/ t) dt
for x > 0, show the following.
a. (lnx)' = l/x, for x > O. b. In x < 0 for 0 < x < 1, In 1 = 0, and In x > 0 for x > 1. c. In(l/x)
= -In x.
d. In(xy) = lnx
+ lny
for x. y > O.
e. As x -+-
00,
In x -+- 00. As x -+- 0+, In x -+- -00.
f. As x -+-
00,
(lnx)/x -+- O. As x -+- 0+, x lnx -+- O.
Exercise 3.7.3. Given
f(x) =
{
-I ~
-1 < x <0, 0< x < It 1 ::: x :5 3.
a. CalcuiateRJ~lf(t)dt,RJ; f(t)dt,RJ: f(t)dt,andRJo
2x
f(t)dt
b. Using the definition of continuity, investigate the continuity of these functions at x = 0 and x = 1. Calculate their derivatives where appropriate. Exercise 3.7.4. Proceed as in the previous exercise, given
-1::: x::s I, 1 < x < 2. f(x) = { 3-x 2 <x< 3.
~
63
The Riemann Integral
Exercise 3.7.5. Given
f (x) == a. Sketch F(x) ==
O x = 0, sin(n/x) 0 < x < 1.
l
RIol: f(t)dt. Hint: f: = 101 + ft.
b. Is F continuous from the right at x
= O? Hint:
== O? Hint: Write
c. Is F differentiable from the right at x
Rf sin (~) dt = R-{ (_t
IF(x)1 < x.
2
)
sin
(7) (- t;) dl
and integrate by parts. d. Suppose
f()
x ==
l O X = o. sin(n/xC!)
0 < x ::: 1. a > O.
What now? Exercise 3.7.6. Let f(x)
==
-x for -1 < x < 1 and f(x
+ 2) = 2f(x)
for all x. a. Calculate
F(x) = R G(x)
==
L:
f(t) dt
Rj2X fCt) dt,
for t in [-1.4]. and
-1
H(x) = R
j
l/2X
-1
f(t) dt.
b. Discuss continuity and differentiability of F, G, and H.
3.. 8
Sequences of Riemann Integrable Functions
What about convergence theorems for sequences of Riemann integrable functions? We would like lim /n = f to imply lim J In == J lim /n. Let's explore this with some exercises.
64
A Garden of Integrals
Exercise 3.8.1. Given
fk(X) =
0 < x < 1/ k, 1/ k < x < 2/ k, 2/ k < x .:::: 1.
kx -k(x 0 2/ k)
{
Exercise 3.8.2. If 71, 72, ••. is an enumeration of the rational numbers in (0, 1), define a sequence of functions {fk} by
jj (x)
=
k
110
x = 7 1: 72, •.. 7k.
for 0 ::::: x < 1.
t
otherwIse,
i
Show that lirnRJo fk(X) dx i:- RJ;(limfk)(X) dx, even though the sequence {fk} is monotonic and uniformly bounded. In general,
Rt
fk(X)dx -R t
f(x)dx
= R t[/k(X) < Rt
f(x)] dx
l/k(x) - f(x)1 dx
< (b - a) sup Ilk
- II <
E(b - a),
[a,b)
provided Ifk - fl < E throughout the interval [a, b]. This would suggest uniform convergence - assuming, of course, the limit function f is Riemann integrable. Theorem 3.8.1 (Convergence for Riemann Integrable Functions). If {fk} is a sequence 01 Riemann integrable juhctiol1S converging un~ormly to the junction f on [a, b], then f is Riemann integrable and RJa f(x) dx = limRJ: Ik(X) dx.
1
is Riemann integrable on [at b]. Let E" > O. From the uniform convergence of fk on [a, b], given E > 0, we have K > 0, so that
Proof We will show that
E
Ik(x) - 4(b _ a) < f(x) < fk{X)
for all x in [a, b], when k > K.
€
+ 4(b _ a)
(1)
The Riemann Integral
65
So J is bounded on [a , b], and because fk is Riemann integrable on [at b] we have a partition P of [a, b] so that Ep(sup Jk -infjj,) 8.x < E/2 (see Theorem 3.3.1). However, from (3.1) we have
LinfJk 8.x - =- < LinfJ 8.x < I:supJ 8.x p 4 p p
::: L sup fk 8.x + ~t
for 1c > K.
p
Lp
That is, (sup f - inf f) 8.x < ~. We may conclude that J is Riemann integrable on [at b], and then by integrating (3.1), the second conclusion follows. 0
Exercise 3.8.3. H.
Given fk(X)
= 1 +~3X3' for 0 <x ::: 1, calculate I' R (1
y;l 10
2x
1 + k 3x 3
d X.
l
b. Calculate R Jo (Riemann's Function) dx. c. Calculate the Riemann integral of the function in Exercise 3.4.1 by defining an appropriate sequence of functions {Jk}. We are now in a position to give (in slightly modified form) an example due to Volterra (1881) of a function with a bounded derivative whose derivative is not Riemann integrable. This example prompted Lebesgue to develop an integration process so that such functions would be (Lebesgue) integrable. Gordon (1994) supplies a complete explanation in his book. Our approach to the example has three stages: construction of the Cantor set, constructi on of a modified Cantor set, and a return to x 2 sin(l / x).
3.9
The Cantor Set (1883)
On a list of famous mathematical objects, the Cantor set would appear early, (I dare say we can all remember our first encounter with it.) It is the origin of many examples and counterexamples in analysis. We begin with these steps. 1. Divide [0, 1] into three equal parts and remove the middle third, the ~). We are left with the closed set Fl [0. tJU{j, 1]. open interval
(t,
=
66
A Garden of Integrals
2. Divide each of the two closed subintervals of FI into three equal parts and remove the middle thirds, the open intervals (~, ~) and (~, ~). The closed set F2 = [0, U [~, ~] u ~] u [~, 1] remains.
II
[1,
3. Continuing in this way, after n steps we have deleted 1 + 21 + ... + 2n- 1 = 2n -1 open intervals and have left the closed set Fn , consisting of 2n closed intervals, each of length 1/3n . The Cantor set C is the intersection of the closed sets Fn:
c = nFn • For comparison, here is an equivalent formulation: The Cantor set is the points of [0, 1.] that have a base 3 expansion without the digit 1. That is, forxeC, x
=
al
a2
a3
3T + 32 + 3 3 + ... ,
where an = 0 or 2.
Example 3.9.1. The Cantor set C has measure zero (Exercise 5.5.2). We will show that C is uncountable. This statement may seem unreasonable. After all, the only obvious memthe countable set of bers of the Cantor set are 0, I, ~. ~, ~, ~, ~, endpoints of the open intervals that we removed. But we have something besides endpoints in this set, for example
t
2 .020202··· = 32
I
2
••• ,
2
+ 34 + 36 + ... =
2/9 1 1- 1/9 = 4'
It is appropriate that Cantor's diagonalization process can be used effectively. Assume that C is countable and make a list of base 3 expansions: Xl
=
X2
= .a21 a 22 a 23 ••.
.alla12 a 13."
0 ann
= 2,
ann
= O.
all = { 2
The Riemann Integral
67
The number x is 110t on our list. We have an uncountable set of measure zero. The Cantor set, a closed set, contains no intervals. It is nowhere dense:
(
3k3+ 1 t 3k3+ n n
2) (a, C
b) C (0 , 1)
for
12
sufficiently large.
Every point of the Cantor set is a limit of points in C (so it is a perfect set) and points not in C. For example, suppose Xo E C = nFn and In is a closed interval of Fn that contains xo. Since the length of In is 1/3'1, we may choose Ii so large, say N, such that for any open interval (a, b) continuing xo, IN C (at b). At least one of the endpoints is different from
xo· Thus, given any open interval about a point of the Cantor set, it contains a different point of the Cantor set and a point not in the Cantor set. The Cantor set is a nowhere dense pelfect set of measure zero.
For Volterra's example, we want a nowhere dense perfect set of positive measure. We will construct a modified Cantor set where we do not remove as much at each stage.
3.10
A Nowhere Dense Set of Positive Measure
Our construction of a modified Cantor set occurs in a series of step~, as follows.
t.
1. Let a I be positive and less than Remove the open interval (a 1 • 1 ad from [0, 1]. We are left with the closed sets F1•1 : [0, all and Fl.2 : [1 - at. 1], each of length al. See Figure l(a). 2. Let a2 be positive and less than tal. Remove the open intervals (a2t al - a2) and (I - al + a2. 1 - a2). We are left with the dosed sets F2,1 = [O.a2], F2,2 = [al- a2,ad. F2,3 = [I-a), I-at +a2], and F2 ,4 = [1 - a2. 1], each of length a2. See Figure l(b). 3. At step Il - 1 we have closed intervals F n- 1,l, ••• Fn-1,211-1, each of length aI/-I. I
4. At step n each of these closed intervals F,l-l,k is divided into two closed intervals and one open interval; see Figure ICc). We have Illlk
= FII -
l, k -
and J",k has length an-l - 2all •
(F",2k-1 U Fn,2k),
68
A Garden of Integrals
(
o
) 1
(
12,1
)
F 2,2
) F 2•4
(
o
1 (b) F 2,h F 2;1.. F 2,], F 2•4
Figure 1. Constructing the modified Cantor set What have we constructed? Let
Let an = [(1/2n) ao - 2al = at - 2a,
1
6.
and l(ll,d =
= 118
a2 - 2a 3 =
+ (1/3 n)]/2. Then
and t(lz.tl
;4 and
t(13,t}
1
6'
+ t(J,,2) = 2 (118)
,
+ t(h,2.) + t(l3,3) + t(h,4) =
4(;4) ,....
The sum of the lengths of the open intervals removed is %+ 2 C118) 4(514) + ... This modified Cantor set has measure In general.
= !.
!.
+
We claim that this modified Cantor set is a nowhere dense perfect set of measure For suppose that pEe. Since .f.CFn.k) = an -+ 0, given any neighborhood of p we have N and K so that p E FN.K C (p - 8, p + 8). In fact, every neighborhood of p contains points of C and points not in C, namely those in (IN+l,K). See Figure 2.
!.
In the next section, we introduce the remarkable Cantor/unction, developed as the uniform limit of a sequence of continuous functions.
69
The Riemann Integral
I
I
al
FN + J• 2K- 1 P
IN+l.K
Figure 2.
3.11
FN,K
Cantor Functions
We will construct a sequence {c"} of continuous, monotone increasing functions on [0, 1]. The Cantor function will be the uniform limit of this sequence. Here are the steps:
1.
Cl (0)
= 0,
Cl (x)
=
t,
~ < x < ~,
Cl (1)
= 1,
Cl (x)
is linear
otherwise and continuous,
- 0, C2 () 1 'f '9 1 < ;c < 9' 2 "4 2 1'f 31 < x :5 3' 2 "4 3 1'f '9 7 <x 2 . C2 (0) X -- '4] c2(1) = 1, C2(X) linear othervvise and continuous. 3, After n steps we have the deleted 2" - 1 open intervals. On them Cn assumes the values 1/2" , 2/2" •... 1 (2" - 1)/2" left to right. Cn (0) 0, c"(I) = 1, and cn(x) is linear on the 2" closed intervals and continuous.
=
The Cantor function C is the pointwise limit of this sequence of increas" ing continuous functions. Clearly C(O) = O. C(1) = 1. We claim C is continuous. Show that
c-c11+11 I"
=
1 (3)" ( 1)
---<211+1
-2
1
-3"+1 - 3 .211+1 .
=
Consequently, limcn C uniformly on [0,1]. Furthermore, C' 0 on the u ... , u set of measure 1. intervals (t, ~), (b.~) U (~, The continuous function C is 0 at 0, 1 at 1, but has a derivative equal to almost everywhere. C is a monotone function on [0.1] whose derivative vanishes almost everywhere.
f)
o
Exercise 3.11.1. Calculate the integrals C Hint: Theorem 3 8.1.
Jot C(x) dx and C fol xC(x) dx,
70
A Garden of Integrals
The next section develops Volterra's example of a function with a bounded but not Riemann integrable derivative. As mentioned before, Volterra's ex· ample prompted development of the Lebesgue integral.
3.12
Volterra's Example
Vito Volterra's example dates from 1881. Let C denote a Cantor set of positive measure, say ~. Suppose In,k = (a, b) is one of the open intervals removed in the construction process for C. Let
I
c = sup {x a < x < a ; b • [ (x - a)' sin
a)
=
+a -
c,
= (a, b) by
and define a function fn,k on In,k
(x - a)2 sin (x~a
fn,k(X) =
C~ J o} .
)
a < x < c.
(c - a)2 sin (c~a)
c< X< b
(x-b)2 sin(b':x)
b+a-c
(x - b)2 sin
a
(b":x) b
a+b (x - a)2 sin
2
(x':a)
_(X-b)2
Figure 3.
In •k
The function fn,k is differentiable on I",k. the derivative is bounded by 3 on IIl,k, and
f; ,k
(a + _1) nx
=
f~ k (b - _1 ) nx
=
± 1.
I
We define Volterra's function V on [0, 1] as follows:
Vex)
= { fn.~(x)
X E
In ,k ; 11. k
otherwise.
=
I. 2, ... ,
71
The Riemann Integral
We claim function V is differentiable on [0, 1]. Certainly the open intervals In,k do not pose a problem. Suppose Xo belongs to C and x > XO. If x also belongs to C, then Vex) V(xo) O. Otherwise, x belongs to some In,k = (a, b), and Xo < a < x < b. But
=
Vex) - V(xo)
=
=
Vex) < max{(x -
and
a)'- (b - X)2} < (x - xO)2 t
Vex) - V(xo)
- - - - - <x-xo. X-Xo
=
Thus V' 0 on C. So V is differentiable on [0, 1], and IV' I < 3 on [0, 1]. We claim V'is not continuous on C, a set of positive measure ~, and thus is not Rjemann integrable on the interval [0, 1]. Let Xo E C. If ,1:"0 is an endpoint of some In,k = (a, b), Xo = a or b, we have a sequence of points {a + l/n1f} or {b -l/nx} with V' = ±1, whereas V'(a) = V'(b) = O. If Xo E C and Xo is not an endpoint of l",k, then we have a sequence of endpoints converging to Xo (nowhere dense), and thus xo will again have points arbitrarily close to Xo with V' = ±1, whereas V'{xo) = O. This concludes Volterra's example. So the Riemann integral does not necessarily integrate bounded derivatives. We will show in Chapter 6 that the Lebesgue integral does not have F' (x) dx = F(1) - F(O) in this example. this defect. In fact, L
J;
3.13
Lengths of Graphs and the Cantor Function
Does the graph of the Cantor function have a length? The Cantor function has a nice graph, continuous and nondecreasing. The length would be
where P is any partition of [0. 1]. Because
L
[(a.,;?
+ (aC)2]1/2
<
L(ax + ac) = 2
p
for any partition, the length of the graph of C is at most 2. We claim that the length of the graph of C is actually 2. Let's analyze this claim.
72
A Garden of Integrals
1. At the first stage of the process, we have a horizontal line segment of length and what remains exceeds 2{~): Use the partition PI = ~, I}.
to, i,
t,
2. At the second stage, we have three horizontal line segments of length t+2(~), and what remains exceeds 4{i): Pz = {O, j, ~r it~, I} is the appropriate partition.
l, t,
3. At the third stage, we have seven horizontal line segments of £ength + 2(~) + 4(1,) and what remains exceeds 8(l), and so on.
t
4. Thus, at the 11th stage, the length of the graph of the Cantor function exceeds (j) + 2(l) + 4{i7) + ... + 2n- I (1/3 n) + 2n(1/2n), and this quantity converges to 2. A comp lete description of the Cantor function and other remarkable functions may be found in the beautiful books by Sagan (1994) and by Kannan and Krueger (1996).
3.14
Summary
Two Fundamental Theorems 01 Calculus for the Riemann Integral
If F is differentiable on [a, b], and F' is bounded and continuous almost everywhere on [a, b], then 1. F' is Riemann integrable on [a, b], and
2.
!
R!~"'C F'Ct) dt = F(x) - F(a), where a ::: x ::; b.
If f is bounded and continuous almost everywhere on (a, b] and F(x) R!: I(t)dt, then 1. F is absolutely continuous on [a. b], 2. F is differentiable almost everywhere on [a, b], and
3. F' ==
1 at points of continuity of 1 (almost everywhere).
I Cauchy t Riemann
Figure 4. Cauchy integrability implies Riemann integrability
=
The Riemann Integral
3.15
73
References
1. Bressoud, David. A Radical Approach to Real Analysis. Washmgton. matical Association of America. 1994
Mathe~
2 Darboux, I. Gaston. Ml!:moire sur 1es fonctioDs discontmues. Annales Sci Ecole Normale Superiellre 4'2 (1875) 57-112. 3. Goldberg, Rlchard R. fl..fethods of Real Analysis New York: Wiley, 1964 4. Gordon, Russel1. The Integrals of Lebesgue, Denjoy, Perron and Henstock Providence, R I.: American Mathematical Society, 1994. 5 Kannan, Rangachary, and Carole King Krueger Advanced Analysis Real Line New York: Springer, 1996.
011
tile
6. Ross, KenneUl. ElementalY Analysis: The TheolY of Calculus. New York: Springer, 1980. 7. Sagan, Hans Space-Filling Curves New York: Springer, 1994. 8. Simmons, George Calcllius Gems New York:
McGraw~Hill,
1992.
9. Stromberg, Karl. IntrodZlctzoll 10 Classical Real AnaZvsis Belmont, Calif.' Wadsworth, 1981.
CHAPTER
4
Neglect of mathematics works injury to all knowledge, since he who is ignorant of it cannot knOl'll the other sciences 01" thin.gs of this world. And what is worse. men who a"'e tluts ignorant are unable to perceive their own ignorance and so do not seek a remedy. - Roger Bacon
4.1
Generalizing the Riemann Integral
After Riemann's fonnulation of the integral, various generalizations were attempted. One of the most successful, the so-called Riemann-Stieltjes integ/"al, was obtained by Thomas StieItjes (1856--1894). Stieltjes was trying to model mathematically the physical problem of computing moments for various mass distributions on the x-axis, with masses lni at distances di from the origin (Birkhoff, 1973): "Such a distribution will be perfectly detennined if one knows how to calculate the total mass distributed over each segment Ox [of OX]. This evidently will be an increasing function of x. . .. Accordingly, let cp(x) be an increasing function defined on the interval (a. b) .... It is convenient always to regard cp(b) - cp(a) as the mass contained in the interval (a, b) .... Let us now consider the moment of such a mass distribution wrt [with respect to] the origin. Let us set a = xo, b = X,,, and let us intercalate 12 - 1 values between Xo and xu: .1.0
<
XI
< X2 < ... <
X n- l
<
XI!'
Next, let us talce It numbers ~l. ~21 ••• I~" such that Xk-l ::: ~k ::: Xk· The limit of the sum ~l [CP(XI) - Cp(xo)] + ~2 [CP(X2) - CP(Xl)] + ... + ~/I [cp(x n ) - cp(xlI-J)] will be the moment, by definition. More
75
76
A Garden of Integrals
generally, let us consider the sum
I(st) [CP(Xl) - cp(xo)]
+ I(S2) [q,(Xl) -
+ ... + I(sn) [cp(xn) -
CP(Xl)]
CP(Xn-l)}'
This will still have a limit, which we shall denote by
f:
I(x) dcp(x).U
So calculation of moments for mass distributions has led us to consi4er sums, Riemann-Stieltjes sums, of the form n
L'/(Ck) [CP(Xk) - cp(Xk-l)] = k=l
L I Acp. P
=
These are Riemann sums when cP (x) x. The optimism expressed by StieLtjes - "this win still have a limit" - is what we want to investigate. But frrst, a more formal definition is needed.
Definition 4.1.1 (The Riemann-Stieltjes Integral). (Stieltjes, 1894) Suppose that we have two bounded functions I and cP defined on the interval [a, b], and we have a number A such that for each t: > 0 there exists a positive constant 0 for which n
L
I(Ck) [q,(Xk) - CP(Xk-l)] - A <
E,
k=l
for every partition P of [a, b] whose subintervals have length less than 8. We say I is Riemann-Stieltjes integrable with respect to cP on [a, b], and we write
R-S
t
f(xl dt/>(x) = A.
Exercise 4.1.1.
a. Suppose I is constant on [a, b] with I(x) = k. Then for any partition of [a, b],
L
I Acp = k[cp(b) - cp(a)].
R-S
P
lb
kdcp(x) = k[cp(b) - tjJ(a)].
a
b. Suppose cp is constant on [a, b] with cp(x) = k. Then for any partition of [a, b],
L f;\.t/> = O. P
R-8
t
f(x) dt/>(x)
= o.
77
The Riemann-Stieltjes Integral
4.2 Discontinuities Thinking back to the mass distribution problem that Stieltjes was trying to solve, and the possibility of point masses, suppose that ¢ has a jump discontinuity, and I is "nice."
I
Exercise 4.2.1. Assume
is continuous on [a,b] and
13t a:5: < ~i, ¢(x)=
x=x, X < x
/3
{ /32
That is, ¢ is a step function with one step at a partition of [a, b].
x
x, for a < x < b. Let P
be
x
a. What happens if is a partition point, say = XK for 0 < K < n? In this case, XK-l < CK :'5 = XK :5 CK+l ::: XK+l. Show
L I!J.¢ = [f(CK) -
x
I(~i)] ({3 - {11)
p
+ [f(CK+I) -
I(x)] (/32 - {1)
+ I(X)(/32 -
/3l).
Hint: As the subintervals of P become small, the continuity of I makes the first two terms approach zero, suggesting that the Riemann-Stieltjes integral has the value 1(9:)(/32 - /3l) = ICx) [¢Cx+) - ¢(Jr:-)
J.
b. What if i is not a partition point? Hint: We have XK-l < i < XK and XK-l ::: CK < Xx. Show
L I!J.¢ = [/(CK) -
f(i)] (/32 - {11)
+ 1(..r:)({12 -
{11);
p
and the continuity of I suggests that the R-S integral has the value ICx){¢(i+) - ¢(..i-)].
f:
c. Show that R-S I(x) d¢(x) = l(x)[¢C..i+) - ¢(..i-)J. Generalize to Jl: = a, X = b, and n steps for ¢. The preceding exercise results in the following theorem. Theorem 4.2.1. Suppose that I is continuolls 017 [at b] and that ¢ is a step fimction withjll1nps at Xl, X2t ... in in [a, b]. Then the RiemQI1I1-SIieitjes integral of I with respect to ¢ exists, and I
R-S
f.
a
n
b
I(x) d¢(x) =
L 1(~ik)(¢(X:) - ¢(.i;)L k=l
78
A Garden of Integrals
where rp(a+) - rp(a-) means rp(a+) - rp{a), and rp(b+) - rp(b-) means
rp(b) - rp(b-).
Exercise 4.2.2. Suppose that f(x) = x 2 and rp{x) = [x], the greatest integer function, where 0 ::: x < n. Show that R-S
f.no x d([xD = 12 + 22 + ... + 2
n2
= n{n
+ 1)(2n + 1) . 6
Wbat happens if f and if; have a common point of discontinuity? If
f(x)
=!
1 x t:- ~, 2 x = ~
and
rp(x)
=!
1 0::: x < ~, 3 ~::: x S 1,
both are discontinuous at x = ~. Suppose P is a partition of [0, 1] and ~ is the partition point Xk. Then
~ftlt/> =
f(Ck) [tP
(4)
+ f(Ck+l)
-tP(Xk-l)] [tP(Xk+l) - tP
G)] =
2f(Ck).
If! is not a partition point, Lp f 8.rp = 2f(Ck). Whether! is a partition point or not, the result is the same. We have problems since f(Ck) could be 2 or 1 according as Ck is ~ or is different from ~. The Riemann-Stieltjes integral does not exist. So, in all that follows, we will be assuming that f and rp are real-valued bounded functions on [a, b] with no common point of discontinuity. ~uppose
=x
=
and rp(x) x 2 on [0,1]. Let P be f(c)8.rp = any partition of [0,1], and form the Riemann-Stieltjes sum Ck [x~ - x;_l]. Then,
Exercise 4.2.3.
f(x)
Lp
Lk=l
LXk-dxl- xi-d ~ LCk[.Ji - x~_l] S LXk[xl-xi-tl· Subtracting
L i[x2 - X2-I] from each term, we have '"" Ck[Xk2 L.,
2 " - Xk-l] - '2 3
o.
We conclude R-SJo xd(x2) = ~ = RJ~ x2x dx. l
L
You may wonder: where did i[x~ -x2-I] come from? We have "fundamental theorems" relating Riemann-Stieltjes integrals to Riemann integrals.
79
The Riemann-Stieltjes Integral
4.3 Existence of Riemann-Stieltjes Integrals Theorem 4.3.1. Sztppose I is continuoZls and ¢ is differentiable. with q/ being Riemann integrable, on the interval [a. b]. Then the Riemanl'l-Stieltjes integral 0/ / with respect to ¢ exists, and R-S
f.
b
lb
I(x) dt/J(x) = R a I (x)t/J' (x) dx.
Proof. Since f t/J'is Riemann integrable, we are tryin,f to show that the number A as defined earlier (Section 4.1. I)_equals RJa I(x)t/J'(x) dx. An estimate for E P 18.t/J - R f (x )t/J' (x) dx is needed. However, with tlie observation that 8.¢ = R JXX:-I t/J' (x) dx, by the Fun-
J:
damental Theorem of Calculus for the Riemann Integral (Theorem 3.7.1), we have that, for any partition P,
LI8.¢-R
f.
b
a
p
n
I(x}¢'(x)dx = LR k=l
l
xk
[/(Ck)-/(x)]t/J'(x}dx.
·"A.-l
The uniform continuity of I and the boundedness of ¢' will now complete the argument. 0
...
Exercise 4.3.1.
1
a. Looking at Exercise 4.2.3, calculate R-S J0 xd(.;y;2) in light of the theorem we just proved. h. Calculate R-S J~l xd(x 2}. c. Calculate R-S J~l Ixld(x Ixl). We have a theorem that can be helpful in some evaluations.
Theorem 4.3.2. Suppose I and t/J are boundedfimctio17S with no common discontinuities on the ilitel'vai [a, bJ. and the Riemann-8tieltjes integral 0/ f with respect to ¢ e.:dsts. Then the Riemann-Stieitjes integral of t/J with respect to I e'Cists, and R-S
f.
a
b
t/J(x)dj(x) = f(b)¢(b) - j(a)t/J(a} - R-S
f.
b
f(x) dt/J(x).
Proof. A partition of [a, b] with cs between xs may be viewed as a partition of [a. b] with ;cs between es. Let E > 0 be given. Since by assumption j is Riemann-Stieltjes inte~ grable with respect to t/J, we have a positive 8 so that, for any partition of
80
A Garden of Integrals
fa, bJ whose subintervals have lenrh less than 8, the associated RiemannStieltjes sum is within E of R-S fa f(x) drp(x). Let P be a partition of [a, b] whose subintervals have length less than 8/2. If Co = a and Cn+l = b, then {co, Cl, C2, ••• , Cn , Cn +l} is a partition of [a, b], for Ck :5 Xk :5 Ck+h and Ck - Ck-l < 8. Then n
L rp(Ck)[!(Xk) -
f(Xk-l)]
k=l
- (!Cb)
t
!(X)d
n
- L t;6(Ck)[!(Xk) -
f(Xk-l)]
k=l
because the Riemann-Stieltjes sum L f(Xk)llrp is associated with a partition of [a, b] whose subintervals, [Ck' ck+d, have length less than 8. 0
Exercise 4.3.2.
a. Evaluate R-S J~l xd(lxD. b. Evaluate R-S
J; x 2d([x]).
We now are well positioned to deal with monotonicity of t;6, the motivation for Stieltjes as he modelled mass distributions.
Monotonicity of cp
4.4
We begin with a Fundamental Theorem of Calculus for the R-S integrals.
Theorem 4.4.1 (FTC for Riemann-Stieltjes Integrals). If f is continuous on the interval [a, b]. and t;6 is monotone increasing on [a t b]. then R-S f(x) dt;6(x) exists.
J;
81
The Riemann-Stieltjes Integral Defining a new fimctiol1 F F(x)
011
the interval [a, b) by
= R-S
f
I(t) dq,(t).
then
rp is continuoZls, and 2. F is differentiable at each point where rp is differentiable (almost everywhere), and at sllch a point F' = I rpl 1. F is continuous at any point where
Proof We may assume rp(b) - rp(a) >
(recall Exercise 4.1.1). For existence, we may mimic the arguments of Riemann's Integrability Criteria (Section 3.2.1). We have p inf 16.rp ::; p 16.rp ::; p sup 16.rp since 6.rp is nonnegative, and we may conclude that any two Riemarm-Stieltjes sums for a partition P are within L p (sup f - inf f) 6.rp of each other, a quantity that can be made small due to the uniform continuity of the function I. As for refinements, suppose we add a point i with Xk-l < Ji < Xk. Then
L
L
l[rp(Xk} - rp("i)
inf [Xk-l.Xk]
+ rp(.£) -
{J
L
rp(Xk-l)]
< inf l[rp(Xk) - rp("i)] [.'t I.'t'k]
+
inf .. I{rp(.i) - rp(xk-d. [.'tk-ltX]
As we add points to P, the associated Riemann-Stieltjes sums for the (sup I - inf1)6.rp of each other. For difrefinements are still within ferent partitions, PI, P2, PI U P2 is again a refinement of each and.... Complete the argument for existence. As for continuity, we may assume If I < B on the interval [a t b]. Thu~, recalling Exercise 4. 1. 1(a), IF(y) - F(x)1 = IR-S I(t) drp(t)1 < B Irp(y) - rp(x)j implies continuity of F. Finally, suppose rp is differentiable at a point x in the interval (a. b). Because I is continuous on the interval fa, b],
Lp
J:
min f[¢(:c
~~+~
+ h) .... rp(x)]
%+/z
< R-S
1
I(t) drp(t)
x
< max l[cfJ(x [x,x+lr]
Application of the intennediate value theorem to (x+1z
R-S J't'
f(t) drp(t}
I
+ It) - rp(x)]. yields
= f(c)[rp(x + h) -
rp(x)]
82
A Garden of Integrals
+ It. Thus,
for h sufficiently small and positive, and c between x and x F(x
+ h) h
h
F(x) _ f(x)4J' (x) _ R-S J:+ f(t) d4J(t) _ l(x)4J'(x) h _ I(c) 4J(x
+ h) -
4J(x) _ I (x)4J' (x)
/z
< f(o) [q,(X
+
ht
q,(x) - q,'(x)]
+ !4J' (x) [/(x) -
f(c)lI.
The conclusion follows from differentiability of 4J and continuity of f, at x. Argue when h is small negative. 0
Exercise 4.4.1.
a. With C the Cantor function (Section 3.11), calculate R-S
f
b. Assuming that
f
C(x)dC(x)
and
f
x dC(x).
is continuous and monotone increasing, show that
J.
b
R-S a f(x)df(x) =
4.5
R-S
Euler's Summation
1
1
2" 12(h) - 2" f2(a).
Formu~a
A beautiful application of Riemann-Stieltjes integration is given by the Euler Summation Formula.
Theorem 4.5.1 (The Euler Summation Formula). Suppose continuous on. the interval [0,00). Then
I
and
I'
are
~)
tl(k) = R rNf(x)dx+ f(1) + feN) +R (N(x - [x] I'(x)dx. 11 2 11 2
k=l
As usual, [ ] denotes the greatest integer function.
Ji'
Proof The Riemann-Stieltjes integral R-8 f(x)d([xD exists, and by Theorem 4.2.1 it is equal to L~=2 f(k), since [x] is a step function on [1, N] with jumps at 2,3, ... , N.
The Riemann-Stieltjes Integral
83
On the other hand, application of Theorem 4.3.2 tells us that
R_S!,N f(x)d([x]) = feN) N _ f(l) 1 _ R-S 0
0
= Nf(N) - f(I) - R
by Theorem 4.3.1. Thus
!,N [xJ/,(x)dx
l
(x - [xJ- D = !,N D
R(
!,N ~~Jdf(x)
f'(x)dx
R
f'(x)dx - R
(x -
iN [xlf'(x)dx
=(N-Df(N)-~f(I)-R( f(x)dx
1
1 - feN) - R
2
2
= -- f(1) Solving for
!,N f(x)d([x]) !,N f(x) dx + LNf(k).
+ f(l) + R-S
- Nf(N)
1
k=l
z::f f(k), we have
LNf(k) = R !,N lex) dx + -1 f(l) + -1 feN) k=l
2
1
+R(
2
(x-[X1-Df'(X)dxo
0
Exercise 4.5.1. Do the following series converge or diverge?
"" sin{.JTi)
L....,---' I
n
' " sin(1n IZ) L....,
n
I
•
d " sin(1n Il) an L...., • p > L P 1
n
Exercise 4.5.2. Let f(x) = l/x and form the sequence {I'll}, )In == 1 + + ... + 1/11 -lnl1 Show < J'll < 1. Hint: Tangent lines at 2, 3, ... ,n. The limit y = Jim )'11 is called Eu1er's constant.
!
4.6
t
UUliform Convargence and R-S
~Ultegra'~ioUl
We conclude our treatment of the Riemann-Stieltjes integral with a convergence theorem.
84
A Garden of Integrals
Theorem 4.6.1 (Convergence Theorem for R-S Integrals). Given that {fk} is a sequence oj continuoZls junctions converging uniformly to J on the interval [a, b] and that ¢ is monotone increasing on [a, b]. Then 1. the Riemann-8tieltjes integral of Jk with respect to ¢ exists for all k,
2. the Riemann-8tieltjes integral of J with respect to ¢ eX.ists, and 3. R-S
J:
J(x) d¢(x) = limR-S
J:
J(x) d¢k(X).
Proof R-8
t
A(x) dq,(x) - R-8
< R-8
t
t
I(x) dq,(x)
IA(x) - l(x)1 dq,(x).
0
Exercise 4.6.1. Calculate limR-SJ;/2 (1- (x/n)rd(sin(x)).
4.7
References
L Apostol, Tom. Mathematical Analysis. 2nd edition. Readmg. Mass.: AddisonWesley, 1974. 2. Birkhoff. Garrett. A Source Book in Classical Analysis. Boston: Harvard University Press. 1973.
CHAPTER
5
lebesgUle Measure The scientist does not study natzwe because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful. If nature were not beautiful, it would not be worth Imowing, and if nature were not worth knowing, life would not be worth living. Of course I do not here spealc of that beauty that strikes the senses, the beauty of qualities and appearances, not that I undervalue sZlch beauty, far from it, but it has nothing to do with science; 1 mean that profounder beauty which comes from the hannoniol(s order ofthe parts, and which a pUl"e intelligf},nce can grasp. - Henri Poincare
Vito Volterra's 1881 example ofa function having a bounded derivative that was not Riemann integrable (Section 3.12) prompted Henri Lebesgue to develop a method of integration to remedy this shortcoming: The Lebesgue integral of a bounded derivative returns the original function.
5.1
lebesgue's Idea
Lebesgue's integration process was· fundamentally different from that of his predecessors. His simple but brilliant idea: Partition the range of the function rather than its domain. How does this work? Suppose the function f is bounded on the interval [a, b], say a < f < fJ. Partition the interval [a, ,8], a
= Yo < Yl < Y2 < ... <
),,,-1
< YII =
fJ,
with E/c denoting the points of the interval [a, bJ for which J'k-l ~ f < Yk. We have partitioned the interval [a, b] into disjoint sets Ek with [a, b] = UEk·
85
86
A Garden of Integrals
In each nonempty set Ek, pick a point Ck and form the sum (length of Ek)' Observe that LYk-l . (length of Ek) <
L I(Ck) .
L I(Ck) . (length of E k)
< I:Yk . (length of Ek), and that the absolute value of the difference of any two such sums J
II: I(Ck)' (length of Ek) - I: I(Ck) . (length of Ek)1 is bounded by maxk=l •... ,n(yk - Yk-i)' (b - a). Apparently... What do we mean by the length of Ek? Certainly if Ek is an interval, there's no problem. But suppose, for example, the function I under discussion is the absolute value function on the interval [-I, 1]. Well, Ek could be [Yk-lt Yk) U (-Yk, -Yk-tJ. No problem, then: the length of Ek is 2(Yk - Yk-l)' But then we have Dirichlet's function on the interval [0. 1], one on the rationals and zero on the irrationals. We would have to calculate the length of the rational numbers in [0, 1] and the length of the irrational numbers in [0,1]. The sets Ek may be very complicated. This integral, if it is to have any power, requires us to assign a length, to assign a measure, to a large variety of sets.
5.2
Measurable Sets
Let's think about what to put on a Measuring Wish List. ..
1. We want to measure aU sets of real numbers. 2. The measure of a subset must not exceed the' measure of the set. 3. A point should have measure zero. 4. The measure of an interval should be its length. 5. Translation of a set of real numbers should not alter its measure. 6. The measure of the whole must be the sum of the measures of its parts. Unfortunately, we cannot define a measure that meets all these requirements. For one thing, Wishes 3 and 6 would imply that the measure of any set of real numbers is zero.
87
Lebesgue Measure
Yet linearity of the integral demands Wish 6, and Wish 6 seems sO reasonable (with appropriate modification) that we will back off on the first item on our list. We will keep additivity at the expense of measuring not all - but as it turns out, enough - sets of real numbers. The idea here is that we will define a measure, commonly referred to as the Lebesgue outer measure, that measures all sets of real numbers (not always in an additive fashion); we will discard those sets where additivity fails. Recall that every set of real numbers may be covered by a countable collection of open intervals. In fact, a set has many such coverings, whose lengths are easily calculated: the length of:. a cover will be the sum of the lengths of the open intervals making up the cover. The infimum of this set of extended real numbers is called the Lebesgue outer measure of the set. Formally, for any set A of real numbers, the Lebesgue outer measure of A, written ,uil{A), is given by
5.2.1
The Wish List and Lebesgue Outer Measure
What do we have if we adapt our Wish List to the Lebesgue outer measure? 1. Every set of real numbers has a Lebesgue outer measure. 2. Since any cover of a set will be a cover of a subset, the infimum property yields monotonicity. 3. The empty set is a subset of every set; {a} C (a - E, a + E); and the Lebesgue outer measure of the empty set, or a singleton set, is zero. 4 .. Because (a, b) is a cover for (a, b), f.L*((a, b)) :5 b generally the outer measure of an interval is its length.)
a. (In fact,
5. Translation invariance is not a problem. 6. Additivity remains. The last three points require some explanation. Adapted PI'lsh 4. We will show that ,u*(a, b)) = b -a. If Uh is a cover of (a. b), then Uik, with (a - E. a + E) and (b - E, b + E), is a cover of [a. bJ. We have an open cover of a compact set. By the Heine-Borel
88
A Garden of Integrals
+ (a nk - ank _ 1) + ... + (a n2 - a) (b nk - ank) + (b nk _ 1 - ank _ 1 ) + ... + (b n1 - an1 ) L::.e(h) + 4£.
b ~ a = (b - ank ) < <
That is, b-a is a lower bound for the length of any cover of (a, b), and since the outer measure of (a, b) is the largest lower bound, JL * (a, b)) > b'.!.... a. Thus, the outer measure of the interval (a, b) is simply its length. For unbounded intervals, say (a, 00), the open interval (a, c) is a subset of (a, (0) for every real number c larger than a, and monotonicity completes the argument. Show JL *(1) = f.{I) for other kinds of intervals, such as [a. b) and (-00, b]. Adapted Wish 5. Why is translation invariance not a problem? Because the length of an interval does not change under translation. If A CUIk , then c + A C U (c + h) and
That is, J.L * (c + A) is a lower bound for the length of any cover of A. In other words, f1..*(c + A) :::: JL*(A) by the infimum property. Starting with a cover of c + A, U!k, argue the other direction. Adapted Wish 6. Additivity remains. But is the Lebesgue outer measure additive? Is the Lebesgue outer measure of a collection of disjoint sets the sum of the Lebesgue outer measures of the individual sets? If so, our quest would be over. In 1905, Vitali constructed a Lebesgue nonmeasurable set of real numbers; additivity does not always hold. What we can show, though, is that the Lebesgue outer measure of any countable collection of sets of real numbers (not necessarily disjoint) is at most the sum of Lebesgue outer measures of the individual sets: JL"1I(UA k) < LJ.L*(Ak)'
If JL""(Ak) = 00 for any k, we are done. Assume JL'1o(Ak) < k. We have an open cover Un1kn of each Ak with
JL*(Ak) ~ Lf.(Ikn) < JL*(Ak) 11
+ E/2k,
for
E
00
for all
> O. (infimum property)
89
Lebesgue Measure
The collection of open intervals [kn, for k, 11 == 1,2, ... , is a cover of UAk, and }L~(UAk) ~ LL>~(Ikn) < L}L"'(Ak) +E. k
k
n
Thus (sub additivity) Vitali's example is particularly upsetting. We are so close with this Lebesgue outer measure. What to do? We need an ingenious way around this.
5.3
Lebesgue
Measurab~e
Sets and Call'atheodory
Here is an ingenious jdea: Any set A of real numbers can be decomposed, relative to a set E, into two disjoint sets, its intersections with E and the complement of E, A n E and A n E C • Additivity would suggest that }L't'(A) = }L""(A n E) + }L~(A n EC). Let us then select from the collection of all subsets of real numbers precisely those sets E that interact in this fashion. Such sets E will be called Lebesgue measurable sets.
Definition 5.3.1 (Caratheodory's Measurability Criterion, 1914). A set of real numbers E is Lebesgue measurable if
holds for every set of real numbers A. So we have a criterion for measurability: Select a set E, and check whether it "splits" every set A of real numbers in an additive fashion. If the answer is yes, keep E. Otherwise, discard E. This seems straightforward enough. But we may end up discarding most of our collection of subsets of the real numbers.
