Information Technology Reference
In-Depth Information
Proof. It is easy to see that any element
(
μ
A
,
ν
A
)
∈M
can be presented in the form
(
μ
A
,
ν
A
)
(
0, 1
−
ν
A
)=(
0, 1
)
,
(
μ
A
,0
)=(
μ
A
,
ν
A
)
⊕
(
0, 1
−
ν
A
)
.
If
(
μ
A
,
ν
A
)
∈F
, then
m
((
μ
A
,0
)) =
m
((
μ
A
ν
A
)) +
m
((
0, 1
−
ν
A
))
.
Generally, we can define
m
:
M→
[
0, 1
]
by the formula
m
((
μ
A
,
ν
A
)) =
m
((
μ
A
,0
))
−
m
((
0, 1
−
ν
A
))
,
so
that
m
is an extension of
m
. Of course, we must prove that
m
is a state. First we prove that
m
is additive.
Let
A
=(
μ
A
,
ν
A
)
∈M
,
B
=(
μ
B
,
ν
B
)
∈M
,
A
B
=(
0, 1
)
, hence
((
μ
A
+
μ
B
−
1
)
∨
0,
(
ν
A
+
ν
B
)
∧
1
)=(
0, 1
)
,
μ
A
+
μ
B
≤
1, 1
−
ν
A
+
1
−
ν
B
≤
1.
Therefore
(
)+
(
)=
(
μ
A
,
)+
(
μ
B
,
)
m
A
m
B
m
ν
A
m
ν
B
=
(
μ
A
,0
)
−
(
− ν
A
)+
(
μ
B
,0
)
−
(
− ν
B
)=
m
m
0, 1
m
m
0, 1
=
m
(
μ
A
+
μ
B
,0
)
−
m
(
0, 1
−
ν
A
−
ν
B
)=
=
(
μ
A
+
μ
B
,
ν
A
+
ν
B
)=
(
⊕
)
m
m
A
B
.
Before the continuity of
m
we shall prove its monotonicity. Let
A
≤
B
, i.e.
μ
A
≤
μ
A
,
ν
A
≥
ν
B
.
Then by Theorem 2.1
(
)=
(
μ
A
,0
)
−
(
− ν
A
)=
m
A
m
m
0, 1
=
Ω
μ
A
dP
+
α
(
1
−
Ω
(
μ
A
+
0
)
dQ
−
0
dP
−
α
(
1
−
Ω
(
0
+
1
−
ν
A
)
dQ
)=
Ω
=
+
α
(
−
Ω
(
μ
A
+
ν
A
)
)
Ω
μ
A
dP
1
dQ
.
Therefore
(
)
−
(
)=
+
α − α
− α
−
m
B
m
A
Ω
μ
B
dP
Ω
μ
B
dQ
Ω
ν
B
dQ
−
(
Ω
μ
A
dP
+
α
−
α
Ω
μ
A
dQ
−
α
Ω
ν
A
dQ
)=
=
Ω
(
μ
B
− μ
A
)
− α
Ω
(
μ
B
− μ
A
)
+
α
Ω
(
ν
A
− ν
B
)
dP
dQ
dQ
.
Of course, as an easy consequence o Proposition 2.1 we obtain the inequality
− α
≥
fdP
fdQ
0
Ω
Ω
