A concise introduction to the theory of integration, second edition

is dense in £ 1 (En, BEn , Jln), where Jln denotes the restriction of Jl to BEn . In view of Theorem 3.3. 13, ........, this will follow as soon as we show that l a is in the · I (J.Ln -closure of ICn for each open C En . If == En == E, then there is no problem, since in that case l a 1 is both p-uniformly continuous and Jln-integrable. On the other hand, if C En is open and =I= E, set F == and define m 'Pm ( x ) == 1 - ( 1 + p ( x , F) ) , m > 1 , where p ( x, F) inf { p ( x , y ) : y E F} . Since p ( · , F) is p-uniformly continuous, we see that 'Pm is p-uniformly continuous. In addition, it is easy to check that 0 < m / l a as m ---t oo. Hence, by the Monotone Convergence Theorem, it follows that m r En � l a r En in L 1 ( En, BEn ' Jln) as m ---t By the preceding, we now know that if f E L 1 (M) vanishes identically off En for some n > 1 , then there is a sequence { m } 1 C ICn such that f i i£ 1 (J.L ) � 0 as m ---t oo. At the same time, it is clear, by Lebesgue ' s Dominated Convergence Theorem, that for any f E L 1 (J,t) , ll fn - f ii £ 1 (J.L ) � 0 as n ---t oo, where fn lEn f · D Notice that, when applied to Lebesgue ' s measure on � N , Corollary 3.3. 14 says that for every f E £ 1 (.XR N ) and E > 0 there is a continuous function cp such that cp vanishes off of a compact set and < E. This fact can be interpreted in either one of two ways: either measurable functions are provides a rather not all that different from continuous ones or · £ 1 crude gauge of size. Experience indicates that the latter interpretation is the more accurate one.

I ILl )

'P 'P I 'Pm -

G

_

G

G

G

GC

00 .

'P

AR N I ! - 'P I L l (AJRN ) I I ( AJRN )

60

III Lebesgue Integration Exercises

Let f be a non-negative, integrable function on the measure space (E, B, v) , and define JL(r) == fr f dv for r E B. Show that JL is a finite measure on (E, B) . In addition, show that JL is absolutely continuous with respect to v in the sense that for each E > 0 there is a 8 > 0 with the property that JL(r) < E whenever v(r) < 8. 3.3. 16 Exercise: Let f be a non-negative, measurable function on the measure space (E, B, JL) . If f is integrable, show that (3.3. 17) 3.3. 15 Exercise:

Next, produce a non-negative measurable f on ( [0, 1] , B[o , l ] , A [o , I] ) (.X[o, l ] is used here to denote the restriction of AR to B[o, 1 ] ) such that (3.3. 17) holds but f fails to be integrable. Finally, show that if f is a non-negative measurable function on the finite measure space (E, B, JL), then f is integrable if and only if CX) JL(f > n ) < oo. n =l 3.3. 18 Exercise: Let J be a closed rectangle in � N and f J � � a continuous function. Show that the Riemann integral (R) JJ f(x) dx of f over J is equal to the Lebesgue integral JJ J (x) AJRN (dx) . Next, suppose that f E L 1 (AJRN ) is continuous, and use the preceding to show that

L

:

J f(x) >.RN (dx)

J /limR N (R) JJ{ f(x) dx , where the limit means that, for any E > 0, there exists a rectangle Jf. such that =

j f(x) >.RN (dx) - (R) 1 f(x) dx < f

whenever J is a rectangle containing Jf. . For this reason, even when f is not continuous, it is conventional to use J f(x) dx instead of J f dAR N to denote the Lebesgue of f. 3.3. 19 Exercise: Let (E, B, JL) be a measure space and let {fn}1 be a se­ quence of measurable functions on (E, B) . Next, suppose that {gn}1 C L 1 (JL) and that gn � g E L 1 (JL) in L 1 (JL) . The following variants of Fatou ' s Lemma and Lebesgue ' s Dominated Convergence Theorem are often useful. ( i ) If fn < gn (a.e., JL) for each n > 1, show that

j

j

nlim ---+ CX) fn dJL. ---+ CX) fn djL < nlim ( ii ) If fn � f either in JL-measure or JL-almost everywhere and if I fn i < gn ( a.e., JL) for each n > 1, show that II fn - f II £ 1 (J.L) � 0 and therefore that limn ---+ CX) J fn dJL == J f dJL.

3.3

Convergence of Integrals

61

Let (E, B, JL) be a measure space. A family JC of measurable functions f on (E, B, JL) is said to be uniformly JL-absolutely continuous if for each E > 0 there is a 8 > 0 such that fr I f I dJL < E for all f E JC whenever E B and JL( ) < 8; and it is said to be uniformly JL-integrable if for each E > 0 there is an R < oo such that � / > R l f l dJL < E for all f E JC . I ( i ) Show that JC is uniformly JL-integrable if it is uniformly JL-absolutely con­ tinuous and supIC l l f ii £ 1 (JL ) < oo . /E Conversely, suppose that JC is uniformly JL-integrable and show that it is then necessarily uniformly JL-absolutely continuous and, when JL(E) < oo , that sup / E IC l l f ii £ 1 (JL ) < oo . ( ii ) If sup / E IC J 1!1 1 + 6 dJL < oo for some 8 > 0, show that JC is uniformly JL-integrable. ( iii ) Let {/n}1 C L 1 (JL) be given. If fn � f in L 1 (JL), show that {fn}! U {f} is uniformly JL-absolutely continuous and uniformly JL-integrable. Conversely, assuming that JL(E ) < oo , show that fn � f in L 1 (JL ) if fn � f in JL­ measure and {fn}1 is uniformly JL-integrable. ( iv ) Assume that JL(E) == oo . We say that a family JC of measurable func­ tions f on (E, B, JL) is tight if for each E > 0 there is a E B such that JL( ) < oo and sup / E IC frc I! I dJL < E. Assuming that JC is tight, show that JC is uniformly JL-integrable if and only if it is uniformly JL-absolutely contin­ uous and sup / E IC ll f ii £ 1 (JL ) < oo . Next, show that JC is uniformly JL-integrable if JC is tight and sup / E IC J 1! 1 1 + 6 dJL < oo for some 8 > 0. Finally, suppose that {fn}1 C L 1 (JL) is tight and that fn � f in JL-measure. Show that l l fn - f i i£ 1 (JL ) � 0 if and only if {fn} is uniformly JL-integrable. 3.3. 20 Exercise:

r

r

r

r

Let (E, B, JL) be a finite measure space. Show that fn � f in JL-measure if and only if J l fn - ! I 1\ 1 dJL � 0.

3.3. 21 Exercise:

Let (E, p) be a metric space and {En}! a non-decreasing sequence of open subsets of E such that En / E. Let Jl and v be two measures on (E, BE ) with the properties that JL(En) V v(En) < oo for every n > 1 and J cp dJL == J cp dv whenever cp is a bounded p-uniformly continuous cp for which there is an n > 1 such that cp 0 off En. Show that Jl == v on BE . 3.3. 22 Exercise:

Although almost everywhere convergence does not follow from convergence in measure, it nearly does. Indeed, suppose that {fn}1 is a sequence of measurable, �-valued functions on (E, B, JL) . Given an �-valued, measurable function J, show that (3.3.8) holds, and therefore that fn � f both (a.e., JL) and in JL-measure, if 3.3. 23 Exercise:

00

L1 JL (Ifn - f l

> E) <

oo

for every E > 0.

62

III Lebesgue Integration

In particular, conclude that fn � f (a.e., J-t ) and in J.-t- measure if

L1 l l fn - J IIL 1 (JL) <

00 .

3.4 Lebesgue's Differentiation Theorem.

Although it represents something of a departure from the spirit of this chap­ ter, we return in this concluding section to Lebesgue ' s measure on � and prove the following remarkable generalization of the Fundamental Theorem of Cal­ culus. Namely, we will show if f is any Lebesgue integrable function on � ' then ( cf. the notation introduced in Exercise 3.3.18) . 1 ! J (t) - f ( x ) ! dt == 0 for (Lebesgue) almost every x E � (3.4. 1) hm ' t '\.{ x } I I I t where ( 3.4.1) is to be interpreted as the statement that, for almost every x E � ' there exists, for each E > 0, a 8 == 8 ( x , E) > 0 such that 0

�

r ! J (t) - f( x ) l dt < E whenever i 3 is an open interval with I I I < 8. };_ I I I it X

In other words, except on a set of Lebesgue measure 0, an integrable function can be recovered by differentiating its indefinite integral.

In order to understand the strategy behind our proof, first note that (3.4.1) is completely obvious when f is continuous. Hence, since ( cf. Corollary 3.3.14 ) the continuous elements of L 1 ( �) L 1 (,\R) are dense in L 1 (�), it suffices for us to show that the set g of f E L 1 (�) for which (3.4. 1) holds is closed in L 1 (�) . To this end, we introduce the Hardy-Littlewood maximal function _

(3.4.2)

M f( x )

sup

{};_I I I }{I i f(t) i dt : j } 3

x ,

xE

�'

for f E L 1 (�). Next, for each f E L 1 ( � ) and E > 0 set b,. ( f, E)

==

{x : 1'\.{)imx} };_I II Jt� i f(t) - f(x) i dt } >

t:

.

Clearly, (3.4.1) holds if and only if I D,.(j, E) I e == 0 for every E > 0. Moreover, for any E > 0 and any J, g E L 1 (�): D,. ( j, 3 E ) C { x : M (f - g) > E } U b,. (g, E) U { x : l g( x ) - f( x) ! > E } .

63

3.4 Lebesgue 's Differentiation Theorem

In particular, this means that if g E g, then

where, in the passage to the last line we have used l � (g, E) l e == 0 and Markov ' s inequality. Finally, suppose that { gn }1 C g and that 9n � f in £ 1 (� ) . Then, the preceding line of reasoning leads us to the conclusion that � � ( / , 3E) I e

< nlim ! {M(J - gn) > E} l e ' ---+ CX)

and so we would be done if we knew that ( 3.4.3) With the preceding in mind, we now turn our attention to the analysis of the Hardy-Littlewood maximal function. To begin with, we first note that M f is measurable, and therefore that we can drop the subscript "e" in ( 3.4.3). To see this, observe that an alternative expression for M f is 1 { sup Mf(x) = ,a b E(O , CX)) a + b }( -a ,b) I J(x + t) l dt and that, for each a, b E (0, oo ) and x < y ,

f I J( x + t) l dt - r I J( y + t) l dt < r I J(t) l dt . r ,b) r ,b) J( x -a , y-a]U[x + b, y+ b) J(-a J(-a Hence, by Exercise 3.3. 15, we know that, for each a , b E (0, oo ) , 1 r 1 1 ( x + t) 1 dt E R xER a + b J( -a ,b ) f---+

is uniformly continuous and, therefore, for each a E � ' that {x : Mf(x) > a} ==

{

i

1 : x l f(x + t) l dt > o: U b a + ( -a ,b ) a , b E (O , CX))

}

is open. We next observe that control on the size of M f will follow from control on the one-sided maximal functions M+ f(x) sup � { I J(t) l dt. I J(t) l dt and M_f(x) sup � { h>O J( x -h, x ] h>O J[x , x +h)

64

III Lebesgue Integration

Indeed, for any I 3 x , let h+ and h_ be the lengths of [x, oo ) ni and ( -oo, x] n i , respectively, and note that 0

0

1

0

� r I J(t) i dt = � I J( t ) l dt I J (t) i dt + � r I I I }J I I I J[ x,x + h+) I I I ( x - h_ , x ] < M+ f( x ) + M_ f(x) < M+ f(x) V M_ f( x) ,

�A

�il

and therefore, since it is essentially obvious that M+ f V M_ f < MJ, we have (3.4.4) as

Moreover, by precisely the same argument we used to prove that { Mf > a} is open, we know that the same is true of both {M+ f > a} and {M_ f > a}. Finally, note that M_ f( x) == M+ /( -x) , where /(t) == f( - t) , which means that we really need to learn how to control only M+ f · For this purpose, let f E £ 1 ( �) and E > 0 be given, and consider the function F€ (x) = { i f(t) i dt - EX , X E JR. }( - CX) , X] By Exercise 3.3. 15, Ff_ is uniformly continuous, and, obviously, lim x � ± CX) Ff_(x) == =f OO . Moreover, it is an elementary matter to see that as

Hence, we will see shortly, all that we need is the following wonderfully simple observation. 3.4.5 Lemma (Sunrise Lemma) . * Let F : � � � be a continuous func­ tion with the property that limx � ± CX) F (x) == =fOO . Set G == {x : 3y > x F( y ) > F(x) }. Then G is an open, and each non-empty, open, connected component of G is a bounded interval (a, {3) with F(a) < F({3) .

PROOF: Clearly G is open. Next, suppose that (a, {3 ) is a non-empty, con­

nected, open component of G, and take 1 E (a, {3). If either {3 == oo or {3 < oo and F({3) < F(1) , then there exists a unique x E [1, {3) such that F(x) == F(1) and F( y ) < F(1) for all y > x . But this would mean that, on the one hand, x E G and, on the other hand, F( y ) < F(x) for all y > x , which is impossible. Hence, we now know that {3 < oo and that F( 1) < F ( {3 ) for all 1 E (a, {3) . Finally, from this it is clear that a > -oo and that F(a) < F({3). D * The name derives from the following picture. The sun is rising infinitely far to the right in mountainous (one-dimensional) terrain, F(x ) is the elevation at x, and G is the region in shadow at the instant when the sun comes over the horizon.

65

3.4 Lebesgue 's Differentiation Theorem

Applying Lemma 3.4.5 to the function Ff. , we see that G€ {M+ f > t } is either empty or (cf. the first part of Lemma 2. 1.9) is the union of countably many disjoint, bounded, open intervals ( an , f3n ) satisfying 0 < F€ (f3n ) - F€ ( an ) =

1(O:n ,/3n ) i f(t) i dt - E (f3n - an )

for each n. Hence, either IG € 1 == 0 or, after summing the preceding over n , we arrive at In other words, we have now proved first that and then, after taking left limits with respect to t > 0,

v

In fact, because M_ f( x) == M+ f( ) we also know that (3.4.6) continues to hold when M+ f is replaced by M_ f throughout. Finally, in conjunction with (3.4.4), these leads to -x ,

which means that we have now proved the renowned

Hardy-Littlewood

maximal inequality:

( 3.4.7) Since (3.4.7) certainly implies (3.4.3) , and (3.4.3) was the only missing in­ gredient in the program with which this section began, the derivation of the following statement is complete. 3.4.8 Theorem ( Lebesgue's Differentiation Theorem ) . For any Lebes­ gue integrable function f on �' (3.4. 1) holds. In particular, (3.4.9)

lim _..;_ { j ( t ) = j(x) }

i�{x} I I I

;

for almost every X

E JR.

Before closing this section, there are several comments which should be made. First, one should notice that the conclusions drawn in Theorem 3.4.8 remain true for any Lebesgue measurable f which is integrable on each compact

66

III Lebesgue Integration

subset of JR. Indeed, all the assertions there are completely local and therefore follow by replacing f with f l ( - R , R) , restricting ones attention to x E ( -R, R) , and then letting R / oo. Second, one should notice that (3.4. 7) would be a trivial consequence of Markov ' s inequality if we had the estimate l i M f i i £ 1 ( R ) < 2 II J II£ 1 (R ) · Thus, it is reasonable to ask whether such an estimate is true. That the answer is a resounding no can be most easily seen from the observation that, if ll f ii £ 1 ( R) f= 0, then a f( -r, r ) I J(t) l dt > 0 for some r > 0 and therefore Mf(x) > l xN-r for all x E JR. That is, if f E £ 1 (JR) does not vanish almost everywhere, then Mf is not integrable. (To see that the situation is even worse and that, in general, M f need not be integrable over bounded sets, see Exercise 3.4. 12 below.) Thus, in a very real sense, (3.4. 7) is about as well as one can do. (See Exercise 6.2.27 for an interesting continuation of these considerations.) Because this sort of situation arises quite often, inequalities of the form in ( 3.4. 7) have been given a special name: they are called weak-type inequalities to distinguish them from inequalities of the form li M J II £ 1 (R ) < C I I J I I£ 1 (R ) , which would be called a strong-type inequality. Finally, it should be clear that, except for the derivation of (3.4. 7), the arguments given here would work equally well in JR N . Thus, we would know that, for each Lebesgue integrable f on JR N , 1 I for almost every ) ( X E IR. N f(x) = 0 d J ( 3.4. 10) B lim y y l � { x} I B }B if we knew that c (3.4.1 1) I {Mf > E } l < -E IIJ IIL 1 ( AR N ) ' where M f is the Hardy-Littlewood maximal function Mf(x) �� � I IJ( y ) l dy I and B denotes a generic open ball in JR N . It turns out that (3.4. 11), and therefore (3.4.10), are both true. However, the proof of ( 3.4. 11 ) for N > 2 is somewhat more involved than the one which we have given of (3.4.7)*.

1r

L

Exercises

Define f : JR � [O, oo) so that f(x) == (x( log x) 2 ) - 1 if x E ( o, - 1 ) and f(x) == 0 if x � ( o, 1 ). Using Exercise 3.3. 18 and the Fundamental Theorem of Calculus, check that f E £ 1 (JR) and that - 1 , X E ( 0, - l ) . f ( t) dt = log X 3.4.12 Exercise: e

e

1(O,x)

-

e

In particular, conclude that f( o , r ) M f(x) dx == oo for every r > 0.

* See, for example, E . M . Stein's Singular Integrals and Differentiability Properties of Func­ tions, published by Princeton Univ . Press ( 1 970)

67

3.4 Lebesgue 's Differentiation Theorem

Given f E L 1 (1R) , define the Lebesgue set of f to be the set Leb(f) of those x E � for which the limit in (3.4. 1) is 0. Clearly, (3.4. 1) is the statement that Leb(j)C has Lebesgue measure 0, and clearly Leb(f) is the set on which f is well behaved in the sense that the averages m f; f (t) dt converge to f(x) as j \. {x} for X E Leb(f) . The purpose of this exercise is to show that, in the same sense, other averaging procedures converge to f on Leb(f). To be precise, let p be a bounded continuous function on � having one bounded, continuous derivative p'. Further, assume that p E £ 1 (JR) , J p(t) dt == 1, J l tp'(t) I dt < oo , and lim l t i � CX) p(t) == limt � CX) tp'(t) == 0. Next, for each t > 0, set pf. (t) == t - 1 p ( t - 1 t ) and define f€ (x) = p€ (x - t)f(t) dt, x E lR and f E L 1 (1R). 3.4. 13 Exercise:

J

The purpose of this exercise is to show that (3.4.14) ft: (x) � f(x) as t � 0 for each x E Leb(f) . (i) Show that, for any f E £ 1 (JR) and x E Leb(j),

8lim �0 ; j(x-8,x] f(t) dt. �0 ; J[rx ,x+8) f(t) dt f(x) 8lim v

(ii)

that

=

=

v

Assuming that f is continuous and vanishes off a compact set, first show f€ (x) =

rJ[o, CX) ) p( - t)f(x + d) dt + J[ro ,CX)) p(t)f(x - d) dt,

and then (using Exercise 3.3. 18 and Theorem 1.2.7) verify the following equal­ ities: (3.4.16) and

rJ[o, CX)) p( -t) f(x + Et) dt Jr(o, CX) ) tp' ( -t) ( ..!..tt J[rx ,x+t:t ) ! ( s) ds) dt =

rJ[o, CX)) p(t)f(x - d) dt

(j

)

f(s) ds dt. tp' (t) ..!..tt J( o , CX) ) (x-t:t , x] Next (using Corollary 3.3. 14) argue that ( 3.4.16) and (3.4. 16 ' ) continue to hold for every f E L 1 (1R) and x E Leb(J). (iii) Combining part (i) with (3.4. 16) and (3.4. 16 ' ), conclude that ( 3.4.16 ' )

=

-

r

' J f. (t) dt, for x Leb(J), tp (x) f -f(x) f.lim �O E

==

and, after another application of Exercise 3.3.18 and Theorem 2. 1.7, note that

- J tp' (t) dt J p(t) dt =

=

1.

Chapter IV Products of Measures

4. 1

Fubini's Theorem.

Just before Lemma 3.2.2, we introduced the product (E1 E2 , B1 B2 ) of two measurable spaces (E 1 , B 1 ) and (E2 , B2 ). We now want to show that if Jl i , i E { 1, 2}, is a measure on ( Ei , Bi ), then, under reasonable conditions, there is a unique measure v on (E1 E2 , B1 B2 ) with the property that v(rl X r2 ) == Il l (rl ) JL(r2 ) for all ri E Bi . The key to the construction of v is found in the following function analog of ?T- and A-systems ( cf. Lemma 3. 1.3). Namely, given a space E, we will say that a collection £, of functions f : E � ( ] is a semi-lattice if both j + and f - are in £, whenever f E .C . A sub collection /C C £, will be called an £-system if: ( a ) 1 E /C; ( b ) if f, g E /C and { ! == oo} n {g == oo} == 0, then g - f E /C whenever either f < g or g - f E .C; ( c ) if a , f3 E [0, ) and J , g E /C, then af + (3g E /C; (d) if {fn}1 C /C and fn / f, then f E /C whenever f is bounded or f E .C . x

x

x

x

- oo, oo

oo

The analog of Lemma 3. 1.3 in this context is the following. 4. 1 . 1

Lemma. Let C be a ?T-system which generates the a-algebra

and let £, be a semi-lattice of functions f : E ( ] £-system and l r E /C for every r E C, then /C contains every f E measurable on ( E, B) .

E,

�

- oo, oo .

B

over If /C is an £, which is

PROOF: First note that {r C E : 1 r E /C} is a A-system which contains C. Hence, by Lemma 3. 1.3, 1 r E /C for every r E B. Combined with ( c ) above, this means that /C contains every non-negative, measurable, simple function

on (E, B) . Next, suppose that f E £, is measurable on (E, B). Then both j + and f - have the same properties, and, by ( b ) above, it is enough to show that j+ , f- E /C in order to know that f E /C. Thus, without loss in generality, we assume that f E £, is a non-negative measurable function on (E, B). But in that case f is the non-decreasing limit of non-negative measurable simple functions ; and so f E /C by (d) . D

4.1

Fubini 's Theorem

69

The power of Lemma 4. 1 . 1 to handle questions involving products is already apparent in the following. 4.1.2 Lemma. Let (E1 , 8 1 ) and (E2 , 82 ) be measurable spaces, and suppose that f is an �-valued measurable function on (E 1 x E2 , 8 1 x 82 ). Then for each x 1 E E1 and x 2 E E2 , j(x 1 , ) and f( x 2 ) are measurable functions on (E2 , 82 ) and (E1 , 8 1 ), respectively. Next, suppose that Jli , i E {1, 2}, is a finite measure on ( Ei, 8i ). Then for every measurable function f on ( E1 x E2 , 8 1 x 82 ) ·

·

,

which is either bounded or non-negative, the functions

(E1 , 81 ) and (E2, 82 ), respectively. PROOF: Clearly it is enough to check all these assertions when f is non­ negative. Let £, be the collection of all non-negative functions on E1 x E2 , and define /C to be those elements of £, which have all the asserted properties. It is clear that 1 r1 xr2 E /C for all ri E 8i . Moreover, it is easy to check that /C is an £-system. Hence, by Lemma 4. 1.1, we are done. D

are measurable on

(E 1 , 8 1 , Jl i ) and (E2 , 82 , Jl 2 ) of finite measure spaces, there exists a unique measure v on (E1 x E2 , 8 1 x 82 ) such that 4.1.3

Lemma. Given a pair

Moreover, for every non-negative function

rE

J (4.1.4)

1 xE

f on (E1 x E2 , 8 1 x 82 ),

v(dx ) x ) x , f(x dx l l 2 2 2

=

L2 (L1 j(x 1 , x2 ) M 1 (dx 1 ) ) M2 (dx2 ) L1 (L2 J(x 1 , x2 ) 1-t2 (dx2 ) ) /-ll (dx l ) · =

PROOF: The uniqueness of v is guaranteed by Exercise 3. 1.8. To prove the

existence of v, define

and

70

IV Products of Measures

for r E B l X 82 . using the Monotone Convergence Theorem, one sees that both v1 , 2 and v2 , 1 are finite measures on (E1 E2 , 8 1 82 ). Moreover, by the same sort of argument as was used to prove Lemma 4. 1.2, for every non-negative measurable function on (E1 E2 , 81 82 ): x

x

and

x

x

j j dv2 , 1 l2 (l1 f(x 1 , x2 ) /-l2 (dx2 ) ) /-l 1 (dx 1 ). =

Finally, since VI , 2 (rl X r2 ) == JL(rl ) JL(r2 ) == v2 , 1 (rl X r2 ) for all ri E Bi we see that both v1 , 2 and v2 , 1 fulfill the requirements placed on v. Hence, not only does v exist, but it is also equal to both v1 , 2 and v2 , 1 ; and so the preceding equalities lead to (4.1.4). D In order to extend the preceding construction to measures which need not be finite, we must ( cf. Exercise 4.1. 12) introduce a qualified notion of finiteness. Namely, we will say that the measure JL on (E, B) is a- fi nite and will call (E, B, JL) a a-finite measure space if E can be written as the union of a countable number of sets r E B for each of which JL(r) < oo . Thus, for example, (� N , BRN , AR N ) is a a-finite measure space. 4.1.5 Theorem ( Tonelli's Theorem ) . Let (E1 , 8 1 , JL I ) and (E2 , 82 , JL 2 ) be a-finite measure spaces. Then there is a unique measure v on (E1 E2 , 8 1 82 ) such that v(rl X r2 ) == JL I (rl ) JL(r2 ) for all ri E Bi · In addition, for every non­ negative measurable function f on ( E1 E2 , 8 1 82 ), J f( · , x 2 ) JL2 (dx 2 ) and J j(x 1 , · ) JL I (dx l ) are measurable on (E1 , B 1 ) and (E2 , B2 ) , respectively, and (4.1 .4) continues to hold. PROOF: Choose sequences {Ei, n}� 1 C Bi for i E {1, 2} so that JLi (Ei,n) < oo for each n > 1 and Ei == U� 1 Ei,n · Without loss in generality, we assume that Ei,m n Ei, n == 0 for m f= n . For each n E z + , define JLi , n(ri) == JLi (ri n Ei, n), ri E Bi ; and, for ( m , n ) E z + 2 ' let V(m , n ) on (El X E2 , B l X 82 ) be the measure constructed from JL I,m and JL 2 , n as in Lemma 4. 1.3. Clearly, by Lemma 4.1.2, for any non-negative measurable function f on ( El X E2 , Bl X 82 ), '

x

x

x

x

is measurable on (E 1 , 8 1 ) ; and, similarly, JE 1 j(x 1 , · ) JL I (dx 1 ) is measurable on (E2 , 82 ). Finally, the map r E 8 1 B2 r-----+ E:, n =l v( m , n ) (r) defines a measure vo on (E 1 E2 , 8 1 82 ), and it is easy to check that vo has all the required properties. At the same time, if v is any other measure on (E1 E2 , 81 82 ) for which v(rl X r 2 ) == /L l (rl ) JL 2 (r2 ), ri E Bi , then, by the uniqueness assertion x

x

x

x

x

4. 1

71

Fubini 's Theorem

in Lemma 4.2 1.3, v coincides with v(m , n ) on 81 82 [El,m E2 , n ] for each ( m , n) E z + and is therefore equal to Vo on 8 1 82 . D The measure v constructed in Theorem 4. 1.5 is called the product of Jll times JL2 and is denoted by Jl l Jl2 · x

x

X

x

4.1.6

Theorem (Fubini's Theorem) . Let (EI , 8 1 , Jll ) and (E2 , 82 , JL2 ) be

a-finite measure spaces and f a measurable function on (E1 Then the f is Jll x Jl2-integrable if and only if

x

E2 , 8 1

x

82 ) .

if and only if

Next, set

and

and define fi on Ei , i

E {1, 2},

by XI

if E Al otherwise and

Then fi is an �-valued, measurable function on ( Ei , 8i ) . Finally, if f is Jll x Jl2integrable, then Jli (Ai C) == 0, fi E L 1 (J-Li ), and

( 4. 1.7) for i

E {1, 2}.

72

IV Products of Measures

PROOF: The first assertion is an immediate consequence of Theorem 4. 1.5.

Moreover, since Ai E Bi , it is easy to check (from Lemma 4. 1.2) that fi is an �-valued, measurable function on (Ei , Bi)· Finally, if f is Jl l x JL 2 -integrable, then, by the first assertion, Jli(Ai C) == 0 and fi E L 1 ( J-Li )· Hence, by Theorem 4.1.5 applied to j + and f- , we see that

r 1

E 1 x E2

j(x 1 , X 2 ) ( J.Ll X J.L2 ) (dx 1 X dx 2 )

- r

}A l

=

X E2

� - (x 1 , X 2 )

( J.Ll X J.L2 ) ( dx 1 X dx 2 )

1 1 (L2 j+ (x 1 , x2 ) J.L 1 (dx2 ) ) J.Ll (dxt ) - 1 1 (L2 f (x 1 , x2 ) J.L2 (dx2 ) ) J.L 1 (dx 1 ) -

=

r ft (X ! ) J.L 1 ( dx ! ) i

jA

1

and the same line of reasoning applies to f2 .

D

Exercises

Let (E, B, JL) be a a-finite measure space. Given a non­ negative measurable function f on (E, B) , define r(j) == { (x, t) E E x [O, oo) : t < f (x) } 4.1.8

Exercise:

and f (J )

{ (x, t) E E x [O, oo) : t < f (x ) }. -Show that both r(j) and r(j) are elements of B X B[o ,CX)) and, in addition, that ==

( 4.1.9) In proving measurability, consider the function (x , t ) E E x [0, oo) r-----+ f(x ) - t E ( - oo, oo] ; and get (4. 1.9) as an application of Tonelli ' s Theorem. Clearly (4. 1.9) can be interpreted as the statement that the integral of a non­ Hint :

negative function is the area under its graph.

4. 1 Fubini 's Theorem

73

Let (E1 , 81 , J.-t 1 ) and (E2 , 82 , J.-t 2 ) be a-finite measure spaces and assume that, for i E { 1, 2} , 8i == a(Ei; Ci), where Ci is a 1r-system con­ taining a sequence {Ei, n}� 1 such that Ei == U� 1 Ei , n and J-t i(Ei , n) < oo, n > 1. Show that if v is a measure on (E 1 E2 , 81 82 ) with the property that v(r1 X r2 ) == J.-l 1 (r1 ) J.-l 2 (r2 ) for all ri E Ci , then V == J.-l 1 X J.-l 2 . Use this fact to show that, for any M, N E z + , 4. 1 . 10 Exercise:

x

x

Let (E1 , 8 1 , J.-t 1 ) and (E2 , 82 , J.-t2 ) be a-finite measure spaces. Given r E 8 1 X 82 ' define r(l ) (x2 ) { Xi E El : (x l , x2 ) E r } E B l for X 2 E E2

4. 1 . 1 1 Exercise:

and r(2 ) (x l )

{ X2 E E2

:

(x l , x 2 ) E r } for Xi E El .

Check both that r( i ) (xj) E 8i for each Xj E Ej and that Xj E Ej r-----+ J.-ti ( r( i ) (xj) ) E [O, oo] is measurable on (Ej, 8j) ({i, j} == {1, 2}). Finally, show that JL 1 JL2 (r) = 0 if and only if JLi (r( i ) ( xi ) ) = 0 for ttralmost every xi E Ei ; and, conclude that JL 1 (r(l ) ( x2 ) ) = 0 for tt2 -almost every x 2 E E2 if and only if /L 2 (r(2) ( xt) ) = 0 for IL l -almost every Xi E El . In other words, r E Bl X 82 has J.-t 1 J.-t 2 -measure 0 if and only if �-t 1 -almost every vertical slice (J.-t 2 -almost every horizontal slice) has �-t 2 -measure (J.-t 1 -measure) 0. 4. 1 . 12 Exercise: The condition that the measure spaces of which one is taking a product be a-finite is essential if one wants to carry out the program in this section. To see this, let E1 == E2 == (0, 1 ) and 8 1 == 82 == 8( o, I ) . Define J.-l 1 on ( E1 , 8 1 ) so that �-t 1 (r) is the number of elements in r ( oo if r is not a finite set) and show that J.-li is a measure on (E1 , 8 1 ) . Next, take �-t 2 to be Lebesgue ' s measure A (o, I ) on (E2 , 82 ). Show that there is a set r E 81 82 such that x

x

x

r l (x l , X 2 ) JL 2 (dx 2 ) = 0 for every X l E El

JE2

r

but ( Hint : Try r == {(x, x) : 0 < X < 1}.) In particular, there is no way that the second equality in ( 4.1.4 ) can be made to hold. Notice that what fails here is really the uniqueness and not the existence in Lemma 4. 1.3.

IV Products

74

of Measures

< a < b < oo and a function f : (a, b) E � � with the properties that /( · , x) E C((a, b)) for every x E E and f ( t, ) is measurable on ( E, B) for every t E (a, b), show that f is measurable on ( (a , b) E, B( a, b) B) . Next, suppose that f ( , x) E C 1 ((a , b)) for each x E E, set f'(t, x) == d� f(t, x) , x E E, and show that f' is measurable on ( (a, b) E, B( a , b ) B) . Finally, suppose that Jl is a measure on (E, B) and that there is a g E L 1 (JL) such that I f( t, x ) I V I f' ( t, x ) I < g(x ) for all (t, x ) E (a , b) E. Show not only that JE f( x) JL(dx) E C 1 ((a, b)) but also that f(t, x) J.L (dx) = f ' (t, x) J.L (dx) . 4. 1.13 Exercise: Let (E, B) be a measurable space. Given

-oo

x

·

x

x

x

x

x

·

�l

4.2

·

,

l

Steiner Symmetrization and the Isodiametric Inequality.