Exercise 5.3.1. Show that the empty set and the set of an rea1 numbers satisfy Caratheodory's measurability criterion. If E is Lebesgue measurable. show that the complement of E is Lebesgue measurable.
5.3.1
Intervals
Are intervals Lebesgue measurable? We know that the Lebesgue outer measure of an interval is its length. We want to show that intervals satisfy
90
A Garden of Integrals
Caratheodory's measurability criterion. Of course, for an interval that can be written as the union of two disjoint intervals, length is additive; as it turns out;, this observation will yield measurability of intervals. Because (a, b) = (a, oo)n(-co, b) when a < b - and because we have to start somewhere - we will show that (a , co) is Lebesgue measurable. Actually, we need only show
tJ.*(A) > tJ.*(A n (a, (0»)
+ J.l.*(A n (-co, a]),
since the reverse inequality holds by the subaddifivity of the Lebesgue outer measure. We may also assume tJ.*(A) < co. By the infimum property, we may select a"tight" open cover U1k of A; that is, tJ.*(A) < L l{lk) < tJ.*(A) + li, for li > O. Thus
tJ.*(A
n (a, co») + tJ. *(A n (-co, an < tJ.1«(Uh) n (a,oo») + tJ.*(Ulk) n (-00, a]) < L tJ.'" (Ik n (a, co») + L tJ. * (Ik n (-co, a]) L[l(h < tJ.*(A)
(monotonicity) (subadditivity)
n (a,co») +l(Ik n (-co,an] == El{lk)
+ li.
and we have
J.l.*(A
n (a. (0») + J.l.* (A n (-00, an
~ tJ.*(A).
This is what we wanted to show. Demonstrate the Lebesgue measurability of (-co, b), (a, b), and so on. Intervals are Lebesgue measurable. We have devised a complicated measure, an infinitum of infinite sums, to measure something we already knew the measure of - the length of an interval. Isn't mathematics humbling? We know a few Lebesgue measurable sets - the empty set, its complement, and intervals and their complements. How about unions and intersections of measurable sets? Most important of aU, if {Ek} is a sequence of mutually disjoint Lebesgue measurable sets, is UEk Lebesgue measurable? Furthennore, does additivity hold: tJ.*(UEk) ;:::::: L tJ.*(Ek)? Is the Lebesgue outer measure, when restricted to Lebesgue measurable sets, countably additive? It is time to discuss sigma algebras and investigate why they are important in the context of measure theory.
lebesgue Measure
5.4
Sigma
91 A~gebras
Definition 5.4,1 (A Sigma Algebra). Given a space n, a collection 0 of subsets of is said to be a sigma algebra provided the following properties hold:
n
1. The empty set belongs to this collection.
2. If a set is in this collection, its complement is in the collection. 3. Given any sequence of sets in this collection, their union is in the collection. Caratheodory showed that the collection of subsets of real numbers that satisfy his measurability criterion forms a sigma algebra and, moreover, that the Lebesgue outer measure is countably additive on this sigma algebra. This is the essence of the next theorem, one of the two most important results regarding Lebesgue measure. The other crocial result is Theorem 5.5.1.
Theorem 5.4.1 (CaratheodoryJ 1914). Define the Lebesgue outer measure JJ. * of any set of real numbers E as follows:
~"'(E) where l(h)
= inf{Ll(Ik) lEe Uhf
= bk -
h,
open intervals},
ak, ak. bk are extended real numbers, ak < bk. Then
1. The collection oj sets E of real numbers that satisfy Ca1"atheodory's measurability criterion, JJ.1< (A) = JJ.,j. (A n E) + JJ. ... (A nEe) Jor every set oj real l'lumbe1"s A, J01711S a sigma algebra lvi.
2. The Lebesgue ollter measure is countably additive on M: For any sequence {E k } oj mutually disjoint Lebesgue measurable sets, 00
JLit
(Ur Ek )
=
L JJ. reEk)' 1
Proof Show the first two requirements for a collection of sets to form a sigma algebra are met. The third requirement, that UEk be a member of M, may be demonstrated as follows. Start with an easier problem: The intersection of two measurable sets, and the union of two measurable sets, is measurable. Show that An(EIUE2)=(AnEdU(AnEfnE2)
An (Ef n Ei)
= An (EI
U E 2 )c.
and
92
A Garden of Integrals
Then
n E I ) + tL*(A n ED measurability of EI = tL*(A n E 1) + tL1«(A n ED n E2) + tL*(A nED n En
tL*(A) = J.L*(A
measurability of E2
n E I ) U (A nED n E2) + tL*(A nED n
::: tL*(A
En
sub additivity
= tL"'(A
n (EI
U E 2 »)
+ tL""(A n (EI U E2 )C)
subadditivity
::: tL*(A). The union of two Lebesgue measurable sets is Lebesgue measurable. Complementation shows the intersection of two measurable sets is measurable: (EI n E 2) = (Ef U ED c . Measurability of finite unions and finite intersections follows by induction. As for (UI' Ek), we may assume the Ek are mutually disjoint. In this case,
n E 2) + J.L*(EI
tL*(EI U E2 ) = tL*(EI U E2 )
U E2 )
n En
+ tL*(Ed,
= tL*(E2)
since Eland E2 are disjoint. By induction, we have finite additivity and
tL* (A
n (Uf Ek») = tL* (A n (U{' Ek) n EX) + tL1«A n (Uf Ek) n E~) n-l
= tL*(A
n En) + Lll*(A n Ek) 1
Since
n
> LJ.L*(A
n Ek)
+ tL"\.( An (ur Ekt).
1
independent of 11, we have co
tL"'(A) > LtLT(A
n Ek) + tLT(A n (U'f Ek)C)
1
> J.L*(A
n (ur Ek) ) + tL*(A n (Uf Ekt),
93
Lebesgue Measure
and
Uf Ek
is a member of M,
Uf Ek
is Lebesgue measurable.
We have shown that the collection of sets of real numbers satisfying Caratheodory's measurability criterion, M, is a sigma algebra. The task of showing countable additivity remains. We have /J. * (UEk) ::: L /J.* (Ek) by the subadditivity of Lebesgue outer measure. For the other direction, we have finite additivity, n
00
L/J."'(Ek)
= /J.*(U~Ek) < /J.* (U~Ek) < LtL*(Ek),
1
1
independent of 11, and hence 00
00
L/J.*(Ek) < /J.~ (Uf Ek) < L/J.*(Ek). 1
This completes the proof.
5.5
1
0
Borel Sets
The Lebesgue outer measure is countably additive when restricted to the sigma algebra of sets M satisfying Caratheodory's measurability criterion. For E E M, we will write /J.(E) for /J.""(E). It is to be understood that writing /J. assumes the set is Lebesgue measurable. This measure /J. may be the measure needed for the Lebesgue integral. Recall that the only sets that we mow are measurable (i.e., are in M) are rp, R, intervals, and their complements. Not quite. We have just shown that we can measure countable unions (rational numbers, for example), their complements (irrational numbers), and so on - some very complicated sets. It is time to introduce Borel sets. We have shown that open intervals are Lebesgue measurable; open intervals are in the sigma algebra M. Show that the intersection of all sigma aJgebras that contain a given collection of sets of real numbers is again a sigma algebra. This sigma algebra is said to be generated by the given collection. We give a special name, the Borel sigma algebra, to that sigma algebra generated by the collection of open intervals, and we write B. Since we have shown that open intervals are Lebesgue measurable, evidently B C M. We have, almost unexpectedly, arrived at a very important result.
Theorem 5.5.1. EvelY Borel set of real numbers is Lebesgue measurable
94
A Garden of Integrals
We can measure the Borel sets (we are bard-pressed to find non-Borel Lebesgue measurable sets), and measuring works as expected for measurable sets: When the whole is decomposed into a countable number of disjoint measurable parts, the measure of the whole is tbe sum of the measures of its parts. Here are some exercises in measuring.
Exercise 5.5.1. Show that the following assertions are true. a. p.(l) = 1,(1). Hint: (as b) C (a, b). So p.lI'{Ca s b)) :5 b - a, and,Section 5.2.1.
b. p,(Os 1]) =
L P. (k +1l' k]1) . 00
(
I
c••Singleton sets are measurable and have measure zero. Hint: {a} n(a -lJk,a + 11k). d. Any countable set is measurable and has measure zero. Hint: {at, az.···, ak,. .. } = Uak, and p.(ak) < p. (ak -
2~' ak + 2~).
e. The rationals are measurable and bave measure zero. f. The irrational numbers in [0,1] are measurable. What is their measure? (A set of positive measure containing no intervals.)
g. Any set with p.*(E) == 0 is Lebesgue measurable. Hint: p.*(A) < p.*(A n E) + p.*(A nEe) ::: p."t&(E) + p.*(A) = p,*(A).
In fact, there are uncountable sets of Lebesgue measure zero.
Exercise 5.5.2. Show tbat the measure of the Cantor set is zero.
5.6
Approximati.ng Measurable Sets
We are familiar with the ideas of open sets, closed sets, compact sets, and so on. Do we have relationships between these topological notions and measurability of sets? It turns out that Lebesgue' measurable sets may be closely approximated by open sets "from the outside" and closed sets "from the inside." Theorem 5.6.1. For any set E of real numbers the following statements are equivalent: 1. E is Lebesgue measurable iii th.e sense of CaratheodolY.
95
Lebesgue Measure
2. Given E > 0, we have an open set G containing E so that 1.£ *(G .... E) < E.
3. Given E > O. we have a closed set F contained in E so that 1.£14-(E .... F) < E.
Proof Assume E is measurable. We will show that the measure of E may be approximated by the measure of an open set. By the infimum property, we have an open cover Uh so that I.£(E) ::s I.£(U[k) < I.£(E) + E; and because E is measurable,
If the measure of E is finite, subtraction yields the desired result. Otherwise, let En = En [-lI, n] and argue as before. So statement 1 implies statement
2. Show, using complementation, that open set approximation yields closed set approximation. So statement 2 implies statement 3. As for the third conclusion, suppose we have a closed subset F of E with I.£*(E .... F) < E. It is sufficient to show I.£~(A) > 1.£:4-(A n E) 1.£* (A nEe) with I.£*(A) being finite. We have
+
F) U F)) F») + I.£*(A n F)
.u*(A n E) = 1.£* (A n (E ....
n (E < E + I.£*(A n F) < 1.£* (A
and
1.£lf"(A nEe) = 1.£*( An (E .... F) U
Ft)
n (E .... F)C n Fe) I.£"'(A n FC).
= I.£~(A <
Because closed sets are measurable, 1.£" (A so addition yields
n F) + I.£*(A n F C ) = I.£T(A),
and we have the desired conclusion. Thus statement 3 implies statement 1, and the equivalence of 1, 2, and 3 follows. 0 We return to Borel sets and some exercises "completing" B.
96
A Garden of Integrals
Exercise 5.6.1. 8.
Show that for any measurable set E, we have a Borel set B 1 so that E is a subset of Bl and M(B} - E) = O. Hint: From the previous result we have an open set G n so that M(G n - E) < lIn; B} = nGn.
b. Show that for any measurable set E we have a Borel set B2 so that B2 is a subset of E and JL(E - B2) = O. Hint: B2 = UFn . c. Show that every Lebesgue measurable set (jf real numbers is the union of a Borel set and a set of Lebesgue measure zero. Hint: E = Bl U (E - Bl)' Adjoining the sets of Lebesgue measure zero to the Borel sets creates the Lebesgue measurable sets.
5.6.1
Vitali's Co"ering Theorem
Here is a result by Giuseppe Vitali (1875-1932) that we will find very useful. We begin with a defmition.
Definition 5.6.1 (Vitali Cover). Let A be a nonempty set of real numbers. A Vitali cover of A is a collection of closed intervals of arbitrarily small length that cover each point of A. Thus every point of A is in an arbitrarily small closed interval from this collection. Theorem 5.6.2 (Vitali, 1908). Given a set ofreal numbers A whose Lebesgue
outer measure isftnite and a Vitali cover of this set A. Then given an E > 0, there exists a finite, disjoint collection of closed intervals II, h .... In from this Vitali cover of A so that I
This finite collection can be extended to a countable collection of mutually disjoint, closed intervals It, 12 , ••• , Ill • ••• from this Vitali cover of A so that JL*(A Ik) = O. An outline of the argument for this result may be found in the wonderful book by Stromberg (1981),
ur
97
Lebesgue Measure
5.7
Measurable functions
Recall that Cauchy's integration process was very effective for continuous functions, functions that preserve openness under inverse images. Since every open set of real numbers is the union of a countable collection of disjoint open intervals, and since inverse images behave nicely j-l (U1k) = Uj-l (Ik) - why not try measurement of inverse images of intervals? This was Lebesgue's idea (Section 5.1). Definition 5.7.1 (Lebesgue Measurable Functions). A real-valued function I that is defined on a measurable set E is said to be Lebesgue measurable on E if the inverse images under I of intervals of real numbers are measurable subsets of E. How would we check to see if a function is Lebesgue measurable? Do we have to argue all types of intervals? Exercise 5.7.1. a. Assume 1-1 ((e, 00)) is Lebesgue measurable for every real number e Show thatthe inverse images of [a, 00), (-00, a), (-oo,a], (a,b), (a, bJf [a, b), [a, b], are Lebesgue measurable. Hint: [a,oo) = n(a (1/ k), 00) and 1-1 (fa, 00)) = nj-l (a - (1/ k), 00). b. Show that the following function is Lebesgue measurable on [-1,2]:
x < I, x
= 1,
x>l. lust what kind of functions are Lebesgue measurable?
5.7.1
Continuous Functions Defined on Measurable Sets
We will show that continuous functions defined on measurable sets are Lebesgue measurable. Consider the equation A = 1-1 (e , co)) {x EEl I(x) > e}. If A is empty, we are done. Othenvise, for each x in A, we have ~(x) > 0 so that for z belonging to the interval (x - ~(x). x + ~(x»), IC::) > e:
=
A=
UxeA ((x
-~(x). x + ~(x)) n E) =
(u.teA
(x - 8(x), X + 8(x))) n E.
98
A Garden of Integrals
5.7.2 Riemann Integrable Functions Cauchy integrable functions are Lebesgue measurable functions. How about Riemann integrable functions? We will show that Riemann integrable functions are Lebesgue measurable functions. As a first step, we will suppose I and g are defined on a measurable set E, that I is Lebesgue measurable on E, and that g = f except On a subset Z of E of Lebesgue measure zero (almost everywhere). We claim that g is Lebesgue measurable on E. Consider the following relationships:
I I(x) >
{x EEl g(x) > c} = {x E Z
g(x) > c}
E - Z I I(x) = g(x) > c} U {x E Z I g(x) > I(x) > c} U {x E Z I g(x) > c = I(x)} U {x E Z I g(x) > c > f(x)} {x EEl I(x) > c} U {x E Z I g(x) > c} U {x E
U {x E
If f is Riemann integrable in [a, b], of measure zero, f is continuous on [a,b]-Z. Let g = f on [a, bl- Z and 0 on g is measurable on [a, b]~ and thus I
Z I g(x) > c > I(x)}.
then I's discontinuities form a set [a, b] - Z, and I is measurable on A. Then g =
f
almost everywhere, is measurable on [a, b].
Lebesgue measurable functions may not be Riemann integrable. Exercise 5.7.2. Let
le;<) = Show that
I
Io
1 x rational, x irrational,
O<x
1.
is Lebesgue measurable on [0, 11.
5.7.3 Limiting Operations and Measurability An important property of sequences of measurable functions is that measurability is preserved under many limiting operations. Theorem 5.7.1 (Limiting Operations and Measurability). If {/k} is a sequence 01 Lebesgue measurable functions defined on a measumble set E witlt lim Ik = I pointwise on E then I is Lebesgue measurable on E. J
99
lebesgue Measure For example, if
gk(X)
== SUP{/k(X), fk+l (X), ... } =
Un;::k{X
E
Elfn(x) > e}.
then gk is measurable. If hk(X) == inf{/k(x). /k+l (x), ... } = Un;::k{X E El/n(x) < e}, then hk is measurable. If lim sup Ik = limgk, then {x E Ellimsup Ik(X) < e} = Uk=l {x E Elgk(X) < c}, and so on. However, Riemann integrable functions do not share this property. Limiting operations may not preserve RIemann integrability. For example, Jet '1, '2 .... be any enumeration of the rational numbers in (0, I), and define a sequence {fk} of Riemann integrable functions by
Ik (x)
=! o
1 x = Fl: F2, otherwIse.
••• , 'k,
Then RJo! /k(X) dx = 0, but the pointwise limit of the sequence {fk} is the Dirichlet function, which is not Riemann integrable.
5.7.4 Simple Functions Simple functions have a finite number of values. That is, for a measurable set E, a simple function ¢ can be written as a finite linear combination of characteristic functions: n
rp(x)
=
L CkXEk(X) k=l
where ek are real numbers, Ek are mutually disjoint measurable sets with E = Uk=l Ek, and XEk (x) is the characteristic function
(.) _!
XEk.x -
1 x E Eb Ox¢. Ek.
The reader may show that simple functions are measurable functions.
Theorem 5.7.2 (Approximating Measurable Functions by Simple Functions). For any measurable function f defined on a measurable set E, there exists a sequence of simple fimctions {(PI,} on E so that lim rpk = f
lor all E. is bounded 017 E, then 1im rpk = / uniformly on E. If f is nonnegative, the sequence {rpk} may be constructed so that it is a monotonically increasing sequence.
If I
100
A Garden of Integrals
Proof We follow Lebesgue's idea: partition the range of I. We may assume that I is nonnegative; that is, we have
I
=
111 + I _ 1/1- I 2
2
for If~ f, If I; f measurable and nonnegative. Let[O, 00) = [0, 1) U [1,2) U ... U [11-1, n) U [n, 00), and partition [0,00) into 2n + 2n + ... + 2n + 1 = n2n + 1 disjoint subintervals. Define if>n by
n2n k - 1
if>n(x)
=L
k=l
~XE/lk(X) + nXF/I(x)
.
with Enk
=
I
X E
\lc-l
kl
E - n- < I(x) < - n 2
-
2
I
Fn = [11,00).
Note that Enk = En+12k-l U E n + 12k. The reader may complete the argument. 0
5.7.5
Pointwise Convergence Is Almost Uniform Convergence
Because uniform convergence transfers many nice properties to the limit function, we look for conditions that generate uniform convergence. For sequences of measurable functions, we have a remarkable theorem showing that pointwise convergence is almost uniform convergence. The theorem is due to Dimitri Egoroff. Theorem 5.7.3 (Egoroff, 1911). Suppose {/k} is a sequence o/measurable junctions that converges to a real-valued junction I almost everywhere on the interval [a, b]. Then lor any 8 > 0, we have a measurable subset E 0/ [a, b1 so that J..L(E} < 8 and the sequence {/k} converges unifonnly to I on [a, b] - E.
5.8
More Measureable Functoons
In addition to continuous functions, differentiable functions, monotone functions, and Riemann integrable functions, two other classes of measurable functions - functions of bounded variation and absolutely continuous functions - will be of interest as we develop the Fundamental Theorems of Calculus for Lebesgue integrals.
101
Lebesgue Measure
5.8.1
Functions of Bounded Variation
Camille Jordan (1838-1922) offered the following definition in 1881. Definition 5.8.1 (Bounded Variation). A function is said to be of bounded variation on an interval [a, b1 provided Lp I/(Xk) - I(Xk-I)1 is bounded Xo < Xl < .,. < Xn b. for all possible partitions P of [a, b], a
=
=
Exercise 5.8.1.
a. Show that continuous functions may not be of bounded variation, for example
I(x) Hint: Xk = 2/(2k
= 1x sin(1l'/x) o
+ 1), k =
0 < X < 1, 0 = x.
It 2t ....
b. Show that monotonic functions are of bounded variation. Hint: Lp I/(Xk) - I(Xk-I)1 < I/(b) - l(a)l·
5.8.2
Functions of Bounded Variation and Monotone Functions
Theorem S.S.l (Jordan, 1894). Functions 01 bounded variation are the difference.()f two monotone increasing junctions. Clearly the difference of two monotone functions is a function of bounded variation. Now suppose I is a function of bounded variation on [a, b]. Define a new function V, the variation of I on [at b], by
Vex)
== sup
L I/(Xk) - I(Xk-i)1 p
over all partitions P of [at x], with a < x < b. The function V is certainly monotone increasing. Since I = V - (V - I), we need on]y show that V - I is monotone increasing, that is, for x < y,
Vex) - I(x) ::: V(y) - l(y)
or
I(y) - I(x) ::: V(y) - Vex).
But the variation on [x t y] is at least as large as I/(x) - l(y)1 (trivial partition). It will be shown (Theorem 5.10.1) that monotone functions are differentiable almost everywhere. Thus functions of bounded variation are differentiable almost everywhere. This seems simple enough, the difference of two monotonic functions characterizing functions of bounded variations. What else is known? Kannan and Krueger (1996) offered the following observation.
102
A Garden of Integrals
Example 5.8.1. A function of bounded variation is the difference of two monotone functions, but it need not be monotonic on any subinterval of its domain. This needs a closer look. Let 't. '2, ... be an enumeration of the rational numbers in (0, 1), and let 0 < a < 1. Define I on [0, 1] by
I(x)
=
Io a
k
x = 'k,k x =F 'k.
= 1,2 .... ,
We claim that the total variation of I on the interval [0. l] is 2a/(I - a). Construct a partition P {O, Xl, X2, ..• X2n-I, I}, with the "odds" being {XI. X3, ... x2n-d = {'I. r2, ... rn} and the "evens" irrational numbers in [0,1]. Then
=
I
E l/(xk) -
I
I
I(Xk-t)1 = 2(a l
+ a 2 + ... + an)
and
p
n
= 1,2, ....
Thus V(l) ~ 2a/(l - a). On the other hand, for any partition P {O, Xl, X2, .••• Xn-l, I}. r1 will be in one of these subintervals or will be an endpoint. Regardless, rl will be in exactly one of the (Xk-l, Xk+l), with k = 1.2, ... ,n - 1, and '1 makes a contribution to the variation only if it is an endpoint, an evaluation point of f. Similarly for '2. '3 •... , rn-I· So the worst case occurs When the n -1 rational numbers Tt. r2 • ...• 'n-l occur as partition points. Thus the variation for any partition is bounded by
=
5.8.3 Absolutely Continuous Functions Vito Vitali developed this definition in 1904.
Definition 5.8.2 (Absolutely Continuous Function). A function f on [a, b] is said to be absolutely continuous on [a. b] if, given any € > 0, we can find a positive number 8 such that for any fmite collection of paiIwise disjoint intervals Cab bk) C [a, b], Ie = 1,2, ... ,12, with l)bk -ak) < 8, we have L I/(bk) - l(ak)1 < €.
103
Lebesgue Measure
The stipulation finite may be replaced by finite or countable.
Exercise 5.8.2. Verify the following statements. n. Absolutely continuous functions are uniformly continuous. Hint: Show that 1 is continuous on the interval [a t b]. b. Absolutelycontinuous functions are functions of bounded variation and thus differentiable almost everywhere. Hint: The variation of lover an interval of length 8 is less than €. Partition [a t b] into subintervals of length less than 8. c. Continuous functions are not necessarily absolutely continuous. Hint: Recall Billingsley's function (Section 2.8), a continuous nowhere differentiable function.
d. Differentiable functions are not necessarily absolutely continuous, for example, lex) = x' sin~1f Ix') 0< x< 1. x =0.
I
Hint: Xk
=
)2/(2" + 1), k = 1,2, ....
e. Differentiable functions with a bounded derivative are absolutely continuous. Hint:
f. Absolutely continuous functions are differentiable almost everywhere. If I' = 0 almost everywhere and I is absolutely continuous, then I is constant.
Theorem 5.8.2. If I is absolutely continuous on [a. b] (and thus diffel"en~ liable almost everywhere), and the derivative 01 1 vanishes almost everywhere on [a, b], then I is constant 012 [a, b]. Proof We will show that I(c) = lea) for any c in the interval {a.b]. Because I is absolutely continuous on [at b], given an E > 0 there is a 8 > 0 so that for any finite or countable collection of pairwise disjoint intervals (ab bk) with length Z)bk-ak) < 8, we have L II(bk) - l(ak)1 < E. Let E {x E (a. c) I I'(x) O}, the interval [at c] except for a set of measure zero. For each x E E we have arbitrarily small closed intervals [x, x + /2] for which I/(x + h) - I(x) I < €h because I'(X) O. This collection of closed intervals is a Vitali cover of E (Section 5.6.1).
=
=
=
A Garden of Integrals
104
We have a finite collection of disjoint closed intervals [XI. Xl + hIl, ... , [x n • Xn + hn], ordered as a < Xl < Xl + hI < X2 < X2 + h2 < ... < Xn + h n < c. That is,
(a. c)
= (a, Xl) U [Xl. Xl + hd U (Xl + hI. X2) U [X2. X2 + 11 2] U··· U [xn, Xn
and p.(E - U[Xk. Xk
+ hk])
(a, c) - U[Xk' Xk
+ Ilk]
+ hnl U (xn + h n, c)
< 8, the 8 of absolute continuity. Because C
(a, c) - E) U (E - U[Xk, Xk
+ hkD
and because (a, c) - E is a set of measure zero, we have
p.(a, c) - U[Xk, Xk
+ hk])
= p.(al, Xl) U
(Xl
+ hI. X2) U ... U (xn + !tn, c»)
< 8. Thus
If(e) - f(a)1
+ [f(xn + hn) - f(xn)] + ... + [f(x! + hI) - f(Xl)] + [f(XI) - f(a)]! < (If(e) - f(xn + hn)1 + ... + If(xl) - f(a) I ) + (If(Xl + hI) - f(x!)1 + ... + If(xn + hn) < E + E(h l + h2 + ... + hn ) < E(l + c - a). 0
= I[f(e) -
f(xn + hn)]
f(xlI)l)
Since the derivative of the Cantor function is zero almost everywhere, and C(O) = 0, C(1) = 1, the Cantor function - even though continuous and differentiable almost everywhere - is not absolutely continuous.
5.9
What Does Monotonicity Tell Us?
A remarkable theorem (Theorem 5.10.1, whose proof will appear later) tells us that monotone functions are differentiable almost everywhere. To appreciate that monotonicity implies differentiability almost everywhere requires a more detailed analysis of difference quotients
fey) - f(x) )I-X
The Dini derivates have proved to be invaluable for such analysis.
105
Lebesgue Measure
5.9.1
Dini Derivates of a Function
Suppose I is defined on an interval containing the point x. We are interested in the four quantities illustrated in Figure 1. They are the limits of the difference quotients at x:
I · .I · I · .I
. sup l(y) - I(x) ,x < y < x D + I{x):= lIm h-+O+ y-x D + I(x):= hm tnf I(y) - I(x) ,x < y < x h-+O+
D- I(x) = 11m sup h-+O+
y-x
I + I I I +h
I (y) - I(x) ,x - h < y < y-x
,
h , X
I
D-/(x):= hm tnf I(y) - I(x) x - II < y < x . 11-+0+ Y- x I
The limits always exist (in the extended reals).
:x
Figure 1. Dini derivates Elcercise 5.9.1. Calculate the Dini derivatives for these functions at x
=
Q.
f(x)
b.
f(.~) =
I I
1 x = 0, 0 x :f: 0,
0 x sin(K/x)
g(x)
x=o, x
# o.
=
I
0 x =0, 1 x :f: 0,
hex) =
= o. Ixl.
106
A Garden of Integrals
c. Letting 1(0) = 0,
f(x)
=
I(x) =
I !
x x rationals -3x x irrational,
for x > 0,
x rational, x irrational,
for x < O.
-4.;t
2x
Exercise 5.9.2.
a. A fimction is differentiable at x in (a, b) iffall four derivates are finite and equal. Demonstrate.
b. The Straddle Lemma. Suppose I is differentiable at a point x in [a, b]. Show that for every (£ > 0 there exists a DE (x) > 0 so that if u #- v and x - DE (x) < u < x < v < x + DE' (x), then
I/(v) - I(u) - II (x) (v - u)1 ::: €(v - u).
Theorem 5.9.1. 1[1 is monotone increasing on [a, b], then alllour derivates are nonnegative andfinite almost everywhere. Proof. Because 0 < D+ [ :::; D+ I and 0 < D_ I ::: D-[, it is sufficient to show that D- I and D+ I are finite almost everywhere. Let E = {x E [a, b)ID+ I(x) = oo}, and assume Jl. ... (E) = a > O. We will arrive at a contradiction after an application of Vitali's Covering Theorem (Section 5.6.1). For x E E~
lim sup! l(y) - I(x) : x < y < x h~O+
Then for any constant [(
y-x
we have a sequence Yll
+ hI = +00. ~
x+ so that
l(yn) - I(x) > K. Yn -x That is, {[x I y,,] Ix E E} is a Vitali cover of E. Thus, we have a finite, disjoint collection, [XI. Yl], [X2. Y2] •.•. , [Xn, Yn], so that Il
or
L (Yk - Xk) > ~. k=1
-
lebesgue Measure
107
Then
feb) - I(a)
~
n
L
n
[/(Yk) - I(Xk)] > K
k=l
L (Ylc - Xk) > K~. k=l
By choosing K larger than (2(/(b) - I(a)) )/Ci. we have a contradiction. Complete the argument by showing that D- I is finite almost everywhere.
o 5.10
Lebesgue's Differentiation Theorem
Finally, as promised, Lebesgue's Differentiation Theorem.
Theorem 5.10.1 (Lebesgue, 1904). If I is nondecr'easing f is differentiable almost everywhere.
012
(a b]. then I
The proof may be found in Gordon (1994). We include it here as a testament to human ingenuity.
Proof Since
I
is nondecreasing, the four Dini derivutes are nonnegative, and (by th~ previous theorem) finite almost everywhere. It will be sufficient to show that the four derivates are equal almost everywhere. Because 0 < D+I < D+ I, we will show that the set E {x E (a, b) I D+/ex) < D+ I(x)} has Lebesgue outer measure zero. In fact, we may reduce the problem further with the observation that
=
E
= Up,q rational numbers{X E
(a, b) I D+/(x) < p < q < D+ I(x)}.
It will be sufficient, then, to show that
J.L"'({X
E
(a, b)ID+/(x) < P < q < D+ I(x)}) = 0
for each pair of rational numbers p, q. Denoting this set by Epq, we assume J.L ~ (E pq) = Ci. > 0 for some pair of rational numbers p and q. We will arrive at a contradiction. Given an E > 0, we have an open set 0 containing E pq. (We may as well assume 0 C (a, b), since (a, b) is an open interval containing E pq .) So J.L1fl(E pq ) < Ji.(O) < Ji.'tl(E pq ) + E; that is, J.L(O) < Ci. + E. For x E E pq , where D+/(x) < p, we have intervals [x, y] with y ~ x+ and [fCy) - I(x)]/(y - x) < p. The collection of such intervals fonns a Vitali cover of E pq. TIlUs, we have a finite number of disjoint intervals
108
A Garden of Integrals
from this collection, say [Xl! Yi], [X2. Y2], ... t [XN YN L all belonging to 0 (with X E Epq C 0, 0 open), so that J.L"'(Epq - Uf==l[Xk, YkD < f, and I
N
L
N
[f(yk) - f(Xk)] < P L
k=l
+ f).
(1)
k=1
Now consider the set E pq ~
(Yk - Xk) < PJ.L(O) < p(~
n (Uf=l [Xk. Yk]). Because
= j.L*(Epq) n (uf=dxk, Yk]))
< J.L* (Epq
+ J.L'" (Epq -
(uf==l[xk,Ykl))
we have j.L'" (Epq n (Uf=l[Xkt Yk])) > ~ - E. A point x in this set belongs to E pq and belongs to exactly one of the disjoint intervals [Xk, Yk], with k 1,2, ... IN. We have problems if x = Yb since we want to approach from the right. But there are no such problems if we consider the set Epq n (Uf!=1 (Xb Yk)). Furthermore, deletion of the endpoints does not alter the outer measure. A point in E pq n (Uf=1 (Xb Yk)) belongs to E pq and exactly one of the open intervals (Xkt Yk), say (XK, YK). Again we have intervals [u, v] with v --+ u+, [u,lJ] C (XK, YK), and [f(v)- f(u)]/(v -u) > q. The collection of such intervals forms a Vitali cover of Epq n (Uf!=l (Xk, Yk)). Thus we have a finite number of disjoint intervals from this collection, say [Ult VI], [U2. V2], ... t [UM, VM], so that
=
J.L'" (( E pq n
(Uf=1 (Xk' Yk)))
-
U~=dvk. Uk])
<
€
and M M L[f(Vk) - f(Uk)] > q L(Uk - Uk)· k=1
(2)
k=1
Since f is nondecreasing, and since each [Uk, Vk] C (Xi. Yi) for k = 1, 2, ... ,M and some i = 1t 2, ... IN, M N L[f(Vk) - f(Uk)] < L[f(Yk) - f(Xk)].
k=l
k=l
From equations 1 and 2 we have M
q
E (Vk k=1
Uk) < p(~
+ f).
109
Lebesgue Measure
We observe that
M
/vI
L(Uk -Ilk) =
L J.L([Uk. Uk]) =
k=l
k=I
> p..f:
jJ.
(Ut~dtlk. Uk])
(Epq n (U£'=I (Xb Yk))) - E > a -
2E
Hence, q(a - 2e) < pea + E), and since e was arbitrary qa < pa, or q < p. We have a contradiction to p < q. This completes the proof. D Having an understanding of measurable sets and measurable functions, we are in position to define the Lebesgue integral and discuss Lebesgue integration.
5.11
References
1. Gordon, Russell A. The Integrals of Lebesgue, Denjoy, Perron, and Henstock. Providence, R.I.: American Mathematical Society, 1994. 2. Kannan, Rangachary, and Carole King Krueger. Advanced Analysis on the Real Line. New York: Springer, 1996.
3. Stromgerg, Karl. An Introduction to Classical Real Analysis. Belmont, Calif: Wadsworth, 1981.
CHAPTER
6
The lLebesgue Integra~ As the drill will not penetrate the granite unless kept to the work hou,. after haUl; so the mind will not penetrate the secrets 01 mathematics unless held long and vigorously to the work. As the sun's rays burn only when concentrated, so the mind achieves mastery in mathematics, and indeed in every branch 01 knowledge, only when its possessOP' hurls all his forces upon it. Mathematics, like all the other sciences. opens its doors to those only who knock long and hard. - B. F. Finkel
6.1
Introduction
The culmination of our efforts regarding measure theory, Lebesgue integration, is a mathematical idea with numerous and far-reaching applications. We will confine our remarks to the essential concepts, but the interested reader will be well rewarded by additional efforts.
6.1.1
Lebesgue's Integral
We begin our exploration of Lebesgue's integral (1902) by defining what it means to be Lebesgue integrable. Suppose 1 is a bounded measurable function on the interval [a, b], so a < 1 < {3. Partition the range of I: Cl = Yo < Yl < ... < YIZ = {3, and let E" = {x E [a. b] I Yk-l < 1 < Yk}, for k = 1,2 •...• n. Form the lower sum Lk=l Yk-lIL (E/c) and the upper sum Lk=l YIcJ.L(Ek) (The terms J.L(Ek) make sense because 1 is measurable by assumption.) All such sums are between a(b - a) and {3Cb - a). Now compare the supremum of the lower sums with the infimum of the upper sums over all possible partitions of [a, {3l. If these two numbers are equal, sal A, we say 1 is Lebesgue integrable on [0 b], and we write A = LJa 1 dJ.L. I
111
112
A Garden of Integrals
Proposition 6.1.1. For bou17ded measurable functions on [a, b], the supremum ofthe lower sums equals the infimum ofthe upper sums. The Lebesgue integral always exists. We will prove this proposition in four steps. A
Step 1. Adding a finite set of points to a partition P of [a,.8] does not decrease the lower sum or increase the upper sum for P. (So~caLled "refinements" of [a,.8] generally increase lower sums and decrease upper sums:) Step 2. No lower sum can exceed an upper sum. Hint: LYk-lJL(Ek) < L
Zi-lJL(Ft) < L
Au~
A
ZiJL(Fi) < LSJJL(Gj).
Au~
~
Step 3. The supremum of the collection of all numbers associated with lower sums is less than or equal to the infimum of the collection of all numbers associated with upper sums.
Step 4. Let E > 0 be given. Construct a partition P'" of [a 111 so that < ... < = b, with Yk - Yk-l < e/(b - a), for a = Yo < k = 1,2, .. ., n. Then I
yr
Y;
n
a(b - a) ::: L
n
Yk-lJL(Ek) <
k=1
and
n
0:::
L k=1
L n
(Yk - Yk-l)JL(Ek) <
YkJL(Ek) < .8(b - a),
k=1
L
(b ~ a) JL(Ek)
=
E.
k=1
The lower and upper sums for this partition are within E of each other. Thus the supremum of the lower sums and the infimum of the upper sums are within E of each other, and thus because E is arbitrary these numbers are the same. For a bounded measurable function on an interval [a, b] the Lebesgue integral always exists. Exercise 6.1.1. Let LJ: f dJL be the Lebesgue integral of the bounded measurable function f. Then for any E > a we may construct a partition P of [a, .8] so that the difference of the upper and lower sums for this partition is less than E. Hint: Properties of supremum and infimum.
The lebesgue Integral
6.1.2
113
Young's Approach
Another common approach to the Lebesgue integral is that of William H. Young (1863-1942). Young's method (1905) closely parallels the development of the Riemann integral, where f is assumed only to be bounded and we work with the domain [a, b] instead of the range [a, .8]. Again, we will explore this approach in stages. First, suppose f is a bounded function on [a, hI with ex < f < .8. Partition the domain of f, [a, b], into a finite number of non-overlapping measurable sets Ek, where J.k(Ei n E j) = 0 and Ek f:. rp, a so~caIled measurable partition of [a, b1. Pick a point Ck in the measurable set Ek, and fonn the Lebesgue sums n
L
!(ck)J.L(Ek),
written
L
f(c)J.L(E}.
(1)
p
k=l
These sums satisfy
L i~f fJ.L(E) ::: L f(Ck)J.L(E) p P ::: Lp sup fJ.L(E) ::: .8(b - a). E
a(b - a) <
Now compare the supremum of the lower Lebesgue sums with tile infimum of the upper Lebesgue sums over all measurable partitions of [a. b]. If these numbers are the same, say A, we say f is Lebesgue integrable on [a, b], and we write A = L f dJ.L. For E, a measurable subset of [a, b],
J:
L
Lf
dtJ. "" L
lb
fIE dtJ.,
Before continuing with Young's approach to Lebesgue integrability. SOme exercises will help clarify a few concepts.
Exercise 6.1.2. Demonstrate the following assertions. a. Measurab1e refinements of [a, b1 do not decrease lower sums or in· crease upper sums. Hint: Ek = Fl U ... U FA, all sets nonempty measurable subsets of fa. b], illf f/J,(Ek) < EJ,:
L
inf fl.L(Fj) < Fj
L
sup fJ.L(FJ) < sup fJ.L(Ek). Fj
EJ..
114
A Garden of Integrals
b. No lower sum can exceed an upper sum. Hint: P = U~ Ekl Q = Fj; form nm measurable sets Ek n Fj. Show
Ur L P
inf IJ.L(E) < E
L
inf IJ.L(E n F) =5
PUQ EnF
:5
L
L
sup IJ.L(E
n F)
PUQ EnF
sup IJ.L(F}.
Q
F
c. Let e > 0 be given. If we have a partition of [a, bI with the upper artd lower sums within e of each other, then the supremum of the upper sums and the infimum of the lower sums are within e of each other. The bounded function I is Lebesgue integrable on [a, b].
d. If the bounded function I is Lebesgue integrable on [a, b], then for any e > 0 we have a partition of [a, b] so that the associated upper and lower sums are within e of each other. Hint: The supremum and infimum properties. Recall that I's being bounded did not guarantee existence of the Rie· mann integral - the additional, and sufficient, condition of continuoZls almost everywhere had to be imposed (Theorem 3.6.1). Likewise, in Young's approach an additional and sufficient condition, measurability 01 I, is imposed. We claim that the bounded function I on [a, b] is Lebesgue integrable on [a, b] iff I is Lebesgue measurable. For the bounded (0:: < f < fJ) measurable function I, we can simply use Section 6.1.1. Hint:
LYk-lJ.L(f-1([Yk-bYk»)) < sup Linf/J.L(E) < infLsUp/J.L(F) P
Q Q
P 1
< LYkJ.L(/- ([Yk-lt Yk»)). For the other direction, suppose the bounded function f is Lebesgue integrable on [a l b]. We claim that I is Lebesgue measurable on [al b]. From Exercise 6.1.2, we have partitions Pn of [a, b], with Pn a refinement of Pn-I, so that (sup 1- inf I)J.L(E) < l/n. II Let rpn inf I and Vrn sup I on Pn. Note that rpn-l ~ tPn < Vrn :5 Vrn-l on [a, b]. We have monotone sequences of measurable functions, (tPn), (Vrn) with rpn :5 I :5 1/111 on [a, b]. Thus limrpn = rp :5 f :5 Vr = lim t,l; rp and Vr are measurable functions on [a, b]. We claim rp = Vr almost Vr almost everywhere and measurability of everywhere, in which case I I follows (Section 5.7.2).
=
LP
=
=
115
The Lebesgue Integral
Let
E
= {x
E
[a , b] 11{I(x) - ¢(x) > 0)
= u..
