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Chapter 13 Riemann Integrability
CIIAL'TER
13
RTEbIANN INTEGRABILITY
C o n s i d e r t h e c l a s s i c a l Lebesgue t h e o r e m : a bounded f u n c t i o n o n a r e a l i n t e r v a l i s Ricmann i n t e g r a b l c i f a n d o n l y i f i t s s e t o f p o i n t s o f d i s c o n t i n u i t y h a s (Lebesgue) measure z e r o . While q u i t e a t t r a c t i v e , t h e t h e o r e m h a s an u n s a t i s f a c t o r y f e a t u r e : i t r e q u i r e s f i r s t d e f i n i n g "Lebesgue measure z e r o " ,
a
c o n c c p t o u t s i d e t h e domain o f Riemann i n t e g r a t i o n . T h i s u n s a t i s f y i n g f e a t u r e can be e l i m i n a t e d . s u b j e c t o f m e a s u r e on o u r r e a l i n t e r v a l . Lebcsgue m e a s ure.
I-lerc m e a s u r e means
C o r r e s p o n d i n g t o "measure"
i n t h e same way
t h a t "Riemann i n t e g r a l " c o r r e s p o n d s t o " L e b e s g u e have t h e concept of " c o n t c n t " .
Consider t h e
integral",
we
The t h e o r e m f o r s e t s c o r r e s p o n d -
i n g t o the Lebesgue theorem above r e a d s : a s e t h a s c o n t e n t ( i s q u a d r a b l e ) i f and o n l y i f i t s f r o n t i e r h a s measure z e r o .
But
t h e f r o n t i e r o f a s e t i s c l o s e d , and f o r a c l o s e d s e t , h a v i n g measure z e r o i s e q u i v a l e n t t o having c o n t e n t z e r o .
Thus t h e
theorem can he s t a t e d : a s e t h a s c o n t e n t i f and o n l y i f i t s f r o n t i e r has content zero. The way t o e l i m i n a t e m e a s u r e f r o m L c b e s g u e ' s t h e o r e m i s
now c l e a r .
We r e p l a c e t h e s e t o f p o i n t s o f d i s c o n t i n u i t y o f a
f u n c t i o n f by t h e s a l t u s f u n c t i o n o f f , which g i v e s t h e p o i n t s
of discontinuity.
The s a l t u s f u n c t i o n i s a l w a y s p o s i t i v e a n d
uppersemicontinuous,
and f o r s u c h a f u n c t i o n , h a v i n g Lebesgue 31 5
Chapter 1 3
316
i n t e g r a l z e r o i s e q u i v a l e n t t o h a v i n g Riemann i n t e g r a l z e r o . Thus t h e L e b e s g u e t h e o r e m b e c o m e s : a bounded f u n c t i o n i s Riemann i n t e g r a b l e i f a n d o n l y i f i t s s a l t u s f u n c t i o n h a s Riemann i n t e g r a l zero. T h i s w i l l b e o u r a p p r o a c h t o Riemann i n t e g r a b i l i t y , u s i n g t h e f a c t t h a t f o r an element f o f C " ( X ) ,
6(f) is the gencraliza-
tion of the saltus function of a function.
560. Riemann i n t e g r a b l e e l e m e n t s
We f i r s t d e f i n e Riemann i n t e g r a b i l i t y w i t h r e s p e c t t o a band J o f C l ( X ) .
I n 5 6 1 , we e x a m i n e i t w i t h r e s p e c t t o a
s i n g l e e l e m e n t 1-1 o f C ' ( X ) . Given a b a n d J o f C l ( X ) , an
e l e m e n t f o f C"(X) w i l l b e
s a i d t o b e Riemann i n t e g r a b l e w i t h r e s p e c t t o J i f 6 ( f ) E t J L . We w i l l d e n o t e t h e s e t o f s u c h e l e m e n t s b y
B(J).
Thus X ( J )
=
6 - l (.JL). From ( S O - l l ) , we h a v e :
(60.1)
F o r e v e r y band J o f C ' (X),
3 (J)
(i)
i s an Ma-subspace o f C"(X),
(ii)
c o n t a i n s C(X),
( i i i ) i s closed under t h e operations u ( - ) , a ( - ) ,
a(.).
I t i s e a s y t o s e e from t h e d e f i n i t i o n t h a t i f f C % ( J ) and
Riemann I n t e g r a b i l i t y
a(f) 5
g 5 u ( f ) , t h e n gE.2 (J).
31 7
A c t u a l l y , from ( 5 0 . 1 3 1 , we have
a much s t r o n g e r r e s u l t :
(60.2)
If A,B c X(J), A
In p a r t i c u l a r ,
a(J)
5 f 5 B , and (AA
- VB)EJL,
then f € x ( J ) .
i s Dedekind c l o s e d .
561.
u-Riemann i n t e g r a b i l i t y
Given p E C ' ( X ) , an e l e m e n t f o f C''(X) w i l l be s a i d t o be Riemann i n t e g r a b l e w i t h r e s p e c t t o IJ,o r 1-1-Riemann i n t e g r a b l e , if ( 6 ( f ) ,
14) =
0.
( 6 1 . 1 ) Given p E C ' ( X ) , t h e n f o r e v e r y f E C " ( X ) , t h e f o l l o w i n g a r e equivalent :
'1
f i s Riemann i n t e g r a b l e w i t h r e s p e c t t o 1-1;
2'
f i s Riemann i n t e g r a b l e w i t h r e s p e c t t o C ' ( X )
Fi'
T h i s f o l l o w s from t h e e a s i l y v e r i f i a b l e f a c t t h a t f o r gEC"(X)+, g E ( C J ( X ) I . I y i f and o n l y i f ( 8 , 11-11)
=
0.
Thus t h e s e t o f 1-1-Riemann i n t e g r a b l e e l e m e n t s o f C"(X) precisely by
a (1-1).
a(C'(X) ) .
u
is
We w i l l a l s o d e n o t e t h i s (MI-)s u b s p a c e
Some c h a r a c t e r i z a t i o n s o f
2 (J):
31 8
Chapter 13
( 6 1 . 2 ) Given a b a n d J o f C l ( X ) , t h e n f o r f E C " ( X ) , t h e following are equivalent: 1'
f i s Riemann i n t e g r a b l e w i t h r e s p e c t t o J ;
2'
f i s Riemann i n t e g r a b l e w i t h r e s p e c t t o e v e r y u E L J :
3'
t h e r e e x i s t a u s c e l e m e n t u a n d an k s c e l e m e n t a. s u c h
that R < f 5 u and u - L E J ' :
'4
( ~ ( f ) , u )= ( f , p > = ( u ( f ) , p ) f o r e v e r y ~ E J .
The p r o o f i s i m m e d i a t e .
We n e x t g i v e some c h a r a c t e r i z a t i o n s o f x ( p ) .
For s i m p l i c -
i t y , we t a k e p > 0.
(61.3)
Given u € C ' ( X ) + , t h e n f o r f E C " ( X ) , t h e f o l l o w i n g a r e
equivalent:
1'
f i s p-Riemann i n t e g r a b l e ;
h>f
g
t h e r e e x i s t s u b s e t s A , B o f C(X) s u c h t h a t A < f < B and ( f , ~ = )
'4
SUP
€5A
(g,p)
=
inf ( ~ , L O ; hEB .
t h e r e e x i s t s e q u e n c e s { g n l , { h n } i n C(X) s u c h t h a t
The p r o o f i s s t r a i g h t f o r w a r d .
Riemann I n t e g r a b i l i t y
319
I n t h e c l a s s i c a l t h e o r y o f Riemann i n t e g r a t i o n on a E u c l i d e a n s p a c e , Rieman i n t e g r a b i l i t y i s d e f i n e d b y o n e o f t h e properties 2
0
,
3 O , o r 4'
above, but using a d i s t i n g u i s h e d
family of "step functions" i n place of continuous functions. Among t h e p r o p e r t i e s p o s s e s s e d b y t h e s e s t e p f u n c t i o n s a r e t h e following.
Each s t e p f u n c t i o n i s a f i n i t e l i n e a r c o m b i n a t i o n
(I)
of f u n c t i o n s which can be o b t a i n e d as s p e c t r a l e l e m e n t s o f continuous functions ( c f . 517). ( 1 1 ) Each s t e p f u n c t i o n i s Riemann i n t e g r a b l e .
