First, here is a nice measurable selection theorem.
Lemma 47.2.2Let U be a separable reflexive Banach space. Suppose there is asequence
{u (ω )}
j
_{j=1}^{∞}in U, where each ω → u_{j}
(ω)
is measurable and foreach ω, sup_{i}
∥u (ω)∥
i
< ∞. Then, there exists u
(ω )
∈ U such that ω → u
(ω)
ismeasurable, and a subsequence n
(ω)
, that depends on ω, such that the weaklimit
lim u (ω) = u (ω )
n(ω)→∞ n(ω)
holds.
Proof:Let
{zi}
_{i=1}^{∞} be a countable dense subset of U^{′}. Let h : U →∏_{i=1}^{∞}ℝ be
defined by
∏∞
h(u) = 〈zi,u〉.
i=1
Let X = ∏_{i=1}^{∞}ℝ with the product topology. Then, this is a Polish space with the
metric defined as d
(x,y)
= ∑_{i=1}^{∞}
-|xi−yi|
1+|xi−yi|
2^{−i}. By compactness, for a fixed ω,the
h
(un(ω))
are contained in a compact subset of X. Next, define
-------------
Γ n(ω) = ∪k ≥nh(uk(ω)),
which is a nonempty compact subset of X. Moreover, Γ_{n}
(ω)
is a measurable
multifunction into X.
Next, we claim that ω → Γ_{n}
(ω)
is a measurable multifunction.
The proof of the claim is as follows. It is necessary to show that Γ_{n}^{−}
(O )
defined as
{ω : Γ n(ω)∩ O ⁄= ∅}
is measurable whenever O is open. It suffices to verify this for O a
basic open set in the topology of X. Thus let O = ∏_{i=1}^{∞}O_{i} where each O_{i} is a proper
open subset of ℝ only for i ∈
which is a measurable set since u_{k} is measurable.
Then, it follows that ω → Γ_{n}
(ω )
is strongly measurable because it has compact values
in X, thanks to Tychonoff’s theorem. Thus Γ_{n}^{−}
(H)
=
{ω : H ∩ Γ n(ω) ⁄= ∅}
is
measurable whenever H is a closed set. Now, let Γ
(ω)
be defined as ∩_{n}Γ_{n}
(ω)
and then
for H closed,
Γ − (H ) = ∩nΓ −n (H )
and each set in the intersection is measurable, so this shows that ω → Γ
(ω )
is also
measurable. Therefore, it has a measurable selection g
(ω )
. It follows from the definition
of Γ
(ω)
that there exists a subsequence n
(ω)
such that
g(ω) = lim h(u (ω)) in X.
n(ω)→∞ n(ω)
In terms of components, we have
gi(ω) = lim 〈zi,un(ω)(ω )〉 .
n(ω)→ ∞
Furthermore, there is a further subsequence, still denoted with n
(ω)
, such that
u_{n(ω)
}
(ω)
→ u
(ω)
weakly. This means that for each i,
〈 〉
gi(ω) = n(liωm)→ ∞ zi,un(ω)(ω) = 〈zi,u(ω)〉.
Thus, for each z_{i} in a dense set, ω →
〈zi,u(ω)〉
is measurable. Since the z_{i} are dense,
this implies ω →
〈z,u (ω )〉
is measurable for every z ∈ U^{′} and so by the Pettis theorem,
ω → u
(ω)
is measurable. ■
Now we consider the case of fixed points for simplices.
Suppose f
(⋅,ω)
: S → S for S a simplex. Then from the Brouwer fixed point theorem,
there is a fixed point x
(ω )
provided f(⋅,ω) is continuous. Can it be arranged to have
ω → x
(ω)
also measurable? In fact, it can, and this is shown here. In other words,
if P
(ω)
are the fixed points of f
(⋅,ω)
, there exists a measurable selection in
P
(ω)
.
S ≡
[x0,⋅⋅⋅,xn]
is a simplex in ℝ^{n}. Assume
{xi − x0}
_{i=1}^{n} are linearly independent.
Thus a typical point of S is of the form
∑n
tixi
i=0
where the t_{i} are uniquely determined and the map x → t is continuous from S to the
compact set
{ ∑ }
t ∈ ℝn+1 : ti = 1,ti ≥ 0
.
To see this, suppose x^{k}→ x in S. Let x^{k}≡∑_{i=0}^{n}t_{i}^{k}x_{i} with x defined similarly with
t_{i}^{k} replaced with t_{i}, x ≡∑_{i=0}^{n}t_{i}x_{i}. Then
n n
xk − x = ∑ tk(x − x ), x − x = ∑ t (x − x )
0 i=1 i i 0 0 i=1 i i 0
Say t_{i}^{k} fails to converge to t_{i} for all i ≥ 1. Then there exists a subsequence, still denoted
with superscript k such that for each i = 1,
⋅⋅⋅
,n, it follows that t_{i}^{k}→ s_{i} where s_{i}≥ 0
and some s_{i}≠t_{i}. But then, taking a limit, it follows that
∑n ∑n
x − x0 = si(xi − x0) = ti(xi − x0)
i=1 i=1
which contradicts independence of the x_{i}−x_{0}. It follows that for all i ≥ 1,t_{i}^{k}→ t_{i}. Since
they all sum to 1, this implies that also t_{0}^{k}→ t_{0}. Thus the claim about continuity is
verified.
