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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
.
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