Construction of Brownian motion on the Wiener measure space

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Construction of Brownian motion on the Wiener measure space

著者 Sato Shuichi

著者別表示佐藤秀一

page range 14p.

year 2019‑09‑23

URL http://hdl.handle.net/2297/00055659

Creative Commons : 表示 ‑ 非営利 ‑ 改変禁止 http://creativecommons.org/licenses/by‑nc‑nd/3.0/deed.ja

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CONSTRUCTION OF BROWNIAN MOTION ON THE WIENER MEASURE SPACE

SHUICHI SATO

Abstract.

We give a self contained construction of the Wiener probability space.

1. Introduction

Let W = W d = C 0

¹

) be the collection of continuous

^R

^d -valued functions on the interval

^R⁰

= 0

¹

). In this note we present a self contained construction of the Wiener probability space ( W

^F

), where

^F

is a sigma-algebra of subsets of W and is the Wiener measure. The mathematical theory of the Brownian motion is based on this probability space. We follow the methods of 4], but we intend to make our presentation more specic. (See also 2], 3], 6].)

2. Definition of the Wiener measure

Denition 2.1. We write t = ( t

¹

:::t n )

²

(

^R⁰

) ⁿ if t = ( t

¹

:::t n )

²

(

^R⁰

) ⁿ =

R

0

R

0

( n -fold product) and 0 < t

¹

< t

²

<

< t n . We also write

^R

^d =

^R

d , when Cartesian products of

^R

^d are considered. Dene t : W

^!^R

_nd , t

²

(

^R⁰

) ⁿ , by

t ( f ) = ( t

¹

( f ) ::: t

n

( f ))

where _t

_j

( f ) = f ( t _j ) and

^R

_nd = (

^R

d ) ⁿ =

^R

d

^R

d ( n -fold product). Set

G

t =

^;1

_t ( B ) : B

²^B

(

^R

_nd )

where

^;1

_t ( B ) =

^f

f

²

W : t ( f )

²

B

^g

and

^B

(

^R

_nd ) denotes the Borel class of

^R

_nd . Since

^B

(

^R

_nd ) is a sigma-algebra, obviously we have the following result.

Lemma 2.2.

^G

t is a sigma-algebra for every t

²

(

^R⁰

) ⁿ . We observe the following result on

^G

t .

Lemma 2.3. Let t

²

(

^R⁰

) ^m , s

²

(

^R⁰

) ⁿ with m

n . Let ~ t =

^f

t

¹

:::t _m

^g

s ~ =

f

s

¹

:::s _n

^g

, which are sets of positive numbers, if t = ( t

¹

:::t _m ) s = ( s

¹

:::s _n ).

Suppose that ~ t

~ s . Then

^G

t

^G

s .

Proof. Take ( s ) = ( s

⁽¹⁾

s

⁽²⁾

:::s

⁽

_n

⁾

) satisfying s

⁽¹⁾

= t

¹

, s

⁽²⁾

= t

²

, ::: , s

⁽

_m

⁾

= t _m with some

² ^S

_n (the permutation group). Suppose A

² ^G

_t . Then there exists

²^B

(

^R

_md ) such that A =

^;1

_t (). Let

⁰

=

^R

ⁿ _d

^;

^m . Dene

^;1

(

⁰

) =

^f

( x

^;1⁽¹⁾

:::x

^;1⁽

_n

⁾

) : ( x

¹

:::x n )

²

⁰^g

:

Key Words and Phrases. Brownian motion, Wiener probability space.

The author is partly supported by Grant-in-Aid for Scientic Research (C) No. 16K05195, Japan Society for the Promotion of Science.

This note was essentially nalized in August 2016.