In order to provide an example which displays the power of Fubini ' s The­ orem, we will prove in this section an elementary but important inequality about Lebesgue ' s measure. Namely, we will show that, for any bounded sub­ set r C �N , (4.2. 1) where n N denotes the volume of the unit ball B(O, 1 ) in �N and rad(r) sup { ly; x l : x, y E r } _

is the radius (i.e., half the di amet er) of r. Notice (cf. ( ii ) in Exercise 2.2.3) that what (4.2.1 ) says is that, among all the subsets of �N with a given diameter, the ball of that diameter has the largest volume; it is for this reason that (4.2. 1) is called the isodiametric inequality. At first glance one might be inclined to think that there is nothing to (4.2.1). Indeed, one might carelessly suppose that every r is a subset of a closed ball of radius rad(r) and therefore that (4.2.1) is trivial. This is true when N == 1. However, after a moment ' s thought, one realizes that, for N > 1, although r is always contained in a closed ball whose radius is equal to the diameter of r, it is not necessarily contained in one with the same radius as r. ( For example, consider an equilateral triangle in �2 .) Thus, the inequality 1 r 1e < n N ( 2rad ( r )) N is trivial, but the inequality in ( 4.2. 1) is not! On the other hand, there are many r ' s for which (4.2.1) is easy. In particular, if r is symmetric in the sense that r == - r { -X : X E r}, then it is clear that x E r ===} 2 l x l == l x + x l < 2rad(r) and therefore r c B (O, rad(r) ) . Hence ( 4.2.1) is trivial when r is symmetric, and so all that we have to do is devise a procedure to reduce the general case to the symmetric one.

75

4.2 Steiner Symmetrization and the Isodiametric Inequality

The method with which we will perform this reduction is based on a famous construction known as the Steiner symmetrization procedure. To describe Steiner ' s procedure, we must first introduce a little notation. Given v from the unit (.LV - 1 ) -sphere s N - 1 {x E �N : l x l == 1 }, let L(v) denote the line {tv : t E �}, P(v) the (N - I ) -dimensional subspace {x E �N : x ..l v}, and define S(r; v) {x + tv : x E P(v) and l t l < �f(r; v, x) } , where f(r; v, x) l {t E � : x + tv E r} l e is the length of the intersection of the line + L(v) with r. Notice that, in the creation S(r; v) from r, we have taken the intersection of r with x + L(v) , squashed it to remove all gaps, and then slid the resulting inter­ val along x + L(v) until its center point falls on x. In particular, S(r; v) is the symmetrization of r with respect to the subspace P(v) in the sense that, for each x E P(v), X

(4.2.2)

x + tv E S(r; v)

¢:::::;>

x - tv E S(r; v);

what is only slightly less obvious is that S(r; v) possesses the properties proved in the next lemma. 4.2.3 Lemma. Let r be a bounded element of BRN . Then, for each v E s N - 1 , S(r; v) is also a bounded element of �N , rad (S( r; v) ) < rad ( r ) , and I S(r; v) I == 1 r 1 . Finally, if R : �N � �N is rotation for which L(v) and r are invariant (i.e., R ( L(v) ) == L(v) and R(r) == r), then RS(r; v) == S(r; v) . PROOF: We begin with the observations that there is nothing to do when N == 1 and that, because none of the quantities under consideration depends on the particular choice of coordinate axes, we may and will assume not only that N > 1 but also that v == eN _ (0, . . . , 0, 1 ) . In particular, this means, by Lemma 4. 1 .2, that a

� E IRN - l f(�) � l lr((�, t) ) dt E [0, oo) �

is BR N- 1 -measurable; and therefore, by Exercise 4. 1 .8, because S(r; e N ) is equal to { ( � , t) E �N - 1 X [O, oo) : t < ( ) } U { ( � , t) E �N - 1 X ( - oo, O] : - t < ( ) } ,

f�

we know both that S(; eN ) is an element of BRN and that where, in the final step, we have applied Tonelli ' s Theorem.

f�

76

IV Products of Measures

We next turn to the proof that rad (S (r ; eN ) ) cannot be larger than rad(r) ; and, in doing so, we will, without loss in generality, add the assumption that r is compact. Now suppose that x , y E S(r; eN ) are given, and choose � ' 'fJ E �N - l and s, t E � so that x == ( � , s) and y == ( TJ , t) . Next, set M ± (x) == ± sup{ T : ( � , ± 7 ) E r} and M ± ( y ) == ± sup{ T : ( TJ , ±T ) E r}, and note that, because r is compact, all four of the points x ± (� ' M ± (x)) and y ± == ( TJ , M ± ( y )) are elements of r. Moreover, 2 l s l < M+ (x) - M - (x) and 2 l t 1 < M + ( y ) - M - ( y ) ; and therefore

In particular, this means that 2 2 2 2 2 < == t s t x s IY - l ITJ - �1 + I - l I TJ - �1 + ( l i + 1 l ) 2 2 + + < 111 - �1 + ( ( M ( y ) - M - (x)) V ( M (x) - M - ( y ) ) ) < (I Y + - x - 1 v IX + - y - 1) 2 < 4 rad(r) 2 . Finally, let R be a rotation. It is then an easy matter to check that P(Rv) == R ( P(v) ) and that f(Rr ; Rv, Rx) == f(r ; v, x) for all x E P(v). Hence, S(Rr, Rv) == RS(r, v) . In particular, if r == Rr and L(v) == R (L (v)) , then Rv == ± v, and so the preceding ( together with (4.2.2)) leads to RS( r, v ) == S(r, v). D Theorem. The inequality in (4.2.1) holds for every bounded r C �N . PROOF: Clearly it suffices to treat the case when r is compact and therefore Borel measurable. Thus, let a compact r be given, choose an orthonormal basis {e l , . . . , eN } for �N , set r o == r, and define rn == S(rn - l ; en) for 1 < n < N. By repeated application of (4.2.2) and Lemma 4.2.3, we know that 1 rn1 == 1 r 1 , rad(rN ) < rad(r) , and that Rm rn == rn , 1 < m < n < N, where Rm is the rotation given by Rm x == x - 2(x , em )RNem for each x E �N . In particular, this means that Rm rN == Rm rN for all 1 < m < N, hence - rN == rN , and therefore ( cf. the discussion preceding the introduction of Steiner ' s procedure) 4.2.4

We will now use ( 4.2.1) to give a description, due to F. Hausdorff, of Lebes­ gue ' s measure which, as distinguished from the one given at the beginning of

4.2 Steiner Symmetrization and the Isodiametric Inequality

r

77

Section 2. 1, is completely coordinate free. Namely, we are going to show that, for all C �N , (4.2.5)

1 r 1 e HN (r) inf { L ON rad(C) N ==

:

C a countable cover of

CEC

r} .

H (r)

We emphasize that we have placed no restriction on the sets C making up the cover C. On the other hand, it should be clear that N would be unchanged if we were to restrict ourselves to coverings by closed sets or, for that matter, to coverings by open sets. Directly from its definition, one sees that H N is monotone and subadditive in the sense that and

HN (Qrn) < �HN(rn)

rn

for all { } f' C P (R N ) .

Indeed, the first of these is completely trivial, and the second follows by choos­ ing, for a given E > 0, {Cm }1 so that c U cm and n N (rad ( C )) N < H N m + T m f

rm

(r )

L

CEC m

and noting that

Moreover, because

1r 1 e < L C i e for any contable cover C, the inequality 1 r 1 e < H N (r) is an essentially trivial consequence of ( 4.2.1). In order to prove the opposite inequality, we will use the following lemma. I

CEC

sequence {Bn}! property that

�N

with IGI < oo , there exists a of mutually disjoint closed balls contained in G with the

4.2.6 Lemma. For any open set G in 00

(4.2.7)

PROOF: If G == 0 , there is nothing to do. Thus, assume that G =I= 0 , and set

Go

==

G.

Using Lemma 2. 1.9, choose a countable, exact cover C0 of G0 by

78

IV Products of Measures

non-overlapping cubes Q. Next, given Q E C0 , choose x E JRN and 8 E [0, oo ) so that N i Q == Il [x - 8, xi + 8] and set BQ == B (x , � ) . 1 Clearly, the BQ ' s are mutually disjoint closed balls. At the same time, there is a dimensional constant a N E (0, 1) for which I BQ I > a N I Q I ; and therefore we can choose a finite subset { Bo, I . . . , Bo, no } C { BQ : Q E Co } in such a way that no Go \ U Bo,m < ,BN I Go l where ,BN 1 - ¥- E (0, 1). m= 1 Now set G 1 == G0 \ U�o Bo , m · Noting that G 1 is again non-empty and open, we can repeat the preceding argument to find a finite collection of mutually disjoint closed balls B1 , m C G 1 , 1 < m < n 1 , in such a way that n1 G 1 \ U B 1 ,m < ,BN I G 1 I · m= 1 More generally, we can use induction on f E z + to construct open sets Gt C Gt - 1 and finite collections Bt , 1 , , Bt , n�. of mutually disjoint closed balls B c Gt so that nt I Gt + l l < ,BN I Gt l where Gt+ 1 == Gt \ U Bt ,m ; m= 1 clearly the collection •

•

•

:

{ Bt ,m f E N and 1 < m < n�_} has the required properties. D RN . PROOF: As we have already pointed out, the inequality 1r1e < H N ( r ) is an immediate consequence of ( 4.2. 1 ) . To get the opposite inequality, first observe that HN ( r ) == 0 if 1r1e == 0. Indeed, if 1r1e == 0, then, for each E > 0 we can first find an open G � r with I G I < E and then, by Lemma 2.1.9, a countable, exact cover C of G by non-overlapping cubes Q, which means that 4.2.8 Theorem. The equality in (4.2.5) holds for any set r c

Next, because H N is countably subadditive, it suffices to prove that H N ( r ) < l r le for bounded sets r. Finally, suppose that r is a bounded set, and let G be

4.2 Steiner Symmetrization and the Isodiametric Inequality

79

any open superset of r with I G I < oo . By Lemma 4.2. 7, we can find a sequence {Bn}! of mutually disjoint closed balls B C G for which CX)

I G \ A I == 0 where A == U Bn. 1

Hence, because H N (r) that H N (f) <

<

H N (G)

<

H N (A) + H N (G \ A) == H N (A), we see

L ONrad (Bn) N == L I Bn I == I A I == I G I CX)

1

1

for every open G � r ; and, after taking the infimum over such G ' s, we arrive at the desired conclusion. D Exercises

Using the definition of HN given by the second relation in (4.2.5), give a direct ( i.e., one which does not make use of the first relation in ( 4.2.5)) proof that H N (f) == 0 for every bounded subset of a hyperplane ( cf. (i) in Exercise 2.2.3) of � N .

4.2.9 Exercise:

Chapter V C hanges of Variable

5.0 Introduction.

We have now developed the basic theory of Lebesgue integration. However, thus far we have nearly no tools with which to compute the integrals which we have shown to exist. The purpose of the present chapter is to introduce a technique which often makes evaluation, or at least estimation, possible. The technique is that of changing variables. In this introduction, we describe the technique in complete generality. In ensuing sections we will give some examples of its applications. Let (E1 , B1 ) and (E2 , B2 ) be a pair of measurable spaces. Given a measure M on (E 1 , B1 ) and a measurable map � on (E1 , B1 ) into (E2 , B2 ), we define the pushforward or image �*M == M o � - 1 of M under � by �* M (r) == M ( � 1 (r) ) for r E B2 . Because set theoretic operations are preserved by inverse maps, it is an easy matter to check that �* M is a measure on (E2 , B2 ). 5.0.1 Lemma. For every non-negative measurable function cp on (E2 , B2 ), -

( 5.0.2 ) Moreover, cp E £ 1 ( E2 , B2 , �* M ) if and only if cp o � holds for all cp E L 1 ( E2 , B2 , �* M ) ·

E

L 1 (E1 , B 1 , M ) , and ( 5.0.2 )

PROOF: Clearly it suffices to prove the first assertion. To this end, note that ( 5.0.2 ) holds, by definition, when f is the indicator of a set r E B2 . Hence, it

also holds when f is a non-negative measurable simple function on (E2 , B2 ). Thus, by the Monotone Convergence Theorem, it must hold for all non-negative measurable functions on (E2 , B2 ). D The reader should note that Lemma 5.0. 1 is really a definition more than it is an honest theorem. It is only when a judicious choice of � has been made that one gets anything useful from ( 5.0.2 ) .

5.1 Lebesgue

Integrals vs. Riemann Integrals

81

Exercises

Referring to Theorem 2.2.2, let A be a non-singular N N matrix and TA the associated linear transformation on � N . Show that (TA)*AR N == j det(A) j - 1 AR N .

5.0.3 Exercise:

x

5 . 1 Lebesgue Integrals vs. Riemann Integrals.

Our first important example of a change of variables will relate integrals over an arbitrary measure space to integrals on the real line. Namely, given a measurable �-valued function f on a measure space (E, B, JL) , define the distribution of f under Jl to be the measure Jl t f*Jl on (�, �) . We then have that, for any non-negative measurable cp on (�, �) _

(5.1.1)

L r.p o f(x) J-L(dx) .k r.p (t) J-Lt (dt). =

The reason why it is often useful to make this change of variables is that the integral on the right hand side of ( 5. 1.1 ) can often be evaluated as the limit of Riemann integrals to which all the fundamental facts of the calculus are applicable. In order to see how the right hand side of (5.1. 1) leads us to Riemann in­ tegrals, we will prove a general fact about the relationship between Lebesgue and Riemann integrals on the line. Perhaps the most interesting feature of this result is that it shows that a complete description of the class of Riemann inte­ grable functions in terms of continuity properties defies a totally Riemannian solution and requires the Lebesgue notion of almost everywhere. 5 . 1 . 2 Theorem. Let v be a finite measure on ((a , b], B (a ,b]) , where - oo < a < b < oo , and set 'lj;(t) == v((a, t]) for t E [a, b] ( '1/J (a) == v(0) == 0) . Then, 'ljJ is right-continuous on [a , b ) , non-decreasing on [a, b] , '1/J (a) == 0, and, for each t E (a, b] , 'lj;(t) - 'lj;(t-) == v( {t} ) , where 'lj;(t - ) lim s/ t 'lj;(s) is the left-limit of 'ljJ at t. Furthermore, if cp is a bounded function on [a , b] , then cp is Riemann integrable on [a , b] with respect to 'ljJ if and only if cp is continuous (a. e. , v) on (a , b] ; in which case, cp is measurable on ( (a , b] , B {a ,b] ) and (5. 1.3)

1(a ,� r.p d v = (R) �f[a ,� r.p (t) d'lj;(t) .

(See Exercise 7. 1.31 to learn how to go from a right-continuous, non-decreasing 'ljJ to a measure Jl · )

PROOF: It will be convenient to think of v as being defined on ( [a , b] , B [a ,b]) by v(r) v(r n (a, b] ) for r E B[ a ,b] . Thus, we will do so.

82

V Changes of Variable

Obviously V; is non-decreasing on [a , b] ; and therefore ( cf. Lemma 1.2.20) V; has at most countably many points of discontinuity. Moreover, V;( a) == v(0) == 0, for each s E [a , b ) V;(s) == JL ( (a , s ]) == limt � s JL ( (a, t] ) == limt � s V;(t) , and for each t E (a, b] V;(t) - V;(t - ) == lim s/ t v ((s , t] ) == v( {t} ). Assume that cp is Riemann integrable on [a, b] with respect to V; . To see that cp is continuous ( a.e., v ) on (a , b] , choose, for each n > 1, a finite, non­ overlapping, exact cover Cn of [a , b] by intervals I such that II Cn II < � and V; is continuous at I - for every I E Cn. If � == U� {I - : I E Cn}, then JL(�) == 0. Given m > 1, let Cm, n be the set of those I E Cn such that sup 1 cp - inf1 cp > � . It is then easy to check that CX) CX) {t E (a, b] \ � : cp is not continuous at t} c u n U cm,n · m= l n= l But, by Exercise 1.2.26, 1

and therefore v (U:= l n� u Cm, n) == 0. Hence, we have now shown that cp is continuous ( a.e., v ) on (a, b] . Conversely, suppose that cp is continuous ( a.e., v) on (a , b] . Let { Cn}1 be a sequence of finite, non-overlapping, exact covers of [a , b] by intervals I such that II Cn ll � 0. For each n > 1, define 'Pn (t) == sup1 cp and cpn (t) == inf1 cp for t E I \ {I - } and I E Cn · Clearly, both 'Pn and -cpn are measurable on ( (a, b] , B< a,b] ) . Moreover, inf cp < -cpn < cp < 'Pn < sup cp ( a , b] 1

( a , b]

for all n > 1. Finally, since cp is continuous ( a. e., v) , cp

and so, not only is cp ( (a, b] , B (a , b] ) , but also -

v

== nlim ----+- CX) -cpn == nlim ----+- CX) 'Pn ( a.e., v); ==

limn CX) cpn ( a.e., v) and therefore measurable on ---+-

r -
=

5.1 Lebesgue

83

Integrals vs. Riemann Integrals

as n � oo. From this it is clear both that cp is Riemann integrable on [a , b] with respect to 1/J and that (5 . 1 . 3 ) holds. D We are now ready to prove the main result to which this section is devoted. 5.1.4 Theorem. Let (E, B, JL) be a measure space and f a non-negative, measurable function on (E, B). Then t E (0, oo) � JL(f > t) E [0, oo] is a

right-continuous, non-increasing function. In particular, it is measurable on ( ( 0, oo), B( o , oo ) ) and has at most a countable number of discontinuities. Next, assume that cp E C ( [0, oo)) nC 1 ( (0, oo)) is a non-decreasing function satisfying cp(O) == 0 <

0, and set cp( oo) == lim t � oo cp ( t ) . Then

f r.p o f ( x) J-L( dx) = fo oo r.p ' ( t)J-L(f > t) AIR ( dt) . J( , ) JE Hence, either JL(f > 8) == oo for some 8 > 0, in which case both sides of ( 5.1.5) are infinite, or for each 0 < 8 < r < oo the map t E [8, r] � cp '(t)JL(f > t) is (5. 1.5)

Riemann integrable and

o f(x) J-L( dx) = lim (R) f r.p ' (t ) J-L(f > t) dt. f r.p }E r/8 �oo0 }[8,r] PROOF: It is clear that t E (0, oo) � JL(f > t) is right-continuous and non­ increasing. Hence, if 8 == sup{ t E (0, oo) : JL(f > t) == oo } , then JL (f > t) == oo for t E (0, 8) and t E (8, oo) � JL(f > t) has at most a countable number of

discontinuities. Furthermore, if

h (t ) == Jl ( f > ktl ) for t E ( � , ktl J , k > 0, and n > 1, n

then each hn is clearly measurable on ( (O, oo), B( o , oo ) ) and h (t ) � JL(f > t) for each t E (O, oo). Hence, t E (O, oo) � JL(f > t) is measurable on ( ( 0, oo), B(o , oo ) ) . We turn next to the proof of ( 5.1.5). Since (cf. Exercise 3.3. 18) n

limo f o:�

}( o: , 8]

limo (R) {6 r.p' ( t) dt = r.p( 8) r.p' ( t) dt = o:� l

o:

and therefore

(l

'P

0

f dfJ,

) (lco, oo ) r.p' (t) J-L(f 1\

>

t) AJR (dt )

)

>

r.p (8 ) J-L(f > 8) ,

it is clear that both sides of (5. 1.5) are infinite when JL(f > 8) == oo for some 8 > 0. Thus we will assume that JL(f > 8) < oo for every 8 > 0. Then the restriction of Jl J to B[8 , oo ) is a finite measure for every 8 > 0. Given 8 > 0, set

84

V Changes

of Variable

==

J.t J ((8, t]) for t E [8, oo) and apply Theorem 5.1.2 and Theorem 1.2.7 to see that r dJ.L J = (R) r

j

1{8 < }

,

)

=

1[8,r]

=

1[8,r]

==

cp(8)�-t(8 < f < r) +

j(8, r) J.t(t < f < r) cp' (t) AR(dt)

for each r E ( 8, oo) . Hence, after simple arithmetic manipulation and an application of the Monotone Convergence Theorem, we get

{

1{8 < / < oo }

(

j(8, oo) cp' (t)�-t(t < f < oo) AR(dt)

after r / oo. At the same time, it is clear that f

('P o J -

j( 8, oo)

and, after combining these, we now arrive at

{

1{! > 8 }

('P o f(x) -

j(8,oo ) J.t(f

>

t) cp' (t ) AR(dt).

Thus, (5.1.5 ) will be proved once we show that

r r f(x) f(x) J.L (dx). (8) (dx) ('P ) = J.L

8} 1 0

E

0

But if 0 < 81 < 82 < oo, then 0 < (cp o f - cp (82 ) ) 1 { /> 82 } < (cp o f ­ cp(8 ) ) 1 { / > 8 } , and so the required convergence follows by the Monotone Con­ vergence Theorem. Finally, to prove the last part of the theorem when J.t(f > 8) < oo for every 8 > 0, simply note that I

1

r

J(O

cp ' (t)�-t(f j r/ oo (li, r]

lim 1 (t) J.L (f > t) AR(dt) = 6)0

and apply Theorem 5.1.2. D

>

t) AR(dt) ,

5. 2 Polar

Coordinates

85

Exercises

5 . 1 . 6 Exercise:

Here are two familiar applications of the ideas discussed in

this section. (i) Let 1/J be a continuous, non-decreasing function on the compact interval [a, b] . Show that (5. 1 . 7)

(R) { f o '1/J ( s) d'lj;( ) = (R) { s

1[a , b]

1[1/J (a),'lj;(b)]

f(t) dt, f E C ( [a, bl ) .

Suppose that Jl is a measure on ( �, BR) with the properties that JL(l) < oo for each compact interval I and JL( { t}) = 0 for each t E � - Let q, : � � � be a function satisfying JL( [a, b] ) = q,(b) - q,( a ) for all - oo < a < b < oo . Note that q, is necessarily continuous and non-decreasing, and show that q,*Jl coincides with the restriction of AR to BR [q,(�) J . (ii)

A particularly important case of Theorem cp(t) = tP for some p E (0, oo ), in which case (5. 1 . 5) yields 5 . 1 . 8 Exercise:

5. 1 .4

1s when

(5. 1 .9)

Use (5. 1 .9) to show that I J I P is JL-integrable if and only if CX)

CX)

Ln= 1 nP+ l J-l ( l f l > ! ) + nL= 1 nP- 1 1-l ( l f l > n) 1

< oo .

Compare this result to the one obtained in the last part of Exercise 3.3. 16.

5.2 Polar Coordinates.

From now on, at least whenever the meaning is clear from the context, we will use the notation "dx " instead of the more cumbersome ",\R N ( dx ) " when doing Lebesgue integration with respect to Lebesgue ' s measure on �N . In this section, we examine a change of variables which plays an extremely important role in the evaluation of many Lebesgue integrals over �N . Let s N - 1 denote the unit (N - 1 )-sphere { x E �N : lx l = 1 } in � N , and define : � N \ {0} -----+ s N - 1 by q,(x) = 1�1 · Clearly q, is continuous. Next, define the surface measure AsN on s N - 1 to be the image under q, of N AR N restricted to Bn ( o , 1 )\{ 0 } · Noting that q,(rx) = q,(x) for all r > 0 and x E �N \ {0} , we conclude from Exercise 5.0.3 that { j o if!(x) dx = r N { j o if! (x) dx 1B ( O , r)\{ 0 } 1B (0, 1 )\{ 0 }

-1

V Changes of Variable

86 and therefore that

N r {n j o if!(x) dx = N {sN- 1 j(w) >. gN- 1 (dw) (5.2.1) J J ( O , r)\{ 0 } for every non-negative measurable f on ( S N - 1 , B8N-1 ). In particular, using N--11 - AsN - 1 ( s N - 1 ) to denote the surface area of s N - 1 ' we have that W� is the volume n N of the unit ball B(O, 1) in �N . Next, define w : (O, oo) X s N - 1 � �N \ {0} by w(r, w) = rw. Note that w is one-to-one and onto; the pair (r, w) w - 1 (x) = (lx l , �(x)) are called the polar coordinates of the point x E � N \ {0}. Finally, define the measure RN on ( (0, oo) , B( o , CX) ) ) by RN (r) = fr r N - 1 dr. The importance of these considerations is contained in the following result.

w

5.2.2 Theorem. Referring to the preceding, one has that

ARN = w* ( RN X AsN- 1 ) a

on

BR N \{ 0 } ·

In particular, if f is non-negative, measurable function on

(

(�N , BR N ) ,

then

)

r N j(x) dx = r r N - 1 rsN- j(rw) ).gN- 1 (dw) dr l 1 JR J(O , CX) ) (5.2.3) = rsN- 1 r f(rw)r N - 1 dr AsN- 1 (dw) . J( o , CX)) l PROOF: By Exercise 3.3.22 and Theorem 4.1.5, all that we have to do is check that the first equation in ( 5.2.3 ) holds for every f E Cc ( �N ) {! E C ( �N ) : f 0 off of some compact set}. To this end, let f E Cc( �N ) be given and set F(r) = fn ( o ,r) f(x) dx for r > 0. Then, by (5.2.1), for all r, h > 0:

(

)

=

f(x) dx F(r + h) - F(r) = f jB ( O , r + h)\B (O , r) f o \II ( r, if!(x)) dx B ( O , r + h)\B ( O , r) ( f(x) - f w(r, �(x)) ) dx + B ( O , r + h)\B ( O , r) (r + h) N - r N o(h), sN-l (dw) + o w) \II ( r, f >. N gN- 1 where "o ( h)" denotes a function which tends to 0 faster than h. Hence, F is continuously differentiable on (0, oo) and its derivative at r E (0, oo) is

J

=

i

J

0

5. 2 Polar Coordinates

87

given by r N - 1 fsN- 1 f o 'll (r, w) AsN-1 (dw). Since F(r) -----+ 0 as r '\. 0, the desired result now follows from Theorem 5. 1.2 and the Fundamental Theorem of Calculus. D Exercises

5.2.4 Exercise:

In this exercise we discuss a few elementary properties of

AgN- 1 . ( i ) Show that if r =I= 0 is an open subset of s N - 1 , then AgN- 1 (r) > 0. Next, show that AsN- 1 is rotation invariant . That is, show that if 0 is an N N­ orthogonal matrix and To is the associated transformation on �N ( cf. the paragraph preceding Theorem 2.2.2), then ( To ) * AsN- 1 == AsN- 1 . Finally, use this fact to show that (5.2.5 ( i )) (e, w)RN AsN- 1 (dw) = o sN- 1 and (5.2.5 ( ii)) (e , w ) R N (TJ , w) JRN AsN- 1 ( dw) = n N (e , TJ ) RN S N- 1 for any e , TJ E �N . Hint: In proving these, let e E � N \ {0} be given and consider the rotation Oe which sends e to - e but acts as the identity on the orthogonal complement of e . x

I

I

( ii ) Define

:

[ ��::] ,

and set J.L = * A [o, 2 ,.1 . Given S 1 by (O) = any rotation invariant finite measure v on (S 1 , B81 ), show that v = "�!1 ) J.L· In particular, conclude that ,\8 1 == M · ( Cf. Exercise 5.3. 15 below. ) Hint: Define 0 cos () sin () for () E [0, 21r] , == - sin 0 cos 0 and note that v = _.!._ f f f d f f o To8 (w) v (dw) dO. 27r s1 s1 J[o, 21r) l l ( iii ) For N E z + ' define 3 : [ - 1, 1] X s N - 1 -----+ s N by 1 2 S(p, w) = ( 1 - � ) 2 w ' and let l 2 1 (r) = � p ( - ) AIR A8N- 1 (dp dw) J.L N for r E B[ - 1 , 1 ) X BsN- 1 . Show that AsN == 3* M N · [0, 21r]

0

-----+

[

]

(

£

)

]

[

x

x

88

V Changes of Variable

( Hint: Consider

(

)

{ rN { r 3 (p, w)) J-L N (dp X dw ) dr f ( sN1 1[ - 1 , 1 ] x 1( o, oo ) for continuous f : �N + 1 � � with compact support.) Finally, use this result to show that rsN J ( ( O , w)JRN ) AsN (dw) = WN - 1 r f(p) ( l - p2 ) � - 1 dp 1 1[ - 1 , 1 ] for all (} E s N and all measurable f on ( [- 1, 1] , B[ - 1 , 1 1 ) which are either bounded or non-negative. Perform the calculations outlined in the following. (i ) Justify Gauss ' s trick: 2 r � � r e - dx == rJR2 e - 2 dx == 27r r re - 2 dr == 27r 1( o, oo ) 1R 1 and conclude that for any N E z + and symmetric N N-matrix A which is strictly positive definite ( i.e., all the eigenvalues of A are strictly positive), 5 . 2 .6 Exercise:

(

2

2)

x

(5.2.7) try the change of variable �(x) == TA _ ! x. ( ii ) D efine r ( 1' ) == f( o, oo ) fY - 1 e - t d for 1' E (0, oo). Show that, for any 1' E (0, oo ) , r(1' + 1) == 1'r( 1') · Also, show that r ( � ) == 1r � . The function r( · ) is called the Euler ' s Gamma function. Notice that it provides an extension of the factorial function in the sense that r( n + 1) == n! for integers n > 0. ( iii ) Show that N 2 2{ 7r) WN - 1 = r ( � ) ,

Hint :

t

and conclude that the volume O N of the N-dimensional unit ball is given by N 2 7r f! N == r ( -N + 1 ) ( iv ) Given a , E (O, oo) , show that

/3

. 2

1(O, oo ) t-! exp [ 2 t 2

-a

-

/3-t2 ]

5. 3

89

Jacobi 's Transformation and Surface Measure

Finally, use the preceding to show that

1

[

]

a2 7r ! e - 2 o ,B 2 � fJ t- exp - a t - t dt == a . fJ ( O ,CX)) Hint: Define 'lj;( s ) for s E � to be the unique t E (0, oo ) satisfying s at ! - (3t - ! , and use part ( i ) of Exercise 5. 1 .6 to show that 2 o ,8 2 e (3 2 - s ds. e t - 1 exp - a t - - dt == t a R

1

( O ,CX))

2

2

[

-

]

1

2

5.3 Jacobi's Transformation and Surface Measure.

We begin this section by deriving Jacobi ' s famous generalization to non­ linear maps of the result in Theorem 2.2.2. We will then apply Jacobi ' s result to show that Lebesgue 's measure can be differentiated across a smooth surface. Given an open set G C �N and a continuously differentiable map* x E G r-----+ 4l(x) ==

�(x)

E �N '

we define the Jacobian matrix J4l(x) == �! (x) of 4l at x to be the N N­ matrix whose jth column is the vector x

84>1 8x · J

In addition, we call 8 4l( x ) l det(J4l(x)) l the Jacobian of 4l at x. 5 .3. 1 Lemma. Let G be an open set in � N and 4l an element of C ( G ; � N )

1

whose Jacobian never vanishes on G. Then tit maps open (or BRN -measurable) subsets of G into open or (BRN -measurable) sets in � N . In addition, if r C G with 1 r 1 e == 0, then 1 4l (r) l e == 0; and if r c 4l( G ) with 1 r 1 e == 0, then (r) l e == 0. In particular,

I tlt - 1

* Because we will be dealing with balls in different dimensional Euclidean spaces here , we will use the notation BJR( a , r) to emphasize that the ball is in � N .

V Changes

90

of Variable

PROOF: By the Inverse Function Theorem,t for each x

E

G there is an open neighborhood u c G of X such that 4l r u is invertible and its inverse has first derivatives which are bounded and continuous. Hence, G can be written as the union of a countable number of open sets on each of which 4l admits an inverse having bounded continuous first order derivatives; and so, without loss in generality, we may and will assume that � admits such an inverse on G itself. But, in that case, it is obvious that both 4l and 4l - 1 are continuous. In addition, by Lemma 2.2. 1 , both take sets of Lebesgue measure 0 into sets of Lebesgue measure 0. Hence, by that same lemma, we now know that both 4l and 4l - 1 take Lebesgue measurable sets into Lebesgue measurable sets. D A continuously differentiable map 4l on an open set U C �N into �N is called a diffeomorphism if it is injective (i.e., one-to-one ) and 8clt never vanishes. If 4l is a diffeomorphism on the open set U and if W == 4l( U ), then we say that 4l is diffeomorphic from U onto W . In what follows, for any given set r C �N and 8 > 0, we use r < 6) - { X E �N : I Y - x l < 8 for some y E r} to denote the open 8-hull of r. 5.3.2 Theorem (Jacobi's Transformation Formula) . Let

set in � N and 4l an element of C2 (G; �N ). ishes, then, for every measurable function f measurable on ( G, BR N [G] ) and (5.3.3)

r

J( y ) dy <

jif!(G)

1G f

0

G

be an open If the Jacobian of cit never van­ on ( 4l(G), BR N [4l(G)] ) , f o cit is

4l (x) 84l(x) dx

whenever f is non-negative. Moreover, if 4l is a diffeomorphism on (5.3.3) can be replaced by

( 5.3.4 )

r

jif!(G)

f( y ) dy =

1G f

0

G,

then

4l(x) 84l(x) dx .