1x
CUm
1x E [a,b] 11{In(x) -
E
[a, hI I >/rex) - ",(x) >
~j
cjJlI(X) >
..!..j. 111
It will be sufficient to show that J.L({x E [a,b] 11{In(x) -¢n(x) > 11m} has measure zero. By construction, LJ:(l{In -¢I1)dJ.L < l/n. That is, "!'J.L
m
J.L
(1 (1
x
E
[a , b] Il{In (x) - cjJ/1 (x) >
~l) m
x
E
[a b] I l{IlI (x) - cjJ'l (x) >
2.1) <
I
< ..!:.,
111
n
m
or for all n.
It
The argument is complete. We conclude, for bounded functions on [a. b], o
Riemann integrability iff continuity exists almost everywhere;
G)
Lebesgue integrability iff measurability holds.
6.1.3
And Another Approach
The reader might prefer a blend of the two approaches just sketched. Assume f is a bounded and measurable function in the interval [a, b], and use measurable partitions of [a , b] the domain of f as in Young's approach: L f(Ck)J.L(Ek)· Complete the details.
6.2
IntegrabUuty: Riemann IEnsures Lebesgue
We now have an important result to explore. In all that follows, we will assume f is Lebesgue measurable whenever we refer to the Lebesgue integral of f. Theorem 6.2.1. All Riemann integrablefil17ctions are Lebesgue integrable. and
R
f.b f(x)dx = L t
f dJl.,
116
A Garden of Integrals
We begin by recalling that bounded, continuous almost everywhere functions are measurable. The nonoverlapping intervals [Xk-l Xk] form a measurable partition P of [a, b]. We have j
Linf/fl.x < sup
(I::
inf/J.L(E)) < inf(:E sup IJ.L(E))
pEP
p
E
< Lsup/fl.x p
where 'P is the collection of all measurable partitions of [a. b]. Complete the argument.
Exercise 6.2.1. Show that Dirichlet's function
I(x) =
11o
x :atio~aIJ
x mational.
is Lebesgue integrable on the interval [0. 1]. The problem, posed by Dirichlet and considered by Riemann t of developing an integration process to "integrate" such functions, was solved by Lebesgue approximately 7S years later. We frequently write L I(x) dx
J:
J:
for L I dJ.L. For exampie, consider the function
I( )= x
11 J0fJ.
x = p Jq. p, q relatively prime natural numbers, p < q, x = 0, 1, or x irrational.
We can show that I is Riemann integrable (Exercise 3.4.2) by Theorem 3.6.1, because I is bounded and continuous on the irrationals, that is, almost everywhere. Thus, I is Lebesgue integrable. Finally, because I = 0 almost everywhere, L
/.1
f dlJ. = 0 = R
f
f(x)dx.
Some Riemann integrals can be calculated by using the power of the Lebesgue integral.
Elcercise 6.2.2. D.
Consider the function (Exercise 3.4.1)
I(x) Show that
=! o 1
IJ21~
<. x 'S:. IJ(2n -
1),11 = 1,2, ... ,
otherwIse.
RJo1 I(x) dx = L Jo1 I(x) dx =
ln2.
117
The Lebesgue Integral
b. For 0 < x < 1, use Lebesgue's idea of partitioning the range [0, 1] of -IX with Ek = [(k - 1)/n, kin), k = 1,2, ... ,11 + I, and show 1 L -IX dx 2 Riol x 2 dx by noting that
10
=
2L (k - 1) 11
2
.!. <
L
(k - 1) (2k - 1) n3
n
[Gr -(k / r] k[(k)2 (k - 1)2] < 2L (k)2 < L;;-;:; ;:;.1
= E k/
11
6.2.1
-
-'1-
Nonnegative Unbounded Measurable Functions
We may extend Lebesgue integrability to nonnegative unbounded measurable functions on [a, b] in several ways. A particularly straightforward approach for nonnegative measurable functions I is to work with "truncations" of I: k
f
1Ik
=
0 < f < k, I >k.
Thus the sequence {k f} of bounded measurable functions converges monotonically to f, and we define the Lebesgue integral of I to be the limit of the monotone sequence {L
Lib
I: I
kid J.L } : .
dJ.L
= limL k
a
lb
k I dJ.L.
a
Of course the limit may be infinite, but we are not interested in this case. We wil1 say that I is Lebesgue integrable on [a, b] provided that limk L k I dfJ. is a real number.
I:
Exercise 6.2.3. a. Show that
I
is Lebesgue integrable on [O,lJ and L
gIVen
ll/~
[(x) =
x = 0,
o <x::::;
and so OX k I(x)
=
{
1,
= 0,
1/ -IX
1/ k 2 < x < I,
Ie
0 < x < 1/ 1c2.
101 I
dJ.L = 2,
118
A Garden of Integrals
b. Show that
I
is not Lebesgue integrable on [0, 1], given
f(x)
Ox
={
l/x
= 0,
a< x
< 1.
c. Show that I is unbounded on every subinterval of [0, 1], that I dJi. = 0, given Lebesgue integrable, and that L
J;
I(x) =
6.2.2
{
= p/q.
q
x
o
factors, a < x < I, otherwise, x = 0, x = 1.
I
is
P, q natural numbers, no common
Positive and Negative Measurable Functions
What about the case when the measurable function negative? Observe that
1=I/I+f _111-1 2
and
2
I
is both positive and
If I = If I + I + If 1- f 2
2'
This allows us to write I as the difference and II I as the sum of two nonnegative measurable functions. Each of the integrals
Lib III +f d a
2
and
Ji.
Libi/l-id a
2
J.L
may be calculated by truncation if necessary. If both integrals are finite, we say I is Lebesgue integrable on [a , b]. Note that I is Lebesgue integrable iff If I is Lebesgue integrable.
Exercise 6.2.4. a. For
I(x)
={
-1 1
x ~atio.nal, x matlOnal,
l
I/(x)1 dx = 1 and RJol I(x) dx does not exist. However, 1 1 both L Jo III dJi. and L J0 f dJi. exist. The function I is Lebesgue integrable iff lfl is Lebesgue integrable. This is not true for Riemann show RJo
integrals in general. b. Show that L
J; I dJi. = 0, given f(x) =
{
1/-IX
0< x <1, -1/.j2-x 1 < x < 2.
The Lebesgue Integral
c. Show that
I
119
is differentiable on [0, 11, given
I(x) =
j x 2 sin(n/x2)
0 < x < 1,
lOx = d. For the previous function, show that L Jei L
1
10o
1 - cos
x
Ic (2:) dx ~ L x n
N
o.
I' dji. does not exist. Hint: N
'1t
(N-l)'1t
cos(y) dy. 2y
6.2.3 Arbitrary Measurable Subsets For an arbitrary measurable subset E of [at b], we may define the Lebesgue integral of lover E, written LJE I dji., as LJ; IXE dji., the Lebesgue integral of I times the characteristic function on E, over [a, b]. If I is Lebesgue integrable on [a, b] and g equals I almost everywhere g dji. = on [at b], then we claim g is Lebesgue integrable on [a, b] and L
J;
LJ: I
dji.. (We freguently say "sets of measure zero do not affect Lebesgue integrals.") Suppose I and g are nonnegative, and consider their truncations {k I}, {kg}, The functions k I, kg are bounded measurable functions, Lebesgue integrable; k I = kg almost everywhere on [a, b]; kif:. kg on a set Z of measure zero; and k I = kg on [a, b] - Z. For the measurable partition ([a, b] - Z) U Z of [a, b]. form the Lebesgue sum (1),
+ (kg - k I) (C2)ji.(Z) O· ji.([a, b] - Z) + (kg - k 1)(C2) ·0
(k g _ k I)(cl)ji.([a, b] - Z)
= We may show L
J: e'
g-
k I)
dji.
- O.
= O. Thus
which can be extended with limits. The corresponding result is not necessarily true for Riemann integrals. The reader may compare Dirichlet's function with the identical1y zero function on [0, 1], or Exercise 6.2.3c.
120
6.2.4
A Garden of Integrals
Another Definition of the Lebesgue Integral
We offer an equivalent (and in hindsight more easily generalized) definition of the Lebesgue integral. As before, the explanation has several steps.
Step 1. If t/J is a nonnegative simple function, t/J =
L Ck XEk' where E
= U7 Ek, for Ek mutually disjoint measurable subsets of E and Ck nonnegative real numbers, then
Step 2. Recall Theorem 5.7.2, that if I is a nonnegative measurable function defined on a measurable set E, we have a monotonically increasing sequence of simple functions {t/Jk} converging to I· Then L JE I dJ.L == limLJE t/Jk dJ.L. (This limit is well defined.) A nonnegative measurable function I is said to be integrable on E if L JE I dJ.L is fmite. Step 3. For a measurable function f, I = (III + 1)/2 - (111- 1)/2. If L JE (III + 1)/2 dJ.L and L JE(III- 1)/2 dJL are both finite, then we say
I
is Lebesgue integrable and define L
rJE f dJ.L = L JEr 111-2 I
dJ.L - L
rJ III 2+ f E
dJ.L.
Note: If g = I almost everywhere, and I is Lebesgue integrable, then g is Lebesgue integrable and L JE g dJL ::::: L JE I dJL (with t/J simple, t/J = 0 almost everywhere, et cetera).
6.3
Convergence Theorems
We now discuss some convergence theorems for the Lebesgue integral, one of their strongest attributes. By definition, we have convergence theorems for special situations, simple and truncations: or We want to replace the special measurable functions t/Jk or k I with more general measurable functions. This will result in convergence theorems that are more powerful than those for the Ri~mann integral. Generally spealcing, we seek to weaken the unifonn convergence requirement that is generally associated with the Riemann integral.
121
The Lebesgue Integral
Theorem 6.3.1 (Bounded Convergence). If {fk} is a uniformly bounded
sequence of Lebesgue measurable functions converging pointwise to f almost everywhere on [at b], then limL
1b /k
dJL = L
1b I
dJL = L
1b
(lim/k) dJL.
Proof Define f to be zero wherever lim fk is not f. (Sets of measure zero do not affect the Lebesgue integral.) Because the sequence {fk} is unifonnly bounded, say Ifkl :::: B on [a,b], and limits of measurable functions are measurable, we have that f is bounded and measurable, and thus Lebesgue integrable on [a 1 b]. So L f dIL makes sense, as does L fk dIL· Let € > 0 be given. By Egoroff's Theorem (5.7.3), we have a subset E of [a, b] so that the sequence {fk} converges uniformly to f on [a, b] - E, and IL(E) < € where If - fkl < 2B. Thus,
J:
L
t
L
1& I dJL
t
l/k - II
dJL
Lll!k - fl
dIL
Ik dJL
=
J:
E
2B IL(E)
+L
+ fIL([a. b] -
r
J[a,bl-E
Ifk - fl dIL
E) < 2Bf + f(b - a)
for k sufficiently large, and the argument is complete.
0
Exercise 6.3.1.
a. For
k2
fk =
10
0 < X < 1/ k,
limfk = 0,
otherwise,
limL
11 /k =
limk =
00.
b. Define a sequence {fk} by .f. ( )
Jk
x
= 11o
x
= rl,1'2""1 r k,
otherwise,
where 1'111"2, • " is an enumeration of the rational numbers in (0, 1). 1 Then Dirichlet's function (lim !k) is Lebesgue integrable and L f dJi.. = o The limit of a unifonnly bounded monotonically increasing sequence of Riemann integrable functions is not Riemann integrable in general.
Jo
c. Revisit Exercise 2.5.2. d. Revisit Exercise 3.4 .1.
A Garden of Integrals
122
6.3.1
Monotone Convergence
Uniform boundedness may be replaced with monotonicity. We will now look at the Lebesgue Monotone Convergence Theorem of Beppo Levi. Theorem 6.3.2 (Levi, 1906). If {I,,} is a monotone increasing sequence 01 nonnegative measurable lunctions converging pointwise to the function I on [a, b], then the Lebesgue integral of I e.'Cists and
Lib I diL = limL ib Ik diL. a
a
.
Proof. The function I, being the limit of a sequence of measurable functions, is measurable. Because every nonnegative measurable function is the limit of a nondecreasing sequence of simple functions (Theorem 5.7.2), and because its integral is by definition (Section 6.2.4) the limit of the sequence of integrals of the simple functions, we have:
o:5 tPll < tP12 <
... :5
cfJln :5 .,. , lim tPln = 11. n Lib 11 diL a
o < cfJal < tP22 < ... !:. tP2n :::: .• . , lim cfJ2n n Lib 12 diL a
0:5
tPml :5 tPm2 <
... :::
tPmn
< ... , lim tPmll n
Lib tn diL a
and
=D limLib tPln diL, n
= ED
a
12,
and
limL ib tP2n diL, n
- Iml
a
and
=D limLib tPmn diL,,· .. n
a
Construct a new sequence of simple functions, ~k: " = fiJn. fiJI fiJ2 = max{tP12, tP22} ::: tP12 ::: tPl1 = tPh ~
~
They are nonnegative and fonn an increasing sequence converging pointwise to I: lim ¢k = lim Ik = I· By definition, then, L I d iL D
J:
=
123
The Lebesgue Integlal b ....
limL fa ¢k dJL. However,
since {/k} is a monotone increasing sequence converging to f. We may therefore conclude that limL Ik dJL = L I dJL. 0
f:
J:
Exercise 6.3.2. If {/k} is a monotone sequence of Lebesgue integrable functions on the interval [a. b] converging pointwise to I, 11 dJL < 00, and limL
b
J: Ik dJ..L
If:
is finite, sho}" that I is Lebesgue integrable on [a,
and L fa I dJL = limL f: Ik dJL. Hint: (/1 - Ik)·. Exercise 6.3.3. Calculate the following. n. L
fol x-liZ dx. Hint:
b. L fol
-x In(x) dx. Hint: Ik(X)
c. L
1
0 /.
=
(00~ (1 ~x /2x )11 )
!
1/ Ie < x ~ I, 0< x < 1/ k.
x In(x) 0
3
2
~
dx. Hint: Show
3 2 X /
L.J (1 o l
+ x2)n
=
!
0 X- 1/2
+ x 3/2
XCI
d.
LL ( )n dx. o f. 1 +
f.
LLf.l----x-. ---dx. + (It - l)x](l + l1x)
0
1
0
I
x = 0, 0<
X
< 1.
b1
124
A Garden of Integrals
6.3.2 Sequential Convergence Pierre Fatou (1878-1929) gave us a result that will prove useful in the development of Fundamental Theorems of Calculus for the Lebesgue Integral. It is widely known as Fatou's Lemma. Lemma 6.3.1 (Fatou, 1906). 1f{/k} is a sequence a/nonnegative Lebesgue i";tegrablefimctions converging pointwise almost eveJywhere to / on [a I b}.
then
L f. bIdJ1.
::;: lim infL f. b fk dJ1..
Proof. Here is the basic idea: "lim inf' yields monotone increasing sequences that are amenable to the Monotone Convergence Theorem. We have /1 == inf{/l. /2 •... }; /1 is measurable, where 0 < / 1 < /n for all 12; and
L f.b Continuing,
L
I, dJ1. < inf {L
t t 11 dJ1.. L
h dJ1. •... } .
== inf{/k. /k+l> ... }; and again
L f.b f..g dJ1. < inf! L f.b !k dJ1.. L f.b !k+1 dJ1. •... } . So we have 0
L f.b f..g < inf {L f.b
Ik dJ1.. L f.b Ik+l dJ1. .... } .
The sequences and are nonnegative monotone increasing sequences of real numbers (integrability of /k) and have limits in the extended reals:
limL f.b f..g dJ1. < liminf {L f.b !k dJ1.. L f.b
Ik+l dJ1. .... } .
The Monotone Convergence Theorem 6.3.2 yields
limL f.b f..g dJ1. = L f.b limf..g dJ.£ = L f.b I
dJ1. <: liminfL f.b !k dJ1.. 0
125
The Lebesgue Integral
6.3.3
The Dominated Convergence Theorem
Finally, we have a very powerful convergence theorem that replaces monotonicity with dominance. Theorem 6.3.3 (Lebesgue, 191 0). Suppose {lk} is a sequence 01 Lebesgue integrable functions (fk measurable and L 11kl dJL < 00 for all k) converging pointwise almost everywhere to the function 1 on [a, b]. Let g be Lebesgue integrable so that 11kl < g for all k on the interval [a. b].
J:
Then
f
is Lebesgue integrable L
lb f
012
[a, b] and
dJ1. = limL
lb /k
dJ1..
Proof Since limits of measurab Ie functions are measurable by Theorem 5.7.1 and since sets of measure zero do not affect Lebesgue integrability, we may assume 1 is real-valued and measurable, and that limlk 1 on [a, b]. Construct two monotone sequences {f k} and {I k}, with
=
Le =
inf{fk. Ik+lt···}
and
1k =
sup{lb fk+l • ... }.
Le
Because -g < < Ik ~ f k ~ g, the functions integrable. "Furthermore, since
o < g + Le < g + Le+l
< 2g
and
0 < g-
fk
Le,
f k' and f are
~ g-
1 k+l
< 2g,
Le} and {g- 1 k} are monotone increasing sequences of integrable functions with lim(g + L) = g + 1 and lim(g - 1 k) = g - f·
the sequences {g +
Application of the Monotone Convergence Theorem yields
L
t
(g
+ f)dJ1. = L = L
lb lb
lb f = lb lb Lc
gdJ1.
+L
g dJ1.
+ lim
dJ1.
limL
(g
+ L) d"
dJ1..
= lim J: Lc d 11.· Similarly, L Jab 1 d JL = lim J; f k d 11.. With the observation that Lc < fie < 1 k, and all functions being inte-
That is, L Jab 1 d 11. grab Ie, we have
L Thus L
lb Lc
J: 1
dJ1. < L
lb /k
dJ1. < L
lb h
dJ1.
for all k.
dJL = lim Jab fk dJL, and the proof is complete.
0
126
A Garden of Integrals
Exercise 6.3.4.
r' k;2 x
a. Show limL
10 1 + on [0, 1].
k
Let g
=!
1
b. Evaluate limL k
k3/2x k2
1+ 1
0
g(x)
c. Show limL k
2
x2k
-2
1 +x
1
e. Show L {
2k
x
=
1 1 x In
2
dx
l
= O. Hint.
O x =0 .' O<x<1.
X- l/2
dx = 2.
(L: ~)
d. Evaluate L {
dx = O. Hint. There are many approaches.
2
G)
dx.
dx
=
:2 .
r
x a- 1 1 1 f. Show L 10 1 + xfl dx = a - a + fJ where a, {3 > O. l
6.3.4
Interchanging Land
1
+ a + 2fJ -
1
a
+ 3{3 + "',
J
Suppose {fk} is a sequence of Lebesgue integrable functions defined on [a. b]. If the series L: Ifk I dJL converges, we can show that the series L: fk converges almost everywhere on [a, b] to a Lebesgue integrable function and
J:
By the Monotone Convergence Theorem 6.3.2,
Because E Ifk I is Lebesgue integrable on [a. b], it is thus finite almost everywhere on [a, b]. We have that E fk converges almost everywhere on (a, bl and IE fkl ::: E Ifkl.
127
The Lebesgue Integral
We claim that E fk is integrable on [a, b]. Let gn = E~ Ik. Then Igil I < E Ifkl· We apply the Dominated Convergence Theorem 6.3.3'
6.4
FUr1ldamenta~
Theorems for the Lebesgue Integral
What remains to explore? We have reached the Fundamental Theorems of Calculus for the Lebesgue integral. Now we can resolve the crucial questions: o Does L
J: f' dp,
o When is
6.4.1
= feb) - f(a)?
(L J~"'C f dJL)'
= f?
Properties of the Indefinite Integral
We will begin by examining properties of the indefinite integral. That is, for a Lebesgue integrable function f on the interval [a b], we will define a new function F on [a, b] by I
F(x)
=L
l
x
f d/L
=L
f
!(t)dt,
What are the properties of this function? Is F continuous? Differentiable? .. For the following discussion, assume f is Lebesgue integrable on [a, b]; that is, f is Lebesgue measurable and L Jab III dJL < 00. Let F(x) = L J~"t f(t) dt. We will investigate seven properties of the indefinite integral.
Property I. If f is bounded 011 [a, b], then F is continuolls on [a, b]. (In fact, F is Lipschitz.) Hint: With a < x < y < b, IF(y) - F(x)1
=
L Y
L
f(t)dt < B(y -x)
Property 2. F is continuoZls on [a, b] whether
for
If I <
B on [a, b].
f is bounded 01' not.
128
A Garden of Integrals
Hint: Truncation. Suppose
f
is unbounded and nonnegative. Then
o ~ limL f.b kf
dlL = L
f.b f dlL.
Let e > 0 be given. We have a natural number K so that
L
f.b f dlL
<
-l
f.b kf dJ1. ~ L f.b f dJ1.,
J:
whenk > K.
Now, F K (x) = L K f (t) d t is continuous by Property 1. We have a ~ so that for y E (x - ~ X + ~) n [a, b], the difference K (y) - F K (x) < e. But
IF
J
IF(y) - F(x)1 ::: ]F{y) - FK{y)]
~L
t
+ IFK(y) -
(f - K f)dIL +E+L
I
FK(x)1
t
+ IFK(x) -
F(x)1
(f - K f)dIL < 3l,
where y E (x - ~J X +~) n [at b]. The reader may remove the requirements that I be nonnegative with the usual f (III + f)/2 - (If 1- f)/2. Because F is continuous on [a, b], we may conclude that F is uniformly continuous on [a b].
=
I
Property 3. Given an e > 0, we have a 8 so that if E is any measurable If I dp. < e. subset of [at b] with }L(E) < 8, L
JE
We have replaced the small interval of Property 2 with a small measurable subset of [a, b]. Start by assuming f is bounded on [a, b] so thatlf I < B. Then If I dp. < Bp.(E). Let 8 < e/B. For I unbounded and nonnegative,
LJE L
Lf
dlL = L
~L
L
t
Lf
(f - K f) dlL
+L
(f -
+ L IE K f
K
f) dJ1.
K
dlL dIL·
Complete the argument using the Lebesgue Monotone Convergence Theorem. Property 4• .if F(x) =
J: f(t) dt = 0 fm' all x in [a, b], then f(t) = 0
almost everywhere in [a b]. I
Suppose the set {x E [a, b] I f(x) > O} = U {x E [a, b] I I(x) > lin} has positive measure. Then for some N, we have a closed subset F of
129
The Lebesgue Integral
{x E [a, b1 1 f(x) > 1/ N} (by Theorem 5.6.1), with JJ.(F) > O. Then [a, b] - F = U(ak,bk), so F(b) =
Lib
f dJJ. = L ( f dJJ.
1F
a
+
Ll
f dJ1.
[a.b]-F
=L LldJl.+ L:[L [. IdJl.-L {k Id~'] =L But F(b)
IF( f
dJJ. >
~J1.(F) > o. N
= 0 by assumption. Complete the argument. f is nonnegative, F is nondecreasing on [a, b]. (If I + f)/2 - (If 1- f)/2, F is the difference of
Property 5. Assuming
In fact, since f = two monotone functions. That is, F is a function of bounded variation, and thus F is differentiable almost everywhere on [a, b] by Lebesgue's Differentiation Theorem 5.10.1. So, what is F'?
Property 6. If f is continuous at Xo, a point in [a, b], then F is differentiable at Xo and F'(xo) = f(xo). Review tile Fundamental Theorem 3.7.2 and supply the details of the argument. Hint:
F(xo
+
'i, -
F(xo) - I(xo) =
!
L i~+h
(J - I(xo)) dJl. .
Property 7. Iff is bounded 011 [a, b], with If I < B, then F'(x)
=
f(x)
almost evelywhere. We will show first that L f~"( (F' - f) dJL = 0 for all x in [a. b]. Because F is differentiable almost everywhere, lim1Z[F(x + I/1I) F (x)] = F' (x) for x a point of differentiabil ity of F, almost everywhere. Then F(x + l/n) makes sense on [a, x] if we extend F to [a, x + 1] by F(t) = F(x), for x < t < t + 1. Since sets of measure zero do not affect the Lebesgue integral, we have
L
t
(F'- f)dJl.
=L
t (limn
[F
(r + ,~) - F(t)] - I) dr.
Can we move "lim" outside the integral? What do our convergence theorems demand? We could use the Bounded Convergence Theorem 6.3.1 - provided we have a uniform bound for the sequence (n[F(x + l/n) - F(x)]).
130
A Garden of Integrals
We do. Since
If I <
B,
[ ( + n1) - F(x)]
n F x
l
x 1/n
If I dJ1. < B.
+
~ It . L x
So, invoking Cauchy integrals, we calculate
L
t
+ ,~) -
limn [F (t
- limL
t
n
F(t)] dt
[F (t
t [n . l
- lim[n.c _ lim
+
F (t
n-
F(tY] de
+t/. F(t) dt
-
lim [It . F(x) .
-
F(x) - F(a).
F(t)dt]
H/
x
C
t f..
+ ~) dt -/I·e -II·
C
.!.n - n . F(~n) . .!.] , 11
' F(t) dt]
a<
~n < a + ..!.., 11
Now use Property 4. Exercise 6.4.1. Let f(x) =
Calculate F(x) = L
I
x
x irr~tional,
o x rational.
I: f(t) dt and its derivative. (LI:
We have shown that, for f measurable and bounded, f dJ1.)' = f almost everywhere. We can, in fact, dispose with the requirement that f be bounded.
6.4.2 A Fundamental Theorem for the Lebesgue Integral f is Lebesgue integrable on [a, b], define a fimction f on [a, b] by F(x) = L f dJ1.. Then F is absolutely continuous on [at b] and F' = f almost everywhere on [a, b].
Theorem 6.4.1 (Lebesgue, 1904). Given that
I:
Proof The absolute continuity of F on the interval [a, b1 will follow from Property 3 of indefinite integrals (Section 6.4.1). Given an € > 0, we have a 8 > 0 so that if E is any measurable subset of [at b] with J1.(E) < 8, then LIE If I dJ1. < €.
131
The Lebesgue Integral
Let (ak' bk) be a finite collection of painvise disjoint intervals with length, L(bk - ak), less than 8. Then
So F is absolutely continuous on [a, b] and thus differentiable almost everywhere. We have shown the validity of the second conclusion (F' = f almost everywhere on [a, bD, with the assumption that f is bounded (Section 6.4.1). Since f = [(If I + f)/2] - [(If I - f)/2], the difference of two nonnegative measurable functions, we assume f is nonnegative, and we begin with the sequence of bounded, measurable functions, the truncations {k f} of f. The function L (f - k f) d J1. is nonnegative and nondecreasing on [a I bl. By Lebesque's Differentiation Theorem 5.10.1, this function has a nonnegative derivative almost everywhere, and by Property 7, (L f~"'{ k f dJ1.)' =k f almost everywhere. So, '
f:
(Ll (f - fl dP.)' = (L { = (L { f dP.)' - f
o<
k
(L {
f dp. )' -
k
f dp.)'
k
almost everywhere. Thus (L f: f dJ1.)' - f f) dJ1. > O. On the other hand,
= F' -
f > 0 almost everywhere, and L fax (F' -
L{(F' - fldp. = L{
F' dp.
-L {
f dp..
Now, F is a nondecreasing continuous function (by Properties 2 and 5). So F is Cauchy integrable. Furthermore,
o<
C{
n
[F (t + ,~) - F(t)] dt
[1
= n· C
x+1/11
't
F(t) dt - C
:" F(x) - F(a) =
J.% f
l
dp.,
Q
a 1/11 +
F{t)dt
]
132
A Garden of Integrals
since F(t) = F(x), x < t < x + 1 and F(a) < F(t), for a < t. The sequence {n[F(x + lin) - F(x)]} is a sequence of nonnegative Lebesgue integrable functions whose limit is F' (x) almost everywhere. Applying Fatou's Lemma 6.3.1, we have
L{
lim[n (F (I + !) - F(!))] dl :0 liminfC { n [F (1.+ !) - F(l)] dl
F' dJ1. = C {
:0 F(x) - F(a) = L {
f dJ1..
f:
That is, L (F' - f) dJ1.. :: 0 for all x in [a, b], and thus F' - f = 0 almost everywhere, by Property 4. Demonstrate the case when f is nonpositive. So, F' = f almost everywhere on [a, b]. This completes the proof. 0 Compare the Fundamental Theorems 2.4.1, 3.7.2, and 8.8.1. As forLf: F' dJ1.., it would be wonderful ifLf: F' dJ1.. = F(b)-F(a), but this is not true. Let C denote the Cantor function. Then C' 0 almost everywhere on [0,1.]. Thus,
=
o= 6.4.3
L
f
C' dJ1. < 1 = C(l) - C(O).
The Other Fundamental Theorem
Are there additional conditions we might impose on the derivative that F'dJ1.. = F(b) - F(a)? This question was the impetus would yield L for Lebesgue's development of his integration process.
f:
If F
is a differentiable junction. and the derivative F' is bounded on the interval [a 1 b], then F' is Lebesgue integrable 011 [a, b1 and
Theorem 6.4.2.
L
1.% F' dJ1. =
F(x) - F(a)
for x in the interval [a, b1. Proof. Because F' (x) = lim [ F (x 11,1-+0
+ h/l) - F (X)] hll
133
The Lebesgue Integral
for any sequence of real numbers {h n } convergmg to zero, and because limits of measurable functions are measurable, we may conclude the derivative is measurable, and it is bounded by assumption. Thus F' is Lebesgue inte~ grable on [a, b]. Because limn[F(x + lin) - F(x)] = F'(X), it is natural to consider the sequence of functions {n[F(x + lin) - F(x)]}, with F extended to the interval [a, b + 1] by F(x) = F(b) + F'(b)(x - b), where b < x < b + 1. This sequence of measurable functions is unifonnly bounded: for t < c < t
1
+ -, 11
by the mean value theorem for derivatives and the assumption that the derivative is bounded. We may use the Bounded Convergence Theorem 6.3.1 for our calculation: L{
F'dJL
=L{
limn [F
(I + ,~) - F(t)] dl
=limR {n[F(t+
= limn R f.x = limn R
F
(I + ,~) dl -limn
:J:+I /1I
1
~)-F(t)] dt
F(t) dt - limn R
x
= F(x) The proof is complete.
f.x F(/) dl f.a+l/ll R
FCt) dt
a
F(a).
0
So what remains? We would Jjke to remove the requirement that the derivative is bounded. This cannot be done with the Lebesgue integration process. Recall Exercise 6.2.4. (A solution will appear in Chapter 8.)
6.4.4 The Bounded Variation Condition We showed that if F has a derivative and the derivative is bounded, then L F' dj1. = F(b) - F(a). The assumption that F is differentiable with a bounded derivative implies that F is absolutely continuous on [a, b]; recall Exercise 5.8.2.
J:
134
A Garden of Integrals
Would bounded variation of F be sufficient to have L F(a)? No; for example F()
:c
L
f
J; F' dJ1. = F(b)-
j 1 0<x
=12
< I, 1 < x < 2;
F' = 0 almost everywhere; F'djJ. = 0 < 1 = F(2) - F(O).
The requirement we are looking for is that F is absolutely continuous on
[a, b].
6.4.5
Another Fundamental Theorem of Calculus
Let's look at one more Fundamental Theorem of Calculus for the Lebesgue integral. Theorem 6.4.3 (Lebesgue, 1904). If F is absolutely continuous on [a. b], then F' is Lebesgue integrable, and L{
for x in [a t b].
F' djJ. = F(x) - F(a)
Proof We know that F is of bounded variation on [a, bI. Thus, F = Fl - F2 with Fl. F2 monotone increasing functions and F' exists almost everywhere. Since IF' I =:: F{ + Fi almost everywhere, L
t
IF'I d}J.:<;; L
t
F! djJ. + L
< Fl (b) - Fl (a)
t F~
djJ.
+ F2(b) -
F2(a)
by the proof of Theorem 6.4.1. Thus F' is integrable. Let G(x) = L J~"t F'dJ1.. By Theorem 6.4.1, G is absolutely continuous, so F - G is absolutely continuous and (F - G)' = F' - G' = 0 almost everywhere. By Theorem 5.8.2, F - G is constant. Because G(a) = 0, we have F(x) - F(a)
=L {
F' djJ..
Compare Theorems 2.3.1, 3.7.1, and 8.7.3.
0
135
The Lebesgue Integral
6.4.6
Comments
For the absolute continuity of ¢, refer to Definition 5.8.2. Tbeorem 6.4.4. If f is continuolls and ¢ is absolutely continuous on an interval [a, b], then the Riemann-Stieljes integral of f with respect to ¢ is just the Lebesgue integral of f¢' 011 the interval [at b]:
R-S
t
f(x)d",(x) = L
t
N'dp..
Proof How can we show this? First, since absolutely continuous functions are of bounded variation, the existence ofR-S f(x) d¢(.;r.) may be shown by Theorem 4.4.1. In fact,
J:
(R-S {
= N'
f(t) d", (t»),
almost everywhere. Second, f¢' is Lebesgue integrable on [a, b] and ¢(c), for fe, d] c [a, b], by Theorem 6.4.3. Next we..,have
R-S
t
f(x) dt/>(x) - L < R-S
+
t
t
N'dp.
f(x) dq,(x) -
L f(Ck)tJ.", -
Also,
L f(ek) It/>(Xk) - t/>CXk-l)]- L = <
f:, Lr.f:, L [L
LJed ¢' dJ.L = ¢(d)-
L
t
L f(Ck)tJ.t/>
t
N' dp. .
N'dp.
(f(Ck) - f(t» ""(t) dl]
IfCCk) - f(/)II",'(/)1 dt.
We muy use unifonn continuity of rem 6.4.3) to finish the argument. 0
f
and integrability of ¢' (Theo-
136
A Garden of Integrals
Theorem 6.4.4 is not true if we assume only that f/J is of bounded variation. For example, let f = f/J = C, the Cantor function. Then by Theorem 4.4.1 and Exercise 4.4.1, 1
R-S
6.5
1.o
1
C de =-
but
2'
L
feel
dJ1. '" O.
Spaces
We will now discuss one of the most exciting'and profound applications of the Lebesgue integral: L-p spaces, written LP. We have dealt with various collections of functions - Cauchy integrable functions, absolutely continuous functions, Riemann integrable functions, and many more. In many cases we can define a distance between functions that satisfies all the usual requirements imposed when dealing with real numbers, and which satisfies our intuitive requirements for how distance operates. That is: 1. The distance from any object to itself should be zero. 2. The distance between different objects should be positive. 3. The distance between object 1 and object 2 should be the same as the distance between object 2 and object 1. 4. Finally, just as real numbers satisfy a triangle inequality (that is, la - ella - bl + Ib - cD, the distance between any two objects should not exceed the sum of the distances between each of these objects and a third object.
6.5.1
Metric Space
More fonnally, we have the idea of a metric space, defined as follows. Let X be a nonempty set. A metric p on X is a real-valued function with domain X x X that satisfies the following criteria: 1. p(x,x)
= 0, X EX.
2. p(x, y) :> 0, x, y E X and x =f:. y.
3. p(x,y):;:: p(y,x), x.y 4. p(x, y) ::: p(x, z)
E
X.
+ p(z, y)
(triangle inequality).
137
The Lebesgue Integral
The set X together with the metric p (distance function) is called a metric space, denoted by (X, p). Let's look at some examples of metric spaces.
Example 6.5.1. For real numbers
X,
y, p(x, y)
= Ix - yl : (R, Ix -
yl).
Example 6.5.2. For continuous functions on [0, 1], C [0, 1], p(x, y) =
max
0:::;t'::;1
Ix(t) - y(t)1 : (C[O. 1],
max Ix - YI) .
0'::;/:::;1
In this second iDstance, the reader may show we have a metric space. For example, invoking the triangle inequality, we have
Ix(t) - y(t)1 < Ix(t) - z(t)1 f Iz(t) - y(t)1 ,
for 0 < t < 1,
because x(t), yet), and z(t) are real numbers. Certainly
Ix(t) - y(t)1 < max Iz(t) - x(t)1 09:::;1
+ O:::;t'::;l ma."'{ Iz(t) -
y(t)I,
and the result follows. Or does it? Recall that x, y are continuous functions, x-y is a continuous function, I~ - Y I is a continuous function, and continuous functions assume a maximum on compact sets. Tighten up the argument.
Example 6.5.3. Consider C [0, 1], the same collection as in the preceding 1 example, but with a different metric: p(x. y) = C J0 1."C(t) - y(t)1 dt. Because we are dealing with the Cauchy integral, it is important that Ix (t) - y(t)1 be a continuous function on [0, 1]. It is. 1 Now, if x ¥- y, is C J0 Ix(t) - y(t)1 dt > O? Let's see: Ix (to) - y(to}] > 0, so we have a neighborhood about to .... And the triangle inequality entails
c[
Ix(t) - y(t)1 dt < C [
Ix(t) - z(t)1 dt
+ C[
Iz(t) - y(t)1 dt.
But Ix(t) - y(t)1 < Ix(t) - z(t)J + Iz(t) - y(t)1 holds for all t in [0, 1]. We can use the monotonicity property of integrals. In this example it was important that, given objects in our collection, the difference of such objects belongs to the collection. Again, we write
( C[Q.!J. [
Ix(t) - y(I)1
dl) .
138
A Garden of Integrals
Example 6.5.4. Consider R[O, 1], the collection of Riemann integrable functions on [0,1], with p(x,y) == RJo1Ix(t) - y(t)J dt. Here, we have that x and y are bounded and continuous almost everywhere, x - y is bounded and continuous almost everywhere, Ix - yl is bounded and continuous almost everywhere. So R Ix(t) - y(t)! dt makes sense. If x =I y, is RJ; Ix(t) - y(t)1 dt > O? Suppose x == 0 in [0,1], with 1 Y = 1, t = ~, 0 otherwise. Then x =fi y, but RJo Ix(t) - y(t)1 dt = O. We have a problem, unless we agree to identify functions that are equal almost everywhere. This we will do. The "points" in our space R[Ot 1] are actually classes of functions, the functions in a particular class differing from each other on a set of measure zero. However, we will follow tradition and talk about a function x from the meh'ic space, in this case R[O, 1], when technically speaking, we should talk about a representation.
f;
l
We have (R[O, I}. RJo Ix(t) - yet) I dt).
Example 6.5.5. Consider Ll [0, I}, the collection of Lebesgue measurable functions x on [0, 1.] with L fol lxl dJ.L < 00. The reader can show we have a metric. The metric space (Ll [0, 1], L follx - yl dJL) is also called L1.
Example 6.5.6. Consider L2[O, 1], the collection of Lebesgue measurable functions x on [0, 1} with LJ; Ixl2 dp. < 00. In this instance, we have a metric space with p(x, y)
=
L
f
Ix -
yl2 dJ.L,
Is L
lliX -
yl2 dJ.L::
L
lliX _z12
dJ.L
+
L
II Iz -
yl2 dJ.L
true? This inequality would follow from Ix - yl2 ::s Ix - Zl2 + Iz _ Y12. But first, is x - y a member of L2[O, 1]? That is, given XI y E L2[O, 1] does it follow that L Ix + Yl2 dJL < oo? We calculate
J;
Ix + yl2
<
(lxl + ly\)2
< (2max{lxl, Iyl}?
~ 4max{1x12 ,lyI2} < 4(1xl2 + lyI2).
139
The Lebesgue Integral 1
So, Ix + yl2 is measurable and LJo Ix + yl2 dJL < 00. Thus it makes sense to talk about x - y, x - z, z - y in this space of functions L2[0, 1]. Still, the triangle inequality remains. However,
Ix - yl2
= I."t - Z + Z -
Ix - z + z - yl (Ix - z\ + 12 = Ix - yllx - zl + Ix - ylly - zl· yl2 <
yl)
Thus, we have
Ll' I.~ - yl'
dp. <
JL l' Ix - yllx - %1 +
L{
dp.
Ix - ylly - zl
dp..
Just because x - y, x - z, and y - z are members of L2[0. 1], does this imply that their products are members of L 1 [0, 1]? Why the square root? In an admittedly roundabout fashion, we have been led to some famous inequalities and linear spaces. Here is the most important requirement for a linear space: Having ."t Y as members guarantees that X + y and ax are members (for a a scala1-). As we shall see, there are some inequalities that will be helpful in establishing the triangle inequality for metric spaces. J
6.5.2
Famous Inequalities
We begin with Young's Inequality; the work of William Henry Young (1864-1942).
Theorem 6.5.1 (Young, 1912). For nonnegative nZimbers a. b, aP ab < P
bq
1 1 + - = 1. q P q Proof Fix b > 0, and maximize the function f(a) = ab - aP I p.
+ -,
for 1 < p <
00, -
0
Next we have the HOlder-Riesz Inequality, which we owe to Otto Holder (1859-1937) and Frederic Riesz (1880-1956).
Theorem 6.5.2 (H~lder-Riesz, 1889, 1910). Let p > 1 and q satisfy (II p) + (l/q) = 1. Jfx E LP[O, 1] and y E Lq[O, I], the1l xy E L1[O.i] and
1
L 11XYI dp. <
(I ) 1/ (1 )I/q L 11xIP dp. L 1 1Ylq dp. P
140
A Garden of Integrals
Proof. If x or y
=1=
0 almost everywhere, We have
Conclude that xy E LI [0,1]; integrate; and so on.
0
The Minkowski-Riesz Inequality, to which Hermann Minkowski (18641909) contributed, is more complex.
.
Theorem 6.5.3 (Minkowski-Riesz, 1896, 1910). Let p > 1. LP[O.I], then we have
(L {IX + yiP d/L) I/p ::: (L /.IlxiP d/L) I/p + (L {
lylP
If x, Y
E
d/L) I/p •
Proof. If p = 1, integrate the triangle inequality. If Ix + yiP = 0 almost ev~ erywhere, there is no problem. We will assume p > 1 and fol Ix + yiP dp.. =1= O. Then
LJ.1IX + yiP
dJ,L =
o·
LJ.IIX + YIP-l Ix + y\
dp..
0
::: L
f
Ix + ylp-I Ixl
d/L + L {
I.~ + yIP-I Iyl d/L ..
