Let X be a separable complete metric space and let
(Ω,C,μ)
be a set, a σ algebra of
subsets of Ω, and a measure μ such that this is a complete σ finite measure space.
Also let Γ : Ω →P_{F}
(X )
, the closed subsets of X.
Definition 47.1.1We define Γ^{−}
(S)
≡
{ω ∈ Ω : Γ (ω)∩ S ⁄= ∅}
We will consider a theory of measurability of set valued functions. The following
theorem is the main result in the subject. In this theorem the third condition is what we
will refer to as measurable.
Theorem 47.1.2The following are equivalent in case of a complete σ finitemeasure space. However 3 and 4 are equivalent for any measurable space consistingonly of a set Ω and a σ algebra C.
For all B a Borel set in X,Γ^{−}
(B)
∈C.
For all F closed in X, Γ^{−}
(F)
∈C
For all U open in X,Γ^{−}
(U )
∈C
There exists a sequence,
{σn}
of measurable functions satisfying σ_{n}
(ω)
∈ Γ
(ω)
such that for all ω ∈ Ω,
--------------
Γ (ω) = {σn(ω) : n ∈ ℕ}
These functions are called measurable selections.
For all x ∈ X,ω →dist
(x,Γ (ω))
is a measurable real valued function.
G
(Γ )
≡
{(ω,x) : x ∈ Γ (ω)}
⊆C× B
(X)
.
Proof: It is obvious that 1.) ⇒ 2.). To see that 2.) ⇒ 3.) note that
Γ^{−}
∞
(∪i=1Fi)
= ∪_{i=1}^{∞}Γ^{−}
(Fi)
. Since any open set in X can be obtained as a countable
union of closed sets, this implies 2.) ⇒ 3.).
Now we verify that 3.) ⇒ 4.). For convenience, drop the assumption that Γ
(ω)
is
closed in this part of the argument. It will just be set valued and satisfy the
measurability condition. A measurable selection will be obtained in Γ
(ω)
. Let
{xn}
_{n=1}^{∞}
be a countable dense subset of X. For ω ∈ Ω, let ψ_{1}
(ω )
= x_{n} where n is the smallest
integer such that Γ
(ω)
∩ B
(xn,1)
≠∅. Therefore, ψ_{1}
(ω )
has countably many values,
x_{n1},x_{n2},
⋅⋅⋅
where n_{1}< n_{2}<
⋅⋅⋅
. Now
{ω : ψ = x } =
1 n
{ω : Γ (ω)∩ B (xn,1) ⁄= ∅}∩ [Ω ∖∪k <n{ω : Γ (ω) ∩B (xk,1) ⁄= ∅}] ∈ C.
Thus we see that ψ_{1} is measurable and dist
(ψ1(ω) ,Γ (ω))
< 1. Let
Ωn ≡ {ω ∈ Ω : ψ1(ω) = xn} .
Then Ω_{n}∈C and Ω_{n}∩ Ω_{m} = ∅ for n≠m and ∪_{n=1}^{∞}Ω_{n} = Ω. Let
Dn ≡ {xk : xk ∈ B (xn,1)}.
Now for each n, and ω ∈ Ω_{n}, let ψ_{2}
(ω)
= x_{k} where k is the smallest index such that
x_{k}∈ D_{n} and B
(x , 1)
k 2
∩ Γ
(ω)
≠∅. Thus dist
(ψ (ω),Γ (ω))
2
<
1
2
and
d(ψ2 (ω ),ψ1 (ω)) < 1.
Continue this way obtaining ψ_{k} a measurable function such that
. This
has shown that if Γ is measurable, there exists a measurable selection, σ
(ω )
∈Γ
(ω)
. Of
course, if Γ
(ω)
is closed, then σ
(ω)
∈ Γ
(ω )
. Note that this had nothing to do with the
measure. It remains to show there exists a sequence of these measurable selections σ_{n}
such that the conclusion of 4.) holds. To do this we define for Γ
(ω)
closed and
measurable,
{ ( ) ( )
Γ (ω) ≡ Γ (ω)∩ B xn,2−i if Γ (ω) ∩B xn,2−i ⁄= ∅ .
ni Γ (ω) otherwise.
First we show that Γ_{ni} is measurable. Let U be open. Then
{ω : Γ ni(ω)∩ U ⁄= ∅} = {ω : Γ (ω )∩B (xn,2−i)∩ U ⁄= ∅}∪
[{ω : Γ (ω)∩ B (x ,2−i) = ∅}∩ {ω : Γ (ω)∩ U ⁄= ∅}]
n
= {ω : Γ (ω )∩ B(xn,2−i)∩ U ⁄= ∅} ∪
[(Ω∖ {ω : Γ (ω)∩ B (x ,2−i) ⁄= ∅})∩ {ω : Γ (ω)∩ U ⁄= ∅}],
n
a measurable set. By what was just shown, there exists σ_{ni}, a measurable function such
that σ_{ni}
(ω)
∈Γ_{ni}
(ω )
⊆ Γ
(ω)
for all ω ∈ Ω. If x ∈ Γ
(ω)
, then
-------------
x ∈ B (xn,2− (i+2))
whenever x_{n} is close enough to x. Thus both x,σ_{n}
(i+2)
(ω)
are in B
( −(i+2))
xn,2
and so
|| ||
σn(i+1)(ω)− x
< 2^{−i}. It follows that condition 4.) holds. Note that this had nothing to
do with the measure.
Now we verify that 4.) ⇒ 3.). Suppose there exist measurable selections
σ_{n}
(ω)
∈ Γ
(ω)
satisfying condition 4.). Let U be open. Then
{ω : Γ (ω) ∩U ⁄= ∅} = ∪∞n=1σ−n1(U ) ∈ C.
