77.2. SOME FUNDAMENTAL THEOREMS 2607

and that γ is a P measurable R∞ valued function. Since γ (ω) ∈ Γ(ω) , it follows that foreach n,γ (ω) ∈ Γn (ω) . Therefore, there exists jn ≥ n such that for each ω ,

d (f(XNC (ω)u jn (·,ω)) ,γ (ω))< 2−n.

Therefore, for a suitable subsequence{

un(ω) (·,ω)}

, we have

γ (ω) = limn(ω)→∞

f(XNC (ω)un(ω) (·,ω)

),

for each ω . In particular, for each k

γk (t,ω) = limn(ω)→∞

f(XNC (ω)un(ω) (t,ω)

)k

= limn(ω)→∞

mk

∫ t

lmk (t)

⟨φ rk

,XNC (ω)un(ω) (s,ω)⟩

Vds, (77.2.5)

for each t.Note that it is not clear that (t,ω)→ f

(XNC (ω)un(ω) (t,ω)

)is P measurable, al-

though (t,ω)→ γ (t,ω) is P measurable.Now here is the proof of the theorem.

Proof of Theorem 77.2.10 By assumption, there exists a further subsequence, still denotedby n(ω), such that, the weak limit

limn(ω)→∞

XNC (ω)un(ω) (·,ω) = v(·,ω)

exists in V . Then,

mk

∫ t

lmk (t)

⟨φ rk

,v(s,ω)⟩

Vds

= limn(ω)→∞

mk

∫ t

lmk (t)

⟨φ rk

,XNC (ω)un(ω) (s,ω)⟩

Vds

= γk (t,ω) , product measurable.

Letting φ ∈ D be given, there exists a subsequence, denoted by k, such that mk → ∞ andφ rk

= φ . Recall(

mk,φ rk

)denoted an enumeration of the pairs (m,φ)∈N×D . For a given

φ ∈D denote this sequence by mφ . Thus we have measurability of

(t,ω)→ mφ

∫ t

lmφ(t)⟨φ ,v(s,ω)⟩V ds

for each φ ∈D .Now we will be a little more careful about the countable set D . Iterate the following.

Let φ 1 ̸= 0. Let F denote linearly independent subsets of V ′ which contain φ 1 such that theelements are further apart than 1/5. Let C denote a maximal chain. Thus ∪C is also in F .If W := span∪C fails to be all of V ′, then there would exist ψ /∈W such that the distance of