Let q satisfy (1/ p) + (l/q) = 1, that is, (p - l)q = p, and apply the HOlder-Riesz Inequality (Theorem 6.5.2) to the integrals on the right-hand side. Thus,
LJ.
1
Ix + yIP-I Ixl
dx:::
(
LJ.
1
Ix + yl(P-I)/. d/L)
t(LJ. Ixl Pd/L)* , 1
and
LJ. 1Ix + ylP-Ilyl dx < (LJ. 1Ix + yl(P-Il/. d/L)t (1 LJ. lylP dp.)* . We have
141
The lebesgue Integral That is,
Exercise 6.5.1. Revisit the examples in Section 6.5.1, and show that the following are metric spaces.
a.
(C [0, 1], maxo~t~l
Ix (t) - yet) I).
b.
(c [0, 1]. cJ; Ix(t) -
y(t)1 dt).
c. (R[O,l],RJo1Ix(t) - y(t)1 dt).
d.
(Ll [0, 1], L Jot Ix (t) -
y(t)1 dt).
e.
(L 2[0. 1]. ~L 1.1 Ix -
YI2
dJ.!.).
Among othtr reasons, if x =
It I = -y,
z =
JL Jo
I
Ix - Yl2 djJ.? 0, then LJ; Ix - yl2 djJ. =
Why the square root for L2[o, 1] : p(x, y) =
~ > L Jo Ixl2 djJ. + L Jo lyl2 djJ.. 1
6.5.3
I
Completeness
We have some function spaces with metrics. Are these spaces complete? That is, given a Cauchy sequence of elements of these spaces, do we have convergence to an element belonging to the space?
Example 6.5.7. The metric space (C[O, 1], maxo
0, there is N so that
For each t, Ix" (t) - Xm (t) I < E; in other words, the sequence of real numbers {xn(t)} is a Cauchy sequence. That is, limx,z(t) = x(t). We have a function x on [0, 1]. Is x a member ofC[O, I]? (Is x continuous on [0, I]?)
142
A Garden of Integrals
Show the convergence is uniform: that IXn(t) - x(t)1 < E for all t in [0, 1] when n > N. The function x is continuous by Weierstrass's Theorem. Convergence with this metric is uniform convergence. Note that if Xn = tnfor 0 < t < 1, then _\ 0 · 1lIDX n 1
0 < t < 1, t
= 1.
Is {xn} a Cauchy sequence in (C[O, 1], maxo~t~l Ix(t) - y(t)!)?If so, then by what we have just shown, we have a continuous function x on [0, 1] so that max Ixn{t) - x(t)1 ~ 0; O!:t~l
that is, x(l) = 1, and x = 0, for 0 < t < 1. We have a contradiction: {xn} is not a Cauchy sequence in C[O, 1] with the metric maxo~t~l Ix - y[.
Example 6.5.8. The metric space (C[O, 1], C fol [x(t) - yet)] dt) is not complete. First, let
xn(t)
=
0
Show that C follXn - xm[ dt < 2/11, for m > n. The sequence {xn} is a Cauchy sequence that does not converge to an element of C[O, 11. Claim: No matter what x E C [0,1] is chosen, we do not have convergence.
x
Figure 1.
Ix(t) -
Xn (t)1
143
The Lebesgue Integral
Hint: Assume such an x. Because x is continuous on [0, 1], x is bounded, Ix I < B, B > 1. Let E > 0 be given. See Figure 1.
IXn - (l/.Jt) I dJ.L = LJo1/
1
Note that LJo
n2
n dJ.L = l/n
(R[O, 1], RJ; IX(I) -
Example 6.5.9. The metric space
~ O.
y(t)1
dt)
is not
complete. The reader can complete the details. Hint: Look at Example 6.5.8. It was critical that x is bounded. We conclude these explorations on a more positive note.
Example 6.5.10. The metric spaces
(Li[O, II,L fiX -yl dP.) are complete -
(L2[0' 11,
and
L
fiX -yI2dP.)
the celebrated Riesz-Fischer Theorem.
The reader should check Example 6.5.8.
6.5.4
The Riesz Completeness Theorem
Frederic Riesz gave us a completeness theorem. Theorem 6.5.4 (Riesz, 1907). For p > I, the metric space
is complete. Proof. We prove the result for p = 2. Suppose {xn} is a Cauchy sequence
in the metric space
(L2[0, 1], JL J; Ix - yl' dp. ) and
E
> 0. We have an
N so that for an
12, 112
> N.
Choose 111
so that L
11k
>
Ilk-l
i
1
o
IXn
1
2
-XIII
so that L
i
1
o
I
IXII
dj.L
< -, 2
2
-XII!..
I
1
djJ. < k'
2
1Z
>
Ilk; •.•
144
A Garden af Integrals
In particular, 1
1
2
L /.0 IXnl -xn11 dp.. < 1
L
/.
-xn2 1 dp.. < 2 2 '
1
L
1
2
I Xn 3
0
2'
2
0 !Xnk_l -xnkl /.
1 dp.. < 2k '
We are interested in the subsequence {xn~J. By the HOlder-Riesz Inequality,
L
111
:C.k+l
So Lk=l L
1
-x•• dp. <
Jo1 \Xnk+l -
L
X nk
11
(X. k+ l -xnkl2dp.°VL
/.112 dP. < ;kO
I dp.. < 1, and the series converges.
By the Lebesgue Monotone Convergence Theorem, {L£= 11 X nk +1 -
and
L IX nk +1 -
X nk
I},
I
x nk converges almost everywhere. So J
x n1
+ L (X nk + 1 -
x nk ) =
XnJ
k=l
converges almost everywhere. We have x (t) = lim xnk (t) when the limit exists and 0 otherwise. Is x a member of L2[Ot 1] : LJ; X2 dp.. < oo? Apply Fatou's Lemma. Since
li~ (x nj (t) - xn/c (t))2 = (x(t) - x nk (t))2 J
almost everywhere and 1
L we have
/ .o
(Xlil -
2 1 xn/c) dp.. < k' 2
145
The Lebesgue Integral
So x - x llk is a member of L2[O, 1]. Since XIlIc is also, we may conclude that x is a member of L2[O, l.]. We have succeeded, therefore, in showing that every Cauchy sequence in the metric space ( L 2[0. 1). JL Ix - y 12 djJ. ) contains a subsequence
J;
that converges pointwise, almost everywhere, to a member of L2[O, 1]. It remains to show that the original sequence converges to x in the L2[O, 1] metric. By the Minkowski Inequality we have
L
10
1
1
Ix" - xl2
djJ. <
L
10 Ix. -
X. A
12
djJ.
1
+
L
10 Ix". - xl 2 djJ..
D
Example 6.5.11. Consider the functions Xl, X2, ... illustrated in Figure 2. The sequence {xn} converges at no point of [0,1]. For p > I, with
Ji
fi,jj
2k
< -
1l
< 2k + 1 ,•
1-----
1---
Figure 2. Function sequence for Example 6.5 11
146
A Garden of Integrals
Example 6.5.12. Recall the Cantor set of measure t (Section 3.10). Define a sequence of characteristic functions (Xn) as follows;
+ X[7/12,I].
Xl
=
X2
= X[O,13/72] + X[17/72,S/12] + X[7/12.SS/72] + X[S9/72,I]
X[OrS/I2]
xn={:
on the 2n closed intervals (Fn, 1 I Fn ,2 I each of length 1/2[(1/2n) + (l/3 n )], otherwise.
••••
Fn ,2" ),
We have
R
f
R (I XlI(t)dt = 2n . ~ 10 2
[Xm(t) - Xn(t)) dt =
~
(~+..!..) = ~ +! (~)n n n 2
3
[G)" -GfJ,
2
2
3
•
for m > n, and
limXn(t) = Xe(t), which is the characteristic function on the Cantor set of measure
i.
In this example, we observe first that {Xn} is a Cauchy sequence in the metric space ( L 2[0,
1], JL
J; Ix-y f dfJ.), such that
(2)n ' {I 1 (2)n L 10 (Xn - xe)2 dJ.L <"2 "3 {I
2
L 10 IXn - Xml dp.
1
<"2 "3
for m >
11,
and
Next, observe that {Xn} is a Cauchy sequence in (R[O, I}. Rf~
{I
R 10
1
(2)n
IXn - Xml dt <"2 '3 ·
Ix - y\ dt):
for In > 12, and
limXn(t) -+ Xc (t). where Xc is not Riemann integrable (because it is discontinuous on a set of positive measure). However,
L
10{I IXII ~ xc I dp. = '12 (2)n 3' . l
So Xn converges to X in the metric space (L[O, I]. L fo
Ix - yl dt).
147
The Lebesgue Integral
Exercise 6.5.2. a. Show the special case of the HOlder-Riesz inequality. p = 1, known as Schwarz's Inequality:
L
10' IxYI dp. ~ l
Hint: L Jo (a Ixl
(L
10' x
2
L
10' x2 d~,.JL 10' y2 dfJ.,
for x, y
E L 2[0, 1].
+ lyl)2 dJi.. ::: 0; also,
dlL )
",2
+ 2 (L lo'IXYI
dfJ.) '"
+ L 10' y2 dfJ. > O.
We have a parabola that opens "up" and this has a minimum.
(L 2[0, 1], JL J; Ix - Y12 dIL )' limits
b. Show that in the metric space are unique in the sense that
imply x = y almost everywhere. Hint: Minkowski Inequality.
c.
JL
0,
Show that if J~ IX n - xl dJ.L -+- then the sequence {xn} is a Cauchy sequence in L2[O, 1]. Hint: Minkowski: Xn - X m , Xn - X, 2
Xm- X • l
d. Show that ifL Jo Ix" - XI2 df.L -+ 0, then O. Hint: By the Minkowski Inequality,
L
10' x~ dlL <
L
10' (x. -
X)2
IL Jot x;dJi.. - L Jo
1
dlL +
L
X
2
dJi..l-+-
10' x2 dIL·
Also,
and
L
10\'" -
X)2
dfJ. <
L
10'
<
L
10' {x" -
.",2
dJL -
L
10' x~ dlL
x)2 dp..
148
A Garden of Integrals
e. Show L
II
(x. - x) djJ. < L
II Ix. - xl
dJ.!
L10 Ix. - xl2 JL 10 1
dJ.!
<
1
J2 dJ.!.
Hint: L2 convergence implies Ll convergence on [0. 1]; or
limL
10
1
x. dJ.!
=
Cfol
X
djJ..
L 2 [-n, n] and Fourier Series Write an infinite series L Ck with the sole requirement that L c~ < 00. Now relabel it as (ao/.Ji) + al + hI + a2 + b2 +.,. + ak + bk +"'J with (aa/2) + Ll (a% + b~) < 00. This looks suspiciously like a Fourier 6.6
series (Theorem 2.7.1), doesn't it? We will construct a function x in L2[_1r. rr] so that its Fourier coefficients are precisely ao/2. al, b 1 • a2. b 2 • ' •• and the partial sums converge to x in the metric space
This result is one of the most surprising and astounding in mathematics, but that is the nature of mathematics - beautiful, intriguing:
Now, consider the sequence of L2[-1t t rr] functions
How can we show this is a Cauchy sequence in
149
The Lebesgue Integral
the given metric space? First, consider that for
1: {
[a;
In
> n,
+ ~(ak coskt + bk Sinkl)] - [ ~ + ~(ak coskt +bk Sinkl>] =
l1r: -1r:
(t
ak
r
dt
+ bk sinkt) 2 dt
coskt
n+l
m
=
1f
L (at + bZ) . n+l
Since (afi/2)
+ Lr(a~ + b~)
converges by assumption,
m
L (a~ +bn -+ 0 n+l
By the ruesz-Fischer Theorem (Example 6.5.10), we have a complete metric space. Thus we have a function x belonging to L2[_1f,1f] so that
L
in
1r: [ a; + ~(ak n cos kl + bk sinkl) -
x
]2 dt -70
as
11
That is, by Exercise 6.5.2,
or
Furthermore,
1 l1r: xCt) cos kt dt -L
= ak,
-1r: 1 l1r: x(t)sinktdt = bk, -L 1f -1r:
7r
for k = 0,1,2, ... , and
~
00.
150
A Garden of Integrals
For example,
L f~ X(I) sinml dt =L L:[X(t)- (a;
+ ~akcoskt+bkSinkt)}inmldt
+L f~ [~ + ~(akCaSkl +bkSinkl)}inmldl
f~ [X(I) -
= L
(a;
+ ~ ak cos kt + bk sinkl) ]
sinml dl
O n < m, + I :rcbm 1Z > m. As 11
L
~ 00,
L:
the first term converges to 0:
X(I) - (;:
L
<
+ ~(ak caskl + bk Sinkt)) sinml
i: L
[X(t) - a;
dl
+ ~(ak coskl + bk Sinkl)r dt
11r-n: sin mt dt. 2
Thus we have
L f~ X (I) sinml dt = 1rbm E)Cercise 6.6.1. Given x(t) :
and
bm
=:
1 -L 1t'
1t'
1n: x(t) sinmt dt. -n:
-It I ,-:rc ~ t < 1t', consider the Fourier
series of x, ak
= -1 L 1t'
bk = -1 L 1C
R.
Show t llat
11r -1r
(1t'
-ltD cos kt dt,
for k = 0, 1, ... , and
1n: (:rc -It I) sinkt dt, -n:
. f . " F' t hIS ouner senes 0 x IS
for Ie = 1, 2, .... 00
1t' -
2
4"
+-
If
L..,
n=l
COS(211 -
(
1)t
)2'
2n - 1
151
The Lebesgue Integral Jr2
b. Conclude that -
2
16
+ 2" L rr 00
1
1+
1 34
1
1 1++ 24 34
1
(211 -
2Jr 2
)4 = - . That is, 1 3
1
+ 54 + ... =
rr 4 96
and
rr 4 +-+"'=-. 54 90
1
Exercise 6.6.2. Suppose x(t) = {
~l o-1C< <x X<
a. Show that the Fourier series of x is evaluate for t = 1 and t = 2.
~" 1 rr L.-J 211 -
1
sin(2n - 1)t t and
b. Conclude that
Exercise 6.6.3. Find the Fourier series of
()_I
xt -
(rr - 1)/2
o
extended periodically. Evaluate at t
0 < t < 2Jr, . otherwIse.
= 1 and t = 2.
Beautiful things happen in L2[-rr, rrJ. In 1966 Lennart Carlson showed that for functions in L2[a, b] the Fourier series converges pointlVise, almost everywhere, to the original function.
6.7
Lebesgue Measure in the Plane and Fubini's Theorem
As we constructed Lebesgue measure on R from intervals, we may also conshuct Lebesgue measure on R2 from rectangles, on R3 from ... Such a development may be found many places, but not here. We conclude by stating a result that will be useful later, the theorem of Fubini regarding dOllble integrals.
152
A Garden of Integrals
Theorem 6.7.1 (Fubini).
L[L
If f
is Lebesgue measurable on R2 and
If(X. y)1 dx ] dy
or
.
fR [L If(x. y)1 dY] dx
exists, then f is Lebesgue integrable on R'l and
Exercise 6.7.1. Given f(x, y) = (x 2
f (f 6.8
f(X.Y)dY) dx =
~.
-
y2)/(X 2
f (f
+ y'l), show f(x.Y)dX) dy = -
~.
Summary
Two Fundamental Theorems of Calculus for the Lebesque Integral
If F is absolutely continuous On [a, b], then 1. F' is Lebesgue integrable on [a, b], and 2. L J~'( F'(t) dt = F(x) - F(a), with a
If f is Lebesgue integrable on [a,
bl and
x < b.
F(x)
=LJ: f(t) dt, then
1. F is absolutely continuous on [a, b], and 2. F' =
f
almost everywhere on [a, b].
I Cauchy I Riemann Lebesque
Figure 3. Integrable functions: Cauchy C Riemann
C Lebesgue
The lebesgue Integral
6.9
153
References
1 Apostol, Tom. klathemalical Analysis. Reading, Mass.: Addison-Wesley, 1974. 2. Boas, Ralph. A Primer of Real Functions. Washington: Mathematical Association of America, 1996. 3. Burk, Frank. Lebesgue kfeOSlire and integration: An Introduction New York: Wiley-Interscience, 1998. 4. Goldberg, Richard. Methods of Real Analysis New York: Wiley, 1964. 5. Gordon, Russell A. The Integrals ofLebesgue, Denjoy. Perron. and Hemlock Providence, R.I.: American Mathematical Society, 1994. 6. Kaunan, Rangachary, and Carole King Krueger. AdvancedAllalysis ofthe Real Line. New York: Springer, 1996. 7 Munroe, M. k[easZlre and Integration. Reading, Mass.: Addison-Wesley, 1959. 8. Royden, Halsey. Real Analysis. New York: Macmillan, 1968. 9. Shilov, G., and B.L. Gurevich. Integral, Measure and Derivative: A Unified Approach New York: Dover, 1977. 10. Spiegel, Murray. Real Variables. New York: Schaum's Outline Series 11, 1969. 11. Stromberg, Karl. An Introduction to Classical Real Analysis Belmont, Calif.: Wadsworth, 1981 12. Titchmarsh, E. The TheOlY of FU1!ctions. Oxford University Press, 1939
CHAPTER
7
The
lebesgue-Stie~tjes ~rotegral
Even now there is a very wavering grasp of the true position of mathematics as an element in the history of thought. I will not go so far as to say that to const'1,ct a history of thought without profound study of the mathematical ideas of successive epochs is like omitting Hamlet from the play which is named after him. That would be claiming too much. But it is certainly analogous to cutting Ollt the part of Ophelia. This simile is singularly e.;y;act. For Ophelia is quite essential to the play, she is Ve7Y charming - and a little mad. Let us grant that the pursuit of mathematics is a divine madness of the human. spirit, a refuge fi1;m the goading urgency of contingent happenings.
-
Alfred North Whitehead
The Lebesgue measure weights intervals according to their length: JL ((a. b]) = b - a. We are looldng for different weightings of intervals, trying to find measures different from Lebesgue measure. Of course the general properties of a measure - nonnegative and countable additivity for disjoint sequences of measurable sets - must be retained. We will discuss three of the most common approaches, which differ from each other essentially in their starting points.
7.1
L..S Measures and Monotone Increasing Functions
In Chapter 4 we explored the particularly fruitful, and most fundamental, approach of Thomas Stieltjes, who modelled mass distributions with monotone increasing right-continuous functions. His approach involved the variation of ¢ on an interval (a, b]: ¢(b) - ¢(a). For Lebesgue-Stieltjes measure, we may also begin with the variation of a monotone increasing right-continuous F on the half-open interval (a, b]
155
156
A Garden of Integrals
as F(b) - F(a), in other words the entity ("total mass" to b minus "total mass" to a), but we now proceed to develop a measure. Whereas Lebesgue measure weights the interval (a, b) by b-a, LebesgueStieltjes measure will weight (a, b] by F(b) - F(a). We have a weight function t": t"«a, bJ) FCb) - F(a) and L(q,) O. Since familiarity with the weight function 't' will be helpful, let's examine its properties.
=
7.1.1
Properti~s
=
of the Weight Function
As an exercise, demonstrate each of these properties. Some suggestions are provided.
Property 1. For a < e < b,
t"(a, b])
= F(b) -
F(a) = F(b) - F(e)
+ F(e) -
F(a)
= L(e, b]) + "t«a, en. Property 2. If (ak, bk], 1 ~ k < n, are disjoint and Ui(ak, bk] C (a, b], then L~L(ak,bkn :::: t"«a,b]). Hint: Relabel so a F(b n) - F(al) n
> I)F(bk) - F(ak)] 1
n
= L
L«ak,bk]).
1
Property 3. If (a, b] c Ui(ak. bk], then t"«a, bl) < L~ 't'(Cab bkl). Hint: Assume all intervals are needed and at < a2 < ... =:: an. Then al =:: a < bI, an < b < bn• with ak+1 < bk < bk+l, and
t"(a, b]) = F(b) - F(a) n-l < L [F(a,,+l) - F(ak)]
+ FCbn ) -
F(an)
1
n
< LT«ak,bk]). 1
Propel1y 4. If (ak, b,,], 1 :::: k < then t"«a, b]) = L~ 't'«ak. bk1).
11,
are disjoint and (a, b] = U~(ak, bk],
157
The Lebesgue-Stieltjes lntegral
Property 5. If (a. b] C Uf(ak, bk], then F(b) - F(a) ~ L~ r(ak. bkl). Hint:
+ 8, b] C F(a + 8) <
a. [a
Uf(ak, bk] C Uf(ak, b;), where hk < b;, and
F(a)
+ €.
b. (a, b] = (a, a + 81 U [a
+ 8, b] C (a, a + 81U (U~l(aki'bk) C (a,a +8]) U (U~l(aki'bki])' (Heine-:-Borel).
c. Using Property 3, N
r(a, bJ) < F(a
+ 8) -
F(a)
+ L [F(b kj ) - F(ak)] 1
N
<
€
+L
[F(bk;) - F(ak-i) +
1
':i] -
co
~
L r «ab bkD + 2€. 1
d. F(b) - F(a) -2€ < L~ r(ab bkJ). By the arbitrary nature of E', the conclysion follows.
Property 6. If (a, bl = Uf(ak, bkl, with mutually disjoint intervals, then r(a,b]) = L~r(ak.bkl). Hint: Since (a,b] C U(ak.bk], by Property 5, r(a,b]) < L r((ak,bkJ). For the other direction, U~ (ab bkl c (a, b]. By Property 3, n
L r(ab bk]) ~ r(a, bJ). 1
independent of n. Thus co
L r(Cab bkJ) < r(a. bn· 1
So r is countab1y additive on ha1f-open, half-closed subintervals.
7.1.2 The l-S Outer Measure Knowing how r operates, as the analog of length f.. in Lebesgue measure, we define the outer measure.
A Garden of Integrals
158
Definition 7.1.1 (Lebesgue-Stieltjes Outer Measure). The Lebesgue-Stieltjes outer measure J1.j.., determined by a monotone increasing right-continuous function F ~ for any set A of real numbers, is defined as
J.LF
~ r(ak. bkJ) I A C Uf(ak. bk1 } •
= F(bk) -
F(ak) and r(rp)
= o.
Exercise 7.1.1. Show that JL j.. satisfies the generaJ requirements for an outer measure. a. J1.p(rp) = O.
b. If A
c.
c B, then JL F(A)
< J1. ~ (B) (monotonicity).
wI-- (UAk)
< L J1.:F(Ak) (subadditivity). Hint: Review Section 5.2.1, the Lebesgue Wish List.
7.2 Caratheodory's Measurability Criterion As with Lebesgue measure, from Lebesgue-Stieltjes outer measure we set a measurability requirement. This again is the work of Constantin Caratheodory (1873-1950).
Definition 7.2.1 (Caratheodory's Measurability Criterion). A set of real numbers E will be Lebesgue-Stieltjes measurable if
holds for every set of real numbers A. The next series of exercises demonstrates that intervals are J1.F measurable.
Exercise 7.2.1. a. Show that JLp(a, b]) = F(b) - F(a) = 't(a, b)). Hint: (a, b] C (a, b), so JLj..((a. < '(((a, = F(b) - F(a). For the other direction, use the properties in Section 7.1.1.
bn
bn
b. Show that intervals of the form (-00, b) satisfy Caratheodory's Criterion, J1.}(A) = J1.F (A n (-00, bD + J1.} (A n (b. 00)). Hint: Section 5.3.1.
159
The lebesgue-5tieltjes Integral c. Show that intervals we jJ.F measurable.
We are mimicking our development of Lebesgue measure in Chapter 5. Along those lines, jJ. F measurability of intervals irpplies that Borel sets are likewise Lebesgue-Stieltjes measurable. fudeed, we have a measure space: (R, B, J1.F). A monotone increasing right-continuous F determines a measure (Lebesgue-Stieltjes) J1.F on a sigma algebra of sets that contains the Borel sets, the Borel sigma algebra B. Here are some examples of measuring with J1.F. In the first example, notice that if F is continuous everywhere, then "points" have LebesgueStieltjes measure zero.
Example 7.2.1.
>+ ,~])
(n (a -
JLF({a)) = JLF
= lim JLF ( ( a -
~, a + ~])
= F(a) - F(a-). Example 7.2.2. a. MF(-:;-oo,X]) = MF (U(x -k,x]) = limJ1.F(x -k,x])
= F(x) -lirnx-+-oo F(x) = F(x) - F(-oo).
= F(x-) -
b. jJ.F(-oo, x))
C. J1.F (x, 00))
= F(oo) -
d. J1.F ([XI 00))
=
F(-oo).
F(x).
F(oo) - F(x-).
e. J1.F(R) = F(oo) - F(-oo).
Example 7.2.3. a. J1.F([a,b1)
=
J1.F({a})
+ J1.F(a,b1)
= F(b) - F(a-).
b. J1.F([a, b)) = F(b-) - F(a-).
c. /1.F(a, b)) = F(b-) - F(a).
Exercise 7.2.2. a. Calculate jJ.F(-CO,OO)), jJ.F(-l,lJ), and MF({O}), given
F(x)
=
I
x
x+
1
x < 0, x >0.
160
A Garden of Integrals
h. Calculate J1.F ({rationals in (0, I)}), J1.F = ({irrationals in (0, I)}), and J1.F(C), where C is the Cantor set (Section 3.9), given OX < 0, F(x) = 2x 0 < x < I, { 2 x > 1.
c. Calculate J1.F ([0, 1]) and J.LF (C), given
F(x)
=
O x < 0, Cantor fu nction 0 < x ::::: I, { i x ~ 1.
d. Calculate J.LF (Ca, b]), given F{x)
7.3
={ ~
x < 0, x ~O.
Avoiding Complacency
For many years I took for granted Stieltjes' contributions in the setting of Lebesgue-Stieltjes measure. Recently I read a beautiful book, Measure, Topology, and Fractal Geometry, where I carne across the following example (Edgar, 1990, p. 136). I have modified it slightly for our purposes. Suppose we weight the half-open interval (a, b] by r(a, bJ) = .J~b---a. We then have an outer measure J.L: that will determine a sigma algebra of measurable sets, and thus we have a measure J1.T:. No problem there. But what is the measure of the interval [-1, O]? It turns out that this interval is not JLl: measurable. Certainly J1.; (0, 1 ~ 1, since (0, 1.] is a cover of itself. On the other hand, if (0, ,1] ~ U(ak' bk1, then
n
Thus 1 < L .jbk - ak. By the infimum property, 1 < J1.;(O, 1]). So J1.~(O, 1]) = 1. Show that J1.; ((-I, 01) = 1.
The Lebesgue-Stieltjes Integral
161
For the interval (-I, I], JL~(-1.1]) ~ if [-1, 0] is JL't measurable, then
.J2. By Caratheodory's Criterion,
for aU sets A of real numbers. In particular, if A
.J2 > JL; (-1, IJ) =
JL; (-1, 1]
= (-1, 1], then
n [-1,0]) + JL:((-I, 1) - [-1,0])
= JL;(-l, 0]) + fL;(O.I]) = 2. Of course in hindsight, we can see that we lacked finite additivity, that is, .Jb - a =F .Jb - a + .Jc - a, for a < c < b - so we might have been suspicious.
L.. S Measures and! Nonnegat~ve Lebesgue Integrable Functions Suppose f is a nonnegative Lebesgue integrable function on the reals. Thus, f is nonnegative, Lebesgue measurable, and JR f dJL < 00. Now define a function F on the reals by F(x) = J~oo f dfL. The function F is clearly 7.4
nondecreasijlg, bounded, and absolutely continuous (by Theorem 6.4.1). As before, we may construct a Lebesgue-Stieltjes measure JL f' and we have a sigma algebra that includes the Borel sets. Even nicer, the sigma algebra of JL f measurable sets contains the Lebesgue measurable sets. Let's look at some properties of this measure.
7.4.1
Properties of f.L f
Assuming that F(x) = J~oo f dJL and that f is nonnegative and Lebesgue integrable, we can show that Lebesgue measurable sets are fL f measurable sets. In fact, JL f (E) = L JE f d fL when E is a Lebesgue measurable subset of R.
Property 1. If fL(E) = 0, then E is fLf measurable and JLf(E) Let that
€
= O.
> 0 be given. Because F is absolutely continuous we have a 0 so
Because JL(E) = 0, we have a cover of E by disjoint half-open intervals (ab bk] so that JL (UfCak. bkJ) < o. But then, /..t (U~(ak. bk]) < ~ for all
162
A Garden of Integrals
n. Thus II
J.Lf
(u1(akt bk])
= L[F(bk) - F(ak)] <
E,
1
independent of n. We have jJ.j(E) :s jJ.j (ui(akt bk ]) < E. That is, J.L j (E) = O. But sets of Lebesgue-Stieltjes outer measure zero are LebesgueStieltjes measurable:
o ::: jJ. i
(A
n E)
J.Lj(A) :::: J.Lj(A
< J.Lj(E) = 0,
n Ee).
and
(monot~nicity)
Property 2. If the set E is Lebesgue measurable, then E is LebesgueStieltjes measurable (J1. f)· From Exercise 5.6.1, every Lebesgue measurable set (E) may be written ..... as the union of a Borel set (B) and a Lebesgue measurable set of measure zero (E - B). That is, E = B U (E - B). By Property 1, E - B is jJ. f measurable. Borel sets are jJ. f measurable (Exercise 7.2.1). Thus E is jJ. f measurable and
J.Lf(E)
= J.Lf(B) + jJ.f(E -
B)
= J.Lf(B).
The sigma algebra of J.L f-measurable sets contains M, the sigma algebra of Lebesgue measurable sets. We have a measure space: (R, M, jJ.f).
Property 3. With F continuous, J.Lf«a, b]) = jJ.f(a, b») =L
t
= {
= J.Lf([a, b») = J.Lf([a, b])
fdJ1.
f
djJ..
}(a,b]
E][ample 7.4.1. For any nonnegative Lebesgue integrable function f, define a nonnegative set function, jl, on the Lebesgue measurable sets E by fleE) = LIE f djJ.. Will {l be a measure? We will explore this question in stages.
a. If A c B, show P,(A) < P,(B). b. If {Ek} is a mutually disjoint sequence of Lebesgue measurable sets, show that {l(UEk) = E {leEk). Hint: Monotone convergence (Theorem 6.3.2) and countable additivity. We have a measure on Lebesgue measurable sets: J.L f.
The Lebesgue-Stieltjes Integral
163
c. Note that
il/«a,bj) = L
i:
f dp. -L
i~ f
d?!
= F(b) - F{a) = J.LF (at b]). d. Given that E is Lebesgue measurable, does ill (E) = J.L I CE)? Because ill and J.L I agree on the covering sets, (a, b], and because R can be written as a countable union of pairwise disjoint sets, R = U{n, n + I], it can be shown that we have uniqueness: ilICE) = J.Lj{E), E Lebesgue measurable.
Exercise 7.4.1. a. Calculate J.L/{R), J.LI([-l, 1]), J.Lj({-l,ll), J.Lj{{O}), and J.L({O}), given F(x) = L
b. Calculate J.LI((-l, 1)), J.Lj(-l, with Exercise 7.2.2, given
1]),
c. Calculate ~~ j
(C-1, 1])
F(x)
= L [~f dp..
Ixl> 1,
+2
F(x) = L
d. Given the Gaussian probability density function,
I(x) =
1 e-x ", - 2t, ~21!t
show limx~oo F(x) limx~-oo F{x).
+00.
D, given
-1 < x < 0, -2x + 2 0:5 x < 1, 2x
<x<
and compare with J.L( (-1, 1
0
={
-00
and J.LI({-I, 2)), and compare
OX < 0, (x) = 2 0:5 x ::: 1, { Ox> 1,
I
I(x)
I:"" f dp.,
= 1.
t > 0,
and
F(x)
=L
i~ f
L:
f dp.,
Approximate J.L/(C-l, 1)). Calculate
e. Calculate F given the Cauchy density function,
1 I(x) = -
1
1!l+x
2'
dp..
and
F(x)
= L i~ f
dp..
164
A Garden of Integrals
7.5 l .. S Measures and Random Variables Our final approach begins with a probability space 9, a sigma algebra of subsets of this space :E, and a probability measure P: (9, :E, Pl. Let X be a random variable on this space. That is, X is a real-valued function defined on 9, and the inverse image of the interval (-00, x], X-I (-00, is a member of the sigma algebra :E for every real number x. Borel sets are P measurable. We may calculate the measure of this inverse image: P ( X-I «-00, x]) ) ;" We defme an extended real-valued function F on the reals by
xn,
Fx(x)
= p(X-1«-oo,x])),
the probability distribution function F: R -+- [0, 1].
7.S.1
Properties of Fx
It will help to consider some characteristics of Fx.
Property 1. The function Fx is monotone increasing on R:
x
= n( -oo,X + ~],
X-I «-00, x]) = nX- 1 (( -OO,A
+ !]).
P is a measure, and so on. Property 3. The limit limx-+-oo Fx(x) = 0, since ¢ = n(-oo, -n]: Fx(-oo) = O. Property 4. The limitlimx-++ oo Fx(x) 1.
= 1 since R = U(-oo.n]; Fx(oo) =
Again, we have a monotone increasing right-continuous function Fx on the reals J and thus we may construct a Lebesgue-Stieltjes measure, /.LX, with the sigma algebra of Ji.x measurable sets containing the sigma algebra of Borel sets B. The probability space (R, B, Ji.x) is said to be induced by the random variable X. This concludes our treatment of generating Lebesgue-Stieltjes measures. Now our development of the Lebesgue-Stieltjes integral will proceed smoothly.
165
The Lebesgue-Stieltjes Integral
7.6 The Lebesgue-Stieltjes Integral To construct the Lebesgue-Stieltjes Integral, we will mimic the construction of the Lebesgue integral (Section 6.2.4), with J.LS denoting J.LF, J.L f' or /.Lx respectively. We proceed in four stages.
Step 1. A real-valued function g is said to be /.LS measurable if g-I (-00, x]) - or equivalently g-l (B), B a Borel set - belongs to the appropriate sigma algebra. Since the sigma algebras for J.LF, j1. f, or j1.x contain, respectively, B, M, or E, we will assume g-l (B) belongs to B, M, or B, in that order. Step 2. For nonnegative simple functions ¢, ¢ = Lk=l CkXEk' where E = UE/c, with Ek mutually disjoint, j1.S measurable sets, and Ck nonnegative real numbers, we defme the Lebesgue-Stieltjes integral of ¢ by 11
L ckj1.s(Ek)·
L-S /, ¢ dJ.Ls =
k=l
E
Step 3. If g is a nonnegative j1.S measurable function defined on a /.Ls measurable set E, we may approximate g by a monotonically increasing ¢k d j1.s} is sequence of simple functions {¢k}. Thus, the sequence {L-S monotone increasing and has a limit (extended reals, perhaps). We define
IE
L-S
Ie
= lim L-S
g dj1.s
Ie
¢k dj1.s.
Step 4. If g is a j1.S measurable function, then
Igl +g
Igl-g
2
2
g=---
¢:
and we have monotone increasing sequences of simple functions and ¢; convergin~ respectively to (lgi + g)/2 and (Igl - g)/2. We define L-S
Ie
g d fJ.s
= lim L-S
Ie rf>t
dfJ.s - IimL-S
Ie rf>;;
d fJ.S ,
provided both integrals on the right are finite.
Exel'cise 7.6.1. a. Calculate L-S Ie-I,l] g dj1.F, given F(x) =
I
-1 2
x < a, x >0,
and
g(x) = 2,
-1 < x ::: 1.
166
A Garden of Integrals
h. Calculate L-S f(-l,l] g dJ.LF given l
F(x)
={
x < 0, x ~ 0,
x-I 2x + 2
and
g(x) = 2,
-1:s x ::: 1.
c. Calculate L-S f(-I,I] g dJ.LF, given F(x) =
I
x < 0, x ~ 0,
-1 2
g(x) = 2 + x 2 •
and
-1::: x ::: 1.
d. Calculate L-S f(-I,2] x 2 dJ.L I, given
I
Ox < 0, (x) = 2 0 < x < I, { o x > 1.
e. Determine Fx(x) and calculate L-S f{-l,2l x dJ.Lx, given (0, E, P) = (R 1M, P), with M a sigma algebra of Lebesgue measurable sets, P a probability measure, P(E) == L fE X[O,I] dj.L, and
o X(lO) =
2w 2 - 2w
o 7.7
CJJ
< 0.
0 < CJJ < 1/2, 1/2 < w ::: 1, 1 < w.
A Fundamental Theorem for l-S Integrals
We conclude this chapter with a Fundamental Theorem of Calculus for Lebesgue-Stieltjes integrals. Theorem 7.7.1 (FTC for Lebesgue-Stieltjes Integrals). If g is a Lebesgue measurable/unction on R. I is a nonnegative Lebesgue integrablefunctio12 on R, and F(x) = L f:'oo f dJ.L, then:
1. F is bounded, monotone increasing. absolutely continuolls, and differentiable almost everywhere, and F' = f almost everywhere. 2.
we have a Lebesgue-Stieitjes measure j.L I so that, lor any Lebesgue measurable set E j.L I (E) = L f E I d J.L. and j.L I is absolutely conI
tinuous with respect to Lebesgue measure. 3. L-S
L
g dJ.LI = L
L
gf dJ.L = L
L
gFI dJ.L.
167
The lebesgue-Stieltjes Integral
Proof The first two parts of the conclusion have already been discussed among the properties of J.L f (Section 704.1). As for the last conclusion, we will give only a sketch. Step 1. Consider g = XE for E a Lebesgue measurable set. We have J
L-S
L
XE dJ.Lf
L = L
= L-S
dILf
= J.Lf(E) = Lief dIL
L
XEI dJ.L
=L
L
XEF' dp..
Step 2. For the simple function tP, with Ck > 0, we have tP No problem; linearity of the integral.
=
L~ CkXE/c.
Step 3. For the nonnegative simple function tPk, with 0 < tPk < tPk+b we have g = limtPk. Ivlonotone Convergence Theorem ... and so on. 0
This concludes our treatment of the Lebesgue-Stieltjes integral.
7.8
Reference
1. Edgar, (!Jerald. kleasure, Topology a'ld Fractal Geomehy. New Yorlc: SpringerVerlag. 1990.
CHAPTER
8
He who knows not mathematics and the results of recent scientific investigation dies without knowing truth. - K. H. Schellbach
In this chapter we present a beautiful extension of the Lebesgue integral obtained by an apparently slight modification of the Riemann integration process. Recall that in Section 3.12 we saw functions with a bounded derivative whose derivative was not Riemann integrable. These examples prompted Lebesgue to develop an integration process by which differentiable functions with bounded derivatives could be reconstructed from their derivatives: L
f'
F' dp.
= F(x) -
F(a).
This was a Fundamental Theorem of Calculus for the Lebesgue integral (Theorem 6.4.2). The next step would be to try to remove the "bounded" requirement on the derivative. We want an integration process in which all Lebesgue integrable functions will still be integrable and where differentiability of F guarantees J~'\: F'(t) dt = F(x) - F(a). Denjoy (in 1912) and Perron (in 1914) successfully developed such extensions: see Gordon's book. The Integrals of Lebesgue. Denjoy. Perron. and Henstock (1994). In 1957, Jaroslav Kurzweil utilized a generalized version of the Riemann integral while studying differential equations. Independently, Ralph Henstock (1961) d.iscovered and made a comprehensive study of this generalized Riemann integral. which we will call the Henstoclc-KlIrzlVeil integral or H-K integral. In Gordon's wonderful book all these integrals are fully developed and shown to be equivalent. We will use the constructive approach due to Kurzweil and Henstock because of its relative simplicity.
169
170
8.1
A Garden of Integrals
The Generalized Riemann Integral
Recall the Riemann integration process for a bounded function interval [a, b]:
I
on the
1. Divide [a, b] into a finite number of contiguous intervals.
2. Select a tag Ck in each subinterval [Xk-l, Xk] at which to evaluate I. 3. Form the collection of point intervals consisting of (Cl' [Xo, Xl)), (C2' [Xl. x2D,
... , (c n , [Xn-I, xnD·
4. Calculate the associated Riemann sum,
Lk=l I(Ck)(Xk -
Xk-l).
If we find that these Riemann sums - these numbers - are close to a number A for all collections of point intervals with subintervals of a uniformly small length (Xk - Xk-l) < 8, with 8 constant, then we declare I to be Riemann integrable on [a, b] and write RJ: I(x) dx = A. In the modified Riemann integration process that we are about to describe, the local behavior of f plays a prominent role. Therefore, instead of dividing [a, b] into a finite number of contiguous intervals ofunifonn length and then selecting the tag C in each subinterval at which to evaluate I, we will first examine the points where f is not well behaved. For instance, we will look for jumps or rapid oscillation. Using that information, we will divide [a, b] into contiguous subintervals of variable length. In this way, we will be able to control the erratic behavior of I about a tag c by controlling the size (the length) of the associated subinterval [Il, v], where 1.t < C < v. Generally, if I does not change much about the point c, the length of the associated subinterval lu, v] may be large. But if I behaves erratically (jumps, oscillates, etc.) about c, then the length of the associated subinterval [u, v1needs to be small. This is the key idea; We want a function, a positive function 8(.), on [a, b], that associates small intervals [u, v] about C when f exhibits erratic behavior at c. In particular, C - 8(c) < u < C :5 v < c + 8(c):
[u, v] C
(c -
8(c), C + 8(c»)
and
v - u < 28(c).
The term in the Riemann sum, f(c)(v -u), would be dominated by I(c), 2o(c). We will clarify this process with examples. As always, we are driven by considerations of area, and thus begin with examples whose integrals, because of I'areas," clearly "should be ... "
The Henstoc:k-Kurzweillntegral
171
Example 8.1.1. Suppose f(x) =
ll~o
: ~ ~:
From area considerations, the H-K integral of lover the interval [0,3] should be 6:
H-K
l'
f(x) dx = 6.
