12 G.W. JOHNSON and M.L. LAPIDUS
We finish this section with two lemmas. The first is a somewhat
technical measure-theoretic result which will be used throughout the
paper, most often without explicit mention. The reader should note the
result but may wish to skip the proof at least initially.
LEMMA 0.1. Let n e M(0,t) and suppose that 9 e L. . Let
(0.14) F^y) := / e(s,y(s))dn(s)
1 (0,t)
for any y e C[0,t] for which the integral exists. Then, for every
A 0,
F,(A-1/2x+O
:Ls _ defined and satisfies
(0.15) |F1(A"1/2x+0
N
for m x Lebesgue-a. e. (x,0 e * E. .
PROOF. We first show: (*) For every A 0 and m x Lebesgue-a.e. (x,£),
e(s,A~lyx(s)+0 is defined and satisfies | 0 (s, A-1'2x(s)+£) | . Ile(s,-)||
for |n|-a.e. s.
Let HA : (0,t) x cQ x RN ^ (0,t) x RN be defined by
Hx(s,x,£) =
(s,A-1/2x(s)+0
H is everywhere defined and continuous
and so 6 ° H, is certainly Borel measurable though it need not be every-
where defined. Let
N := ((s,v) e (0,t) x RN
:
e(s,v) fails to be defined or e(s,v) is
defined but |e(s,v)| || e (s, •) 1^}.
that e is defined and satisfies |e(s,v)| _ ||eCs,-)|| ^or InI x Lebesgue-
a.e. (s,v); i.e., N is |n| x Lebesgue-null. Let A 0 be given. Note
that to establish (*), it suffices to show that H~ (N) is |n| x m x
-1 N
Lebesgue-null. Accordingly, we section H. Qj) at (s,£) e (0,t) x R
:
[H^1(^)](S^)
= {x CQ :
(s,X,OeH^1(^)}
=
{x:(s,A"1/2x(s)+0
N}
= {x: A "1/2x(s)+£
e
N(S)]
= {x:x(s) e
A1/2
[ N
( s )
-£]},
where N^ : = {u e RN : (s,u) e N].
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