PROOF: We first note that (5.3.4 ) is a consequence of (5.3.3) when cit is one-to­

one. Indeed, if 4l is one-to-one, then the Inverse Function Theorem guarantees that 4l - 1 E C 2 (4l(G) ; �N ) . In addition, Hence we can apply (5.3.3) to 4l - 1 and thereby obtain

1G t

f fl- (x) c5 +(x) dx < r 0

jif! ( G )

J( y ) ( c5f!J ) · - 1 ( y ) c5f!J - 1 ( y ) dy = r 0

jif! ( G )

J( y ) dy ;

See, for example, W. Rudin ' s Principles of Mathematical Analysis, McGraw Hill ( 1976) .

5. 3

Jacobi 's Transformation and Surface Measure

91

which, in conjunction with (5.3.3) , yields (5.3.4) . We next note that it suffices to prove (5.3.3) under the assumptions that G is bounded, 4l on G has bounded first and second order derivatives, and 8 4l is uniformly positive on G. In fact, if this is not already the case, then we can choose a non-decreasing sequence of bounded open sets Gn so that 4l r Gn has these properties for each n > 1 and Gn / G. Clearly, the result for 4l on G follows from the result for 4l r Gn on Gn for every n > 1. Thus, from now on, we assume that G is bounded, the first and second derivatives of 4l are bounded, and 8clt is uniformly positive on G. Let Q == Q ( c; r ) == flf [ci - r, ci + r] C G. Then, by Taylor 's Theorem, there is an L E [0, oo ) (depending only on the bound on the second derivatives of 4l) such that (Cf. Section 2.2 for the notation here.) At the same time, there is an M < oo (depending only on L, the lower bound on 8 4l, and the upper bounds on the first derivatives of 4l) such that

Hence, by Theorem 2.2.2, 1 4l(Q) I < 84l(c) I Q ( O, r + Mr 2 ) I == (1 + Mr) N 84l(c) I Q I . Now define JL(r) == fr 84l(x) dx for r E BRN [G] , and set v == 4l*JL• Given an open set H C 4l (G), use Lemma 2.1.9 to choose, for each m E z + , a countable, non-overlapping, exact cover Cm of 4l - 1 (H) by cubes Q with diam( Q) < ! . Then, by the preceding paragraph,

where CQ denotes the center of the cube Q. After letting m � oo in the preceding, we conclude that I H I < v(H) for open H C 4l(G) ; and so, by Exercise 3. 1.9, it follows that (5.3.5)

l r l < v(r) for all r E BR N [4l(G)] .

Starting from (5.3.5) , working first with simple functions, and then passing to monotone limits, we now conclude that ( 5.3.3 ) holds for all non-negative, measurable functions f on (4l(G) , BR N [4l(G)]) . D As an essentially immediate consequence of Theorem 5.3.2, we have the following.

V Changes of Variable

92 5.3.6 Corollary. Let

morphism. Set

G be an open set in JRN and cit E C 2 ( G; JRN )

M� (r) =

l 8+(x) dx

for r E

BR N [ G ] .

Then clt*Jlif! coincides with the restriction Aif!(G) of AR N to particular,

f E L 1 (cit ( G), BRN [cit ( G)] , Aif!(G) )

�

a diffeo­

BR N [clt ( G )] .

In

f � E L 1 ( G, BR N [G] , Jlif!) ' 0

in which case (5.3.4) holds.

As a mnemonic device, it is useful to represent the conclusion of Corollary 5.3.6 as the statement f( y ) dy == f o �(x) 8�(x) dx when y == clt(x). We now want to apply Jacobi ' s formula to show how to differentiate Lebes­ gue 's measure across a smooth surface. To be precise, assume that N > 2 and say that M c RN is a hypersurface if, for each p E M, there exists an r > 0 and a three times continuously differentiable F : BR N (p, r) � lR with the properties that N (p, r) n M == { y E BR N (p, r) : F(y) == o } B R (5.3.7) and I VF( y ) l -1- 0 for any y E BR N (p, r) , where \7 F(y ) [F , 1 ( y ), . . . , F , N ( Y ) ] is the gradient of F at y and, once again, we have used the notation F ,j to denote the partial derivative in the direction ej . Given p E M, the tangent space Tp ( M ) to M at p is the set of v E JR N for which there exists an E > 0 and a twice continuously differentiable curve 1 on ( - E , E ) into M such that 1(0) == p and i' (O) == v . ( If 1 is a differentiable curve, we use i' (t) to denote its derivative � at t.) Canonical Example: The unit sphere s N - 1 is a hypersurface in ]R N . In fact, at every point p E s N - 1 we can use the function F(y) == IYI 2 - 1 and can identify Tp (s N - 1 ) with the subspace of v E JRN for which v - p is orthogonal to p. 5 .3.8 Lemma. Every hypersurface M can be written as the countable union of compact sets and is therefore Borel measurable. In addition, for each p E M, Tp ( M ) is an (N - I)-dimensional subspace of R N . In fact, if r > 0 and F E C3 ( BR N (p, r) ; R) satisfy (5.3.7) , then, for every y E BR N (p, r)nM, T y ( M ) coincides with the space of the vectors v E JRN which are orthogonal to \7 F(y) . Finally, if, for r c M and p > 0, (5.3.9) r(p) { y E ]RN : :3p E M (y - p ) l_ Tp ( M ) and I Y - PI < p} } ,

5.3 Jacobi 's Transformation and Surface Measure

93

PROOF: To see that M is the countable union of compacts, choose, for each p E M, an r(p) > 0 and a function Fp so that (5.3.7) holds. Next, select a countable subset {Pn}! from M so that 00

M C U BR N (Pn, � ) , with rn 1

==

r(pn)·

Clearly, for each n E z+ ' Kn { y E BR N (Pn, � ) : F( y ) = 0 } is compact ; and M == U� Kn. Next, let p E M be given, choose associated r and F, and let y E BR N (p , r) n M. To see that VF( y ) l_ Ty (M) , let v E Ty (M) be given and choose t > 0 and 1 accordingly. Then, because 1 : ( - t , t ) � M, Conversely, if v E �N satisfying (v , VF( y )) R N == 0 is given, set V(z) = v -

( v , VF(z)) R N

F(z) V' 2 z F IV' ( ) l for z E BR N (p, r) . By the basic existence theory for ordinary differential equations,* we can then find an t > 0 and a twice continuously differentiable curve 1 : ( - t , t ) � BR N (p, r) such that 1(0) == y and i' (t) == V (1(t)) for all t E ( - t , t ) . Clearly i' (O) == v , and it is an easy matter to check that Hence, v E Ty (M) . To prove the final assertion, note that a covering argument, just like the one given at the beginning of this proof, allows us to reduce to the case when there is an r > 0 and an a three times continuously differentiable F : M (2r) � � such that I VF I is uniformly positive and M == {x E M (2r ) : F(x) == 0} . But in that case, we see that : x E r and r(p) = x + t VF(x) lt p , l < F(x) l i V' and so the desired measurability follows as an application of Lemma 2.2 . 1 to the Lipschitz function

{

}

D * See Chapter 1 of E. Coddington and N . Levinson's Theory of Ordinary Differential Equa­ tions, McGraw Hill ( 1 955) .

V Changes of Variable

94

We are, at last, in a position to say where we are going. Namely, we want to show that there is a unique measure AM on (M, BR N [M] ) with the property that (cf. (5.3.9)) ( 5.3. 10)

1 - AR N (r(p) ) AM (r) == plim �O 2p for bounded r E BR N with r c M.

Notice that, aside from the obvious question about whether the limit exists at all, there is a serious question about the additivity of the resulting map r � AM (r) . Indeed, just because r1 and r2 are disjoint subsets of M' it will not be true, in general, that r1 (p) and r2 (p) will be disjoint. For example, when M == s N - 1 and p > 1 ' r1 (p) and r2 (p) will intersect as soon as both are non-empty. On the other hand, at least in this example, everything will be all right when p < 1; and, in fact, we already know that (5.3. 10 ) defines a measure when M == S N - 1 . To see this, observe that, when p E ( 0, 1) and M == S N - 1 , r(p) =

{

y :

1 - p < IYI < 1 + p and , : ,

E

}

r '

and apply (5.2.3) to see that

where the measure AsN- 1 is the one described in Section 4.2. Hence, after letting p � 0, we see not only that the required limit exists but also gives the measure AsN - 1 . In other words, the program works in the case M == s N - 1 , and, perhaps less important, the notation used here is consistent with the notation used in Section 4.2. In order to handle the problems raised in the preceding paragraph for general hypersurfaces, we are going to have to reduce, at least locally, to the essentially trivial case when M == �N- 1 {0}. In that case, it is clear that we can identify M with �N- 1 and r(p) with r X ( -p, p) . Hence, even before passing to a limit, we see that x

In the lemmata which follow, we will develop the requisite machinery with which to make the reduction. p E

M there is an open neighborhood U of 0 in �N- 1 and a three times continuously differentiable injection (i.e. , one-to-one) \}1 U -----+ M with the properties that p == w(O) and, for each E U, the set { \}1 , 1 ( ) , . . . , \}1 , N 1 ( ) } forms a basis in T w ( u) ( M) . 5.3. 1 1 Lemma. For each :

u

_

u

u

5.3 Jacobi 's

Transformation and Surface Measure

95

PROOF: Choose r and F so that (5.3.7) holds . After renumbering the coor­

dinates if necessary, we may and will assume that F , N (P) =I= 0. Now consider the map Y1 - P1 E �N . YN - 1 - P N - 1 F( y ) Clearly, cit is three times continuously differentiable. In addition,

[ 1

J cit (p) == IRVN-

]

0 F , N (P ) '

where v == [F , 1 (p) , . . . , F , N _ 1 (p) J . In particular, 8 clt (p) =I= 0, and so the In­ verse Function Theorem guarantees the existence of a p E (0, r] such that cit r BR N (p, p) is diffeomorphic and clt - 1 has three continuous derivatives on the open set W cit ( BR N (p, p)) . Thus, if U

{u E � N - 1 : (u, O) E W },

then U is an open neighborhood of 0 in �N - 1 , and is one-to-one and has three continuous derivatives. Finally, because \}1 takes its values in M, it is obvious that

At the same time, as the first (N - 1) columns in the non-degenerate matrix Jclt - 1 (u) , the \}1 ,j (u) ' s must be linearly independent. Hence, they form a basis in Tw (u) ( M) . D Given a non-empty, open set U in �N - 1 and a twice continuously differen­ tiable injection \}1 U � M with the property that :

{ w , 1 , . . . , w , N - 1 } forms a basis in Tw(u) ( M ) for every u E U, we say that the pair ( \}1 , U) is a coordinate chart for M. 5.3. 12 Lemma. Suppose that (w, U) is a coordinate chart for M, and define (5.3. 13)

V Changes of Variable

96

Then 8 '}1 never vanishes, and there exists a unique twice continuously differ­ entiable n : U ------+ s N - 1 with the properties that n(u) l_ Tw ( u ) (M) and *

det [ W , 1 . . . W , N - 1 n(u) T ] = 8W(u)

for every u

E

U.

Finally, define

�(u, t) == W(u) + tn(u) T

for (u, t)

E

U X �-

Then

8 w(u, 0) == 8 w(u) , u E U, and there exists an open set U in �N such that � r U is a diffeomorphism, u == { u E �N - 1 : (u, 0) E U} , and w(U) == �(U) n M. In particular, if x and y are distinct elements of \}1 ( U), then { x} (p) n \}1 ( U) is disjoint from { y } (p) n w(U) for all p > 0. PROOF: Given a u E U and an n E s N - 1 which is orthogonal to Tw ( u ) (M) , { \}1 1 ( u), . . . , \}1 , N - 1 ( u) , nT } is a basis for �N and therefore det [ w , 1 (u) . . . W , N - 1 (u) nT ] =f- 0. Hence, for each u E U there is precisely one n(u) E s N - 1 n Tw ( u ) (M) with the property that det [ W , 1 (u) . . . W , N - 1 (u) n(u) T ] > 0. To see that u E U � n(u) E s N - 1 is twice continuously differentiable, set p == W(u) and choose r and F for p so that (5.3.7) holds. Then n(u) = ± 1 ���=� 1 , and so, by continuity, we know that, with the same sign throughout, n( w ) = ± \7 F(w(w)) I V' F(W(w)) l for every w in a neighborhood of u. Turning to the function \}1 , note that 8�(u, 0) 2 = (det [ W , 1 (u) . . . W , N - 1 (u) n(u) T J r W , 1 (u)T w , 1 ( u) . . . w , N - 1 ( u) n ( u) T ] == det [ \}1 , N - 1 ( ) T n(u) = 8W(u) 2 . = det (( ( W , i (u) , : ,j (u)) R N (5.3. 14)

,

U

[

)) �]

* Below, and throughout , we use A T to denote the transpose of a matrix A. In particular , if A is a row vector, then A T is the corresponding column vector, and vice versa.

5. 3 Jacobi 's Transformation and Surface Measure

97

Hence, (5.3. 14) is proved. In particular, this means, by the Inverse Function Theorem, that, for each u E U, there is a neighborhood of (u, 0) in RN+ 1 on which \}1 is a diffeomorphism. In fact, given u E U, choose r and F for p == w(u), and take p > 0 so that

'VF ( W ( w )) Then, because n (w) = ± j F ( w ( )) j for all w E BR N- 1 (u, p) , w V'

F ( �(w, t) ) == ± t ! VF ( w(w)) ! + E(w, t) for (w, t) E BR N- 1 (u, p) ( - � , � ) , where ! E(w, t) l < Ct2 for some C E (0, oo ) . Hence, by readjusting the choice of p > 0, we can guarantee that \}1 r BIRN - 1 ( u, p) X ( - p, p) is both diffeomorphic and satisfies x

In order to -prove we must find an open U in �N such - the final -assertion, that w(U) == w(U) n M and w r U is a diffeomorphism. To this end, for each u E U, we use the preceding to choose p( u) > 0 so that: BJRN - 1 ( u, p( u) ) C U and � r BJRN - 1 ( u, p( u) ) X ( - p( u) ' p( u) ) is both a diffeomorphism and satisfies Next, choose a countable set { un}1 C U so that CX)

U == U BJRN- 1 (un , if- ) where 1 and set

Pn

== p (un) ;

n

Un == U BJRN -1 (urn , e; ) and Rn == P 1 1\ . . . 1\ Pn · 1 To construct the open set U, we proceed inductively as follows. Namely, set E 1 == '¥- and k1 == u 1 X [ - E 1 , E 1 ] · Next given Kn , define En + 1 by

( {I

2En+ 1 = Rn + l 1\ En 1\ inf W ( ) - � ( t) and take

U

W,

I

:

U

W,

E Un + l \ Un , ( t) E kn ,

)

and l u - w l > R';+ 1 } ,

V Changes of Variable

98

Clearly, for each n E z +-' Kn is compact, Kn c Kn + 1 ' Un X ( - En, En) c K-n , and 8 '}1 never vanishes on Kn. Thus, if we can show that En > 0 and that \}1 f Kn is one to one for each n E z + ' then we can take [; to be the interior of u � Kn. To this end, first- observe - that there is nothing to do when n == 1. Furthermore, if En > 0 and \}1 f Kn is one to one, then En + 1 == 0 is possible only if there exists a u E BR N-1 (Un + 1 , Pn + 1 ) and -a (w, t) E BRN- 1 (Um , Pm ) X ( - pm , Pm ) for some 1 < m < n such that w(u) == w(w, t) and l u - w l > R�+ 1 • But, because this would mean that \}1 ( w) == \}1 ( u) and therefore, since \}1 is one to one, would R�+ 1 • Hence, En + 1 > 0. Finally, lead to the contradiction that 0 == ! w - u l > - to see that \}1 f Kn + 1 is one to one, we need only check that (u, s) E (Un + 1 \ Un) X [ - En + 1 , En + 1 ] and (w, t) E Kn ===} �(u, s) =/= �(w, t) . But, if l u - w l < R�+ 1 , then both (u, s) and ( t) are in BR N- 1 (un+ 1 , Pn + 1 ) ( -Pn + 1 , Pn + 1 ) and � is one to one there. On the other hand, if l u - w l > R�+ 1 , then l �(u, s) - � ( , t) l > l w(u) - �( , t) l - l sl > 2En + 1 - En + 1 == En + 1 > 0. D w,

x

w

w

5 . 3 . 1 5 Lemma. If (w, U) is a coordinate chart for M and

ment of BR N with

r c w(U) ,

then (cf.

(5.3. 13))

r a bounded ele­

(5.3. 16) in Lemma 5.3.12. Since r is a compact subset of the open set G == �(U), E == dist ( r, GC) > 0. Hence, for p E (0, E), PROOF: Choose U and \}1

as

which, by (5.3.4 ) and Tonelli ' s Theorem, means that

Since w - 1 (r ) == � - 1 (r ) is compact, 8� f w - 1 ( r ) is uniformly bounded and continuous. In particular, by Lebesgue ' s Dominated Convergence Theorem and (5.3. 14) ,

f _!_ 2p }( - , ) 8�(u, t) dt � 8�(u, 0) = 8W(u) p p

5.3 Jacobi 's

Transformation and Surface Measure

99

boundedly and point-wise for u E U. Hence, after a second application of Lebesgue ' s Dominated Convergence Theorem, we arrive at (5.3. 16). D 5.3. 1 7 Theorem. Let M be a hypersurface in � N . Then there exists a unique

measure AM on (M, BRN [M] ) for which (5.3. 10) holds. In fact, AM (K) < oo for every compact subset of M. Finally, if ( \}1 , U) is a coordinate system for M and f is a non-negative, BRN [M]-measurable function, then (cf. (5.4. 13))

(5.3.18)

r

lw ( U)

f(x) AM (dx) = r f W(u)8W(u) )..R N - 1 (du). lu 0

PROOF: For each p E M, use Lemma 5.3. 11 to produce an r(p) > 0 and a coordinate chart ( \}1 Up ) for M such that BRN (p, 3r(p) ) is contained in ( cf. Lemma 5.3. 12 ) w(Up ) · Next, select a countable set {Pn}! C M so that P'

CX) M C U BRN (pn , rn), where rn r(pn), 1

n- 1 Mn == (BRN (pn , rn) n M) \ u Mm for n > 2. 1 Finally, for each n E z + , define the finite measure Jln on (M, BRN [M] ) by

and set (5.3.19) Given a compact K c M, choose n E z + so that K c u� BRN (pm , rm ), and set r == r 1 1\ · · · 1\ rn and E == j . It is then an easy matter to check that, for any pair of distinct elements x and y from K, either l x - Y l > r, in which case it is obvious that {x}(E) - -n { y }(E) == 0 , or l x - Y l < r, in which case both {x}(E) and { y }(E) lie in W m (U m ) for some 1 < m < n and, therefore, the last part of Lemma 5. 3.12 applies and says that { x} (E) n { y } (E) == 0 . In particular, if r E BRN is a subset of K and rm == r n Mm , then, for each p E [0, E), {r1 (p ) , . . . , rn (P) } is a cover of r( p ) by mutually disjoint measurable sets. Hence, for each 0 < p < E: n ARN ( r (p ) ) == ARN ( rm (P) ) . 1 P

L

V Changes of Variable

100

At the same time, by Lemma 5.3. 15,

In particular, we have now proved that the measure defined in (5.3.19) satisfies (5.3. 10) and that AM is finite on compacts. Moreover, since (cf. Lemma 5.3.8) M is a countable union of compacts, it is clear that there can be only one measure satisfying (5.3. 10) . Finally, if ( \}1' U) is a coordinate chart for M and r E BR N with r c \}1 ( U) is bounded, then (5.3.18) with f == l r is an immediate consequence of (5.3.16). Hence, (5.3. 18) follows in general by taking linear combinations and monotone limits. D The measure AM produced in Theorem 5.3.1 7 is called the surface measure on M. Exercises

5.3.20 Exercise: In the final assertion of part ( ii ) in Exercise 5.2.4 and again in ( i ) of Exercise 5.2.6, we tacitly accepted the equality of 1r , the volume 02 of

the unit ball BR 2 (0, 1 ) in �2 , with the half-period of the sin and cos functions. We are now in a position to justify this identification. To this end, define 4l : G (0, 1) (0, 27r) ----+ �2 (the here is the half-period of sin and cos) by 4l(r, 8) == (r cos O, r sin O)T. Note that 4l(G) == B(O, 1) \ {x E B(O, 1) : x 1 == 0} and therefore that 1 4l(G) I == 0 2 . Now use Jacobi ' s Transformation Formula to compute 1 4l(G) I . 1r ,

x

1r

Let M a hypersurface in �N . Show that, for each p E M, the tangent space Tp (M) coincides with the set of v E � N such that 5.3.21 Exercise:

1. lmo dist(p +�2�v, M) e�

<

00 .

Given v E Tp (8G) , choose a twice continuously differentiable associated curve 'Y, and consider � r-----+ ! P + �v - 'Y ( �) I· Hint :

Given r > 0, set s N - l (r) == { X E �N : l x l == r } , and observe that s N - l (r) is a smooth region. Next, define � r : s N - l � s N - l (r) by � r (w) == rw ; and show that AsN-l (r) == r N - l (� r )*AsN- 1 . 5 .3.22 Exercise:

5.3.23 Exercise:

then AM (r) > 0.

Show that if r =I= 0 is an open subset of a hypersurface M,

5. 3 Jacobi 's Transformation and Surface Measure

10 1

In this exercise we introduce a function which is intimately related to Euler ' s Gamma function. ( i ) For (a, {3) E ( 0 , oo ) 2 , define

5.3.24 Exercise:

B (a, /3) = f u - 1 ( 1 - u) f1 - 1 du. 1(0, 1 ) 0

Show that

(a)r(/3) B ( a, {3) = rr(a + {3) where r( · ) is the Gamma function described in ( ii ) of Exercise 5.2.6. ( See part ( iv ) of Exercise 6.3.18 for another derivation.) The function B is called the Beta function. Clearly it provides an extension of the binomial coefficients in the sense that

(

m+n+ 1 m+n B (m + 1 , n + 1 ) m for all non-negative integers m and n.

)

Think of r(a) r({3) an integral in (s, t) over (O, oo) 2 , and consider the map uv 2• E oo) ( u , v ) E (0 , oo) (0 , 1 ) (0 , u( 1 v) as

Hint:

f--+

x

( ii ) For ,\ > � show that

[

]

where {cf. part ( ii ) of Exercise 5.2.6) W N _ 1 is the surface area of s N - 1 . In particular, conclude that

Use polar coordinates and then try the change of variable 4l{r) == 1 � r2 • ( iii ) For ,\ E { 0 oo ), show that 2

Hint :

,

and conclude that, for any N E z + ,

e ) � - 1 d�

WN . WN - 1 (- 1,1) Finally, check that this last result is consistent with part ( iii ) of Exercise 5.2.4.

1

(1 -

=

102

V Changes of Variable

Let U be a non-empty, open subset of JR N - 1 and f E C3 (U ; JR) be given, take 5.3.25 Exercise:

M = { (u, f(u)) : u E U} and W(u) =

[!�)] ,

u E U.

That is, M is the graph of f. (i) Check that M is a hypersurface and that (U, w) is coordinate chart for M which is global in the sense that M == w(U ) . (ii) Show that

(((w w ) IRN )) 1
,j

.

1

+ v J T V J,

Given a non-zero, row vector v, set A == l + v T v and e == 1�1 . Note that ( 1 + l v l 2 ) e and that Aw == w for w ..l e. Thus, A has one eigenvalue equal to 1 + l v l 2 and all its other eigenvalues equal 1. (iii) From the preceding, arrive at Hint: Ae ==

JM

.M fu

a formula which should be familiar from elementary calculus. Let G be a non-empty, open set in JRN , and assume that M {x E G : F(x) == 0} f= 0 and that F , N never vanishes on M. (i) Set U == {u E JR N - 1 : (u, t) E M for some t E JR}, and show that there exists an f E C3 (U ; JR) with the property that F (u, f(u)) == 0 for all u E U. In particular, M is the graph of f. (ii ) Define \}1 from f as in Exercise 5.3.25, and conclude that {

. M = { cpFi V F I o W (u) du . lu I , N I JM 5.3.26 Exercise: _

Again let G be a non-empty, open set in JRN and F E C3 (G; JR) , but this time assume that F , N vanishes nowhere on G. Set T == Range(F) and, for t E T, Mt == { x E G : F(x) == t} and Ut == {u E �N - 1 : (u, t) E Mt } · ( i ) Define cJl on G by 5 .3.27 Exercise:

4l ( x ) ==

XN- 1 F( x)

103

5.4 The Divergence Theorem

and show that � is a diffeomorphism. As an application of Exercise 5.3.26, show that, for each t E T, Mt is a hypersurface and that, for each cp E Cc (G) : Fl cpl\7 u, = o � - 1 (u, t) du, t E T. t) M d (

.. , l�(G) I F, N I Mt In particular, conclude that

1

1RN -1

is bounded and Br-measurable. (ii) Using Theorems 4.1.6 and 5.3.2, note that, for cp E Cc(G),

1

1;

o � - 1 ( y ) dy cp (x) dx = f jfJ!(G) , N I G = l�(G) ( u , t) I ; I o � - 1 (u , t) du dt . N After combining this with part (i) , arrive at the following (somewhat primitive) version of the co-area formula

)

£ (lN -l

(5.3.28) Take G == {x E �N : X N -1- 0}, F == lxl for x E G, and show that (5.2.3) can be easily derived from ( 5.3.28). (iii)

5 .4 The Divergence Theorem.

Again let N > 2. Perhaps the single most striking application of the con­ struction made in the second part of Section 4.3 is to multi-dimensional inte­ gration by parts formulae. Namely, given a non-empty open G in � N , we will say that the region G is smooth if its boundary 8G is a hypersurface in �N and, for each p E 8G, the number r > 0 and F (p, r) � � in ( 5.3. 7) can be chosen so that, in addition to ( 5.3. 7), :

BRN

(5.4.1) Notice that if G is a smooth region, then, for each p E 8G, there is a unique n(p) E s N - 1 n Tp ( 8G)_l_ with the property that, for some E > 0: GC if � E ( 0 ' E ) (5.4.2) p + � n (p ) E G if � E ( - t:, O).

{

V Changes of Variable

104

For obvious reasons, n (p) is called the outer normal to 8G at p. Notice that x E 8G � n (x) E s N - 1 is locally ( i.e., on each compact ) Lipschitz continuous, since if r and F are chosen for p so that (5.3.7) and (5.4. 1) hold, then (cf. Lemma 5.3.8) X E BJRN (p, r) n aG. n(x) = I \1 F(x) ' F(x) l Y' Our main goal in this section will be to prove that if f E Cc (� N ; �) ( i.e., vanishes off some compact subset ) and G is a smooth region, then for every w E sN- 1 : d

r f(x + �w) dx = r f(x) (w, n (x)) R N Aaa ( dx). � la e = o la c Observe that (5.4.3) is another way of seeing that surface measure is in truth the derivative of ARN across the surface. Indeed, since there is no requirement that f be differentiable, it must be Lebesgue ' s measure which is absorbing the derivative on the left hand side of ( 5.4.3) . The key to (5.4.3) is contained in the following lemma. 5 .4.4 Lemma. Let G be a smooth region. Then, for each p E 8G, there is a coordinate chart (w, U) for 8G and an > 0 with the properties that: U is bounded, p E w(U), and the associated map (5.4.5) (u, t) E U U X ( -€ , € ) � �(u, t) W(u) + tn(w(u)) T E � N d

(5.4.3 )

f.

is a diffeomorphism such that both � and � - 1 have bounded, continuous first and second order derivatives, and, for each u E U,

w(u, t) E G if and only if t E (- € , 0 ) .

(5.4.6)

Furthermore, given such a coordinate chart, (5.4.3 ) holds for every continuous � with compact support in (i.e., vanishing off a compact subset f : �N of) w(U) . PROOF: To prove the first part, use Lemma 5.3. 1 1 to choose some coordinate -----t

chart (4l, W) for 8G with p-E 4l(U) ,- assume that W is connected, and define associated n (w) , w E W, W, and 4l as in Lemma 5.3. 12. Clearly, n (w) == ± n ( 4l ( w)) , with the same choice of sign for every w E W. Hence, if necessary after replacing w by W' == { w : ( w 1 , . . . ' - W N - 1 ) E w} and cit by we may and will assume that n( w) ==- n ( 4l ( w)) for all w E W. In particular, - because 4l(W) n 8G == 4l(W) and 4l(w, t) -E G- for sufficiently small strictly negative t ' s, we know that, for each ( w, t) E W, 4l( w, t) E G if and only if t < 0.

5.4 The Divergence Theorem

105

Finally, choose t >- 0 so that BR N (p, 2 t) c 4- (U) , set U == � - 1 ( BR N -(p, t)n8G) , and take \}1 and \}1 to be, respectively, the restrictions of � and � to U and U == U X (-t, t). Turning to the second part of the lemma, set n == \}1 ( U) , and define p( y ) = W ( --J, - 1 ( y h , . . . , --J, - 1 ( y ) N - 1 ) and n( y ) = n(p( y ) ) for y E n. Given w E s N - 1 ' set

Wn (Y ) == (w, n( y )) R N and Wn ( Y ) == W0 ( y ) n (y ) , y E fl ; and, given f E C(�N ; �) with compact support K c n, set r ==dist ( K, nC ) . Then, by the translation invariance of Lebesgue ' s measure, for every � E �: �( � )

fa f(x + �w) dx - fa f(x) dx = � 1 (�) + � 2 (�),

where and

�2 (� )

L ( 1 c (x - �wn (x) ) - 1 a (x) ) f(x) dx.

In order to prove that (5.4.3) holds, we will show that

( ) � � 1 0 lim = ---+ 0 � e � 2 (� ) = f(x) wn (x) A (dx); lim aa e ---+ O �

(5.4.7)

1

G

and we will begin with the second, and easier part of ( 5.4. 7) . To this end, we use Theorem 5.3.2 and Fubini ' s Theorem to write � 2 (� ) = where g(� , u)

fu

g( � , u) du,

1� 1 < r,

- 1 ) [1 ( - oo ,O) (t - �wn (W(u)) ) - 1 ( - oo ,o) (t) ] f ( --J, (u, t) ) 8--J, (u, t) dt. ( - t: , t:

By elementary reasoning, one sees that, as � ---t 0,

V Changes of Variable

106

uniformly and boundedly for

.hm � 2 (x) e---. o �

=

u E U; and so (cf. (5.3. 18))

j f(u)wn ( W(u) ) 8W(u) .XR N- 1 (du) 1 =

u

aa

f(x) wn(x) Aaa (dx).

(5.4. 7) is more involved. To handle it, we introduce the D( y ) = q, - 1 ( Y ) N = (y - p( y ), n ( y ) ) JRN ' y E n. Observe that if p E f! n 8G, then ( cf. Exercise 5.3.20) . D(p + tv) - D (p) == 0 if v E Tp (8G) hm t---. o t 1 if v == n(p). The first part of notation

{

Hence,

( 5.4.8 )

V D(p) == n(p)

for p E

f!

n 8G.

Next, set

E( y , � ) == D (y - �wn ( Y ) ) - D( y - �w)

for

( y , �) E K ( - r, r). x

Clearly,

x - �w E G but x - �wn ( Y ) � G x - �w � G but x - �wn ( Y ) E G

===} ===}

0 < D (x - �wn (x) ) < E(x, � ) E(x, � ) < D(x - �wn (x)) < 0.

Thus,

� � � ( � ) � < ll f ll u AR N (r( � ) ) where r( � ) {X E K I D (x - Wn(x)) I < I E(x, � ) I }. =

In order to estimate

:

ARN ( r( � )) ' first observe that, for some cl < oo ,

Hence, since p ( x - �w0(x))

==

p(x),

which, together with ( 5.4.8 ) , leads to the existence of a c2 But, since w - w0(x) that

..l

<

00

for which

n(x) , this, in conjunction with Taylor 's Theorem, says

107

5.4 The Divergence Theorem for some

C < oo.

In other words, we have now shown that

r( � ) c { x E K : I D (x - �wn(x)) i < ce } C q, ( { (u, t) E fJ : l t - �wn (W(u)) l < ce } ) · Finally, set

and use Theorem

5.3.2 to conclude that

D

which is more than enough to prove the first part of (5.4.7) .

G be a smooth region in JR N and U an open neighbor­ hood of G. Then (5.4.3 ) holds for every f E Cc(U; JR) . In particular, if G is bounded, then ( 5.4.3 ) holds for every f E C(U; JR) . 5.4.9 Theorem. Let

PROOF: We first observe that, without loss in generality, we may assume that

U == JRN and that f E Cc (JR N ; lR). Indeed, when G is unbounded but f E Cc(U; JR), choose a compact set K C U so that f vanishes off K; and when G is bounded, choose K so that G C K and K C U. Next, in either case, let E denote the distance between K and UC and define j JR N � lR so that -

0

:

f(x) = [ ( � dist (x, UC ) ) 1\ 1 J f(x)

and j

0 off U.

It is then clear that j E

for x

Cc(lR N ; JR) .