Let's see why this is so. For any ordinary Riemann partition of [0.3] with all the tags Ck different from x = 1.
However, x = 1 may be the tag of one interval, say (I, [Xk-l. xkD, or it may be the tag of two adjacent subintervals, (1. [Xk-b 1]), (1, [1, Xk+lD. The difference of such a sum and the number 6 is bounded in absolute value by 1/(1) - 21 (Xk - Xk-l), 1/(1) - 21 (Xk+l - Xk-l), respectively, So, given E > 0, the difference between any ordinary Riemann sum and the number 6 will be bounded by € if the length of the subinterval(s) containing the tag x = 1, Xk - Xk-l (Xk - Xk-l, Xk+l - Xk), is less than
€/ {4[ 1/(1)1 + 2]}: 1/{l) -
21 (X/c -
€
Xk-l)
< [1/(1)1
+ 2] . 4[ 1/(1)1 + 2]
€
=
'4'
or
1/(1) - 21 (Xk+l < ([ If(I)1
- X/c-l)
+ 2J . 4[ If(1') I + 2J) + ([ 1/(1)1 + 2J . 4[ If(:)1 + 2J)
E
2' If we can guarantee that for any partition of [a b] the subinterval containing x = 1 has length less than E/ {4[ 1/(1)1 + 2]}, then we will conclude that the integral has value 6. Obviously we could require all the subintervals to have a length less than E/ {4[ 1/(1)1 + 2]}; and this is the usual Riemann integral when 0 is a constant, in this case, E/{4[1/(1)1 + 2]}. As we have discussed however, the only subintervals of importance are those that contain the point of discontinuity, x == 1. Thinking of 8 as a I
A Garden of Integrals
172 function, we have
If this function 0(') forces the subinterval of any partition that contains x = 1 to have length less than €o/ {4[ 1/(1)1 + 2]}, the integral would have value 6.
Exercise 8.1.1. Show H-K!g I(x) dx = 3, given I(x) Hint:
O(x) =
I
€o/
=
I
4 x = O. I x # O.
{2[ 1/(10)1 + I]}
x=
O.
.x -:F O.
You will have noticed that both of these functions were integrated by ordinary Riemann techniques. We are trying to get used to the idea of a variable 0(.). Let's try another example.
Example 8.1.2. Suppose f(x) =
Io
l/x
x
= 1, !.l .... ,
otherwise.
Even though I is unbounded and thus not Riemann integrable, I is Lebesgue integrable (f = 0 a.e.), and area considerations suggest thatHK!; f(x) dx =
O. Of course the only way a Riemann sum will be different from zero is if at least one tag Ck belongs to the set jt ... }; and such a tag could be the tag of two contiguous subintervals. Let €o > 0 be given. Then for any partition of [a, b],
{I.!.
II: I(ck)(xk - Xk-l) - 01 = L I(ck)(xk - Xk-l). If Ck
#
l/n for a11n, then this Riemann sum is zero. If Ck = I/j, then
In that case, let's require O(Ck)
= 0 (~) = . €o • +2 . } } ·2J
173
The Henstock-Kurzwellintegral
Since a partition may assume only a finite number of values from the set 1, and a tag may be the tag of two subintervals, the difference between any associated Riemann sum and zero will be bounded by
i, t, ... ,
It appears that ~(x)
==
f./ n2 n+2 1
1
x = l/n, rt
= L 2, ...
I
otherwise
would work.
Exercise 8.1.2. Show that H-K!~ I(x) dx = 0, given that I is the Dirichlet function,
f(x) = Hint:
~(x)
1~
= 1f./2n +2
x rational, otherwise.
X = J'n., n = 1,2, ... , otherwise.
1
Example 8.1.3. Consider the integral of a familiar function, f(x) = x 2 , with 0 < x < 1. Then C
1
Thus H-K!ol
1
o
I (;t) dx
lex) dx
=R
11 I 0
should be
(x) d x
11
= L0i dJ.L = -.31
t.
Now, f changes least about 0 and most about 1. Our function 8(·) should reflect this behavior. We want to approximate the area under the curve between Xk-J and Xk by Cf(Xk - xk-d. That is,
since we know the answer. Furthermore, the total error,
174
A Garden of Integrals
needs to be made small. But 2
Xk-l (Xk - Xk-l)
<
(XZ-l +
=
3) 3"1 (3 Xk-Xk-l
< X~ (Xk
Then Cl(Xk - Xk-l) Ci(Xk - Xk-l) -
XkXk-l
3
+
xz)
(Xk - Xk-l)
- Xk-l).
-l (xl- x:- 1)
is between
X~(Xk - Xk-l) and CZ(Xk - Xk-l) - x~_l (Xk - Xk-l).
That is, the individual errors are bounded by
and the cumulative error is bounded by n
L:
28(Ck)(1
+ Ck)(Xk -
Xk-l),
k=l
which we want to be less than Would it work to set 8(x)
f£.
= e/[2(1 + x)]?
Let's see:
Yes. Comment: Consider
8(0) = e/2 > e/4 = 8(1). Notice that 8(·) is a decreasing function. Thus the largest value of 8(·), where the function changes least, is x = 0; the smallest value of 8(.), where the function changes most, is x = 1. The subintervals of such a partition decrease in length as we move from left to right. All of these probablys and maybes may be making us uneasy. In the next section we will tighten up our arguments and give precise definitions.
The Henstock-Kurzweil Integral
8.2
175
Gauges and 8-fine Partitions
To begin this discussion, we must establish some tenninology.
[a, b] -+ R+, where B(t) > 0 and a < is called a gauge on [a, b].
o A positive function B :
o
IS
t :::
b,
A tagged partition of [a, b] is a finite collection of point intervals (Ck' [Xk-I. Xk]) where 1 < k < n, Xk-l ::: Ck :::: Xb and a = Xo < Xl < X2 < ... < Xn = b. We call Ck the tag of the interval [Xk-I. Xk].
o Given a gauge B(·) on [a, b] and a tagged partition of [a, b], we say the
tagged partition is B-fine if Ck-B(Ck) < Xk-l < Ck ~ XIc < Ck+B(Ck), where 1 < k ::: IZ. To say we have a B-fine partition of [a. b] means we have a tagged partition of [a, b] satisfying Ck - B(Ck) < Xk-I < Ck :5 Xk < Ck + B(Ck) for each subinterval [Xk-l, xkl. Note that Xk - Xk-l < 2B(Ck), where 1
Lemma 8.2.1 (Cousin, 1895). If B is a gauge on [a, b] (i.e., a positive function on [a, b]), then there exists a B-jine partition of [a. b]. In fact, injinitely many exist. Cousin's argument uses nested intervals. Suppose we do not have a Bfine partition of [a, b]. Then either [a, (a + b)/2] or [(a + b)/2. b] does not have a B-fine partition, say lab b I ]. Repeat this bisction method. We have a sequence of nested intervals whose lengths approach zero. Consequently, we have a point C in the interN section; that is, C E [al/,bll ] for all 11. Because B(c) > 0, we have a natural number N so that c -B(c) < aN ~ C < bN < C+B(c). But then we have a B-fine partition of [aN, bN] consisting of the tagged partition (c. [aN. bND, a contradiction. Note: If 0 < B{x) < a1 (x) on [a, b], then any a-fine partition of [a, b] is a BI -fine partition of [a, b]. Why?
176
8.3
A Garden of Integrals
HoOK Integrable Functions
We are now in a position to defme the new integral.
Definition 8.3.1 CHenstock-Kurzweil Integral). A function I on the interval [a, b] is said to be Henstock-Kurzweil (H-K) integrable on [a, b] if there is a number A with the following property: For each E > 0 there exists a gauge (positive function) 8e (·) defined on [a, b] such that for any 8e -fine partition of [a, b], with Ck - 8e (Ck) < Xk-l < Ck < Xk < Ck + 8E (Ck). for 1 < k < n, we have n
L
I(Ck)(Xk - Xk-l) -
A <
E.
k=l
We write H-KJ: I(x) dx = A. As always, to say I is H-K integrable will mean that the integral exists and is finite. The sum L ICc) 6.x will be called an H-K Sllm when we are discussing the H-K integral.
Exercise 8.3.1. Show that a function I has at most one H-K integral. Assume that H-KJ: I(x) dx = Ai, where i = 1,2, and show that Al = A 2 • Hint: 8(x)
= min{8 1 (x), 82 (x)}, where 8i
is associated with Ai.
Just what kindspffunctions are H-K integrable? We have discussed a few examples. A more thorough and systematic treatment will now be given.
8.3.1
Step Functions
Step functions are H-K integrable.
Example 8.3.1. Suppose the step function s is defined by at
sex) =
{
f3 a2
if a < x < c, if x = c, if C < x
We will show that s is H-K integrable and
H-K
E
f.b s(x) dx =
O
a)
+ 0<2(b -
c).
A possible difficulty arises at x = c, where s may be discontinuous. Let > 0 be given. We will define a gauge 8e (·) on [at b] so that for any 8e -fine
177
The Henstock-Kurzweil Integral
partition of [a, b] the tag of the subinterval containing C - say, [Xk-l, Xk] - is c. That is, (Ck. [Xk-l. xkD = (c, [Xk-l, xkD Define
8e(x) =
1 t Ix8-- ci
~f x i: C
I
If x = c,
with 8 to be determined. We claim C = Ck' Otherwise, Ck i: c and Ic -- ckl < 28 E (Ck) < ICk -- cl, a contradiction. Now form the associated H-K sum (s(c) == fJ) for a 8c fine partition
[a, b]:
+ fJ(Xk -Xk-l) + a2{b -Xk) aI (c -- a) + a2{b -- c) + (fJ -- al)(c -- Xk-l) + (fJ -- (2)(xk -- c).
LS(Ck) ll.x = al(xk-l -a)
=
Thus,
IES(Ck)ll.x -- [al(c --a)
+ a2(b -- c)]1
<
1.8 -- ad, 281O (c)
We want this last expression to be less than E. Let
8,.(c)
=
E
2(1 fJ -- Cl!ll + IfJ -- Cl!2! + 1) -
E
2(sum of jumps
+ 1)'
That is, our gauge 8100 on [a, b] is given by
8( ) E X
-1 -
t Ix -- CI
E/[2(sum of jumps + 1)]
if x i: C, if x = c.
We have shown that the step function is H~K integrable on [a, b] and
H-K
J.b sex) dx = "1 (c - a) + "2(b -
c).
Next, a more complicated step function.
Example 8.3.2. Let a step function s be defined by
fJo al
sex)
= fJI a2
11,.
=
if x a, if a < x < c, if x = c, if C < x
=
We will show that s is H-K integrable on [at b] and
H-K
J.b s(';t) dx = "I(C - a) + ",(b -
c),
178
A Garden of Integrals
Discontinuities of s may occur at a, c, or b. As in the previous example, the technique is to defme a gauge that forces a, c, and b to be tags of their associated subintervals. Let E > 0 be given. Trial and error suggests an appropriate gauge on [a, b] as o,,(x)
=! t t
ffiin{lx - ai, Ix - bl, Ix - cl} min{ (c - a). (b - c), o}
if x ¥- a, b, c, if x = a or b or c.
with 0 to be determined. We claim that a is the tag of [a, Xl]; b is the tag of [Xn-l, b]; and C is the tag of [Xk-l, xk1 for some k, where 2 < k < II - 1, for any o,,-fine partition hf [a, b]. For example, to show a is the tag of [a, xd when a = Cl, assume otherwise. Then Cl - o,,(Ct) < a, so CI - a < o,,(Ct). If Cl is not b b, or c, then Cl - a < O,,(CI) < ~(Cl - a), a contradiction. If CI then b - a < o,,(b) = t(b - c) and a > c. Finally, if Cl = c, then C - a < OE (c) < (c - a). We have shown that (CI, [a, Xl]) = (a, [a, Xl]). The reader may consider the case (c n • [Xn-l l bD = (b, [Xk-lt b]). As for c, can it belong to the subinterval [a. xd? If so, then C - a < 20,,(a) < C - a. Can c belong to the subinterval [Xu-I. b]? If so, then b-c < 20,,(b)::: b-c. Thus C belongs to [Xk-l,Xk], where 2 < k < 11-1. The tag Ck of [Xk-lt Xk] is not a or b. If Ck is not c, then [c - ckl < 2o(Ck) < ICk - ci. The tag of [Xk-lt Xk] must be c. F onn the associated H -K sum:
=
t
L S(Ck) 6.x = al (C - a) + a2(b - c) + (fJo - al)(Xl - a) + (fJI - al)(C - Xk-l)
+ (fJI
- 0!2)(Xk - c)
+ (P2 -
(2)(b - Xn-l).
Thus
+ a2(b - C)]I [Po - ad . 2oe(a) + [,81 - all· 2oe(C) + [,81 - Ct!21 . 2oe(C) + [,82 - a21· 20" (b).
II:S(Ck) 6.x - [al(c -a) <
This last expression is less than
0=
E
if we choose E
2(sum of jumps
+ 1)'
The Henstock-lCurzweil Integral
179
For the gauge 8E O on [a, b] defined by
~ min{lx - ai, Ix -
i
bl, Ix - cl}
8E (x) = { 1 . { 2' mm (c - a), (b - c),
E
2(sum of jumps +1)
jf
x
= a,
b or c,
the associated H-K sum satisfies
We have shown that this step function is
H-K
f.b s(x) dx
='
H~K
IZ, (c - a)
integrable on [a, b] and
+ IZ2(b -
c).
Extend these arguments to a general step function on [a, b]. So~
a finite number of discontinuities does not pose a problem. However, we know that step functions are Riemann and Lebesgue integrable. Furthermore,
R
f.b s(x) dx = L f.b s d/L = H-K f.b s(x) dx.
Have we anything new? This question will be answered in Sections 8.7.1 and 8.7.3. But first we have some preliminary work to do.
8.3.2
Riemann Integrable Functions
Riemann integrable functions are H-K integrable, and the integrals have the same value. Suppose f is Riemann integrable on. [a, b]. Let e > 0 be given. We have a positive.constant 8 so that if P is any partition of [a, b] with Xk -Xk-l < 8, then .
Lf(Ck)(.~k -Xk_') -R f.b f(x)dx P
< E.
a
Define our gauge 8,; by 8E (x) = 8/2 for all x in [a, b]. Suppose we take any 8E -fine partition of (a, b], the existence of which is guaranteed by Cousin's Lemma 8.2.1. Then Zj - Zj-l < 28(cj) < 8, and consequently
180
A Garden of Integrals
The function f is H-K integrable and H-K!: f(x) dx = R!: f(x) dx. Of course this result makes the step function example of Section 8.3.1 trivial.
Example 8.3.3. The function f(x) = {
llf
0< x< I, X =0,
is H-K integrable on [0, 1] and H-K!~ f(x) dr = 2. How shall we demonstrate this? We begin by looking at the fluctuations. Because of the rapid change in f about zero, we want a small interval about zero. The area under the curve between Xk-l and Xk is given by 2.JXk - 2JX k-1. Thus we want
1
- ( X k - Xk-l) ~
.JCk
2,JXk - 2JX k-l
for Ck > O.
If Cl = 0, then O(Xl - 0) ~ 2.JXl and 2.JX'i < 2J8(x) < ~/2, where we assume E < 4. So, 8(0) e2 /16. Otherwise, for k = 2 •...• 11, routine manipulation shows
=
181
The Henstoclc-Kurzweillntegral
Thus, the cumulative error for the interval [Xl, 1] is bounded by
~ 28(Ck) L....J 3/2 (Xk - Xk-l), k=2 ck which we want to be less than E/2. We have this if 28(Ck)/C;/2 < E/2; 2 that is, if 8(Ck) < E/4cZ/ for k = 2, ... ,11. This suggests
8(x) =
! !
E2/16 (E/4)X3/2
x = 0,
°<
X
< 1.
\Vill this gauge work? Let E > 0 be given (we may assume 8( ) £ X =
E2/16 (E/4)X3/2
°
~
< 4), with
x = 0,
<
X
< 1.
Let P be any Be-fine partition of [0, 1]. (Once again, its existence is guaranteed by Cousin's Lemma 8.2.1.) Consider the collection of associated point intervals: (CI, [0, Xl]), (C2, [Xl, X2]), ... , (C II , [XIl-I, XnD. We claim that for the first interval, [0, Xl], the tag Cl must be zero. 2 2 If Cl is greater than zero, then Be(c.) = (E/4)ci/ , and Cl - (E/4)c:/ must be less than zero. But C1 _
~c3/z =' C1 ( 4 4
€c:'Z)
4
1
> C1 (4 - €) > O. -
4
So the tag of the first subinterval is zero. Thus, n
k=1 n
- L !(Ck)(Xk -
Xk-l) - 2
k=2
n
<
2../Xl + L
1
r,:;-(Xk -Xk-l)
k=2 '" Ck
-2(.JXk - JXk-d
182
A Garden of Integrals
So f is H-K integrable on [0, 1] and H-KJri f(x) dx = 2. This function is not Riemann integrable, although it is Lebesgue integrable.
Example 8.3.4. A function that is unbounded on every subinterval of [0, 1], with
H-K
f
I(x) dx = L
f
I dJ.L =
o.
As in Exercise 6.2.3, let
f(x) =
Io
q x = p(q; p, q relatively prime natural numbers, otherwIse.
Recall Exercise 8.1.2 and Example 8.1.2.
Example 8.3.5. Suppose that f is zero almost everywhere on [a, b]. Then f is H-K integrable on [a, b] and
H-K
t
I(x)dx
=L
f
I dJ.L
= O.
Let E > 0 and E = {x E [a, b] I f(x) '# O}. Because E has measure zero, we can cover E by an open set G so that J1. (G) < E. To start, suppose we have a Riemann partition of [a, b]. The only contribution to this Riemann sum will occur when a tag Ck belongs to E C G. How do we estimate f(Ck)? All we mow is that f(Ck) :/= O. The idea is to partition E into disjoint measurable sets, anyone of which would yield an estimate for f. One way to do this is to observe that
E = {x
E
[a, b] I f(x)
= U{x E [a, b]
f.
O}
= {x E [a, b] Ilf(x)1 f. O}
In - 1 < II(x)1
~ n}.
Each of the mutually disjoint sets En, {x E [a, b] 111- 1 < If(x)1 ~ n}, is a subset of E, which has measure zero. The set En is measurable, and JJ.(En) = O. We have an open set Gn covering En with J1.(G n ) < E/n2" (recall Section 5.6 on approximating measurable sets). Now we need a gauge. If a tag Ck belongs to [a, b] - E, then f(ek) = 0 and we would have no contribution to a Riemann sum. In this case,- 8e (Ck) does not matter. Otherwise, Ck belongs to E and we have a unique N so that Ck E EN. Since Il(ek) I > N - 1, which is potentially very large, we need the subintervals associated with such Ck, [Xk-l, Xk], to be small. Recalling that
The Henstodc-Kurzweil Integral
183
EN is a subset of the open set G N, and considering that J.L (G N) is less than E/ N2 N , is there some way to force the nonoverlapping subintervals [Xk-l. Xk] to be a subset of G N and thus to have a total length not exceeding
E/ N2N? Since [Xk-l. XkJ C (Ck - 8(Ck). Ck + 8(Ck)), is there some way to force (Ck - 8(Ck), Ck + 8(Ck)) to be in GN? Yes. Let 8(Ck) be the distance from Ck to points not in GN. That is, let 8 (Ck) equal the distance from Ck to points in the complement of the open set GN . We are ready to define the gauge:
I
8E (x)
=
X E
{ distance from x to the complement of Gn
[a, b] - E.
for x Il
E
Ell,
= 1,2 •....
Consider a 8£-fine partition of [a b]. Only tags in En, for some n, will contribute: En) •...• Ellm • Because If I < ilion E"i' the total contribution is bounded by I
1ttJ.L(Gn1)
+ 1l2J.L(Gn2 ) +"·+nkJ.L(Gn,,,) < LIlJ.L(Gn ) <
Thus, H-KJ: fex) dx
E.
= O. Look at Example 8.3.4.
We offer two additional observations. First, if C is the Cantor function, l C' dJ.L. then C' = 0 almost everywhere and H-K!Ol C'(x) dx = 0 = L Second, almost everywhere equal functions have the same jntegral.
fo
Exercise 8.3.2. Suppose f is H-K integrab1e on [a. b] and g equals f almost everywhere on [a. b]. Demonstrate that g is H-K integrable on [a. b] and that
H-K
lb
g(x) dx = H-K
lb
I(x) dx.
Hint: g = (g - f) + f; g - f is H-K integrable by Example 8.3.5; and by the same example we have H- K [g(x) - f(x)] dx = O. Now use linearity of the integral.
f:
Is the H-K integral an extension of the Lebesgue integral? We have some preliminary work to do before we can answer this question.
8.4
T~le
Cau(hy Criterion for H... ~( Integrability
As with other integration processes, Cauchy-type conditions are useful. An H-K sum is a Riemann sum associated with a 8E -fine partition.
184
A Garden of Integrals
Definition 8.4.1 (Cauchy Criterion for H-K Integrability). A function f on the inteIVal [a. b] is H-K integrable on [at bJ iff for every f > 0 we have oE-fine partitions PI and P2 of [a. bJ with the associated H-K sums within f of each other.
> 0 be given. If f is H-K integrable on [a, b], then we have a gauge OE(') on [a. b] so that the H-K sum of any oE-fine partition of [a, hI
Proof Let
f
is within f/2 of the H·K integral of f. Thus any two such sums will be within f of each other. For the other direction, assume that for any f > 0 we can determine a gauge OE(-) so that for any two o·fine ~artitions of [a, b] the H-K sums are within f of each other. Then for f = 1I n, we have a positive function 8n (.) so that for any two ~n·fine partitions the associated H-K sums are within 1/11 of each other and an+l < On on [a, b]. Consider the sequence of the H·K sums
L
f(cI)/)qX,
~
L fCC
2
)6.2 X,
••. ,
~
L f(c
n
)6. n x, ... ,
h
with Pn a on·fine partition of [a, b1. We claim the sequence
is a Cauchy sequence of real numbers. To show this, we compare
Pn
Pm
with 111 > 11. The key is the observation that by our construction process Om < an. so a am-fine partition is a an-fine partition of [a, b]. We have two H-K sums for a on-fine partition of [a, b]. For m > 11,
and we have a Cauchy sequence. Let A = lim LP1, f(cn)lj..nx. Then for each 11 1 <-. 11
185
The Henstock-Kurzweil Integral
Let E > 0 be given and choose N so that 1/ N < partition of [a, b]. Then
L fCc) 6.x P
A <
L
E.
Let P be any oN-fine
fCc) 6.x - LfCC N )6.Nx
P
PN
1 I <-+-<2E N N ' since P and PN are oN-fine partitions of [a, b]. Thus f is H-K integrable on [a, b] and H-Kf: f(x) dx = A.
0
Another way to look at H-K integrability involves approximation above and below. We will explore this by way of an example.
Example 8.4.1. Suppose for each E > 0 we have H-K integrable functions ep and 1/1 on [a, b] so that ep < f < 1/1 and H-K Then
f
lb
",(xl dx - H-K
lb
"'(x) <
€.
is H-K integrable on [a, b].
Before we demonstrate this, notice that since ep and 1/1 are H-K integrable, we have ot/J- and ol/r-fine partitions of fa, b], ep < f < 1/1 on [a, b] so that
L ep(e) 6.x - H-K 1¢(x) dx b
P~
a
and
L l/1(d)6.y - H-K l P~
b
1/I(x) dx
a
are less than E. Define a gauge 0(') by oCx) = min{8 tP Cx), 0l/r(x)} and let P be any 0fine partition of fa, bJ. Then a o-fine partition of [a, bJ wil1 be a 0tP- and
A Garden of Integrals
186 81/r-fine partition of [a, b], and
H-K
t
",(x) dx -
€
L ",(e) box < L f(c) b.x < L t(c) b.x p p <
p
a
f.b ",(x) d.~ + • < H-K f.b
Any two H-K sums for a o-fine partition of [a, b] will be within 3E" of each other. By the Cauchy Criterion (Definition 8.4.1), f is H-K integrable on [a, b]. We conclude this discussion with a final property.
Example 8.4.2. If f is H-K integrable on [a, bl, then f is H-K integrable on every closed subinterval [e, d] of [a b]. Suppose a < e < d < b. (The reader may consider the other cases.) The idea here is to construct two 8e -fine partitions of [a, b] that are identical on [a, c] and [d, b], and then use Cauchy's Criterion. Let E > 0 be given. Because f is H-K integrable on [a, b], we have a gauge 01£ on [a, b1. Thus we have a gauge 8e on (a, c], [c, d], and [d, b]. By Cousin's Lemma 8.2.1), we have 8e-fine partitions of [a, c], [e, dl, and [d,b]. Let Pac, P;d' P:d , Pdb denote such 8e-fine partitions with PcId' P;d any two oc;-fme partitions of [c, d]. We have I
Pde =
{(Clt
[a, xtl), ... I (eN, [XN-l. cD},
P;d = {(db [c, zi]).···, (dM [Zl.c-l' dD}, P;d = {(eJ, [c , zf]),··· (eL. [zt-l' dD}, Pdb = {(fI, (d, Yl]), ... , (fK, [YK-I. cD}· t
I
Notice that PaeUP;dUPdb and PacUP:dUPdb are tw08 e -fmepartitions of [a, b]. By the Cauchy Criterion, the difference of the associated H-K sums is less than E". But these H-K sums are identical on [a, c] and [d, b]. Thus the difference of these H-K sums is a difference of any two 8e -fine partitions on [c, d] that is less than E". Using the Cauchy Criterion again we conclude that f is H-K integrable on [c t d].
187
The Henstock-Kurzwellintegral
8.5
Henstock's Lemma
Another usefuJ result, due to Ralph Henstock, tells us that good approximations over the entire interval yield good approximations over unions of subintervals. Lemma 8.5.1 (Henstock, 1961). Suppose f is H-K integrable on [a. b], and for E" > 0 let Be be a gauge on [a, bJ so that if we have a Be-fine partition of [a, b], then
Suppose F l , F2 , .•• F J is a finite collection of 1'l0110verlapping (no common interiOl~ closed subintervals of la, b]. with Yj E Fj C (y j-8 e (yj), Yj + Be (y j )), where 1 :::: j < J. Then I
and J
{ ; !(yj)l(Fj)-H-Kh,!(X)dX <2(. where l (Fj) denotes the length of the subinterval Fj. Proof The set [a b] - u{ Fj is a finite collection of open intervals. Adjoin their endpoints. For this finite collection of closed subintervals, K 1, K 2 • ... , KN. of [a. b]. f is H-K integrable (Example 8.4.2). Choose a gauge Bn < 8" so that for TJ > 0 and a 811 -fine partition of /(", call it K(P,,), we have I
L K(Pn )
f(x) 6,x - H-K
1
f(x) dx < ; .
K"
The partitions K(P.), K(P2), ... K(PN), together with Fli F2 • ... , FJ fonn a B,,-fine partition of [a, b]. (See previous result.) We have Lhe case as illustrated in Figure I.
188
A Garden of Integrals
1
a
K..
I
2
F2
b
Figure 1. Partition of [a, b] Then
t
If(YJ)l(Fj)-H-KL f(X)dxj J
-
+ f(y2)l(F2~ + ... + j(YJ)l(FJ)
[f(Yl)l(Fd
L
+
f(x) 6.x
+
K(PI)
-H-K[
+f
jK1
+
L
f(x) box
K(P2)
f
f(x)dx
JFI
f(x)dx
+ {
+(
E
L
f(x) 6.x]
K(PN)
lF2
f(x)dx
f(x)dx
+ .. , +
+ '" + f
lK2
r
jFJ
f(x)dx
f(X)dX]
lKN
E(H-K Ix. f(X)dX) - t, L~/(X)b.:C) N
<
+ " .+
+L
H-K
n=l
f
j Kn
f(x) dx -
L
f(x) dx <
E
+ N ; = E + 11.
K(PII )
By the arbitrary nature of 11. the first inequality is valid. For the second equality. select from the subintervals Fl. F2 •.. . , FJ those for which f(yj)l(Fj) - H-K!F. f(x) dx > O. Thus J
o<
Lf(yj)l(Fj)-H-K ( f(x)dx= lFJ
LI
I <E.
Otherwise, f(Yj)l(Fj) - H-K!FJ f(x) dx < 0, and
- Lf(Yj)l(Fj)
+ H-K
L. J
All together,
The argument is complete.
0
f(x)dx = L
I I <E.
The Henstack-I
8.6
189
Convergence Theorems for the H-K Integral
\Ve now establish a convergence theorem for H-K integrals.
8.6.1
Monotone Convergence
Theorem 8.6.1 (H-K Monotone Convergence). Suppose that {fk} is a monotone sequence ofH-K integrablefimctions on [a, b] converging pOintwise to f on fa, b]. Then f is H-K integrable on [a, b] iff the sequence {H-Kj: fk(X) dx} is bounded on [a, b]. In this case,
H-K
f.b f(x) dx = limH-K f.b fk(x) dx.
Proof Assume the sequence {!k} is monotone increasing: Ji < ... < fk < fk+l < f on [a, b]. If ! is H-K integrable, then H-KJ: fk(X) dx <
H-KJ: f(x) dx for all k, and the monotone increasing sequence {H-KJ: fk(X) dx}, being bounded above by H-KJ: f(x) dx, converges. Now, assume the monotone increasing sequence
{H-KJ: fk(X) dx}
is
bounded above and converges to A: limH-KJ: !k(X) dx = A. We will show that H-KJ: f(x) dx
= A; that is,
f is H-K integrable on [a, b] to
A. Let E > 0 be given. For k > K, we have 0 < A-H-KJ: fk{X) dx < E. By the assumption of H-K integrability of fk' we have a gauge 8: for fic so that
for every 8:-fine partition of [a, b]. Let x be any point in [a, b]. Because lim fk (x) = f(x), we have a natural number n(x) > K so that If(x) - fk(X)1 < E whenever lc ~ n(x) > K, a natural number associated with each point of [a, b]. The function 8E (x) = B~(."I:) is a gauge on [a, b]. Suppose P is a Be-fine partition of [a, b]. We will show that the difference of the associated H-K sum and the number A can be made arbitrarily small.
190
A Garden of Integrals
I
Lf(c)Ax~A < L p
I
f(Ci) AXI - L fn{ct) (Ci) AXi
'=1
i=1 I
+
L {fn(C;)(Ci) AXI - H-K (
. fn(cj) (X) dxl
JAx,
1=1 I
+
LH-K (
_ fn(c,l) (x) dx - A .
JAx,
i=1
The fIrst term on the right is dominated by "Ef=1 If(c;) - fn(Ci) (c;)1 AXil which is less than E(b - a), For the third term, recall that the sequence {fk} is monotone increasing. The natural numbers n(cd, n(c2), .. .• n(c1) are each greater than or equal to K. Then
so
The second term remains:
The natural numbers 'I(Cl), n(c2), ... t n(cI) may not be distinct. Those that are the same correspond to the same partition, and we use Henstock's Lemmai For example, if n(cil) n(ci2) n(cil)' then
=
= ... =
The second term is dominated by "E E/2k = E. We have shown that for this oE-fIne partition of [a, b) the difference between the associated H-K sum for the function f and the number A is bounded by e(b - a) + e + E.
191
The Henstock-Kurzweil Integral
This is what it means to say that I is H-K integrable on [a, b] and that the H-K integral of I equals A:
H-K f.b f(x) dx The argument for
11
> ... >
Ik
= limH-K f.b A (x) dx. >
Ik+l
> ... >
I
is similar.
D
We note that pointwise convergence of Ik to I may be replaced by Ik convergence pointwise to I almost everywhere on [a, b1. We define I to be zero whenever pointwise convergence fails.
8.6.2
Dominated Convergence
We have the Henstock-Kurzweil counterpart to the Lebesgue Dominated Convergence Theorem 6.3.3. Theorem 8.6.2 (H-K Dominated Convergence). Suppose {/k} is a sequence 01 H-K integrable/unctions on [a, b] converging pointwise to I 017 [a, b]. Ifwe Have H-K integrable/unctions ¢ and 1/1 sllch that ¢ < Ik < 1/1 1m" all k, then I is H-K integrable and
H-K f.b f(x) dx
= limH-K f.b A (x) dx.
The proof may be found in Gordon (1994) or Bartle (200 I).
8.7
Some Properties of the H-K Integral
It is almost time to discuss Fundamental Theorems of Calculus for the HK integral. Before doing so, however, we will explore some more of its properties.
8.7.1
Extension of Lebesgue
We begin by showing that Lebesgue integrable functions are H-K integrable. In fact, the H-K integral is an extension of the Lebesgue integral. Liberal use of the convergence theorems faci1itates the argument. Theorem 8.7.1. Suppose I is Lebesgue integrable on [a, b]. Then H-K integrable 011 [a, b] and
H-K f.b f(.') dx
= L lb f dp..
I
is
192
A Garden of Integrals
Proof. Our argument has three parts. We begin by recalling that Section 8.3.1 showed that step functions are Lebesgue and H-K integrable and the integrals are equal.
Part 1. Simple functions are H-K integrable, and the H-K and Lebesgue integrals have the same value. Let G be an open subset of [a , b). It can be written as a countable union of disjoint intervals: G UIk. Defme a sequence of step functions {fk} by fk = Xli + XI2 + ... + Xlk' Then fk is H-K integrable and limfk = XG· Note that if x E [a,b] -G, then fkCX) = a for all k and limfk(x) O. If x E G, then we have a natural number N so that x ¢. II U··· U IN-I, X E IN, x ¢ IN +1, .... That is,
=
=
= fN-l (x) = 0 , 1 = fN+l (x) = ...
fleX) = ... fN(X) =
I
and
lim fk(X) = 1. By the Monotone Convergence Theorems (H-K and Lebesgue), since a < fk < 1 on [a , b] the characteristic function on Gis H-K and Lebesgue integrable, and
H-K
f.b XG(x)dx = limH-K f.b fk(X)dx = limL f.b fk dp. = L f.b XG dp..
Next, let E be a measurable subset of [a, b]. We may cover E with a monotone decreasing sequence of open sets {Gk} such that limj.L(Gk) = /-LeE) (by Theorem 5.6.1). Again, fk = XG k defines a monotone sequence ofH-K and Lebesgue integrable functions converging pointwise to XE, with a
H-K
f.b XE(X) dp. = limH-K f.b fk(x) dx = limL f.b /k dp. = L f.b XE dp..
Finally, a simple function is by definition a linear combination of characteristic functions on measurable sets. Thus, H-K
{b (simple function) dx = L {b (simple function) dj.L.
Ja
,Ja
193
The Henstock-KurIweil Integral
Part 2 Assume 1 is a nonnegative Lebesgue integrable function on [a, b]. Then we have a monotone sequence {¢k} of simple functions converging pointwise to 1 on [a, b] (Theorem 5.7.2). Now,
H-K
lb
¢k(X)
dx = L
lb
¢k
dp. < L
lb I
dp. <
00.
Application of the Monotone Convergence Theorems yields
H-K
lb
lb lb
I(x) dx = IimH-K
= limL
¢k(X) dx
¢k
dp.
=L
lb I
dp..
Part 3. Apply Part 2 to the nonnegative Lebesgue integrable functions
1/1+1 2
1/1-1
and
2
.
0
So Lebesgue integrable functions are Henstock-Kurzweil integrable and the integrals have the same value. Are there functions that are H-K integrable and not Lebesgue integrable? We will show very shortly that there are.
8.7.2 Recovering Functions via Differentiation We now prove a theorem that tells us that t for differentiable functions, the HK integral "recovers" the function from its derivative. Recall the Lebesgue integral requirement that the derivative be bounded (Theorem 6.4.2). Theorem 8.7.2. IfF is a differel1tiableftmct~on on Ihe interval [at b], then the derivative 01 F, F'J is H-K integrable 017 [a. b Jand
H-K { ' F'(f) dl
= F(x)
- F(a).
Proof Let € > 0 be glVen. For each point x in [a, b] we have a positive number 8f (x) so that for u < v, x - 8fi(.'\:) < U ~ x ~ v < x + 8e (x), and
IF(v) -
F(u) - F' (x)(v -
by the Straddle Lemma (Exercise 5.9.2).
u)1 < €(v -
u)
194
A Garden of Integrals
We have a gauge oe{')' Let P be a oe-fine partition of [a, b]. Then by Cousin's Lemma 8.2.1 we have
with Ck -
fot 1
<::
k
<::
OE{q) <
Xk-l <:: Ck
::5
Xk
<
Ck
+ Oe (Ck),
n. Thus
L F'(Ck) nx - [F{x) -
F(a)]
p
= <::
Lp (F'(Ck)(Xk - Xk-l) EL flXk = Eeb -a).
[F(Xk) - F(Xk-l)I)
So we have that F' is H-K integrable and H-K !~'C F' (t) dt = F(x) - F(a) for x in the interval [a b]. 0 J
We have shown that every derivative is H-K integrable, bounded or not.
8.7.3 H-I(, but not Lebesgue, Integrable We will demonstrate that an H-K function need not be Lebesgue integrable. Consider 2 F(x) = { x sin(1l'/ X2) 0 < X <:: 1,
o
x =0.
Then
F'(x) =
j
0 < x ::5 1, l O x = o.
2x sinC1l'/x2) - 21l'/x cOS(1l'/X2)
So F' is H-K integrable and H-K!Ol F'(x) dx = F(l) - F(O) = O. However, P' is not Lebesgue integrable. Recall that a function is Lebesgue integrable iff its absolute value is Lebesgue integrable. The function 2x sin(1l' / x 2 ) (when 0 < x ::5 1) and 0 (when x = 0) is continuous on
[0, 1]. Exercise 8.7.1. If F' was Lebesgue integrable, we could conclude that
L
f 2; leos C)I
Show that this is not the case, Hint:
Xn
dx <
00.
= Jl/{2n.
+ 1).
195
the Henstock-I
The H-K integral is indeed an extension of the Lebesgue integral. Can we weaken the hypothesis that F be differentiable? The continuous function 2..{i has a derivative of 1/ ..{i for x -:f: O. In Example 8.3.3 the H-K integral of the function 1/..{i (when 0 < x ~ 1) and 0 (when x = 0) was shown to exist and to have value 2.Jl.
Exercise 8.7.2. Show that the continuous function F(x) = 2..{ij with a derivative F' defined by 1/..{i for x > 0 and 0 when x = 0 satisfies F(x) = H-K!; F'(t) dt. Maybe F does not have to be differentiable at all points. On the other hand, the Cantor function C is a continuous function, differentiable almost everywhere (C' = 0 almost everywhere), and (by Example 8.3.5)
C(J) - e(O)
= 1,= 0 = L
f
c' d,..
= H-K
f
C' d,...
A set of measure zero, with regards to the derivative, still causes problems. It turns out we can successfully deal with a countable set of problem points. We have a Fundamental Theorem of Calculus for the H-K integral - in fact, as we shall see, we have two fundamental theorems to examine.
8.7.4
Fundamental Hwl( Theorem
Theorem 8.7.3 (FTC for the H-K Integral).
qF
is continuous on [a, b] and if F is difJerentiable all [a, b] with at most a countable number of e"'Cceptional points, then F' is H-K integrable all [a, b J and
H-K
f.'
for each x
F'(t)dt = F(x) - F(a)
E
[a, b].
Proof At the exceptional points, say aI, a2 • ... , ale we define F' to be zero: F'(ak) ::: 0, with k = 1,2, .... Let E > 0 be given. We now construct a gauge on [a, b]. I ••• ,
For x
-:f: Qb
we have a positive number
IF(v) -
8:
F(u) - F' (x)(v
0:
(x) so that
-tt)1 < E{V -
0:
u)
whenever x (x) < u ~ x < v < x + (x), by the Straddle Lemma (Exercise 5.9.2). Because F is continuous at ak, we have a 0; (ak) so that
A Garden of Integrals
196
whenever ak -0; (ak) < x < ak + o~(ak). Define the gauge OE(X) on [at b] by 8 ( ) = ex
I 0;
0; (x) (ak)
x:f: at. a2 • ... , x
= ak.
(differentiability) (continuity)
This gauge will work for [a, x]. By Cousin's Lemma 8.2.1 we have a oe-fine partition of [at x]:
.
If none of the tags Ck is an exceptional point al. a2, .•• t ak, •.. , argue as in Theorem 8.7.2. Otherwise, suppose the tag Ck is the exceptional point aN. If Ck tags only one subinterval, (Ck. [Xk-h xkD, then for F'CCk) = F'(aN) = 0 we have IFCXk) - F(Xk-l) - F'(Ck)(Xk - Xk-l)1 = IF(Xk) - F(Xk-l)1 <
2(2:+2 ) = 2:+ 1 '
But if Ck tags two subintervals, either (Ck [Xk-2, xk-!l) and (Ckt [Xk-l t Xk]), or else (Ck' [Xk-ll xkD and (Ck, [Xk' xk+d), we could have I
and
to estimate. In this case, we have
The worst case would be if each tag occurs as an exceptional point. But we have only a finite number of tags, so the total contribution from such exceptional tags must be less than L E./2 N E.. For the tags that are not exceptional points,
=
and
197
The Henstoclc-Kurzweil Integral
We have for any DE-fine partition P of [a. xJ
F(x) - F(a) -
L F'CCk) llXk
< Eeb - a) + E.
P
Thus, F' is H-K integrable and H-KJ~l: F'(t) dt = F(x) - F(a).
0
Compare Theorems 2.3.1, 3.7.1, and 6.4.2.
8.7.5
Sawtooth Functions
We offer one example of a so-called SlI1l1tooth function. The reader may construct other examples; these examples are informative. We construct a function F on [0, 1] as follows.