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We see that f

²

^;1

_s (

^;1

(

⁰

)) if and only if

( s

¹

( f ) ::: s

n

( f ))

²

^;1

(

⁰

) which means that

( s

⁽¹⁾

( f ) ::: s

⁽n⁾

( f ))

²

⁰

: The denition of

⁰

implies that this is equivalent to

( s

⁽¹⁾

( f ) ::: s

⁽m⁾

( f ))

²

: This can be rewritten as

( _t

¹

( f ) ::: _t

_m

( f ))

²

which is equivalent to f

²

^;1

_t (). Thus A =

^;1

_t () =

^;1

_s (

^;1

(

⁰

)), and hence A

²^G

s , since

^;1

(

⁰

)

²^B

(

^R

_nd ).

Denition 2.4. ^Let

^G⁽

ⁿ

⁾

⁼

t

^2(R⁰⁾ⁿ^G

t and

^G

=

¹

_n

⁼¹^G⁽

ⁿ

⁾

.

Lemma 2.5.

^G

is an algebra of subsets of W .

Proof. We easily see that W

²^G

. Let A

²^G

. Then there is n 1 such that A

²^G

t

for some t

²

(

^R⁰

) ⁿ . Since

^G

t is a sigma-algebra (Lemma 2.2), we have A ^c

² ^G

t

and hence A ^c

²^G

. Suppose that AB

²^G

. Lemma 2.3 implies that AB

²^G

t for some t . Thus A

B

² ^G

t by Lemma 2.2. Collecting results, we see that

^G

is an algebra.

Let

^j

x

^j

= (

²¹

+

²

_d )

¹

⁼

²

be the Euclidean norm of x = (

¹

::: d )

² ^R

^d . Dene g ( txy ) = 1

(

^p

2 t ) ^d exp

; j

y

^;

x

^j²

2 t

xy

²^R

^d t > 0 :

We have the following formulas:

Lemma 2.6.

Z

Rd

g ( txy ) dy = 1

Z

Rd

g ( sax ) g ( txb ) dx = g ( s + tab ) :

Denition 2.7. Dene t on

^G

t , t

²

(

^R⁰

) ⁿ , by t ( A ) =

Z

g ( t

¹

0 x

¹

) g ( t

²^;

t

¹

x

¹

x

²

) :::g ( t n

^;

t n

^;1

x n

^;1

x n ) dx

¹

:::dx n

with A =

^;1

_t (), t = ( t

¹

:::t n ),

²^B

(

^R

_nd ).

Lemma 2.8. ^Let t be as in Denition 2 : 7. Then, t denes a probability measure on the sigma-algebra

^G

t .

Proof. If t is well-dened on

^G

t , it is easy to see that t is a probability measure.

Suppose that A =

^;1

_t () =

^;1

_t (

⁰

) with

⁰ ²^B

(

^R

_nd ). We show that =

⁰

. To see this, we notice that t is a surjection from W onto

^R

_nd . Thus the relation

^\

t ( W ) = t ( A ) =

⁰^\

t ( W ) implies =

⁰

. It follows that t is well-dened on

^G

t .

To show that t is a probability measure, we have to prove

(1) t ( W ) = 1

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(2) if A _k

²^G

_t , k = 1 2 ::: , and A _j

^\

A _k =

, j

⁶

= k , then t (

¹

_k

⁼¹

A k ) =

^X¹

k

⁼¹

t ( A k ) :

Obviously, we have part (1), since W =

^;1

_t (

^R

_nd ). To prove part (2), let A _k =

^;1

_t ( k ) with k

²^B

(

^R

_nd ). We note that

^;1

_t ( j

^\

k ) = A j

^\

A k =

if j

⁶

= k . This implies that j

^\

k =

, since t is a surjection. Thus the countable additivity of t follows from the denition of t with the countable additivity of Lebesgue integral.

We note that if A =

^;1

_t (), t > 0, t

^f

f

²

W : f ( t )

²

^g

= t ( A ) =

Z

(

^p

2 1 t ) ^d exp

; j

x

^j²

2 t

dx:

(2.1)

Lemma 2.9. Let A

² ^G

. Then we can dene on

^G

by ( A ) = t ( A ) with t

²

n

¹

(

^R⁰

) ⁿ satisfying A

²^G

t .