E

U

Moreover, since

r i(x + �w) dx = r f(x + �w) dx for 1�1 < �2 . la la it is obvious that (5.4.3) for f is equivalent to (5.4.3) for f. In view of the preceding, we now assume that f E Cc (JR N ; JR) ; and we start the proof of (5.4.3) by observing that when f vanishes off the compact set K and K n {)G == 0 , then E == dist ( K, 8G) > 0 and

fa f(x

+

�w) dx =

Hence, in this case,

L l a (x - �w)f(x) dx = fa f(x) dx

(5.4.3 ) is essentially trivial.

for all

1�1 < E.

108

V Changes of Variable

To complete the proof, we must show how to reduce to the situations for which the result is already proved. Thus, let f E Cc (l� N ; �) be given and choose a compact K off which f vanishes. As we noted above, there is nothing more to do if K n 8G == 0. Thus, we assume that K n 8G =I= 0. Using the compactness of K n 8G and Lemma 5.4.4, we now choose coordinate charts { ( w m , Urn) }� of the sort described in that lemma, points {Pm }1 C K n 8G, and numbers {rm }1 C (O, oo) so that BJR N (Pm , 4rm) C � m ( Um) for each 1 -< m -< n and n

K n 8G C U BJR N (Pm , rm) 1

Set

·

n

H == �N \ U BJR N (pm , rm) ,

1

and define

'1/Jo ( x ) == dist ( x, HC ) and '1/Jm ( x ) == dist ( x, BJR N (pm , 3rm) C ) for 1 < m < n. It is then clear that each '1/Jm is a non-negative, continuous function and that n

s (x) - L 'l/Jm (x) > r 1 1\ · · · 1\ Tn > 0 for all x E �N . 0 Hence, if

x x '1/J )f( ( m fm ( x ) = s x ) ' x E IR. N and 0 < m < n, ( ) then each fm is continuous, f == 2:� fm , fo vanishes off a compact subset of 8G C , and, for 1 < m < n, fm vanishes off of a COmpact subset of � ( U ) Prn

In particular, this means that

1

Prn

·

1

d f(x + t:,w) dx = n d fm( x + t:,w) dx L e c dC c 0 d o = e =o e �

�

n

= L r fm (x) (w , n ( x )) JR N A ac (dx) = r f(x ) (w, n(x )) JR N Aac (dx). 0

lac

lac

D

Although ( 5.4.3 ) is the basic fact which we wanted to prove in this section, it does not present the result in the form which is most frequently encountered in applications. To see how to pass from ( 5.4.3 ) to the expression which we are after, assume that f is continuously differentiable in a neighborhood of G, and ( cf. Exercise 4. 1 . 13 ) note that ( 5.4.3 ) becomes

r ('V f ( x ) , w) JR N dx = r f (x) (w , n(x)) JR N Aac (dx) .

lc

lac

109

5.4 The Divergence Theorem

Next, suppose that F : � N � � N is once continuously differentiable and that F 0 off a compact set. By applying the preceding with f == Fk and w == ek and summing over 1 < k < N, we obtain

(5.4. 10)

{ div F (x) dx = { ( F ( x ) , n(x)) JR N Aaa (dx) , lac

lc

where div F �� Fk,k is the divergence of F. The formula in sufficiently important to warrant our stating it as a theorem.

(5.4.10)

is

5.4. 1 1 Theorem (Divergence Theorem) . Again let G be a smooth region

in � N and U an open neighborhood of G. If F : U � �N is continuously differentiable and either G is bounded or F 0 off of a compact subset of U, then { div F (x) dx = { ( F (x) , n(x)) JR N Aaa (dx) . _

lac

lc

Before dropping this topic, we will give some examples of the way in which The Divergence Theorem is used in the analysis of partial differential equations. Let � = �f" 1 8�::! denote the standard Laplacian. The following vari­ ant on The Divergence Theorem provides one of the keys to the analysis of equations in which � appears. 5.4. 12 Theorem (Green's Identity) . Let G be a smooth region in �N and

U an open neighborhood of G. If u and v are twice continuously differentiable �-valued functions on U and either G is bounded or u has compact support in U, then (5.4.13)

fa u �v dx - fa v �u dx 1 u ov Aac (dx) - 1 ==

where

d f(x

of (x)

�

On

ac

�

ac

un

ou A ac (dx ) , v� un

+ � n( x ) ) �=O = (V' f(x) , n(x)) JR N

denotes differentiation in the direction (cf. ( 5.4.2 ) ) n. PROOF: Simply note that u �v - v �u == div (u \7 v - v \7u) , and apply The Divergence Theorem. D In order to extract information from Green ' s Identity, one must make judi­ cious choices of v for a given u. For example, one often wants to take v to be the fundamental solution g given by g(O) == oo and

( 5.4.14)

- log l x l { g( x ) - l xi 2 - N _

2 N>3

if N == if

for X E �N \ {0} .

V Changes of Variable

1 10

Note that g and I Vg l are integrable on every compact subset of JRN . In addi­ tion, the following facts about g are easy to verify: if N == 2 -x N (5.4. 15) flg(x) == 0 and l x i \7g(x) == (2 - N)x if N > 3

{

on

JR N \ {0}.

Our first application allows us to solve the Poisson equation

flu == - f.

5 .4. 16 Theorem. Set CN == 21r or (N - 2) W N - 1 (cf. Exercise 5.2.6) depending on whether N == 2 or N > 3. Given f E c; ( lR N ; lR) ' define UJ on JRN by (cf.

(5. 1 . 14))

=

r g(x - y )f( y ) dy . }JRN CN Then U J E C2 ( JRN ; lR ) and flu! == -f . U J (X)

_2._

PROOF: First observe that another expression for UJ ( · ) is JJRN g( y )f( · - y ) dy , and it is clear ( cf. Exercise 4. 1 . 13) from this latter expression not only that U J E C2 ( 1RN ; JR) but also that flu, (x) == JJRN g ( y )flf(x - y ) dy . Thus, all that we need to do is check that JRN g( y )flf(x - y ) dy == - eN f(x). Fix x and choose R > 1 so that f 0 off of B(x , R - 1) . For 0 < r < R, set Gr == B(O, R) \ B(O, r) . Then

r N g ( y ) b.. f (x - y ) dy = rlim r g ( y ) b.. f (x - y ) dy ; � O R } Jar and, by Green ' s Identity and (5.4. 15) , for each 0 < r < R ( cf. Exercise 5.3.22) : r g ( y ) b.. f (x - y ) dy Jar of (x - y ) AsN- 1 ( r) (dy ) � == g( y ) f(x ) (y ) Aaar (dy ) y n n 0 0 8Gr 8Gr f o g( ) (x - y ) >. a B ( o, r) (dy ) y = a p 8B ( O , r)

1 1

1

+

i N-1

og ( y ) A B(O , r) (dy ) f(x - y ) 0p a

S (r ) of (x + rw)AsN- 1 (dw) == - r N - 1 gN- 1 g(rw) 0p g o 1 N + rw) +r (rw)AsN- 1 (dw), gN- 1 f(x 0p where g denotes differentiation in the outward radial direction and we have P used Exercise 5.3. 17 together with the fact that, for Gr , :n == - gp on s N - 1 (r). But r N - 1 g(rw) � 0 uniformly as r � 0 and ( cf. (5.4. 15) ) if N == 2 -1 g _ o 1 N r -p ( rw) == - (N - 2) if N > 3. o

1

i

{

5. 4

The Divergence Theorem

111

After combining this with the preceding, we now see that lim

r g( y ) f}.j(x - y ) dy

r �O Ja r

{ j(x + rw) ).. s N- 1 (dw) = - cN f(x) . = - WCN rlim �O sN1 N- 1 l

0

Our second application of Green ' s Identity will be to harmonic functions. A function u E C2 (G ; �) is said to be harmonic in G if �u == 0. Notice that if N == 1 and u is harmonic on (a, b) and continuous on [a, b] , then u(x) == �:=:�u(a) + ��=:Ju(b) for x E [a , b]. In particular, u ( atb ) is precisely the mean of the values that u takes on 8(a, b). We will now use Green ' s Identity to derive the analogous fact about harmonic functions in higher dimensions. 5.4. 17 Theorem ( The Mean Value Property ) . Suppose that u is an har­

monic element of C2 (G ; �) . Then, for each

B(x, R) c G,

x

E

G

and

R

>

0

satisfying

rsN u(x + Rw) AsN- 1 (dw) . WN - 1 l - 1 PROOF: Without loss in generality, we will assume that x == 0. Set ( cf. (5.4. 14) ) gn (x) == g(x) - g (Re 1 ) where e 1 == ( 1 , 0, . . . , 0) E s N - 1 . Then, by Green ' s Identity applied to the functions u and 9R in the region ( cf. the proof of the preceding ) Gr (5.4. 18)

u(x) =

1

0 = { (gR ( Y ) f}. u( y ) - u( y ) i}. gR ( Y ) ) dy Jar = - RN - 1 {sN- 1 u(Rw) Q:pR (Rw)AsN- 1 ( dw) l OU (rw)>.. s N- 1 (dw) - r N - 1 gN- 1 gn (rw) !l up

l

+ r N - 1 {sN- 1 u(rw) Q:pR (rw)).. s N-1 (dw), l

where we have used the same notation as in the preceding proof. Note that the first term on the right equals ( cf. (5.4. 15) )

l

CN u(Rw) AsN-1 , (dw) NWN - 1 S 1 the second term tends to 0 as r � 0, while the third term tends to - cN u(O) .

D

V Changes of Variable

1 12

Exercises

5.4. 19 Exercise: Let

u

be a twice continuously differentiable function in a neighbor hood of the closed ball B ( x, T ) , and assume that �u < 0 in B ( x , T ) . Generalize the Mean Value Property by showing that (5.4.20) u(x) > 1 { u(x + rw) .\gN- 1 (dw)

WN - 1 1s N- 1

Next, show that (5.4. 18 ) and (5.4.20) yield, respectively, (5.4.21 (i) ) u(x) = 1 N r u(x + y ) dy

O N T 1B ( x, r )

and

u(x) > 0 1 N { u(x + y ) dy . N T 1B ( x, r) Using (5.4. 21 (ii) ) , argue that if G is a connected open set in �N and E C2 (G; �) satisfies �u < 0, then u achieves its minimum value at an x E G if and only if u is constant on G. This fact is known as the strong minimum (5.4.21 (ii) )

u

principle. 5.4.22 Exercise: Let G be a bounded, smooth region in the plain �2 . In addi­

tion, assume {)G is a closed curve in the sense that there is a 1 with the properties that

( i ) Show that

E C 2 ( [0, 1); �2 )

t E [0, 1 ) � 1(t) E 8G is an injective surjection, 1 ( 0) == lim 1 ( t) , i' ( 0) == tlim t/ 1 / 1 i' ( t) , i(O) == tlim / 1 i(t) , and li' (t) l > 0 for t E [0, 1 ) .

r cp ( ( ) Aaa ( d( ) = r cp I'( t) li' (t) I dt 1aa 1[o, 1 ] for all bounded measurable cp on 8G. ( ii ) Let n(t) denote the outer normal to G at 1(t), check that n(t) == ± li' (t) l - 1 (i'2 (t) , -i'1 (t) ) , with the same sign for all t E [0, 1 ) , and assume that 1 has been parameterized so that the plus sign is the correct one. Next, suppose that h E C2 ( G ( P ) ; �) for some p > 0, and define u == �� and == - �Z . Next, set f == u + A and r(t) == 1 1 (t) + A12 (t) . Show that r J( l' (t)) t (t) dt = v=� r r�h] ( ( ) d( . ( 5.4.23 ) 1a 1[o, 1 ] A particularly important case of (5.4.23) is the one when �h == 0; in which 0

v

v

case f is a complex analytic function and (5.4.23) leads to the famous Cauchy

integral theorem.

5. 4

The Divergence Theorem

113

G, a non-empty, open set in JRN , and F E C3(G ; JR) be given, and assume that l \7 F l never vanishes in G. Next, set T == Range ( F) and, for t E T, define Mt == { x E JRN F ( x) == t}. Check that, for each t E T, Mt is a hypersurface. Further, by combining the result obtained in Exercise 5.3.27 with the localization technique used in the proof of Theorem 5.4.9, show that, for each c.p E Cc(G) , t E T � JMt c.p d>.. Mt E lR is bounded and measurable and that the co-area formula ( 5.3.28 ) continues to hold. 5.4.24 Exercise: Let

:

Chapter VI Soine Basic Inequalities

6. 1 Jensen, Minkowski, and Holder.

There are a few general inequalities which play a central role in measure theory and its applications. The ones dealt with in this section are all conse­ quences of convexity considerations. A subset C C � N is said to be convex if (1 - t)x + ty E C whenever x, y E C and t E [0, 1] . Given a convex set C C � N , we say that g : C � � is a concave function on C if

g ( (1 - t)x + ty )

> (1 - t)g(x) + tg( y ) for all

x, y E C and t E [0, 1] .

Note that g is concave on C if and only if { (x, t) E C x � : t < g(x) } is a convex subset of � N + l . In addition, use induction on n > 2 to see that

for all n > 2, { y 1 , . . . , Yn} C C and {a 1 , . . . , an} C [0, 1] with E � ak == 1. The essence of the relationship between these notions and measure theory is contained in the following. 6. 1 . 1 Theorem ( Jensen's Inequality ) . Let C be a closed, convex subset of

�N , and suppose that g is a continuous, concave, non-negative function on C. Let (E, B, JL) be a probability space and F : E � C a measurable function on (E, B) with the property that IFI E £ 1 (JL) . Then EC and (See Exercise 6. 1.9 for another derivation.)

6.1

Jensen, Minkowski, and Holder

1 15

PROOF: First assume that F is simple. Then F == L:�= O Yk l rk for some n E z + , Yo , . . . , Yn E C and cover {ro , . . . , rn} of E by mutually disjoints elements of B. Hence, since I:� JL(rk) == 1 and C is convex, F dJL == I:� YkJL(rk) E C and, because g is concave,

JE

Next, let F be general. Choose and fix some element y0 of C, and let be a dense sequence in c. Given m E z + ' choose Rm > 0 so that

and nm E z + so that C n B(O, Rm ) c U�"\ B ( Yk, ! ) . Define rm, O I F( � ) I > Rm } , and use induction to define

rm, £ = � E E \

{

for 1 < f <

JE

nm . Finally, F m dJL E C and g

set

£ -1

!Jo rm, k : F(� )

Fm

==

E

B ( Yt , ,!. )

L:�Tno Yk l rm , k .

(L Fm dp) > L g

0

==

{ yk}1

{� E E :

}

Then, by the preceding,

F m dp

Moreover, it is easy to see that II I F m - F l I I L 1 (J.L ) 0 as m � oo . Thus, because C is closed, we now see that F dJL E C. At the same time, because g is continuous, g o F m � g o F in JL-measure as m � oo . Hence, by Fatou ' s Lemma, for each

m E

z+ .

JE

----7

We now need to develop a criterion for recognizing when a function is con­ cave. Such a criterion is contained in the next theorem. 6. 1 . 2 Lemma. Suppose that C C

�N is an open convex set, and let C be its Moreover, if g is continuous on C and g E C 2 (C) ,

closure. Then C is convex. then g is concave on C if and only if its Hessian matrix

is non-positive definite ( i.e., all of the eigenvalues of H9 (x) are non-positive) for each x E C. 0

VI Some Basic Inequalities

116

PROOF: The convexity of C is obvious. In order to prove that g is concave on C if H9 (x) is non-positive definite at every x E C, we will use the following simple result about functions on the interval [0 , 1] . Namely, suppose that u is continuous on [0 , 1] and that u has two continuous derivatives on ( 0, 1). Then u( t) > 0 for every t E [0, 1] if u(O) == u(1) == 0 and u"(t) < 0 for every t E (0, 1). To see this, let t > 0 be given and consider the function uf. - u - tt(t - 1). Clearly it is enough for us to show that uf. > 0 on [0 , 1] for every t > 0. Note that uf. (O) == uf. (1) == 0 and u� ( t) < 0 for every t E ( 0, 1). In particular, if uf. ( t) < 0 for some t E [0, 1], then there is an E (0, 1) at which uf. achieves its absolute minimum. But this is impossible, since then we would have that u� ( ) > 0. { The astute reader will undoubtedly see that this result could have been derived as a consequence of the strong minimum principle in Exercise 5.4.19 for N == 1.) Now assume that H9 ( x ) is non-positive definite for every x E C. Given x , y E C , define u(t) == g({1 - t)x + ty ) - (1 - t)g(x) - tg( y ) for t E [0, 1] . Then u(O) == u(1) == 0 and 0

s

s

0

u" (t) = (y - x , H9 ((1 - t)x + ty ) ( y - x ) ) R N < 0 for every t E {0, 1). Hence, by the preceding paragraph, u > 0 on [0, 1] ; and so g({1 - t)x + ty ) > ( 1 - t)g(x) + tg( y ) for all t E [0, 1] . In other words, g is concave on C, and, by continuity, it is therefore concave on C. To complete the proof, suppose that H9 ( x ) has a positive eigenvalue for some x E 6 . We can then find an w E s N - l and an t > 0 such that (w, H9 (x)w ) R N > 0 and x + tw E 6 for all t E ( - t , t ) . Set u(t) == g( x + tw) for t E ( - t , t ) . Then u"(O) == (w, H9 (x)w) R N > 0. On the other hand, 0

+ u( -t) - 2u{O) u(t) " (O) == 1. u ' t----.. 0 t2 liD

and, if g were concave,

( )

2u{O) == 2u t -2 t

(

)

2g 21 (x + tw) + 21 (x - tw) > g ( x + tw) + g( x - tw) == u(t) + u( - t),

==

from which would we would get the contradictory conclusion that

0. D

6 . 1 . 3 Lemma. Let A =

u"(O)

<

[ � : ] be a real symmetric matrix. Then A is non­

positive if and only if both a + c < 0 and ac > b2 . In particular, for each 2 0 -o: and ( x , y ) E [O, oo) 2 r-----+ a E {0, 1), the functions (x, y) E [O, oo) r-----+ x y 1 ( X 0 + yo: ) are continuous and concave. 1

a:

6. 1

117

Jensen, Minkowski, and Holder

PROOF: In view of Lemma 6. 1.2, it suffices for us to check the first assertion. To this end, let T == a + c be the trace and D == ac b2 the determinant of A. Also, let A and Jl denote the eigenvalues of A. Then, T == A + Jl and D == AJl. If A is non-positive and therefore A V Jl < 0, then it is obvious that T < 0 and that D > 0. If D > 0, then either both A and Jl are positive or both are negative. Hence if, in addition, T < 0, A and Jl are negative. Finally, if D == 0 and T < 0, then either A == 0 and Jl == T < 0 or Jl == 0 and A == T < 0. D -

6.1.4 Theorem (Minkowski's Inequality) . Let !I and /2 be non-negative,

measurable functions on the measure space [1 , oo ) ,

( E, B, JL ) .

Then, for every p E

PROOF: The case when p == 1 follows from (3.2.13). Also, without loss in generality, we assume that ff and ff are JL-integrable and that !I and !2 are [0, oo ) -valued. Let p E ( 1, oo ) be given. If we assume that JL ( E ) == 1 and we take a == ! , then, by Lemma 6. 1.3 and Jensen ' s inequality,

l (it + h ) p d/1 l [ (fit + (f�t] ! dJ1 ! < [ (l Ji dJ1 ) + (l g dJ1 ) ] [ (l fi dJ1) i + (l g dJ1) i r =

a

a

More generally, if JL ( E ) == 0 there is nothing to do, and if 0 < JL ( E ) < oo we can replace Jl by JL(E) and apply the preceding. Hence, all that remains is the case when JL ( E ) == oo . But if JL ( E ) == oo , take En == {!I V !2 > � } , note that JL ( En ) < nP J ff dJl + nP J ff dJl < oo , apply the preceding to JI , !2 , and Jl all restricted to En , and let n ---t oo . D 6. 1.5 Theorem (Holder's Inequality) . Given p E (1, oo ) define the Hol­ der conjugate p' of p by the equation 1p p1, == 1. Then, for every pair of

+

non-negative, measurable functions !I and !2 on the measure space

for every p E

( 1, oo ) .

(E, B, JL ) ,

VI Some Basic Inequalities

118

PROOF: First note that if either factor on the right hand side of the above inequality is 0, then !I f2 == 0 (a.e. , JL) , and so the left hand side is also 0. Thus we will assume that both factors on the right are strictly positive, in which case, we may and will assume in addition that both Jf and f� are JL-integrable and that !I and f2 are both [0, oo )-valued. Also, just as in the proof of Minkowski ' s inequality, we can reduce everything to the case when JL(E) == 1. But then we can use apply Jensen ' s Inequality and Lemma 6. 1.3 with a == � to see that I

l ft h dj1 l ( fi) o (f{ ) l - o dj1 < (l Jf dj1) o (l J{ dj1y -o (l Jf d11) (l J{ d11) ? D =

*

Exercises

6 . 1 . 6 Exercise: Here are a few easy applications of the preceding.

(i) Show that log is a continuous and concave on every interval [E, oo ) with E > 0. Use this together with Jensen ' s inequality to show that for any n E z + ' Jl i , . . . , Jln E (0, 1) satisfying L: � = I Jlm == 1, and ai , . . . , an E (0, oo ) , n

n

m= l

m= I

II a�m < L Jlm am .

In particular, when Jlm == � for every 1 < m < n this yields ( a I a ) :n < � L:� = l am , which is the statement that the arithmetic mean dominates the 1

·

geometric mean. (ii) Let n E z + ,

·

·

n

and suppose that !I , . . . , f are non-negative, measurable functions on the measure space ( E, B, Jl ) . Given PI , P E ( 1, oo ) satisfying L:� = I P� == 1, show that n

. . .

2,

,

n

Minkowski ' s and Holder ' s inequalities are in­ timately related and are both very simple to prove. Indeed, let !I and f2 be bounded, non-negative, measurable functions on the finite measure space (E, B, JL) . Given any a f= 0, observe that 6. 1. 7 Exercise: When p

==

6. 2

The Lebesgue Spaces

119

from which it follows that

2 r h 12 df-l < t r tt df-l + �t r Ji df-l JE JE JE

for every t > 0. If either integral on the right vanishes, show from the preceding that JE f1 f2 dJl < 0. On the other hand, if neither integral vanishes, choose t > 0 so that the preceding yields

(6. 1.8) Hence, in any case, (6.1.8) holds. Finally, argue that one can remove the restriction that f1 and f2 be bounded, and then remove the condition that JL (E ) < oo . In particular, even if they are not non-negative, so long as ft and fi are JL-integrable, conclude that f1 f2 must be JL-integrable and that (6. 1.8) continues to hold. Clearly (6.1.8) is the special case of Holder ' s inequality when p == 2. Because it is a particularly significant case, it is often referred to by a different name and is called Schwarz's inequality. Assuming that both ft and fi are JL­ integrable, show that the inequality in Schwarz ' s inequality is an equality if and only if there exist ( a , {3) E �2 \ {0} such that af1 + /3!2 == 0 ( a.e. , JL ) . Finally, use Schwarz ' s inequality to obtain Minkowski's inequality for the case when p == 2. Notice the similarity between the development here and that of the classical triangle inequality for the Euclidean metric on � N . 6. 1.9 Exercise: There is a "better" proof of Jensen ' s inequality which is

based on the following geometric fact. Namely, if g is a continuous, concave function on the closed, convex subset C in � N , then, for each p E C, there is a v E �N such that the line L { ( x, g (p) + (v, x - p) R N ) : x E �N } lies above 6 { (x, t) E C x � : t < g(x) } in �N + l . That is, g(x) < g (p ) + (v, x - p)R N for every x E C. Assuming this fact*, give another derivation of Jensen ' s Inequality. ( Hint: Take p == J F dJL. )

6.2 The Lebesgue Spaces.

I I £ 1 (JL)

and the space L 1 ( JL ) . We are now In Section 3.2 we introduced ready to embed L 1 ( JL ) into a one-parameter family of spaces. Given a measure space ( E, B, JL ) and a p E [1, oo ) , define ·

* When C is the closure of its interior and g is smooth, this fact is an easy consequence of Taylor 's Theorem. In general, it can be seen as an application of the Hahn-Banach Theorem for JR N .

120

VI Some Basic Inequalities

for measurable functions ( E, B) define

f on

(E, B) . Also, if f is a measurable function on

II J IIL= ( J.t ) = inf { M E [0 , oo] : I f I < M (a.e. , J-L) }. Obviously, as p varies 11 / II Lv(J.L ) provides different estimates on the size of f as it is "seen" by the measure Jl · Although information about f can be gleaned from a study of 1 1 / II L P(J.L ) as p changes (for example, spikes in f will be emphasized by taking p to be large) , all these quantities share the same flaw as 11 / IIL l (J.L) : they cannot detect properties of f which occur on sets having JL-measure 0. Thus, before we can hope to use any of them to get a metric on measurable functions, we must invoke the same subterfuge which we introduced at the end of Section 3.2 in connection with the space £ 1 (JL) . Namely, for p E [1, oo] , we denote by LP (Jl) == LP (E, B, JL) the collection of equivalence classes [/]'"'-� (cf. Remark 3.2.14) of �-valued, measurable functions f satisfying ll f ll £v (J.L ) < oo, and, once again, we will abuse notation by using f to denote its own equivalence class 1-L

[/]

1-L

I"V .

Note that, by

(3.2.13 ) and Minkowski ' s inequality,

(6.2. 1) for all p E [1, oo ) , /1 , /2 E LP (Jl) , and a , f3 E �- Moreover, it is a simple matter to check that (6.2. 1) continues to hold when p == oo. Thus, each of the spaces LP (Jl) is a vector space. In addition, because of our convention and Markov's inequality (Theorem 3.2.8) , 11 / II Lv (J.L ) == 0 if and only if f == 0 as an element of LP (JL) . Finally, (6.2. 1) allows us to check that l l /2 - /I IILP (J.L ) satisfies the triangle inequality and, together with the preceding, this shows that it determines a metric on LP (JL) . Thus, when {/n }1 U {/} C LP (JL) , we often write fn � f in LP (Jl) when we mean ll fn - f i ! Lv (J.L ) � 0. The following theorem simply summarizes obvious applications of the results in Sections 3.2 and 3.3 to the present context. The reader should check that he sees how each of the assertions here follows from the relevant result there. 6.2. 2 Theorem. Let

and

f, g E LP (Jl) ,

(E, B, JL)

be a measure space. Then, for any p E

[1, oo]

Next suppose that {In } ! C LP (Jl) for some p E [1 , oo] and that I is an �­ valued measurable function on (E, B) . (i) If p E [1, oo ) and fn � f in LP (JL) , then In � I in JL-measure. If fn � f in L CX) (JL) , then fn � f uniformly off of a set of JL-measure 0. (ii) If p E [ 1 , oo ] and In � f in JL-measure or ( a.e. , JL ) , then 11 / II L P(J.L) < lim n---+ CX) II fn II LP(J.L) . Moreover, if p E [1 , oo ) and, in addition, there is a g E LP (Jl) such that Ifn i < g ( a.e. , Jl) for each n E z+ ' then In � f in LP(JL) .

6.2

The Lebesgue Spaces

121

(iii) If p E [ 1 , oo] and limm � oo II In - lm iiLP(JL ) == 0, then there is an I E LP ( Jl ) such that In � f in LP (JL) . In other words, the space LP ( Jl ) is

SUPn > m

complete with respect to the metric determined by II · IIL P(JL ) · Finally, we have the following variants of Theorem 3.3. 13 and Corollary 3.3. 14. ( iv ) Assume that JL(E) < oo and that p, q E [1 , oo ) . Referring to Theorem 3.3. 13, define S as in that theorem. Then for each I E LP ( Jl ) n L q (Jl) there is a sequence { 'Pn }1 C S such that 'Pn � I both in LP ( Jl ) and in Lq (JL) . In particular, if is generated by a countable collection C, then each of the spaces LP (Jl), p E [1 , oo) , is separable. ( v ) Let ( E, p) be a metric space, and suppose that Jl is a measure on ( for which there exists a non-decreasing sequence of open sets En satisfying JL(En) < oo for each n > 1 . Then, for each pair p, q E [1, oo) and I E LP ( Jl ) n Lq (JL) , there is a sequence {cpn }1 of bounded p-uniformly continuous functions such that 'Pn 0 off of En and 'Pn � I both in LP (Jl) and in Lq (Jl) . The version of Lieb ' s variation on Fatou ' s Lemma for £P-spaces with p =I= 1 is not so easy as the assertions in Theorem 6.2.2. To prove it we will need the following lemma.

B

E, B ) E /E

6.2.3 Lemma. Let

p

E ( 1 , oo) , and suppose that {In } ! C LP (Jl) satisfies

supn > 1 ll ln iiLP(JL ) < oo and that In � 0 either in JL-measure Then, for every g E LP (JL) , l /n l p - 1 191 dJl == 0 == nlim l ln l l g l p - 1 dJL. nlim

J

or (a.e. , JL) .

J

� oo � oo PROOF: Without loss in generality, we assume that all of the In 's as well as g are non-negative. Given b > 0, we have that

J R:- 1 g dJ-L = }r{ Jn < 8g } ��- 1 g dJ-L + }r{ fn > 8g } ��- 1 g df-l < 8P - 1 II g ll iv (JL ) + r

}{ Jn >8 2 }

�� - 1 g dJ-L + r

}{ g <8 }

�� - 1 g dJ-L.

Applying Holder ' s inequality to each of the last two terms, we now see that

J ��- 1 g df-l < 8p- 1 II g ll iv (tt)

Since, by Lebesgue ' s Dominated Convergence Theorem, the first term in the final brackets tends to 0 as n ---t 0, we conclude that nl!__.�

J ��- 1 g df-l < 8P- 1 ll g l l iv (tt) + �� II fn ll ivttt ) II l { g < 6 } gll LP(tt)

for every b > 0. Thus, after another application of Lebesgue ' s Dominated Convergence Theorem, we get the first result upon letting b � 0.

VI Some Basic Inequalities

122

To treat the other case, apply the preceding with l!:,- 1 and gP- 1 replacing In and g, respectively, and with p' in place of p. D 6.2.4 Theorem (Lieb) . Let (E, B, JL) be a measure space, p

{ In } ! U { I } C LP (JL) . If SUPn> 1 ll ln ii L P ( JL ) < or (a.e., JL) , then

oo

E [1 , oo ) ,

and and In -----+ f in JL-measure

(6 . 2 . 5) and therefore ll ln - I II L P ( JL ) -----+ 0 if ll ln i i L P ( JL ) � II I I I LP ( JL ) · PROOF : The case when p == 1 is covered by Theorems 3.3.5 and 3.3. 12, and so we will assume that p E ( 1 , oo ) . Given such a p, we first check that there is a KP < oo such that (6.2.6) l l b i P - l a i P - l b - a l P I < Kp ( l b - a i P- 1 I a l + l a l p - 1 l b - a ! ) , a , b E �. Since (6.2.6) clearly holds for all a , b E � if it does for all a E � \ {0} and b E �' we can divide both sides of (6.2.6) by l a i P and thereby show that (6.2.6) is equivalent to

l l ci P - l - I e - l i P I < Kp ( l c - l l p - 1 + l c - 1 1 ) ,

cE

�.

Finally, the existence of a Kp < oo for which this inequality holds can be easily verified with elementary consideration of what happens when c is near 1 and when l ei is near infinity. Applying (6.2.6 ) with a == ln ( x ) and b == l ( x ) , we see that

pointwise. Thus, by Lemma 6.2.3 with In and g there replaced by In - f and I , respectively, our result follows. D We now turn to the application of Holder's inequality to the £P-spaces. In order to do so, we first complete the definition of the Holder conjugate p' which, thus far, has only been defined ( cf. Theorem 6. 1 .5) for p E ( 1 , oo ) . Thus, we define p' == oo or 1 according to whether p == 1 or oo . Notice that this is completely consistent with the equation p! + p1, == 1 used before. 6.2. 7 Theorem. Let (E, B, JL) be a measure space. (i) If I and g are measurable functions on (E, B) , then for every p E [1 , oo ]

( 6.2.8 ) In particular, if I

E £P (Jl)

and g

E LP (Jl) , '

then f g E

L 1 (JL) .

6. 2

(ii) If p E

123

The Lebesgue Spaces

[1, oo ) and f E £P (J-L ) , then

(6.2.9) In fact, if function

II!II L P(J.L )

>

0,

then the supremum in

(6.2.9)

is achieved by the

p - l sgn o f f l l . g= ll f ll i,tl' )

(iii) More generally, for any f which is measurable on ( E, B) ,

(6.2. 10) if p == 1 or if p E a-finite.

(1, oo )

and either J-L( I J I >

8)

<

oo

for every 8 >

0 or J-L

is

PROOF: Part (i) is an immediate consequence of Holder ' s inequality when p E ( 1, oo ) . At the same time, when p E { 1, oo } , the conclusion is clear without any further comment. Given (i) , (ii) is easy. When p == 1, (iii) is obvious; and, in view of (ii) , the proof of (iii) for p E (1, oo ) reduces to showing that, under either one of the stated conditions, IIJI ILP(J.L) == oo implies that the right hand side of (6.2.10) is infinite. To this end, first suppose that J-L( I f l > 8) < oo for every 8 > 0. Then, for each n > 1, the function Moreover, if

II 'l/Jn II LP' (J.L )

II J II Lv (J.L )

-----+

oo .

then, by the Monotone Convergence Theorem, ' Thus, Since II f'l/J n II L (J.L) == II 'l/Jn II pLP ' (J.L ) , we see that ==

oo , •

1

Finally, suppose that J-L is a-finite and that J-L( I J I > 8) == oo for some 8 > 0. Choose { En } ! C B so that En / E and J-L ( En ) < oo for every n > 1. Then it is easy to see that limn-+ II f 9n I I L (J.L) == oo when 00

Since

1

IIYn ii LP' (J.L) < 1, this completes the proof.

