722 CHAPTER 26. INDEPENDENCE

Thus K is contained in

G ≡ {A ∈ σ (X i : i ∈ I1) : P(A∩B) = P(A)P(B)} .

Now G is closed with respect to complements and countable disjoint unions. Here is why:If each Ai ∈ G and the Ai are disjoint,

P((∪∞i=1Ai)∩B) = P(∪∞

i=1 (Ai∩B))

= ∑i

P(Ai∩B) = ∑i

P(Ai)P(B)

= P(B)∑i

P(Ai) = P(B)P(∪∞i=1Ai)

If A ∈ G , P(AC ∩B

)+P(A∩B) = P(B) and so

P(AC ∩B

)= P(B)−P(A∩B) = P(B)−P(A)P(B)

= P(B)(1−P(A)) = P(B)P(AC) .

Therefore, from the lemma on π systems, Lemma 9.3.2 on Page 243, it follows

σ (X i : i ∈ I1)⊇ G ⊇ σ (K ) = σ (X i : i ∈ I1) .■

Definition 26.4.5 When X is a random variable with values in a Banach space Z,the notation E (X) means

∫Z X (ω)dP where the latter is the Bochner integral. I will use

this notation whenever convenient. E (X) is called the expectation of X or the expectedvalue of X. I will sometimes also use E as the name of a Banach space, but it should beclear from the context which is meant.

Recall Lemma 10.16.1

Lemma 26.4.6 Let f ,g be nonnegative measurable nonnegative functions on a measurespace (Ω,µ). Then

∫f gdµ =

∫∞

0∫[g>t] f dµdt =

∫∞

0∫

∞

0 µ ([ f > s]∩ [g > t])dsdt.

Corollary 26.4.7 If { fi}mi=1 are nonnegative measurable functions, it follows from in-

duction that ∫ m

∏i=1

fidµ =∫

∞

0· · ·∫

∞

0µ (∩m

i=1 [ fi > ti])dt1 · · ·dtm

Proof: The case of n = 2 was just done. So suppose true for n ≥ 2. Then from thiscase and induction,

∫ n+1

∏i=1

fidµ =∫

∞

0

∫[ fn+1>tn+1]

n

∏i=1

fidµdtn+1 =∫

∞

0

∫ n

∏i=1

X[ fn+1>tn+1] fidµdtn+1

=∫

∞

0

∫∞

0· · ·∫

∞

0µ(∩n

i=1[X[ fn+1>tn+1] fi > ti

])dt1 · · ·dtndtn+1

=∫

∞

0· · ·∫

∞

0µ (∩n

i=1 [ fn+1 > tn+1]∩ [ fi > ti])dt1 · · ·dtn+1

=∫

∞

0· · ·∫

∞

0µ(∩n+1

i=1 [ fi > ti])

dt1 · · ·dtn+1 ■