Proof. Suppose that A =

^;1

_s () =

^;1

_t (

⁰

) with s

²

(

^R⁰

) ^m , t

²

(

^R⁰

) ⁿ and

²^B

(

^R

_md ),

⁰ ²^B

(

^R

_nd ). We show that

Z

g ( s

¹

0 x

¹

) g ( s

²^;

s

¹

x

¹

x

²

) :::g ( s m

^;

s m

^;1

x m

^;1

x m ) dx

¹

:::dx m

=

^Z

0

g ( t

¹

0 x

¹

) g ( t

²^;

t

¹

x

¹

x

²

) :::g ( t n

^;

t n

^;1

x n

^;1

x n ) dx

¹

:::dx n : If m < n , put =

^R

ⁿ _d

^;

^m , s = ( s

¹

:::s _m s _m + 1 s _m + 2 :::s _m + n

^;

m ).

Then

²^B

(

^R

_nd ), s

²

(

^R⁰

) ⁿ ,

^;1

_s () =

^;1

_s

( ) and

Z

g ( s

¹

0 x

¹

) g ( s

²^;

s

¹

x

¹

x

²

) :::g ( s _m

^;

s _m

^;1

x _m

^;1

x _m ) dx

¹

:::dx _m

=

Z

g ( s

¹

0 x

¹

) g ( s

²^;

s

¹

x

¹

x

²

) :::g ( s _n

^;

s _n

^;1

x _n

^;1

x _n ) dx

¹

:::dx _n : So, we may assume that st

²

(

^R⁰

) ⁿ and

⁰ ²^B

(

^R

_nd ).

Let ~ s

^\

^~ t =

^f

s

⁽¹⁾

:::s

⁽

k

⁾^g

=

^f

t

⁽¹⁾

:::t

⁽

k

⁾^g

, (1) <

< ( k ), (1) <

<

( k ) with

²^S

n . We show that () = (

⁰

) = ;

^R

ⁿ _d

^;

^k for some ;

²^B

(

^R

_kd ).

Let ; =

^f

( x

¹

:::x k ) : ( x

¹

:::x k x k

⁺¹

:::x n )

²

() for some ( x k

⁺¹

:::x n )

²^R

ⁿ _d

^;

^k

^g

;

⁰

=

^f

( x

¹

:::x k ) : ( x

¹

:::x k x k

⁺¹

:::x n )

²

(

⁰

) for some ( x k

⁺¹

:::x n )

²^R

ⁿ _d

^;

^k

^g

: If x = ( x

⁰

x

⁰⁰

)

²^R

_kd

^R

ⁿ _d

^;

^k , let k ( x ) = x

⁰

. Then ; = k ( ()), ;

⁰

= k ( (

⁰

)). We can show ; = ;

⁰

as follows. Let ( x

¹

:::x k )

²

;. Then ( x

¹

:::x k x k

⁺¹

:::x n )

²

() for some ( x k

⁺¹

:::x n )

²^R

ⁿ _d

^;

^k . Thus

( x

¹

:::x k x k

⁺¹

:::x n ) = ( y

⁽¹⁾

:::y

⁽

k

⁾

y

⁽

k

⁺¹⁾

:::y

⁽

n

⁾

)

with some ( y

¹

:::y k y k

⁺¹

:::y n )

²

. Since s (

^;1

_s ()) = , there exists f

²

A such that

( f ( s

¹

) :::f ( s k ) f ( s k

⁺¹

) :::f ( s n )) = ( y

¹

:::y k y k

⁺¹

:::y n ) :

(5)

Therefore

( x

¹

:::x k x k

⁺¹

:::x n ) = ( f ( s

⁽¹⁾

) :::f ( s

⁽

_k

⁾

) f ( s

⁽

_k

⁺¹⁾

) :::f ( s

⁽

_n

⁾

)) : Since

^;1

_s () =

^;1

_t (

⁰

), we also have

( f ( t

¹

) :::f ( t k ) f ( t k

⁺¹

) :::f ( t n ))

²

⁰

: Thus ( f ( t

⁽¹⁾

) :::f ( t

⁽

_k

⁾

) f ( t

⁽

_k

⁺¹⁾

) :::f ( t

⁽

_n

⁾

))

²

(

⁰

) : Since

( x

¹

:::x k ) = ( f ( s

⁽¹⁾

) :::f ( s

⁽

_k

⁾

)) = ( f ( t

⁽¹⁾

) :::f ( t

⁽

_k

⁾

))

it follows that ( x

¹

:::x k )

²

;

⁰

. This proves ;

;

⁰

. Similarly, we also have ;

⁰

;.

