This material follows [?], [?] and [?]. More can be found on this subject in these
references. Here H will be a separable real Hilbert space.
Definition 60.9.1Cylinder sets in H are of the form
{x ∈ H : ((x,e1) ,⋅⋅⋅,(x,en)) ∈ F }
where F ∈ℬ
(ℝn)
, the Borel sets of ℝ^{n}and
{ek}
are given vectors in H. Denote thiscollection of cylinder sets as C.
Proposition 60.9.2The cylinder sets form an algebra of sets.
Proof:First note the complement of a cylinder set is a cylinder set.
C
{{x ∈ H : ((x,e1),⋅⋅⋅,(x,en)) ∈ F} }
= x ∈ H : ((x,e1),⋅⋅⋅,(x,en)) ∈ FC .
Now consider the intersection of two of these cylinder sets. Let the cylinder sets be
{x ∈ H : ((x,e1),⋅⋅⋅,(x,en)) ∈ E},
{x ∈ H : ((x,f ),⋅⋅⋅,(x,f )) ∈ F }
1 m
The first of these equals
{x ∈ H : ((x,e ),⋅⋅⋅,(x,e ),(x,f ),⋅⋅⋅,(x,f )) ∈ E × ℝm }
1 n 1 m
and the second equals
{x ∈ H : ((x,e1),⋅⋅⋅,(x,en),(x,f1),⋅⋅⋅,(x,fm )) ∈ ℝn × F }
Therefore, their intersection equals
{x ∈ H : (m(x,e1)n,⋅⋅⋅,(x,en),(x,f1) ,⋅⋅⋅,(x,fm ))
∈ E × ℝ ∩ ℝ × F} ,
a cylinder set.
Now it is clear the whole of H and ∅ are cylinder sets given by
{x ∈ H : (e,x) ∈ ℝ }, {x ∈ H : (e,x) ∈ ∅}
respectively and so if C_{1},C_{2} are two cylinder sets,
C1 ∖ C2 ≡ C1 ∩ CC2 ,
which was just shown to be a cylinder set. Hence
C ∪ C = (CC ∩ CC)C ,
1 2 1 2
a cylinder set. This proves the proposition.
It is good to have a more geometrical description of cylinder sets. Letting A be
a cylinder set as just described, let P denote the orthogonal projection onto
span
(e1,⋅⋅⋅,en )
. Also let α : PH → ℝ^{n} be given by
α (x ) ≡ ((x,e1),⋅⋅⋅,(x,en)).
This is continuous but might not be one to one if the e_{i} are not a basis for example. Then
consider α^{−1}
(F)
, those x ∈ PH such that
((x,e1),⋅⋅⋅,(x,en)) ∈ F.
For any x ∈ H,
((I − P )x,ek) = 0
for each k and so
((x,e1),⋅⋅⋅,(x,en)) = ((Px,e1),⋅⋅⋅,(P x,en)) ∈ F
Thus Px ∈ α^{−1}
(F )
, which is a Borel set of PH and
x = Px + (I − P)x
so the cylinder set is contained in
−1 ⊥
α (F) +(P H)
which is of the form
(Borel set of P H )+ (P H )⊥
On the other hand, consider a set of the form
⊥
G + (PH )
where G is a Borel set in PH. There is a basis for PH consisting of a subset of
{e1 ⋅⋅⋅,en}
. For simplicity, suppose it is
{e1⋅⋅⋅,ek}
. Then let α_{1} : PH → ℝ^{k} be given
by
α1(x) ≡ ((x,e1),⋅⋅⋅,(x,ek))
Thus α is a homeomorphism of PH and ℝ^{k} so α_{1}
(G )
is a Borel set of ℝ^{k}.
Now
α −1(α1(G )× ℝn−k) = G
and α_{1}
(G )
× ℝ^{n−k} is a Borel set of ℝ^{n}. This has proved the following important
Proposition illustrated by the following picture.
PICT
Proposition 60.9.3The cylinder sets are sets of the form
⊥
B + M
where M is a finite dimensional subspace and B is a Borel subset of M. Furthermore, thecollection of cylinder sets is an algebra.
Lemma 60.9.4σ
(C)
, the smallest σ algebra containing C, contains the Borel setsof H,ℬ
(H )
.
Proof: It follows from the definition of these cylinder sets that if f_{i}
(x)
≡
(x,ei)
, so
that f_{i}∈ H^{′}, then with respect to σ
(C)
, each f_{i} is measurable. It follows that every
linear combination of the f_{i} is also measurable with respect to σ
(C)
. However, this set of
linear combinations is dense in H^{′} and so the conclusion of the lemma follows from
Lemma 58.4.2 on Page 6264. This proves the lemma.
Also note that the mapping
x → ((x,e1),⋅⋅⋅,(x,en))
is a σ
(C)
measurable map. Restricting it to span
(e1,⋅⋅⋅,en)
, it is Borel measurable.
Next is a definition of a Gaussian measure defined on C. While this is what it is called, it
is a fake measure in general because it cannot be extended to a countably additive
measure on σ
(C)
. This will be shown below.
