Here is an interesting lemma about weak convergence.
Lemma 60.4.1Let μ_{n}converge weakly to μ and let U be an open set with μ
(∂U )
= 0.Then
nl→im∞ μn(U) = μ(U) .
Proof: Let
{ψk }
be a sequence of bounded continuous functions which decrease to
X_{U}. Also let
{ϕk}
be a sequence of bounded continuous functions which increase to X_{U}.
For example, you could let
ψk (x) ≡ (1 − k dist(x,U ))+ ,
ϕ (x) ≡ 1 − (1− kdist(x,UC))+ .
k
Let ε > 0 be given. Then since μ
(∂U )
= 0, the dominated convergence theorem implies
there exists ψ = ψ_{k} and ϕ = ϕ_{k} such that
∫ ∫
ε > ψd μ− ϕd μ
Next use the weak convergence to pick N large enough that if n ≥ N,
∫ ∫ ∫ ∫
ψdμn ≤ ψdμ + ε, ϕdμn ≥ ϕdμ − ε.
Therefore, for n this large,
[∫ ∫ ]
μ (U ),μn(U) ∈ ϕdμ− ε, ψd μ+ ε
and so
|μ(U)− μn (U)| < 3ε.
since ε is arbitrary, this proves the lemma.
Definition 60.4.2Let
(Ω,ℱ,P )
be a probability space and let X : Ω → E be a randomvariable where here E is some topological space. Then one can define a probabilitymeasure, λ_{X}on ℬ
(E)
as follows:
λX (F) ≡ P ([X ∈ F ])
More generally, if μ is a probability measure on ℬ
(E)
, and X is a random variabledefined on a probability space, ℒ
(X )
= μ means
μ (F) ≡ P ([X ∈ F ]).
The following amazing theorem is due to Skorokhod. It starts with a measure, μ on
ℬ
(E)
and produces a random variable, X for which ℒ
(X )
= μ. It also has something to
say about the convergence of a sequence of such random variables.
Theorem 60.4.3Let E be a separable complete metric space and let
{μn}
be asequence of Borel probability measures defined on ℬ
(E)
such that μ_{n}converges weakly toμ another probability measure on ℬ
(E)
. Then there exist random variables, X_{n},Xdefined on the probability space,
([0,1),ℬ ([0,1)),m)
where m is one dimensionalLebesgue measure such that
ℒ (X ) = μ,ℒ (X ) = μ , (60.4.8)
n n
(60.4.8)
each random variable, X,X_{n}is continuous off a set of measure zero, and
X (ω) → X (ω) m a.e.
n
Proof: Let
{ak}
be a countable dense subset of E.
Construction of sets in E
First I will describe a construction. Letting C ∈ℬ
(E)
and r > 0,
Cr ≡ C ∩ B (a ,r),Cr ≡ B (a ,r)∩ C ∖Cr ,⋅⋅⋅,
1r 1 2( n−1 r2) 1
Cn ≡ B (an,r)∩ C ∖ ∪k=1Ck .
Thus the sets, C_{k}^{r} for k = 1,2,
⋅⋅⋅
are disjoint Borel sets whose union is all of C. Of
course many may be empty.
Cr(ns(iiznde)ex of the {ak} it is close to)
Now let C = E, the whole metric space. Also let
{rk}
be a decreasing sequence of
positive numbers which converges to 0. Let
Ak ≡ Erk1,k = 1,2,⋅⋅⋅
Thus
{A }
k
is a sequence of Borel sets, A_{k}⊆ B
(a ,r)
k 1
, and the union of the A_{k}
equals E. For
(i ,⋅⋅⋅,i )
1 m
∈ ℕ^{m}, suppose A_{i1,⋅⋅⋅
,im} has been defined. Then for
k ∈ ℕ,
Ai ,⋅⋅⋅,i k ≡ (Ai ,⋅⋅⋅,i )rm+1
1 m 1 m k
Thus A_{i1,}
⋅⋅⋅
,i_{m}k⊆ B
(ak,rm+1)
, is a Borel set, and
∪ ∞k=1Ai1,⋅⋅⋅,imk = Ai1,⋅⋅⋅,im. (60.4.9)
(60.4.9)
Also note that A_{i1,}
⋅⋅⋅
,i_{m} could be empty. This is because A_{i1,⋅⋅⋅
,imk}⊆ B
and also, the construction shows the Borel sets, A_{i1,}
⋅⋅⋅
,i_{m} are disjoint. Note that to get
A_{i1,⋅⋅⋅
,imk}, you do to A_{i1,⋅⋅⋅
,im} what was done for E but you consider smaller sized
pieces.