Thus we have ; = ;

⁰

.

Next we show that () = ;

^R

ⁿ _d

^;

^k . Let ( x

¹

:::x k x k

⁺¹

:::x n )

²

;

^R

ⁿ _d

^;

^k . Then, since ( x

¹

:::x _k )

²

;

⁰

= k ( (

⁰

)) and t is surjective from

^;1

_t (

⁰

) to

⁰

, we can nd f

²

^;1

_t (

⁰

) such that

( x

¹

:::x k ) = k (( f ( t

⁽¹⁾

) :::f ( t

⁽

_k

⁾

) f ( t

⁽

_k

⁺¹⁾

) :::f ( t

⁽

_n

⁾

))) with ( f ( t

¹

) :::f ( t k ) f ( t k

⁺¹

) :::f ( t n ))

²

⁰

:

Since s

⁽

_k

⁺¹⁾

:::s

⁽

_n

⁾²

= ^~ t , f can be chosen so that f ( s

⁽

_k

⁺¹⁾

) = x k

⁺¹

:::f ( s

⁽

_n

⁾

) = x n for any x k

⁺¹

:::x n

²^R

d . Since f

²

^;1

_s () also and s

⁽

j

⁾

= t

⁽

j

⁾

, 1

j

k ,

( x

¹

:::x k x k

⁺¹

:::x n ) = ( f ( s

⁽¹⁾

) :::f ( s

⁽

_k

⁾

) f ( s

⁽

_k

⁺¹⁾

) :::f ( s

⁽

_n

⁾

)) with ( f ( s

¹

) :::f ( s k ) f ( s k

⁺¹

) :::f ( s n ))

²

:

This implies that ;

^R

ⁿ _d

^;

^k

(). The reverse inclusion is obvious. Also we have (

⁰

) = ;

^R

ⁿ _d

^;

^k .

If ~ s

^\

^~ t =

, by the arguments above we have =

⁰

=

^R

_nd provided that

^;1

_s () =

^;1

_t (

⁰

) (

⁶

=

,

⁰⁶

=

).

We note that

Z

g ( s

¹

0 x

¹

) g ( s

²^;

s

¹

x

¹

x

²

) :::g ( s n

^;

s n

^;1

x n

^;1

x n ) dx

¹

:::dx n

=

Z

⁽⁾

g ( s

¹

0 x

^;1⁽¹⁾

) g ( s

²^;

s

¹

x

^;1⁽¹⁾

x

^;1⁽²⁾

)

:::g ( s n

^;

s n

^;1

x

^;1⁽

_n

^;1)

x

^;1⁽

_n

⁾

) dx

¹

:::dx n : We show that

(2.2)

Z

⁽⁾

g ( s

¹

0 x

^;1⁽¹⁾

) g ( s

²^;

s

¹

x

^;1⁽¹⁾

x

^;1⁽²⁾

)

:::g ( s n

^;

s n

^;1

x

^;1⁽

_n

^;1)

x

^;1⁽

_n

⁾

) dx

¹

:::dx n

=

Z

;

g ( s

⁽¹⁾

0 x

¹

) g ( s

⁽²⁾^;

s

⁽¹⁾

x

¹

x

²

) :::g ( s

⁽

_k

⁾^;

s

⁽

_k

^;1)

x _k

^;1

x _k ) dx

¹

:::dx _k : To see this we rst note that

Z

Rd^(1);1

^(1);1Y

i

⁼⁰

g ( s i

⁺¹^;

s i x

^;1⁽

i

⁾

x

^;1⁽

i

⁺¹⁾

) dx

^;1⁽¹⁾

:::dx

^;1⁽

^(1);1)

= g ( s

⁽¹⁾

0 x

¹

)

(6)

with s

⁰

= 0, x

⁰

= 0,

^;1

(0) = 0. We consider this integral when (1) 2.