D

For reasons which will become clearer in the next section, it is sometimes useful to consider the following slight variation on the basic £P-spaces. Namely,

VI Some Basic Inequalities

124

(E I , 81 , J.1 1 ) PI , P2 E [ 1 , oo ) . let

and ( E2 , 82 , J.1 2 ) be a pair of a-finite measure spaces and let Given a measurable function f on (E1 x E2 , 8 1 x 82 ) , define

[l2 (l1 ! J(x 1 , x2 ) 1P1 JL 1 (dx 1 ) ) * tt2 (dx2 )]

l l f ii £(P t ·P2 l ( J.t 1 , J.1 2 )

pI2

,

and let £ (P I , p2 ) ( J.-t1 , J.-t 2 ) denote the mixed Lebesgue space of �-valued, 81 x 82 -measurable f ' s for which ll f ll £ c PI ·P2)(J.L I ,J.L 2 ) < oo . Obviously, when p 1 == p == P2 , II J II L c PI . P2 ) (J.L I ,J.L 2 ) == ll f ii £P(J.L I X J.L 2 ) and £ (P I , P2 ) (J.-t i , J.12 ) == LP(J.-t l x J.12 ) ·

6.2. 11

Lemma. For all f and g which are measurable on

and all a,

f3 E � '

(E1

x

E2 , 81

x

82 )

l l af + f3g ii £(PI ·P2) (J.L I ,J.L 2 ) < l a l l lf ll £cPI . P2) (J.L I ,J.L 2 ) + l /31 I IYI I £(PI . P2) (J.L I ,J.L 2 )

(6.2. 12)

Moreover, if {fn}! U { ! } C £ ( P I ,P2 ) (J.-t l , J.12 ), fn � f (a.e., J.-l l X J.12 ) , and I fn i < g (a.e. , J.-l l x J.12 ) for each n > 1 and some g E £ (P I ,P2 ) (J.-t i , J.12 ), then II fn - f II £(PI , p 2 ) (J.L I ,J.L 2 ) � 0. Finally, if J.-l l and J.-t 2 are finite and g denotes the class of all 'ljJ 's on E 1 x E2 having the form E� =l 1ri , m ( · I)'Pm ( · 2 ) for some n > 1 , { 'P m } f C L00 ( J.-t 2 ), and mutually disjoint r1 , 1 , . . . , rl, n E 8 1 , then, for every measurable f E £ (P I ,p2 ) ( J.-t1 , J.-t 2 ) and E > 0, there is a 'ljJ E g such that

II ! - '¢ 11 £(PI . P 2 ) (J.L I ,J.L 2 ) < E.

PROOF: Note that (6.2. 13)

Hence the assertions in (6.2.12) are consequences of repeated application of Minkowski's and Holder's inequalities, respectively. Moreover, to prove the sec­ ond statement, observe ( cf. Exercise 4. 1 . 1 1 ) that for J.-t 2 -almost every x 2 E E2 , fn ( · , x 2 ) � f ( · , x 2 ) ( a. e. , J.-t 1 ) , I fn ( · , x 2 ) I < g ( , x 2 ) ( a. e. , J.-t 1 ) , and g ( , x 2 ) E £P I (J.-t 1 ). Thus, by part (ii) of Theorem 6.2.2, ·

for J.-t 2 -almost every x 2

E

E2 .

for J.-t 2 -almost every x 2

E

E2 and, by (6.2. 13) with g replacing J ,

In addition,

·

6. 2

125

The Lebesgue Spaces

Hence the required result follows after a second application of ( ii ) in Theorem 6.2.2. We turn now to the final part of the lemma, in which the measures Jll and Jl 2 are assumed to be finite. In fact, without loss in generality, we will assume that they are probability measures. In addition, by the preceding, it is clear that, for each f E £ (P I ,p2 ) (JL I , Jl 2 ), Thus, we need only consider f ' s which are bounded. Finally, because Jl l x is also a probability measure, Jensen ' s inequality and (6.2. 13) imply that

JL 2

Hence, it suffices to show that, for every bounded measurable f on ( E1 x E2 , B1 x B2 ) and E > 0, there is a 'ljJ E g for which II! - 'l/J IIL q(J.L 1 xJ.L 2 ) < E. But, by part ( iv ) of Theorem 6. 2.2, the class of simple functions having the form n

L am lrl , m r2 ,m m=l with ri,m E Bi is dense in Lq(JL 1 JL 2 ). Thus, we will be done once we check X

'ljJ ==

x

that such a 'ljJ is an element of g. To this end, we use the same technique as we did in the final part of the proof of Lemma 3. 2.3. That is, set I == ( {0, 1 } ) and, for 'IJ E I, define r1,17 == n � =l ri � ) where r < o ) rC and r< 1 ) r. Then n

n

'¢ (x1 , x 2 ) == L am L 'TJmlr1 , 11 ( ) lr2 , m (x 2 ) == L lr1 ,11 (xi ) cp11 ( 2 ) , m=l 17 EI 17 E I x

xi

where

n

'P11 == Since the

L 'T/m amlr2 , m m= l

·

r1 , 17 ' s are mutually disjoint, this completes the proof. D

For our purposes, the most important fact that comes out of these consid­ erations is the following continuous version of Minkowski's inequality.

(Ei , Bi , Jli), i E { 1 , 2} , be a-finite measure spaces. Then, for any 1 < P I < P2 < oo and any measurable function f on ( E1 E2 , B l X 82 ), I I J II £(PI ·P2) (J.L I ,J.L2 ) < II J II £(P2.Pl ) (J.L 2 ,J.L l ) •

6.2. 14 Theorem. Let

x

PROOF: Since it is easy to reduce the general case to the one in which both Jll and Jl 2 are finite, we may take them to be finite. In fact, without loss in generality, we will assume, from the outset, that they are probability measures.

VI Some Basic Inequalities

126

'¢ rl ,m

Let g be the class described in the last part of Lemma 6.2. 11. Given == 's which is an element of Q , note that, since the L:� == 2::: � for any r E [0, oo ) and are mutually disjoint, 2::: � E � - Hence, by Minkowski ' s inequality for p == � ,

l r l , Tn ( . 1) 'Pm ( . 2 ) I am l r1 m l r l am l r l r1 , m , a1 , ... , an PI 1 P2 I 'I/J I L

[L1 � l rl , = (X 1 ) l cpm l i�2 (tt2 ) /Ll (dx1 )]

1

Pl

2 P [L1 (L2 � l r1 . = (x l ) cpm (x2 ) JL2 (dx2 ) ) � JL1(dx1 )] I '¢ I L(P2 ,p I ) (J.L2 1 ) . f (E1 E2 , B1 B2 ). f l f i £ (J.L ,J.L ) 1 2 6.2.11, {'¢n}1 C 1 1/Jn -l 'l/Jfni £-
Pl

=

, j.L

==

Hence, we are done when the function is an element of Q. To complete the proof, let be a measurable function on x x Clearly we may assume that < oo . Using the last part of � g so that Then, Lemma choose and by Jensen ' s inequality, it is easy to check that � in Jll x Jl 2 -measure. Hence, without loss in generality, therefore that � we will assume that � (a.e., Jl x In particular, by Fatou ' s Lemma and Exercise 4.1.11, this means that

E2 ;

for JL 2 -almost every x 2 E and so, by the result for g and another application of Fatou ' s Lemma, the required result follows for D

f.

The following result is typical of the way in which one applies Theorem

6.2. 14.

6.2

127

The Lebesgue Spaces

(E and (E2 , 82 , J.1 2 ) be a pair of a-finite mea­ , 8 , J.-L ) 1 1 l sure spaces, and suppose that K is a measurable function on ( E1 E2 , 8 1 82 ) 6.2. 15 Theorem. Let

x

x

which satisfies

M1 X2 E E2 I l K ( · , x2 ) 1 Lq(JL 1 ) sup

<

oo

and

for some q E [1 , oo ) . Define

M2 XI E El I K (x1 , · ) 1 Lq(JL2 ) sup

<

oo

/Cj (x1 ) J{E2 K (x1 , x2 ) J (x2 ) J.t2 (dx2 )

(6.2. 16)

=

f (J.-L2 ) . Then for each p

for E Lq '

E [1, oo ] satisfying

� � + � - 1 > 0,

(6.2.17) PROOF: First suppose that (i) of Theorem 6.2.7,

r == oo

and therefore that p

==

q ' . Then, by part

and so (6.2. 17) is trivial in this case. Next, suppose that p == 1 and therefore that q == r . Noting that < we can apply Theorem 6.2. 14 to obtain

I KJI I £r (JL 1 )

I KJI I £< I ,r)(JL2 ,JL 1 ) , I KJI I £r (JL I ) I KJ I £ ( l ,r)(JL2 ,JL I ) I KJI I £ (r, l )(JL I ,JL2 ) 1 l2 (l1 I K (x1 , x2 ) J (x2 W J.l l (dx l ) ) J.t2 (dx) r I l K ( . , X 2 ) 1 L r( J.II ) I f ( x2 ) 1 J.l 2 (dx2 ) < Ml l f l u ( JL 2 )· JE2 [1 , ) p (1, ). ( q, ) ;. (0, 1 ) q. (1 - )p' g L r ' ( J.-L 1 ), (6. 2. 12) , I Y KJ I L 1 (JL I ) I Y KJI I L 1 (JL I X JL2 ) < I I K I Q Jl l £(r,p)(JL 1 ,JL2 ) I Y 1 K i l - o" £ (r1 , p1 ) (JL I ,JL2 )" <

<

r

=

=

Finally, the only case remaining is when r E oo and E a that r E oo , set a == Then, a E and E we have, by the second inequality in that

oo

==

Noting Given

<

Next, observe that

=

1

[l2 (l1 IK (x1 , x2 ) l ar ! J (x2 W J.ll (dx l ) ) J.l2 (dx2 )] Mfl l f l v'(�t2 )· �

P

<

VI Some Basic Inequalities

128

At the same time, since p < r and therefore r ' < p' , we can apply Theorem 6.2. 14 to see that < Hence, we find that by the same reasoning as we just applied to < Combining these two, we arrive at < 1; for all E Lr 1 ( J-LI ) with < and so ( 6.2. 17 ) now follows from part (iii) of Theorem 6.2.7. D

l g I K I 1 -o: l £
6. 2. 18 Corollary. Let everything be as in Theorem 6.2. 15, and, for measur­

able

f : E2

-----+

�' define

and

Next, let p E [ 1, oo ] satisfying ! ! ! 1 . Then r q

== + p

-

�

+�

> 1 be given and define

r

E [ 1 , oo ] by

( 6.2. 19 ) LP In particular, K maps LP (�-t 2 linearly into Lr (�-t 1 ). In fact, f Lr ( J-LI ) is the unique continuous mapping from LP (�-t2 finto £LrP (J-t(�-t21)) whoseKJrestriction to LP ( �-t 2 n Lq ( �-t 2 is given by the map K in ( 6. 2. 16 ) . PROOF: == == q', £ z+ . == (J-t ) fl f f n P 2 n, n ] [ I f(6.2.n 17L)P (J-t2 ) n L 00 (J-tI K2 )I ( I fn i , 6. 2. 20) ' fn LP (J-t2 ) n Lq (J-t2 ). < q' for

E

E

(�-t 2 ) . �

)

E

I

)

)

)

If r oo , and therefore p there is nothing to do. Thus, we will assume that r and therefore p are finite. be given, and set Let E for n E Because 0 E cf. Exercise p and E Hence, by applied to and I

j

, ) ( x1 x2 i l f( x2 ) l M 2 (dx2 ) K { x2 : I J (x2 )l < n}

r

I

In particular, by the Monotone Convergence Theorem, this proves both parts of ( 6.2. 19 ) . Furthermore, if E P ) and E � ' then

J, g £ (�-t2 a, {3 K (af+f3g) == a Kf+ {3 Kg AK(f) n AK(g). 0, )C )C AK( AK( g ! Lr ( �-t 1 K Kf == Kf on

Thus, since both a mapping into

and have �-t1-measure we now see that, as is linear. Finally, it is obvious that for ),

6. 2

129

The Lebesgue Spaces

J-L J-L ) ) Lq f E (J-L2 ) n Lq (J-L2 )· ( ( 2 2 ( L L (J-L ) (J-L ) r 2 {fn }1 ) I l K/ - K' / I £r (J.I t ) < }�--•� I l K/ - KJn I £r (J.I t ) I JC (f - fn) I Lr( 1 ) < M;!: M; - ; n---+ I J - fn i LP(J.I t ) 0. D I

I

£P n Hence, if /(' is any extension of /( r £P as a continuous, linear mapping from P to I , then with the same choice of as above

== limCX) n ---+

limCX)

J.L

=

Exercises

J-L ) f E Lq1 (J-L) nLq2 (J-L) ,

6.2.20 Exercise: Let ( E, B,

given. If

be a measure space and 1 show that for any t E (0, 1)

< QI < Q2 < oo be

1 t 1-t where - == - + Pt QI Q2

(6.2.21)

--

(

Note that 5. 2. 21) says that p �

- I f I LP (J..L) log

is a concave function of

�.

6.2.22 Exercise: The following exercises give some insight into the £P-spaces

in various situations. is a (i) If ( E, B, is a probability space, show that p E [1 , oo ] � non-decreasing function for any measurable on E, B ) . (ii) Let E == z + and define on B == P ( E ) by == 1 for all E z + . In this case show that p E (1 , oo ] � is non-increasing for every on E. (iii) Let ( E, B, be a measure space and : E � � is a B-measurable function. Show that limp/ CX) Further, assuming either that �-t E oo or that oo, show that == limp / CX) ( iv ) Let ( Ei , Bi ) , i E { 1 , 2}, be measurable spaces, and suppose that is a a-finite measure on E , B . Using part (iii) , show that, for every measurable is function on ( EI x E2 , BI x B , the function X I � measurable on ( EI , BI ) · In particular, we could have defined for all P I , P2 E [1 , oo] .

J-L)

J. .L f v ) i L ( l ( f J-L I J I Lv (J..L) J-L ( {n}) nf J-L) I J I Loo (J..L) < l ffi Lv (J..L) · ( )< I J I L l (J..L) < l f/-Li L2 P(J..L) · I J I Loo (J..L) ) ( 2 2 f 2) I J (xi,I J I ·£2c)v1l .LPoo2 ) ((J.J..L.L21 ),J..L2 ) J-L ) p E (1, oo ) f C E (0, oo)

6.2. 23 Exercise: Let ( E, B,

L (J-L)

of P for some for which there exists a

(6.2.24)

( i ) Set

v(r)

be a measure space, g a non-negative element , and is non-negative, B-measurable function such that

MU > t) < ct Jr{f> t } g dJ.L,

== fr g d�-t for

rE

t E (O, oo) .

B, note that (6.2.24) is equivalent to

J-L( f > t) < Tc v (f > t) ,

tE

(0, oo

)

,

VI Some Basic Inequalities

130

and use (5. 1 .4) ( cf. Exercise 5. 1 .8) to justify

I I I iP(J.I)

=p

r

J( o, oo)

tp - 1 M U > t) dt

Finally, note that conclude that

l f l iv �l (J.t)

l f l iv� l ( v)"

tP - 2 v(f > t) dt = C-p1 P J( o, oo) = J fP - 1 g dJ.-l, and apply HOlder ' s inequality to

< Cp

{

(6.2.25) (ii) Under the condition that

I J I Lv (JL)

<

oo ,

it is clear that (6. 2.25) implies

(6.2.26)

f fn f

Now suppose that M(E) < oo . After checking that (6.2.24) for implies (6.2.24) for 1\ R, conclude that (6.2.26) holds first with replacing and then, after R / oo , for itself. In other words, when M is finite, (6.2.24) always implies (6.2.26) . (iii) Even if M is not finite, show that (6.2.24) implies (6.2.26) as long as M ( > E ) < oo for every E > 0. Hint: Given E > 0, consider M E == M r > E } ] , note that (6.2.24) with M implies itself with ME , and use (ii) to conclude that (6.2.26) holds with ME in place of M· Finally, let E � 0.

fn f

f

f

B[{f

6.2.27 Exercise: Recall the Hardy-Littlewood maximal function

Mf

defined in (3.4.2) , and observe that the definition extends, without change, to any measurable on � which is integrable on compacts. Show that

f

l i MJI I Lv (R) I J I Lv (R) , P E ) f E £P � (�) , f fE ECc£P(�) C Cc(�) , ( ) {fn}1 fn � f in LP � l i MJI I LP(R) < n oo l i Mfn i Lv (R) ·

(6.2.28)

Hint : When

handle general that

<

2p p- 1

( 1 , oo and

( ).

use (3.4.7) together with (iii) of Exercise 6.2.23. To so use v in Theorem 6.2.2 to find ( ) , and apply Fatou ' s Lemma to see that lim ---+

6.3

Convolution and Approximate Identities

131

6.3 Convolution and Approximate Identities.

We will use the ideas of the last section to develop in this section an im­ portant notion of multiplication for functions on � N ; and, because the only measure involved will be Lebesgue ' s, we will use the notation LP(� N ) instead of the more cumbersome LP (

AR N ) .

oo]

6.3. 1 Theorem ( Young's Inequality ) . Let p and q from [1, satisfying ! + ! > 1 be given, and define r E [1 , by ! == p! + ! - 1 . Then, for each p q r q E LP (� N ) and E Lq (� N ) , the complement of the set

oo] f g (6.3.2) A(f, g) {x N l) f (x - y) i i g (y) i dy < oo } has Lebesgue measure 0. Furthermore, if A( J, g ) J ( wh en ) ) d f(x g y y y N { R (6.3.3 ) f * g (x) 0 otherwise, then f * g g * f and ( 6.3.4 ) Finally, the mapping ( J , g ) LP (�N ) Lq (� N ) f*g Lr (�N ) is bilinear. PROOF: r oo. r . A(J, g ) A(g , f) f f g * * g [ 1 oo ) g Lq (�N ) , K ( y) g (x - y ) x , y �N . ( ) I L q( R N ) l g i L q( R N ) < oo ; , ) ( · N I K ) I L q( R I l K y yERN xERN 6.2. 18, A (J , g ) AK(f) . f f * g /C 0 LP(�N ) , A (f, g )C f Lr (� N ) ( 6.3.4 ) LP(�N ) f f * g g Lq (�N )f; g. D ( 6.3.3 ) f * g LP (� N ) f gg. Lq (�N ) r oo ) p 1 p r p Given � N , define Th f for functions f on �N by Th f ( x ) � N and p [1 , oo]. Then Th is an isometry on LP (�N ) for every ( fMoreover, ifp [ 1 oo ) and f LP(�N ) , then E

IR

:

x

E

==

E

E

�

x

We begin with the observation that there is nothing to do when == Thus, we will assume throughout that and therefore also p and q are all finite. Next, using the translation invariance of Lebesgue ' s measure, first note that == == and then conclude that Finally, given qE , and E Obviously, for set x , == E == sup

sup

x

,

·

==

and, in the notation of Corollary == and == In particular, for each E has Lebesgue measure and is linear for each E holds. In addition, E � E and therefore the bilinearity assertion follows after one reverses the roles of and described in is called the convolution of with The quantity f In applications, the most useful cases are those when E and or == (and therefore E where either == q' (and therefore == == q) . To get more information about the case when == q' , we will need the following. 6.3.5 Lemma. x + h) . E

( 6.3.6 )

hE

==

hE

,

E

E

VI Some Basic Inequalities

132

PROOF : The first assertion is an immediate consequence of the translation invariance of Lebesgue ' s measure. Next, suppose that p [1 , oo ) is given. If g denotes the class of LP ( �N ) for which 6 . 3 . 6 holds, it is clear that Cc ( � N ) C Q . Hence, by v in Theorem 6.2.2, we will know that g LP ( � N ) as soon as we show that g is closed in C g and suppose that � in LP ( �N ) . LP ( � N ) . To this end, let Then

E

( )

E f () fn f

=={ fn}1

h---+<0 I Th f - f(i LP(Rfn)N ) h==---+20l lfi Tnh -!J-I Lv (RINL)P(RN ) + h---+0 I Th fn - fn I LP(RN) + l fn - f i LP(RN ) lim

lim

lim

�0

as

n ---t

oo .

D

E [1 , ] f E LP (�N ) , and g E LP' (�N ) . Then Th (f *g) == (Th f) * g == f * (ThgN) for all h E �N . Moreover, f * g is uniformly continuous on � and (6.3.8) Finally, if E ( 1 , ) then (6 .3. 9) f* g(x) == O. PROOF: ARfN*. g (6.3.8) E [1 , ) ( 6 . 3 8) ( 6 . 3 6) I Th (f *g ) - J * g l u == I (Th j - f) * g l u < I Th j - JI I LP(RN ) l g i LP' (RN ) == f g lhl E Cc ( � N ) f (6 3. 9) g E LP' (�N ) 6.3. 7 Theorem. Let p

p

oo ,

oo ,

lim

l x 1 ---+ CX)

for that

.

The first assertion is again just an expression of translation invariance Further, is a simple application of Holder ' s inequality. To see is uniformly continuous, first suppose that p oo . Then, by and . , �0

as and in this 0; and when p oo , simply reverse the roles of argument. To prove the final assertion, first let be given, and define gf to be the class of for which . holds. Then it is easy to check that Cc ( � N ) C g, . Moreover, by (6.3.8) , one sees that gf is closed in LP' ( � N ) . Hence, just as in the final step of the proof of Lemma 6.3.5, we conclude that g1 == LP' ( �N ) . Next, let LP' ( � N ) be given and define 1{9 to be the class of LP ( � N ) for which 6 . 3 . 9 is true. By the preceding, we know that Cc ( � N ) C 1{9 . Moreover, just as before, 1{ 9 is closed in LP ( �N ) ; and therefore 1{ 9 LP ( � N ) . D ---t

fE ==

g( E )

6.3. 10 Remark: Both Theorem 6.3.3 and Theorem 6.3.7 tell us that the

convolution of two functions is often more regular than either or both of its factors. An application of this fact is given in Exercise 6.3.26 below, where one sees how it leads to an elegant derivation of Lemma 2. 1. 15.

6. 3

Convolution and Approximate Identities

133

The next result can be considered as another example of the observation made in the preceding Remark.

C 1 ( �N ) , and assume that g as well as g , 1 , . . . , g , N are elements of LP' ( l� N ) for some p E [1 , oo ] . (Recall that g ,i g;i . ) Then f * g E C 1 (� N ) for every f E £P(�N ), and

6.3. 1 1 Lemma. Let g E

* g) f g J ( B aX i - *

(6.3. 12)

·

,t ,

1 < i < N. -

-

PROOF: Let w E s N - 1 be given. If p' E [1, oo ) , then, by Theorem 6.3.7, Ttw ( f * g) - f * g == f * (Ttw g - g) for every t E � - Since

1 (

- g( y ) w, t [0, 1 )

Ttw 9 (Y)

_

and, by Theorem 6.2. 14,

" v

g ( y + s tw ) ) R N d s

r (w,V'g( - + stw) - V'g( · )) R N ds 1[0, 1 ) LP1 (R N ) < r Tstw (w, V' g ) R N - ( w, V' g) JRN LP' (RN ) ds ----+ Q J[o, 1 ) as t � 0, the required result follows from (6.3.4) . On the other hand, if p' then

Ttw 9 (Y) - g( y )

t

=

== oo ,

r (w, V'g( y + stw) ) JRN ds ----+ (w, V'g( y )) JRN J[o, 1 )

boundedly and point-wise, and therefore the result follows, in this case, from Lebesgue's Dominated Convergence Theorem. D The preceding result leads immediately to the conclusion that the smoother g is the smoother is f * Y · More precisely, given a multi-index a == (a 1 , . . . , a N ), where the a i ' s are non-negative integers, define l a l == L: f ai and

Then, as an immediate corollary to Lemma 6.3. 1 1 , we see that if g E and ao: g E £P' ( � N ) for some p E [1 , oo ] and all a ' s, then

C CX) ( �N )

(6.3. 13) for every f E £P(� N ). We next turn our attention to the case when g E here is the one which follows.

£ 1 ( �N ) .

The main result

VI Some Basic Inequalities

134

9t (

L 1 (JR N )

(

and t > 0, define ) = t -Ng t - 1 ) . Then E L 1 (JR N ) and J gt dx = J g dx . In addition, if J g dx = 1, then for every p E [1 , oo ) and f E LP(� N ) : lim (5.3. 15) = 0.

6.3. 14 Theorem. Given g

9t

E

·

·

t�O I ! * 9t - ! I LP(RN )

PROOF: We need only deal with the last statement. Assume that J g dx = 1 . Given f E LP(� N ), note that, for almost every X E �N ,

{ ( f( x - y ) - f ( x ) ) gt ( ) dy { ( f( x - ty) - f(x) ) g ( y ) dy . *t J N N R JR t (x , y ) (J (x 6.2. 14, ) I ! * gt - JI I LP(RN ) < l '��t l £( l ,p)(AJRN ,AJRN ) { l l t f - J I L P(R N ) I g ( y ) i dy < l � t l i L < P . 1 l( >.a N , >.a N ) J N R I l T- ty !-JI I LP(RN ) < 2 I J I LP (RN ) , (6.3.6) . D 6.3. 14, g L 1 (� N ) J g dx 1 , {gt : t an approximation of the identity, g B(O, 1 ) . the volume under the graph 9t 1 t B(O, t) . f g ( X ) - f (x )

Hence, if w

Y

=

=

ty) - J(x) g(y) , then, by Theorem

=

r- y

=

.

Since we now see that the result follows from the above by Lebesgue ' s Dominated Convergence Theorem and

For reasons which ought to be made clear by Theorem if E and > 0} is called an approximate = the corresponding family identity. To understand how an approximate identity actually carries out consider the case when is non-negative and vanishes off of Then of continues to be as � 0 while the base of the graph is restricted to Hence, all the mass is getting concentrated over the origin. Combining Theorem 6.3. 14 and (6.3. 13) , we get the following important approximation procedure.

C CX) (JR N ) n L 1 (JR N ) with JR N g( x ) dx = 1 be given. In addition, let p E [1 , oo ) and assume that aa g E LP' ( R N ) for all E N N . Then, for each f E £P(JRN ), f f in LP(JR N ) as t � 0; and, for all 6.3. 16 Corollary. Let g

E

a

9 * t t f * 9t has bounded, continuous derivatives of all orders and ( ! * 9t ) J * ( a0 Yt ) , NN . -----+

aa

> 0,

=

a E

Exercises

J, g L 1 (JRN ), j f * g( x ) dx j f(x ) dx j g( x ) dx .

6.3. 1 7 Exercise: Given

show that

E

=

6.3. 18 Exercise: Given a family {ft : t E (0, oo ) } C family is a convolution semigroup if fs + = Is

L 1 (� N ) , we say that the and all t E (0, oo ) .

t * ft

s,

Here are four famous examples of convolution semigroups, three of which are also approximate identities.

6. 3

Convolution and Approximate Identities

135

� (

( i ) Define the Gauss kernel r ( x ) = (21r) - exp _ l xt ) for X E JRN . Using the result in part ( i ) Exercise 5.2.6, show that JR N 1' ( x ) dx == 1 and that 1' v's * 1'v't == 1'v'8+I

t E ( 0, oo ) . Clearly this says that the approximate identity { 1' v't : t E (0, oo ) } is a convolu­ for

s,

tion semigroup of functions. It is known as either the heat flow semigroup or the Weierstrass semigroup.

( ii ) Define v on � by

Show that JR v(�) d� == 1 and that

(6.3. 19) Hence, here again, we have an approximate identity which is a convolution semigroup. The family { vt 2 : t > 0} , or, more precisely, the probability mea­ sures A r-----+ JA Vt 2 ( �) d�, play a role in probability theory, where they are called the one-sided stable laws of order � . Hint: Note that for 'TJ E (0, oo )

-f--e , and use part ( iv ) of Exercise 5.2.6. ( iii ) Using part ( ii ) of Exercise 5.3.24, check that the function P on �N given

try the change of variable � (�)

==

by

N+l 2 2 ' X E �N ' (1 + lxl ) - -

P (x )

==

2

WN

-

has Lebesgue integral 1 . Next prove the representation

(6.3.20) Finally, using (6.3.20) together with the preceding parts of this exercise, show that

(6.3.21)

Ps * Pt

==

Ps+ t ,

s,

t E (0, oo ) ;

and therefore that { Pt : t > 0} is a convolution semigroup. This semigroup is known as the Poisson semigroup among harmonic analysts and as the Cauchy semigroup in probability theory; the representation (6.3.20) is an example of how to obtain one semigroup from another by the method of sub­ ordination.

136

VI Some Basic Inequalities

Yo:g (xlR) (r(a)[O, oo)) - 1 x0 - 1 e- x9o: (x)x 00. o: JR 9o: ( x ) dx 1. ( r( a ,B) f t o: - 1 ( 1 - t ) !3 - 1 dt ) ga + f3 (X ) , * ( ) go: g{3 X - r(a) r(,B) j(O,l) a

(iv) For each > 0, define and (cf. (ii) in Exercise 5.2.6)

==

:

� ==

so that

==

for

>

if x < 0 Clearly,

Next, check that +

_

and use this, together with Exercise 6.3. 17, to give another derivation of the formula, in (i) of Exercise 5.3.24, for the Beta function. Clearly, it is also shows that > 0} is yet another convolution semigroup, although this one is not an approximate identity.

{go: : a

6.3.22 Exercise: Show that if Jl is a finite measure on then for all f E Cc(RN ) the function f * Jl given by

f * J-L(X) =

JRN

and

p

E [1 , oo] ,

{R f(x - y) J-L(dy) , X E JR.N JN

is continuous and satisfies

(6.3.23 ) Next, use (6.3.23 ) to show that for each p E [1, oo) there is a unique continuous map JC JL : £P(JR N ) � LP(JRN ) such that JC JL f == f * Jl for f E Cc(� N ). Finally, note that (6.3.23 ) continues to hold when f * Jl is replaced by JC JL f, but that KJL J need not be continuous for every f E £P(JR N ). 6.3.24 Exercise: In many applications it is extremely important to have

compactly supported approximate identities. (i ) Set

1 1 ( [ dx ) 1 2 ] r CN ln( 0 , 1 ) 1 - 1X 1) (xp ) { CN [� 1 _jx l2 ] XX � B(O, B(O, 1 ) exp

==

and define

=

p

exp

if if

E

Show that E C00 (1R N ) . ( ii ) Use the preceding to show that if F is a closed subset of JR N and G is an open subset satisfying dist ( F, GC ) > 0, then there is an 'TJ E C00 ( �N ) such that lp < 'TJ < la . Such a function 'TJ is sometimes called a bump function. (iii) Show that for each pair p , q E [1 , oo) and every f E £P(JR N ) n L q ( �N ) there exists a sequence 1/Jn E cgo ( RN ) such that 1/Jn � f both in LP ( �N ) and in Lq ( R N ) . In particular, cgo ( JR N ) is dense in LP(JR N ) for every p E [1 , oo).

6. 3

137

Convolution and Approximate Identities

( iv ) Let G be an open subset of JRN . When

N > 2, we showed in Theorem

u

5.4. 17 that every E C2 (G) which is harmonic on G satisfies the Mean Value Property ( 5.4. 18 ) for balls B(x, R) whose closures are in G. Moreover, as pointed out in the paragraph preceding that theorem, the Mean Value Property is a triviality when N == 1 . In this exercise, prove the converse of the Mean Value Property. Namely, show that if E C(G) satisfies ( 5.4. 18 ) whenever B(x, R) c G, then E C00 (G) and is harmonic on G. The proof can be C G, then the Mean accomplished in two steps. First show that if B(x, Value Property implies that cf. part ( i ))

u

(

u

2t)

f (G) f}.j (x) 02 t� lrsN -1 (f(x tw) - f(x) ) AsN - l(dw) . x ( 6.3.25 ) , f ( 5.2.5 ) . u r (x) N u(x) l r l l� , r - r - { y - x, y r} ,

and conclude from this that and x E G,

( 6.3.25 )

u u

=

� £N

Hint : To prove

E c oo

E c oo (G) . Second, show that for any

lim t� O

+

in a two place Taylor expansion around

expand

and use the relations in 6.3.26 Exercise: Let

E BR have finite Lebesgue measure and set < where � == X : E

l _r * lr . Show that and that == 1r1 . Use these observations, together with Theorem 6.3.7, to give another proof of Lemma 2. 1 . 15.

u(O)

6.3.27 Exercise: Define the o--finite measure

�-t(r) [ � dx =

JL

(

==

)

)

on (0, oo , B( o , oo ) by

for r E B ( o ,oo) ,

and show that Jl is invariant under the multiplicative group in the sense that r = r E (0, oo) , J( o, oo) J( o , oo) and

f(ax) �-t (dx) f(x) �-t (dx), a { f (_!_) M ( dx ) { f(x) M ( dx ) X J J [O, oo] . : (O, oo ) f f g, x { A11 (j, g) {x (O, oo) : f l g ( y ) l JL(dy ) oo } , ( ) J =

( O , oo)

for every B( o ,oo) -measurable ble, JR-valued functions and ==

E

( O ,oo)

Next, for B( o ,oo) -measura­

�

set

( o ,oo)

Y

<

VI Some Basic Inequalities

138

J f ( �) g ( y ) J.t ( dy ) { f g (x ) f. f g g •

when

( O ,oo )

==

0

and show that • ! + ! ! > 0 then 1p r ' q ==

•

x E All- ( !, g)

otherwise,

In addition, show that if p, q E [1 , oo ] satisfy

f () g

for all E LP Jl and E L q (JL) . Finally, use these considerations to prove the following one of G.H. Hardy ' s many inequalities :

[Jf xl+a (Jf cp(y) dy) dx] ) ) P

1

( o ,oo)

�

( o , x)

-<

p a

(Jf (yycpl+(ya) t dy) cp.

