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Proof. First
(
)=
(
(
)) =
((
)) =
m x
R
m
x
R
m
1, 0
1.
= ∅
(
)
(
)=(
)
If A
B
, then x
A
x
B
0, 1
, hence
(
)=
(
(
)) =
((
(
)
(
)) =
m x
A
B
m
x
A
B
m
x
A
x
B
=
m
(
x
(
A
)) +
m
(
x
(
B
)) =
m x
(
A
)+
m x
(
B
)
.
(
)
(
)
Finally, A n
A implies x
A n
x
A
, hence
m x
(
A n
)=
m
(
x
(
A n
))
m
(
x
(
A
)) =
m x
(
A
)
.
Proposition 3.2.
Let x :
σ ( J ) →F
be an observable, m :
F→ [
0, 1
]
be a state.
Define
F : R
[
0, 1
]
by the formula
(
)=
(
((
)))
F
u
m
x
, u
.
Then F is non-decreasing, left continuous in any point u
R ,
(
)=
(
)=
lim
u
F
u
1,
lim
F
u
0.
u
→−
Proof. If u
<
v , then
((
)) =
((
))
((
))
((
))
x
, v
x
, u
x
u , v
x
, u
,
hence
(
)=
((
))
(
((
))) =
(
)
F
v
m
, v
m
x
, u
F
u
,
F is non decreasing. If u n
u , then
x
((
, u n
))
x
((
, u
))
,
hence
F
(
u n
)=
m
(
x
((
, u n
)))
m
(
x
((
, u
))) =
F
(
u
)
,
F is left continuous in any u
R . Similarly u n
implies
x
((
, u n
))
x
((
,
)) = (
1, 0
)
.
Therefore
F
(
u n
)=
m
(
x
((
, u n
)))
m
((
1, 0
))) =
1
for every u n
, hence lim u
F
(
u
)=
1. Similarly we obtain
u n
= ⇒−
u n
,
hence
m
(
x
((
u n ,
u n
)))
m
(
x
(
R
)) =
1.
Now
1
=
lim
n
F
(
u n
)=
lim
n
m
(
x
((
u n ,
u n
))) +
lim
n
F
(
u n
)=
=
1
+
lim
n
F
(
u n
)
,
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