Similarly,

Z

Rd⁽m^+1);⁽m^);1

⁽

m

Y^+1);1

i

⁼

⁽

m

⁾

g ( s i

⁺¹^;

s i x

^;1⁽

_i

⁾

x

^;1⁽

_i

⁺¹⁾

) dx

^;1⁽

⁽

_m

⁾⁺¹⁾

:::dx

^;1⁽

⁽

_m

^+1);1)

= g ( s

⁽

_m

⁺¹⁾^;

s

⁽

_m

⁾

x m x m

⁺¹

) for 1

m

k

^;

1, where this is considered when ( m + 1) ( m ) + 2, and

Z

Rnd^;⁽k⁾

n

^Y^;1

i

⁼

⁽

k

⁾

g ( s _i

⁺¹^;

s _i x

^;1⁽

_i

⁾

x

^;1⁽

_i

⁺¹⁾

) dx

^;1⁽

⁽

_k

⁾⁺¹⁾

:::dx

^;1⁽

_n

⁾

= 1 : This integral is considered when ( k )

n

^;

1. Let us denote by F ( x

¹

:::x _n ) the integrand of the left hand side of (2.2). Then the integral on the left hand side of (2.2) equals

Z

;

Z

Rnd^;k

F ( x

¹

:::x k x k

⁺¹

:::x n ) dx k

⁺¹

:::dx n

!

dx

¹

:::dx k : (2.3)

Collecting results above, we see that the inner integral is equal to

g ( s

⁽¹⁾

0 x

¹

) g ( s

⁽²⁾^;

s

⁽¹⁾

x

¹

x

²

) :::g ( s

⁽

_k

⁾^;

s

⁽

_k

^;1)

x k

^;1

x k ) : Using this in (2.3), we get (2.2).

In the same way, we have (2.4)

Z

⁽⁰⁾

g ( t

¹

0 x

^;1⁽¹⁾

) g ( t

²^;

t

¹

x

^;1⁽¹⁾

x

^;1⁽²⁾

)

:::g ( t n

^;

t n

^;1

x

^;1⁽

_n

^;1)

x

^;1⁽

_n

⁾

) dx

¹

:::dx n

=

Z

; 0

g ( t

⁽¹⁾

0 x

¹

) g ( t

⁽²⁾^;

t

⁽¹⁾

x

¹

x

²

) :::g ( t

⁽

k

⁾^;

t

⁽

k

^;1)

x k

^;1

x k ) dx

¹

:::dx k : From (2.2) and (2.4), it follows that

Z

⁽⁾

g ( s

¹

0 x

^;1⁽¹⁾

) g ( s

²^;

s

¹

x

^;1⁽¹⁾

x

^;1⁽²⁾

)

:::g ( s n

^;

s n

^;1

x

^;1⁽

_n

^;1)

x

^;1⁽

_n

⁾

) dx

¹

:::dx n

=

Z

⁽⁰⁾

g ( t

¹

0 x

^;1⁽¹⁾

) g ( t

²^;

t

¹

x

^;1⁽¹⁾

x

^;1⁽²⁾

)

:::g ( t n

^;

t n

^;1

x

^;1⁽

_n

^;1)

x

^;1⁽

_n

⁾

) dx

¹

:::dx n since ; = ;

⁰

, s

⁽

_j

⁾

= t

⁽

_j

⁾

for 1

j

k , and hence

Z

g ( s

¹

0 x

¹

) g ( s

²^;

s

¹

x

¹

x

²

) :::g ( s n

^;

s n

^;1

x n

^;1

x n ) dx

¹

:::dx n

=

Z

0

g ( t

¹

0 x

¹

) g ( t

²^;

t

¹

x

¹

x

²

) :::g ( t n

^;

t n

^;1

x n

^;1

x n ) dx

¹

:::dx n : This implies that is well-dened on

^G

.

