Now here is a definition of what it means for a set of measures to be tight.
Definition 60.3.1Let Λ be a set of probability measures defined on the Borel sets of atopological space. Then Λ is “tight”if for all ε > 0 there exists a compact set, K_{ε}suchthat
μ([x ∕∈ K ]) < ε
ε
for all μ ∈ Λ.
Lemma 60.2.3 implies a single probability measure on the Borel sets of a separable
metric space is tight. The proof of that lemma generalizes slightly to give a simple
criterion for a set of measures to be tight.
Lemma 60.3.2Let 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 existsa finite collection of balls,
{B(ai,r)}
_{i=1}^{m}such that
( -------)
μ ∪mi=1B (ai,r) > 1− ε
for every μ ∈ Λ.
Proof: If Λ is tight, then there exists a compact set, K_{ε} such that
μ (K ε) > 1− ε
for all μ ∈ Λ. Then consider the open cover,
{B(x,r) : x ∈ Kε}
. Finitely many of these
cover K_{ε} and this yields the above condition.
Now suppose the above condition and let
----------
Cn ≡ ∪min=1B (ani,1∕n)
satisfy μ
(Cn )
> 1 − ε∕2^{n} for all μ ∈ Λ. Then let K_{ε}≡∩_{n=1}^{∞}C_{n}. This set
K_{ε} is a compact set because it is a closed subset of a complete metric space
and is therefore complete, and it is also totally bounded by construction. For
μ ∈ Λ,
∑ ∑
μ (KCε ) = μ(∪nCCn) ≤ μ(CCn) < ε-= ε
n n 2n
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
from topology which are interesting for their own sake.
Theorem 60.3.3Let H be a compact metric space. Then there exists a compactsubset 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
otherwise the 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}
_{i=1}^{m1}. Letting H_{i}^{1}≡B
(hi,1)
, it follows H_{i}^{1} is also a compact metric space and so
there exists a 1/2 net for each H_{i}^{1},
{ i}
hj
_{j=1}^{mi}. Then taking the intersection of B
( i 1)
hj, 2
with H_{i}^{1} to obtain sets denoted by H_{j}^{2} and continuing this way, one can obtain compact
subsets of H,
{ i}
H k
which satisfies: each H_{j}^{i} is contained in some H_{k}^{i−1}, each
H_{j}^{i} is compact with diameter less than i^{−1}, each H_{j}^{i} is the union of sets of
the form H_{k}^{i+1} which are contained in it. Denoting by
{ i}
H j
_{j=1}^{mi} those sets
corresponding to a superscript of i, it can also be assumed m_{i}< m_{i+1}. If this is not so,
simply add in another point to the i^{−1} net. Now let
{ i}
Ij
_{j=1}^{mi} be disjoint
closed intervals in
[0,1]
each of length no longer than 2^{−mi} which have the
property that I_{j}^{i} is contained in I_{k}^{i−1} for some k. Letting K_{i}≡∪_{j=1}^{mi}I_{j}^{i},
it follows K_{i} is a sequence of nested compact sets. Let K = ∩_{i=1}^{∞}K_{i}. Then
each x ∈ K is the intersection of a unique sequence of these closed intervals,
{ k}
Ijk
_{k=1}^{∞}. Define θx ≡∩_{k=1}^{∞}H_{jk}^{k}. Since the diameters of the H_{j}^{i} converge to 0 as
i →∞, this function is well defined. It is continuous because if x_{n}→ x, then
ultimately x_{n} and x are both in I_{jk}^{k}, the k^{th} closed interval in the sequence
whose intersection is x. Hence, d
(θxn,θx )
≤ diameter(H_{jk}^{k}) ≤ 1∕k. To see the
map is onto, let h ∈ H. Then from the construction, there exists a sequence
{ k }
Hjk
_{k=1}^{∞} of the above sets whose intersection equals h. Then θ
( ∞ k)
∩i=1Ijk
= h.
■
Note θ is maybe not one to one.
As an important corollary, it follows that the continuous functions defined on any
compact metric space is separable.
Corollary 60.3.4Let H be a compact metric space and let C
(H )
denote thecontinuous functions defined on H with the usual norm,
||f||∞ ≡ max {|f (x)| : x ∈ H }
Then C
(H )
is separable.
Proof: The proof is by contradiction. Suppose C
(H )
is not separable. Let ℋ_{k}
denote a maximal collection of functions of C
(H)
with the property that if
f,g ∈ℋ_{k}, then
||f − g||
_{∞}≥ 1∕k. The existence of such a maximal collection of
functions is a consequence of a simple use of the Hausdorff maximality theorem.
Then ∪_{k=1}^{∞}ℋ_{k} is dense. Therefore, it cannot be countable by the assumption
that C
(H )
is not separable. It follows that for some k,ℋ_{k} is uncountable. Now
by Theorem 60.3.3 there exists a continuous function, θ defined on a compact
subset, K of
[0,1]
which maps K onto H. Now consider the functions defined on
K
Gk ≡ {f ∘θ : f ∈ ℋk }.
