60.7. THE SUBMARTINGALE CONVERGENCE THEOREM 1977

and also for these ω,

∑n

∑k(Xn (ω) ,ek)

2H < ∞.

It follows from the estimate 60.7.18 that for ω not on a suitable set of measure zero, S (ω)defined by 60.7.17,

S (ω)≡∞

∑k=1

Yk (ω)ek

makes sense. Thus for these ω

S (ω) = ∑l(S (ω) ,el)el = ∑

lYl (ω)el ≡∑

l∑n(Xn (ω) ,el)H el

= ∑n

∑l(Xn (ω) ,el)el = ∑

nXn (ω) .

This proves the theorem.Now with this theorem, here is a strong law of large numbers.

Theorem 60.7.5 Suppose {Xk} are independent random variables and E (|Xk|) < ∞ foreach k and E (Xk) = mk. Suppose also

∞

∑j=1

1j2 E

(∣∣X j−m j∣∣2)< ∞. (60.7.19)

Then

limn→∞

1n

n

∑j=1

(X j−m j) = 0 a.e.

Proof: Consider the sum∞

∑j=1

X j−m j

j.

This sum converges a.e. because of 60.7.19 and Theorem 60.7.4 applied to the randomvectors

{X j−m j

j

}. Therefore, from Lemma 59.7.4 it follows that for a.e. ω,

limn→∞

1n

n

∑j=1

(X j (ω)−m j) = 0

This proves the theorem.The next corollary is often called the strong law of large numbers. It follows immedi-

ately from the above theorem.

Corollary 60.7.6 Suppose{

X j}∞

j=1 are independent random vectors, λ Xi = λ X j for alli, j having mean m and variance equal to

σ2 ≡

∫Ω

∣∣X j−m∣∣2 dP < ∞.

Then for a.e. ω ∈Ω

limn→∞

1n

n

∑j=1

X j (ω) = m