The set function will extend to the Wiener measure on the sigma-algebra

generated by

^G

.

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3. Countable additivity of the Wiener measure

We rst prove countable additivity of on the algebra

^G

, which along with a result from the measure theory will imply what we want to show.

Proposition 3.1. Let be as in Lemma 2 : 9. Then is countably additive on the algebra

^G

.

It is easy to see that this follows from the next result.

Proposition 3.2. Let be as in Lemma 2 : 9. Let A n

²^G

, n = 1 2 ::: . Suppose that A n

^#

. Then lim n

^!1

( A n ) = 0.

To prove Proposition 3.2 we shall show the following.

Assertion 1. Suppose that A _n

²^G

, n = 1 2 ::: , A _n

^#

and lim n

^!1

( A _n )

⁶

= 0.

Then lim n

^!1

A _n

⁶

=

.

To prove this we assume that A n =

^;1

_t

_n

( n ) for some t n

²

(

^R⁰

) ^N

ⁿ

and n

²

B

(

^R

^N _d

ⁿ

), where N n = card~ t n . We may also assume that

~ t n =

t

⁽¹

ⁿ

⁾

t

⁽²

ⁿ

⁾

:::t

⁽

_a ⁿ

⁽

_n

⁾⁾

₂ ¹ _n ₂ ² _n ::: n ₂ ² _n ⁿ

where

^f

a ( n )

^g

is a strictly increasing sequence of positive integers, which can be seen by Lemma 2.3. Let

~ t _n =

t

⁽¹

ⁿ

⁾

t

⁽²

ⁿ

⁾

:::t

⁽

_n ⁿ

⁾

₂ ¹ _n ₂ ² _n ::: n ₂ ² _n ⁿ

~ t _n =

1 2 ⁿ ₂ ² _n ::: n ₂ ² _n ⁿ

which are subsets of ~ t n . We note that

j

¹

~ t j =

j

¹

~ t _j (3.1)

and M n = card(~ t _n )

3 n 2 ⁿ

^;1

.

To prove the assertion we need the following.

Lemma 3.3. For any > 0, there exists > 1 such that

1

X

n

⁼¹

^f

f

²

W : ! n ( f ) > 2

^;

ⁿ⁼

³

^g

<

where

! n ( f ) = max

^fj

t ( f )

^;

s ( f )

^j

: 0 < t

^;

s

2

^;

ⁿ st

²

t ^~ _n

^g

: In proving this we apply the following formula.

Lemma 3.4. ^Let ⁰ < s < t <

¹

,

²^B

(

^R

^d ). Then

^f

f

²

W : _t ( f )

^;

_s ( f )

²

^g

=

Z

(

^p

2 ( 1 t

^;

s )) ^d exp

; j

x

^j²

2( t

^;

s )

dx:

(8)

Proof. Dene a continuous function h :

^R²

_d

^!^R

^d by h ( xy ) = y

^;

x . Let

⁽

st

⁾

( f ) = ( s ( f ) t ( f )) :

Then, we note that

f

²

W : t ( f )

^;

s ( f )

²

^g

= f

²

W :

⁽

_st

⁾

( f )

²

h

^;1

()

: Therefore

^f

f

²

W : t ( f )

^;

s ( f )

²

^g

=

⁽

st

⁾

(

^;1⁽

_st

⁾

( h

^;1

()))