In ( 6 1 . 5 ) we generalize the classical procedure t o t h e p r e s e n t c a s e i n t h e f o r m o f a c o n d i t i o n f o r p-Riemann i n t e g r a b i l i t y i n terms of s p e c t r a l elements o f continuous f u n c t i o n s . F i r s t , h o w e v e r , we m u s t a n s w e r t h e f o l l o w i n g q u e s t i o n .
Are t h e
s p e c t r a l e l e m e n t s o f c o n t i n u o u s f u n c t i o n s Riemann i n t e g r a b l e ( b y o u r d e f i n i t i o n ) ; a n d i f n o t , a r e s u f f i c i e n t l y many Riemann integrable t o enable us t o carry out the classical process? The q u e s t i o n was a n s w e r e d b y Loomis
[32].
Indeed, he proved a
more g e n e r a l r e s u l t , w h i c h r e d u c e s , i n o u r c o n t e x t , t o t h e
f o 1l o w i n g :
(61.4) Theorem (Loomis).
Given p E C ' { X ) ,
then f o r every fEC"(X),
the following are equivalent:
'1
f i s p-Riemann i n t e g r a b l e ;
2'
a l l b u t a c o u n t a b l e number o f t h e s p e c t r a l e l e m e n t s o f
f a r e p-Riemann i n t e g r a b l e .
Chapter 1 3
320
Proof.
We c a n assume p > 0 , IIui1
=
1, and 0 < f < n(X).
F o r t h e r e m a i n d e r o f t h e p r o o f , we w i l l d e n o t e n(X)
s i m p l y b y I]
and e a c h s p e c t r a l e l e m e n t e f ( X ) s i m p l y b y e ( 1 ) . Assume 1' h o l d s . (6(e(x)),p) e s t a b l i s h 2'.
2
E
We show t h a t f o r e v e r y E > 0 ,
f o r o n l y a f i n i t e number o f A ' s , w h i c h w i l l S p e c i f i c a l l y , we p r o v e t h e
Then
We show t h e e q u i v a l e n t i n e q u a l i t y
Note f i r s t t h a t f A X n n - f A A n - l n we w o r k w i t h t h i s l a t t e r f o r m .
= (fAXnn -
An-ln)'
;
Riemann I n t e g r a b i l i t y
This i s t h e f i r s t i n e q u a l i t y i n ( i i ) .
T h i s i s t h e second i n e q u a l i t y i n ( i i ) , s o we have ( i i ) , and
with it, ( i ) .
I t follows f r o m ( i ) t h a t f o r n = l,.-.,m,
321
Chapter 1 3
32 2
(Here we have used t h e a s s u m p t i o n t h a t f - hence e a c h g n i s Riemann i n t e g r a b l e ; c f . 4'
i n (61.2).)
Then
Since l E = l ( 6 ( e ( ~ n ) ) , 2 ~ )m E , t h i s g i v e s u s m < 1 / ~ and , comp l e t e s t h e p r o o f o f t h e Lemma.
Now assume 2'
holds.
I t i s c l e a r t h a t t h e proof of
F r e u d e n t h a l ' s s p e c t r a l theorem (17.10) can be m o d i f i e d ( f o r o u r f ) s o t h a t t h e s p e c t r a l e l e m e n t s a r e u-Riemann i n t e g r a b l e . Thus f i s i n t h e norm c l o s u r e o f j ; ! ( u ) .
Since
x(u)
i s norm
c l o s e d , f E j;! ( p ) .
QED We can now g i v e o u r g e n e r a l i z a t i o n o f t h e c l a s s i c a l cond i t i o n f o r Riemann i n t e g r a b i l i t y .
( 6 1 . 5 ) Given p E C ' ( X ) + , t h e n f o r f € C t t ( X ) , t h e f o l l o w i n g a r e equivalent:
' 1
f i s u-Riemann i n t e g r a b l e ;
2'
t h e r e e x i s t s e q u e n c e s { g n } , {h,}
satisfying:
Riemann I n t e g r a b i l i t y
323
t h e g n ' s and h n ' s a r e l i n e a r combinations o f
(i)
u-
Riemann i n t e g r a b l e s p e c t r a l e l e m e n t s o f e l e m e n t s
of C ( X ) ; gl 5 g 2
(ii)
2 * *-* f<-< * . * I
( i i i ) supn(gn,v)
Proof.
(f,v)
S u p p o s e 1' h o l d s .
t o s a t i s f y 4'
(*)
=
h2 < h1 , =
infn(hn,p).
Choose s e q u e n c e s { g ; } , { h , l }
of (61.3).
For e a c h n , t h e r e e x i s t g n , h n , l i n e a r combinations o f p -
Riemann i n t e g r a b l e s p e c t r a l e l e m e n t s o f e l e m e n t s o f C ( X ) , s u c h t h a t gn 5 g;
,
hn
2
hA
and
[ I g n - gA[19 IIhn - hA[I
5
l/n.
I n e f f e c t , gn c a n b e o b t a i n e d b y t h e m o d i f i c a t i o n o f t h e proof of (17.10)
p o i n t e d o u t a t t h e end of t h e p r e c e d i n g p r o o f ,
a n d s i m i l a r l y f o r hn ( t h e a p p r o x i m a t i n g s t e p e l e m e n t s i n ( 1 7 . 1 0 ) can b e chosen above t h e g i v e n a ) . Applying (17.12) and ( 6 0 . 1 ) , w e can assume, i n a d d i t i o n , t h a t t h e s e q u e n c e {g,} descending.
They t h u s s a t i s f y ( i ) a n d ( i i ) o f 2'.
f o l l o w s f r o m 4' T h a t 2'
i s a s c e n d i n g and t h e sequence {hn} i s
i n ( 6 1 . 3 ) a n d t h e norm c o n t i n u i t y o f
i m p l i e s 1'
f o l l o w s f r o m 5'
(iii) p.
i n (61.3). QED
Remark. p-Riemann i n t e g r a b i l i t y i n o r d i n a r y a n a l y s i s i s d e f i n e d f o r f u n c t i o n s on X C''(X)a.
- i n our context, f o r elements of
I t i s easy t o v e r i f y t h a t the space of functions
w h i c h a r e p-Riemann i n t e g r a b l e i n t h e o r d i n a r y s e n s e i s
Chapter 13
324
2 (u),
precisely
562. Riemann n e g l i g i b l e e l e m e n t s
We r e t u r n t o a g e n e r a l band J o f C ' ( X ) . R(J)
n J'
We w i l l c a l l
t h e s e t o f Riemann n e g l i g i b l e ( w i t h r e s p e c t t o J )
e l e m e n t s , an d w e w i l l d e n o t e i t b y f X ( J 1 . w i l l also write
,TX
I f J = C'(X )
I.r'
we
(1-1).
( 6 2 . 1 ) Given a b a n d J o f C ' ( X ) , t h e n f o r f € C " ( X ) , t h e f o l l o w i n g are equivalent: 1'
fCh%(J);
2'
u ( f ) ,k(f)EJ1:
'3
u(/fl)EJ'.
S u p p o s e 1' h o l d s , t h a t i s , 6 ( f ) E J '
Proof. f - 6(f), f
w e h a v e 2'. 3'
holds.
+
6(f)EJL. T h a t 2'
h'?, (J)
Then
Since f - 6cf) < tcf) < u(f) < f + 6(f),
i m p l i e s '3
f o l l o w s from (49.12).
Th en, a g a i n by ( 4 9 . 1 2 1 ,
u ( f ) , t ( f ) E J L , hence G(f)EJ'. fE
and fEJ'.
(u(f)I,
T h a t fEJ'
Suppose
l t ( f ) IEJL, hence
is trivial.
Thus
. QED
(62.2)
C o r o l l a r y 1.
Given a b a n d J o f C ' ( X ) ,
t h e n f o r fEC"(X),
Riemann Integrability fE Jvx (J) if and only if
1 f( -<
325
u for some usc element u in J L .
I t follows , f $ ( J ) is the Riesz ideal of C " ( X )
generated by the
usc elements in (J'),.
Proof.
Suppose g € l X ( J ) and 0 5 If1 5 /gl. Then u(lf/) 2
~ ( / g / ) .Since,
by '3
hence, again b y '3 ideal.
above,u(jgj)EJL, it follows u(lfl)EJL.
above, fcJY*x(J).