�

( o ,oo)

E (0, oo , p E [1 , oo , and non-negative, B ( o ,CX)) -measurable Hint : To prove everything except Hardy ' s inequality, simply repeat the argu­ ment used in the proof of Young ' s Inequality. To prove Hardy ' s result, take for all

a

f(x) = (�) l [l ,oo) (x) g (x) = x 1 - ;cp(x), I ! g i Lv (JL) < I J I L 1 (JL) I Y I Lv (JL) · 0:

P

and use

•

and

Chapter VII A Litt le Abstract Theory

7. 1 An Existence Theorem.

In Chapter II we constructed Lebesgue ' s measure on � N , and in ensuing chapters we saw how to construct various other measures from a given mea­ sure. However, as yet, we have not discussed any general procedure for the construction of measures ab initio; it is the purpose of the present section to provide such a procedure. The basic idea behind what we will be doing appears to be due to F. Riesz and entails the reversal, in some sense, of the process by which we went in Chapter III from the existence of a measure to the existence of integrals. That is, we will suppose that we have at hand an integral and will attempt to show that it must have come from a measure. Thus, we must first describe what we mean by an integral. Let E be a non-empty set. We will say that a subset L of the functions f : E � is a lattice if f 1\ g and f V g are both elements of E L whenever f and g are. Given a lattice L of �-valued functions, we will say that L is a vector lattice if the constant function 0 is an element of L and L is a vector space over �- Note that if L is a vector space of �-valued functions on E, ther1 it is a vector lattice if and only if j + f V 0 E L whenever f E L. Next, given a vector lattice L, we will say that the mapping I : L � � is an integral on L if ------+

( a) I is linear,

I is non-negative in the sense that I(f) > 0 for every non-negative f E L, ( c ) I (fn) � 0 whenever {fn }� C L is a non-increasing sequence which

(b)

tends (point-wise) to 0.

Finally, we will say that the triple I == (E, L, I) is an integration theory if � and I is an integral on L. L is a vector lattice of functions f : E ------+

7. 1 . 1 Examples: Here are three situations to which the preceding notions

apply. ( i ) The basic model on which the preceding definitions are based is the one which comes from the integration theory for a measure space ( E, B, Jl ) . Indeed, in that case, L == £ 1 (JL) and I(f) == J f dJL.

VII A Little Abstract Theory

140

(ii) A second basic source of integration theories is the one which comes from finitely additive functions on an algebra. That is, let A be an algebra of subsets of E and denote by L (A) the space of simple functions I : E � � with the property that {I == a } E A for every a E � - It is then an easy matter to check that L (A) is a vector lattice. Now let Jl : A � [0, oo ) be finitely additive

in the sense that

JL ( rl U r2 ) == JL ( rl ) + JL ( r2 )

for disjoint

r1 , r2 E A.

Note that, since JL(0) == JL(0 U 0) == 2JL(0) , JL(0) must be 0. Also, by proceeding in precisely the same way as we did (via Lemma 3. 2 .3 ) in the proof of Lemma 3.2.4 and then in the proof of Lemma 3.2. 1 1 , one can show that

I E L (A) � I( l ) -

a ERange(/)

is linear and non-negative. Finally, observe that I cannot be an integral unless Jl has the property that

( 7. 1 . 2 )

JL ( rn)

� 0 whenever

{rn }� C A decreases to 0 .

On the other hand, if ( 7. 1 . 2 ) holds and {In }� C L(A) is a non-increasing sequence which tends point-wise to 0, then for each E > 0, Thus, in this setting, ( 7. 1 . 2 ) is equivalent to I being an integral. (iii) A third important example of an integration theory is provided by the following abstraction of Riemann ' s theory. Namely, let E be a compact topo­ logical space, and note that C(E; �) is a vector lattice. Next, suppose that I : C(E; �) � � is a linear map which is non-negative. It is then clear that I I(I) I < C l l l ll u ,E , I E C(E ; � ), where C == I ( l ) and ll l l lu , E supx E E l l (x) l is the uniform norm of I on E. In particular, this means that I I(In) - I( J ) I < C l l ln - l ll u ,E � 0 if In � I uniformly. Thus, to see that I is an integral, all that we have to do is use Dini's Lemma (cf. Lemma 7. 1.23 below) which says that In � 0 uniformly on E if {In }� C C(E; �) decreases point-wise to 0.

Our main goal will be to show that, at least when 1 E L, every integration theory is the sub-theory of the sort of theory described in (i) above. Thus, we must learn how to extract the measure Jl from the integral. At least in case (ii) above, it is clear how one might begin such a procedure. Namely, A == {r c E : l r E L(A) } and JL(r) == I( l r ) for r E A. Hence, what we are attempting to do in this case is tantamount to showing that Jl can be extended as a measure to the a-algebra a ( A ) generated by A. On the other hand, it is not so immediately clear where to start looking for the measure Jl in case (iii) ;

7.1

An Existence Theorem

141

the procedure which got us started in case (ii) does not work here since there will seldom be many r C E for which l r E C(E ; �) . Thus, in any case, we must learn first how to extend I to a larger class of functions f : E � � and only then look for Jl. Our extension procedure has two steps, the first of which is nothing but a rerun of what we did in Section 3.2, and the second one is a minor variant on what we did in Section 2. 1 . 7. 1 .3 Lemma. Let

(E, L , I)

be an integration theory, and define L u to be the class of f : E � ( - oo , oo ] which can be written as the point-wise limit of a non-decreasing sequence { 'Pn} r C L . Then L u is a lattice which is closed under non-negative linear operations and non-decreasing sequential limits (i.e., {fn}r C Lu and fn / f implies that f E Lu )· Moreover, I admits a unique extension to L u in such a way that I ( fn) / I(f) whenever f is the limit of a non-decreasing {fn}r C L u . In particular, for all J, g E L u , - oo < I(f) < I ( g ) if f < g and I(af + {3g) == ai ( f) + {3I (g ) for all a , {3 E [0, oo ) .

PROOF: The closedness properties of L u are obvious. Moreover, given that an extension of I with the stated properties exists, it is clear that that extension is unique, monotone, and linear under non-negative linear operations. Just as in the development in Section 3.2 which eventually led to The Mono­ tone Convergence Theorem, the proof ( cf. Lemma 3.2.6) that I extends to L u is simply a matter of checking that the desired extension of I is consistent. Thus, what we must show is that when '¢ E L and { 'Pn }r C L is a non-decreasing sequence with the property that 'ljJ < limn� 'Pn point-wise, then I( '¢ ) < limn� CX) I ( 'Pn). To this end, note that 'Pn == '¢ - ( '¢ - 'Pn) > '¢ - ( '¢ - 'Pn) + , ( 'ljJ - 'Pn) + � 0, and therefore that (X)

> nlim �CX) I( 'Pn)

limCX) I ( ( '¢ - 'Pn)+) == I( '¢ ). I( '¢ ) - n�

As we said before, once one knows that I is consistently defined on L u , the rest of the proof differs in no way from the proof of The Monotone Convergence Theorem (cf. Theorem 3.3. 2) . D Lemma 7. 1 .3 gives the first step in the extension of I. What it provides a rich class functions which will be used to play the role that open sets played in our construction of Lebesgue ' s measure. Thus, given any f : E � � ' we define (7. 1.4 )

I( ! ) = inf { I ( cp ) :

cp

E L u and

f < cp } ·

(We use the convention that the infimum over the empty set is + oo. ) Clearly I(f) is the analog here of the outer measure r � 1r1e in Section 2. 1. At the same time as we consider I, it will be convenient to have (7. 1.5)

I( ! ) = sup { - I ( cp ) :

cp

E L u and - cp <

f} ;

VII A Little Abstract Theory

142

the analog of which in Section 2 . 1 would have been the interior measure

r

f----*

lrh

sup

{ I F I : F is closed and F c r}.

(In keeping with our convention about the infimum, we take the supremum over the empty set to be - oo. ) 7. 1 . 6 Lemma. For any JR.-valued function

f on E

I(f) < I(J),

(7 . 1 . 7)

and if a E [0, oo ) if a E ( -oo, O] .

== ai(f) and I( a f) == ai(f) I( a f) (7. 1 .8) I(af) == ai(f) and I(af) == ai(f) -2 Moreover, if ( J , g) : E � JR. then

f
(7. 1 .9)

and, when

( J , g)

(7. 1 . 10)

===}

I(J) < I(g) and I(f) < I(g),

takes values in

JR2 (cf. Section 3.2), .-..

( I( J ), I(g) ) E i2 ( I(J), I(g) ) E i2

===} ===}

I(f + g) < I(f) + I(g) I(f + g) > I(f) + I(g) .

Finally,

f E Lu

(7. 1 . 1 1)

===}

I(f) == I(f) == I(f).

PROOF: To prove (7. 1 . 7) , note first that it suffices to treat the case in which I(f) > -oo and I(f) < oo and then that, for any ( 'ljJ) E L� with < f < + 'ljJ > 0 and therefore I( + I( 'ljJ) == I( + 'ljJ) > 0. Both (7. 1 .8) and (7. 1.9) are obvious, and, because of (7. 1.8) , it suffices to prove the first line in (7. 1 . 10) . Moreover, when I(f) 1\ I(g) > - oo, the required result is easy. On the other hand, if I(f) == -oo and I(g) < oo, then we can C L u so that f < choose U and < for each E z + , g < and I( 'ljJ) < oo. In particular, f + g < + 'ljJ for all E z + , and so I(f + g) < limn -H:x) I( + I( 'ljJ) == -oo. Finally, suppose that f E L u . Obviously, I(f) < I(f) . At the same time, if C L is chosen so that E L C L u and / f, then (because < f for each E z + ) == -(

'¢, cp

, cp cp

cp)

{ {'¢} n}1 cp '¢ ,

{-cp'Pnn)} � 'Pn

'Pn

'Pn )

I'Pn(cpn)

'Pn I(f) > n ---+ - I ( - cpn) n ---+ I ( cpn) CX)

==

limCX)

-n

n

-cpn

n

lim

-cp

==

I(f). D

n

7.1

An Existence Theorem

143

� == ==

From now on, we will use 9R(I) to denote the class of those f : E which I(f) I( J) , and we define i : 9J1(I) JR so that I(f) i(J) Obviously ( cf. (7 . 1 .8) )

�

==

JR for I(f) .

==

E 9J1(I) ===} af E 9J1(I) and i ( a) ai(J) , a E JR. Finally, let £ 1 (I) denote the class of JR-valued f E 9J1(I) with i( J ) E JR. 7. 1 . 13 Lemma. If f : E JR, then f E L 1 (I) if and only if, for each E > 0, there exist cp, 1/J E L u such that -cp < f < 1/J and I( cp) + I( 1/J) < E. In particular, f E £ 1 (I) ===} j + E £ 1 (I). Moreover, L 1 (I) is a vector space and i is linear on L 1 (I). Finally, if {fn } ! C L 1 (I) and 0 < fn / f, then f E 9J1(I) and 0 < I(fn) / I(f) . PROOF: First suppose that f E L 1 (I). Given E > 0, there exists (cp, 1j;) E L� f

(7. 1 . 12)

� -

-

< 1/J,

<

such that -cp < f I(f) < -I(cp) + � ' and I(1j;) I(f) + � ; from which it is clear that I( cp) + I( 1/J) < E. Conversely, suppose that f : E JR and that, for some E E (O, oo) , there exists (cp, 1j;) E L� such that -cp < f < 1/J and I ( cp) + I ( 1/J) < E. Because I ( cp) 1\ I ( 1/J) > - oo, - oo < I ( 1/J) < E - I ( cp) < oo and - oo < I(cp) < E - I(1j;) < oo. In addition, -I(cp) < I(f) < I(f) < I(1j;) . Hence, not only are both I(f) and I(f) in JR, but also I(f) - I(f) < E, which completes the proof of the first assertion. Next, suppose that f E L 1 (I) . To prove that j + E L 1 (I) , let E E (O, oo) be given, and choose (cp, 'lj;) E L� so that -cp < f < 1/J and I(cp) + I(1j; ) < E. Next, note that -cp - cp /\ 0 E L u , cp + cp V 0 E L u , and that cp - < j + < '¢+ . Moreover, because cp + 1/J > 0, it is easy to see that -cp - + 7j; + < cp + 1/J, and therefore we know that ----t

==

==

To see that £ 1 (I) is a vector space and that i is linear there, simply apply (7. 1 .8) and (7. 1 . 10) . Finally, let {fn }1 be a non-decreasing sequence of non­ negative elements of L 1 (I), and set f limn-H:x) fn · Obviously, limn � CX) i( Jn ) < I(f) < I(f) . Thus, all that we have to do is prove that I(f) < limn � CX) i ( Jn ) · To this end, set h 1 !1 , hn fn - fn - 1 for n > 2, and note that each hn is a non-negative element of L 1 (I) . Next, given E > 0, choose, for each m E z + , 1/Jm E Lu so that hm < 1/Jm and I(1/Jm ) < i(hm ) + 2 - m E. Clearly, 1/J L:� 1/Jm E L u and f < 1/J. Thus, I(f) < I(1j;) . Moreover, by Lemma 7. 1 .3 and the linearity of i on L 1 (I) ,

==

I(1j;)

==

==

==

(�1 ) = �L..-t1

nlim � CX) I L..-t 1/Jm

nlim � CX)

1 (1/Jm )

n limCX) '""""' i(hm ) + E < n---+ L..-t

1

limCX) i ( Jn ) + E. D n---+

== (E, be an integration theory. == (E, L 1 (I) , i ) is again an integration theory, C £ 1 (I), and i agrees

7. 1 . 14 Theorem (Daniell) . Let

Then i

==

I

L, I)

L

VII A Little Abstract Theory

144

with I on L. Moreover, if {fn } � C L I (I) is non-decreasing and fn / J, then f E L I (I) if and - only if- sup n f ( x ) < oo for each x E E and supn i ( Jn ) < oo , in which case I(fn) / I ( f ) .

P ROOF: In order to prove the first assertion, note that, by Lemma 7. 1. 13, Moreover, by (7. 1 . 1 1) , L C LI (I) L I (I)- is a vector lattice and i is linear there. and I r L == I r L; and, by (7. 1.9) , I(f) > I(O) == 0 if f > 0. Finally, if {fn}1 C L I (I) and fn / J , then, by the last part of Lemma 7. 1 . 13 applied to {fn - fi } l , f - fi E 001 (I) ,

f (fn) == f ( JI ) + f (fn - fi ) / f ( JI ) + f ( J - fi ) · But, by (7. 1 . 10) , f == !I + (f - !I ) E 001 (I) and i ( J ) == i ( JI ) + i ( J - /I ) · Hence, the last assertion is now proved. In addition, if {fn}1 C L I (I) and fn � 0, then I (fn) � 0 follows from the preceding applied to { - fn} 1, which completes the proof that I is an integration theory. D We are now ready to return to the problem, raised in the discussion following Examples 7. 1 . 1 , of identifying of the measure underlying a given integration theory (E, L, I) . For this purpose, we introduce the notation a(L) to denote the smallest a-algebra over E with respect to which all of the functions in the vector lattice L are measurable. Obviously, a (L) is generated by the sets {f > a} as f runs over L and a runs over �. 7. 1 . 15 Theorem (Stone) . Let ( E, L, I) be an integration theory, and assume that 1 E L. Then

(7. 1 . 16) the mapping (7. 1 . 17) is a finite measure on ( E , a (L I (I) ) ) , a (L I (I) ) is the completion of a(L) with respect to Jli , L I ( JLI ) == £ I (I) , and (7. 1 . 18) Finally, if ( E, B, v) is any finite measure space with the properties that L C LI (v) and I ( f ) == f dv for every f E L, then a ( L ) C B and v coincides with Jli on a(L) .

JE

PROOF: Let 1i denote the collection described on the right hand side of equa­ tion (7. 1. 16) . Using Theorem 7. 1 . 14, one can easily show that 1i is a a-algebra over E and that r E 1i � Jli (r) - f (1r) E [0, oo )

7.1

An Existence Theorem

145

defines a finite measure on ( E, 1t) . Our first goal is to prove that

To this end, for given f 9n

:

E

------+

� and a E � ' consider the functions

[n(l - l /\ a)] /\ 1 ,

n E ll + .

If I E L 1 (I) , then each 9n is also an element of L 1 (I) , 9n / 1{/> a } as n � oo , and therefore 1{/> a } E L 1 (I) . Thus, we see that every I E L 1 (I) is 1t-measurable. Next, for given I : E � [0, oo ) define

L 1 (I) U L 1 (E, 1i, JLI), then (cf. the preceding and use linearity) In E £ 1 (I ) n £ 1 (JLI ) , In / I, and so I E L 1 (E, 1t, I) n L 1 (JLI) and i(l) == I dJl i.

If I E

JE

Hence, we have now proved (7. 1 . 19) . Our next goal is to show that

== 1t == a ( £ 1 (I) ) . Since L C £ 1 (I) and every element of £ 1 (I) is 1t-measurable, what we know so far is that a(L) c a ( L 1 (I) ) c 1t. a(L) JL I

(7. 1 .20)

Thus, to prove (7. 1.20) , all we need to do is show that

(7. 1 .21 ) JL I , then there exist A, B E 1t such that A C r C B and E 1t But if r and I(1A) == Jli(A) == Jli(B) == I(1 n ) . Hence, we can choose sequences {cpn }1 {'l/Jn}1 from Lu nL 1 (I) so that - cpn < 1 A < 1r < 1n < 'l/Jn , - I(cp n ) / I(1 A ) , and I( 'l/Jn ) � i(1n) . Further, after replacing 'Pn and 'l/Jn by 'P 1 1\ 1\ 'Pn and 'l/; 1 1\ 1\ 'l/Jn , we may and will assume that each of these sequences is non­ increasing. Next, take cp == limn---+ CX) 'Pn and 'ljJ == limn---+ CX) 'l/Jn . Note that cp and 'ljJ are in L 1 (I), -cp < 1 r < 'ljJ, and -i ( cp) == i ( 'ljJ) . In particular, since -i(cp) < I (1r) < I (1r) < i('lf;) , this means that 1r E L 1 (I) and therefore that r == { 1r > 0} E a ( L 1 (I) ) . To prove that r E a(L) JL1 , first note that every element of Lu is a(L)-measurable. Hence, both cp and 'ljJ are a(L)-measurable, and so both the sets C == { cp < 0} and - D == {-'ljJ > 1 } are elements of a (L) . Finally, from -cp < 1r < 'ljJ and -I( cp) == I( 'ljJ ) , it is easy to check that < -cp - < 1r < 1v < 'ljJ and therefore that JL(D \ C) JL== JL(D) - JL(C) - < 1c I('l/J) + I(cp) == 0, which completes the proof that r E a(L) I . Hence, we have now proved ( 7. 1. 21 ) and therefore ( 7. 1 .20 ) . ·

·

·

·

·

·

VII A Little Abstract Theory

146

We have now completed the proof of everything except the concluding as­ sertion of uniqueness. But if L C L I (v), then obviously a(L) C B. Moreover, if J f dv == I(f) for all f E L, then we can prove that v f a(L) == J-LI f a(L) as follows. Namely, it is clear that a(L) is generated by the 1r-system of sets r of the form

r == { fi > a I , , ft > al }, where f E z+ , { ai , . . . , at} C � ' and {fi , . . . , ft} C L . Thus, by Exercise 3. 1.8, we need only check that v agrees with J-LI on such sets r. To this end, define ·

·

·

]

[

9n == n I min < k
j

j

v ( r ) == nlim ---+ CX) gn dJ-LI == J-LI (r) . D ---+ CX) gn dv == nlim With the preceding result, we are now ready to handle the situations de­ scribed in ( ii ) and ( iii ) of Examples 7. 1 . 1 .

7. 1 . 22 Corollary ( Caratheodory Extension ) . Let A be an algebra of sub­

sets of E and suppose that J-L : A [0, oo ) is a finitely additive function (cf. ( ii ) in Examples 7. 1 . 1) with the property that (7. 1 .2) holds. Then there is a unique finite measure jl on (E, a(A)) with the property that jl coincides with J-L on A. PROOF: Define L(A) and I on L(A) as in ( ii ) of Examples 7. 1 . 1 . It is then an easy matter to see that a(A) == a (L(A)) . In addition, as was shown in ( ii ) of Examples 7. 1 . 1 , I is an integral on L(A) . Hence the desired existence and uniqueness statements follow immediately from Theorem 7. 1. 15. D ------+

Before we can complete ( iii ) in Examples 7. 1 . 1 , we must first prove the lemma alluded to there. 7. 1 . 23 Lemma ( Dini's Lemma) . Let

{fn }�

be a non-increasing sequence of non-negative, continuous functions on the topological space E. If fn � 0, 0 uniformly on each compact subset K C E (i.e., ll fn llu , K then fn 0.) PROOF: Without loss in generality, we assume that E itself is compact. Let E > 0 be given. By assumption, we can find for each x E E an n(x) E z+ and an open neighborhood U(x) of x such that fn ( x ) ( Y ) < E for all y E U(x) . Moreover, by the Heine-Borel Theorem, we can choose a finite set {x i , . . . ' X L } c E so that E == u � I U(x£). Thus, if N(E) == n(xi ) v . . . V n(xL), then fn < E as long as n > N(E) . D ------+

------+

Given a topological space E, let Cb (E; �) denote the space of bounded continuous functions on E and turn Ch (E; � ) into a metric space by defin­ ing IIY - f llu, E to be the distance between f and g. We will say that A :

7. 1

An Existence Theorem

147

is a non-negative linear functional if A is linear and A(f) > 0 for all f E Cb ( E; [0, oo ) ) . Furthermore, if A is a non-negative linear functional on Ch (E; JR), we will say that A is tight if it has the property that for every 8 > 0 there is a compact K8 C E and an A8 E (0, oo ) for which

Cb (E; JR)

�

lR

I A(J) I < A8 ll f llu, K6 + 8 ll f ll u, E

for all

f E Cb (E; lR) .

Notice that when E is itself compact, then every non-negative linear functional on Ch (E; JR) is tight. 7. 1.24 Theorem ( Riesz Representation ) . Let E be a topological space, �

set B == a ( Cb (E; JR) ) , and suppose that A : Ch (E; �) � is a non-negative linear functional which is tight. Then tbere is a unique finite measure Jl on (E, B) with the property that A(J) == JE f dJL, f E Ch (E; JR) .

PROOF: Clearly, all that we need to do is show that A ( fn ) � 0 whenever {fn}! C Cb (E; JR) is a non-increasing sequence which tends (pointwise) to 0. To this end, let E > 0 be given, set 8 == + ll f1 ll u,E , and use Dini ' s Lemma to 1 2 choose an N(8) E z+ so that l l fn ll u , K6 < � for all > N(8) , where K8 and 2 6 A8 are the quantities appearing in the tightness condition for A. Then, for > N(8), ! A(fn) l < E. D n

n

The importance of Theorem 7. 1 .24 is hard to miss. Indeed, it seems to say that it is essentially impossible to avoid Lebesgue ' s theory of integration. On the other hand, it may not be evident how one might use Corollary 7. 1 .22. For this reason we close this section with a typical and important application of Caratheodory ' s Theorem. Namely, for each i from a non-empty index set J, let (Ei , Bi , Jli ) be a probability space. Given 0 =I= S C J, set Es == fl i E S Ei , and use lis to denote the natural projection map from E E3 onto Es . Finally, let � stand for the set of all non-empty, finite subsets F of J, and denote by BJ the a algebra over E generated by sets of the from

(7. 1 . 25)

rF

II F 1

(rriE F ri)

'

F E � and

ri E Bi , i E F.

Our goal is to show there is a unique probability measure ( E, BJ ) with the property that

(7. 1 .26)

JL ( rF )

==

JL

ni E J Jl i

on

IT Jli (ri) iE F

for all choices of F E � and ri E Bi , i E F. We begin by pointing out that uniqueness is clear. Indeed, the generating class described in (7. 1 .25) is obviously a 1r-system, and therefore ( cf. Exercise 3. 1 .8) the condition in (7. 1 .26) can be satisfied by at most one measure. Sec­ ondly, observe that there is no problem when J is finite. In fact, when J has

VII A Little Abstract Theory

148

only one element there is nothing to do at all. Moreover, if we know how to handle J ' s containing n E z + elements and J == {i 1 , . . . , in + 1 }, then we can take f1� + 1 Jli m to be the image (cf. Lemma 5.0. 1) of (cf. Lemma 4. 1 .3)

under the mapping X

Ei n + l �

n+ 1 ( xi l ' · · · ' Xi n + l ) E II Ei m · 1

Thus, assume that J is infinite. Given F E � ' use /-L F to denote niEF Jli · In order to construct J.L, we first introduce the algebra A

U li p 1 ( BF) where Bp FE�

II Bi iEF

and note that BJ is generated by A. Next, observe that the map J.L : A � [0, 1] given by

J.L( A ) == J.L F ( r ) if A == II F 1 (r) for some F E � and r E BF is well-defined and finitely additive. To see the first of these, suppose that, for some r E Bp and r ' E BF ' , Il p 1 ( r ) == Il p } ( r ') . If F == F' , then it is clear that r == r ' and that there is no problem. On the other hand, if F c F', then one has that r ' == r X E F'\F and therefore (since the Jli ' s are probability measures) that

/-LF' ( r' ) == J.L F ( r ) II Jli (Ei ) == J.L F ( r ) . iEF'\F That is, J.L is well-defined on A. Moreover, given disjoint A and A' from A, choose F E J so that A, A' E II F 1 ( BF) ' note that r == li p ( A ) is disjoint from r ' == IJ p , ( A') , and conclude that

J.L ( A u A' ) == J.L F ( r u r' ) == J.L F ( r ) + J.L F ( r ' ) == J.L( A ) + JL ( A' ) . In view of the preceding paragraph and Corollary 7. 1 .22, all that remains is to show that if {An } ! is a non-decreasing sequence from A and n� An == 0, then J.L(An ) � 0. Equivalently, what we need to know is that if {An} ! C A is non-decreasing and, for all n E z + and some E > 0, J.L( An ) > E, then n� An =I= 0. Thus, suppose that such a sequence is given, choose {Fn }1 so that An E li p� ( BFn ) for each n E z + , and set S == J \ U� Fn. Without loss in generality, we assume that Fn C Fn + 1 for all n . Under the condition that

7.1 JL

An Existence Theorem

149

( An) > E > 0 for all E z + , we must produce a sequence {am}! such that a I E E p1 , am E Epm \Fm_ 1 for m > 2, and ( ai , . . . , an) E Il pn ( An) for all E z + . In fact, given such a sequence {am} 1 , observe that by taking a E E so that Il p1 ( a) == a I , Ilpm \Fm_ 1 ( a) == am for m > 2, and (when S =I= 0) lis ( a) is an arbitrary element of Es, one gets an element of n� An. n

n

To produce the am ' s, first choose and fix, for each i E J, a reference point e i E Ei . Next, for each n E z + , define .Pn : E pn � E so that if i E Fn if i E J \ Fn , and set fn == lA n o -Pn. Obviously,

f < JL ( An) =

1EFn fn (XFn ) ILFn (dxFn ) ·

FUrthermore, if, for each m E z + , we define the sequence functions on E pm so that 9m , m (xFm ) == fm (xFm ) and

gm , n (xF,. )

=

{

{ gm , n

:

n

>

m} of

fn (xF,. , Y Fn \F.,. ) ILFn \F.,. ( Y Fn \ F.,. )

JE Fn \Fm when > m, then 9m , n + I < 9m , n and gm + l, n (XF,. , Y F,.+1 \F.,. ) JL F,.+1 \F.,. ( dy F,.+ 1 \F.,. ) gm , n (XF,. ) = E pm +1 \Fm for all 1 < m < Hence, 9m limn ---+ CX) 9m , n exists and, by the Monotone n

1

n.

-

Convergence Theorem,

J Finally, since limCX) { fn (xFn ) ILFn ( dxFn ) = nlim p( An) > E, n ---+ CX) ---+ }FE Fn . E F1 there exists an a I E E p1 such that 9I ( a I ) > E. In particular ( since 9I < !I ) , this means that ai E Il p1 ( AI )· In addition, from (7. 1 .27) with m == 1 and X p1 == ai , it means that there exists an a2 E Ep2 \F1 for which 92 ( ai , a2 ) > E and therefore ( since 92 < !2) ( ai , a2) E Il p2 ( A2) · More generally, if ( ai , . . . , am ) E E pm and 9m (ai ' . . . ' am ) > E, then ( since 9m < fm) ( ai ' . . . ' am) E Il pm ( Am) and, by (7. 1.27) , there exists an am + I E Epm + 1 \Fm such that 9m + I ( ai , . . . ' am ri ) > E. Hence, by induction, we are done and we have proved the following theorem.

{ g 1 (xF1 ) ILF1 ( dXF1 ) h

=

7. 1 . 28 Theorem. Let J be an arbitrary index set and, for each i E J, let

(Ei , Bi , Jli ) be a probability space. If E == fl i E J Ei and BJ is the a-algebra over E generated by the sets rF in (7. 1 . 25) , then there is a unique probability measure JL on ( E, BJ ) satisfying (7. 1.26) .

150

VII A Little Abstract Theory

The existence result proved in Theorem 7. 1 .28 plays an important role in probability theory, where it becomes a ubiquitous tool with which to construct infinite families of mutually independent random variables. 7. 1 . 29 Remark: As the preceding discussion demonstrates, the version of

Caratheodory ' s theorem which we have proved here is sufficient to handle the construction of non-trivial measures which are finite. Moreover, by an obvious localization procedure, one can reduce problems involving a-finite situations to finite ones. On the other hand, when confronting problems leading to measures which are not a-finite, the results proved in this section are insufficient, and one must work harder. For an excellent account of both the methodology ( as well as reason ) for handling such problems, see L.C. Evans and R. Gariepy's Measure Theory and Fine Properties of Functions, published by the Studies in Advanced Math. Series of CRC Press ( 1991) . Exercises

7. 1 .30 Exercise: Assume that E is a metric space. Show that the Borel field

BE coincides with a ( Cb(E; � )) . Next, assume, in addition, that E is locally compact ( i.e. , every point x E E has a neighborhood whose closure is compact ) and show that BE == a ( Cc ( E; �) ) ( the space of f E C ( E; �) which vanish off some compact subset of E ) . 7. 1 .31 Exercise: Let 1/J be a bounded, right-continuous, non-decreasing func­

tion on �. Set 1/J (±oo) == lim x� ± CX) 1/J(x ) . (i) For each cp E Ch ( �; � ) , show that

I(cp)

i

lim ( R ) cp d1j; E � R� CX) [ - R , R]

exists. (ii) Show that cp E Ch ( �; � )

I( cp) E � is a non-negative, linear functional. Further, check that, for each R > 0, r-----+

where A == 1/J( oo) - 1/J( -oo) and E(R)

==

A - (1/J(R) - 1/J( -R) ) .

(iii) As a consequence of ( ii) and Theorem 7. 1 .24, show that there is a unique

measure J-l'lf; on ( �, BR ) with the property that �-t ( ( -oo, x] ) == 1/J(x) - 1/J( -oo) for all x E �. Notice that the mapping 1/J r-----+ J-l 'lf; inverts the map discussed in Theorem 5.4.2.

7.2 Hilbert

Space and the Radon-Nikodym Theorem

151

7. 2 Hilbert Space and the Radon-Nikodym Theorem.

In Exercise 6. 1. 7 we saw evidence that, among the £P-spaces, the space £ 2 is the most closely related to familiar Euclidean geometry. In the present section, we will expand on this observation and give an application of it. Throughout (E, B, JL ) will be a measure space. By part ( iii ) of Theorem 6.2.2, the space L 2 ( JL ) is a vector space which becomes a complete metric space when we use I I ! - g ii £2 ( J.L ) to measure the distance between f and g. In addition, if we define (7.2 . 1) then ( f, g ) L 2 ( J.L ) is bilinear ( i.e. , it is linear as a function of each its entries ) and ( cf. (6.2.9) ) , for f E L 2 ( JL ) , (7.2.2)

(J , f) l 2 ( J.L ) = sup { ( !, g ) £2 ( 11 ) : g E L 2 ( J..L ) with ll g ll £2 ( 1' )

I I ! I I L 2 ( J.L ) ==

1

}

<1 .

