34.4. THE IMPLICIT CASE 1185

∩nm(ω)<r<p

{ω :

⟨Bgr−

m

∑i=1⟨Bgr,ei⟩Bei,gr−

m

∑i=1⟨Bgr,ei⟩ei

⟩= 0

}As earlier, these sets

{Em+1

p}∞

p=1 are disjoint and measurable. As before, let nm+1 (ω) = p

where ω ∈ Em+1p . Then from the algorithm, em+1 (ω)≡

X∪pEm+1p

(ω)gnm+1(ω)−∑

mi=1⟨Bgnm+1(ω),ei

⟩ei

Dm,

where Dm = ⟨B(gnm+1(ω)−∑

mi=1⟨Bgnm+1(ω),ei

⟩ei),

gnm+1(ω)−∑mi=1⟨Bgnm+1(ω),ei

⟩ei

⟩1/2

Thus the ek (ω) are all measurable into W thanks to the algorithm. However, they all havevalues in V. Thus if φ ∈V ′, let φ n→ φ in V ′ where φ n ∈W ′.

⟨φ ,ek (ω)⟩V ′,V = limn→∞⟨φ n,ek (ω)⟩V ′,V = lim

n→∞⟨φ n,ek (ω)⟩W ′,W

which is the limit of measurable functions. By the Pettis theorem, this shows ek is measur-able into V also.

To verify the assertion on right continuity, the same kind of argument holds. We sup-press the dependence on ω . Consider first e1. Begin by considering n1 (t)

E1k ≡ {t : ⟨B(t)gk,gk⟩ ̸= 0}∩∩ j<k

{t :⟨B(t)g j,g j

⟩= 0}

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

k aredisjoint. Then

n1 (t)≡{

1 if t /∈ ∪kE1k = /0

k if t ∈ E1k

If t ∈ E1k , then from the definition, ⟨B(t)gk,gk⟩ ̸= 0 and k is the first index for which this

is nonzero. Let tl ↓ t. Then by continuity, for all l large enough, ⟨B(tl)gk,gk⟩ ̸= 0. What of⟨B(tl)g j,g j

⟩for j < k? By assumption,⟨

B′ (t)g j,g j⟩≤ kg j (t)

⟨B(t)g j,g j

⟩and so, letting Kg j (t) =

∫ t0 kg j (s)ds,

ddt

(e−Kg j (t)

⟨B(t)g j,g j

⟩)≤ 0

e−Kg j (tl)⟨B(tl)g j,g j

⟩≤ e−Kg j (t)

⟨B(t)g j,g j

⟩= 0

Thus one obtains right continuity of t→ n1 (t) and for E1k , there is an interval [t, t+δ )⊆E1

k .From the algorithm,

e1 (t)≡X∪kE1k(t)

gn1(t)⟨Bgn1(t),gn1(t)

⟩1/2