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hence
(
−
)=
m
f
,1
f
dP
,
(4)
Ω
for any measurable
f
:
Ω
→
[
0, 1
]
.
Now take our attention to the second term
m
(
0,
f
+
g
)
in the right side of the equality (1). First
define
M
:
S→
[
0, 1
]
by the formula
M
(
A
)=
m
(
0, 1
−
χ
A
)
.
As before it is possible to prove that
M
is a measure. Of course,
(Ω)=
(
)=
α ∈
[
]
M
m
0, 0
0, 1
.
S→
[
]
Define
Q
:
0, 1
by the formula
m
(
0, 1
−
χ
A
)=
α
Q
(
A
)
.
As before, it is possible to prove
(
−
)=
α
m
0, 1
f
fdQ
,
Ω
for any
f
:
Ω
→
[
0, 1
]
measurable, or
(
)=
α
Ω
(
−
)
m
0,
h
1
h
dQ
,
(5)
for any
h
:
Ω
→
[
0, 1
]
,
S
-measurable. Combining (1), (4), and (5) we obtain
m
(
A
)=
m
((
μ
A
,
ν
A
)=
m
((
μ
A
,1
−
μ
A
)) +
m
((
0,
μ
A
+
ν
A
))
=
Ω
μ
A
dP
+
α
(
−
Ω
(
μ
A
+
ν
A
)
)
1
dQ
.
A simple consequence of the representation theorem is the following property of the mapping
P
−
α
Q
:
S→
R
.
Proposition 2.2.
S→
[
]
α
Let
P
,
Q
:
0, 1
be the probabilities mentioned in Theorem 2.1,
is the
corresponding constant. Then
(
)
− α
(
)
≥
P
A
Q
A
0
for any
A
∈S
.
Proof. Put
B
=(
0, 0
)
,
C
=(
χ
A
,0
)
. Then
B
≤
C
, hence
m
(
0, 0
)
≤
m
(
χ
A
,0
)
. Therefore
α
=
m
(
0, 0
)
≤
m
(
χ
A
,0
)=
P
(
A
)+
α
(
1
−
Q
(
A
))
.
Theorem 1.1 is an embedding theorem stating that every IF-events algebra
F
can be embedded
to an
d
MV-algebra
M
. Now we shall prove that any state
m
:
F→
[
0, 1
]
can be extended to a
state
m
:
M→
[
0, 1
]
([63]).
Theorem 2.2.
Let
M⊃F
be the MV-al
ge
bra constructed in Theorem 1.1. Then every state
F→
[
]
M→
[
]
m
:
0, 1
can be extended to a state
m
:
0, 1
.
