Finitely additive integral representations

F o r t h e r e s u l t s i n t h i s s e c t i o n t h e e x t r a s t r u c t u r e c o n t a in e d in a s s u m p tio n A3 i s u s e d . A lth o u g h th e y c o u ld be p ro v e d w ith o u t A3, th e f u r t h e r c o m p lic a tio n s and t h e s p a r s i t y o f c o n v in c in g a p p l i c a t i o n s f o r th e more g e n e r a l form s a r e good enough r e a s o n s f o r n o t d o in g s o h e r e . The e f f e c t o f A3 i s t o c o n v e r t t h e " lo w e r s e m i - c o n t i n u i t y " a s s u m p tio n A4 i n t o

one o f " c o n t i n u i t y ” .

LEMMAo I f C s a t i s f i e s A l, A3 and A4 th e n {h > 3} € F(K) / o r every h i C and 3 > 0 .

P r o o f . Suppose K € K i s g iv e n . Choose an h* € C t a k i n g th e v a lu e 1 on K (u s e S to n e ’s c o n d i t i o n ) , and th e n o b s e rv e t h a t

{h > 3) n K = { (1 + 3)h * \h < 1} n X , w hich i s a / ( - s e t by v i r t u e o f A4. □

Now t o s e e w hat s o r t o f c o n d i t i o n i s n eed ed t o o b ta in a f u l l

r e p r e s e n t a t i o n , l e t us b e g in w ith th e s im p le c a s e w here 1 € C 0 In o r d e r t h a t th e two l i n e a r f u n c t i o n a l s T and y* s h o u ld a g re e on C i t i s n e c e s s a r y and s u f f i c i e n t t h a t T ( l ) = y * ( l ) (= sup{y(X ) : K € K} ) .

[Compare t h i s w ith th e u s u a l t i g h t n e s s c o n d i tio n o f P r o h o r o v .] T h is means t h a t f o r any £ > 0 t h e r e s h o u ld e x i s t a ( K su ch t h a t

T( h) > T (1) - £ f o r e v e r y h > 1v . By means o f A3 we can f u r t h e r re d u c e K

£

t h i s t o th e c o n d i tio n t h a t T ( h ' ) < £ f o r e v e ry h r 5 1 w hich v a n is h e s on

K^ . With su ch m o tiv a tio n th e f o llo w in g d e f i n i t i o n becom es somewhat more n a t u r a l . We sa y t h a t K e x h a u s ts T on C i f 3 to e v e ry h € C and

£ > 0 3 th e r e e x i s t s a K i K such t h a t T ( h r) < e w henever h > h r 6 C

and h 1 = 0 on K . F o r e x a m p le , t h i s c o n d i tio n i s s a t i s f i e d i f , f o r any

h i C , {h > 0} i s c o n ta in e d i n some K-s e t 0 l cf „ th e c o n tin u o u s f u n c t i o n s o f com pact s u p p o r t on a l o c a l l y com pact s p a c e . ]

THEOREM 5 J o Assume A l, A2j A3., A4 and A5 o f S e c tio n 33 and a ls o th a t K i s a ( 0 , U/ , (]/) p a v in g on X . Tfoen a n e c e ssa ry and s u f f i c i e n t

c o n d itio n f o r th e r e to e x i s t a K -re g u la r f i n i t e l y a d d i ti v e measure r e p r e s e n tin g T i s t h a t K e x h a u s ts T on C and

T( h) = lim T( h A n) f o r e v e ry h € C . ( 5 .1 )

f t - * »

TTze r e p r e s e n tin g measure i s th e one d e term in e d by ( 3 . 1 ) .

P r o o f . T h ro u g h o u t t h i s p r o o f , y r e f e r s t o th e f i n i t e l y a d d i t i v e m easure c o n s t r u c t e d i n Theorem 4 . 1 .

To p ro v e s u f f i c i e n c y , su p p o se t h a t K e x h a u s ts T on C . G iven £ > 0 and h , l e t K b e th e K -s e t d e te rm in e d by th e e x h a u s tio n p r o p e r t y . By ( 5 .1 ) and ( 3 . 5 ) , t h e r e i s no l o s s o f g e n e r a l i t y i n assu m in g t h a t h 5 1 . Choose an > 1^ and a n a t u r a l number n f o r w hich n ^T(ft^) < £ .

Also define -1 n

k

=

n

y

1 %-1 being members

K.

=

z

K

n

{h

>

i/n}

for £ = 0, 1, ..., n+1 and set

Notice that

K

=

K

z> ... z> Z = 0 all of these sets 0 — — n+1

K (by the Lemma), and that

k

+ n 1 >

h

on

K .

K -

K

The simple function fe is less than

h

. We show that

>

T(h) -

3e , from which it follows that y*(/z) =

T(h)

. The idea is to approximate

k

with a C-function

h*

>

k

, and then use the exhaustion condition applied to

h\ h*+n ^h,

to prove that

T{h)

~

T(h*)

. So choose

' ' '

n

for which

T [ h < ]i

(isf.) + e and set

h*

=

n~

Z ^ • Notice

£ £=1

that

h* > k

, so that 7z* + n > /z on

K

. Note also that

K

TÜi*)

< £ + n Z y(^-) = e + • The C-function 7z\ £=1

7z*+n ^Tz, therefore satisfies the conditions for the exhaustion property, whence

T(h)

< T

7 z \

/z*+n 1/z1 + i7h*+n~^h.

<

e + 2 W ) + n 1T(/zJ < y*(fc) + 3e .

Now for the necessity. Given /z € C and e > 0 , choose

k - Y

ot.1

< h

for which

V*(k) > T(h) -

£ , where the

K.'s

are

£=1

'L Ki

t'

pairwise disjoint

K -

sets. Put

K -

U

K.

, and consider any

h r < h

which vanishes on

K

.

V

N" K2.

Observe that

h* + k S h

, so that m ) = y*(fc)

>

]i3k(hr)

+

V*(k)

(superadditivity) >

T(h')

+

T(h)

- e .

Thus the exhaustion property is satisfied. Clearly (5.1) is just a restatement of (3.5) when y represents

T

, so it is also a necessary

condition. □

We shall use this theorem in two different ways. The more obvious of these is simply to derive finitely additive representation theorems. The classical Markoff-Alexandroff representations fall into this category (see the examples at the end of this section). On the other hand, only slight modifications are needed to convert Theorem 5„1 into a theorem on

a-additive and T-additive representations. This more important application will be dealt with in Section 6.

Before giving the examples of the direct use of Theorem 5.1 though, we should mention an equivalent functional analytic formulation of the

exhaustion property. This is sometimes a more readily interpretable condition.

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