Now we verify that 4.) ⇒ 5.). Let F
(ω)
≡dist
(x,Γ (ω ))
. Then letting U be an open
set in [0,∞), F
(ω )
∈ U if and only if d
(x,σ (ω))
n
∈ U for some σ_{n}
(ω)
. Let
h_{n}
(ω )
≡ d
(x,σ (ω))
n
. Then h_{n} is measurable and F^{−1}
(U )
= ∪_{n=1}^{∞}h_{n}^{−1}
(U)
∈C. This
shows that for all x ∈ X, ω →dist
(x,Γ (ω))
is measurable and this proves
5.).
Now we verify that 5.) ⇒ 4.). We know dist
(x,Γ (⋅))
is measurable and we
show
{ω : Γ (ω )∩U ⁄= ∅}
∈C whenever U is open and then use 3.) ⇒ 4.). Since
X is separable, there exists B
(xi,ri)
such that U = ∪_{i=1}^{∞}B
(xi,ri)
. Then
{ω : Γ (ω) ∩U ⁄= ∅} = ∪∞i=1 {ω : Γ (ω)∩ B (xi,ri) ⁄= ∅}
= ∪∞ {ω : dist(x,Γ (ω)) < r} ∈ C.
i=1 i i
Therefore, 5.) ⇒ 4.) as claimed.
Now we must prove 5.) ⇒ 6.). We note that ω →dist
(x,Γ (ω))
is measurable and
x →dist
(x,Γ (ω))
is continuous. Also, the graph of Γ,G
(Γ )
is given by
G (Γ ) = {(ω, x) : dist(x,Γ (ω )) = 0}.
We wish to show that
(ω,x)
→dist
(x,Γ (ω))
is product measurable because then G
(Γ )
,
being the inverse image of
{0}
will be product measurable. Let
{xk}
be a countable
dense set in X and let
ϕ (ω,x) ≡ dist(x ,Γ (ω))
k n
where n is the first index such that x ∈ B
(xn,2−k)
. Then ϕ_{k}
(ω,x)
→dist
(x,Γ (ω))
due
to the continuity of x →dist
(x,Γ (ω))
and so we must argue that ϕ_{k} is product
measurable. On
En ≡ Ω × (B (xn,2− k)∖∪m <nB (xm,2−k)),
ϕ_{k}
(ω,x )
= dist
(xn,Γ (ω))
. Thus, on this set, ϕ_{k} equals a measurable function of ω and
does not depend on x on E_{n}. It follows that there are measurable simple C measurable
functions, s_{m}
(ω )
which increase pointwise to dist
(xn,Γ (ω))
on E_{n}. Thus s_{m}
(ω)
X_{En}
(x)
increases to ϕ_{k}
(ω,x)
on E_{n} showing that ϕ_{k}X_{En} is product measurable with respect
to C× σ
(τ)
since E_{n} is a measurable rectangle with respect to C and σ
(τ)
.
Therefore, ϕ_{k} is product measurable and so
(ω,x)
→dist
(x,Γ (ω))
is also product
measurable.
It remains to prove 6.) ⇒ 1.). This follows from Theorem 46.0.19.
being a complete measure space. Note that
without this assumption we could not draw the conclusion desired. This required
consideration of the measure. The following theorem is like part of the above but without
an assumption that Γ
(ω )
is closed.
Theorem 47.1.3The following are equivalent for any measurable space consistingonly of a set Ω and a σ algebra C. Here nothing is known about Γ
(ω)
other thanthat is a nonempty set.
For all U open in X,Γ^{−}
(U )
∈C
Γ − (U ) ≡ {ω : Γ (ω)∩ U ⁄= ∅}
There exists a sequence,
{σn}
of measurable functions satisfying σ_{n}
(ω)
∈ Γ
(ω)
such that for all ω ∈ Ω,
Γ (ω)-= {σ-(ω) : n-∈-ℕ}
n
These functions are called measurable selections.
Proof: First 1.) ⇒ 2.). A measurable selection will be obtained in Γ
(ω)
. Let
{xn}
_{n=1}^{∞} be a countable dense subset of X. For ω ∈ Ω, let ψ_{1}
(ω)
= x_{n} where n is the
smallest integer such that Γ
(ω )
∩ B
(xn,1)
≠∅. Therefore, ψ_{1}
(ω)
has countably many
values, x_{n1},x_{n2},
⋅⋅⋅
where n_{1}< n_{2}<
⋅⋅⋅
. Now
{ω : ψ1 = xn} =
{ω : Γ (ω)∩ B (xn,1) ⁄= ∅}∩ [Ω ∖∪k <n{ω : Γ (ω) ∩B (xk,1) ⁄= ∅}] ∈ C.
Thus we see that ψ_{1} is measurable and dist
(ψ1(ω) ,Γ (ω))
< 1. Let
Ωn ≡ {ω ∈ Ω : ψ1(ω) = xn} .
Then Ω_{n}∈C and Ω_{n}∩ Ω_{m} = ∅ for n≠m and ∪_{n=1}^{∞}Ω_{n} = Ω. Let
Dn ≡ {xk : xk ∈ B (xn,1)}.
Now for each n, and ω ∈ Ω_{n}, let ψ_{2}
(ω)
= x_{k} where k is the smallest index such that
x_{k}∈ D_{n} and B
( )
xk, 12
∩ Γ
(ω)
≠∅. Thus dist
(ψ2(ω),Γ (ω))
<
12
and
d(ψ (ω ),ψ (ω)) < 1.
2 1
Continue this way obtaining ψ_{k} a measurable function such that