=

Z

h

^;1⁽⁾

g ( s 0 x

¹

) g ( t

^;

sx

¹

x

²

) dx

¹

dx

²

=

Z

R

2d

( x

²^;

x

¹

) g ( s 0 x

¹

) g ( t

^;

sx

¹

x

²

) dx

¹

dx

²

=

Z

Rd

Z

g ( s 0 x

¹

) g ( t

^;

sx

¹

x

¹

+ x

²

) dx

²

dx

¹

=

Z

Rd

g ( s 0 x

¹

) dx

¹^Z

g ( t

^;

s 0 x

²

) dx

²

=

Z

g ( t

^;

s 0 x

²

) dx

²

=

Z

(

^p

2 ( 1 t

^;

s )) ^d exp

; j

x

²^j²

2( t

^;

s )

dx

²

where

denotes the characteristic function of . Proof of Lemma 3 : 3. Note that

(3.2)

^f

f

²

W : ! n ( f ) > 2

^;

ⁿ⁼

³

^g

=

0

<t st

^;

s

²²^~

t

_n^;ⁿ

f

²

W :

^j

t ( f )

^;

s ( f )

^j

> 2

^;

ⁿ⁼

³

^g

: This in particular implies that the set on the left hand side is in

^G

t

n ^G

(see the proof of (3.3) below). Since

^f

f

²

W :

^j

t ( f )

^;

s ( f )

^j

> c

^g

=

Z

j

x

^j

>c 1

(

^p

2 ( t

^;

s )) ^d exp

; j

x

^j²

2( t

^;

s )

dx which follows from Lemma 3.4, by (3.2) we have

^f

f

²

W : ! _n ( f ) > 2

^;

ⁿ⁼

³

^g ^X

0

<t st

^;

s

²²^~

t

_n^;ⁿ

Z

j

x

^j

>

²^;ⁿ⁼³

1 (

^p

2 ( t

^;

s )) ^d exp

; j

x

^j²

2( t

^;

s )

dx

X

0

<t

^;

s

²^;ⁿ

st

²^~

t

_n

C d

^p

t

^;

s 2 ⁿ⁼

³

^;1

exp

;

²

2( t

^;

s )2

²

ⁿ⁼

³

d

(9)

where C _d = ^d

³^p⁼²^p²

and to get the last inequality we have used the estimate

Z

j

x

^j

>c 1 (

^p

2 t ) ^d exp

; j

x

^j²

2 t

dx

d

X

i

⁼¹

Z

j

v

i^j

>c=

^p

d 1

p

2 t ^exp

;

v

²

_i 2 t

dv i

d

p

2 td

p

c ^exp

;

c

²

2 dt

with c > 0. Therefore,

^f

f

²

W : ! n ( f ) > 2

^;

ⁿ⁼

³

^g ^X

0

<t

^;

s

²^;ⁿ

st

²^~

t

_n

C d 2

^;

ⁿ⁼

²

2 ⁿ⁼

³

^;1

exp

;

²

2 2

^;2

ⁿ⁼

³

d 2

^;

ⁿ

M _n

²

C d 2

^;

ⁿ⁼

⁶

^;1

exp

^;

²

2 ⁿ⁼

^3;1

d

^;1

(9 = 4) C d n

²

2

²

ⁿ

^;

ⁿ⁼

⁶

^;1

exp

^;

²

2 ⁿ⁼

^3;1

d

^;1

: Thus

1

X

n

⁼¹

^f

f

²

W : ! n ( f ) > 2

^;

ⁿ⁼

³

^g

(9 = 4) C d

1

X

n

⁼¹

n

²

2

¹¹

ⁿ⁼

⁶

exp

^;

2 ⁿ⁼

^3;1

d

^;1

!

^;1

and hence taking large enough depending on , we get the conclusion.

Choose

⁰

> 0 so that lim t

n

( A n ) >

⁰

. Applying Lemma 3 : 3, take

⁰

> 1 such that

1

X

n

⁼¹

t

n^f

f

²

W : ! n ( f ) > 2

^;

ⁿ⁼

³

⁰^g

<

⁰

= 3 : Choose a compact set

⁰

_n

n such that

t

n

(

^;1

_t

_n

( n

ⁿ

⁰

_n )) <

⁰

3

^;

ⁿ

(see 5, p. 48, Theorem 2.18]). There exist a compact set

⁰⁰

_n in

^R

^N _d

ⁿ

such that

⁰⁰

_n

⁰

_n and

^;1

_t

_n

(

⁰

_n )

^\^f

f

²

W : ! n ( f )

2

^;

ⁿ⁼

³

⁰^g

=

^;1

_t

_n

(

⁰⁰

_n ) : (3.3)

This can be shown as follows.