Thus J v x ( J ) is a Riesz
The rest of the Corollary is clear.
Remark. Note that, since h?$?(J)= X ( J )
n
QED JL,
,qx(J) is norm
closed.
It follows from (62.1) that Riemann negligibility can be defined first, and then used to define Riemann integrability. Specifically, define g to be Riemann negligible if u(lgl)EJL, then (since 6(f) is a positive usc element) define f to be Riemann integrable if 6 ( f ) is Riemann negligible.
(As
we
pointed out in the introduction to the present chapter, this eliminates the mention of Lebesgue negligibility in Lebesgue's characterization of Riemann integrability. By definition X ( J )
=
6-'(JL).
Contained in the above dis-
cussion is the fact that actually X ( J )
=
&-lWX(J)).
This
gives us:
(62.3) Given a band J of C' (X), let H be the Riesz ideal
C h a p t e r 15
326
g e n e r a t e d Oy Sa(J)). Then (i)
S-'(H)
=
g(H
1
) , a n d H i s t h e s m a l l e s t Riesz i d e a l of
Clt(X) w i t h t h i s p r o p e r t y ; ( i i ) jZ(HI)
=
J ( J ) , and H
1
i s t h e l a r g e s t b a n d o f C1(X )
with t h i s property
The b a n d H 1 g e n e r a t e d by S ( ( R ( J ) )
(that is, the order clos-
u r e o f t h e above H) i s o f c o u r s e t h e smallest band o f C"(X) s u c h t h a t S-'(H1) t h a n J'
=
T h i s band may b e s t r i c t l y s m a l l e r
X(J).
- equivalently,
H
I
may b e s t r i c t l y l a r g e r t h a n J .
an example, l e t X b e a f i n i t e s e t .
Then Cll(X)
=
As
C(X), s o
S ( C t l ( X ) ) = 0 , s o f o r e v e r y band J o f Cll(X), H = 0 , s o H1
=
0.
We e m p h a s i z e ( i i ) a b o v e . : f o r e v e r y band J o f C l ( X ) , t h e r e
i s a l a r g e s t b a n d ( i t c o n t a i n s J ) w i t h t h e same Riemann i n t e g r a b l e e l e m e n t s i n C"(X).
We c l o s e t h i s 9 w i t h a n i n t e r e s t i n g r e s u l t .
(62.4)
Giv e n a band J o f C t ( X ) , e v e r y Riemann i n t e g r a b l e e l e -
ment o f C1l(X)d i s Riemann n e g l i g i b l e :
g(J) n
Proof.
C"(X)d
c&?,(J).
Consider f €$(J)
we show f E J'. a ( f ) = 6 ( f ) EJ'.
By ( 5 4 . 1 2 ) ,
n C 'l (X )d,
L(f)
=
and we c a n t a k e f > 0;
0 , hence u ( f )
Thus 0 < f < u ( f ) € J',
=
u(f) -
whence f EJ'. QED
327
Riemann I n t e g r a b i l i t y
563. Riemann i n t e g r a b i l i t y and Riemann s u b s p a c e s
Let J be a band o f C'(X) Then
J(J),
=
a n d I its d u a l band i n
Cll(X).
J ( J ) / J ~ % ( J ) A. r e a s o n a b l e e x p e c t a t i o n i s t h a t I n ( 6 3 . 2 ) , we l i s t
R ( J ) , = R(1) t h e Riemann s u b s p a c e o f I . some common b a n d s f o r which t h i s i s t r u e . p r i n c i p a l b a n d s {C'(X)
Among them a r e t h e
l ~ € C ' ( X ) l , which a r e t h e o n e s d e a l t w i t h
LJ i n ordinary integration theory.
We w i l l s e e below, however,
t h a t f o r a g e n e r a l band o f C'(X),
t h e e x p e c t a t i o n need n o t be
born o u t . First,we record:
( 6 3 . 1 ) For e v e r y band J o f C'(X),
w i t h d u a l band I ,
nJ),= R ( I ) -
Proof. f - R(f)EJ'.
Suppose f E J ( J ) .
Then 6 ( f ) E J ' ,
I t follows a ( f ) , = f I
=
so u ( f ) - f ,
u ( f ) I , whence f I E R ( I )
(57.6). QED
( 6 3 . 2 ) L e t J b e a band of Cl(X),
w i t h d u a l band I .
f i e s any o f t h e f o l l o w i n g c o n d i t i o n s , t h e n X ( J ) ,
If J satis-
= R(1):
( i ) J i s a p r i n c i p a l band; (ii) J i s vaguely c l o s e d - e q u i v a l e n t l y , I i s a u-band;
Chapter 1 3
328
( i i i ) Jd i s v a g u e l y c l o s e d - e q u i v a l e n t l y , I i s a n R-band; (iv)
J i s v a g u e l y d e n s e i n C'(X) - e q u i v a l e n t l y ,
U(1)
=
Cll(X).
I n e a c h c a s e , i t o n l y r e m a i n s t o show t h a t
Proof. R(1) c a ( J ) , .
Assume ( i ) h o l d s , t h a t i s , J
=
C'(X)
f o r some u E C ' ( X ) .
I-r
By ( 2 7 . 4 ) , t h e r e e x i s t s e q u e n c e s { ( g ) I , n P I ( h n ) l - r l i n C(X) s u c h t h a t ( g ) + f , ( h ) + f . LJ n1-1 n1-1
Consider f E R ( 1 ) .
To show t h i s , we r e p l a c e Cgn3,thn3 by s e q u e n c e s
C1711> k i t h t h e d e s i r e d p r o p e r t y . -
g1
=
-
gl,
then s e t hl
= hlvgl.
h e prmceed i n d u c t i l e l y .
We h a v e (gl)l-l
= (gl)p,
and (E1I-r 1 = (hlll-lv(Fl)l-l = ( h l ) l - l v ( g l ) u = ( h l l V .
-
-
g l , - ..,gm, El,..-,En
Ti
< * - *
n Set
{En},
have been chosen s o t h a t
of course,
Assume
El z - -+ -gn* <<
5Kl a n d (Ek)l-l = =
Set
( g k ) l - l , (Ek)l-l = ( h ) ( k = l , . . . , n ) k1-1 (gn+lVzn)AEn. C o n t i n u i n g i n t h i s f a s h i o n , we
.
f -<
obtain (*), N o w s e t R = Vn g n' u = An hn '
T h e n R-< u a n d R = f = u . 1-1 1-1
To e s t a b l i s h t h a t f E J ? j ( J ) I , we n e e d o n l y show t h a t uE%'(J). T h i s f o l l o w s from u - I! EJ'
( s i n c e (u
-
a) I
=
0 ) a n d 3'
in
(61.2). Assume ( i i ) h o l d s .
Then by ( 5 9 . 5 ) a n d ( 6 0
C(X), c 2 ( J ) , . Assume ( i i i ) h o l d s , a n d c o n s i d e r f E R ( 1 ) .
We c a n assume
Riemann I n t e g r a b i l i t y
We show t h a t i n t h i s c a s e , n o t o n l y i s f i n
0 < f < ll(X)I.
but is actually i n
je(J),,
329
J ( J ) . By ( 5 7 . 6 ) , t h e r e e x i s t a n Rsc 2 f 5 uI.
element k and a u s c element u such t h a t k I (uV0)A
l l ( X ) ; t h e n (u,) I = f a l s o .
S e t El
= ( k V 0 ) A U ( X ) I (by
a s s u m p t i o n , l l ( X ) l i s a n Rsc e l e m e n t ) ; t h e n ( R 1 ) I
R I E I , s o we a c t u a l l y have R1 We t h u s have R1 - R1 € I d =
=
I t follows
hence f € X ( J ) .
, ' J
Assume ( i v ) h o l d s .
Then i t f o l l o w s e a s i l y from t h e f a c t
t h a t p r o j , mgps C ( X ) MU-isomorph c a l l y o n t o C(X), again, R(I) c
But
f also.