Step 1. Define F as
FeO) = 0, F
F
('21)
= 1. F
1Z) ( 11+1 F(1)
(2)3
= 1-
1 (3)4
2'
F
1
=1--+···+(-1) n+l 2
= 1-
(1) n
11
'2 + 3' ...
, ...
= ln2,
and let F be linear otherwise.
Step 2. F is continuous on [0, 1] so ]im.l:-+l- F(x) = ]n2. Step 3. F is differentiable except at
on (0. ~). on (~, ~), on (~. ~),
2
-3 F'ex)
=
4
(_1)11+1 (11
t. ~ .... .Il/(n + 1) •.... Thus
+ 1)
on
(n - l)/1Z,lZ/(ll + 1)) •....
Step 4. F' is H-K integrable on [0, 1] and H-K!Ol F'(x) dx = F(1) F(O) = In2. Step 5. F' is not Lebesgue integrable on [0, 1].
198
A Garden of Integrals
How did we reach that final step? Assume F' is Lebesgue integrable. Then IF'I is Lebesgue integrable, and 1
L
IF'I df.L >
0
/.
/.IJI(n+I)
L a
= 2 (~) 2
+3
IF'I d,u
(!)6 + 4 (~) + ... + 12
(n
+ 1) ( n(n 1+ 1) )
III
= 1+ 2
+ -3 + ... + 11-.
We have a contradiction. This concludes our treatment of recovering a function from its derivative using the H-K integral. How about recovering a function from its H-K integral using differentiation?
8.8
The Second Fundamental Theorem
Before proving the Second Fundamental Theorem of H-K integrals, we will discuss some properties of the function F(x) = H-KJ: I(t) dt, for a ::s x < b, where we assume that I is H-K integrable on [a, b].
8.8.1
Some H-K Properties
Property 1. F is continuous on [a, b]. Let € > 0 be given and select any point c in (a, b). We will show that F is continuous at c. Because I is H-K integrable on [a. b], we have a gauge 8E (·) so that for any 8c:-fine partition of [a, b1,
Define a new gauge 8(·) on [a, b] by
8(x)
= i min {!Ix - cl, 8e(x)}
l min {8 (c). c - a, b - c, €/(l/(c)1 + 1)} E
if x -:F c, if x = c.
For any 8-fine partition of [a, b], c is a tag. For x in (c - 8(c), c + 8(c)) - that is, Ix -cl < 8(c) - application of Henstock's Lemma to (c, [x, cD
The Henstock-I
199
and (c I [c IX]) implies
If(c)(c - x) - (FCc) - F(x)) I ~
f
If(c)Cx -
f.
I
c) - (FCx) - F(c») ~
and
That is,
IF(x) - F(c)1 <
E
+ If(c)llx - cl
<
f
E
+ If(c)l· If(c)1 + 1
< 2€.
Thus, if Ix - cl < 8(c), then IF(x) - F(c)1 < 2E, so F is continous The reader may consider c = a and c = b.
Property 2.
If f
is nonnegative, F is nondecreasing
For a < x < y < b, F(y) - F(x) are nonnegative.
Property 3. Let that
E
If f
E
= H-Kf:
f(t) dt. All the H-K sums
is continuous at cECa, b), then F is differentiable at c.
f
> 0 be given. Because
f(c) -
at c.
< f(t) < fCc)
+E
is continuous at c, we have a 8(c) so
It -
for
cl < 8e (c), t
E
[a, b].
For h > 0,
h(f(c) -
E)
(c+1I
< H-K
lc
f(t) dt ::: h(f(c)
+ E).
That is,
(c+h -€h < H-K lc [f(t) - f(c)] dt < EIt, and
F(c
+ h) h
F(c) _ f(c)
=
H-K (c+/l [f(t) - f(c)] dt
h
~
E.
It
Complete the argument by considering h < 0, c = a, c = b, and show that 1. if c E [a, b) and limx - c + fex) exists, then F has a right-hand derivative at c equal to f(c+);
200
A Garden of Integrals
2. if C E (a, b] and limx_c- f(x) derivative at c equal to f(c-).
=
f(c-) exists, then F has a left-hand
In fact, if we remove the requirement that f is continuous or has a leftor right-hand limit, it is still true that F' = f almost everywhere. This is the Second Fundamental Theorem.
8.8.2
Second Fundamental Theorem for H-I<
Theorem 8.8.1 (Second FTC for the H-K Integral). Given that f is H-K integrable on the interval [a. b], define afimction F on [a, b] by F(x) == H-Kf~'t f(t) dt. Then
1. F is continuous on [a. b], 2. F' 3.
== f
almost evelywhere on [a b], and I
f is Lebesgue measurable.
Proof We have shown that F is continuous (Section 8.8.1). To show that F' = f almost everywhere on [a, b1, we will show that the Dini derivate D + F equals f almost everywhere. Let E be the set of points in [a, b] for which D+ F is not f. First we will show that E has measure zero. For x in the set E, F(y) - F(x) , I 1m sup ---..;;;;...;;....-.......;.....;.. h-f'O+ Y- x is not f. If that upper limit is greater than f(x), then for every h > 0 we have a number Yh in [x, x + h] n [a, b], so that
F(Yh) - F(x) x
Yh -x
> f(x)
+ TJXI
with TJx a positive number. That is, F(y;~) - F(x) - I(x)(ih - x) > T/x(yj. - x). However, if that upper limit is less than I(x), then for every h > 0 we have a number Yh in [x, x + h] so that
F(Yh) - F(x) x < I(x) - TJx. Y/z -x with T/x a positive number. That is, for x in the set E and every h > 0 we have a number Yi~ in (x. x + Il] n [a, b) so that
IF(yt) -
F(x) - l(x)(yZ -
x)1 > TJx(yh -
x).
201
The Henstock-Kurzweillntegral
For x in the set E, we have a sequence of points Yh in [a, b) converging to x from the right so that IF(y:) - F(x) - f(x)(Yh - x)1 > TJX(Yh -x), for TJ:c positive. With each point x in E we associate a positive number TJ."t. Then
We will show that En = {x EEl TJ."t > lin} has measure zero for 12 = I, 2, ... ; thus E will have measure zero. Fix 11. Let E be greater than zero. We have a gauge 8",(.) so that if P is any 8-fine partition of [a, b], then E
< -. 11.
The collection of closed intervals ([x, yhl Ix E Ell! X < Yi~ < x + h < x + 8e {x)} forms a Vitali cover of En. We have a finite collection {[Xl, y;1]1 ... , [XM, y;M]} of disjoint closed intervals, Xl, " ' , XM points in En so that M
J.L*(En} -
L (y;,m - Xm) <
t!.
711=1
That is, M
L
(y:m - Xm)
> J.L*(En} -
t!.
m=l
But
Apply Henstock's Lemma 8.5.1. We conclude M
:1 > L ?E
Tm
v· {i/l f(x",} {y;m - xm} - H-K J~ J(t) dt
::em
711=1
L If(x m ) (y;'''' -xm ) - [F (y;'m) - F(xm )] I > L TJ.'(m (yftm - Xm) > !:..(J.L*(E E).
=
ll ) -
IZ
Thus J.L"t:(EIl ) < 3E and En is Lebesgue mensurable since p.T(ElI) = O.
202
A Garden of Integrals
Again, let E denote the points x in (a, b] for which F_(x)
F_(x) == lim inf{ F(y) - F(x) ,x - h < Y <
Y-x
h-+o+
If F-Cx) > f(x) , then for It > 0 we have
F
(Yh):t' -
FCx)
Yh -x
>
Yh < x
x}
~
f(x). That
:f; f(x).
in (a, b], so that
+ Tlx.
f()
x
for Tlx a positive number. That is, F
(Yh) -
(Yh -
F(x) -
x) f(x) < Tlx
(Yh -
for y~ -x < O.
x) < 0
However, if F-(x) < f(x), then
(Yh)x -
F(x)
F(x) - f(x)
(Yh -
F
Yh -x
(Yi) -
That is, F
<
f() X
-
Tlx.
x) > -Tlx (Yh
-
x) > O. In short,
IF (Yh) - F(x) - f(x) (Yh - x)! > -Tlx (Yh - x) = Tl;r; (x - yiz) . As before, En = {x EEl Tlx > lIn} and E == UEn • Fix n. The col-
lection of intervals {[Yh x] I x E En, X - 8e (x) < X - h < x - Yi~ < x} forms a Vitali cover of En. We have a finite collection {[y~l, Xl] [y:M XM]} of disj oint closed subintervals of (a , b], with L~=l (Xm - y~m) > p.*(En) - E. But I
I "
[Yi: m , xm] C
• I
(xm - 8e(xm), Xm
I
+ 8eCx m»).
Applying Henstock's Lemma 8.5.1, we have
~
-
It
M > '"' f(xm)
~ m-l
(Xm - y:m) - H-K 1~ f(t) dt Y'''''" h
M
= L !f(xm) (Xm - y~m) -
[F(x m) - F
(YZ m )] I
m=l
> Tlx
(Xm -
yZm) > .!:.(p."'CE n) n
e).
Thus p."F(E'I) < 3E. By the arbitrary nature of E, we have that p.* (En) = 0, so Eu is Lebesgue measurable and p.(En) = O.
203
The Henstock-Kurzweil Integral
Show that the other two derivates equal sequently F' = I almost everywhere.
I
almost everywhere, and con-
The third conclusion - that I is Lebesgue measurable - follows, since differentiability of F almost everywhere implies that F is continuous almost everywhere and thus (Section 5.7.2) F is measurable. Define F for x > b by F(x) = F(b). Then In(x) = II [F (x + (lIn)) - F(x)1 is a measurable function. Thus, lim/n(x) = F'(x) almost everywhere on [a, b1 is a measurable function (Theorem 5.7.1). Since F' = I almost everywhere, I is measurable (Section 5.7.2). 0 Compare Theorems 2.4.1,3.7.2, and 6.4.1. This concludes our discussion of the H-K Integral.
8.9
Summary
Two Fundamental Theorems 01 Calculus for the Henstock-Kurzweil Integral
If F is continuous on [a, b1 and F is differentiable with at most a countable number of exceptions on [a, b], then
1. F' isJIenstock-Kurzweil integrable on [a. b] and 2. H-KJ~"'C F'(t) dt
=
F(x) - F(a), a <x ~ b.
If I is Henstock-Kurzweil integrable OIl [a, then 1. F is continuous on [a,
2. F' = 3.
I
f
b1 and F(x)
= H-KJ~"'C I(t) dt
b1,
almost everywhere on [a,
b1. and
is measurable.
I Cauchy I Riemann Lebesque I-Ienstock-Kurzwei1
Figure 2. C eRe L c H-K integrable functions.
204
8.10
A Garden of Integrals
References
1. Bartle. Robert. A Modern Theory of Integration. Graduate Studies in Mathematics, Vol. 32. Providence. R.I.: American Mathematical Society, 2001. 2. DePree, John; and Charles Schwartz. Introduction to Analysis. New York: Wiley, 1988. 3. Gordon, Russell. The integrals of Lebesgue, Denjoy, Perron, and Henstock Graduate Studies in Mathematics, Vol. 4. Providence, R.I.: American Mathematical SocielYs 1994. 4. McLeod, Robert The Generalized Riemann Integral. Carus Monographs, No. 20. Washington: Mathematical Association of America, 1980. 5. Yee, Lee, and RudolfVybomy. The Integral: An EasyApproach, after Kurzweil and Henstock. Cambridge University Press, 2000.
CHAPTER
9
lhe Wiener
~ntegrai
What science c(;ln there be more noble, more e.xcellent, more llseful/or men, more admirably high and demonstrative than this of mathematics? - Benj amin Franklin
In the preceding chapters, the integrals under discussion were defined on sets of rea] numbers. So the domains of integration have consisted of real numbers. In contrast, the Wiener integral has as its domain of integration the space of continuolls junctions on the inter·val [0, 1] that begin at the origin. A cQJltinuous function now plays the role of a real number. With the Wiener integral path replaces pOint: these are lntegrals over sets of continuous functions, integrals over "paths." Hence the terminology path integral. The approach, as in the development of the Lebesgue integral, is threefold: 1. We begin by defining a measure on special subsets of our space of continuous functions. 2. We extend this measure to an appropriate sigma algebra. 3. With a measure in place, we develop an integration process leading to the Wiener integral.
9.1
Bro\lvnian Motion
The story begins in Scotland in the 1820s with a botanist named Robert Brown (discoverer of nuclei of plant cells) who was studying the erratic motion of organic and inorganic particles (pollen and ground silica, respectively) suspended in liquid. Neighborlng particles experienced unrelated motion, movement was equally likely in any direction, and past motion had no bearing on future motion.
205
206
A Garden of Integrals
Despite numerous experiments that varied heat and viscosity, a satisfactory explanation was not forthcoming. Decades later a theoretical explanation was put forward by Albert Einstein (1905) and Marian Von Smoluchowski (1906); the eventual experimental verification by Jean Perrin (1909) resulted in his receiving the 1926 Nobel prize in physics. Consider a collection of particles with initial displacement zero. Einstein argued that the number of particles per unit volume around position x at time t (in other words, the particle density at x at time t) is given by (2rc t)-1/2 e-x2 / 2t , which is a solution of the diffusion equation. Moreover, the mean square displacement from the beginning position is proportional to the elapsed time t. Consider a single particle, replacing density by probability. Starting from position x = 0 at time t = 0, the probability that the particle will be found between a and b at time t is written P (a < xCt) < b I xeO) = 0). We calculate this probability by
P(a < x(t) < b I x(O) = The mean square displacement calculated by
E(x')
= (21rt)-l!'
(2Jrt)-l!'l e-X'!Zt dx.
the expected value of x 2
L:
and the particle density, (2rct)-1/2 e-x equation
au
b
0) =
2
x'e-X'!2' dx /2t,
-
(1)
may be
= t,
is a solution of the diffusion
a2 u at ="2 ax 2 · 1
Thinking of a particle starting at the origin at time zero, we are assessing its chances, its probability, of passing through the window [a, b) at time t. position
b "count" this particle "ignore" this particle --~--~~~~~~---------7----------------- rime
Figure 1. Particle probability
207
The Wiener Integral
Exercise 9.1.1. a. Using equation 1, calculate numerically the probability that a particle at time t = ~ passes through the windows [-1,1), [-2,2), [-3,3), assuming x (0) = O. Hint: As the window increases in size, we should count more particles, and for the largest window [-00, (0), we should count all particles. b. Show P (- 00 < x(t) < = 1, for t > O.
00
0) = (2rrt)-1/2 J~ e- x7"/2t dx
I x(O) =
We would expect the particle to wander up or down (so to speak) with equal probability. That is, the expected space coordinate x at time t, for a particle beginning at the origin, should be zero. Readers with some familiarity with probability will understand that the expected value of x, E(x), should be zero.
Exercise 9.1.2. a. Show E(x) = (2rrt)-1/2 J~ xe-."C integration by parts.
2
/2t
dx = O. Hint: Symmetry or
h. Show E(x) = (2rr(t - tl))-1/2 J~oo xe-(:C-~1)2/2(I-rl) dx = ~b for t > t12 assuming X(tl) = ~I' Hint: If the particle is at location ~1 at tl, its expected location t - tl seconds later is ~I' There is nothing special about starting at x = 0 at t = O. Here is anotber thought about the previous exercise. If we started with a collection of particles at the origin at time zero, we would expect the particles to spread out - to disperse away from zero - although most should remain close to zero, the expected value. The reader may recall that the variance is a measure of the dispersion away from the expected value; in this setting, it's away from zero. Fonnally, the variance of position x, written as E( (x
- E(x)f), equals E(x 2 ) since E(x) = O.
Exercise 9.1.3. a. Show E(x 2 )
= (2rrt)-1/2 J~ x2e-x2/2t dx = t.
h. Assuming x(t!) = ~b show E(x 2 ) = t - fl. Hint: The expected value of the squate of the displacement, the so~called mean squar'e displacement, is proportional to the elapsed time.
208
A Garden of Integrals
Exercise 9.1.4. Calculate numerically
P (
-~ < x (~) < ~ I x (0) = 0)
1
p ( __ < x
.../2-
P (- .;;
,
(~) < ~ I x(O) = 2.../2
0) '
~ x (U < .;; I x(O) = 0). and
P (-1 :5 x(1) < 1 I x(O) = 0). 2
Exercise 9.1.5. Show that the function (2rct)-1/2 e-x /2t is a solution of the diffusion equation Ut = iu,x,x. Yes, the distribution of the particles' position, the distribution of x at time t, starting from the origin (x(O) = 0), is a normal distribution with expected value 0 and variance t.
9.1.1
Wiener's Explanation
In the 1920s Norbert Wiener developed a mathematical explanation of Brownian motion, a staggering achievement given the complexity of the physical phenomenon. Wiener wrote: "In the Brownian movement. .. it is the displacement of a particle over one interval that is independent of the displacement of the particle over another interval" (Collected Works, 1976, p.459). That is, the quantities (random variables)
X(tl) - X (0). x (t2) - X (II) •...• X(tn) - X(tn-l) vary independently, 0 < 11 < 12 < ... < tn < 1, and are normally distributed with mean zero and variance tk - lk-l, with k = 1,2, ... 111, respectively. We have this mathematical model of Brownian motion:
1. x(O)
= 0 (all particles begin at the origin).
2. x(·) is continuous for 0 ::: t :5 1 (erratic, but continuous paths). 3. The random variable x(t) -xes), which is the change in position over the time interval t - s, has a nonnal distribution with mean zero and variance t - s, for 0 < s < I < 1. We write
P(a < x(t) -x(s} <
b) =
(21«t -s)r
1 Z /
lb e-e2/z('-·)d~.
209
The Wiener Integral
4. X(tl) - x(O) •... x (tlZ ) - X(tll-I) are independent random variables for 0 < tl < ... < tn < 1. I
From these intuitive concepts we now develop Wiener measure. Note that in what follows
K(x. t) = (2rrtr 1/. exp (-
~;).
for 0 < t < I.
Construction of the Wiener fVleasure
9.2
Our construction of Lebesgue measure on the space of real numbers R I began by assigning a measure (length) to intervals of real numbers: .e([a, b») == b - a. For Wiener measure, our space will be the continuous functions on [0. 1] beginning at the origin, Co, with the sup norm:
Ilxll = max Ix(t)l. O.:st.:sl
The norm defines a topology, sometimes called the metric topology on Co. A set {x E Collix - Xo II < r} is called an open ball with center Xo and radius r. A subset E of Co is called an open set if for each x in E there exists r > 0 so the {yilly -xii < r} C E. In Section 9.3 we develop these ideas. When we assign a measure to the "quasi-interval" of continuous functions, the set
{X(·)
Co
E
I at
< x(tJ) < bl. 0 < tl < I},
what do we have? It is the probability of a Brownian partic]e starting at the origin and passing through the window [all b t ) at time tt, where the position at time tl, x (tl), satisfies al < X(tl) < bl . Figure 2 illustrates this probability. We write WIl (
{x(·) E Co
I at
< x(td < bl. 0 < tl < I})
==
Lf.b (211" l
t l)-1/2 exp (-
9.2.1
gt ) dg l .
2tI
al
Borel Cylinders with One Restriction
More generally, we may replace the interva1 [al. b I ) with any Borel set B 1 in R I , dg] denoting Lebesgue measure in RI; see Figure 3. We write Btl
= {x(·)
E Co
I X(tI)
E BI, BI a Borel set in R
1
,O <
11 <
I}
210
A Garden of Integrals
position
"ignore"
time
Figure 2. The window [at. bl) and
Wll (Btl) = L ( d ~I K(~I, tl). JBl
The set of continuous functions Bt 1 is called a Borel cylinder with one restriction. Exercise 9.2.1. Fix 0 < II < 1. The collection of all Borel sets in Rl is a sigma algebra of subsets of R 1 , Bl. To each Borel set B 1 in R I, we correspond the Borel cylinder in Co,
Show that this collection of Borel cylinders is a sigma algebra of subsets of Co, Bf1' with Wtl (Bt1) = LIBl d~lK(~I.tl) as the measure; d~1 is Lebesgue measure.
position
time
Figure 3. Borel cylinder, one restriction
211
The Wiener Integral
Borel sigma algebras of R 1 (Rl.Bl,d~d
Borel sigma algebra of Co (Co. Bli ' WI 1 )
Figure 4. Borel sigma algebras
9.2.2
Borel Cylinders with Two Restrictions
Now suppose 0 < t1 < t2 < 1. Consider the quasi-interval of continuous functions {x(·) E Co I al < X(tl) < bIt U2 ::: x (t2) < b2}.
(Think of Brownian particles beginning at the origin and passing through the windows [Ui, bi) at times ti.) This situation is depicted in Figure 5.
1
Figure 5. Borel cylinder. two resb.ictions Assign a w-measure of
Why this measure? It is detennined as follows. Given the probabiJity of a Brownian particle starting at the origin at time zero and passing through the window [al. b t ) at time tl and given, furthermore, that it has passed through this window at time t1, let us calculate its probability of passing through the window [U2' b2) at time t2 > tl.
212
A Garden of Integrals
1
Figure 6. The kith subrectaIlgle First, we partition [all bd x [a2' b2) into (2R)2 subrectangles. We write
bl - al • b2 - a2 _ 2R
2R
I:::..
-
1
.1:::.. 2·
Now, what is an appropriate contribution for the kith subrectangle? See Figure 6. The probability of the particle passing through the kth segment at time tl is approximately K{al + k1:::.. 1, tl)l:::..l. Given that the particle is at the kth segment at time t1. the probability that it is at the ith segment at time t2 is approximately
Thus the total contribution of that kth segment would be 2"-1
K(al
+ k1:::.. 1 , tl)l:::..l
L
K(a2
+ i 1:::..2 -
al - k1:::.. 1 • t2 - tl)1:::..2.
i=O
Summing oVer k, we have 2"-12/-1
L L
K(al
+ kl:::..b tl)l:::..lK(a2 + i 1:::..2 -
al - kl:::..b t2 - tl)1:::..2
k=o ;=0
~
In Figure 7 the contribution is
~1
A
= al + k1:::.. 1, ~2 = a2 + i1:::..2:
/{ (~11 tl) /{ (€2 - €1. t2 - tl) 1:::..§11:::..€2.
213
The Wiener Integral
~--~------~~------~--------r-~----~---i
€1=al+k6.1 €2 = a2 + i 6.2 K (€t. tl) K (€2 - €l. t2 - tl) 6.€16.€2 Figure 7. Shaded areD. is the contnbution of the kith segment Next, as before, we replace [all b l ) x [a2. b2) with an arbitrary Borel set B in R2 . That is, for the Borel cylinder 2
we have the measure
where d(~l. ~2) denotes Lebesgue measure in R2. The set of continuous functions Btl t2 is a Borel cylinde1" with two restrictions. Example 9.2.1. Fix 0 < tl < t2 < 1. The collection of all Borel cylinders in Co whose base is a Borel set in R2 is a sigma algebra of subsets of Co. The procedure is clear. Fix 0 < tl < t2 < ... < til < 1. To a Borel set Bn in Rn, correspond the Borel cyJinder Btl t2 ...tll in Co. We have
and we assign a measure, Wt\/2 .. tll(BtI12 ...I,,)
=
Lf .. ·JBIl{ [((~1.tz)K(~2 -~1.'2 .,. K(~n - ~"-l t tIl
-il)
- tn-l)d(~t. ~21 "" ~n).
214
A Garden of Integrals
(Co. B71 ...tn•Wtl ...tn)
(Rn. r;n. d(~b ~2t ..•• ~n»)
Figure 8. Borel cylinders and sigma algebras of subsets This collection of Borel cylinders in Co is a sigma algebra of subsets of Co, B~lt2 ...tll; see Figure 8. Let's calculate the w-measure of some Borel cylinders in Co.
E){ercise 9.2.2. a. Suppose Btl Btlt2
= {xC·) ECo I XCtl) ER I , 0
E
I (XCtl),X(t2))
Co
E
R2, 0 <
Since Co = Btl = Bt[t22 we want Wtl (Btl) that this is in fact the case. b. Show that for 0 < Wtlt2 t 3(
{x(·)
=
E
tl
<
t2
<
t3
and tl
<
t2
< I}.
= Wtl 12 (Btl 12) = 1. Show
< 1,
Co I (X(tI), X(t2) , XCt 3)) E [al. b l ) x Rl x [a3, b3)} )
Wtlt3(
{xC,)
E
Co I (X(tl).X(t3))
E
[altbl) x [a 3 ,b3)}).
Hint: Fubini's Theorem. c. Show that for 0 < tl <
t2 <
t3 ::: 1,
(Cbapman-Kolmogorov Equation).
9.2.3 The Sigma Algebra
B[O,l]
What are we to do with all these collections of Borel cylinders and sigma algebras? Suppose we form the collection of all Borel cylinders with one
215
The Wiener Integral
restriction, Btl' as t1 varies through (0, 1], a collection of sigma algebras of subsets of Co. Now form. the collection of all Borel cylinders with two restrictions, tl, t2, where 0 < tl =F t2 :s 1. This gives us a collection of sigma algebras that contains the sigma algebras of one restriction. Continuing in this way, form the collection of all such sigma algebras with n restrictions, tIt t21 •.. t tn, and so forth. 'When we gather together all these collections, we have a large collection of finitely generated sigma algebras of subsets of Co and the associated probability measures. Let B[O,I] denote the smallest sigma algebra generated by this collection of finitely generated sigma algebras. This large collection of sigma algebras has now been replaced by one sigma algebra, B[O,l]. Can we similarly replace the large collection of associated probability measures with one probability measure on B[O,I]? That is what Wiener accomplished - a measure on Co [0, 1], a measure on an infinite-dimensional space.
9.3
Wiener's Theorem
.
Wiener takes us from the space of continuous functions to a probability space. Theorem 9.3.1 (Wiener, 1922). Let Co denote the space of continuoZls functions x(·) 01'1 [0, 1] with x(O) = 0 and IIxll = maxO::st::Sl Ix(t)l. Fo/' o < tl < t2 < ... < tn < 1 and BTl a Borel set in R n, form the Borel cylinder
and its associated pl'Obability measure W/ 1t2" til: Writ':!",!"
tll) =
(Bllt2 ...
Lf· .. JBIff [((~1' tl) •.• [(~1I - ~'I-il
tn - tn-t>d(~l t ~21 ••. ~1l)' I
These probability measures may be extended uniquely to a p1'Obabi/ity measure, the Wiener meaSll1'e f1.w. 011 the sigma algebra generated by the collection of all finitely restricted sigma algebras of subsets of Co. B[O,I].
A proof may be found in Yeh (1973). What kinds of sets are in B[0,1]? Recall that with Lebesgue measure J.l on ~ Lebesgue measurable sets E may be covered "tightly" by opeD sets
216
A Garden of Integrals
0, where /-l(O - E) < E. They may also be approximated tightly from the inside by closed sets F, where /-l(E - F) < E.
In fact, we have Borel sets so that UFk C E
c nOk
and
I with Fk closed and Ok open. Borel sets and Lebesgue measurable sets are
closely related. Do we have similar results for Wiener measure? We will show that 8[0,1] 8(Co), where B(Co) denotes the sigma algebra of Borel sets in the topological space Co with metric the sup norm.
=
Example 9.3.1. Note that Co = {x(.) E Co I X(tI) E RI, 0 < II::: I}
tP = {x(·)
E
and
Co I 0 < X(II) < O}
are members of 8[0,1]. Consider the functions of Co satisfying Ix(t)1 ::: fJ, for 0 < t ::: 1 and fJ > 0: {x(·) E Co I -fJ ::: x(t) :s fJ. 0 < t < I}. We will show that this subset of Co is actually a member of 8 0,11, even though it has an uncountable number of t restrictions. This appears to be a Borel cylinder- (-fJt fJ) is a Borel set in R 1 - but we have an uncountable number of t restrictions. Select a countable dense subset {tl' t2 • ..• } of [0, 1], and define a sequence {Sn} of Borel cylinders of Co as follows:
82 = {x(·) E Co
I -fJ < X(tl) I -fJ :s X(tl)
fJ}. ::: fJ, -fJ :s X(12)
8 n = {x(·)
I -fJ ::: XCtk)
<
8 1 = {x(·)
E
E
Co
Co
~
fJ. k = 1,2, ..
<
of
fJ}. n},
Clearly 8n is a Borel cylinder in Co, a member of 8[0,1], and 8 1 :J 82 :J ... :J 8n :J .... We claim that
S. =
n n
!X(.) E
m=l
=
!X(.)
m=l
Co l-fJ - ! < X(tk) ~ fJ. k = 1.2 •...• nj
ECo l-fJ < X(tk) < fJ + ~,
k = 1,2 .... ,12\.
217
The Wiener Integral
x
x
If E Sn, then is continuous and -f3 < X(tk) < f3, where k = 1,2, ... ,11. Thus we have -{3 - 11m < -{3 < X(tk) < {3 for all m and k = 1. 2, ... ,11. That is,
XE
n
IX(')
Co 1-{3 -
E
~ < X(tk) ~ f3,
;k = 1,2, ....
l1j.
~ < X(tk) ~ {3,
;k = 1,2, .. .
,n} .
m.
m=1
Assume
XE
n
IX(')
E
Co I -{3 -
m
m=l
Then x belongs to Co, and -{3 -11m < X(tk) < {3, for k = I.2, ... ,n. We have that .;t belongs to Co, and -{3 :::: X(tk) < {3, for Ie 1,2 •... ,'1. That is, x is a member of SII' As a countable intersection of members of 13[0,11, 8 n is a member of 13[0,1]. We claim that
=
n 00
{x(·) E Co
I -f3
< x(t) :::: {3, 0 :::: 1 < I}
=
S".
n=1
Certainly this set is a subset of Sn, where n = I, 2, .... So it is a subset of n~=I Sn. Assume E n~=l Su. Then is continuous on [0, 1], and -{3 :::: .i(t;) :::: {3, for i = 1,2, ... and -f3 < x(·) < {3, on a dense subset of [0,1]. On the other hand, if ¢ {xC,) E Co I -{3 < x(t) :::: f3r 0 < t < I}, where are we? Because is continuous, we have a point to in (0, 1] with x(to) > f3 + ex or ,,i(to) < -f3 - a for some a > O. Furthermore, we have a sequence {t,Z ,,} from our dense subset of [0, 1], with tnk. -+ to. By continuity of X, we have x(t",.) -+ £(to). That is, (tn,J > {3. For M > nk, x ¢: SM. Consequently x rfi n~=lslI' Thus
x
x
x x
Ix
I
~i E
{x(·) E Co I -{3 < x(t) < {3, 0
We conclude that {xO E Co I -f3 < x{t) < f3. 0 < t < I} belongs to 13[0,1]. The so-called closed baH center x == 0 and radius {3. We have sets with an uncountable number of restnctions, in the finitely generated sigma algebra 13[0,1].
218
A Garden of Integrals
Example 9.3.2. We claim that the set {x(.) 8[0,1]. If we can show that
V {x(.)
Ilxl <,8} =
{x(.) e Co
E
Co Ilxl <,8} belongs to
e Co Ilxl ::0 ,8 -
!} ,
l/n
{x(.) belongs to
8[0,1],
Co Ilxl ::::,8 - lin, 0
E
by the previous example.
Suppose ..~ E {x(.) E Co Ilxl < ,8}. Then Ixl < ,8 for all t in [0,1]. Since x is continous and it assumes maximum and minimum values on [0,1], we have -,8 < "~(~n) ::: x(t) < x(iM ) < ,8. That is, we have a natural number M so that ,8 > 1I M and
-,8 + ~ < x(t) < ,8 -
~,
for 0 :::: t < l.
Put another way,
{x(.) e Co Ilxl < fJ - ~} .
Xe Clearly,
{x(.)
e Co
!}
Ilxl <,8 -
c {x(.) e Co
Ilxl !S fJ}.
Exercise 9.3.1. Show the set {x(.) E Co Ilx(t) -xo(t)1 < E, 0
E
Co Ilx(t) - xo(t)1 < =
E,
0:::: t < I}
U {x(.) e Co Ilx(t) -
XO(I) I
€ -
II
=
un n
{x(.) E Co
I Ix (tk) -xo(tk)1
!I <
k
E -
.!.I. n
The reader may conclude that
{x(·)
E
Co IlIxll ::::,8}, {x(.) and
{x(.)
E
E
Co
IUxll < ,8},
Co III.t - xoll <
E}
are members of B[O,11. (Recallllxll = maxo::::t::::llx(t)I.) The set {x(.) E Co Illx - xoll < E} is called the open ball with center Xo of radius E.
219
The Wiener Integral
9.3.1
Open Sets and Open Balls
Every open set in a separable metric space is a countable union of open bal1s. We have shown the open ball in Co with center Xo and radius E, {x(.) E Co Illx - xoll < E}, is a member of B[O.I]. If we can show that Co with the sup metric is a separable metric space, we may conclude that every open set in Co is a member of B[O,l]. That is, we may conclude that B(Co) C B[0,1].
Example 9.3.3. Form the collection of all polygonal functions on [O,1J. These are linear on [(k - 1) In, kill], for k = 1, 2, ... , IZ, vanish at t = 0, and assume rational values at kin. We will show that this countable collection of "points" is a dense subset of Co. Pick an arbitrary open ball in Co, {x(.) E C Ilix - xoll < E}. Now construct a polygonal function x belonging to this ball as follows. Let E > 0 be given. Because Xo is uniformly continuous on [0,1], there exists 0 > 0 so that Ixo(t) -xes)1 < E/S whenever It -sl < 0 for all t, x in [0, 1]. Choose n so that 1I It < O. We have 0 < l/n < 2/n < .,. < n/11 = 1. So Ixo(u) - xo(v)1 < E/S whenever we have ll. v E [(k -1)/17,lcln], for k = 1,2, ... ,11. In each subinterval (k - 1)1 n , k I 1Z], pick a rational number J'k and define the polygonal function .x so that x(k/n) = xo(rk), with x(O) = O. For t E [(k - l)ln, kin], we calculate
Thus
Iii -
xol/ <
E.
220
A Garden of Integrals
We have shown that every open ball contains a "polygonal function." Suppose G is an open set in Co and Xo belongs to G. We have € > 0 so that {x(.) E Co [llx - xoll < €} c G. Select a polygonal function y so that lIy-xoll < €/3 and E/3 < r < E/2, for r a rational number. Then
{xO
E
Co [ Ilx -
yll
<
r}
C
{x(.)
E
Co [llx - xoll <
€}.
The open set G in Co is a union of countably many open balls with centers in the countable collection of polygonal functions and rational radii. We have shown that every open set in the metric space Co with sup norm is a member of B[0,11. We write B(Co) C B[0,11. Example 9.3.4. We will show that the Borel cylinder
is a member of B(Co). Given nk=1 {xO E Co I ak < X(tk) < bk}, we will define a projection Ptk on Co by Ptk (x) = X(tk). This projection Ptk is a continuous function from Co to R 1. That is, for Ilx - yll < 8, we have
Thus
is a Borel set in Co. We conclude that B[O,lJ C B(Co ). Since we have already shown that B(Co) C B[O.I]. we may conclude that B[O,l] = B(Co). We have achieved an understanding of Wiener measurable sets. The next step is measurable functions.
9.4
Measurable Functionals
Definition 9.4.1 (Wiener Measurable Functional). A real-valued function F defined on Co is a Wiener measurable functional if it is measurable with respect to B[O,I]. That is. F- I (c, 00)) must be a member of B[O,l] for all real numbers c.
The Wiener Integral
221
Example 9.4.1. For x
Co and 0 < t1 < 1, define F[x] = xltd. We will show that F is a Wiener measurable functional on Co. E
Because F assigns to every function in Co its value at t1, we have that
F- 1 ((e , 00»)
= {xO E Co
1
F[x]
= {x(·)
1
X(t1) E (c,co)}.
E Co
E (c , oo)}
This is a Borel cylinder with one restriction, a member of B[O,l].
Example 9.4.2. For x
Co and 0 < tl < 1, define F[x] Wiener measurable functional on Co? E
= IX(tl)l. Is
F a
We have two possibilities. If C < 0, then
F- 1 ((e, (0»)
= {x(·)
E
Co 1 e < IX(t1)1 < co} = Co.
If C > 0, then
F-1((e,oo))
Ie <
= {x(·)
E
Co
= {x(·)
E
Co 1 -co < X(tI) < -c}
U {x(·) E Co
IX(tl)1 < co}
< X(tI) < oo}.
1e
We have Borel cylinders, members of B[O,11. So F is a Wiener measurable functional on Co. Co, suppose F[x] = X2(tl), where 0 < tl < 1. Show that F is a Wiener measurable functional on Co.
Exercise 9.4.1. For x
E
How far can we take this approach? Consider F[x] with x E Co and 0 < tl < t2 < 1. We have
F- I ((e,
co»)
= {x(·)
E
Co
I c < x 2 (td IX(t2)1
If e = 0, then X2(t2) Ix (t2) 1 > 0 unless X(tl) = 0, X(t2) X2(t) = 0, X(t1) = X(t2) = O. Thus, ... There must be a better way.
= X 2 (tl) IX(t2)1. < oo}.
#- 0,
X(tl)
#- 0,
Example 9.4.3. Suppose the functional F defined on Co depends on only a finite number of fixed values of t. Assume F[x] = I (x (fl), ...• x (In»), and I is a real-valued continuous function on Rn. We will show that F is then a Wiener measurable functional on Co:
F- 1 ((e , co»)
= {x(·) = {x(·) = {x(·)
F[x] > c}
E
Co
1
E
Co
I I(X(tI)""
E Co
I (X(t1) ....
I
I
X
x (t,,») >
c}
(tn ») E 1-1 (c.
co»)}
222
A Garden of Integrals
Because I is continuous on Rn J then 1-1 (c, 00)) is an open set, a Borel set, in Rn , say Bn. Then
F- 1(c, 00)) = {x(·) eCo I (X(tl), ... ,X(tn )) e B n }, a member of 8[0,IJ. So F is a Wiener measurable functional on Co. We have explored a sigma algebra of Wiener measurable subsets of Co, 8[0,11, the Borel sets Co with sup nOIlD, and Wiener measurable functionals on Co, F[x]. Turning now to the Wiener integraUtself, we shall mimic the development of the Lebesgue integral.
9.5
The Wiener Integral
With a measure space, (Co 8[0,1] t iLw) and measurable functions - Wiener measurable functionals on Co, F[xJ - only the integral remains. Given an element x(·) in Co, we can assign a real number F[x]. In fact, we can assign F[x] to each path x(·) in Co. Thinking of this path integral as an averaging process, we will assign the number F[x] to a collection of paths that are "close" (so to speak). Then we average over all such collections. We are now in position to define the Wiener integral. Symbolically, we write I
(
lco
F[x] diLw.
As with the Lebesgue integral, we will define the Wiener integral for characteristic functionals, simple functionals, limits of simple functionals, and so on.
Example 9.5.1. Suppose F[x] is the characteristic functional on a measurable subset C of Co, where C e 8[0,1]. We will define the Wiener integral of the characteristic functional F [x]. First, we observe that ifx(·) e C, then F[xJ = 1 and if x ¢ C, then F[xJ = O. Note too that F[xJ is a Wiener measurable functional on Co:
F-1(-00,c])=
¢ Co-C { Co
ifc < 0, ifO
Then feo F[x] diLw should be iLw(C). We define it as such:
(
lco
Xc [x] diLw
== iLw(C),
223
The Wiener Integral
JViener simplefimctionals are linear combinations of characteristic functionals by definition.
Example 9.5.2. Suppose t/J Ck members of B[O,t].
= L~ akXcJ..' with Co = U':Ck, CI n Cj =
t/J,
We define the Wiener integral of t/J by
How do we deal with more complicated functionals?
°
Example 9.5.3. Assume F[x] = X2(tl), with < tl :5 1, as suggested by Figure 9. The functional F weights each path in Co by the square of its value at at t1. We shall detennine
x
y
o
1 time
Figure 9. Wiener simple functional Following the ideas of Lebesgue, we will construct a monotone sequence of\lliener simp1e functionals converging to F[x]. Recall that F[x] is Wiener measurable (Exercise 9.4.1).
Step 1. Consider the case [0,00) = [o,~) U [~, 1) U [1, (0). We have
C II
= {X(.) E Co I 0 :::: X2(tl)
C12 =
C1
!
x(·) E Co
<
H'
I~ < x'(ll) < 11.
= {.t(.) E Co 11
< X2 (tl)}.
and
224
A Garden of Integrals
Then Cll , ClZ , and C 1 are disjoint members of B{O,l] with union Co. Define the simple functional FIfx] by
1
Ft[x] = O· XCu [xl + 2" . XC12[X]
+ 1· Xc. [x].
Repeat this process.
Step
11.
Cnk =
Consider the case [0,00) = Uk:lo-lCnk U Cn with
I
x(·) E Co
k k+ 2n < X2(tl) < 2n
I
I) '
o< k
< n2n - I,
and
Cn = {xO E Co 111 < X2(tl)}' The simple functional on Co is defined by n2/1-1
F,.[x]=
L
k 2nXCnk[x]+nXCn[x],
k=O The reader may show that 0 ::::: Fn[x] ::::: Fn+Ilx] on Co, and that for 2 11 > x (td then 0::: FIx] - Fn[x] 1/2n. The Wiener measurable functional F[x] is the limit of a monotone sequence of nonnegative Wiener measurable simple functionals on Co:
The Wiener Integral
225
1 1 -.;n 1
+ L -.Jkfi!f
(
-.Jek+l)/211 00
+L
(11 -
..fii
+L < L
i:
-co
nIe) - ~~ K(~l' tl)d~l ] 2
~f)J(~l' tl)d~1
(n - ~~)K(~l' tl)d ~1
HK(~l' tl)d~l'
The reader may show that the four integrals with square roots as limits converge to zero as 11 becomes large. For example, k/2n - ~f < 1/211 on the interval
(..jk/2 n, ..j(k + 1)/2n )
nated by 1/2" J~ J(~ll tl) ., ., r 00 e-x- d x < I/?_ae-Q!- . Ja!
dr
L~~O-l[ ... ]
and the sum
As for
LJ:JnCn -
is domi-
~r)K(~l' tl) d~b use
Thus,
We define
1
F[x] dJ1.w
==
liml FII[x1 dJ1.w.
