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~ [ ( 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 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 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
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
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 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
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, 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
~
L
< 2 (b €- a ) 6.x + (sup rb] La, €
f - f) inf
[a,b}
L
sup f -inff
6.x >---!i..2lb.;;;;a)
€
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 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
.!.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
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 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
=
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 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
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; 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])
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)
~l) m
x
E
[a b] I l{IlI (x) - cjJ'l (x) >
2.1)
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 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 ... }; and again
L f.b f..g dJ1. < inf! L f.b !k dJ1.. L f.b !k+1 dJ1. •... } . So we have 0 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
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
(Xk) - t/>CXk-l)]- L =
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
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 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.
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 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)
+ ,~])
(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)]
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
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 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)
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 )
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 0 be given and choose N so that 1/ N < partition of [a, b]. Then
L fCc) 6.x P
A
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 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
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 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
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
-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}
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 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)
{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 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 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
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
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 (; 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
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
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
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
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 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
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 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
