59.1. RANDOM VARIABLES AND INDEPENDENCE 1859
provided N is large enough. Thus it follows µ is inner regular on ∪∞i=1Fi. Why is it outer
regular? Let Vi ⊇ Fi such that µ (Fi)+ ε/2i > µ (Vi) and
µ (∪∞i=1Fi) =
∞
∑i=1
µ (Fi)>−ε +∞
∑i=1
µ (Vi)≥−ε +µ (∪∞i=1Vi)
which shows µ is outer regular on ∪∞i=1Fi. It follows F contains the π system consisting of
open sets and is closed with respect to countable disjoint unions and complements, and soby the Lemma on π systems, Lemma 12.12.3, F contains σ (τ) where τ is the set of opensets. Hence F contains the Borel sets and is itself a subset of the Borel sets by definition.Therefore, F = B (X) .
One can say more if the metric space is complete and separable. In fact in this case theabove definition of inner regularity can be shown to imply the usual one.
Lemma 59.1.9 Let µ be a finite measure on a σ algebra containing B (X) , the Borel setsof X , a separable complete metric space. Then if C is a closed set,
µ (C) = sup{µ (K) : K ⊆C and K is compact.}
It follows that for a finite measure on B (X) where X is a Polish space, µ is inner regularin the sense that for all F ∈B (X) ,
µ (F) = sup{µ (K) : K ⊆ F and K is compact}
Proof: Let {ak} be a countable dense subset of C. Thus ∪∞k=1B
(ak,
1n
)⊇C. Therefore,
there exists mn such that
µ
(C \∪mn
k=1B(
ak,1n
))≡ µ (C \Cn)<
ε
2n .
Now let K =C∩ (∩∞n=1Cn) . Then K is a subset of Cn for each n and so for each ε > 0 there
exists an ε net for K since Cn has a 1/n net, namely a1, · · · ,amn . Since K is closed, it iscomplete and so it is also compact since it is complete and totally bounded, Theorem 7.6.5.Now
µ (C \K)≤ µ (∪∞n=1 (C \Cn))<
∞
∑n=1
ε
2n = ε.
Thus µ (C) can be approximated by µ (K) for K a compact subset of C. The last claimfollows from Lemma 59.1.8.
Definition 59.1.10 A measurable function X : (Ω,F ,µ)→ Z a topological space is calleda random variable when µ (Ω) = 1. For such a random variable, one can define a distri-bution measure λ X on the Borel sets of Z as follows.
λ X (G)≡ µ(X−1 (G)
)This is a well defined measure on the Borel sets of Z because it makes sense for every Gopen and G ≡
{G⊆ Z : X−1 (G) ∈F
}is a σ algebra which contains the open sets, hence
the Borel sets. Such a measurable function is also called a random vector.