(10)

Let h :

^R²

_d

^!^R

^d , h ( xy ) = y

^;

x as above. Let B ( r ) =

^f

x

²^R

^d :

^j

x

^j

r

^g

. Then

f

²

W : ! _n ( f )

2

^;

ⁿ⁼

³

⁰^g

=

0

<t st

^;

s

²²^~

t

_n^;ⁿ

f

²

W : h ( _t ( f ) _s ( f ))

²

B (2

^;

ⁿ⁼

³

⁰

)

^g

=

0

<t

^;

s

²^;ⁿ

st

²^~

t

_n

f

²

W : ( t ( f ) s ( f ))

²

h

^;1

( B (2

^;

ⁿ⁼

³

⁰

))

^g

=

0

<t

^;

s

²^;ⁿ

st

²^~

t

_n

f

²

W : t

n

( f )

²

( n

⁰

st )

^g

=

^;1

_t

_n

0

B

@

0

<t st

^;

s

²²^~

t

_n^;ⁿ

( n

⁰

st )

1

C

A

for some ( n

⁰

st )

²^B

(

^R

^N _d

ⁿ

) which is closed and can be written as ( n

⁰

st ) =

h

^;1

( B (2

^;

ⁿ⁼

³

⁰

))

^R

^N _d

ⁿ^;2

with some = st

²^S

N

n

. Thus we can take

⁰⁰

_n =

⁰

_n

^\

0

B

@

0

<t

^;

s

²^;ⁿ

st

²^~

t

_n

( n

⁰

st )

1

C

A

: Next, we show

n

\

j

⁼¹

^;1

_t

_j

(

⁰⁰

_j )

⁶

=

n = 1 2 ::::

(3.4)

To prove this we rst observe that A n =

^\

ⁿ

j

⁼¹

A j =

^\

ⁿ

j

⁼¹

^;1

_t

_j

( j ) =

^\

ⁿ

j

⁼¹

^;1

_t

_j

( j

ⁿ

⁰

_j )

^;1

_t

_j

(

⁰

_j

ⁿ

⁰⁰

_j )

^;1

_t

_j

(

⁰⁰

_j )

0

@

n

j

⁼¹

^;1

_t

_j

( j

ⁿ

⁰

_j )

1

A

0

@

n

j

⁼¹

^;1

_t

_j

(

⁰

_j

ⁿ

⁰⁰

_j )

1

A

0

@

n

\

j

⁼¹

^;1

_t

_j

(

⁰⁰

_j )

1

A

: We next note that

^;1

_t

_j

(

⁰

_j

ⁿ

⁰⁰

_j ) =

^;1

_t

_j

(

⁰

_j )

ⁿ

^;1

_t

_j

(

⁰⁰

_j )

^f

f

²

W : ! j ( f ) > 2

^;

^j=

³

⁰^g

: Thus

0

@

n

\

j

⁼¹

^;1

_t

_j

(

⁰⁰

_j )

1

A

t

n

( A n )

^;^X

ⁿ

j

⁼¹

t

j

^;1

_t

_j ^;

j

ⁿ

⁰

_j

^;^X

ⁿ

j

⁼¹

t

j

^;1

_t

_j ^;

⁰

_j

ⁿ

⁰⁰

_j

>

⁰^;^X

ⁿ

j

⁼¹

⁰

3

^;

^j

^;^X

ⁿ

j

⁼¹

t

j^f

f

²

W : ! j ( f ) > 2

^;

^j=

³

⁰^g

>

⁰^;

⁰

= 2

^;

⁰

= 3 =

⁰

= 6 > 0

Construction of Brownian motion on the Wiener measure space