1184 CHAPTER 34. GELFAND TRIPLES AND RELATED STUFF

+

∥∥∥∥∥Bxk−∞

∑i=1⟨Bxk,ei⟩Bei

∥∥∥∥∥+∥∥∥∥∥ ∞

∑i=1⟨Bxk,ei⟩Bei−

∞

∑i=1⟨Bx,ei⟩Bei

∥∥∥∥∥≤ ε +

∥∥∥∥∥ ∞

∑i=1⟨B(xk− x) ,ei⟩Bei

∥∥∥∥∥ ,the term ∥∥∥∥∥Bxk−

∞

∑i=1⟨Bxk,ei⟩Bei

∥∥∥∥∥equaling 0 by 34.4.18. From 34.4.21 and 34.4.19,

≤ ε +∥B∥1/2

(∞

∑i=1|⟨B(xk− x) ,ei⟩|2

)1/2

≤ ε +∥B∥1/2 ⟨B(xk− x) ,xk− x⟩1/2 < 2ε

whenever k is large enough, the second inequality being implied by 34.4.19. Therefore,

Bx =∞

∑i=1⟨Bx,ei⟩Bei

in W ′. It follows that

⟨Bx,x⟩= limk→∞

⟨k

∑i=1⟨Bx,ei⟩Bei,x

⟩= lim

k→∞

k

∑i=1|⟨Bx,ei⟩|2 ≡

∞

∑i=1|⟨Bx,ei⟩|2

Now consider the measurability assertion on the ei. Consider first e1. Begin by consid-ering n1 (ω)

E1k ≡ {ω : ⟨B(ω)gk,gk⟩ ̸= 0}∩∩ j<k

{ω :⟨B(ω)g j,g j

⟩= 0}

As explained above, B(ω) = 0, if and only if E1k = /0 for all k. Also note that these E1

k aredisjoint and F measurable. Then

n1 (ω)≡{

1 if ω /∈ ∪kE1k = /0

k if ω ∈ E1k

Then n1 (ω) is clearly measurable because it is constant on measurable sets. Then from thealgorithm,

e1 (ω)≡X∪kE1k(ω)

gn1(ω)⟨Bgn1(ω),gn1(ω)

⟩1/2

Thus e1 (ω)= 0 if ω /∈∪kE1k . Also e1 (ω) is measurable because ω→ n1 (ω) is measurable.

Thus e1 has constant values on measurable sets. So suppose ni (ω) is measurable for i≤m.Then define Em+1

p ≡{ω :

⟨Bgp−

m

∑i=1

⟨Bgp,ei

⟩Bei,gp−

m

∑i=1

⟨Bgp,ei

⟩ei

⟩̸= 0

}∩{ω : nm (ω)< p}