= f.
f < u l , w i t h ( u , ) ~= f . I 1
=
S e t u1 =
XCJ)
(not j u s t
B(J
(52.3) that,
,
I)
QED
The r e m a i n d e r o f t h i s s e c t i - n i s d e v o t e d t o a n example t o show t h a t f o r a g e n e r a l band J o f C l ( X ) , w i t h d u a l band I ,
X(J), #
For c l a r i t y , we f i r s t c o n s t r u c t a n example i n
R(1).
o r d i n a r y f u n c t i o n t h e o r y , t h e n show t h a t we a c t u a l l y h a v e a n example i n C ' l ( X ) . Let w = t h e f i r s t i n f i n i t e o r d i n a l ; o1 = t h e f i r s t u n c o u n t a b l e o r d i n a l ; t h e s e t o f o r d i n a l s { B 16 < w ) ;
N
=
fi
= the s e t of ordinals
{ B I B 5 01;
IN1 = t h e s e t o f o r d i n a l s t a l a < w l l ;
-
N1
=
t h e s e t o f o r d i n a l s Icxlci
We e n d o w m and
ml
5
with t h e order topology.
p a c t Hausdorff. Let X =
N1
Z =
m1
Q
WEjl
=
all.
fl
x x
X
(product topology);
w) IJ (wl W)
x
m);
U ( ~ 1W)
They a r e t h e n com-
330
So
Chapter 13
Z is a closed subset of X, and Q c Z. Consider the function f on Q with value 1 on the points of
N1x w
and 0 on the points of w1
x
istic function of the closed set W, istic function of the open set X \ ( w ,
Let u be the character-
IN.
w,
x x
and
m).
il
the character-
Then u is upper-
semicontinuous, R is lower semicontinuous, and
Note that we do not have
2 u ; we show there exist no
R
R lower-
semicontinuous and u uppersemicontinuous such that R < u and ( * ) is satisfied.
u(wl,n)
=
Suppose such a pair R,u exists.
0 for all n E N .
Now
Since u is uppersemicontinuous, i t
follows that limsupa+w u(a,n) < 0 for all n E N . Hence, since 1 N1 is uncountable, there exists a. such u(a,n) < 0 for all c1L c1 0
and nEl?i.
By assumption, R < u , so we also have
R(a,n) < 0 for all a
2
a.
and n C N . R being lowersemicontinuous,
< 0 for all a > a o , which conthis gives in turn that R ( a , w ) -
tradicts the definition of R . The above construction has been carried out in identify this with C r r ( X ) a .
k"(X).
X is dispersed, so C " ( X )
=
It f o l l o w s u above is a usc element of C " ( X )
(37.12).
an Rsc element. R l Q = RI,
Let I be the band R"(Q)
so by ( * ) , f E R ( I ) ,
.
Then ulQ
=
Cll(X). and k u I and
but it is easily shown, using the
above argument, that there is no gEC"(X) such that ( k ( g ) ) I (u(g)),
= f.
Now
=
Riemann I n t e g r a b i l i t y
331
5 6 4 . Examples
Consider t h e c a s e J
=
C(X)
u'
and we c a n t a k e p
2
We
0.
have no s p e c i a l comment t o make a b o u t x ( p ) beyond t h e s t a t e m e n t i n ( 6 3 . 2 ) , b u t we t a k e t h i s o c c a s i o n t o remark on s t a n d a r d We r e l a t e i t t o o u r
Riemann i n t e g r a t i o n i n f u n c t i o n t h e o r y .
a p p r o a c h by - a s u s u a l - i d e n t i f y i n g L"(X)
with CII(X)u
" f u n c t i o n " will mean an e l e m e n t o f C l f ( X ) a ) .
I t i s e a s y t o show
(SO
t h a t t h e s e t o f f u n c t i o n s which a r e Riemann i n t e g r a b l e ( w i t h respect to
u)
i n the standard sense i s precisely
X ( U ) ~ ,and
the
s e t o f t h o s e which a r e Riemann n e g l i g i b l e i n t h e s t a n d a r d s e n s e is precisely
y x ( ~ =)$ ~ X(p)
CII(X)u.
And r e m a r k a b l y , by ( 6 2 . 4 ) .
Thus, by ( 6 3 , 2 ( i ) ) , t h e s p a c e o f Riemann i n t e g r a b l e f u n c t i o n s modulo t h e s p a c e o f Riemann n e g l i g i b l e o n e s c a n be i d e n t i f i e d w i t h t h e Riemann s u b s p a c e
) - otherwise s t a t e d , LJ This i s t h e Caratheodory
R(C"(X)
w i t h t h e Riemann s u b s p a c e o f L m ( p ) . theorem. We l o o k a t o t h e r e x a m p l e s .
(64.2)
X ( C ' (X)) = C(x) = .R(C"(X)).
332
Chapter 1 3 Proof.
( C * ( X ) ) ' = 0 , s o I ( c ~ ( x ) )= 6
-1
(0) = c(x).
We
have a l r e a d y n o t e d t h e s e c o n d e q u a l i t y i n 5 5 9 ( c f . ( 4 6 . 3 ) ) . QED
We have remarked e a r l i e r ( 5 5 9 ) t h a t t h i s g e n e r a l i z e s t h e c l a s s i c a l theorem f o r a r e a l i n t e r v a l : a f u n c t i o n f i s RiemannS t i e l t j e s i n t e g r a b l e w i t h r e s p e c t t o e v e r y f u n c t i o n o f bounded v a r i a t i o n i f and o n l y i f f i s c o n t i n u o u s .
The above i s a s t r o n g s t a t e m e n t .
For f EC"(X) t o be
Riemann i n t e g r a b l e w i t h r e s p e c t t o e v e r y a t o m i c Radon m e a s u r e , f must l i e i n C(X); i t i s n o t enough t h a t i t be c o n t i n u o u s on X .
The o n l y r e s u l t we have t o o f f e r a t p r e s e n t on x ( C l ( X ) d ) i s a c h a r a c t e r i z a t i o n due t o C . Goffman f o r t h e c a s e where X i s
m e t r i c and d e n s e - i n - i t s e l f ( o r a l c o m m u n i c a t i o n ) .
( 6 4 . 4 ) Lemma.
First, a
I f X i s a compact m e t r i c s p a c e , a p o s i t i v e u s c
e l e m e n t u l i e s i n Cii(X)u i f and o n l y i f ( u , x ) = 0 f o r a l l b u t
a c o u n t a b l e number of
XIS.
Riemann I n t e g r a b il i t y Proof.
333
The s u f f i c i e n c y f o l l o w s from ( 5 4 . 1 6 ) .
Comversely,
s u p p o s e U E C " ( X ) ~and t h a t ( u , x ) > 0 f o r a n u n c o u n t a b l e number of x ' s .
Then t h e r e e x i s t s 1 > 0 and a c l o s e d u n c o u n t a b l e s u b > 1 €or a l l xEZ 0'
s e t Zo o f X ,
such t h a t ( u , x )
take X
S i n c e X i s compact m e t r i c , Z o c o n t a i n s a c l o s e d
=
1.
d e n s e - i n - i t s e l f subset 2.
For s i m p l i c i t y ,
L e t e be t h e c h a r a c t e r i s t i c e l e m e n t
Then ( e , x ) < ( u , x ) f o r a l l x , s o t h e Isomorphism theorem
of Z .
gives us e < u. L e t I be t h e band o f C"(X) g e n e r a t e d by e , and J i t s d u a l band i n C'(X).
Then J i s t h e v a g u e l y c l o s e d band g e n e r a t e d
by Z ( c f . (51.10) and ( 3 6 . 7 ) ) , h e n c e , b y ( 3 7 . 1 0 ) , c o n t a i n s a d i f f u s e measure ~ > 0 . But t h e n ( e , u ) > 0 , s o ( u , ~ )> 0 , c o n t r a d i c t i n g u EC'l(X)u. QED
S i n c e f o r e v e r y f EC"(X), 6 ( f ) i s a p o s i t i v e u s c e l e m e n t , t h e sbove g i v e s us o u r v e r s i o n o f Goffman's t h e o r e m :
( 6 4 . 5 ) I f X i s compact m e t r i c , t h e n f o r f E C " ( X ) , i f and o n l y i f ( 6 ( f ) , x )
=
f Ex(C'(X)d)
0 f o r a l l b u t a c o u n t a b l e number o f
X I S .
Goffman's s t a t e m e n t i s a s f o l l o w s :
(64.6)
(Goffman).
A bounded r e a l f u n c t i o n on a c l o s e d i n t e r v a l
i s Riemann i n t e g r a b l e w i t h r e s p e c t t o e v e r y p u r e l y n o n a t o m i c
3 34
Chapter 13
measure if and only if its set of points of discontinuity is countable.