Construction of intervals depending on the measure
Next I will construct intervals, I_{i1,}
⋅⋅⋅
,i_{n}^{ν} in [0,1) corresponding to these A_{i1,⋅⋅⋅
,in}. In
what follows, ν = μ_{n} or μ. These intervals will depend on the measure chosen as
indicated in the notation.
∞ ∞
ν (Ai⋅⋅⋅,i ) = ∑ ν (Ai⋅⋅⋅,i ,k) = ∑ m (Iν ) = β − α,
1 m k=1 1 m k=1 i1⋅⋅⋅,im,k
the intervals, I_{i1}
⋅⋅⋅
,i_{m},j^{ν} being disjoint and
Iν = ∪∞ Iν .
i1⋅⋅⋅,im j=1i1⋅⋅⋅,im,j
These intervals satisfy the same inclusion properties as the sets
{Ai ,⋅⋅⋅,i }
1 m
. They
are just on [0,1) rather than on E. The intervals I_{i
1⋅⋅⋅
,imk}^{ν} correspond to the
sets A_{i1⋅⋅⋅
,im,k} and in fact the Lebesgue measure of the interval is the same as
ν
(Ai ⋅⋅⋅,i ,k)
1 m
.
Choosing the sequence
{rk}
in an auspicious manner
There are at most countably many positive numbers, r such that for ν = μ_{n} or
μ,ν
(∂B (ai,r))
> 0. This is because ν is a finite measure. Taking the countable union of
these countable sets, there are only countably many r such that ν
(∂B (ai,r))
> 0 for
some a_{i}. Let the sequence avoid all these bad values of r. Thus for
F ≡ ∪ ∞m=1 ∪∞k=1 ∂B (ak,rm )
and ν = μ or μ_{n},ν
(F)
= 0. Here the r_{m} are all good values such that for all k,m,∂B
(ak,rm)
has μ measure zero and μ_{n} measure zero. The next claim is illustrated in the
following picture. In the picture, A represents one of those A_{i1,⋅⋅⋅
,im} and A_{1} and A_{2} are
two of the sets A_{i1,⋅⋅⋅
,imk} which partition A_{i1,⋅⋅⋅
,im}.
PICT
Claim 1: ∂A_{i1,}
⋅⋅⋅
,i_{k}⊆ F. This really follows from the construction. However, the
details follow.
Proof of claim:Suppose C is a Borel set for which ∂C ⊆ F. I need to show
∂C_{k}^{ri}∈ F. First consider k = 1. Then C_{1}^{ri}≡ B
(a1,ri)
∩ C. If x ∈ ∂C_{1}^{ri}, then B
(x,δ)
contains points of B
(a1,ri)
∩ C and points of B
(a1,ri)
^{C}∪ C^{C} for every δ > 0. First
suppose x ∈ B
(a1,ri)
. Then a small enough neighborhood of x has no points of
B
(a1,ri)
^{C} and so every B
(x,δ)
has points of C and points of C^{C} so that x ∈ ∂C ⊆ F by
assumption. If x ∈ ∂C_{1}^{ri}, then it can’t happen that
||x − a1||
> r_{i} because then there
would be a neighborhood of x having no points of C_{1}^{ri}. The only other case to consider
is that
||x− ai||
= r_{i} but this says x ∈ F. Now assume ∂C_{j}^{ri}⊆ F for j ≤ k − 1 and
consider ∂C_{k}^{ri}.