Note that (J , g ) L 2 ( J.L ) plays the same role for L 2 ( JL ) that the Euclidean inner product plays in �N . That is, by Schwarz ' s inequality ( cf. Exercise 6. 1 .7) ,

can be thought of as the angle between f and g in the plane spanned by f and g. Thus, we say that f E L 2 ( JL ) is orthogonal or perpendicular to S C L 2 ( JL ) and write f ..l S if (J , g ) L 2 ( J.L ) == 0 for every g E S. A topological vector spaces with this kind of structure is known as a Hilbert space. A distinguishing feature of Hilbert spaces is the property proved in the following lemma. Indeed, the reader might want to observe that, although we restrict our attention to £ 2 -spaces, the result proved depends only on com­ pleteness and the existence of a bilinear map for which (7.2.2) holds; hence, it is a property possessed by all Hilbert spaces. 7.2.3 Lemma. Let F be a closed linear subspace of L 2 ( JL ) . Then, for each

g E L 2 ( JL ) , there is a unique f E F for which (7.2.4)

In fact, the unique f E F which satisfies (7.2.4) is the unique f E F such that (g - f) ..l F. That is, f is the perpendicular projection of g onto F.

152

VII A Little Abstract Theory

(7.(7. 22. 4.4),) (g - f) 1/J t I I g - f - t'ljJ II �2 (J.I) I g - f I i2 (J.I) - 2t (g - J, 'ljJ) L2 (J.I) + t2 I 'ljJ I i2 (J.I) t (g - f) (g - J, 1/J ) £2 (JL1/J) f 1/J l g - 1/J I �2 (JL) I Y - JI I �2 (JL) + 2 (g - f, f - 1/J ) L2 (JL) + I / - 1/J I �2 (JL) I Y - f i �2 (JL) + I ! - 1/J I �2 (JL) > I Y - JI I �2 (JL) • f; !I , ! i {1, 2}, (!2 ( fi) g 2 II !2 - !Il l £ 2 ( JL ) f {fn}! C (7. 2.4) l g - fn II £2 (J.I) a { l g - cp l £2 (J.t) : cp } L 2 (JL) { fn}1 cp, 1/J £2 (JL) , (7. 2 . 5) I 'P + 1/J I �2 (JL) + I 'P - 1/J I �2 (JL) 2 l cp i �2 (JL) + 2 I 1/J I �2 (JL) · cp g - fn 1/J g - fm (7. 2 . 5),

if and only if ..l F. and observe that, for any

ff

P ROOF: We first check that E F satisfies To this end, suppose that E F satisfies E F, the function E � 1----+

= has a minimum at == 0. Hence, by the first derivative test, we see that == 0 for every l_ F, E F. Conversely, if E F and then, for any E F, ==

==

Thus, we have now proved the equivalence of the two characterizations of and, as a consequence of the second characterization, uniqueness is easy. ..l F, E F and Indeed, if E then - !I ) l_ F and == 0. therefore In view of the preceding, it remains only to prove that there is an E F for holds. To this end, choose F so that which inf

-----+

Since that

EF .

and therefore F are complete, all that we have to do is show is Cauchy convergent. In order to do this, note that for any

E

==

Taking

and

==

in

==

we obtain

2 ( l g - fn l i2 (J.I) - a2 ) + 2 ( l g - fm l i2 (J.I) - a2 ) , ln �lm 2 2 N z+ � fn g > N. D > N, l fn - fm i £2 (JL) < 7.1. 3 (6. 2 . 8) . Let A : L 2 (JL) R be a linear mapping. Then (7.2 . 7) I A I { A(cp) : cp L2 (J.L) and l cp i £2 (J.t) < 1 } < if and only if there is an h £ 2 (JL) for which (7. 2 .8) A(cp) (h, cp) L2 (J.t) ' cp L2 (J.L) ; in which case h is uniquely determined by (7. 2 . 8 ) and l h i £ 2 ( JL ) is the supremum in (7. 2 . 7 ). <

where we have used the fact that Now let E 0 be given, choose n and conclude that

E F in order to get the last inequality.

E

so that E for m,

for

n

The result obtained in Lemma is a basic existence assertion from which a great many other existence results follow. Perhaps the single most important such result is the following sharpening of the 7. 2.6 Theorem (F. Riesz) .

sup

-----+

E

oo

E

=

E

7. 2 Hilbert

Space and the Radon -Nikodym Theorem

153

PROOF: Everything except the existence of f when (7.2. 7) holds is clear. To prove this existence, first note that (7.2. 7) implies that

and therefore that A is continuous on £ 2 (JL) . Hence, F { cp E £ 2 (JL) : A( cp) == 0} is both linear and closed. Moreover, either F == £ 2 (JL) , in which case we may take h == 0, or there is a g � F. In the latter case, choose f E F so that k (g - f) ..l F and note that A( k ) =I= 0. In addition, observe that, for any cp E £ 2 (JL) , A ( cp ) A

(

and therefore that

(k,

-

���j k E F.

)

Hence,

(

)

A ( cp ) A ( cp ) I l k II 2 k = 0, k, =

and so we can take

D Following J. von Neumann, we will now use Theorem 7.2.6 to derive an important property about the relationship between measures. Namely, given two measures Jl and v on the same measurable space (E, B) : ( a) we say v dominates Jl and write Jl < v if JL(r ) < v(r) , r E B; ( b ) we say Jl is absolutely continuous with respect to v and write Jl << v if JL ( r ) == 0 for all r E B with v(r) == 0; ( c ) we say Jl and v are singular and write Jl ..l v if there is a � E B with the property that JL(�) == v ( �C ) == 0. Obviously, both domination and absolute continuity express a relationship between Jl and v, the former being a much stronger statement than the latter. In contrast, singularity is a statement that the measures have nothing to do with one another and, in fact, live on different portions of E. The result alluded to above comes in two parts and applies to Jl ' s which are finite and v ' s which are a-finite. The first part says that Jl can be written (in a unique way) as the sum of a measure Jla which is absolutely continuous with respect to v and a measure Jla which is singular to v. The second part tells us that there is a unique non-negative f E L 1 (v) with the property that (7.2.9)

!La (r) =

l f dv,

rE

13.

In particular, if Jl itself is absolutely continuous with respect to v, then Jl == Jla and so (7.2.9) holds with Jl in place of Jla ·

154

VII A Little Abstract Theory

The key to von Neumann ' s proof of these results is the observation that everything can be reduced to consideration Jl ' s which are dominated by v, in which case the existence of f becomes a simple application of Theorem 7.2.6. 7.2. 10 Lemma. Suppose that (E, B, v ) is a a-finite measure space and that

Jl is a finite measure on (E, B) which is dominated by v. Then there is a unique [0, !]-valued h E L 1 (v) with the property that

l r.p dJ.L l r.ph dv

(7.2. 1 1)

=

for every B-measurable cp : E

----t

[0, oo] .

PROOF : Since we can write E as the union of countably many, mutually dis­ joint, B-measurable sets of finite v-measure, we assume, without loss in gen­ erality, that v is finite on (E, B) . But, in that case, L 1 (JL) C L 2 (v) and the linear mapping r.p E L 2 (v) � A (r.p) -

I 'P I £2 (v) ·

l r.p dJ.L E R

satisfies I A (cp) l < v(E) ! Hence, by Theorem 7.2.6, there is an h E L 2 (v ) such that (7.2. 1 1) holds for every cp E L 2 (v) and, therefore, for every bounded B-measurable function. We now want to show h (which is determined only up to a set of v-measure 0) can be chosen to take its values in [0, 1] . To this end, set An == { h < - � } and Bn == { h > 1 + � } for n E z+ . Then, by (7. 2. 1 1 ) ,

from which we conclude that v ( An) == v ( Bn) == 0 , n E z+ , and therefore that v( h < 0) == v( h > 1) == 0. In other words, we may assume that h takes its values in [0, 1 ] , and, clearly, once we know this, (7.2. 1 1 ) for all non-negative, B­ measurable cp 's is an easy consequence of the Monotone Convergence Theorem. Finally, the uniqueness assertion is trivial, since (7. 1 . 1 1) determines the v­ integral of h over every r E B. D The first part of the next theorem is called Lebesgue's Decomposition Theorem, and the second part is the Radon-Nikodym Theorem. 7. 2. 1 2 Theorem. Suppose that (E, B, v) is a a-finite me&Sure space, and let

Jl be a finite measure on (E, B) . Then there is a unique measure Jla < Jl on (E, B) with the properties that Jla << v and Jla ( JL - Jla) l_ v. In addition, there is a unique non-negative f E L 1 ( v ) for which (7.2.9) holds. In particular, Jl << v if and only JL ( r ) == fr f dv, r E B, for some non-negative f E L 1 ( JL ) .

7. 2

Hilbert Space and the Radon -Nikodym Theorem

155

PROOF: We first note that if JL (r) == fr f dv, r E B , for some f E L 1 ( v ) , then ( cf. Exercise 3.3. 15) JL << v and ( cf. Exercise 3.2. 17) f is necessarily unique and ( cf. the preceding ) non-negative as an element of L 1 ( v ) . Next, we prove that there is at most one choice of J.la · To this end, suppose that JL == J.la + J.la == JL� + JL� , where JLa and JL� are both absolutely continuous with respect to v and both JLa and JL� are singular to v. Choose E , �' E B so that

v ( � ) == v ( � ' )

and set A == � U �' . Then any r E B,

JLa (r)

==

and J.la ( �C ) == JL� ( � ' C ) == 0,

== 0

v( A)

== J.la ( AC) == JL� ( AC) == 0; and therefore, for

JLa ( A n r) == JL ( A n r ) == JL� ( A n r ) == JL � (r) .

Hence, J.la == JL� . To prove the existence statements, we first use Lemma 7.2. 10, applied to JL and JL + v, to find a B-measurable h : E [0, 1] with the property that -----t

l

=

for all non-negative, B-measurable cp ' s. It is then clear that

l cp( 1 - h) d l
(7.2. 13)

=

first for all cp E £ 1 (JL) n £ 1 ( v ) and then for all non-negative, B-measurable cp ' s. Now set � == { h < 1} and JLa (r) == JL ( � n r ) , r E B. Since

v ( E C ) = v(h = 1) = f

J{h= 1 }

h dv = f

J{h= 1 }

( 1 - h) dJ.L = o,

it is clear that JL - JLa is singular to v. At the same time, if f then, by (7.2. 13) with cp == ( 1 - h) - 1 1rnE ,

J.La(r)

=

J.La (r n E)

=

(1 - h) - 1 hlE ,

L cp( 1 - h) dJ.L £ f dv =

for each r E B. D Given a finite measure JL and a a-finite measure v, the corresponding mea­ sures JLa and JLa are called the absolutely continuous and singular parts of JL with respect to v. Also, if JL is absolutely continuous with respect to v, then the corresponding non-negative f E L 1 (v) is called the Radon­ Nikodym derivative of JL with respect to v and is often denoted by � . The choice of this notation is explained by part ( ii ) of the Exercise 7. 2. 14 which follows.

156

VII A Little Abstract Theory Exercises

7. 2 . 14 Exercise: Suppose that C is a countable partition of the non-empty

set E, and use B to denote a(C) . ( i ) Show that f : E � R is B-measurable if and only if f is constant on each r E C. Also, show that the measure v is a-finite on (E, B) if and only if v(r) < oo for every r E C. Finally, if Jl is a second measure on (E, B) , show that Jl << v if and only if JL(r) == 0 for all r E C satisfying v(r) == 0. ( ii ) Given any measures Jl and v on (E, B) and a B-measurable, v-integrable f : E � [0, oo ) , show that JL(r) == fr f dv for all r E B implies that, for every r E C, f ���� on r if v(r) E (0, oo ) and JL(r) = 0 if v(r) = oo .

( iii ) Using the preceding, show that, in general, one cannot dispense with the

assumption in Theorem 7. 1 . 13 that v is a-finite.

7.2 . 1 5 Exercise: Readers with good memories may be disturbed by the ap­

parent difference between the notions of absolute continuity used here and that used earlier in Exercise 3.3. 15. To allay such concerns, check that, as long as Jl is finite, Jl << v implies that for every E > 0 there is a 8 > 0 with the property that JL(r) < E whenever r E B and v(r) < 8 . (Because v is not assumed to be a-finite, you should not use the Radon-Nikodym Theorem here.) 7. 2 . 16 Exercise: The purpose of this exercise is to take a closer look at

the relationship 'ljJ � Jl,p (established in Exercise 7 . 1 .30) taking a bounded, right-continuous, non-decreasing 'ljJ on � into a finite measure Jl,p on ( �, BR ) . ( i ) Given a measure space (E, B, JL) and an x E E, one says that x is an atom of (E, B, JL) if JL(r) > 0 for every r E B which contains x. Further, one says that (E, B, JL) is non-atomic if it has no atoms and that (E, B, JL) is purely atomic if JL(r) == 0 for every r E B which does not contain any atoms of ( E, B, Jl) . Assuming that { x} E B for every x E E, show that x is an atom of (E, B, JL) if and only if JL ( {x} ) > 0, and, when Jl is a-finite, conclude that the set A == { x E E : Jl ( { x} ) > 0} of atoms is countable. In particular, if { x} E B for every x E E and Jl is a-finite, show that (E, B, JL) is purely atomic if an only if there is a countable set S such that JL(r) ==

L

x E r nS

JL ( {x} ) ,

r E B.

( ii ) Given a bounded, right-continuous, non-decreasing 'ljJ on � ' show that 'ljJ is

continuous if and only if Jl,p is non-atomic. Next, using either Lemma 1 .2.20 or the preceding, show that the set D( 'ljJ) of x E � for which 'lj;(x) - 'l/J(x - ) > 0 is countable, and define the discontinuous part of 'ljJ to be the function 'l/Jd : � � � given by t E ( - CX) ,x] n D ( 'f/; )

( 'lj;(t) - 'lj;(t-) ) ,

X

E �-

7. 2

Hilbert Space and the Radon-Nikodym Theorem

157

Show that 'l/Jd is a non-negative, bounded, right-continuous, non-decreasing function and that the continuous part 'l/Jc 'ljJ - 'l/J( -oo) - 'l/Jd of 'ljJ is a bounded, continuous, non-decreasing function. Finally, say that 'ljJ is purely discontinuous if 'ljJ == 'l/J ( -oo) + 'l/Jd , and check that 'ljJ is purely discontinuous if and only if Jl,p is purely atomic. ( iii ) We now want to see how to characterize Jl,p << AR in terms of 'l/J. For this purpose, we say that the bounded, right-continuous, non-decreasing function 'ljJ is absolutely continuous if, for every E > 0, there is a 8 > 0 such that 00

L ( 'l/J (bm) - 'l/J ( am )) < E

m= l

whenever { ( am, bm ) }1 is a sequence of mutually disjoint, open intervals which satisfy E:= l (bm - am ) < 8. Obviously, every absolutely continuous 'ljJ is uniformly continuous. Show that, in fact, 'ljJ is absolutely continuous if and only if Jl,p << AR . In this connection, we say that 'ljJ is singular if Jl,p ..l AR . ( iv ) The preceding considerations lead to the following decomposition of a bounded, right-continuous, non-decreasing function 'l/J. Namely, let Jl,P,a and Jl,P,a be the absolutely continuous and singular parts of Jl,p with respect to AR , and define the absolutely continuous part 'l/Ja and singular part 'l/Ja of 'ljJ by 'l/Ja( x ) == Jla ( (-oo, x] ) and 'l/Ja ( x ) == Jla ( (-oo, x] ) for x E �- Further, let 'l/Ja,c and 'l/Ja,d be the continuous and discontinuous parts of 'l/Ja . Obviously, (7.2. 17)

'l/J == 'l/J (

- 00

) + 'l/Ja + 'l/Ja,c + 'l/Ja,d ·

Show that the decomposition in (7.2. 17) is canonical in the sense that if 'ljJ == 'l/J ( -oo) + 'lj; 1 + 'lj;2 + 'ljJ3 , where, for each 1 < i < 3, 'l/Ji is right-continuous, non­ decreasing, and tends to 0 at -oo, 'lj; 1 is absolutely continuous, 'lj;2 is continuous but singular, and 'lj;3 is purely discontinuous, then 'l/J 1 == 'l/Ja, 'l/J2 == 'l/Ja ,c , and 'l/J3 == 'l/Ja,d · 7.2 . 1 7 Exercise: It is clear that absolutely continuous non-decreasing and purely discontinuous non-decreasing functions exist. But are there any contin­ uous, singular, non-decreasing functions? In order to show that there are, we will now describe the Cantor-Lebesgue function. Referring to Exercise 2 . 1 .20, recall that the closed set Ck is the union of 2 k disjoint closed intervals and that [0, 1] \ ck is the union of 2 k - 1 disjoint open intervals ( ak,j ' bk,j ) ' 1 < j < 2 k ' where we have ordered these so that bk ,j < ak,j + l for 1 < j < 2 k - 1. Next, set ak ' o == -oo, bk ' o == 0, a k ' 2 k == 1 , bk ' 2 k == oo, and define 'l/Jk : � � [0, 1] so that: ( a) 'l/Jk is constant on each of the intervals [ak,j , bk , j ] , 0 < j < 2 k ; ( b ) 'lfJk (bk,j ) == 2 - k j for 0 < j < 2 k and 'l/Jk is linear on each of the intervals [bk,j , ak , j + l ] , 0 < j < 2 k .

158

VII A Little Abstract Theory

Notice that each 'l/J k is continuous and non-decreasing from � onto [0, 1] . In addition, check that i l '¢k +1 - '¢k l lu , IR = 2 �k ; and conclude from this that '¢k converges uniformly on � to a continuous, non-decreasing 'ljJ : � [0, 1] with the property that Jl'lf; (C) == 1 . At the same time, by Exercise 2 . 1 . 20, AR (� \ C) == 0, and therefore 'ljJ is singular. -----+

Notation Description

Notat ion (a.e. , J-t)

a+ &

a

-

( v , w )!RN lim An & n-+(X) -

To be read almost everywhere with respect to J-t The positive and negative parts of The inner product of v and w in

a

E

§3.3

JIR

§ 1 .2

JJRN

n := l U n > m An and limit inferior lim An u :=l n n > m An of the sets { An } r n-+(X) The limit superior

E 3. 1 . 12

-

BE (a, r) BE B [E' ] J-L

-

B

Seet

BlRN

-

The ball of radius r around a in the metric space E The Borel a-algebra over the topological space E

§3. 1

The restriction of the a-algebra B to the subset E'

§3. 1

The completion of the a-algebra B with respect to the measure

The a-algebra of Lebesque measurable subsets in JIRN The product a-algebra generated by sets of the form

§2. 1

rl

r2 for ri E Bi C ( E ) & C( E; F )

§3. 1

J-t

X

§3.2

The spaces of continuous JIR-valued and F-valued func­ tions on E The space of bounded, continuous, JIR-valued functions on the topological space E.

Cc (G)

The space of f

E

C (G) with compact support in G.

The space of f : G

II C II

�

JIRN with

n

continuous derivatives

The mesh size of the collection C

§1.1

To be read C2 is a refinement of C 1

§1.1

The least common refinement of C 1 and C2

§1.1

The length of the interval I and the change 1/; ( I+ ) 1/; ( I - ) of 'lj; over I The Jacobian of c . . Ab rev1ation tor

ar

The boundary

§ § 1 .2 & 1 .3 §5.3

cp a:

ax 1

ala I oN . . . ax N

1

r \ f of the set r

159

§5.2

Notation

160

The restriction of the function f to the set r

J rr

f (x+ ) & f (x - )

The right and left limits of f at x

1 1 / l l u & 11 / l l u , E

The uniform norms of the function f & f f E

f l\ g & f V g 1 1 / I I L v ( J.L ) ( J , g) £ 2 ( J.L ) I I f I I £ ( P l ,p 2 ) ( J.L l ,J.L2 ) J' & J'u

==

J.L

0

The minimum & maximum of f and

g

The p-Lebesgue norm of f with respect to the measure J.L The inner product of f and g in L 2 (J.L )

q, - 1

The mixed (P 1 , P2 ) -Lebesgue norm with respect to the

pair of measures (J.L 1 , J.L 2 )

The collections of all closed subsets and all countable unions of closed subsets

The image of the measure J.L under the map

tP

The exterior or outer Lebesgue measure of the set r

rc

The complement of the set r

r

The interior of the set r

-

The closure of the set r

r

rC6

)

� & �6 lr

I - & I+ IJRN

1

/ dx

Jtl>

LP (J.L) £P (JRN ) L (PI ,P2 ) (J.L 1 ' J.L 2 )

§5.2 §2. 1 (5.3.3 )

1r1e

0

§3 . 2 & §5.2 §7.2

The convolution of the functions f and g

f*Y p* J.L

§ 1.2

E lR

The open 6-hull of the set r

§5. 1 §2. 1

§5 .3

The collections of all open sets and all countable intersections of open sets The indicator or characteristic function of the set r

§2. 1 §3.2

The left and right end points of the interval I The identity matrix on

lR N

The Lebesgue integral of f over r

§2.2 E

-

BJRN

The Jacobian matrix of tP The Lebesgue space of f with 1 1 / I I L P ( J.L )

§5.3 < oo

§3 .2 & §5.2

The Lebesgue space of L P ( AJRN ) Mixed norm Lebesgue space

§6 .2

161

Notation

Lebesgue's measure restricted to E

AE J.L 1 J.L

X

J.L 2

<<

l/

dJ.L -

E

-

BJRN

The product of the measures J.L 1 and J.L 2 .

§3.4

To be read J.L is absolutely continuous with respect to

v.

§7.2

The Radon-Nikodym derivative of J.L with respect to

v

§7.2

dv

J.L l_ V

Equivalence relation determined by the measure J.L·

§3.2

J.L

The completion of the measure J.L ·

§3 . 1

N

The set of non-negative integers.

VF

The gradient of the function F.

§5.3

n(x)

The outer normal at the point x.

§5.3

P (E )

The surface area of the sphere and the volume of the unit ball in JR N

The power set (set of all subsets ) of E The set of all rational numbers in

lR

The extended real line

-...

2 The extended plain JR with the two points ( oo - oo ) and ( - oo

,

oo

E. 5 . 2 . 6 §3. 1

Q JR2

1

§7.2

v.

J-L r-v

W N - 1 & O, N

(R)

To be read J.L is singular to

lR

§3.2 ,

) removed

§3. 2

The Riemann-Stieljes integral of cp with resp ect to 1/J

§ 1.2

sgn(x )

The signum of x

§ 1 .2

a(E ; C)

The a-algebra over E generated by the collection C

cp(x) d'ljl (x)

sN- 1 TA Tx

B(C) z & z+

E lR

The unit sphere in

lR

§3 . 1

N

The linear transformation determined by the matrix A

§2.2

The tangent space at x

§5.3

The set of selections for the collection C

§1.1

The set of all integers and its subset of positive integers

Index

A

absolutely continuous, 60 for measures , 1 53 for non-decreasing function, 1 57 part of a measure, 1 55 uniformly, 6 1 absolutely continuous part for non-decreasing functions, 1 57 algebra, 34 almost everywhere, 53 convergence, 53 approximate identity, 1 34 compactly supported , 136 atom , 1 56 atomic non, 1 56 purely, 156 axiom of choice, 27 B

ball, 25 volume of, 88 volume of in JR N , 86 Beta function, 101 , 1 36 Borel a-algebra or field over E , 35 Borel measurable , 35 Borel-Cantelli Lemma, 39 c

Cantor set , 28 Cantor-Lebesgue function, 157 Caratheodory's Extension Theorem, 146 co-area formula, 103 , 1 13 concave function, 1 14 Hessian criterion, 1 1 5 continuous part of non-decreasing function, 1 57 convergence J.L-almost everywhere, 53 in J.L-measure, 54 metric for, 6 1

convex set, 1 1 4 convolution , 131 for the multiplication group , 137 Young's Inequality for, 1 3 1 convolution semigroup , 134 Cauchy or Poisson, 135 Weierstrass, 135 coordinate chart , 95 global, 102 countably additive, 36 cover, 1 exact, 3 , 24 non-overlapping, 1 , 24 D

Daniell 's Theorem, 143 diameter, 74 of rectangle, 1 diffeomorphism, 90 Dini 's Lemma, 146 discontinuous part of non-decreasing function, 1 56 distribution function of f in computation of integrals, 1 29 distribution of f , 8 1 i n computation of integrals , 83 divergence, 109 Divergence Theorem , 109 E

Euclidean invariance of Lebesgue ' s measure, 3 1 extended real line, 4 1 F

Fatou ' s Lemma, 52, 57 Lieb's version, 54, 58, 1 22 finitely additive, 140 Fubini 's Theorem, 71 function bump , 136

Index function ( continued) concave, 1 14 harmonic, 1 1 1 measurable, 43 modulus of continuity of, 10 fundamental solution, 109 Fundamental Theorem of Calculus, 9 G

Gamma function , 88 Gauss kernel, 135 gradient , 92 Green's Identity , 109 H

Hardy's inequality, 138 Hardy-Littlewood maximal function, 62, 66 maximal inequality, 65, 130 harmonic function, 1 1 1 mean value property of, 1 1 1 generalized, 1 1 2 heat flow or Weierstrass semigroup , 135 Hessian matrix, 1 1 5 Hilbert space, 1 5 1 Holder's Inequality, 1 1 7 Holder conjugate, 1 1 7, 1 22 hyperplane, 33 hypersurface , 92 I image of a measure, 80 injective, 90 integrable, 49 function the space L 1 (J.L) , 49 uniformly, 6 1 integration by parts, 9 isodiametric inequality , 74 J

Jacobi's Transformation Formula, 90 Jacobian , 89 matrix, 89 Jensen ' s Inequality, 1 14, 1 1 9

1 63 L

.X-system , 35 Laplacian, 109 lattice, 1 39 vector, 1 39 integral on, 1 39 integration theory for, 139 lattice operations, 7 £-system , 68 Lebesgue integral exists, 47 notation for, 43 Lebesgue measurable, 23 Lebesgue set , 67 Lebesgue spaces L 1 (J.L) , 49 LP (J.L) , 1 19 mixed £ (PI (J.L l , J.L 2 ) , 1 24 Lebesgue's Decomposition Theorem , 1 54 Lebesgue ' s Dominated Convergence Theo­ rem, 53, 57 Lebesgue's measure, 23 Euclidean invariance of, 3 1 existence of, 26 Hausdorff's description, 76 interior, 142 notation for, 23, 36, 85 outer or exterior , 2 1 notation for , 2 1 of a parallelapiped , 33 limit of sets, 39 limit inferior , 39 limit superior , 39 linear functional non-negative, 14 7 tight , 1 47 Lipschitz continuous, 30 Lipschitz constant , 30 L 1 (J.L) , 49 £PI (J.L l , J.L 2 ) , 1 24

, p2 )

, p2

M

Markov 's Inequality , 47 Mean Value Property, 1 1 1 , 137 measurable function , 43 criteria for, 50 indicator or characteristic, 43 Lebesgue integral of, 46 simple, 43

1 64

Index

measurable ( continued ) (function continued ) (simple continued ) Lebesgue integral of, 43 map , 4 1 measurable space, 36 product of, 4 1 measure, 36 finite, 36 probability, 36 product , 7 1 arbitrary number o f factors, 149 pushforward or image of, 80 a-finite, 70 surface, 100 zero, 53 measure space, 36 complete, 37 completion of, 36 finite, 36 probability, 36 a-finite, 70 mesh size, 3 Minkowski's Inequality, 1 1 7 continuous version, 125 modulus of continuity, 10 monotone class, 3 8 Monotone Convergence Theorem, 5 1 more refined , 4 N

non-atomic, 156 non-measurable set, 28 non-overlapping, 1 0

one-sided , stable law of order open 8-hull, 90 outer normal, 104

�,

1 35

p parallelepiped , 33 volume of, 33 1r-system , 35 Poisson equation , 1 10 Poisson or Cauchy semigroup , 135 polar coordinates , 86 power set , 34 purely atomic, 1 56

purely discontinuous non-decreasing function, 1 57 pushforward of a measure, 80 R

radius, 74 Radon-Nikodym derivative, 1 55 Theorem, 154 rectangle, 1 cube, 24 diameter ( diam) of, 1 volume (vol ) of, 1 Riemann integrable, 3 1/J-Riemann integrable, 8 with respect to 1/J, 8 with respect to 1/J , 8 1 sum, 3 lower, 4 upper, 4 Riemann integral , 4 Riemann-Stieltjes integral, 8 Riesz Representation Theorem for L 2 , 152 for measures, 14 7

s Schwarz's inequality, 1 19 semi-lattice, 68 a-algebra, 34 generated by, 35 signum function (sgn) , 16 singular for measures, 1 53 part of a measure, 155 part of non-decreasing function, 1 57 singular part for non-decreasing functions, 1 57 smooth, 103 regions, 103 sphere, 75, 85 surface area of, 86, 88 surface measure, 85 Steiner symmetrization procedure, 75 Stone's Theorem, 144 strong minimum principle, 1 1 2 subordination, 135 surface measure, 100 of graphs, 102 symmetric, 74

165

Index T

tangent space, 92 tensor product, 4 1 tight family of functions , 6 1 linear functional, 14 7 Tonelli's Theorem, 70 u

uniform norm, 6, 140

v

variation (Var) , 12 bounded, 12 negative variation (Var ) 1 2 positive variation (Var + ) , 12 _

,

y

Young's Inequality, 1 3 1 Young's inequality for the multiplicative group , 138

Solution to Selected Problems

1 . 1 .9: To prove Theorem 1 . 1 .8, simply note that Lemma 1 . 1 . 7 says that

lim II C II �O

and that lim

II C II �O

To prove that

U (f; C ) == inf U (f; C) c

£( ! ; C) == sup £(!; C) . c

f V g is Riemann-integrable if f and g are, observe that

a' V b' - a V b < (a' - a) V (b' - b) < Ia' - al + lb' - bl for any a, a', b, and b' E �Thus, for any C,

U ( f v g; C ) - .C ( f v g; C ) from which it is clear that addition, since

<

[u (f; C) - .C(f; C) ] + [u (g; C) - .C(g; C) ] ;

fVg

is Riemann-integrable if

f

and

g

are. In

for every C and � E B(C), the corresponding inequality is obvious. One can handle f 1\ g by either applying an analogous line of reasoning or simply noting that f 1\ g == - ( - f) V (-g) . Finally, the linearity assertion follows immediately from the linearity of the approximating Riemann sums. &

1 . 1 . 10

1 . 1 . 12: Given

C and E > 0, set

C(E) == {I E C :

sup f - inf j > E } . I

I

Assume that f is Riemann-integrable, and let exists a 8 > 0 such that

E > 0 be given.

Then there

U (f ; C) - C(J; C) < E 2 whenever

I I C II < 8. Thus, if II C II < 8, then

E L

(sup f - i�f f) vol(J) < L (sup f - i� f ) vol(J) == U ( f; C) - £( ! ; C) < E 2

vol(J) <

I E C ( E)

L

IEC

as

long

as

II C II < 8.

I

IE C ( E)

I

1

Solution to Selected Problems

167

Conversely, suppose that for each t > 0 there is a C€ for which

L

vol(J) <

t.

lE C e ( E )

Then

U( f; Cf. ) - C (J; Cf. ) s p f - i� f f vol( I) + s p f - i �f f = lE C e \ Ce ( E ) � lE C e ( E ) � < 2 ll f l l u vol( J) vol( J) + t

L( L

)

lE C e ( E )

L (

L

) vol( I)

lE C e \ C e ( E )

2 ll f ll u t + t i J I == t ( 2 ll f ll u + I J I) � o t � 0. Hence, infc U (f; C) == supc £(!; C) and so f is Riemann-integrable. Finally, suppose that f is a bounded function on J which is continuous at all but a finite number of points, a 1 , . . . , am . Given t > 0, choose non-overlapping <

as

cubes Q 1 , . . . , Qm so that at is the center of Q l and vol(Q£) < � for each 1 < f < m. Next, observe that f is continuous on J J \ (U� Q£) and therefore that

- f(x) : x, y E J and I Y - x l < 8 } < t for some 8 > 0. Moreover, J admits a finite exact cover C by non-overlapping sup {f (y )

�

�

cubes of diameter less than 8, and clearly C

C u { Q 1 n J, . . . , Q m n J}

is a finite, non-overlapping cover of J with the property that

L

vol(Q) <

{Q E C:sup Q f - i nf Q / > E }

1 .2.28: Let 1/J

m

L vol (Qt) < t. l =l

C(J).

What we must show is that, for each there is a 8 > 0 such that E

S (1/J; C ' ) - S(1j;; C) < whenever

II C' II

and

t > 0,

t

< 8. To this end, suppose that

where J - == ao < < an == minl < m < n ( am - am- 1 ) so that ·

w..p ( 8 )

C

sup

·

·

J+ .

Next, given

{ I 'I/J (t) - 7/J (s) l : s, t

E

t > 0,

choose 0 < 8 <

J and I t - s l < 8 }

<

2: .