13
RTEbIANN INTEGRABILITY
C o n s i d e r t h e c l a s s i c a l Lebesgue t h e o r e m : a bounded f u n c t i o n o n a r e a l i n t e r v a l i s Ricmann i n t e g r a b l c i f a n d o n l y i f i t s s e t o f p o i n t s o f d i s c o n t i n u i t y h a s (Lebesgue) measure z e r o . While q u i t e a t t r a c t i v e , t h e t h e o r e m h a s an u n s a t i s f a c t o r y f e a t u r e : i t r e q u i r e s f i r s t d e f i n i n g "Lebesgue measure z e r o " ,
a
c o n c c p t o u t s i d e t h e domain o f Riemann i n t e g r a t i o n . T h i s u n s a t i s f y i n g f e a t u r e can be e l i m i n a t e d . s u b j e c t o f m e a s u r e on o u r r e a l i n t e r v a l . Lebcsgue m e a s ure.
I-lerc m e a s u r e means
C o r r e s p o n d i n g t o "measure"
i n t h e same way
t h a t "Riemann i n t e g r a l " c o r r e s p o n d s t o " L e b e s g u e have t h e concept of " c o n t c n t " .
Consider t h e
integral",
we
The t h e o r e m f o r s e t s c o r r e s p o n d -
i n g t o the Lebesgue theorem above r e a d s : a s e t h a s c o n t e n t ( i s q u a d r a b l e ) i f and o n l y i f i t s f r o n t i e r h a s measure z e r o .
But
t h e f r o n t i e r o f a s e t i s c l o s e d , and f o r a c l o s e d s e t , h a v i n g measure z e r o i s e q u i v a l e n t t o having c o n t e n t z e r o .
Thus t h e
theorem can he s t a t e d : a s e t h a s c o n t e n t i f and o n l y i f i t s f r o n t i e r has content zero. The way t o e l i m i n a t e m e a s u r e f r o m L c b e s g u e ' s t h e o r e m i s
now c l e a r .
We r e p l a c e t h e s e t o f p o i n t s o f d i s c o n t i n u i t y o f a
f u n c t i o n f by t h e s a l t u s f u n c t i o n o f f , which g i v e s t h e p o i n t s
of discontinuity.
The s a l t u s f u n c t i o n i s a l w a y s p o s i t i v e a n d
uppersemicontinuous,
and f o r s u c h a f u n c t i o n , h a v i n g Lebesgue 31 5
Chapter 1 3
316
i n t e g r a l z e r o i s e q u i v a l e n t t o h a v i n g Riemann i n t e g r a l z e r o . Thus t h e L e b e s g u e t h e o r e m b e c o m e s : a bounded f u n c t i o n i s Riemann i n t e g r a b l e i f a n d o n l y i f i t s s a l t u s f u n c t i o n h a s Riemann i n t e g r a l zero. T h i s w i l l b e o u r a p p r o a c h t o Riemann i n t e g r a b i l i t y , u s i n g t h e f a c t t h a t f o r an element f o f C " ( X ) ,
6(f) is the gencraliza-
tion of the saltus function of a function.
560. Riemann i n t e g r a b l e e l e m e n t s
We f i r s t d e f i n e Riemann i n t e g r a b i l i t y w i t h r e s p e c t t o a band J o f C l ( X ) .
I n 5 6 1 , we e x a m i n e i t w i t h r e s p e c t t o a
s i n g l e e l e m e n t 1-1 o f C ' ( X ) . Given a b a n d J o f C l ( X ) , an
e l e m e n t f o f C"(X) w i l l b e
s a i d t o b e Riemann i n t e g r a b l e w i t h r e s p e c t t o J i f 6 ( f ) E t J L . We w i l l d e n o t e t h e s e t o f s u c h e l e m e n t s b y
B(J).
Thus X ( J )
=
6 - l (.JL). From ( S O - l l ) , we h a v e :
(60.1)
F o r e v e r y band J o f C ' (X),
3 (J)
(i)
i s an Ma-subspace o f C"(X),
(ii)
c o n t a i n s C(X),
( i i i ) i s closed under t h e operations u ( - ) , a ( - ) ,
a(.).
I t i s e a s y t o s e e from t h e d e f i n i t i o n t h a t i f f C % ( J ) and
Riemann I n t e g r a b i l i t y
a(f) 5
g 5 u ( f ) , t h e n gE.2 (J).
31 7
A c t u a l l y , from ( 5 0 . 1 3 1 , we have
a much s t r o n g e r r e s u l t :
(60.2)
If A,B c X(J), A
In p a r t i c u l a r ,
a(J)
5 f 5 B , and (AA
- VB)EJL,
then f € x ( J ) .
i s Dedekind c l o s e d .
561.
u-Riemann i n t e g r a b i l i t y
Given p E C ' ( X ) , an e l e m e n t f o f C''(X) w i l l be s a i d t o be Riemann i n t e g r a b l e w i t h r e s p e c t t o IJ,o r 1-1-Riemann i n t e g r a b l e , if ( 6 ( f ) ,
14) =
0.
( 6 1 . 1 ) Given p E C ' ( X ) , t h e n f o r e v e r y f E C " ( X ) , t h e f o l l o w i n g a r e equivalent :
'1
f i s Riemann i n t e g r a b l e w i t h r e s p e c t t o 1-1;
2'
f i s Riemann i n t e g r a b l e w i t h r e s p e c t t o C ' ( X )
Fi'
T h i s f o l l o w s from t h e e a s i l y v e r i f i a b l e f a c t t h a t f o r gEC"(X)+, g E ( C J ( X ) I . I y i f and o n l y i f ( 8 , 11-11)
=
0.
Thus t h e s e t o f 1-1-Riemann i n t e g r a b l e e l e m e n t s o f C"(X) precisely by
a (1-1).
a(C'(X) ) .
u
is
We w i l l a l s o d e n o t e t h i s (MI-)s u b s p a c e
Some c h a r a c t e r i z a t i o n s o f
2 (J):
31 8
Chapter 13
( 6 1 . 2 ) Given a b a n d J o f C l ( X ) , t h e n f o r f E C " ( X ) , t h e following are equivalent: 1'
f i s Riemann i n t e g r a b l e w i t h r e s p e c t t o J ;
2'
f i s Riemann i n t e g r a b l e w i t h r e s p e c t t o e v e r y u E L J :
3'
t h e r e e x i s t a u s c e l e m e n t u a n d an k s c e l e m e n t a. s u c h
that R < f 5 u and u - L E J ' :
'4
( ~ ( f ) , u )= ( f , p > = ( u ( f ) , p ) f o r e v e r y ~ E J .
The p r o o f i s i m m e d i a t e .
We n e x t g i v e some c h a r a c t e r i z a t i o n s o f x ( p ) .
For s i m p l i c -
i t y , we t a k e p > 0.
(61.3)
Given u € C ' ( X ) + , t h e n f o r f E C " ( X ) , t h e f o l l o w i n g a r e
equivalent:
1'
f i s p-Riemann i n t e g r a b l e ;
h>f
g
t h e r e e x i s t s u b s e t s A , B o f C(X) s u c h t h a t A < f < B and ( f , ~ = )
'4
SUP
€5A
(g,p)
=
inf ( ~ , L O ; hEB .
t h e r e e x i s t s e q u e n c e s { g n l , { h n } i n C(X) s u c h t h a t
The p r o o f i s s t r a i g h t f o r w a r d .
Riemann I n t e g r a b i l i t y
319
I n t h e c l a s s i c a l t h e o r y o f Riemann i n t e g r a t i o n on a E u c l i d e a n s p a c e , Rieman i n t e g r a b i l i t y i s d e f i n e d b y o n e o f t h e properties 2
0
,
3 O , o r 4'
above, but using a d i s t i n g u i s h e d
family of "step functions" i n place of continuous functions. Among t h e p r o p e r t i e s p o s s e s s e d b y t h e s e s t e p f u n c t i o n s a r e t h e following.
Each s t e p f u n c t i o n i s a f i n i t e l i n e a r c o m b i n a t i o n
(I)
of f u n c t i o n s which can be o b t a i n e d as s p e c t r a l e l e m e n t s o f continuous functions ( c f . 517). ( 1 1 ) Each s t e p f u n c t i o n i s Riemann i n t e g r a b l e .
