60.7. THE SUBMARTINGALE CONVERGENCE THEOREM 1975

Proof: Let Fn ≡ σ (X1, · · · ,Xn) . Consider Sn ≡ ∑nk=1 Xk.

E (Sn+1|Fn) = Sn +E (Xn+1|Fn) .

Letting A ∈Fn it follows from independence that∫A

E (Xn+1|Fn)dP ≡∫

AXn+1dP

=∫

Ω

XAXn+1dP

= P(A)∫

Ω

Xn+1dP = 0

and so E (Xn+1|Fn) = 0. Therefore, {Sn} is a martingale. Now using independence again,

E (|Sn|)≤ E(∣∣S2

n∣∣)= n

∑k=1

E(X2

k)≤

∞

∑k=1

E(X2

k)< ∞

and so {Sn} is an L1 bounded martingale. Therefore, it converges a.e. and this proves thetheorem.

Corollary 60.7.3 Let {Xk} be a sequence of independent real valued random variablessuch that E (|Xk|)< ∞,E (Xk) = mk, and

∞

∑k=1

E(|Xk−mk|2

)< ∞.

Then ∑∞k=1 (Xk−mk) converges a.e.

This can be extended to the case where the random variables have values in a separableHilbert space.

Theorem 60.7.4 Let {Xk} be a sequence of independent H valued random variables whereH is a real separable Hilbert space such that E (|Xk|H)< ∞,E (Xk) = 0, and

∞

∑k=1

E(|Xk|2H

)< ∞.

Then ∑∞k=1 Xk converges a.e.

Proof: Let {ek} be an orthonormal basis for H. Then {(Xn,ek)H}∞

n=1 are real valued,independent, and their mean equals 0. Also

∞

∑n=1

E(∣∣∣(Xn,ek)

2H

∣∣∣)≤ ∞

∑n=1

E(|Xn|2H

)< ∞

and so from Theorem 60.7.2, the series,∞

∑n=1

(Xn,ek)H