1976 CHAPTER 60. CONDITIONAL, MARTINGALES

converges a.e. Therefore, there exists a set of measure zero such that for ω not in this set,∑n (Xn (ω) ,ek)H converges for each k. For ω not in this exceptional set, define

Yk (ω)≡∞

∑n=1

(Xn (ω) ,ek)H

Next define

S (ω)≡∞

∑k=1

Yk (ω)ek. (60.7.17)

Of course it is not clear this even makes sense. I need to show ∑∞k=1 |Yk (ω)|2 < ∞. Using

the independence of the Xn

E(|Yk|2

)= E

( ∞

∑n=1

(Xn,ek)H

)2

= E

((∞

∑n=1

∞

∑m=1

(Xn,ek)H (Xm,ek)H

))

≤ lim infN→∞

E

((N

∑n=1

N

∑m=1

(Xn,ek)H (Xm,ek)H

))

= lim infN→∞

E

(N

∑n=1

(Xn,ek)2H

)

=∞

∑n=1

E((Xn,ek)

2H

)Hence from the above,

E

(∑k|Yk|2

)= ∑

kE(|Yk|2

)≤∑

k∑n

E((Xn,ek)

2H

)and by the monotone convergence theorem or Fubini’s theorem,

= E

(∑k

∑n(Xn,ek)

2H

)= E

(∑n

∑k(Xn,ek)

2H

)

= E(

∑n|Xn|2H

)= ∑

nE(|Xn|2H

)< ∞ (60.7.18)

Therefore, for ω off a set of measure zero, and for

Yk (ω)≡∞

∑n=1

(Xn (ω) ,ek)H ,

∑k|Yk (ω)|2 < ∞