In ( 6 1 . 5 ) we generalize the classical procedure t o t h e p r e s e n t c a s e i n t h e f o r m o f a c o n d i t i o n f o r p-Riemann i n t e g r a b i l i t y i n terms of s p e c t r a l elements o f continuous f u n c t i o n s . F i r s t , h o w e v e r , we m u s t a n s w e r t h e f o l l o w i n g q u e s t i o n .
Are t h e
s p e c t r a l e l e m e n t s o f c o n t i n u o u s f u n c t i o n s Riemann i n t e g r a b l e ( b y o u r d e f i n i t i o n ) ; a n d i f n o t , a r e s u f f i c i e n t l y many Riemann integrable t o enable us t o carry out the classical process? The q u e s t i o n was a n s w e r e d b y Loomis
[32].
Indeed, he proved a
more g e n e r a l r e s u l t , w h i c h r e d u c e s , i n o u r c o n t e x t , t o t h e
f o 1l o w i n g :
(61.4) Theorem (Loomis).
Given p E C ' { X ) ,
then f o r every fEC"(X),
the following are equivalent:
'1
f i s p-Riemann i n t e g r a b l e ;
2'
a l l b u t a c o u n t a b l e number o f t h e s p e c t r a l e l e m e n t s o f
f a r e p-Riemann i n t e g r a b l e .
Chapter 1 3
320
Proof.
We c a n assume p > 0 , IIui1
=
1, and 0 < f < n(X).
F o r t h e r e m a i n d e r o f t h e p r o o f , we w i l l d e n o t e n(X)
s i m p l y b y I]
and e a c h s p e c t r a l e l e m e n t e f ( X ) s i m p l y b y e ( 1 ) . Assume 1' h o l d s . (6(e(x)),p) e s t a b l i s h 2'.
2
E
We show t h a t f o r e v e r y E > 0 ,
f o r o n l y a f i n i t e number o f A ' s , w h i c h w i l l S p e c i f i c a l l y , we p r o v e t h e
Then
We show t h e e q u i v a l e n t i n e q u a l i t y
Note f i r s t t h a t f A X n n - f A A n - l n we w o r k w i t h t h i s l a t t e r f o r m .
= (fAXnn -
An-ln)'
;
Riemann I n t e g r a b i l i t y
This i s t h e f i r s t i n e q u a l i t y i n ( i i ) .
T h i s i s t h e second i n e q u a l i t y i n ( i i ) , s o we have ( i i ) , and
with it, ( i ) .
I t follows f r o m ( i ) t h a t f o r n = l,.-.,m,
321
Chapter 1 3
32 2
(Here we have used t h e a s s u m p t i o n t h a t f - hence e a c h g n i s Riemann i n t e g r a b l e ; c f . 4'
i n (61.2).)
Then
Since l E = l ( 6 ( e ( ~ n ) ) , 2 ~ )m E , t h i s g i v e s u s m < 1 / ~ and , comp l e t e s t h e p r o o f o f t h e Lemma.
Now assume 2'
holds.
I t i s c l e a r t h a t t h e proof of
F r e u d e n t h a l ' s s p e c t r a l theorem (17.10) can be m o d i f i e d ( f o r o u r f ) s o t h a t t h e s p e c t r a l e l e m e n t s a r e u-Riemann i n t e g r a b l e . Thus f i s i n t h e norm c l o s u r e o f j ; ! ( u ) .
Since
x(u)
i s norm
c l o s e d , f E j;! ( p ) .
QED We can now g i v e o u r g e n e r a l i z a t i o n o f t h e c l a s s i c a l cond i t i o n f o r Riemann i n t e g r a b i l i t y .
( 6 1 . 5 ) Given p E C ' ( X ) + , t h e n f o r f € C t t ( X ) , t h e f o l l o w i n g a r e equivalent:
' 1
f i s u-Riemann i n t e g r a b l e ;
2'
t h e r e e x i s t s e q u e n c e s { g n } , {h,}
satisfying:
Riemann I n t e g r a b i l i t y
323
t h e g n ' s and h n ' s a r e l i n e a r combinations o f
(i)
u-
Riemann i n t e g r a b l e s p e c t r a l e l e m e n t s o f e l e m e n t s
of C ( X ) ; gl 5 g 2
(ii)
2 * *-* f<-< * . * I
( i i i ) supn(gn,v)
Proof.
(f,v)
S u p p o s e 1' h o l d s .
t o s a t i s f y 4'
(*)
=
h2 < h1 , =
infn(hn,p).
Choose s e q u e n c e s { g ; } , { h , l }
of (61.3).
For e a c h n , t h e r e e x i s t g n , h n , l i n e a r combinations o f p -
Riemann i n t e g r a b l e s p e c t r a l e l e m e n t s o f e l e m e n t s o f C ( X ) , s u c h t h a t gn 5 g;
,
hn
2
hA
and
[ I g n - gA[19 IIhn - hA[I
5
l/n.
I n e f f e c t , gn c a n b e o b t a i n e d b y t h e m o d i f i c a t i o n o f t h e proof of (17.10)
p o i n t e d o u t a t t h e end of t h e p r e c e d i n g p r o o f ,
a n d s i m i l a r l y f o r hn ( t h e a p p r o x i m a t i n g s t e p e l e m e n t s i n ( 1 7 . 1 0 ) can b e chosen above t h e g i v e n a ) . Applying (17.12) and ( 6 0 . 1 ) , w e can assume, i n a d d i t i o n , t h a t t h e s e q u e n c e {g,} descending.
They t h u s s a t i s f y ( i ) a n d ( i i ) o f 2'.
f o l l o w s f r o m 4' T h a t 2'
i s a s c e n d i n g and t h e sequence {hn} i s
i n ( 6 1 . 3 ) a n d t h e norm c o n t i n u i t y o f
i m p l i e s 1'
f o l l o w s f r o m 5'
(iii) p.
i n (61.3). QED
Remark. p-Riemann i n t e g r a b i l i t y i n o r d i n a r y a n a l y s i s i s d e f i n e d f o r f u n c t i o n s on X C''(X)a.
- i n our context, f o r elements of
I t i s easy t o v e r i f y t h a t the space of functions
w h i c h a r e p-Riemann i n t e g r a b l e i n t h e o r d i n a r y s e n s e i s
Chapter 13
324
2 (u),
precisely
562. Riemann n e g l i g i b l e e l e m e n t s
We r e t u r n t o a g e n e r a l band J o f C ' ( X ) . R(J)
n J'
We w i l l c a l l
t h e s e t o f Riemann n e g l i g i b l e ( w i t h r e s p e c t t o J )
e l e m e n t s , an d w e w i l l d e n o t e i t b y f X ( J 1 . w i l l also write
,TX
I f J = C'(X )
I.r'
we
(1-1).
( 6 2 . 1 ) Given a b a n d J o f C ' ( X ) , t h e n f o r f € C " ( X ) , t h e f o l l o w i n g are equivalent: 1'
fCh%(J);
2'
u ( f ) ,k(f)EJ1:
'3
u(/fl)EJ'.
S u p p o s e 1' h o l d s , t h a t i s , 6 ( f ) E J '
Proof. f - 6(f), f
w e h a v e 2'. 3'
holds.
+
6(f)EJL. T h a t 2'
h'?, (J)
Then
Since f - 6cf) < tcf) < u(f) < f + 6(f),
i m p l i e s '3
f o l l o w s from (49.12).
Th en, a g a i n by ( 4 9 . 1 2 1 ,
u ( f ) , t ( f ) E J L , hence G(f)EJ'. fE
and fEJ'.
(u(f)I,
T h a t fEJ'
Suppose
l t ( f ) IEJL, hence
is trivial.
Thus
. QED
(62.2)
C o r o l l a r y 1.
Given a b a n d J o f C ' ( X ) ,
t h e n f o r fEC"(X),
Riemann Integrability fE Jvx (J) if and only if
1 f( -<
325
u for some usc element u in J L .
I t follows , f $ ( J ) is the Riesz ideal of C " ( X )
generated by the
usc elements in (J'),.
Proof.
Suppose g € l X ( J ) and 0 5 If1 5 /gl. Then u(lf/) 2
~ ( / g / ) .Since,
by '3
hence, again b y '3 ideal.
above,u(jgj)EJL, it follows u(lfl)EJL.
above, fcJY*x(J).
