12 R.H. CAMERON AND D.A. STORVICK

EXAMPLE: Let

0 if x e DV

(2.9) F(x) = , _

1 if x e C - D'

Then since D is a scale invariant null set,we have F(x) «*1 , and hence Fes .

Clearly F is also an element of S . By Theorem 2.1,if F were an element of S ,

it would be continuous (B.P.A.) s-almost everywhere on C [a,bj , and it would there-

v v

fore be zero s-almost everywhere on C [a,b] , since it is zero for all elements of D

This contradicts the fact that it is unity s-almost everywhere and so F/S and

hence S f S D S .

Now finally, we shall show that S'/s . From Proposition 2 of [8], when v = 1

we have that #?'/#?, and if a is the measure generated by the unit mass concentrated

in an element v of L - BV , then a e 771 - #?' .

Let

b b ^

F(x) s J expCi J v(t)dx(t))da(v)=exp{i J vQ(t)dx(t)]eS .

T

a a

L2

By Propositions I & II of [8], we have F/S' and the lemma is proved.

LEMMA 2.7- If F,GeS and F(x)=G(x) almost everywhere on CV[a,b] , then

F(x) ? G(x) .

PROOF. By definition of S , there exist *1 ,*p e 171 such that

_ ~

v b

(2.10) F(x)=f expU £ J v (t)dx (t)3dM-. (v) .

TVrQ

-h i J=l a

J D

2U

and

v b

(2.11) G(x) = J exp{i Z J v.(t)dx.(t)]dnp(v) .

l£[a,b ] H a J '

Thus F(x) is almost everywhere equal to both the right member of (2.10) and the right

member of (2.11). Since Fes cs , it follows from Theorem 2.1 in [k], that H and

—*

w

-+

H are identical. Thus F(x) = G(x) follows from (2.10) and (2.11), and the Lemma is

proved.

REMAJRK: If F e S , then the values of F on

DV[a,b]

determine the values of F

s-almost everywhere on C [a,b] • , and conversely, the values of F s-almost everywhere

on C [a,b] determine the values of F everywhere on D [a,b].

LEMMA 2.8. If Fes there exists F eS such that F « F on C [a,b].