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In-Depth Information
X
∞
.If
σ
-algebra
A
on topological space
P
0
and
{P
1
θ
,θ ∈ Θ}
are Baire mea-
X
∞
then for each of them exists the only continuation
[6] of the probability measure on (
sures [6] on compactum
X
∞
,
A
). As a result we can consider
P
0
and
{P
1
θ
,θ ∈ Θ}
X
∞
,
A
θ ∈ Θ
to be defined on (
). From perpendicularity for every
∃A
0
(
θ
, ∃A
1
(
θ
,A
0
(
θ
,A
1
(
θ
∈A
A
0
(
θ
∩ A
1
(
θ
∅,
)
)
)
)
,
)
)=
X
∞
\ A
0
(
X
∞
\ A
1
(
P
0
(
θ
,P
1
(
θ
.
)) = 0
)) = 0
X
∞
, A
It is easy to see that perpendicular measures on (
) are perpendicular
on (
).
If we suppose that for every
X
∞
,
A
θ ∈ Θ
the set
A
0
(
θ
)
∈
A
and
A
0
(
θ
) is a closed
set in Tychonoff product then the set
A
0
=
θ∈Θ
A
0
(
θ
)
X
∞
\ A
0
aremeasurablesetsin
is closed. Consequently, the set
A
0
and
A
.
X
∞
is a topological space with countable basis. Then [8] a
Tychonoff product
=
F
X
∞
,
A
support
S
of the measure
P
0
on (
) exists, i.e.
S
, where intersection
of sets is such that every
F
is closed,
P
0
(
F
)=1and
P
0
(
S
)=1.Asforevery
θ ∈ Θ
we have
P
0
(
A
0
(
θ
)) = 1 and
A
0
(
θ
)isclosed,then
A
0
=
θ∈Θ
A
0
(
θ
)
⊇
F
=
S.
As a result
A
0
)) = 1.
As every subset in
P
0
(
is closed then every cylindrical set is also closed.
We have already proved that if
X
X
∞
, A
P
0
is perpendicular in (
)toall
P
1
θ
,
A
0
(
θ
)isclosedfor
θ ∈ Θ
, then there is a measurable
A
0
of
A
that for every
X
∞
\ A
0
)=0.
θ ∈ ΘA
0
∩ A
1
(
θ
)=
∅
and
P
0
(
Let
B
1
,B
2
, ...
be a decreasing sequence of cylindrical sets in which
B
n
is
defined by all possible vectors from
X
n
, standing on as the first
n
elements of
the sequences of
A
0
.Then
n→∞
B
n
=
lim
B.
It is clear that
A
0
⊆ B
.Weprovethat
B ⊆ A
0
.Let
ω ∈ B
. Then for every
n
the set
A
(
n
),
A
(
n
)
⊆ A
0
, is defined as a set of all sequences in
A
0
for which
their first
n
elements coincide with the first
n
elements of
ω
. One can see that
A
(
n
)
=
∅
and
A
(
n
)
,n
=1
,
2
, ...,
is a decreasing sequence.
Projection
π
n
of the sequences of
A
0
to the first
n
elements of the sequences
X
n
is continuous as the inverse image of any open set
D
n
of
is the set (
D
n
×
X
∞
)
∩ A
0
. This set is open in topological space
A
0
andclosedinTychonoff
X
∞
. Then the inverse image of any closed set is closed in
X
∞
and
product
π
−
1
n
X
∞
,where
(
ω
n
)=
A
(
n
) - is a closed set in
π
n
(
ω
)=
ω
n
.Then
=
∞
A
A
(
n
)
n
=1
