Information Technology Reference
In-Depth Information
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
)
,
→
∞