Co
Co
In this example, F[x] = X2(tl), the Wiener integral is evaluated as an elementary Lebesgue integral.
Example 9.5.4. Given F[x]
1
= x 2 (tdlx(t2)1. calculate
x2(tl)lx(t2)1
for 0 < t1 < t2 < 1.
djJ. w ,
Co
First we partition R 1 : (-00, -11) U [ -n,
-11 + 2~ )
U .. · U [11 -
2~' .11) U ~l, co).
Let
C;; = Cllk
=
{x(·) E Co I -co < XCtl) < -n}, x(.) E Co
1
Ie ::S x(tt) < 11
I2
C,; = {x(·) E CO III < ;r(tt)}·
lc+l1 2"
I
226
A Garden of Integrals
We have partitioned Co into a finite collection of disjoint Borel cylinders, and if we replace X(ti) with ~l we have a partition of Rl into a finite collection of disjoint Borel sets in Rl: B;;, Bnk,andB;t, respectively. Repeat the process for X(t2), letting
D; = {xC,) Dni =
E Co
x(·) E Co
1
D;t = {xC·)
E
I-co < x (t2) < -12},
I-z2.
.+ 11 ,
< X(t2) < _1- n 2
n -
Co I n < X(t2)};
with corresponding disjoint Borel sets in Rl: G;;, Gni, G;t. Together, we have a partitioning of R 2 = R 1 X R 1 into disjoint Borel sets that correspond to disjoint Borel cylinders in B[O,11. The diagram in Figure 10 indicates the appropriate regions.
;2 B;
n G;
B; n G;
B,,,,n G~
n
B;
n Gill
BnJ. n
B,i n Gill
Gill
-n
B;
n G;;
B"kn
G;;
B;
-n
n G;
1J
Figure 10. Partition of R1 x RI Adding (vertically) these approximations and taking limits we obtain L
L: d~2 L~ d~1~~1~2IK1K2 + L: d~2 Ld~lW~2IK1K2 L
+L for KIK2
L:
d~
fa dM?I~2IK1K2'
= K(~l,tl)K(~2 -~l,t2 -td, yielding L
J': d~2 J': d~1~~1~2IK(~1' Il)K(~2
-
~l' 12 -
1\).
The Wiener Integral
227
Routine (though lengthy) calculations will yield
Jeo FIJ [xl dJ,Lw = L
lim (
1 1 ~rl~2IK(~1. tl)K(~1.-~l' 00
00
-00
-00
t2- t l)
d~ld~2.
From these last two examples, {
Jeo
X
2
(tl)dJ,Lw =
Ll°O ~?K(~l,tl)d~l -00
and x 2(td
(
Jeo
Ix (t2) I dJ,Lw
=L
L: f:'
s~I~IK(~I. t,)K(h -
SI.t. -
tl)dSld~2'
The Wiener integrals have reduced to Lebesgue integrals of the same form.
9.6
functionals Dependent on a Finite Number of t Values
Theorem 9.6.1. Suppose a jimctionaZ F[x] defined on Co depends on a finite number of fi.;r.ed t values, 0 < t1 < t2 < ... < til < 1 sZlch that F[x] = f(x(tl). x (t2) • .... x(tn »), where f is a real-vallledjil17ction continuolls Oil RII Then F is a Wiener measurable functiol1al 011 Co. and {
Jeo
FIx] dJ,Lw = (
Jeo
= L
f(x(tl), ...• X (tn ») d/Lw
L: -L: dSn ..
dS!/(Sl ..... Sn)K(~I.tl)
... /(~u - ~/l-1. til - tn-I)
whenever the last integral exists. If the Wiener jimctiol1a/ depends continuoZlsly on only a finite '1umber of t values, it may be evaluated as an ordinalY Lebesgue integral. Proof. We will sketch the proof. As discussed in Example 9.4.3, F[x] is a Wiener measurable functional OD Co:
228
A Garden of Integrals
We can mimic the development of the Lebesgue integral. Assume 1 is nonnegative and continuous. This is true for characteristic functionals, simple functionals, and limits of monotone sequences {Fm} of simple functionals. (The reader may partition R n into nonoverlapping Borel rectangles. On each of the associated Borel cylinders, calculate the infimum of 1 times the Wiener measure of the as~ociated Borel cylinder.) To conclude, we calculate
{
leo
F[x] dJ-Lw
=(
leo
=
I(x(td ..... X (tn)) dJi.w
f··· fRfl I(~II ~21
••• •
= lim m
(
leo
Fm[x] dJ-Lw
~n)K(~11 td
... K(~n - ~n-ll tn - t,,-I) d~ld~2 ... d~nl by monotone convergence. For (111- 1)/2, and so on.
Exercise 9.6.1. Calculate
1
continuous, proceed as
feo F[x(·)] dJ-Lw
(I + 1/1)/2,
for the following functionals F.
a. F[x(,)] = X(tI), 0 < tl < 1. b. F[x(.)] c. F[x(o)]
= X 2 (tI), 0 < tl = X2(tI) [x(t2)1,
< 1. 0 < tl < t2 ::::: 1.
d. Show
{ xn (tl) dJ-Lw
leo e. Show
=
!
feo X(tl)X(t2) dJ-Lw =
nl2
0
1 . 3 . . . . . (II - 1) t 1
n odd, n even.
tI, 0 < tl < t2 < 1.
f. F[x(-)] = X(t2) - X(tI), 0 < tl < t2 < 1. Hint: J-Lw ({x E CO[X(t2) x(ttl <,Bn = J-Lw({x E CO[(X(tl),X(t2)) : (~11~2) E B2 with ~2~I < ,B} = L f B2 f(~2 -~l)K(~I' tl)K(~2 -~I. t2 -tI) d(~11 ~2)' Let ZI = ~h Z2 = ~2 - ~}, and use Fubini's Theorem. g. F[x(.)] = (X(t2) - X(tI))2, 0 < tl < t2 ::: 1. h. Show
feo (x (t2) - X(tl))4 dJ-Lw
i. F[x(.)]
=
= 3(t2 -
t1)2, 0 < tl < t2 < 1.
(x (t2) - X(tl))n, 0 < tl < t2 ~ 1.
j. Show feo (x(t2)-x(td)(x(t4)-X(t3)) d/.L w t4 < 1.
= 0, 0 <
tl < t2 < 13 <
229
The Wiener Integral
9.6.1
More Interesting Functionals
What if F depends on more than a finite number of values of t?
Example 9.6.1. Suppose F[x] = Cfo1Ix(t)1 dt. The continuous function x (.), the path x (-), is to be weighted by the integral of its absolute value. Let
FN[X]
1 N
=
L N
x
(k) N
•
k=l
an approximation to C f~ Ix(t)1 dt. (Realize that
is a continuous function of R N .) By Example 9.4.3, FN[X] is a Wiener measurable functional on Co, and by Theorem 9.6.1, [
J~
FN[X]dJ.Lw =
~1°O d~N"'l°O d~l(]~d+"'+I~NI)K(~lltl) N
-00
-00
... K(~N - ~N-l. tN - tN-I).
We want
fco F[x] dJ.Lw =
1.
limN
fco FN[X] dJ.Lw. We have
1 N
1
o Ix(t)1 dt - N { ; x
=
(kN )
( L [kiN Ix(t)l(k-I)/N N
x
( Ie" )
k=l
<
LN J(k-1)IN {kiN
Ix(r)l- x
kiN L [ k=l J{k-I)IN
xC!") -
-)~ N
(k) N
dt
dt
k=l
(~ )
11
<
X
~
dt.
Because x (-) is uniformly continuous on [0 1], this last sum may be made arbitrarily small. That is, limN FN [x] = F[x]. The function F[x] is a nonnegative, Wiener measurable functional on Co as a limit of Wiener measurable functions FN on Co. As for evaluating the I
230
A Garden of Integrals
Wiener integral of F, using Fubini's Theorem 6.7.1, we have
Lo (c [lx(r)ldr)
(Lo Ix(t)1
dJ1.w = L {
dJ1.w) dt
(L: I~I K(~. -r)d~) gO)'
= L{
dt
l/
= L { ( g-r ') dt = Evaluating as a limit, we have
and
lim { FN[X] dx = lim N
leo
N
Exercise 9.6.2. Calculate
f2 V-;
leo F[xJ dJLw
(t Vfk.~) Ii k=l
N
for the following cases (0 < t <
1). 8.
F[x] = L I~ Ix(,r)l m dr, 111 = 1,2, ... .
b. F[xJ = L J~ X 2m - 1 (r)d 7:, In = 1,2, .. .. c. F[x] = L I~ x2tn(7:)d7:, m = 1,2, ... .
Exercise 9.6.3. Suppose
F[x] = exp ( -L FN[X]
f.'
= exp (- ~
x 2 (t)d-r)
Ex2 (k ~)).
and
The Wiener Integral 3.
231
Argue Wiener measurability on Co of FN[X] and F[x].
b. Because IFNlx]l :::: 1, use the Bounded Convergence Theorem 6.3.1 to conclude that
Lo exp (
-L
L X
2
(T)dT) dJ.Lw
1 1 00
limL N
tk
•••
-00
X
with
00
K(~11
exp
-00
(-~N kt ~;) =1
tl) ... K(~n - ~n-l' tn - tn-l)d~l ... d ~IZI
= k t / N.
Exercise 9.6.4. Given F[x] = exp ( -L J~ V(xC'r))d-r), where 0 < lvlV~
V(~)
<
E R I , with V continuous on RI. Proceeding as above,
Show that -
1
F[x] dJ.Lw
= lim N
Co
(
2rr -t N
)-N/2 . L
t
exp ( - N (N; [ where ~o
1 1 00
00
...
-00-00
sk - ~k-I
V(~,,) + ('= t/ N
)2]) d~l
... d~nl
= o.
Note that
is a discrete approximation of
But Vex) + (dx/ dt)2 is the Hamiltonian of a particle of mass 1 in a field V. The Feynman integral (next chapter) involves the Lagrangian.
232
A Garden of Integrals
9.7
Kac's Theorem
As the reader has observed, the Wiener integrals are often the limits of fairly obvious discrete integrals, just as Riemann integrals often may be evaluated as limits of discrete sums. As with the Riemann integral, this arduous process sometimes may be avoided. We have a remarkable theorem due to Mark Kac that relates the evaluation of a class of Wiener integrals to solutions to appropriate integral equations. Essentially, he shows that a very complicated Wiener integral is related to a solution of an integral equation. The proof of this theorem may be found many places, including Kac (1959). As a tribute to his genius, we conclude our treatment of the Wiener integral with Kac's proof of this theorem. Theorem 9.7.1 (Kac, 1949). Suppose that V(n is a continuous nonnegative bounded function on R 1. Then
Ie"
exp (-
f.' V(x(~J)d') dp.,. I: H(x.t» dx.
o < t :s I,
= L
where H satisfies the integral equation H(x, t)
+ Lf.t dr: o
1 d~K{x -~, 00
t-
-00
Recalling Exercise 9.1.5, note that for V
1
=
1')V(~)K(~J 1:) = == 0 we have
r 1dJ.Lw = 1 K(x, t) dx, 00
JeD
K(x. t).
where Kr =
-00
1
2: Kxx.·
Proof Assume 0 < V(~) < M for all ~ in R 1 • There are five stages to this proof.
Step 1. By Exercise 9.6.4, Wiener measurability ofexp (-Lf~ V(x(1:»)dr:) is assured. Furthermore, for 0 < L f~ V (x (1'») d r: < M t we have
exp
(-L f.
V(X(~»d~) = F-o(-l)k
tOOL [
fot V(x(1:»d1' Jk k!
233
The Wiener Integral
and application of the Bounded Convergence Theorem implies that
fea exp
(-L LV(x(~»)dr) d/l=
Step 2. Let Ho(x, t)
t
k=o
w
(~1t r
lco
c.
(LJ.'0 V (X (-r)) dr)k dJ-Lw.
= K(x, t) = (2"t}-1/2 exp (-x 2 /2t)
and define
Llao dr[OO d~K(x-~,t-7:)V(~)Hk-l(~''t'). t
Hk(X,t)
=
-00
Show
i:
Hk(.<, t) dx =
and
L Hint:
L:
~I La (L
l'
Ho(x,t)dx = L
V(X(r»)dt) k d/l- w ,
i:
K(x,t)dx
=
1,
k = 1,2, ...
t > O.
(L!~ V(X(7:))d7:)2 = 2!L!~ d7:2!;2 d7:1 V(X(7:1»)V(X(7:2))'
Step 3. Estimating Hk(X, t). we have
0::: Hl (x. t) = L ~
t
lo
d't'
[00 d~K(x -~,
t - r)
V(~)Ho(~, 7:)
-00
MtHo(x. t).
(by 9.2.2)
In general,
Ie
= 1,2 .....
234
A Garden of Integrals
Step 4. Define H(x, t) = Lk=O(-l)k Hk(X, t), which converges for all x and t > 0, since we have IH(x, t)1 < eMt Ho(x, t). Step 5. Then 00
Hex,t) = Ho(x,t)
+ L(-l)kHk(X,t) k=l
= Ho(x,t)
+ L:(-l)k L I.' dr: 00
k=l
0
[:00 d~K(x -~, t = Hoex, t) - L
~)V(~)Hk-l (~, r)
J.' dt
L: d~K(x ~,t ~)V(~) (,t.(-l)kHk(~' ~)) J.' L: -
= Ho(x, t) - L
dt
-
d~K(x -~, t - ~)V(~)H(~, ~).
That is,
Hex, t)+L
I.Eo d"C 1-(00 d~K(x-~, t-r:)V(~)H(~, r:) = K(x, t),
t > 0,
00
and
for all x with t > O.
0
The requirement that V be bounded on R 1 may be removed by truncation arguments.
9.8
References
1. Kac, Mark Probability and Related Topics i1Z Physical Sciences. London: Interscience Publishers, 1959. 2. Wiener, Norbert. Collected Works Vol. I. Cambridge, Mass.: ]\IUT Press, 1976. 3. Yeb, J Stochastic Processes and the Wiener httegral. New York: Marcel
Dekker, 1973.
CHAPTER
10
[Mathematics] . .. there is no study in the world which brings into more - J. 1. Sylvester harmonious action all the faculties of the mind.
10.1
~lJ1troduction
In the 1920s Norbert Wiener introduced the concept of a measure on the space of continllolls junctions. As you recall from Chapter 9, this idea arose from his attempts to model the Brownian motion of small particles suspended in a fluid. In the 1940s Richard Feynman (1918-1988) developed an integral on the same space of continuous functions in his efforts to model the quantum mechanics of very small particles such as electrons. To succeed, Feynman's path integral approach to quantum mechanics had to be consistent with SchrOdinger's Equation.
10.1.1
Schrodinger's Equation
A frequent correspondent of Albert Einstein, Erwin SchrOdinger (18871961) made a series of brilliant advances in quantum theory and the general theory'of relativity. Our topic here is his breakthrough wave equation l discovered in 1925. Suppose a particle of mass m is at position Xo at time t = 0 with a potential V(xo). The particle may move to position x at time t. The Heisenberg Uncertainty Principle sets accuracy limits on the detennination of position x at time t. From physical considerations, then, we assign a probability to each path from Xo at time t = 0 to x at time t. This probability is P(t. x) = l¥r(t, x)1 2 , where t/I is a complex-valued quantity called the probability amplitude.
235
236
A Garden of Integrals
SchrOdinger's Equation, which is the partial differential equation
81/1
at =
i h 82 1/1 2m 8x 2
i -
"iz Vy"
with -00 < x < 00, t > 0, and h = 1.054 x 10-27 erg~sec, is satisfied by the probability amplitude y, with 1/1(0, x) = f(x). Because P(t, x) 11/I(t,x)1 2 is a probability, we want
=
1
2
h!r(t x)1 dx = 1,
10.1.2
I
for t 2: O.
Feynman's Riemann Sums
Feynman's path integral approach exploits the idea of Riemann sums. Suppose we have a nonnegative continuous function f on the interval [at b] and we are faced with the task of determining the area of the region between the x-axis and the graph of f for x between a and b. Roughly, we can say the area is the sum of all the ordinates-the sum of all the! s. In practice, we take a finite subset of the ordinates, equally spaced, and calculate the sum. Take another set of ordinates, equally spaced but closer together, and form another sum. Generally, as we take more and more ordinates (equally spaced, closer and closer together) and compute their sum, these sums will not approach a limit. Clearly, La:sx.5b f(x) doesn't make sense. But what if we assign a weight to each ordinate before summing? We can assign an appropriate weight h, a normalizing constant reflecting the spacing between consecutive ordinates. Now we have L !(xk)·h, and such sums have a limit, the sowcalled Riemann integral of the function ! over the interval [a, b]. We write
"/(x!)
+ hl(xi) + ... + hl(x.) ~ R
t
I(x) dx.
It is this appropriate weight, this normalizing constant, that we will be trying to determine for each possible path. To each possible path, Feynman postulated a probability amplitude, the squared magnitude of which was to be the probability for that particular path. All paths contribute, and each path contributes an equal amount to the total amplitude, but at different phases. The phase of the contribution from a particular path is proportional (the normalizing constant) to the action along that path. The action for a particular path is the action for the corresponding classical system. What does all this mean? Let's look at an example.
237
The Feynman Integral
10.2
Summing Probability Amplitudes
Suppose a small particle of mass m is at location x (0) = Xo at time O. It moves to location x(t) = x at time t in the presence of a potential Vex), along a path x('r), where 0 < -r < t. See Figure 1. position
~-"7
(0, xo)
(t. X(/»)
time
Figure 1. Path (r, x(r») We begin by isolating several key pieces of this puzzle: o
g
The Lagrangian is the difference between the kinetic energy and the potential energy, or ~m~i2 - Vex). The action A along this path x(-r) is a functional given by the integral of the Lagrangian along this path: A. [x (.)] = J~ [~mi2(-r) - V(x(r))] d-r.
o
The phase of the contribution from a particular path is proportional to the action along that path e(i/tl)A[x(.)], where tz = 1.054 X 10-27 •
o
The probability amplitude for this path is proportional to its phase, KeCi/n)A[x(.)], where [( is the nonnalizing constant, the same constant for each path.
a The total probability amplitude 1/1 is the sum of the individual probability amplitudes over all continuous paths from (0, xo) to (t, x(t»):
1/1(0, xo); (t, x») =
KeCi/n)A[x(.)] .
oil connecting contiouous pnths
This function 1/1 is to be a solution of Schr~dinger's Equation, and the probability of going from Xo at time 0 to x at time t is 11/112. We apply the Riemann sum analogy as suggested by Feynman.
238
A Garden of Integrals
10.2.1
First Approximation
Divide the time interval [0, t] into It equ~ parts, Xk = x(ktln), with o ~ k ~ 11. Now replace continuous paths with polygonal paths, using XII = x(t) = x and tk = ktill. See Figure 2. position
Figure 2. Polygonal approximation
10.2.2
Second Approximation
Approximate the action A[x(·)] along the polygonal path
A[x(ol] =
J.' [~mx2('l - V(x(r)) ] dr
.. ~ [~m (Xk ~/:k_l)2 - V(Xk-ll] =
~ [2(~n) (Xk - Xk_ll
2 -
V(xk-ll
G) G)]
0
The phase of the contribution from this polygonal path is
Summing over all polygonal paths, we arrive at the Feynman integral approximation of the sum of all probability amplitudes over all continuous paths,
L
1111 connecting continuous paths
Ke(i/fi)A[x(.)1.
The Feynman Integral
239
This approximation is l/rn (0. xo); (t,x))
= fa dXIl-I···fa ·exp
dXIK
~ {~[2(~n) (Xk -Xk_tl V(Xk-tlC)]} 2
-
Note: We obtain all polygonal paths as Xl,X2, ..• ,Xn-l vary over R; Xo and x" = x are fixed. Taking the limit, we have
as the solution of SchrOdinger's Equation, al/r
at =
Hz i 2m l/r.u - 'Ii V l/r,
with -
00
< x <
00, t
> O.
What kind of integral is this? The integral is highly oscillatory and is not absolutely convergent when V is real. What does the limit mean? Is it possible to choose a normalizing constant K that will guarantee a limit in some sense? Would a K that yields a limit also give us a solution to Schrodinger's Equation? In Feynman's words, "to define such a normalizing constant seems to be a very difficult problem and we do not know how to do this in general terms."
10.2.3
The Normalizing Constant
Feynman goes on to suggest that the normalizing constant K is given by
K
=
)n/2 m ( 2rrift(t /n)
By employing the principle of superposition, that is, l/r(t, x)
= fa t(O,xo); (t,x))f(xo)dxo,
with l/r(O, x)
=
f(x).
and incorporating this normalizing constant K. we finally have Feynman's solution to what we shaH call Schrodinger Problems A and B.
240
A Garden of Integrals
SchrDdinger Problem A
Problem A assumes that the potential V is O. In this case.
1/I(t, x) =
r"
l~m (i"'i~t/n) fa dXn-I""' fa dxo . exp { 2t.~~n) ~(Xk - Xk-l)2} !(Xo) ,
for
Xn
= X,
solves
fA a2 1/1 at = 2m ax 2 '
a1/1
1/1(0, x) = [(x),
with with
-00
fa 1/1 (t, 1
< x < oo,t > 0.
X)[2 dx = 1. t > O.
SchrDdinger Problem B
Problem B asssumes V
i- O. In this case
solves with
-00
< x < oo,t > 0,
1/1(0, x) = [(x),
In the remainder of this chapter we will explain what is meant by "solves" in the context of SchrOdinger Problem A and make some general comments about Problem B.
10.3 A Sinnple Example It is time for an example.
The Feynman Integral
241
Example 10.3.1. Suppose a particle of mass m at location x(O) == Xo at time 0 moves to location xCt) = Xn at time t in the absence of a potential field. In other words, V = O. We claim that
",(t, x) =
li~ C1fi~t/Il,r/2 f·· L ~
n-l
. exp
where
Xn
(21l~~n) ~(Xk - Xk_l)2) dXl dx, ... dXn-l.
= x, solves 2
a""= i 7l a "" at 2m -ax-2
with - 00 < x < 00, t > O.
Let a = m/[2fz(t/n)]. To evaluate this integral, we must integrate expressions of the form
L
{i [a(x _U)2 + bey -
exp
U)2]} duo
Our approach has three parts.
Part One. Show
1
00
iax2
e
d
_I¥i
x -
-00
Hint: Consider the contour integral
. ., f
for a > O.
I
a
cfic eiaz2 dz,
for a > O.
Write e az- = 1l (x, y) + i v (x y), and show that II, v satisfy the CauchyRiemann equations, U x = Vy and u y = -V x • Continuity of u, v and their partial derivatives show eia ;:2 is an analytic function. Consequently, with a suitable contour C to be described, cfic eiaz'J. dz = O. I
2
Let C be the contour for ei az indicated in Figure 3. Then
o = ~ ei • z ' =
dz
fo Bei ax' dx + foB e; o(B+;,)' i dy + isO ei° O+1)'.' (1 + i) dx.
Using
.Jf =
(1
+ i)/.../2, we conclude that 00
o /.
. = 1 +i ~ eiax2 d x -.
2
2a
242
A Garden of Integrals (B,B)
(0, 0)
410--_ _~---~
(B,O)
Figure 3. Contour C Thus
Part Two. Completing the square, show
J.
du exp i [a{x - u)2
+ b{u -
y)2] =
R
[1ri e[iab/(a+b)](::c-y)2.
y~
Conclude
1fi
ei (a/n)(xn-xo)2
[11/{1l-1)]a
.
All together,
fa dXll-l ... fa dXl exp (a 1fi
1fi
2a (3/2)a
[(Xl - XO)2+
(X2 -
1fi
Xl?+'"
+ (xn
- XIl _1)2])
ei (a/n)(x,,-xo)2
[1l/{11 - l)}a
'h
WIt
m a = 2tz(t/n)"
243
The Feynman Integral
Part Three. For XlI =
XJ
show
The limit is trivial. Exercise 10.3.1. Show that
'" {(0, x0); (I, x»)
J
= 2n:~ 'II 1 e(i m/2li)[(x-xo)'l '1
is a solution of the SchrOdinger Equation (V = 0), with - co < x <
00, t
> O.
Ifwe add the boundary condition that 1/1(0, x) = I(x), from the principle of superposition and from Exercise 10.3.1 we reach a formal conclusion:
1fr(t, x) = lljr((O,XQ); (t , x))/(xo) dxo,
for ljr(O, x)
= I(x),
with
That is,
'" (t , x)
= (:: ht )
I/'l
e (/ m/2lit)(x-%0)' f (;<0) dxo,
for '" (0, x)
= f(.~).
The reader may verify that formally
However, the integral may not malce sense unless some restrictions are imposed on I.
244
A Garden of Integrals
10.4
The Fourier Transform
The Fourier transfonn is an important method of solving constant coefficient partial differential equations. Such methods will be helpful in solving SchrOdinger Problem A. Let us briefly review such techniques.
Definition 10.4.1 (Fourier Transfonn). Suppose that f is a complex-valued If(x) I dx < 00. In other words, f belongs to function on R, where L~(R). Then the Fourier transform of f, :F(f), is given by
JR
:F(f)(y) = (2re)-1/2
L
e-i:cy f(x)
Exercise 10.4.1. Let f(x) = e-I.-c l , with
dx.
< x < +00. Using the Lebesgue Dominated Convergence Theorem 6.3.3, show that :F(f)(y) = (2re)-1/22(1 + y2)-1. Note: :F(f)(y) belongs to La(R). -00
Exercise 10.4.2. Let 1(X)
={~
-1 ::: x ::: I, otherwise.
Show that
:F(f)(y) = (2.11")-1/2 2 sin(y) . y
Note: :F(f) is not a member of La(R), but is a member of L~(R). Aside: Show that eiz / z is analytic for z "I- 0, and conclude that the contour integral if; eiz / z dz = 0 when C is the contour; see Figure 4.
-r
r
Figure 4. Contour C
JR
Show that sin(y)/y dy = re. By the way,
t -00
sin(k) =
Ie
1
00
-00
sin(x) dx = x
1 (SiD(X»)2 = t 00
-00
dx
x
-00
(Sin(k»)2 . k
245
The Feynman Integral
Exercise 10.4.3. Let f(x)
==!
e-
x
o
x
~ 0,
x < O.
Show that F(f)(y)
== (27f)-1/2 ( 1 -
£y ) .
1 + y2
This Fourier transform is complex-valued.
Exercise 10.4.4. Let f(x) = e-x2 /2, where -00 < x < +00. Argue that F(f)(y) is a differentiable function of y and show (F(f)(y)' + yF(f)(y) = 0, with F(f)(O) = 1. Solve this differential equation, and y2 conclude that F ( e- x2 / 2 ) (y) = e- /2. Note that in the preceding exercise y2F
Exercise 10.4.5. Show F
(e{-S/2}X
2
)
(e-
x2
/
2) belongs to
L~(R).
(y) = s-1/2 e - y 2/ 2S , for s > O. Hint:
F'. Exercise 10.4.6. We want to replace real s in Exercise 10.4.5 with complex z having positive rea] part. That is, we want (2". )-1/'
L
e-ixy.(-.I')X'- dx = Z-l/'.-y'/2% , 2
for z = a + ib, a > O. Show that (27f)-1/2 JR e-iXY e(-Z/2)x. dx and z-1/2 e- y2/2Z are analytic (Cauchy-Riemann equations). Conclude
F
(e
C- Z/ 2)X 2 )
(y)
= z-1/2 e-y2/2Z
for Re z > O.
10.5 The COl1volution Product We wilJ need a special product of functions. the convolution product, to solve Schr6dinger Problem A.
Definition 10.5.1 (Convolution Product). Suppose of La(R). The convolution product of f with g, f
(I * g)(x) == (27f)-1/2!a f(y)g(x
1
and g are members * g is
- y) dy.
246
A Garden of Integrals
Does this product make sense? Yes: by the HOlder-Riesz Inequality (Theorem 6.5.2),
L
11(y)g(x - y)1 dy <
1
(
L1112
1/2
dx )
(
.
L
2
) 1/2
Ig1 dx
Suppose 1 and g are members of Lb(R). Does the convolution product * g still make sense? We see that
L(L
If(y)g(x - y)1 dX) dy =
= =
L (L L (L L11(Y)1 LIg(u)1 If(Y)1
Ig(x - y)1 dX) dy
If(Y)1
Ig(u)1 dU) dy
dy
du <
00.
Thus f(y)g(..r: - y) belongs to Lb (R). By Fubini's Theorem, J~ f(y)g(x - y) dy exists for almost all x and is integrable. So f * g makes sense for 1 and g members of L~ (R).
10.6 The Schwartz Space For Schr6dinger Problem A we will also find useful the inverse Fourier transforms, and the SchrOdinger requirement 11/1(t, x)1 dx = 1, for t > 0, suggests pleasant behavior for large values of x. This brings us to the Schwartz space.
2
iR
Definition 10.6.1 (Schwartz Space). The Schwartz space S consists of those complex-valued functions on R,
1 : R ~ C, such that
1.
1
has derivatives of all orders.
2.
f
and all its derivatives decrease to zero as lim
Ixl-+oo
for all nonnegative integers
Ixl m D n l(x)
In
Ix I ~ 00:
= 0
and n.
It can be shown that the Schwartz space is a dense subspace of L~ (R) for p > 1. The Fourier transform on the Schwartz space has many beautiful properties.
247
The Feynman Integral
Exercise 10.6.1. a. Show that F(f)(y) = (21l')-1/2 fR e- ixy f(x) dx for f
E S has
derivatives of all orders. In particular, show that (F(f)(y») (11) = (-on F(x n f)(y). Show also that F (fen») (y) iy)" F(f)(y)·
= (__
b. Show F maps S into S, or F(S) c S.
10.6.1
Plancherel's Theorem
We have a useful theorem due to Michel Plancherel, whose proof can be found in Weidmann (1980, p. 292). The theorem has five parts.
1. The Fourier transfonn is a one-to-one linear map of the Schwartz space to itself. In fact, the inverse Fourier transform of F, rl, is given by
r
1
(f)(x) = (21C)-1/2/a eixy f(y)dy 1
for f a member of the Schwartz space S. Note that F- (f)(y)
=
F(f)(-y)· 2. For all members of the Schwartz space S, a dense subspace of L~ (R),
:r- 1 can
3. The Fourier transform F and its inverse the Schwartz space S to a11 of La(R). If then for r > 0,
Fr(f)(y) = (21f)-1/2
f
be extended from is a member of La(R).
L
e-ixy f(x)dx
is a member of La (R). The mean-square limit - which is called the Fourier transform of f and is also written F(f) (y) - exists and defines a function in La (R):
IIF(f)(y) - Fr(f)(Y)112 -70 as r That is. for
f
E
-T
co.
La (R),
r F(f)(y) == lim (21C)-1/21 e-i:cy I(x) dx r-l-c.::l
==
-r
(21l')-1/2/a e- ixy f(x) dx.
248
A Garden of Integrals
Similarly,
:;:-l(f)(x)
=
lim (2Jr)-1/2 (r eixy ICy) dy r~CO J-r
== (2Jr)-1/2
L e
ixy
l(y) dye
4. For all members I of L~ (R), the Fourier and inverse Fourier transforms are bounded linear operators with 11.r(/) 112 = 11/112 = IIr 1 (/)1I 2. 5. For I and g members of L~(R), the convolution product the property
(2rr)-1/2 I
*g =
has
:;:-1 (F(/) . F(g)) = F (r i (/)·:;:-1 (g») .
Moreover, the inner product
(I, g)
I *g
(I, g) == iR I g dx satisfies
= (F(/) • .r(g») =
(:;:-l(/),:;:-l(g)).
Exercise 10.6.2. Using Plancherel's Theorem, show that
for Re z > O. Hint: By Exercise 10.4.6,
for Re z > O. As an application of PlanchereI's Theorem, we showed (Exercise 10.4.2) that the Fourier transform of
f(X) = {
~
-1 -< x -< 1,
otherwise,
is given by
F(/)(y) = (2rr)-1/2 2 sin(y) .
y So I is a member ofL~(R)nLb(R), and F(/) is a member ofL~(R)\L~(R).
The Feynman Integral
249
We have
L: 11' .
r~~ (2n)-1/2
;:-1 (F(/)(Y») =
=
lim (2n)r-+oo
e'XY
r-+oo
=
-1
Y
hm
r
l l
-r
•
elXY
r
dy
(elY - e-iY) • dy zy
sin (x
+ l)Y) -
sin (x -l)y) d
y
-r
y
-1 < x < 1,
x
o
I
.
r-+oo
I !2
={
2 sin(y)
-r
= (2:rr)-1 lim = (27l')
ei:x:y F(/)(y) dy
= ±l '
otherwise,
almost everywhere.
Exercise 10.6.3. We showed in Exercise 10.4.1 that the Fourier transform of e-I.tl is given by {2:rr)-1/22(1 + y2)-1. Show that e-I:x:1 = (2:rr)-1/2 lim
-
1
-
:rr
1
00
-00
r-+oo
l'
+ y2)-leixy dy
-r
cos(xy) d
1+y
(2:rr)-1/22(I
2
y.
In particular, for x = 0,
.!.l
1- rr =
co
_00
1 d 1 + y2 Y
.!.rrr-oo lim [tan- 1 (r) -
tan- 1 (-r)]
Exercise 10.6.4. Suppose that I(x) = x/(l +x2), where -00 < x <
I Show that I
a. Show that
is not a member of Li:(R).
b.
is a member of L~(R).
c. Calculate F(/)(y). Hint: See Figure 5;
+00.
fcz/(l + z2)e-izy dz.
d. Is y2F(/)(y) a member of L~ (R)? Exercise 10.6.5. Suppose I(x) = 1/(1 Calculate .1'(/).
+ x 2 ),
for
-00
< x <
+00.
250
A Garden of Integrals
y
y>O
Figure 5. Contour C
10.7
Solving Schrodinger Problem A
Using Fourier transforms, let's solve (heuristically) the SchrOdinger Equation in Problem A in the form (V = 0),
a 1/1 itt a21/1 -=--, 1/1(0, x)
for-oo<xO,
2m ax 2
at
=
f(x),
with
fa 11/I(t, x)1
2
dx = I, t > O.
Assume t(t. x) has a Fourier transform:
:F(1/I(t, x))(y) = If 1/1 and
F
f
(21l')-1/2fa e-
ixY 1/I(t.
x) dx.
are nice enough, then
ea~) = :t F, F (~:~) = _y2 F(",) , F(t/!{O, x»)(y) = F(f)(Y),
and thus
a:F -in 2:F at = 2m y .
We conclude that :F(1/I) =
2 e(-ifz!2m)y t :F(f).
Then
1/I(t, x) = ;:-1 (e(-ifl/2m)y1 , :F(f))(x)
* ;:-1 (:F(f)){x) /[2di{t/m)1 * f(x)
= ;:-1 (e(-ifl/2m)y2 t ){x) =
ittt)-1/2 e-:C 2
-
( 2m
= (21f~11t) 1/2
fa
.
eUm/1Jit)(;<-xo)2
I
(xo) dxo·
The Feynman Integral
251
Note how the thlrd line recalls Ex.ercises 10.4.6 and 10.6.2. \Ve have several problems. For instance. c What does
:F- 1 (e{-itz / 2m)y'2 t ) mean when Re z = O? Note that
e-iA/2my21 is not f)
in Lb(R).
Are there any requirements on
f
besides
fR If(X)1 2 dx = I?
Ignoring these issues for the present, let's compare this ex.pression (via Fourier transforms) with Feynmants expression (via Riemann sums) as a solution of the SchrOdinger Equation, for -
1/1(0, x)
=
f(x),
with
00
L
< x <
00,
t > 0,
2
11/I(t. x)1 dx = 1. t > O.
That is, we compare
1/I(t,x)
=
. 1/2 (2rcIn) z'At
L I
i In .., exp -(x-xo)-
2ȣ
R
I
f(xo)dxo
with Feynman's expression
1/I(t. x) = lim ( n
. In(/ )
2rc I 'A t
)n12 J. dX -1 ...
. !
Il
lm
·exp
71
R
f.
R
dxo
n
2n(t/1I)~(Xk-Xk-l)
2
I
f(xo),
where x" = x, '!/F(O, x) = f(x). The common features of the integrands are encouraging.
10.7,1
Finding the Right Space of Functions
The requirement that fR 11/I(t.x)1 2 dx = 1 for t > 0 suggests that we are looking for a solution of the SchrOdinger Equation in the space of complexvalued square integrable functions on R - the Hilbert space L~(R). On the other hand, we have differentiability requirements: o What does it mean to say we have a solution of SchrOdinger' s Equation
in L~(R)? o Is there a dense subspace ofLa(R) where differentiability makes sense?
252
A Garden of Integrals
o Do the elements of this subspace (and their derivatives) satisfy growth
restrictions as x
-7
±oo?
We want a dense subspace of L~(R) consisting of smooth functions that decay sufficiently fast. The Schwartz space is our space.
10.8 An Abstract Cauchy Problem We are going to reformulate the SchrOdinger problem as an abstract Cauchy problem, an approach developed by Einar Hille ,and Ralph S, Phillips in the 1940s and 1950s. Our discussion follows Fattorini (1983), Goldstein (1985), and Johnson and Lapidus (2000). Thmk ofu(t) as the state ofa physical system 'U at time t. Think of the time rate of change of u (t), u' (t), as a function A of the state of the system 1l. We are given the initial datum u(o) I. We have u'(t) A[u], with u(O) = I. Thus the Schrodinger Equation for V 0,
=
81/1 i 11 82 1/1 - 2m 8x 2 '
for -
at -
1/1(0, x) = I(x),
with -
00
00
=
<x <
< x <
=
00, t
00,
> 0,
L
2
11/I(t. x)1 dx = I, t > 0,
can now be reformulated as a differential equation in the Hilbert space L~ (R). We have u' (t) = A[u(t)], u(o) =
for t > 0;
f
The differential operator A, with domain an appropriate dense subspace of L~ (R), is given by the differential expression d2
fit
2m dx 2 ' while u(o)(x)
10.8.1
= I(x)
and ll(t)(x)
= 1/I(t
1
x).
Defining the Abstract Cauchy Problem
In the abstract Cauchy problem we have u' (t) = A[u(t)].
u(O) = f, with the following two conditions:
for t > 0,
The Feynman Integral
253
1. The differential operator A is a linear operator from a Banach space X to itself. The domain of A is a dense subspace of X. 2. The solution of this differential equation '£l'(t) = A[u(t)] for t > 0 is a function u (t) continuously differentiable for t > O. By Condition 2 we mean that 1L
'( t ) = l'1m --..,;..-......;....---..,;.. u(t + /z) - '/.l(t) h--.O h
exists and is continuous in the norm of X. The function u(t) belongs to the domain of A, with
u(t · 11m
h~O
+ /z)h -
u(t)
-
A[ ()1 1t
t
X
0 =.
The Cauchy problem is well posed if the following two conditions are satisfied: 1. We have a dense subspace of X such that for any member IlO of this dense subspace we have a solution 11' (t) = A[u(t)] with u(O) = uo.
2. We have a nondecreasing, nonnegative function B{t) defmed for t > 0 such that, for any solution u ofu'(l) = A[u(t)],
lIu(t)lIx < B(t) lIu{O)lIx,
for t > O.
10.S.2 Operators on a (ompleJ( Hilbert Space It will be helpfuf to consider sOlne general comments regarding operators on a complex'Hiibert space 1i.. Th'e domain of a linear operator T - that is, D(T) - is a subspace. We have T(O!lXl +0!2X2) = ell T Xl +0!2Tx2 for all scalars all 0!2 and all elements Xl, X2 in D(T). The following terminology applies.
1. Bounded, A linear operator T in 1i. is bounded if there exists a constant C ::: 0 so that IITIII < C 11/11 for all f in the domain of T 2. Continuolls. Saying a linear operator is continuoZls at I in D{T) means that for every sequence {!,,} in D(T) for which II!', - III -+ 0 it follows that liT!" - Tfll ~ 0,
254
A Garden of Integrals
3. Bounded ifjContinuolis. A linear operator T is bounded iff it is continuous. If T is bounded, liTI I < C II I II, then II TIn - TIll = II T (In - I) I <
Cil/n - III·
But ifT is not bounded, we have {In} in D(T), so that I TIn II > nll/nll. For gn = (IJnll/nIDln, Ilgnll-+ 0, but IITgnll > 1.
4. Norm. The norm of T,
II Til, is
defined by for
I
a member of D(T).
5. Unitary. A bounded linear operator U of 'H. into C is unitalY iff U is isometric, (UI. UI) = (I. I), and onto. For example, the Fourier transform :F is a unitary operator. by Plancherel's Theorem.
In all that follows, 'H. will be the Hilbert space of complex-valued Lebesgue measurable functions defined on R, such that II 12 is Lebesque integrable: 1/12 dx < 00. In this space the inner product of functions I and g is defined by
JR
(/,g)
= J. Igdx. D R
11/112 = (I. I) =
fa 1/1 dx. 2
We follow tradition by denoting this space as L~(R). Convergence of In to I means that
IIfn -
!II
=
(L Ifn - fl2
dx )
1/2
--->- 0 as
11
--->-
ClO.
The space L~ (R) is a complete metric space (see Chapter 6).
10.9
So~ving in the Schwartz Space
Example 10.9.1. Given Problem A, where V =
°in the SchrOdinger Equa-
tion,
at 1/1(0, x)
2 in a 1/1 --
2m 8x 2
= I(x).