Thus J v x ( J ) is a Riesz
The rest of the Corollary is clear.
Remark. Note that, since h?$?(J)= X ( J )
n
QED JL,
,qx(J) is norm
closed.
It follows from (62.1) that Riemann negligibility can be defined first, and then used to define Riemann integrability. Specifically, define g to be Riemann negligible if u(lgl)EJL, then (since 6(f) is a positive usc element) define f to be Riemann integrable if 6 ( f ) is Riemann negligible.
(As
we
pointed out in the introduction to the present chapter, this eliminates the mention of Lebesgue negligibility in Lebesgue's characterization of Riemann integrability. By definition X ( J )
=
6-'(JL).
Contained in the above dis-
cussion is the fact that actually X ( J )
=
&-lWX(J)).
This
gives us:
(62.3) Given a band J of C' (X), let H be the Riesz ideal
C h a p t e r 15
326
g e n e r a t e d Oy Sa(J)). Then (i)
S-'(H)
=
g(H
1
) , a n d H i s t h e s m a l l e s t Riesz i d e a l of
Clt(X) w i t h t h i s p r o p e r t y ; ( i i ) jZ(HI)
=
J ( J ) , and H
1
i s t h e l a r g e s t b a n d o f C1(X )
with t h i s property
The b a n d H 1 g e n e r a t e d by S ( ( R ( J ) )
(that is, the order clos-
u r e o f t h e above H) i s o f c o u r s e t h e smallest band o f C"(X) s u c h t h a t S-'(H1) t h a n J'
=
T h i s band may b e s t r i c t l y s m a l l e r
X(J).
- equivalently,
H
I
may b e s t r i c t l y l a r g e r t h a n J .
an example, l e t X b e a f i n i t e s e t .
Then Cll(X)
=
As
C(X), s o
S ( C t l ( X ) ) = 0 , s o f o r e v e r y band J o f Cll(X), H = 0 , s o H1
=
0.
We e m p h a s i z e ( i i ) a b o v e . : f o r e v e r y band J o f C l ( X ) , t h e r e
i s a l a r g e s t b a n d ( i t c o n t a i n s J ) w i t h t h e same Riemann i n t e g r a b l e e l e m e n t s i n C"(X).
We c l o s e t h i s 9 w i t h a n i n t e r e s t i n g r e s u l t .
(62.4)
Giv e n a band J o f C t ( X ) , e v e r y Riemann i n t e g r a b l e e l e -
ment o f C1l(X)d i s Riemann n e g l i g i b l e :
g(J) n
Proof.
C"(X)d
c&?,(J).
Consider f €$(J)
we show f E J'. a ( f ) = 6 ( f ) EJ'.
By ( 5 4 . 1 2 ) ,
n C 'l (X )d,
L(f)
=
and we c a n t a k e f > 0;
0 , hence u ( f )
Thus 0 < f < u ( f ) € J',
=
u(f) -
whence f EJ'. QED
327
Riemann I n t e g r a b i l i t y
563. Riemann i n t e g r a b i l i t y and Riemann s u b s p a c e s
Let J be a band o f C'(X) Then
J(J),
=
a n d I its d u a l band i n
Cll(X).
J ( J ) / J ~ % ( J ) A. r e a s o n a b l e e x p e c t a t i o n i s t h a t I n ( 6 3 . 2 ) , we l i s t
R ( J ) , = R(1) t h e Riemann s u b s p a c e o f I . some common b a n d s f o r which t h i s i s t r u e . p r i n c i p a l b a n d s {C'(X)
Among them a r e t h e
l ~ € C ' ( X ) l , which a r e t h e o n e s d e a l t w i t h
LJ i n ordinary integration theory.
We w i l l s e e below, however,
t h a t f o r a g e n e r a l band o f C'(X),
t h e e x p e c t a t i o n need n o t be
born o u t . First,we record:
( 6 3 . 1 ) For e v e r y band J o f C'(X),
w i t h d u a l band I ,
nJ),= R ( I ) -
Proof. f - R(f)EJ'.
Suppose f E J ( J ) .
Then 6 ( f ) E J ' ,
I t follows a ( f ) , = f I
=
so u ( f ) - f ,
u ( f ) I , whence f I E R ( I )
(57.6). QED
( 6 3 . 2 ) L e t J b e a band of Cl(X),
w i t h d u a l band I .
f i e s any o f t h e f o l l o w i n g c o n d i t i o n s , t h e n X ( J ) ,
If J satis-
= R(1):
( i ) J i s a p r i n c i p a l band; (ii) J i s vaguely c l o s e d - e q u i v a l e n t l y , I i s a u-band;
Chapter 1 3
328
( i i i ) Jd i s v a g u e l y c l o s e d - e q u i v a l e n t l y , I i s a n R-band; (iv)
J i s v a g u e l y d e n s e i n C'(X) - e q u i v a l e n t l y ,
U(1)
=
Cll(X).
I n e a c h c a s e , i t o n l y r e m a i n s t o show t h a t
Proof. R(1) c a ( J ) , .
Assume ( i ) h o l d s , t h a t i s , J
=
C'(X)
f o r some u E C ' ( X ) .
I-r
By ( 2 7 . 4 ) , t h e r e e x i s t s e q u e n c e s { ( g ) I , n P I ( h n ) l - r l i n C(X) s u c h t h a t ( g ) + f , ( h ) + f . LJ n1-1 n1-1
Consider f E R ( 1 ) .
To show t h i s , we r e p l a c e Cgn3,thn3 by s e q u e n c e s
C1711> k i t h t h e d e s i r e d p r o p e r t y . -
g1
=
-
gl,
then s e t hl
= hlvgl.
h e prmceed i n d u c t i l e l y .
We h a v e (gl)l-l
= (gl)p,
and (E1I-r 1 = (hlll-lv(Fl)l-l = ( h l ) l - l v ( g l ) u = ( h l l V .
-
-
g l , - ..,gm, El,..-,En
Ti
< * - *
n Set
{En},
have been chosen s o t h a t
of course,
Assume
El z - -+ -gn* <<
5Kl a n d (Ek)l-l = =
Set
( g k ) l - l , (Ek)l-l = ( h ) ( k = l , . . . , n ) k1-1 (gn+lVzn)AEn. C o n t i n u i n g i n t h i s f a s h i o n , we
.
f -<
obtain (*), N o w s e t R = Vn g n' u = An hn '
T h e n R-< u a n d R = f = u . 1-1 1-1
To e s t a b l i s h t h a t f E J ? j ( J ) I , we n e e d o n l y show t h a t uE%'(J). T h i s f o l l o w s from u - I! EJ'
( s i n c e (u
-
a) I
=
0 ) a n d 3'
in
(61.2). Assume ( i i ) h o l d s .
Then by ( 5 9 . 5 ) a n d ( 6 0
C(X), c 2 ( J ) , . Assume ( i i i ) h o l d s , a n d c o n s i d e r f E R ( 1 ) .
We c a n assume
Riemann I n t e g r a b i l i t y
We show t h a t i n t h i s c a s e , n o t o n l y i s f i n
0 < f < ll(X)I.
but is actually i n
je(J),,
329
J ( J ) . By ( 5 7 . 6 ) , t h e r e e x i s t a n Rsc 2 f 5 uI.
element k and a u s c element u such t h a t k I (uV0)A
l l ( X ) ; t h e n (u,) I = f a l s o .
S e t El
= ( k V 0 ) A U ( X ) I (by
a s s u m p t i o n , l l ( X ) l i s a n Rsc e l e m e n t ) ; t h e n ( R 1 ) I
R I E I , s o we a c t u a l l y have R1 We t h u s have R1 - R1 € I d =
=
I t follows
hence f € X ( J ) .
, ' J
Assume ( i v ) h o l d s .
Then i t f o l l o w s e a s i l y from t h e f a c t
t h a t p r o j , mgps C ( X ) MU-isomorph c a l l y o n t o C(X), again, R(I) c
But
f also.
