60.3 Tight Measures
Now here is a definition of what it means for a set of measures to be tight.
Definition 60.3.1 Let Λ be a set of probability measures defined on the Borel sets of a
topological space. Then Λ is “tight” if for all ε > 0 there exists a compact set, Kε such
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 60.3.2 Let E be a separable complete metric space and let Λ be a set of Borel
probability measures. Then Λ is tight if and only if for every ε > 0 and r > 0 there exists
a finite collection of balls,
i=1m such that
for every μ ∈ Λ.
Proof: If Λ is tight, then there exists a compact set, Kε such that
for all μ ∈ Λ. Then consider the open cover,
Finitely many of these
and this yields the above condition.
Now suppose the above condition and let
1 − ε∕
for all μ ∈
Λ. Then let Kε ≡∩n=1∞Cn.
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
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 60.3.3 Let H be a compact metric space. Then there exists a compact
,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
Since H is compact, it is totally bounded. Therefore, there exists a 1 net for H
Letting Hi1 ≡B
it follows Hi1
is also a compact metric space and so
there exists a 1/2 net for each Hi1,
Then taking the intersection of B
to obtain sets denoted by Hj2
and continuing this way, one can obtain compact
subsets of H,
which satisfies: each
is contained in some Hki−1,
is compact with diameter less than i−1,
is the union of sets of
the form Hki+1
which are contained in it. Denoting by
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
closed intervals in
each of length no longer than 2
which have the
property that Iji
is contained in Iki−1
for some k
. Letting Ki ≡∪j=1miIji,
it follows Ki
is a sequence of nested compact sets. Let K
each x ∈ K
is the intersection of a unique sequence of these closed intervals,
. Define θx ≡∩k=1∞Hjkk.
Since the diameters of the Hji
converge to 0 as
this function is well defined. It is continuous because if xn → x,
are both in Ijkk,
closed interval in the sequence
whose intersection is x.
To see the
map is onto, let h ∈ H
. Then from the construction, there exists a sequence
of the above sets whose intersection equals 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 60.3.4 Let H be a compact metric space and let C
continuous functions defined on H with the usual norm,
Proof: The proof is by contradiction. Suppose C
is not separable. Let
denote a maximal collection of functions of C
with the property that if
The existence of such a maximal collection of
functions is a consequence of a simple use of the Hausdorff maximality theorem.
is dense. Therefore, it cannot be countable by the assumption
is not separable. It follows that for some
is uncountable. Now
by Theorem 60.3.3
there exists a continuous function, θ
defined on a compact
Now consider the functions defined on
Then Gk 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
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 60.3.5 Let Λ =
n=1∞ be a sequence of probability measures defined on
where E is a separable complete metric space. If
Λ is tight then there exists a
probability measure, λ and a subsequence of
n=1∞, still denoted by
that whenever ϕ is a continuous bounded complex valued function defined on
Proof: By tightness, there exists an increasing sequence of compact sets,
for all μ ∈ Λ. Now letting μ ∈ Λ and ϕ ∈ C
and so the restrictions of the measures of Λ to Kn are contained in the unit ball of
Recall from the Riesz representation theorem, the dual space of C
space of complex Borel measures. Theorem
on Page 1327
implies the unit ball
is weak ∗
sequentially compact. This follows from the observation
is separable which is proved in Corollary
and leads to the
fact that the unit ball in C
is actually metrizable by Theorem 15.5.5
. Therefore, there exists a subsequence of Λ,
such that their
converge weak ∗
to a measure, λ1 ∈ C
. That is, for every
ϕ ∈ C
By the same reasoning, there exists a further subsequence
such that the
restrictions of these measures to
converge weak ∗
to a measure λ2 ∈ C
Continuing this way,
Here the jth sequence is a subsequence of the
. Let λn
denote the measure in
to which the sequence
diagonal sequence. Thus this sequence is ultimately a subsequence of every one of the
above sequences and so μn
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 λp to the Borel sets of Kn equals λn.
Proof of claim: Let H be a compact subset of Kn. Then there are sets, V l open in
Kn which are decreasing and whose intersection equals H. This follows because this is a
metric space. Then let H ≺ ϕl ≺ V l. It follows
Now considering the ends of this inequality, let l →∞
and pass to the limit to
Then passing to the limit as l →∞,
Thus the restriction of λp,λp|Kn to the compact sets of Kn equals λn. Then by inner
regularity it follows the two measures, λp|Kn, and λn are equal on all Borel sets of Kn.
Recall that for finite measures on separable metric spaces, regularity is obtained for
It is fairly routine to exploit regularity of the measures to verify that λm
a Borel subset of Km.
Note that ϕ →∫
is a positive linear functional and so
0. Also, letting ϕ ≡
Define for F a Borel set,
The limit exists because the sequence on the right is increasing due to the above
observation that λn = λm on the Borel subsets of Km whenever n > m. Thus for
n > m
be a sequence of disjoint Borel sets. Then
the last equation holding by the monotone convergence theorem.
It remains to verify
for every ϕ bounded and continuous. This is where tightness is used again. Then as noted
because for p > n,λp
and so letting
the above is obtained. Also,
∞ < M
First let n
be so large that 2M∕n < ε∕
2 and then pick k
large enough that the above
expression is less than ε. ■
Definition 60.3.6 Let E be a complete separable metric space and let μ and the
sequence of probability measures,
defined on ℬ
for every ϕ a bounded continuous function. Then μn is said to converge weakly to μ.