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(i)
P
(
1
,0
Ω
)=[
1, 1
]
,
P
(
0
,1
Ω
)=[
0, 0
]
,
Ω
Ω
=(
Ω
)=
⇒
((
⊕
)) =
(
)+
(
)
(ii)
A
B
0
,1
P
A
B
P
A
P
B
,
Ω
=
⇒
(
)
(
)
(iii)
A
n
A
P
A
n
P
A
.
It is easy to see that the following property holds.
P
(
,
P
(
Proposition 2.1.
Let
P
:
F→J
,
P
(
A
)=[
A
)
A
)]
. Then
P
is a probability if and only if
P
,
P
:
F→
[
0, 1
]
are states.
Hence it is sufficient to characterize only the states ([4], [5], [54]).
Theorem 2.1.
F→
[
]
S→
[
]
For any state
m
:
0, 1
there exist probability measures
P
,
Q
:
0, 1
α ∈
[
]
and
0, 1
such that
m
((
μ
A
,
ν
A
)) =
Ω
μ
A
dP
+
α
(
1
−
Ω
(
μ
A
+
ν
A
)
dQ
)
.
Proof. The main instrument in our investigation is the following implication, a corollary of
(ii):
∈F
+
≤
=
⇒
(
)=
(
−
)+
(
+
)
f
,
g
,
f
g
1
m
f
,
g
m
f
,1
f
m
0,
f
g
.
(1)
S→
[
]
(
)=
(
χ
A
,1
− χ
A
)
We shall define the mapping
P
:
0, 1
by the formula
P
A
m
.
Let
∈S
∩
= ∅
χ
A
+
χ
B
≤
− χ
A
)
(
χ
B
,1
− χ
B
)=(
)
A
,
B
,
A
B
. Then
1, hence (
χ
A
,1
0, 1
. Therefore
P
(
A
)+
P
(
B
)=
m
(
χ
A
,1
−
χ
A
)+
m
(
χ
B
,1
−
χ
B
)=
=
m
((
χ
A
,1
−
χ
A
)
⊕
(
χ
B
,1
−
χ
B
)) =
=
m
(
χ
A
+
χ
B
,1
−
χ
A
−
χ
B
)=
m
(
χ
A
∪
B
,1
−
χ
A
∪
B
=
P
(
A
∪
B
)
.
∈S
(
=
)
Let
A
n
n
1, 2, ...
,
A
n
A
. Then
χ
A
n
χ
A
,1
−
χ
A
n
1
−
χ
A
,
hence by (iii)
P
(
A
n
)=
m
(
χ
A
n
,1
−
χ
A
n
)
m
(
χ
A
,1
−
χ
A
)=
P
(
A
)
.
(
Ω
)=
(
χ
Ω
−
χ
Ω
)=
((
)) =
S→
[
]
Evidently
P
m
,1
m
1, 0
1, hence
P
:
0, 1
is a probability
measure.
Now we prove two identities. First the implication:
∈S
∈
[
]
,
A
i
∩
A
j
= ∅(
=
)=
⇒
A
1
, ...,
A
n
,
α
1
, ...,
α
n
0, 1
i
j
n
i
=
1
α
i
χ
A
i
,1
−
n
i
=
1
α
i
χ
A
i
)=
n
i
=
1
m
(
α
i
χ
A
i
,1
−
α
i
χ
A
i
)
.
m
(
(2)
It can be proved by induction. The second identity is the following
0
≤
α
,
β
≤
1
=
⇒
m
(
αβχ
A
,1
−
αβχ
A
)=
α
m
(
βχ
A
,1
−
βχ
A
)
.
(3)