= f.
f < u l , w i t h ( u , ) ~= f . I 1
=
S e t u1 =
XCJ)
(not j u s t
B(J
(52.3) that,
,
I)
QED
The r e m a i n d e r o f t h i s s e c t i - n i s d e v o t e d t o a n example t o show t h a t f o r a g e n e r a l band J o f C l ( X ) , w i t h d u a l band I ,
X(J), #
For c l a r i t y , we f i r s t c o n s t r u c t a n example i n
R(1).
o r d i n a r y f u n c t i o n t h e o r y , t h e n show t h a t we a c t u a l l y h a v e a n example i n C ' l ( X ) . Let w = t h e f i r s t i n f i n i t e o r d i n a l ; o1 = t h e f i r s t u n c o u n t a b l e o r d i n a l ; t h e s e t o f o r d i n a l s { B 16 < w ) ;
N
=
fi
= the s e t of ordinals
{ B I B 5 01;
IN1 = t h e s e t o f o r d i n a l s t a l a < w l l ;
-
N1
=
t h e s e t o f o r d i n a l s Icxlci
We e n d o w m and
ml
5
with t h e order topology.
p a c t Hausdorff. Let X =
N1
Z =
m1
Q
WEjl
=
all.
fl
x x
X
(product topology);
w) IJ (wl W)
x
m);
U ( ~ 1W)
They a r e t h e n com-
330
So
Chapter 13
Z is a closed subset of X, and Q c Z. Consider the function f on Q with value 1 on the points of
N1x w
and 0 on the points of w1
x
istic function of the closed set W, istic function of the open set X \ ( w ,
Let u be the character-
IN.
w,
x x
and
m).
il
the character-
Then u is upper-
semicontinuous, R is lower semicontinuous, and
Note that we do not have
2 u ; we show there exist no
R
R lower-
semicontinuous and u uppersemicontinuous such that R < u and ( * ) is satisfied.
u(wl,n)
=
Suppose such a pair R,u exists.
0 for all n E N .
Now
Since u is uppersemicontinuous, i t
follows that limsupa+w u(a,n) < 0 for all n E N . Hence, since 1 N1 is uncountable, there exists a. such u(a,n) < 0 for all c1L c1 0
and nEl?i.
By assumption, R < u , so we also have
R(a,n) < 0 for all a
2
a.
and n C N . R being lowersemicontinuous,
< 0 for all a > a o , which conthis gives in turn that R ( a , w ) -
tradicts the definition of R . The above construction has been carried out in identify this with C r r ( X ) a .
k"(X).
X is dispersed, so C " ( X )
=
It f o l l o w s u above is a usc element of C " ( X )
(37.12).
an Rsc element. R l Q = RI,
Let I be the band R"(Q)
so by ( * ) , f E R ( I ) ,
.
Then ulQ
=
Cll(X). and k u I and
but it is easily shown, using the
above argument, that there is no gEC"(X) such that ( k ( g ) ) I (u(g)),
= f.
Now
=
Riemann I n t e g r a b i l i t y
331
5 6 4 . Examples
Consider t h e c a s e J
=
C(X)
u'
and we c a n t a k e p
2
We
0.
have no s p e c i a l comment t o make a b o u t x ( p ) beyond t h e s t a t e m e n t i n ( 6 3 . 2 ) , b u t we t a k e t h i s o c c a s i o n t o remark on s t a n d a r d We r e l a t e i t t o o u r
Riemann i n t e g r a t i o n i n f u n c t i o n t h e o r y .
a p p r o a c h by - a s u s u a l - i d e n t i f y i n g L"(X)
with CII(X)u
" f u n c t i o n " will mean an e l e m e n t o f C l f ( X ) a ) .
I t i s e a s y t o show
(SO
t h a t t h e s e t o f f u n c t i o n s which a r e Riemann i n t e g r a b l e ( w i t h respect to
u)
i n the standard sense i s precisely
X ( U ) ~ ,and
the
s e t o f t h o s e which a r e Riemann n e g l i g i b l e i n t h e s t a n d a r d s e n s e is precisely
y x ( ~ =)$ ~ X(p)
CII(X)u.
And r e m a r k a b l y , by ( 6 2 . 4 ) .
Thus, by ( 6 3 , 2 ( i ) ) , t h e s p a c e o f Riemann i n t e g r a b l e f u n c t i o n s modulo t h e s p a c e o f Riemann n e g l i g i b l e o n e s c a n be i d e n t i f i e d w i t h t h e Riemann s u b s p a c e
) - otherwise s t a t e d , LJ This i s t h e Caratheodory
R(C"(X)
w i t h t h e Riemann s u b s p a c e o f L m ( p ) . theorem. We l o o k a t o t h e r e x a m p l e s .
(64.2)
X ( C ' (X)) = C(x) = .R(C"(X)).
332
Chapter 1 3 Proof.
( C * ( X ) ) ' = 0 , s o I ( c ~ ( x ) )= 6
-1
(0) = c(x).
We
have a l r e a d y n o t e d t h e s e c o n d e q u a l i t y i n 5 5 9 ( c f . ( 4 6 . 3 ) ) . QED
We have remarked e a r l i e r ( 5 5 9 ) t h a t t h i s g e n e r a l i z e s t h e c l a s s i c a l theorem f o r a r e a l i n t e r v a l : a f u n c t i o n f i s RiemannS t i e l t j e s i n t e g r a b l e w i t h r e s p e c t t o e v e r y f u n c t i o n o f bounded v a r i a t i o n i f and o n l y i f f i s c o n t i n u o u s .
The above i s a s t r o n g s t a t e m e n t .
For f EC"(X) t o be
Riemann i n t e g r a b l e w i t h r e s p e c t t o e v e r y a t o m i c Radon m e a s u r e , f must l i e i n C(X); i t i s n o t enough t h a t i t be c o n t i n u o u s on X .
The o n l y r e s u l t we have t o o f f e r a t p r e s e n t on x ( C l ( X ) d ) i s a c h a r a c t e r i z a t i o n due t o C . Goffman f o r t h e c a s e where X i s
m e t r i c and d e n s e - i n - i t s e l f ( o r a l c o m m u n i c a t i o n ) .
( 6 4 . 4 ) Lemma.
First, a
I f X i s a compact m e t r i c s p a c e , a p o s i t i v e u s c
e l e m e n t u l i e s i n Cii(X)u i f and o n l y i f ( u , x ) = 0 f o r a l l b u t
a c o u n t a b l e number of
XIS.
Riemann I n t e g r a b il i t y Proof.
333
The s u f f i c i e n c y f o l l o w s from ( 5 4 . 1 6 ) .
Comversely,
s u p p o s e U E C " ( X ) ~and t h a t ( u , x ) > 0 f o r a n u n c o u n t a b l e number of x ' s .
Then t h e r e e x i s t s 1 > 0 and a c l o s e d u n c o u n t a b l e s u b > 1 €or a l l xEZ 0'
s e t Zo o f X ,
such t h a t ( u , x )
take X
S i n c e X i s compact m e t r i c , Z o c o n t a i n s a c l o s e d
=
1.
d e n s e - i n - i t s e l f subset 2.
For s i m p l i c i t y ,
L e t e be t h e c h a r a c t e r i s t i c e l e m e n t
Then ( e , x ) < ( u , x ) f o r a l l x , s o t h e Isomorphism theorem
of Z .
gives us e < u. L e t I be t h e band o f C"(X) g e n e r a t e d by e , and J i t s d u a l band i n C'(X).
Then J i s t h e v a g u e l y c l o s e d band g e n e r a t e d
by Z ( c f . (51.10) and ( 3 6 . 7 ) ) , h e n c e , b y ( 3 7 . 1 0 ) , c o n t a i n s a d i f f u s e measure ~ > 0 . But t h e n ( e , u ) > 0 , s o ( u , ~ )> 0 , c o n t r a d i c t i n g u EC'l(X)u. QED
S i n c e f o r e v e r y f EC"(X), 6 ( f ) i s a p o s i t i v e u s c e l e m e n t , t h e sbove g i v e s us o u r v e r s i o n o f Goffman's t h e o r e m :
( 6 4 . 5 ) I f X i s compact m e t r i c , t h e n f o r f E C " ( X ) , i f and o n l y i f ( 6 ( f ) , x )
=
f Ex(C'(X)d)
0 f o r a l l b u t a c o u n t a b l e number o f
X I S .
Goffman's s t a t e m e n t i s a s f o l l o w s :
(64.6)
(Goffman).
A bounded r e a l f u n c t i o n on a c l o s e d i n t e r v a l
i s Riemann i n t e g r a b l e w i t h r e s p e c t t o e v e r y p u r e l y n o n a t o m i c
3 34
Chapter 13
measure if and only if its set of points of discontinuity is countable.