Cri ≡ B (ak,ri)∩ C ∖∪kj−=11Crij
k ( ( )C)
= B (ak,ri)∩ C ∩ ∩kj−=11 Crji (60.4.10)
Consider x ∈ ∂C_{k}^{ri}. If x ∈int
(B (ak,ri)∩ C)
(int≡ interior) then a small enough ball
about x contains no points of
(B (ak,ri)∩ C )
^{C} and so every ball about x must contain
points of
( )
∩k−1(Cri)C C = ∪k−1Cri
j=1 j j=1 j
Since there are only finitely many sets in the union, there exists s ≤ k − 1 such that every
ball about x contains points of C_{s}^{ri} but from 60.4.10, every ball about x contains points
of
(Crsi )
^{C} which implies x ∈ ∂C_{s}^{ri}⊆ F by induction. It is not possible that
d
(x,ak)
> r_{i} and yet have x in ∂C_{k}^{ri}. This follows from the description in 60.4.10. If
d
(x,ak)
= r_{i} then by definition, x ∈ F. The only other case to consider is that
x
∕∈
int
(B (ak,ri)∩ C )
but x ∈ B
(ak,ri)
. From 60.4.10, every ball about x contains
points of C. However, since x ∈ B
(ak,ri)
, a small enough ball is contained in B
(ak,ri)
.
Therefore, every ball about x must also contain points of C^{C} since otherwise,
x ∈int
(B(ak,ri) ∩C )
. Thus x ∈ ∂C ⊆ F by assumption. Now apply what was just
shown to the case where C = E, the whole space. In this case, ∂E ⊆ F because
∂E = ∅. Then keep applying what was just shown to the A_{i1,⋅⋅⋅
,in}. This proves the
claim.
From the claim, ν
(int(Ai1,⋅⋅⋅,in))
= ν
(Ai1,⋅⋅⋅,in)
whenever ν = μ or μ_{n}. This is
because that in A_{i1,⋅⋅⋅
,in} which is not in int
(Ai1,⋅⋅⋅,in)
is in F which has measure
zero.
Some functions on [0,1)
By the axiom of choice, there exists x_{i1,}
⋅⋅⋅
,i_{m}∈int
(Ai1,⋅⋅⋅,im)
whenever
int(Ai1,⋅⋅⋅,im) ⁄= ∅.
For ν = μ_{n} or μ, define the following functions. For ω ∈ I_{i1,}
⋅⋅⋅
,i_{m}^{ν}
Z ν (ω ) ≡ x .
m i1,⋅⋅⋅,im
This defines the functions, Z_{m}^{μn} and Z_{m}^{μ}. Note these functions have the same values
but on slightly different intervals. Here is an important claim.
Claim 2 (Limit on μ_{n}):For a.e. ω ∈ [0,1),lim_{n→∞}Z_{m}^{μn}
(ω)
= Z_{m}^{μ}
(ω)
.
Proof of the claim: This follows from the weak convergence of μ_{n} to μ and Lemma
60.4.1. This lemma implies μ_{n}
(int(Ai1,⋅⋅⋅,im))
→ μ
(int(Ai1,⋅⋅⋅,im))
. Thus by the
construction described above, μ_{n}
(Ai1,⋅⋅⋅,im )
→ μ
(Ai1,⋅⋅⋅,im)
because of claim 1 and the
construction of F in which it is always a set of measure zero. It follows that
if ω ∈int
( μ )
Ii1,⋅⋅⋅,im
, then for all n large enough, ω ∈int
( μn )
Ii1,⋅⋅⋅,im
and so
Z_{m}^{μn}
(ω)
= Z_{m}^{μ}
(ω)
. Note this convergence is very far from being uniform.
Claim 3 (Limit on size of sets, fixed measure): For ν = μ_{n} or μ,
ν
{Zm }
_{m=1}^{∞} is
uniformly Cauchy independent of ν.
Proof of the claim: For ω ∈ I_{i1,}
⋅⋅⋅
,i_{m}^{ν}, then by the construction, ω ∈ I_{i1,⋅⋅⋅
,im,im+1⋅⋅⋅
,in}^{ν}
for some i_{m+1}
⋅⋅⋅
,i_{n}. Therefore, Z_{m}^{ν}
(ω)
and Z_{n}^{ν}
(ω)
are both contained in A_{i1,⋅⋅⋅
,im}
which is contained in B
(aim,rm)
. Since ω ∈ [0,1) was arbitrary, and r_{m}→ 0, it follows
these functions are uniformly Cauchy as claimed.
Let X^{ν}
(ω)
= lim_{m→∞}Z_{m}^{ν}
(ω)
. Since each Z_{m}^{ν} is continuous off a set of measure
zero, it follows from the uniform convergence that X^{ν} is also continuous off a set of
measure zero.
Claim 4: For a.e. ω,
lim X μn (ω) = Xμ (ω ).
n→ ∞
Proof of the claim: From Claim 3 and letting ε > 0 be given, there exists m large
enough that for all n,