Then G_{k} is an uncountable set of continuous functions defined on K with the property
that the distance between any two of them is at least as large as 1∕k. This contradicts
separability of C
(K )
which follows from the Weierstrass approximation theorem in which
the separable countable set of functions is the restrictions of polynomials that involve
only rational coefficients. ■
Now here is Prokhorov’s theorem.
Theorem 60.3.5Let Λ =
{μn}
_{n=1}^{∞}be a sequence of probability measures defined onℬ
(E)
where E is a separable complete metric space. If Λ is tight then there exists aprobability measure, λ and a subsequence of
{μn}
_{n=1}^{∞}, still denoted by
{μn}
_{n=1}^{∞}suchthat whenever ϕ is a continuous bounded complex valued function defined onE,
∫ ∫
lim ϕdμn = ϕdλ.
n→∞
Proof: By tightness, there exists an increasing sequence of compact sets,
{Kn }
such
that
1-
μ(Kn ) > 1− n
for all μ ∈ Λ. Now letting μ ∈ Λ and ϕ ∈ C
(Kn )
such that
||ϕ||
_{∞}≤ 1, it follows
||∫ ||
|| ϕdμ|| ≤ μ(Kn ) ≤ 1
Kn
and so the restrictions of the measures of Λ to K_{n} are contained in the unit ball of
C
(Kn)
^{′}. Recall from the Riesz representation theorem, the dual space of C
(Kn )
is a
space of complex Borel measures. Theorem 15.5.5 on Page 1327 implies the unit ball
of C
(Kn )
^{′} is weak ∗ sequentially compact. This follows from the observation
that C
(Kn )
is separable which is proved in Corollary 60.3.4 and leads to the
fact that the unit ball in C
(Kn )
^{′} is actually metrizable by Theorem 15.5.5 on
Page 1327. Therefore, there exists a subsequence of Λ,
{μ1k}
such that their
restrictions to K_{1} converge weak ∗ to a measure, λ_{1}∈ C
(K1)
^{′}. That is, for every
ϕ ∈ C
(K1 )
,
∫ ∫
lim ϕd μ1k = ϕd λ1
k→∞ K1 K1
By the same reasoning, there exists a further subsequence
{μ2k}
such that the
restrictions of these measures to K_{2} converge weak ∗ to a measure λ_{2}∈ C
(K2 )
^{′} etc.
Continuing this way,
μ11,μ12,μ13,⋅⋅⋅ → Weak ∗ in C (K1)′
μ21,μ22,μ23,⋅⋅⋅ → Weak ∗ in C (K2)′
μ31,μ32,μ33,⋅⋅⋅ → Weak ∗ in C (K3)′
..
.
Here the j^{th} sequence is a subsequence of the
(j − 1)
^{th}. Let λ_{n} denote the measure in
C
(Kn)
^{′} to which the sequence
{μnk}
_{k=1}^{∞} converges weak∗. Let
{μn}
≡
{μnn}
, the
diagonal sequence. Thus this sequence is ultimately a subsequence of every one of the
above sequences and so μ_{n} converges weak∗ in C
(Km )
^{′} to λ_{m} for each m. Note that this
is all happening on different sets so there is no contradiction with something converging
to two different things.
Claim:For p > n, the restriction of λ_{p} to the Borel sets of K_{n} equals λ_{n}.
Proof of claim: Let H be a compact subset of K_{n}. Then there are sets, V_{l} open in
K_{n} which are decreasing and whose intersection equals H. This follows because this is a
metric space. Then let H ≺ ϕ_{l}≺ V_{l}. It follows
Now considering the ends of this inequality, let l →∞ and pass to the limit to
conclude
λn(H ) ≥ λp(H ).
Similarly,
∫ ∫
λ (H ) ≤ ϕ dλ = lim ϕdμ
n Kn l n k→∞ Kn l k
∫ ∫
= lk→im∞ ϕldμk = ϕldλp ≤ λp (Vl).
Kp Kp
Then passing to the limit as l →∞, it follows
λ (H ) ≤ λ (H ).
n p
Thus the restriction of λ_{p},λ_{p}|_{Kn} to the compact sets of K_{n} equals λ_{n}. Then by inner
regularity it follows the two measures, λ_{p}|_{Kn}, and λ_{n} are equal on all Borel sets of K_{n}.
Recall that for finite measures on separable metric spaces, regularity is obtained for
free.
It is fairly routine to exploit regularity of the measures to verify that λ_{m}
(F)
≥ 0 for
all F a Borel subset of K_{m}.Note that ϕ →∫_{Kn}ϕdλ_{n} is a positive linear functional and so
λ_{n}≥ 0. Also, letting ϕ ≡ 1,
1
1 ≥ λm (Km ) ≥ 1− m-. (60.3.7)
(60.3.7)
Define for F a Borel set,
λ(F ) ≡ lim λn (F ∩ Kn ).
n→∞
The limit exists because the sequence on the right is increasing due to the above
observation that λ_{n} = λ_{m} on the Borel subsets of K_{m} whenever n > m. Thus for
n > m