394 CHAPTER 14. SOME EXTENSION THEOREMS

Then P0 is well defined because of the consistency condition on the measures νJ . P0 isclearly finitely additive because the νJ are measures and one can pick J as large as desiredto include all t where there may be something other than Mt . Also, from the definition,

P0 (Ω)≡ P0

(∏t∈I

Mt

)= ν t1 (Mt1) = 1.

Next I will show P0 is a finite measure on E . After this it is only a matter of using theCaratheodory extension theorem to get the existence of the desired probability measure P.

Claim: Suppose En is in E and suppose En ↓ /0. Then P0 (En) ↓ 0.Proof of the claim: If not, there exists a sequence such that although En ↓ /0,P0 (En) ↓

ε > 0. Let En ∈ EJn . Thus it is a finite disjoint union of sets of RJn . By regularity of themeasures νJ , there exists a compact set KJn ⊆ En such that

νJn (πJn (KJn))+ε

2n+2 > νJn (πJn (En))

Thus

P0 (KJn)+ε

2n+2 ≡ νJn (πJn (KJn))+ε

2n+2

> νJn (πJn (En))≡ P0 (En)

The interesting thing about these KJn is: they have the finite intersection property. Here iswhy.

ε ≤ P0(∩m

k=1KJk

)+P0

(Em \∩m

k=1KJk

)≤ P0

(∩m

k=1KJk

)+P0

(∪m

k=1Ek \KJk

)< P0

(∩m

k=1KJk

)+

∞

∑k=1

ε

2k+2 < P0(∩m

k=1KJk

)+ ε/2,

and so P0(∩m

k=1KJk

)> ε/2. Now this yields a contradiction, because this finite intersection

property implies the intersection of all the KJk is nonempty, contradicting En ↓ /0 since eachKJn is contained in En.

With the claim, it follows P0 is a measure on E . Here is why: If E = ∪∞k=1Ek where

E,Ek ∈ E , then (E\∪nk=1Ek) ↓ /0 and so

P0 (∪nk=1Ek)→ P0 (E) .

Hence if the Ek are disjoint, P0(∪n

k=1Ek)= ∑

nk=1 P0 (Ek)→ P0 (E) . Thus for disjoint Ek

having ∪kEk = E ∈ E ,

P0 (∪∞k=1Ek) =

∞

∑k=1

P0 (Ek) .

Now to conclude the proof, apply the Caratheodory extension theorem to obtain Pa probability measure which extends P0 to a σ algebra which contains σ (E ) the sigmaalgebra generated by E with P = P0 on E . Thus for EJ ∈ E , P(EJ) = P0 (EJ) = νJ (PJE j) .