168

Solution to Selected Problems

If II C 'II < 8 and A is the set of those I' E C ' for which there is an I E C with I' C I, then for each I' E B = C \ A there is precisely one m E {1 , . . . , n - 1} for which am E I' . In particular, because C ' is non-overlapping, B has at most n elements. Moreover, if I' E B, am E I' , and we use L(I') and R(I') to denote I' n [am - 1 , am J and I' n [am , am + 1 J , respectively, then 0

0

C V C' == A U U { L(I' ) , R ( I' ) }. l'EB Hence,

S ('l/J ; C' ) - S( 'lj;; C ) < S ( 'l/J ; C' v C ) - S('lj;; C ) == L ( I � L (J') 'l/J I + � � R(J') 'l/J I ) < 2nwV; ( 8 ) < E. l'EB Finally, by ( 1.2. 14 ) , ( 1 .2. 16 ) , and ( 1.2.17 ) , the analogous result for var_ and var + follow immediately. Now, suppose that 'ljJ E C 1 ( J ) . For n E z + and 1 < m < n , set am ,n == J- + 7: �J, and use the Mean Value Theorem to choose �m ,n E ( am ,n , am +1,n ) so the cp( am +1,n ) - cp( am ,n ) == cp'(�m ,n ) �J . Then

n- 1 + == (R) ' (�m ) ,n '""" 1/J Var + ( 'lj; ) == nlim �� n m� =O �J

and similarly for Var _ ( 'ljJ) and Var ( 'ljJ) . 1 . 2.29: Let 'lj; 1 and

1

J

cp' ( t ) + dt,

'lj;2 be given. Then, for any c < s < t < d ,

'lj; 1 ( t ) - 'lj; 1 ( s ) and 'lj;2 ( t ) - 'lj;2 ( s ) are non-negative. Hence, for any c < s < t < d and any non-overlapping exact cover C of [s, t] , + 'I/J2 ( t ) - 'I/J2( s ) = L AJ 'l/J2 > L ( AJ'l/J ) ; lEC JEC since both

and so 'l/J2 ( t ) - 'lj; 2 ( s ) > 1/J+ ( t ) - 1/J+ ( s ) . Since � 1 1/J1 - � 1 1/J - == � 11/J 2 - �I'l/J+ for any interval I C J , we now see that 'lj;2 - 1/J + and 'lj; 1 - 1/J - are non-decreasing.

0 and 'lj; ( t ) == t sin ( ; ) , t E ( 0, 1] . Clearly 'ljJ is continuous on [0, 1] . On the other hand, if tn == 2n� 1 for n E N and Cn consists of the intervals [0, tn J , [tn , tn - 1 J , . . . , and [t 1 , to J , it is easy to check that S ('lj;; Cn) diverges as n ----t oo at least as fast as the harmonic series does. Thus, 'ljJ has unbounded variation on [0, 1] . 1 . 2.30: Define 1/J ( O )

==

Solution to Selected Problems

169

To handle the second part of the exercise, consider the function if t E (0, 1 ] \ { � } if t == � . Clearly var ('lj; ; [0, 1 ] ) == 2. On the other hand, for any cp E C ( [0, 1 ]) , one has that R ( cp l'l/J; C, �) == 0 as long as � is not an endpoint of any I E C. Thus, (R) f[o, I) cp (x) d'lj;(x) == 0 for every cp E C ( [0, 1 ]) .

2 . 1 . 20: Part (i) is obvious, since Ck is the union of 2 k mutually disjoint intervals of length 3- k . Furthermore, the first assertion in part (ii) is clear,

and the second one follows by induction on f. To prove the characterization of c�_, note that a is the left hand endpoint of one of the intervals I making up Ct if and only if ak ( a ) E {0, 2} for 0 < k < f, ak ( a ) == 0 for k > f, in which case the associated right hand endpoint is a + 3- l . Thus, X E c�_ if and only if ak ( x ) == ak ( a ) , 0 < k < f, for such an a and there is a k > f such that ak ( x ) f= 0. Once one has these facts, the remainder of (ii) is easy. Moreover, as a consequence of (ii) , we know that the subset n � 1 Ct of C is isomorphic to A which, in turn, differs from {0, 2}N by a countable set. Hence, the uncountability of C follows from that of { 0, 2 }N. 0

0

2. 2.4: Thinking of the vk ' s as column vectors, let [ VI . . . v N J denote the N x N-matrix whose ith column is vi . It is then clear that P ( v 1 . . . VN ) is the image of Q0 under [v 1 . . . v N ] ; and, therefore, Lebesgue would as­ sign P (VI , . . . , v N ) volume equal to the absolute value of the determinant of [vi . . . vN ] ·

In order to connect this with the classical theory, it is helpful to observe first that

where v[ denotes the row vector representation of vi and ( vi , vi ) R N is the inner (i.e. , "dot" ) product of vi and Vj · Next, we work by induction on N. Clearly there is no problem when N == 1. Next, assume that the two coincide for some N E z + and let VI , . . . ' V N + I E �N + I be given. Since there is no prob­ lem in the linearly dependent case, we will assume that vI , . . . , v N + I are lin­ early independent. Choose an orthogonal matrix 0 so that Tovi , . . . , TovN E H (ei , . . . , eN ) . Then, the Lebesgue volume of P ( VI , . . . , v N ) thought of as a

Solution to Selected Problems

170

subset of H ( VI , . . . , v N ) is the same as that of P (Tovi , . . . , Tov N ) thought of as a subset of H (ei , . . . , e N ) and is therefore equal to the square root of

Hence, all that remains to check is that

where u == V N + I w and w is the perpendicular projection of V N + I onto H ( VI , . . . , v N ) . But, since the determinant is a linear function of any one of its columns and because w E H ( VI , . . . , v N ) , we see that -

det

[ (( ==

( vi , vi ) R N

det

)) 1
( vi , VI ) R N ( v 2 , VI ) R N

==

( det [vi . . . V N + I ] ) ( vi , V N ) RN (v 2 , V N ) R N

2

==

( det [vi . . . V N u] )

2

0 0

set of all r C E with the property that, for every E > 0 there exist a closed F c r and an open G � r such that JL( G \ F) < E . To prove the opposite inclusion, we first check that It is clear that A C A is a a-algebra. Obviously, 0 E A, and r E A ===} rC E A. Moreover, if {r n } � c A, r == u � rn , and E > 0 is given, choose { Fn }� c � and {Gn } � c <5 so that Fn c rn c Gn and 3. 1 . 9: Denote by

A the

BE JL ·

n

E z+ .

G == U� Gn, A == U� Fn , and An == U� =I Fm , E z + . Then, G E <5, {An }� c �' A c r c G, An / A, and JL(G \ A) < � · Hence, after choosing E z+ so that JL ( A \ An) < � ' we see that r E A and therefore that A is a a-algebra. We next check that C A, and, in view of the preceding paragraph, this comes down to checking that � C A. But if r E � and we set Fn = x E: p ( x, rC ) > ! }, then Fn / r and so J..t ( Fn) / J..L ( r). Hence, for given E > 0, n

Next, set

n

BE

{

Solution to Selected Problems

171

we can take find an n E z + such that JL ( r \ Fn ) = JL ( f ) - JL ( Fn ) < t ; and therefore we can take G = r and F == Fn . Finally, suppose that rl c r c r2 , where rl , r2 E BE and JL ( r2 \ rl ) = 0. To see that r E A, let t > 0 be given, and use the preceding to find F E � and G E <5 so that F C f1 , f2 C G,

JL ( r1 \ F ) < 2t , and JL ( G \ r2 ) < 2t .

Clearly,

F c r c G and JL (G \ F ) < t .

3. 1. 1 1 : To see that

a} )

JL ( 8Q) = 0, it suffices to check that JL ( {x E �N Xi = :

0 for every 1 < i < N and a E �; and, by translation invariance, this comes down to checking that JL ( { x E Q 0 : Xi = a} ) = 0 for 1 < i < N and a E [0, 1 ] . But, if p = JL ( { x E Q 0 : Xi = a } ) , then, by translation invariance, =

JL ( { X E Q o : Xi = ,8 } ) = JL ( { ( ,8

for every ,B E

1

- Q

) ei + X : X E

[0, 1] . Thus, n

=

JL(Q o ) > L JL ( m =O

{

x

E

Qo :

Xi =

Qo and Xi = } ) = p Q

: }) > (

for every n E z + ; and therefore p = 0. Next, to see that JL ( [O, m,\] N ) = m N JL ( [O, ,X] N ) for note that

m

E

n

+ l)p

z+

and

,\ E (0, oo ) ,

and so, by the preceding and translation invariance,

From the preceding, we now know that JL ( [O, q] N ) = q N for any q E Q n ( O , oo ) ( Q is used to denote the rational numbers) . Thus, if r is any element of ( 0, oo ) , then by choosing {qn }1 C Q so that Qn � r, we see that Combining this with translation invariance, we conclude that JL ( Q) AJR N ( Q) for every cube Q in �N ; and, since every open set G in �N is the countable union of non-overlapping cubes, it follows immediately, from what we already know, that JL(G) AJRN (G) for all open G ' s in � N . Thus, since, for every R E (0, oo ) , the set of all open subsets of ( - R, R) N is 1r-system which generates BRN [( - R, R) N ] , we now know that JL coincides with AnN on BR N [( - R, R) N ] for every R E ( O , oo ) and therefore on the whole of BRN · ==

==

172 An C rn

Solution to Selected Problems

3. 1 . 12: (i) For m

for every

zz+ +, . Am nn> m rn ·

E

set Hence

E

n

r · U n n n> m z+ .

Then

==

m

(ii) Set B == Then B � lim n E Hence, if J;(B 1 ) < oo , then

Am / n---+ rn lim

n---+ rn rn c Bn and

CX)

CX)

and

for every

E � JL (rn) ( J-L (U� rn ) J-L (n---+ rn) n---+ J-L (Bn) m---+ n> m J-L (rn) 0. 2. Srnn r{1,n+ l ,. r 2 , n -J-L (rl rn+l ) ) ( F r - L - l ) card (F) JL (r F n ( r n r n +d ) ca d l F JL r ) ( ) L 0-#FCSn - 1 0-#FCSn - 1

(iii) If

<

oo ,

limCX)

then ==

<

00

and so cf. part (ii) ) :

< limCX)

limCX)

"' �

==

3. 1 . 14: Part (i) is clear. To prove (ii) , we work by induction on

set by

. . , n } , assume the result for some n conclude that u . . . u

==

u

(

n

> Thus, > and, after replacing is equal u

(

n{ FCS,n+ln}CF+ I

rd (F) JL (rF ) , ca 1 ) ( 0#FCSn+ 1 L

where we have used

and part (i) in order to get the second line. (iii) In this case,

(mn ) m. (nn�m) .

card ( r ) ; n .1 == ( n - m ) ! Since there are . . . , n } , card . . . , n } with exactly m elements, we now see that

JL(r) F C { 1, F C { 1,

and, for each non-empty == ' ' sets

=

(r F )

.

Solution to Selected Problems

173

J.-L is a finite measure on ( E, B) , it suffices to check that it is countably additive. Thus, let { rn } c; C B be a sequence of mutually disjoint 3.3. 15: To see that

n

sets, and define

Fn == f L= 1 lrm for n E z + . Then Fn / f l r where r == U� rn ; and so, by the Monotone Convergence Theorem, n r f dv = L JL ( fn ) · JF dll = lim n JL ( r) = rr fdv = nlim L n ---. cx: :> ---. 1 1r 1 To check the absolute continuity assertion, suppose that there were an > 0 such that for every n .z + there exists a r n B for which v (r n) < 2 n and m

ex:>

CX)

E

m= l

rn

E

E

> E . Set r == limn ---. CX) rn · Since J.-L ( E ) < oo , one would know ( cf. (ii) of Exercise 3. 1. 12) that J.-L ( r ) > E . On the other hand, by the Borel-Cantelli Lemma (cf. (iii) of Exercise 3. 1. 12) , one would also have that v ( r ) and therefore fr f both vanish. Since J.-L (r ) == fr f dv, this is clearly a contradiction.

J.-L(rn)

dv

-

> ,\ ) � 0 as ,\ therefore, by the Lebesgue Dominated Convergence Theorem,

J.-L (I f l

3.3. 16: Clearly, by Markov ' s Inequality,

f 1 J 1 dJL 1{ 1 / 1>>.. }

------+

o

as

.x

------+

� oo ;

and

oo .

But, by a second application of Markov ' s inequality, this proves (3.3. 1 7) . To produce the required example, simply consider the function f (x ) == x ( l -1log x ) , x E [0, 1 ] ( oo when x == 0) , -

1 and note that l { x E [0, 1 ] : f ( x ) > ,\} 1 == x(,\) , where ,\x(,\) == ( 1 - log x(,\) ) - . Finally, if f is non-negative and measurable and one defines r n == {f E (n - 1, n] } for n E .z+ , then

CX)

when 2::: � J.-L ( f >

r

1{ / < N}

n)

<

oo ;

and, when

N

N

J.-L ( E ) < oo ,

N

f djL = L r f djL > L (n - l ) JL ( fn ) > L nJL( fn ) - JL ( E ) r 1

N

>

1n

1

1

L J.-L ( f > n) - 2J.-L ( E )

1 for every N E z + , from which the required result follows after one lets N /

oo .

Solution to Selected Problems

174

Without loss in generality, we assume that all the 9n ' s are �-valued and that, in part (i) , fn < 9n, and, in part (ii) , I fn i < 9n , everywhere. To prove part (i) , first choose a subsequence { fn ' } of {fn} so that

3.3. 19:

J

J

---+ CX) fn dJl , nlim ' -----+- CX) fn ' dJl == nlim and second choose a subsequence {gn" } of {gn' } so that Yn " � g (a.e., JL). Then, by Fatou ' s Lemma applied to hn 9n - fn > 0, g dJl - nlim fn dJl == lim hn " dJl > lim hn " dJl -----+> CX) n -----+- CX) n -----+- CX) == [g - n,�im-----+- CX) fn" ] dJl > g dJl - nlim ---+ CX) fn dJL. Given the preceding, the proof of part (ii) in the case of almost everywhere convergence is exactly the same as the derivation of Lebesgue ' s Dominated Convergence Theorem from Fatou ' s Lemma. Namely, one sets hn == l fn - J l and observes that hn < gn + g. The case of convergence in measure is then handled in precisely the same way as it was in the proof of Theorem 3.3. 1 1 . 3.3.20: (i) If /C is uniformly JL-absolutely continuous and M sup / E K ll f iiL l (J.L ) < oo , choose, for a given E > 0, 8 > 0 so that sup / E K fr I l l dJl < E whenever r E B satisfies JL(r) < 8 and choose R E (0, oo ) so that -% < 8. Then, by Markov ' s Inequality, sup / E K JL( I f l > R) < 8 and so sup / E K � / I > R I ll dJl < E. Next, suppose that /C is uniformly JL-integrable and define A(R) supK { J I dtt for R E (0, oo ) . I / E 1{ 1/ I > R } Clearly supK f I! I dtt < Rtt (r) + A(R) / E 1r for any r E B and R E (0, oo ) . Hence, if, for given E > 0, we choose R E (0, oo ) so that A(R) < � ' then fr I! I dJl < E for all f E /( and r E B with JL(r) < 2� ; and so /C is uniformly JL-absolutely continuous. In addition, when JL(E) < oo , by choosing R E (0, oo ) so that A(R) < 1, we see that II ! II L l (J.L ) < RJL(E) + 1 < oo for all f E /C. (ii) Note that, for any R E (0, oo ) , { J I dtt < R - o 1!1 1 + 6 dJL < R - 8 1!1 1 + 6 dJL. I 1{ 1/ I > R } 1/ I > R (iii) Suppose that fn � f in L 1 (JL). Clearly f E L 1 (JL) and supn EZ+ ll !n i i L l (J.L ) < In addition, for given E > 0, choose m E z + so that ll !n - ! I I L l (J.L ) < � for > m ; and, using Exercise 3.3. 15, choose 8 > 0 so that n < m 1fr I fn i dtt < � 1 <max

J

J

J

II

J

l

00 . n

J

II

J

J

J

1 75

Solution to Selected Problems

for all r E B with J.t ( r ) < b , and note that fr I l l d�-t < � whenever J.t ( r ) < Hence, for this choice of 8 > 0,

8.

for all r E B with J.t ( r ) < b. Conversely, if �-t(E) < oo , fn f in J.t-measure, and {fn }� is uniformly I-t-integrable, note that f is J.t-integrable, and ( cf. (i) above and Exercise 3.3. 15) choose, for a given t > 0, a b > 0 so that ------+

for all n E z + . Hence , if rn = J.t ( rn) < b for all n > m , then

{ l fn - f l > �-'tE) } and

m

is chosen so that

(iv) Tightness enables one to reduce each of these assertions to the finite

measure situation, in which case they have already been established. 3.3.22: Clearly the general case follows immediately from the case in which

�-t(E) V v(E) < oo; and therefore we will assume that J.t and v are both finite. Thus, by Exercise 3. 1 .8, all that we have to do is check that �-t(G) == v(G) for every open subset of E. Let G be an open set in E, and define

cp ( x ) =

(

dist ( x, GC ) 1 + dist ( x, GC )

)

:n 1

X

'

E

E and E z + ' n

where dist ( x, r ) inf { p ( x, r ) : y E r } is the p-distance from X to r. One then has that 'Pn is uniformly p-continuous for each n E z + and that 0 < 'Pn(x) / 1 a ( x ) as n oo for each x E E. Thus, by the Monotone Convergence Theorem, J.t ( G) == nlim 'Pn dj.t == nlim ---+ 'Pn dv == v( G). ------+

---+ CX)

J

CX)

J

4. 1 . 13: Without loss in generality, we will assume that a == 0 and that b == 1 . To prove the first assertion, define, for n > 2 ,

cpn ( t, X ) =

{�

( � X) '

if if

( t, x ) E ( t, x ) E

( kn 1 , � ) X E ( 1 - � , 1 ) X E.

Solution to Selected Problems

1 76

Clearly each VJn is measurable on ( (0, 1) x E, B( o , l ) x B) , and f is the point-wise limit of { VJn } �. Thus, f is itself measurable. Thrning to the second assertion, note that, for each t E ( 0, 1 ) , f' ( t , ) is the point-wise limit of functions which are measurable on (E, B). Thus, since f' ( , x) is continuous for each x E E, the measurability of f' follows from the first part. To prove the last assertion, let t E (0, 1) be given, and suppose that { tn } � C (0, 1 ) \ { t } is a sequence which tends to t. Set In == [t, tn] or [tn, t] according to whether tn > t or tn < t, and define ·

·

fn ( tn, x) - f(t, x) E z + and E E. ' tn - t f'(t, x) and l cpn( x) l I In l - l ( R) r f ' (s, x) ds < g (x)

VJn (x) Then

VJn(x)

�

=

n

=

x

}I n

x E E. Hence, by Lebesgue ' s Dominated Convergence Theorem, JE f ( tn, X) J.L ( dx) - JE f ( t, x ) J.L ( dx) r cpn(x) ( dx) r f' (t, x) J.L (dx); J.L tn - t JE JE from which we see that t E (0, 1) JE f(t, x) J,t( dx) is not only differen­ tiable at t but also that its derivative is equal to the J,t-integral of f ' ( t, ) . for every

=

�

�

·

Finally, given the preceding, the continuity of the derivative is simply another application of Lebesgue ' s Dominated Convergence Theorem.

5.2.4: (i) First, let r be a non-empty open subset of s N - l ' and set G == { X E B ( O, 1) \ {0} : �(x) E r} . Then, because � maps B ( O, 1) \ {0} continuously

s N G is a non-empty open subset of �N ; and therefore AsN ( r) NARN ( G) > 0. Next, let 0 be an orthogonal matrix and denote by R the corresponding rotation To on � N . Then � o R Ro � on �N \ {0} , l n ( O , l) o R == l n ( o,l ) , and R* ARN ARN ; and therefore the asserted invariance of As N follows directly from its definition. To prove the asserted orthogonality relations, define for each e E � N \ {0} the rotation Re : �N �N by 2( , x)RN Re x - X - e 2 e ' X E m N . lel onto

-I ,

==

-1

==

-1

�

�

Then, by rotation invariance,

==

Solution to Selected Problems

177

Similarly, for '11 ..l e ,

rsN- ( e , w)RN ( 7J , W )JRN AsN- l ( dw) = r (e , R� w) R N (7J , R� w ) RN AsN-l ( dw) 1 1 1sN- 1 = - rsN- 1 ( e , w)RN ( 7J , W )JRN AsN-l (dw) ; 1 and therefore, for any '11 E � N ,

Finally, if { e 1 , . . . , e N } is the standard orthonormal basis for � N , then, again by rotation invariance,

and so

to denote the rotation determined by the matrix Oo described in the hint, we see, by rotation invariance and Tonelli's Theorem, that (ii) Using

Ro

(

)

r f dv = 21 {g r f Ro(w) dO v( dw) 1S 1 1 1 1[0,2 1r] 7r

0

for any non-negative, Borel measurable f on 8 1 . Next, for fixed w E 81 , choose 'TJw E [0, 2 7r ) so that and observe that

{ f Ro(w) dO = { f (cos ( 'T7w + 0) , sin ( 77w + 0)) dO 1[0, 27r] 1[0 ,27r] f (cos 0, sin 0) dO + r f ( cos ( 21r + 0) , sin ( 21r + 0)) dO = 1[0, 7Jw ] [71w , 2 7r] = f f (cos O, sin O ) dO = f 1 f dJ.L . 1[0, 2 1r] 1S o

1

Solution to Selected Problems

178

Combined with the preceding, this now show that

f

be a non-negative, continuous function on �N with compact support. Then, by Tonelli ' s Theorem and Theorem 5.2.2, (iii) Let

N 2 p ) f (S (p,w) ) dp Ag N -1 (dw) dr J 1 - p2 ) � F( p) dp ( J [- 1 , 1) x SN - 1

( O,CX) )

(1 -

2 -1

-1

[- 1 , 1 )

where

F(p)

J

dy Hence,

2 p Jx S N - 1 ) � f(S(p, w) ) dp A8N - 1 (8w) dr (1 -

[- 1 , 1)

-1

(R) [-8( y ) , 8 ( y)]

where, for each

(0 , 1)

y E �N \ { 0 } ,

is chosen so that

1/;y (p) == (

PIYI 1 - p2 ) 2

,

p

E

[ -8( y ), 8( y )] , and 8( y ) E

p

== 0 for 8( y ) < < 1 .

Solution to Selected Problems

(

Since, by part i) of Exercise 5 . 1 . 6,

(R)

J f(y , 'l/Jy (p)) d'l/Jy (P)

=

[- 8 ( y ) , 8(y)]

=

y { J(o,oo)

for each

E � N \ {0} , we now see that

TN

(R)

179

J

['l/ly { -8( y )) , 'l/ly (8(y))]

f(y ,a)da

, ( J y JJR.l e7) de7 r

x s N -1 r 1 j ( z)dz r r j ( rw)>. sN (dw) ) dr ; ( J(o,oo) JsN f }JRN +

[- 1 , 1 )

TN

=

=

from which the desired result follows easily by taking ·

1J( I z l ) g ( 1 � 1 )

dr

to be of the form

5 .2.6: (iv) Define 1/J as in the hint and observe that 1

( )2

1/J s and therefore that

dd

==

1

-'lj;(s) 2 s

s+

( s 22 a a ) 2 + 4 ,B

1

sE�

------

==

1

2a

­

Thus, by Exercise 5. 1 .6, for any R E (0, oo ) :

e2;13 j

[ '1/J( - R ) � , 'ljJ ( R ) � ] 1

and so

1

1

==

(R)

=

_!_ (R)

C�

[ - R,R]

exp

2 [ -oh - � ] dt 2 2 /3 [- a '¢( s) - 1 ] d1j;(s) exp

2 e1 8 2a

�

'lj;( s) 2

�

2(s a ) 2 ds 1 2 2 ea /3 1 /3 t ] dt ! [-oh t a J [ - R, R]

exp

[ 'l/J ( - R ) � , '1/J ( R ) � ) 1

1+

s

+ 4 ,B

1

=

1

==

e - 8 2 ds ; e - 8 2 ds ,

[- R,R]

[ - R ' R]

and, after differentiating with respect to ,B and applying Exercise 4. 1. 13, one also has that

2 2 ,3 ] o: f3 e2 J t 2 [- a t - -t dt ,8 1 e - 8 2 ds . 1 1

['l/J( - R ) � , '1/J ( R ) � )

exp

==

3

Clearly the desired results follow when one lets R / oo .

[- R, R]

Solution to Selected Problems

180

5.3. 2 1 : Choose r > 0 and F : BJRN (p, r) � lR so that (5.3. 7) holds. Without

loss in generality, we will assume that r is taken so that F has bounded second order derivatives on BJRN (p, r) . In particular, since F vanishes on BJRN (p, r ) n M, there is a C < oo with the property that I F( y ) l < Cdist (y, M) for all y E BJRN (p, � ) . Hence, 1.

1m

e� o

dist (p + �v) "2

�

===}

Conversely, if v

E

1. F (p + �v) - F (p) < oo "2

< oo ===} 1m

e� o

( v, \7F (p)) JR N == 0

Tp (M) and 1 :

dist (p + �v, M) <

(

�

===}

-E, E ) �

v E Tp (M).

M is chosen accordingly, then

IP + �v - ! ( � ) I == 0 �2 )

(

as � � 0.

5.3.24: (i) By Tonelli's Theorem,

r (a) r (,B) = JJ Sa- l tf3- l e- s - t ds dt. (O,CX)) 2 X

Now let � be the mapping suggested in the hint, and observe that � maps (O, oo) x (0, diffeomorphically onto (O, oo) 2 ) and that 8� , == Hence,

(u v ) u . 1 1 ) ( f3+t l l l v 0 s v u dudv f3u ,B s u sae dt = J J d )) ( ) t = J J ( ( r (a) r ( ) (O, ) (O,CX)) (0, 1 ) CX) 2 = �f(O, CX)) ua+f3- t e -u du �f(0, 1 ) v0- 1 ( 1 - v ) f3- 1 dv = r(a + ,B)B(a,,B). 1)

X

(ii) By Theorem 5.2.2,

Moreover, if � is the map in the hint, then, by Exercise 4. 1 .6, the integral on the right is equal to

Hence, since ( cf. (iii) in Exercise 4. 2 . 16 ) follows.

N

WN- l = ;(;)

,

the desired result

181

Solution to Selected Problems A second approach to the same computation is to first observe that

= J(O,oo) tA - 1 [-t ( l + l x l 2 )] dt l A t ( { e - t l x l 2 dx ) dt et r(.X) { ( 1 + l x l 2 ) - A dx = { }JRN }JRN J(o , oo) = J(o{ ,oo) tA - l ( � ) � e - t dt = tr � f (,\ - N ) · r( ,x ) ( 1 � I X ! 2 ) >,

exp

and therefore that

2

6.2.22: (i) Let 1 < p <

q

< oo and set inequality applies and yields

a ==

� - Then, since J,t(E)

1, Jensen ' s

==

(ii) First note that, by Holder 's inequality or direct computation,

for any measure Jl and any 1 < p < q < oo . Second, note that, for the particular measure under consideration, II J II L oo ( JL ) < II J I I LP( JL ) for any p E [1, oo ] . (iii) Since there is nothing to do when JL E) == 0, we will assume that JL E) E (0, oo ] . To prove the first assertion, suppose that 0 < M < II J II L oo ( JL ) and set r == { I l l > M} . Then JL(r) > o, and so

(

(

p ---+ 00

p ---+ 00

In other words, limp---+ oo II J IILv (JL) > M for every M E [0 , II J II L oo (JL) ) , and so limp ---+ oo II J II LP(JL ) > II J II L oo (JL ) · To prove the second assertion, assume that JL E) < oo or ll f ii£ 1 (JL ) < oo , and note that

(

II f II LP( J.I ) < (II f II £'X' ( J.I ) M ( E) * ) (\ (I I f II �=t ) II f I I t ( J.I ) ) . (iv) Suppose that ( E 1 , B 1 ) is a measurable space and that ( E2 , B2 , Jl 2 ) is a a-finite measure space. Choose { E2 , n } � C B2 so that E2 , n / E2 and Jl 2 ( E2 , n ) < oo , E z + . Given a measurable f on ( E1 E2 , B 1 B2 ) , observe

that

n

x

x

l i f (x l , · 2 ) 11 L= (J.I) = nl�n�J f(x l , · 2 ) 1E2 , n ( 2 ) 11 L= (J.I) == n ---+ oo oo I I f (X 1 ' . 2 ) E2 ( 2 ) I I LP ( JL ) •

lim lim ---+ p

1

'

n

•

.

Solution to Selected Problems

182

Thus, since, by Lemma 4. 1 . 2,

is

B1 -

measurable for each

n

E

z + and p

E [1,

) so is

oo ,

{An }o B

C E > 0 and a sequence n > E but v n < 2 - . Hence, if B limn-H:x) An , then (cf. such that part (ii) of Exercise 3. 1 . 12) B > E while, by the Borel-Cantelli Lemma (cf. part (iv) of Exercise 3. 1 . 1 2 ) , v(B) 0. 7. 2. 1 5 : Suppose not. Then there exists an

== (JL(A ) ) == xEE (E, B, JL ) {x} E AB JL JL(E) An == {xA E E : JL ( {x} ) > � } An nJL(E) . A JL(AC) == 0 JL JL(r) == JL(r n A) == L JL(r n {x} ) == L JL( {x }). JL(An )

7.2. 16: (i) Assume that

for every and that is a a-finite. In proving that the set of atoms is countable, it is easy to reduce to the case when is finite. Thus, we will assume that < oo . Next, , and observe that card(An) < set Hence, since is the union over n E z + of the s, it follows that must be countable. Moreover, when is purely atomic, and therefore '

xEA

Conversely, if there is a countable S such that

JL(r) == L JL ({x} ) , r E BR , A == { x E S JL ( { x}) > 0 } r n A == 0 JL(r) == ( JL1/J ({X} ) == JL1/J ( (X - E, X] ) == ( (X) - 'l/J (X - E) ) == 'l/J (X) - 'l/J (X - ) . == { 'l/J(x) - 'lj;(x-) > 0, JL }) ( x 'lj; (x) - 'l/J(x- ) JL'l/J JL1/J A A == 'l/J( x ) -'l/J( ) == ('l/J(t) - 'lj; (t-)) == 'l/Jd (x) . JL ( {t}) == x E r nS

then

:

===}

o.

is the set of atoms, and it is clear that

ii) Clearly,

lim

€� 0

lim 'ljJ

€� 0

Hence, x is an atom if and only if in which case . In particular, is non-atomic if and only if 'ljJ is continuous. On the other hand, if is purely atomic and is its set of atoms, then D( 'ljJ) and - oo

tE D ( 'lj; ) n ( - CX) , x )

t E D ( 'lj; ) n ( - CX) , x )

183

Solution to Selected Problems Conversely, if V; is purely discontinuous, then

JL� ( ( where

- oo ,

x] ) ==

tED( � ) n(

(V;(t) - V;(t - ) ) == v ( ( - oo , x] ) , x E JR, - oo , x]

v is the purely atomic measure given by v(r) ==

L (V; (x) - V;(x - ) ) ,

r E BR ·

xED( � )nr

Hence, since C == { ( - oo , x] : x E 1R} is a 1r-system which generates BR and JL� r c u {JR} == v r c u {JR} , it follows that JL� == v on BR . (iii) First suppose that

/L,p

JL� << AR · Clearly JL� is then non-atomic and therefore

( ( a, b) ) = 'ljJ(b) - '1/J( a ) for all a < b. Next, set f = 'Z,.; . Given E > 0, choose R > 0 so that J{ />R} f dx < � and take 8 == 2� . If { ( am , bm ) } is a sequence of mutually disjoint intervals with L:( bm - am ) < 8 , then

( L

r 1 dx L Jr(am , bm ) n{ f R} < R ( bm - am ) + r f dx < <

)

€,

j{J > R}

where, in the passage to the last line, we have used the fact that the intervals are mutually disjoint. Conversely, suppose that V; is absolutely continuous. Given r E BR with AR (r) == 0, choose, for a given E > 0, 8 > 0 so that

Next, choose an open set G =:) r so that ,\R (G) < 8 , and find (cf. Lemma 2. 1. 10) a sequence { ( am, ,Bm )} of mutually disjoint open intervals so that G == U( am, bm ) · Then L:( bm - am ) == AR ( G ) < 8 and so

JL� << AIR . (iv) Set JL�1 and v2 JL�2+�3 • Clearly, JL� + v2 , << AR , v2 ..l AJR , and therefore JL�,a while v2 == JL�,a· In particular, this means that V; I V;a and that V;2 + V;3 V;a . Finally, because V;2 is continuous and V;3 is purely discontinuous, this second equality means that V;a d '¢3,d V; 3 , from which 'l/J2 'ljJa, is trivial. Hence

VI

==

VI

==

==

==

VI

VI

==

==

,

==

c

==

==

Solution to Selected Problems 2 - k note first that 7. 2 . 1 7: To prove that = 6 , and that r�\ 184

l 'l/Jk + 1 - 'l/Jk l u,R

'l/Jk ck

-

'lfJk+ 1 JR. \ ck r

=

('l/Jk x) (�) k x x 3- k ] 1 k + 3-k k1- 1 ] k 1 (�)- k - 1 x (3- - , 23- - ) 'l/Jk+ 1 (x) 2

Second, check that

for

==

E [0,

and that

if X E [0, if X E

=

Finally, combine these two to complete the computation. � 'ljJ uniformly on � and, there­ Given the preceding, it is obvious that fore, that 'ljJ is a continuous, non-decreasing function with 'l/;( - oo ) = 'l/;(0) = 0 and 'ljJ 1) = 'ljJ ( oo ) = 1 . Moreover, 'ljJ is constant on each of the intervals [a ,j b ,j , k E N and 0 < j < = 0 for every k , and so Hence, JL ,p (� \ JL,p (� \ = 0 also.

( k , k C)]

'l/Jk

2k .

Ck )