990 CHAPTER 27. ORLITZ SPACES
which converges to zero as i→ ∞. Therefore, if the Fi are disjoint and F = ∪∞i=1Fi, let
Sm ≡ ∪mi=1Fi so that Sm ↑ F. Then since XSm →XF in EA (Ω) and L is continuous,
λ (F) ≡ L(XF) = limm→∞
L(XSm)
= limm→∞
m
∑i=1
L(XFi) =∞
∑i=1
λ (Fi) .
Next observe that λ is absolutely continuous with respect to µ. To see this, supposeµ (F) = 0. Then if t > 0, ∫
Ω
A(
XF (x)t
)dµ = 0 < 1
for all t > 0 and so ||XF ||A = 0. Therefore, λ (F)≡ L(XF) = 0.It follows by the Radon Nikodym theorem there exists v ∈ L1 (Ω) such that
L(XF) = λ (F) =∫
Fvdµ.
Therefore, for all s ∈ S,L(s) =
∫F
svdµ. (27.2.19)
I need to show that v is actually in LÃ (Ω) . If v = 0 a.e., there is nothing to prove so assumethis is not so. Let u be defined by.
u(x)≡
{Ã(
r|v(x)|||L||
)/ v(x)||L|| if v(x) ̸= 0
0 if v(x) = 0.(27.2.20)
for r ∈ (0,1) . Now let
Fn ≡ {x : |u(x)| ≤ n}∩{x : v(x) ̸= 0} (27.2.21)
and defineun (x)≡ u(x)XFn (x) . (27.2.22)
I claim ||un||A ≤ 1. It is clear that since µ (Ω)< ∞, un ∈ EA (Ω) . If ||un||A > 1, Then for ε
small enough,||un||A− ε > 1
and so, by convexity of A and the fact that A(0) = 0,
1 <∫
Ω
A(|un (x)|||un||A− ε
)dµ ≤ 1
||un||A− ε
∫Ω
A(|un (x)|)dµ
and so, letting ε→ 0+ and using 27.1.6 and convexity of A as in the proof of the preceedingproposition,
||un||A ≤∫
Ω
A(|un (x)|)dµ ≤∫
Fn
A(
rÃ(
r |v(x)|||L||
)/
r |v(x)|||L||
)dµ
≤ r∫
Fn
A(
Ã(
r |v(x)|||L||
)/
r |v(x)|||L||
)dµ ≤ r
∫Fn
Ã(
r |v(x)|||L||
)dµ
≤ r1||L||
∫Fn
u(x)v(x)dµ =r||L||
∫Ω
unvdµ. (27.2.23)