Definition 60.9.5Let Q ∈ℒ
(H,H )
be self adjoint and satisfy
(Qx, x) > 0
for all x ∈ H,x≠0. Define ν on the cylinder sets, C by the following rule. For
{ek}
_{k=1}^{n}an orthonormal set in H,
ν ({x ∈ H : ((x,e1),⋅⋅⋅,(∫x,en)) ∈ F})
≡ ---------1---------- e− 12t∗θ∗Q−1θtdt.
(2π)n∕2(det(θ∗Q θ))1∕2 F
where here
∑n
θt ≡ tiei.
i=1
Note that the cylinder set is of the form
θF + span(e1,⋅⋅⋅,en)⊥ .
Thus if B + M^{⊥}is a typical cylinder set, choose an orthonormal basis for M,
{ek}
_{k=1}^{n}and do the above definition with F = θ^{−1}B.
To see this last claim which is like what was done earlier, let
((x,e1),⋅⋅⋅,(x,en)) ∈ F.
Then θ
((x,e1),⋅⋅⋅,(x,en))
= ∑_{i}
(x,ei)
e_{i} = Px and so
x = x − Px + Px = x− P x+ θ((x,e1),⋅⋅⋅,(x,en ))
∈ θF + span (e1,⋅⋅⋅,en)⊥
Thus
{x ∈ H : ((x,e1),⋅⋅⋅,(x,en)) ∈ F } ⊆ θF + span(e1,⋅⋅⋅,en)⊥
To see the other inclusion, if t ∈ F and y ∈span
(e ,⋅⋅⋅,e )
1 n
^{⊥}, then if x = θt, it
follows
t= (x,e )
i i
and so
((x,e1),⋅⋅⋅,(x,en))
∈ F. But
(y,ek)
= 0 for all k and so x + y is in
{x ∈ H : ((x,e1),⋅⋅⋅,(x,en)) ∈ F} .
Lemma 60.9.6The above definition is well defined.
Proof:Let
{f }
k
be another orthonormal set such that for F,G Borel sets in ℝ^{n},
A = {x ∈ H : ((x,e1),⋅⋅⋅,(x,en)) ∈ F}
= {x ∈ H : ((x,f1),⋅⋅⋅,(x,fn)) ∈ G}
I need to verify ν
(A)
is the same using either
{fk}
or
{ek}
. Let a ∈ G. Then
∑n
x ≡ aifi ∈ A
i=1
because
(x,fk)
= a_{k}. Therefore, for this x it is also true that
((x,e1),⋅⋅⋅(x,en))
∈ F. In
other words for a ∈ G,
( n n )
∑ (e ,f)a ,⋅⋅⋅,∑ (e ,f) a ∈ F
i=1 1 i i i=1 n i i
Let L ∈ℒ
n n
(ℝ ,ℝ )
be defined by
∑
La ≡ Ljiai,Lji ≡ (ej,fi).
i
Since the
{ej}
and
{fk}
are orthonormal, this mapping is unitary. Also this has shown
that
LG ⊆ F.
Similarly
L ∗F ⊆ G
where L^{∗} has the ij^{th} entry L_{ij}^{∗} =
(f,e )
i j
as above and L^{∗} is the inverse of L because L
is unitary. Thus
F = L (L∗(F)) ⊆ L (G) ⊆ F
showing that LG = F and L^{∗}F = G.
Now let θ_{e}t ≡∑_{i}t_{i}e_{i} with θ_{f} defined similarly. Then the definition of ν
(A)
corresponding to
{ei}
is
1 ∫ − 1t∗θ∗Q−1θ t
ν(A) ≡ ----n∕2------∗----1∕2 e 2 e edt
(2π) (det(θeQθe)) F
Now change the variables letting t = Ls where s ∈ G.
( ) ( ) ( )
det θ∗fθe det(θ∗eθf) = det θ∗fθeθ∗eθf = det θ∗fθf = 1.
This proves the lemma.
It would be natural to try to extend ν to the σ algebra determined by C and obtain a
measure defined on this σ algebra. However, this is always impossible in the case where
Q = I.
Proposition 60.9.7For Q = I, ν cannot be extended to a measure defined onσ
(C )
whenever H is infinite dimensional.
Proof:Let
{en}
be a complete orthonormal set of vectors in H. Then first note that
H is a cylinder set.
H = {x ∈ H : (x,e1) ∈ ℝ}
and so
1 ∫ − 1t2
ν(H ) = √2-π e 2 dt = 1.
ℝ
However, H is also equal to the countable union of the sets,
An ≡ {x ∈ H : ((x,e1)H ,⋅⋅⋅,(x,ean)H) ∈ B (0,n)}
where a_{n}→∞.
1 ∫ − 1|t|2
ν (An) ≡ (√---)an e 2 dt
2π ∫B(n0,n)∫ n
≤ (√--1)a- ⋅⋅⋅ e−|t|2∕2dt1⋅⋅⋅dta
2π n −n − n n
( ∫n −x2∕2 )an
= -−ne√-----dx
2π
Now pick a_{n} so large that the above is smaller than 1∕2^{n+1}. This can be done because for
no matter what choice of n,