'
for with
00
L11/I(t,
<x <
00.
x)1 2 dx
t > 0,
= I,
t>
0,
we will solve the corresponding abstract Cauchy problem u'(t) = A[u(t)], with u(o) = I. Suppose I is a member of the Schwartz space S, a dense
255
The Feynman Integral
subspace of L~(R), operator
fR If(x)1 2dx
= 1, and the domain of the differential
is the Schwartz space S. This is a lengthy exploration, and we will break it irl
Palt One. Assuming u is a member of S, the Fourier given by
three parts. I
[lsform of 11 is
F(u(t)(x)) (y) = (27r)-lnl e-ixYu(t)(x) dx.
Proceeding as in Section 10.7, we have
-i'li
I
(F(u») = F(u') = F(A[ll]) = _y2 F(u), 2m
with F(u(O)(x») = F(f(x)). Solving this differential equation yields F(u) = e(-if,/2m)tY 'J. F(/). Because e(-if./2m)ty 2 F(/) is a member of S, we may conclude that u(t)(x) =
;:-1
(e(-ifl/2m)y'lt F(f)(y»)
(x),
with u(O)(x) = I{x).
Also, u(t)(x) is a member of S, the domain of A, by Plancherel's Theorem (Section 10.6.1). The reader should verify that
solves the SchrOdinger Equation with V rem, 1/I(t, x) is a member of S. That is,
= O. Hint: By Plancherel's Theo-
is a member of S. Show that
a'l/l ::: (2Jl')-1/2
at
2
a
1/12 ::: (2Jl')-1/2
ax
J. J.
R
R
-ifi
y2 ei.'Y:Y
[eC-ifl/2mh2t F(/)(Y)] dy
2m
_y 2 eixy [e(-iIl/2m»)'2 t :F(/)()')] dy.
and
256
A Garden of Integrals
Furthermore, 1/1(0, x)
= :F- 1 (:F(/) (y)) (x) = I{x)
and
L
11/I{t,x)1 2 dx = (",,(t,x), ",,(t,x)}
=
(rl
(e(-i t, /2m)y2 tF(f»). (rle<-tM2m)y2t:F(f»))
= (e(-ih/2m)y2, :F(f), e(-ifJ/2m)y 2t :F(f») = (:F(f), :F(f)} = (f, I)
LI/(x)1
=
2
dx = 1
for t 2:: O.
Part Two. Next we will verify that, given
l.i(t)(X) = :;:-1 (e(-if,/2m)y2 r:F(f)(y») (x), u solves the abstract Cauchy problem in the sense of our definition (Section 10.8.1). Assume I is a member of S. By Plancherel's Theorem, u is a member of S and we can calculate
u(t
+ h) -
u(t)
h eC-i11/2m)Y2CI+II) _ e(-ifJ/2m)y'J,
= :;:-1
h
(
)
:F(f)(y)
=:;:-1 (e<-itl/2m)Y2(t+h/2).
(-2isin(fmY2~)) .~ y2 . :F(f)(Y») 2m 2
-1Ly2l! 2m 2
'-:::0 r l (eC-iA/2m)y21 . -i'h. y2 :F(f)(y») 2m
=
rl
-i'h. ) ( 2m y2 :F(u(t» =
r l ( F(A[u(t)]) )
-
A[u(t)].
Also, for any solution v,
(A[v1, v) = (:F(A[vD, F(v») = (-i'h. y2 :F(v). :F(v))
2m
-i'h. = -2
m
Thus Re (A[v], v) = O.
f. y 2 1:F(v)1 2 dy. R
The Feynman Integral
257
Because
d dt
IIvll2 =
2Re (dV dr' v )
= 2Re (A[v], v) = 0,
JR
1/12 dx = 1, for t > O. So for I we conclude that IIvll2 = Ilv(O)f = a member of 5, :F- 1 (e(-ifz/2m)y2 r :F(/») (x) solves the abstract Cauchy problem.
Part Three. How does this solution compare with our heuristic solution in Section 10.7? Recall that we arrived at
2) m )1/21 ex.p (im (21fitzt 2tzt (x - xo) 1(·"0) dxo. R
We need an integral representation for :F- 1 (eC-ifl/2m)y2t :F(f)(y») (x). From Plancherel's Theorem (Section 10.6.l) and Exercise 10.6.2 we know that if f is a member of (R), then for Re z > 0,
L'a
rl
(e(-zL.;2)y2 :F(/)(Y») (x)
= rl
2
(:F(Z-1/2 e-X /2Z)Cy) . :F(/)(Y)) (x)
--~, (",_)-1/2 Z -1/2 e-x'l/2z..~:c I( )
= (21r)-1/2
fa
e-(x-;co)2/2Z f(xo)
dxo.
for Re z > O. Select a sequence Zn -)0 (itz/m)t for t > 0, with Re (Zn) > O. Because :F- 1 is a bounded linear operator on the complete nonned linear space (R) (by Plancherel's Theorem), it follows that
La
rl
(e(-ZII/2)y2 :F(f)(Y»)
-)0
rl
(e(-ifl/2m)y2 r :F(f)(Y»)
(x)
in L~ (R), and we have subsequence {z"A:} such that pointwise
almost everywhere in x (Theorem 6.5.4). Assuming I is a member of L~ (R) n L~ (R), the Lebesgue Dominated
258
A Garden of Integrals
Convergence Theorem 6.3.3 gives us
r l (e(-fft/2m)y2 t :F(f)(Y») (x) = lir;:--l (e(-Zllk/ 2)Y'l . :F(f)(Y») (x) - liFFl (:F (z;;;/2e-x2/2ZRIc ) (y) . :F(f)(y») (x)
* f(x)
-
lirz;;;'/2e-X"J./2Znk
-
l~ (2.1rZn ,J-1/21 e-{x-xo)2/2Znk I(xo) dxo
-
(
~
) 1/2
2.1rztlt
J.
e(im/2'At)(x-xo)2
R
/
(xo) dxo l
I
almost everywhere in x. Note that for Re z > 0, Z =F 0, le-(x-xof / 2Z < 1, with
I
a member of
rl
(e(-ih/2m)y2 t :F(f)(y»)
= (
m )
2.1r i'h t
1/21.
(x) e(im/2ttt)(:c-xo)2/(x ) dx
ROO,
almost everywhere in x, with the assumption that f is a member ofLa(R)n La (R). (The integral is an ordinary Lebesgue integral.) Since S c La (R) n La (R), we may conclude that the solution of the SchrOdinger Equation (Section 10.1.1) or the solution of the abstract Cauchy problem (Section 10.8.1) bas the two representations,
u(t)(x) = ;:--1
=
(e(-ifz!2m)y2 t :F(/)(Y»)
(x)
(2::"1) 1/21 exp (:: (x - Xo)' )
f(Xo} dXQ
almost everywhere in x, for finS. (The integral is an ordinary Lebesgue integral.)
10.9.1
Extending the Solution of Problem A
Let's try an extension of what we mean by a "solution" of Schrodinger Problem A, based on Johnson and Lapidus (2000, pp. 166-168).
E)cample 10.9.2. We will enlarge the domain of A to a subspace ofLa(R) containing S.
The Feynman Integral
For
f
259
a member of S we showed that
A[/}
i11 = ?_111
d
2
I =;=-1 (-itI ) -?_y2 F(f)('I) _m
-d ? x-
(x).
However, the operator :F- 1 (-i11/2m)y2 :F(/) (y)) is defined when y2 F(/) is a member of L~(R) or when (l + y2)F(/) is a member of L~(R). So we extend the operator A:
-= ;:-1 (-Uz/2m)y2 F(f))
A[f]
with domain D(A) = {g E L~(R) 1 (1 second derivative of f>
+ y2)F(g)
E L~(R)}.
The ordinary
~~~, is being replaced by the operator
with the requirement that (1 + y2)F(/) is a member ofL~(R). The domain D(A) is called the Sobelev space of order 2. So how does this work? We have :F (f(X
where III h(e ihy as h -;.. O. So
f(x
+ I~ 1)1
<
f(X)) (y)
lyl
+ 'iz - f(x)
< Jl
= :, (ejyh -
+ y2
J)F(f)(y),
< 1 + y2 and
-7;:-1 (iyF(f)(y))
(x)
II h(eiylz-l) -? iy in L~(R).
We have a subsequence {h n } converging to 0 so that [f(x + h n ) l(x)}1 h n converges pointwise almost everywhere in x as h n -? O. If I is a member of S, then f'ex) = ;::-1 (iYF(f)(y)) (x). Similarly, f"(x) = F- 1 (iyiyF(/)(y)) (x). Sometimes we say that F- 1(iy:F(/)(y)) and :F- 1( - y2 F(/) (y)) f O) and 1(2), respectively - are distributed derivatives, or weak derivatives, or L2 derivatives, and so on. They agree almost everywhere with the ordinary derivatives I' and I" when I has such derivatives and they make sense as operators on a dense subspace of L~ (R) that contains S, the so-called Sobelev space: {g E L2(R)
I (1 + y2) :F(g)
E L2(R}
= D(A).
260
A Garden of Integrals
We note that for J in the domain of A and g an infinitely differentiable functi On of compact support,
(J(1). g}
= {Fl (iy.r(J)(y»), g} = (iy.r(J) (y), .r(g)} = -(.r(J)(y), iy.r(g)} = _(/.Fl(iy.r(g)(y»)) = -(/.8')
and
(J(2). g}
= (rl ( - y2 .r(J)(Y)) , g} = (J. Fl( - y2 .r(g)(y»))"=
(J, gil).
U
That is, the "integration by parts formulae,
fa J(l)(x)g(x)dx = - fa f(x)g'(x)dx fa J(2) (x)g(::c) dx = fa J(x)g" (x) dx,
and
hold for all infinitely differentiable functions g of compact support. Thus u(t)(x) = F- 1 (e(-iIi!2tn)y2 r .r(J)(y)) (x) makes sense as a solution of the abstract Cauchy problem (Section 10.8.1)for {g E L~(R) 1(1 + y2).r(g) E LE (R) }, a subspace of LE (R) that contains the Schwartz space S. Extending the operator A as previously indicated, we have
F(A[g1)
' 11
= _l2Y
2
.r(g).
111
Exercise 10.9.1. Show that u(t)(x) = F- 1(e(-itJ/2m)y2 t .r(J)(y») (x) for J a member of {g E L~(R) I (1 + y2)F(g) E LE(R)} solves the abstract Cauchy problem (Section 10.8.1). Hint: By Plancherel's Theorem (Section 10.6.1),
u(t
+ .6.t)(x) -
u(t)(x)
.6.t
=
e(-ilJ/2m)y2(t+At) _ e(-iIi!2tn)y2 t ) FI ( .6.t . .r(J)(y) (x)
-if1y 2.r(1l(t») ) = F1(.rA[u(t)]) ~ Fl ( 2m since
.r(u(t») = e(-ill/2m)y2 t F(J)(y) and y 2.r(u(t») = e(-ifl/2m)y2 r ),2 .r(J)(y) E LE(R),
= A[u(t)],
The feynman Integral
because
I
E
261
{g E L~(R) I (l + yl)F(g)
e LE(R)} Also, for any solution
v, l
'f, (A[v], v) = (.r(A[vD • .r(v») = --2
J.
In
Re (A[vl, v)
= 0,
R
y
2
1.r(v)1 1
dy;
and
= ({v, V})f = 2Re (v', v) = 2Re (A[vl. v) = O. The element IIvlli is a constant: Ilv(t)lI~ = IIv(O)II~ = 1. (IIvll~)'
Compare this solution of the abstract Cauchy problem
rl for
(e(-ifz/2m) yl t F(/)(Y)) (x)
a member of {g E L~(R)
I
I (1 + y2).r(g) e L~(R)}
(2:~hl) III fa exp
G:
with
(x - xo)') f(xo) dxo.
Mimic the argument of Example 10.9.1, starting with Part Three and I a member of Lb(R) n LE(R). Use Lb(R) to apply the Lebesgue Dominated Convergence Theorem 6.3.3. Conclude that u(t)(x)
= rl
(e(-11I/2m)y2 t F(/)(Y») (x)
171
= ( 2n:if,t
) 1/2
J. ( m
I
a member of
almost everywhere in x I for
R exp
i 2ht (x -
xo? )
I(xo) dxo
(The integral is all ordinary Lebesgue integral.)
Example 10.9.3. We wanf to remove the requirement that ~~OO.
I
'
be a member
Note first that the integral
(2::"SI' fa = (,m I
=
exp
G~: (x -
XO)2) f(xo) dxo
)1/2 e(im/lt,t)x 2 . lim (2n:)-1/2
ftt
(/:t)
r-+oo
r
e<-imx/fit}Xo[eimxij/lfll I(xo)]dxo
J-r
1/2 eimx2/2tlt • F(e(lm(.)2/ 1tll 1(,))
(;~ x)
262
A Garden of Integrals
makes sense in L~(R) ("mean"), as does :F- 1 (e(-ih/2m)y2 t :F(f)(y)). Now truncate. Given
fr(x)
== { f(x) Ixl
then fr is a member of Lb(R)
o
< ': otherwise,
n L~(R), and
u(t)(x) = p i (e(-iht/2m)y2 :F(f)(y») (x)
= lim p I (eC-ila/2m)y2t :F(fr ley») (x) r~oo
= r~oo lim ( In) .tz 1/2 2TC I t
i ( R
m ) 1/2 [ = ( 2TCint JR exp for f a member of {g E L~(R)
10.9.2
(
i nz exp ~(x - xo) 2nt
im 2tLt (x - xo)
(by 10.6.1)
2) fr(xo) dxo
2) f(xo) dxo.
I (1 + y2):F(g)
E L~(R)}.
A Theorem
In Examples 10.9.1, 10.9.2, and 10.9.3, as well as Exercise 10.9.1, we developed the following theorem. Theorem 10.9.1. Given the abstract Cauchy problem u'(t) = A[u], with u (0) = f the solution u is given by I
u(t)(x) = p i (e(-it&/2m)y2 t :F(f)(y)) (x)
= (2Jr~hJ 1/2
i
exp
G~ (x -
xo)' ) f(xo) dxo
almost everywhere in x, for the following conditions:
it, d 2
.
1. A[f] = 2m dx 2 f. D(A) IS the Schwartz space, f and u are members 01 the Schwartz space, and the "integral" is an ordinary Lebesgue integ7·al. 2. A[/] = p I
(-itz y2 :F(f)(y»), 2m
{g E L~(R) I (1 + y2):F(g) E L~(R)}, f is a member of Lb(R) n D(A), and the "integ74al" is an ordinalY Lebesgue integral. D(A)
=
The Feynman Integral
263
3. A[f] = :F- 1 ( ,tfz y2 F(f)(Y)) , _Tn
=
D(A) {g e LE(R) I (1 + y2):F(g) E LE(R)}, D(A), and the integral "mean" is limJ~r'
10.10 10.10.1
f
is a member Df
Solving Schrodinger Problem B Prelude to Problem B
We are almost ready to tackle Schrodinger Problem B, for V observe that for f in L~(R),
rl
#-
O. First,
(e(-ifl/2rn) ylt :F(f)(y)) (x) =
(
111
2rritzt
) 1/2
f.
R exp
(im ?) 2fzt (x - xo)- f(xo) dxo.
=
with "meann integrals: JR limroo+ CXl J~r' Secondly, the L2 interpretation solves the abstract Cauchy problem (Section 10.8.1) with V = 0 via Fourier transforms, observing the requirement that f be a member of
{g e L~(R) I (1
+ y2):F(g)
e
La (R)}
and fR Ifl2 dx = 1. More ex.plicitly, define a family of mappings {F (t)}, for t ::: 0, on LE (R) into L2(R) by F(O)f = f and
(F(t)f)(x) = :;=-1 (e(-itZ/2m)y2 t :F(f)(y)) (x)
= (, 1~fi ) 1/2 ... rrz It
f. (;:2 R
ex.p
.... nt
(x - XO)2) f(xo) dxo .
Then
F(DFG)! = FG) (r
l (e(-1t./2m)y2(1/2}F(f)))
= F- 1 ( e(-i/./2m) ,2 (I 12) F(F- 1 (e(-iA / 2m)Y'(I/2) F(f)) ) )
=r
1 (e C- Zfl/2m)y'J. t :F(f))
= F(t)f.
264
A Garden of Integrals
Generally, F(t)f = (F(t/ll))n f. and thus
which is Feynman' s solution of SchrOdinger Problem A - assuming that f is a member of {g E L~(R) I (l + y2).r(g) E L~(R)} and JR Ifl2 dx = 1. SchrOdinger Problem B remains, the situation with V ~ 0, nonconstant coefficients.
10.10.2
Trotter's Contribution
In 1964, Edward Nelson put all the pieces together using the Trotter Product Formula (Trotter 1958, 1959). A proof may be found in Johnson and Lapidus (2000, pp. 200-201).
Theorem 10.10.1 (Trotter Product Formula). If A, B, and A + B generate (Co) contraction semigroups T. S, and U on a Hilbert space 1-f., then
.~ (T G) S G))" f
= U(t)f,
for f a member of 'H.. Trotter's product formula will provide a solution of Schrodinger Problem B, but first we have many items to define and discuss. These considerations lead to another method of solving abstract Cauchy problems, without the requirement that we have constant coefficients. For this approach, semigroups of linear operators, the main references are Goldstein (1985) and Johnson and Lapidus (2000).
10.10.3
Semigroups of Linear Operators
Assume a linear map T(t) maps f(x), the solution at time 0, to the solution at time t. We write T(t)f = u(t). Thus T(l
+ s)f = u(t + s) = T(t)u(s) = T(t)T(s)u(O)
= T(t)T(s)f
The Feynman Integral
265
(we hope), and T(O)f = f. We have a collection of linear operators in the Hilbert space L~(R) so that T(O) = 1 and T(t + s) = T(t)T(s). These ideas lead us to the notion of a semigroup of linear operators. Our discussion begins with the approach laid out by Martin Schechter (2001). Suppose we want to solve the differential equation li' (t)
= Ali(t).
u(O)
= 1l0.
for t > 0, where A and 110 are constants. Of course u(t) = etAuo is a solution for t ~ O. As before, we will abstract this problem to a new setting. Suppose we have a Banach space X, and to each real variable t we assign a function u (t) in X. We can think of A as a mapping of X into X. That is, for u in X, Au belongs to X. We interpret u' (t) = Au (t) as
u (t
+ h)lz -
u (t)
A () ut
-
-40
as
h
0
-4.
In this new setting, can we make sense out of et A when A is an operator? Naturally, we try erA
= 1 + ~tA + ~t2 A2 + ... 21
l!
'
and with u(t) a member of X, we have
u(t)
= e'''I/(O) = (I + :!tA +
;t
A' + ... ) 1/(0).
Thus
u(t + /z) -lI(t) e(t+h)A --.....;.....-......-.;.., = Iz h and
u'(t)
=
lim
elLA
11-+0 (
-1)
h
erA
u(O) =
etAll(O)
(eitA
-1) lz
etA 1£ (0) ,
= AetAu(O) = Au(t).
More questions suggest themselves: o
Can we justify these manipulations?
o Does
(1 + frtA + ~t2 A2 + ... ) u(O)
o Does e(l+h)A
=
etA. e'lA
=
t1rA • e( A
converge to an element of X?
make any sense?
266
A Garden of Integrals
Suppose A is a bounded linear operator in X and thus continuous. Then for N > M and t > 0, N
L
1
M'
Defining etA by I for t > O. Next,
1
N
L k' tkllA.ll k
k k' t Ak <
-4
0 as M t N ~
00.
M'
+ I! t A + i! t 2 A2 + ... I
IIA
e
J1
makes sense, and
II et A II
< et 11 A II
1
00
= '"' _hk-1 Ak L..J k' t l'
and ehA -
lz
I
- A
1 ehliA II - I k 1 k! /zk-l UAll - = It
IIAU ----T 0
as h
-4
O.
= etA. ehA = ehA • etA.
Is
2
Thus u'{t) = AetAu{O). But we still need to discuss whether e(t+Ia)A it true in some sense that
I: ~!(l +h)kAk = (I: ~!lmAm) (I: >nAn) = (I: ~!hnA") (I: ~!tmAm)? F or real numbers,
(x
~
+ yl
{x + y)k k!
k!
k
= '"'
~ m!(k - nz)! (
=
xn) (
~ n!
xmyk-m ym)
~ m!
How does tills fare in our setting? We have
.
and
The Feynman Integral
267
Thus
LN m=O
m
LN
In
_t Am. _1 An ml 0 11! 11=
LN k
=0
1
-(t k!
+ hl Ak
m,n::,N
n+m>N
< n,m-s.N
1Z+m>N N
= '"' L.,
m=O
~
_1 tmllAllm.
m!
N
N
n=O
k=O
'"' ~hnIIAnn- '""' ~(t+h)kIIAllk L., III L., Ie!
etliAllehllAIl _ e(t+h)IIAII =
o.
This is very encouraging. Unfortunately, in our particular application A is not bounded; A is a differential operator. However, loosely speaking (A as a limit of bounded linear operators) linearity of A on a dense subspace of X is good enough.
10.10.4
Serrligroup Terminology and a Theorem
Before we proceed, it is necessary to make some definitions concerning a family T = {T(t) : 0 < t < oo} of everywhere defined bounded linear operators from a Banach space X to itself.
Definition 10.10.1 (Semigroup). The family T is called a (Co) semigroup iff 1. T{t
+ s) = T(t)T{s)J, with tiS >
O. for all J in X.
2. T(O)J = J.
3. The map from [0.00) to X. t X for each J in X.
~
T(t)J is continuous in the nonn of
The third criterion is caned the strong operator continuity of the semigroup: It - sl small implies II T(t)J - T(s)J Ilx = I (Tet ) - T(s)) f Ilx small.
268
A Garden of Integrals
Definition 10.10.2 (Contraction Semigroup). The family T is called a (Co) contraction semigroup if, in addition, a fourth criterion holds: 4. IIT(t)fll::: IIfll for all t > 0 and all f in X.
Definition 10.10.3 (Generator ofa Semigroup). The generator A ofa (Co) semi group T is defined by A[f]:E lim .!.[T(t)f - f] = lim (T{t) - l)f t-+O+ t t-+O+ t
= T'(O)j,
where the domain of A, D(A), consists of those elements of X for which this limit exists. A (Co) semigroup with generator A solves the abstract Cauchy problem, in the sense we defined it in Section 10.8.1. We can malce sense out of
u(t) = etAuo. Theorem 10.10.2. Let {T(t)} be a (Co) semigroup on the Banach space X with generator A, and let f be a member of the domain of A. That is, A[f]
==
lim T(t)f - f =
t
t-+O+
~T(t)f dt
1_ = t-O
T'(O)f
exists. Then the domain of A is a dense subspace of X, and u(t) == T(t)f is the unique continuollsly differentiable soilltion on [0,00) of the abstract Cauchy problem for t ::: 0, u'(t) = A[u(t)],
u(O) =
f
Proof. Details can be found in Goldstein (1985) and in Johnson and Lapidus (2000). We argue only existence, in six steps. Step I. u(O)
== T(O)f =
f.
Step 2. The domain of A is a linear subspace of X, and A is a linear operator on this subspace of X. (By definition we have linearity of T on the Banach space X.) Step 3. The domain of A is a dense subspace of X. Let f be a member of X. Form the Riemann-type integrall/t J~ T(s)f ds, for t > O. These Riemann-type sums converge in the norm of X. This integral makes sense because s -+ T(s)f is a continuous map form [0,00) to X by assumption.
269
The Feynman Integral
Let h be greater than
T(h)
o. Then
Glot
(~ 10' T(s)f dS) -
T(s)f
dS)
It
t1 11h 10[t T(h +s)J ds - h1 10[t T(s)f ds t1 -?
11h I +
t h
I
I
[h T(u)J du - h1 10 T(s)J ds }
T(t)J - J as h
-?
0+ .
t
Thus lIt J~ T(s)J ds belongs to the domain of A, and
On the other hand,
~ I.t T(s)f ds -? T(O)J = t
J as t
-?
0+.
0
Given an arbitrary element J of X, we have an element in the domain of A, 1/ t J~ T(s) J ds arbitrarily close for t sufficiently small. The domain of A is a dense subspace of X.
Step 4. The operator T(t) maps the domain of A into A, and T(t)A[J] = A[T(t)J] for J in the domain of A. Suppose f belongs to the domain of A. We show that Tet)/ belongs to the domain of A:
A[T(t)J]
=
lim T(h)T(t)J - T(t)J
h
11-+0+
=
lim T(t) (T(/Z) - I)J h-+O+
h
= T(t)A[J]. We know that
lim (T(/Z) 11-+0+ h
I) f = A[J]
because by assumption J belongs to the domain of A. Therefore the limit exists and A[T(t)/] = T(t)A[J]. The family T commutes with its generator A
270
A Garden of Integrals
Step 5. For any interval [a, b] of [0, (0), we have a constant M such that IIT(t)11 < M for all t in the interval [a, b]. Because T is a (Co) semigroup. s -7 T(s)f is a continuous function from [0, (0) to X for a fixed f E X. So S -7 IIT(s)fll is a continuous function from [0, (0) to R for a fixed
f EX. We have a constant MI, so that IIT(s)fll < MI with a ;:::: s < b, for each f EX. By the Uniform Boundedness Principle. we have a constant M so that II T(s) II < M for a <s < b.
=
= T(t)f
Step 6. We claim u(t)
is a solution of Ui(l) A[u(t)] for any f in the domain of A. That is, for any member f of the dense domain of A, we have a solution, T(l)f, of the differential equation u' = A[u]. The argument is as follows. Let f belong to the domain of A, and suppose t > O. Then lim T(l
+ It)f -
T(t)f
=
h
h--.O+
lim T(h) - I T(t)f h--.O+ h
= A[T(t)f],
because T(t)f belongs to the domain of A by Step 4. Furthermore.
lim T(t)f - T(t - h)f h
=
h-+O+
lim T(t _ h) (T(h) - I)f h
= T(t)A[f],
h-+O+
because f belongs to the domain of A by assumption. For the last equality, observe that
T(t - h) (T(h) - I)f - T(t)A[f]
x
h < T(I -II) { (T(II\-
<M
(T(II\- 1)1
1)1 - A[flJ x + I (T(I - II) - T(I») A11111 x
+ II(T(I - h) -
- All]
T(I») A 11111 x .
x As t --)0. T(t)A[f] is a continuous function from [0, (0) to X, this allows us to estimate (T(t - h) - T(t))A[f] By definition,
I
Ilx'
lim (T(h) - I)f = A[f], 11--.0+ h because f belongs to the domain of A. Since by Step 4 T(t)A[f] A[T(t)f]. we may conclude that (T(t)f)' = A[T(t)f].
The Feynman Integral
271
Thus u(t) == T(t)f is a solution of the differentiatial equation 11'(t) = Arll(t)]. In fact, because (T(t)f)' = T(t)A[f], the function u(t) is continuously differentiable for t > O. 0 We have shown that ll(t) == T(t)f is a solution of the abstract Cauchy problem according to our definition (Section 10.8.1).
10.10.5
Some I\lotes on Our Solution
Note 1.
T(I)! -
! == ==
for
f
l' :, l'
(T(s)!)ds
==
l'
A[T(s)J] dx
T(s)A[J] ds
in the domain of A.
Note 2 Suppose the sequence {fn} of members of the domain of A converges to f; that is, suppose IIfn - fllx -? O. Likewise, say the sequence {Aj;z} converges to g, so IIAfn - gllx ~ O. We claim that f belongs to the domain of A and that Af = g. By Step 5 we have IIT(t)1I < M for 0::: t .:::: T. To show that f belongs to the domam of A, we fonn the quotient
T(l)f - f t
= lim T(t)fn
1t
1t = - 1t
- fn = lim ~
t
t
T(s)A[J,,]ds
0
1 = -1 T(s) IimA{J,,]ds t o t
T(s)g ds.
0
We can do so because
-1
1t
t
0
[T(s)A[fn] - T(s)g]ds
x
I.t T(s)[A[J,,] - g]ds t
= -1
0
1 {t
< Mt
10
X
IIA[J,,]- gllx ds
< lItI II A [J,,] - g II x .
Thus lim Tet)f - f = lim ~ r-+o+ t t-+O+ t
t
10
T(s)gds = T(O)g = g.
Hence f belongs to the domain of A, and Af closed operator on the Banach space X.
= g.
We refer to A as a
272
A Garden of Integrals
Note 3. The solution u(t) == T(t)f is the unique solution to the abstract Cauchy problem u'(t) = A[u(t)], for t > 0 and Il(O) = f for f in the domain of A.
10.10.6 Applying the Theorem To aid our understanding of Theorem 10.10.2, we will mimic the argument for a specific T. For f a member of L~(1?), define for t > 0,
T(O)f
=!
We proceed as before. Step 1. By Plancherel's Theorem (Section 10.6.1). T is a linear operator from L~(1?,) into L~(1?,). Step 2. Again by Plancherel's Theorem,
II T(t)f 112 = ('r-1 (e(-ifz/2m)y2 t :F(f») r l I
(e(-ifr./2m)y2 t :F(f») )
= (e(-itl/2m)y'-t :F(f). e(-i#Z/2m)y2 t :F(f)) = (:F(f), :F(f») =
IIfl1 2 ,
We have that T is bounded: IIT(t)1I = 1 for all t > O. We have a family T = {T(t) : 0 < t < oo} of everywhere defined bounded linear operators in L~(1?,). Step 3. We calculate
(T(t)T(s)f)
= T(t) (rl (e(-ifl/2m)y2 :F(f»)) S
(rl (e(-th/2m)," 1'(J»)) )
=
r
1 ( e(-tA/2m),2, l'
=
r
1(e(-ifl/2m)y2{r+s):F(f»)
= T(t + s)f.
Step 4. PlancherePs Theorem gives us
T(t
+ 6.t)f =
T(t)T(6.t)f
A-;:!:O r
l
=
r l (e(-ifA/2m)yl(I+At) :F(f»)
(e(-ifl/2tn)y2 t :Fef») = T(t)!
273
The Feynman Integral
"Ve have a (Co) contraction semi group T. The generator of T is given by
A[f] == lim T(nf - f , t~O+
t
with the domain of A being those elements of L~ ('R.) for which this limit makes sense. However (again invoking Plancherel's Theorem), lim
rl
=
lim
t~O+
eC- i lr./2m)y2 t ((
r1
t ....... O+
=
r
for (1
-1) ) :F(f)
t
e(-ifl/2m)y2
(
[
(
.~.
?t)
sm _10 y-2m 2
'h ? t -y-2m 2
2
.
]
~ ~ :F(f)
)
2m 2
) 1 -itt ( 2m y2 :F(f)
+ y2):F(f)
in L~(R). Yes, A [fl
for D(A) = {g
E
L~(R)
= :r' (
2::
I (1 + y2):F(g)
y2 FU) )
E
L~(R)}.
Step 5. The domain of A is a (dense) subspace of L~(R). It contains the Schwartz space. Step 6. We noted that
A[T(t)f] =;:-1
(-itl 2m
y2 :F(T(t)f))
=
:r' (2:>2 F (:r 1 (e(-lA/2m,,>, F(!)) ) )
=
rl (-tn y2 e(-it.!2tn)y2 t :F(f)) 2m
=
rl (e<-itl/27n)y2 t • -i'h y2:F(f») = 2m
and so on. Thus for f a member of {g
1.l(t)(x)
=r
E
L~(R)
T(t)A[f],
I (1 + y2):F(g)
E
1(e(-itJ/2m)y2 t :F(f)y) (x)
= ( n~
21rlt1t
) 1/2
J. exp ( i ~l (x - XO)2) f(xo) R
211t
dXQ
L~(R)},
274
A Garden af Integrals
(almost everywhere in x, "mean" integral) solves the abstract Cauchy problem
~~
f.
= A[u] with u(O) =
Step 7. For
f
a member of La(R), for t > 0,
(T(t)f) (x)
== r l = (
(e(-ifz I 2m)y2 1:F(f)(y))
m
2rri'ht
) 1/2
r
Ja
(x)
e( im /2ttt)(x- xo)2
f(x ) dx 0
0,
and T(O)f = f. Thus the family T is a (Co) contraction semigroup on La(R), and the generator of T is A.
10.10.7 Problem B and the Trotter Product Because SchrOdinger's Equation
81/t = ~ 8 1/t _ ~V1/t 2
2m 8x 2
8t
'II
and the operator A dealt with
we are led to the operator of multiplication, B = (-ij'h)V. Define
(S{t)f}(x)
= e(-llfi)V(x)1 • f(x) D
for t > 0, and S(O)f = f for f a member ofL~(R), with V a real-valued Lebesgue measurable function.
1. S is a linear operator from L~(R) into L~(R). 2. IIS{t)fIl 2 =
Ilfll2, S
3. Set
+ s)f =
4. Set
+ ~t)f =
is bounded, and IIS(t)11 = l.
S(t)S(s)f. S(t)S(~t)f = e(-tlr!2m)V(I+dt) • f -+-dl-+O S(t)!.
We have a (Co) contraction semigroup S. The generator B of S is given by
B[f]
=0
lim S(t)f - f = -i Vex)! t 'h.
1-+0+
The Feynman Integral
275
The operator B is linear, and its domain, a subspace ofLa(R), consists of the elements of La(R) that, upon multiplication by V, remain a member of LE(R). The domain of B contains those infinitely differentiable functions of compact support, a dense subspace of L~(R). We have two contraction semigroups on L~(R). For f a member of
L~ (R), TCt)f = :F- 1 (e(-ih/2m)y2 t .1=') with generator
A[f] '" r l (2:>'F(f)) , and DCA) = {g E L~(R) I (l + y2).1='(g) E La(R)}. We have S(t)f e(-i/fl)Vt • f, for V a real-valued Lebesgue measurable function with generator -i
TV. f,
B[f] =
and D(B)
= {g E L~(R) I V . g E L2(R)}.
Form T(t/n)S(t/n)f:
(T G) s G) J ) = T
=
G)
(eHlt')yf!ln)
(2"'i~lln)
r/. L
dxo exp
m ) 2rr itl(t / 1Z)
. exp
J) (Xl)
(2tz~~1!) (Xl - xo)·)
tzi V(xo) ~) J(xo)
. exp (
=(
(Xl)
1/2
1.
dxo
R
(!..tz (~(XI - ~"CO)2 2t / n
V(xo)~)) f(xo) 11
almost everywhere in XI' In general, for f E La (R),
([T G) s GH J) (x.. ) = (
In
2rritl{t /11)
)"/21.R
l. III
n "'\'
_tlZ
1
dXn-l'"
. exp ( ?'h( / ) L)Xk -
1.
R
dxo
• 2 x'c-d -
'-t " ) V(xk-d f(xo) L
~
""
L.,
"Ill
276
A Garden of Integrals
defines a function in L~(R), with V a real-valued Lebesgue measurable function. Consider the operator A + B with domain D(A) n D(B). o Is this operator the generator of a (Co) contraction semigroup? E)
Is D(A)
n D(B)
a dense subspace of L~(R)?
T. Kato (1951) showed that, in particular, if the real-valued Lebesgue measurable function V is in Li(R), then D(B) ::> D(A), and the operator A+B is the generator of a (Co) contraction seroigroup. Apply the Trotter Product FOImula (Theorem 10.10.1):
([r G) s G)]" f) _
(
m
2tri1z{t/n)
(x)
)n/2 [ dXn-l'" JR
( dxo
JR
im ~ 2 i t ~ ) . exp ( 2'h{t/n) ~(Xk - Xk-l) - h n ~ V(Xk-l) f(xo) is a member of L~(R), say y,n(t,x), almost everywhere in x = XII, with "mean" integrals. Thus we have a function in L~(R), y,(t. x), and L~(R) convergence. That is,
l1 1frn(t,X) - y,(t,x)11 2
-70
h
h
as n -700.
So, lim
n-+oo (
111
2tri'h(t/n)
)n/2
R
. 111
exp
(
2'h(t/n)
dXn-1 . . .
n
~(Xk -
R
dxo
2
Xk-l) -
l.
t
n
h 11 ~ V(Xk-l)
)
I(xo)
solves Schrodinger Problem B under the assumption that 1 is a member of
fR
V is a member of L~(R), and 1/12 dx = 1. Compare with Feynman's solution to Schrodinger Problem B. This concludes our treatment of the Feynman integral.
The Feynman Integral
10.11
277
References
1. Fattorini. Hector. The Cauchy Problem Reading, Mass.. Addison-Wesley, 1983. 2. Feynman, Richard P., and Albert R. Hibbs. Quantum lv/echanlcs and Path Integrals New York: McGraw-Hill, 1965. 3. Goldstein, Jerome. Semigroups of Linear Operators and Applications. Oxford University Press, 1985. 4. Johnson, Gerald, and Michel Lapidus. The Feymnan Integral and Feynman 's Operational Ca/cll/us. Oxford University Press, 2000 5. Kato, T. Fundamental properties of Hamiltonian operators of SchrOdinger types. Transactions of the American i'vlathematical Society 70 (1951) 195211.
6. Nelson, Edward. Feynman integral and SchrDdinger equation. JU1l1'llai of~Math ematical Physics 5 (1964) 332-343. 7. Schechter, Martin. Principles of Functional Analysis. Graduate Studies in Mathematics, Vol. 36. Providence, R.I : American Mathematical Society, 2001. 8. Trotter. Hale F. Approximation of semigroups of operators. Pacific Journal of Mathematics 8 (1958) 887-920. 9. - - . On the plOduct of semigroups of operators. Proceedings of the American klathemgtical Society 10 (1959) 545-51. 10 Weidmann, Joachim. Lineal' Operators in Hilbert Spaces. New York: SpringerVerlag, 1980
No matter how far we go into the future there will always be new things happening, new information coming in, new worlds to e.;r.plore, a constantly expanding domain of life, conscioZlsness, and memory_ -
Freeman Dyson
Abbey, Edward, quoted, 45 absolutely continuous function, 102 Archimedes, 6, 29 axiom of Eudoxus, 5 Bacon, Roger, quoted, 75 Billingsley, Patrick, continuous nowhere differentiable function, 43 Borel cylinder, 210 Borel sets, 93 Borel sigma algebra, 93 bounded convergence theorem, 121 bounded variation, 101 Bronowski, Jacob, quoted, 45 Brown, Robert, 205 Cantor set, 65 Carath~odory,
Constantin, 158 Carath~odory's measumbility criterion, 89 Carles on, Lennart, 21, 15] Cauchy, Augustin-Louis, 11, 29 Cauchy critenon for H-K integrability, 184 Cauchy integral, 33 Chapman-Kolmogorov equatIon, 214 Cousm's lemma, 175 Darboux integrability criteria, 51 Dini derivatives, 104 Dirichlet, Gustav Peter Lejeune, 41 Dirichlet's convergence theorem, 41 dommated convergence theorem, 125 Dyson, Freeman, quoted, 278 Egoroff's theorem, 100 Einstein, Albert, 206 Eudoxus,4 Eudoxus' axiom, 5
Euler's summation fonnula, 82 Fatou, Pierre, 124 Fatou's lemma, 124 Fennat, Pierre de, 29 Feynrnnan,FUchard, 26,235 Finkel, Benjamin Franklin, quoted, 111 Fourier transform, 244 Fourier, Joseph, 40 Franklin, Benjamin, quoted, 205 Fubini's theorem, 152 Fundamental Theorem of Calcul.us for the Cauchy integral, 44 for the Henstock-KUIZWeil integral,203 for the Lebesgue integral, 152 for the Lebesgue-Stieltjes integral, 166 for the Riemann integral, 72 for the FUemann-Stieltjes integral, 80 gauge, 175 Henstock, Ralph, 19, 169 Henstock-Kurzweil integral, 176 Hille, Einar, 252 Hippocrates, lune of, 2 H-K dominated convergence theorem, 191 H-K monotone convergence theorem, 189 Holder, Otto, 139 Holder-FUesz inequality, 139 Jacobi, Carl Gustav Jacob, quoted, 29 Jordan, Camille, I 01 Kurzweil, Jaroslav, 19, 169
279
280 Lagrange's identity, 30 Lebesgue, Henri, 15 Lebesgue dominated convergence theorem, 125 Lebesgue measurable functions, 97 Lebesgue monotone convergence the~ orem, 122 Lebesgue outer measure, 87 Lebesgue-Stielges measure. 18 Leibniz, Gottfried, 8 Levi, Beppo, 122 McShane, Edward James, quoted, 1 measurable function, 17 Minkowski, Hennann, 140 Minkowslci-Riesz inequality, 140 monotone convergence theorem, 122 Muldowney, Patrick, 21 Newton, Isaac, 8 Perrin, Jean. 206 Phillips. Ralpb Saul, 252 Poincare, Renn, quoted, 85 Riemann, Bernhard, 12 Riemann integrability criteria, 37 Riemann-Stiel~es integral, 76 Riemann-Stieltjes sums, 14 Riesz, Frederic, 139
A Garden of Integrals
Riesz completeness theorem, 143 Riesz~Fischer theorem, 143 Schellbach, Karl Heinnch, quoted, 169 SchrOdinger. EIWin, 235 SchrOdinger's equation, 236 Schwartz's inequality, 147 sigma algebra, 91 Smoluchowslci, Marian von, 206 Stieltjes, Thomas, 14, 31, 75 Stirling, JaTDes, 31 Sylvester. James Josepb, quoted, 235 tag, 175 tagged partition, 175 Trotter product formula, 264 Vitali, Giuseppe, 96 Vitali cover, 96 Volterra, Vito, 13, 70 von Smoluchowslci, Marian, 206 Wallis, John, 30 Weiner, Norbert, 22, 208 Weiner measurable functional, 220 Whitehead. Alfred North, quoted, 155 Young, Wilham Henry. 113 Young's inequality. 139
