61.3. TIGHT MEASURES 1991

Lemma 61.3.2 Let E be a separable complete metric space and let Λ be a set of Borelprobability measures. Then Λ is tight if and only if for every ε > 0 and r > 0 there exists afinite collection of balls, {B(ai,r)}m

i=1 such that

µ

(∪m

i=1B(ai,r))> 1− ε

for every µ ∈ Λ.

Proof: If Λ is tight, then there exists a compact set, Kε such that

µ (Kε)> 1− ε

for all µ ∈ Λ. Then consider the open cover, {B(x,r) : x ∈ Kε} . Finitely many of thesecover Kε and this yields the above condition.

Now suppose the above condition and let

Cn ≡ ∪mni=1B(an

i ,1/n)

satisfy µ (Cn) > 1− ε/2n for all µ ∈ Λ. Then let Kε ≡ ∩∞n=1Cn. This set Kε is a compact

set because it is a closed subset of a complete metric space and is therefore complete, andit is also totally bounded by construction. For µ ∈ Λ,

µ(KC

ε

)= µ

(∪nCC

n)≤∑

nµ(CC

n)< ∑

n

ε

2n = ε

Therefore, Λ is tight.Prokhorov’s theorem is an important result which also involves tightness. In order to

give a proof of this important theorem, it is necessary to consider some simple results fromtopology which are interesting for their own sake.

Theorem 61.3.3 Let H be a compact metric space. Then there exists a compact subset of[0,1] ,K and a continuous function, θ which maps K onto H.

Proof: Without loss of generality, it can be assumed H is an infinite set since otherwisethe conclusion is trivial. You could pick finitely many points of [0,1] for K.

Since H is compact, it is totally bounded. Therefore, there exists a 1 net for H {hi}m1i=1 .

Letting H1i ≡ B(hi,1), it follows H1

i is also a compact metric space and so there exists a 1/2

net for each H1i ,{

hij

}mi

j=1. Then taking the intersection of B

(hi

j,12

)with H1

i to obtain sets

denoted by H2j and continuing this way, one can obtain compact subsets of H,

{H i

k

}which

satisfies: each H ij is contained in some H i−1

k , each H ij is compact with diameter less than

i−1, each H ij is the union of sets of the form H i+1

k which are contained in it. Denoting by{H i

j

}mi

j=1those sets corresponding to a superscript of i, it can also be assumed mi < mi+1.

If this is not so, simply add in another point to the i−1 net. Now let{

Iij

}mi

j=1be disjoint

closed intervals in [0,1] each of length no longer than 2−mi which have the property that