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In-Depth Information
First it can be proved by induction the equality
1
q
βχ
A
,1
1
q
βχ
A
)=
(
−
(
βχ
A
,1
− βχ
A
)
qm
m
∈
holding for every
q
N
. Therefore
1
q
βχ
A
,1
1
q
βχ
A
)=
1
q
m
m
(
−
(
βχ
A
,1
−
βχ
A
)
p
q
βχ
A
,1
p
q
βχ
A
)=
p
q
m
(
−
(
βχ
A
,1
− βχ
A
)
m
,
hence (3) holds for every rational
α
. Let
α
∈
R
,
α
∈
[
0, 1
]
. Take
α
n
∈
Q
,
α
n
α
. Then
α
n
χ
A
αχ
A
,1
−
α
n
χ
A
1
−
αχ
A
.
Therefore
m
(
αβχ
A
,1
−
αβχ
A
)=
lim
n
m
(
α
n
βχ
A
,1
−
α
n
βχ
A
)=
→
∞
=
(
βχ
A
,1
− βχ
A
)=
α
(
βχ
A
,1
− βχ
A
)
lim
n
→
∞
α
n
m
m
,
β
=
hence, (3) is proved, too. Particularly, if we give
1, then
(
αχ
A
,1
−
αχ
A
)=
α
(
χ
A
,1
−
χ
A
)
m
m
.
Let
f
:
Ω
→
[
0, 1
]
be simple,
S
-measurable, i.e.
n
i
=
1
α
i
χ
A
i
,
A
i
∈S
(
i
=
1, ., , ,
n
)
,
A
i
∩
A
j
=
∅
(
i
=
j
)
.
f
=
Combining (2), (3), and the definition of
P
we obtain
n
i
=
1
m
(
α
i
χ
A
i
,1
−
α
i
χ
A
i
)=
(
f
,1
−
f
)=
m
n
i
=
1
α
i
m
(
χ
A
i
,1
− χ
A
i
)=
=
n
i
=
1
α
i
P
(
A
i
)=
=
fdP
,
Ω
hence
(
−
)=
m
f
,1
f
fdP
,
Ω
Ω
→
[
]
Ω
→
[
]
S
for any
f
:
0, 1
simple. If
f
:
0, 1
is an arbitrary
-measurable function, then
(
)
there exists a sequence
f
n
of simple measurable functions such that
f
n
f
.
Evidently,
1
−
f
n
1
−
f
. Therefore
m
(
f
,1
−
f
)=
lim
n
m
(
f
n
,1
−
f
n
)=
lim
n
f
n
dP
=
fdP
,
→
∞
→
∞
Ω
Ω
