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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

.