Let f
(⋅,ω)
: S → S be continuous such that ω → f
(x,ω )
is measurable. When
doing f
(⋅,ω)
to a point ∑_{i=0}^{n}t_{i}x_{i}, one obtains another point of S denoted
as ∑_{i=0}^{n}s_{i}
(ω)
x_{i}. The coefficients s_{i} must be measurable functions. This is
because
∑n
ω → f (x,ω) = si(ω )xi
i=0
and the left side is measurable so it follows the right is also. Now as noted above, the
map which takes a point of S to its coefficients is continuous and so each s_{i} is measurable
as a function of ω. Note that if x is replaced with x
(ω)
, with ω → x
(ω )
measurable, the
same conclusion can be drawn about the s_{i}
(ω)
. This is because, thanks to the
continuity of f in its first argument, the function on the left in the above is
measurable.
Label x_{j} as p_{j} where p_{0},
⋅⋅⋅
,p_{n} are the first n + 1 prime numbers. Thus the vertices of
S have been labeled. Next triangulate S so that all simplices have diameter less than ε. If
[y0,⋅⋅⋅,yn ]
is one of these small vertices, each is of the form ∑_{i=0}^{n}t_{i}x_{i} where t_{i}≥ 0 and
∑_{i}t_{i} = 1. Define r_{k}
(ω)
≡ s_{k}
(ω )
∕t_{k} if t_{k}> 0 and ∞ if t_{k} = 0. Thus this is a measurable
function.
E ≡ ∩ [ω : r (ω) ≤ r (ω )], F ≡ E ,F ≡ E ∖E ,⋅⋅⋅,F ≡ E ∖ ∪n−1E
k j⁄=k k j 0 0 1 1 0 n n i=0 i
Then p
(yi,ω)
will be the label placed on y_{i}. It equals
∑n
p(yi,ω ) ≡ pkXFk (ω )
k=0
obviously a measurable function. Note also that this new method of labeling does not
contradict the original labels placed on the vertices x_{i}. This is because for x_{i},t_{i} = 1 and
all other t_{j} = 0 so the only ratio that is finite will be s_{i}∕t_{i}. All others are ∞ by definition.
As for the vertices which are on the k^{th} face
[x0,⋅⋅⋅,ˆxk,⋅⋅⋅,xn]
, these will be labeled
from the list
{p0,⋅⋅⋅,pˆk,⋅⋅⋅,pn}
because t_{k} = 0 for each of these and so r_{k}
(ω)
= ∞. By
the Sperner’s lemma procedure described above, there are an odd number of simplices
having value ∏_{i≠k}p_{i} on the k^{th} face and an odd number of simplices in the
triangulation of S for which the product of the labels on their vertices equals
p_{0}p_{1}
⋅⋅⋅
p_{n}≡ P_{n}. We call this the value of the simplex. Thus if
[y0,⋅⋅⋅,yn]
is one
of these simplices, and p
(yi,ω )
is the label for y_{i}, a measurable function of
ω,
n∏ n∏
p (yi,ω) = pi ≡ Pn
i=0 i=0
For ω ∈ F_{k}, what is r_{k}
(ω )
? Could it be larger than 1? r_{k}
(ω )
is certainly finite
because at least some t_{j}≠0 since they sum to 1. Thus, if r_{k}
(ω)
> 1, you would
have s_{k}
(ω)
> t_{k}. The s_{j} sum to 1 and so some s_{j}
(ω)
< t_{j} since otherwise, the
sum of the t_{j} equalling 1 would require the sum of the s_{j} to be larger than 1.
Hence r_{k}
(ω)
was not really the smallest so ω
∕∈
F_{k}. Thus r_{k}
(ω)
≤ 1. Hence
s_{k}
(ω)
≤ t_{k}.
Let S denote those simplices whose value is P_{n} for some ω. In other words, if
{y ,⋅⋅⋅,y }
0 n
are the vertices of one of these simplices, and
∑n
ys = tsixi
i=0
then for some ω, r_{ks}
(ω )
≤ r_{j}
(ω)
for all j≠k_{s} and
{k0,⋅⋅⋅,kn}
=
{0,⋅⋅⋅,n}
. There are
finitely many of these simplices, so S ≡
{ n }
∏
F2 ≡ ω ∕∈ F1 :i=0p(yi,ω ) = Pn : [y0,⋅⋅⋅,yn] = S2
Continue this way obtaining disjoint measurable sets F_{j} whose union is all of Ω. The
union is Ω because every ω is associated with at least one of the S_{i}. Now for ω ∈ F_{k} and
[y0,⋅⋅⋅,yn ]
= S_{k}, it follows that ∏_{i=0}^{n}p
(yi,ω)
= P_{n}. For ω ∈ F_{k}, let b
(ω )
denote the
barycenter of S_{k}. Thus ω → b
(ω)
is a measurable function, being constant on a
measurable set. Thus we let b
(ω )
= ∑_{i=1}^{m}X_{Fi}
(ω)
b_{i} where b_{i} is the barycenter of
S_{i}.
Now do this for a sequence ε_{k}→ 0 where b_{k}
(ω)
is a barycenter as above.
By Lemma 47.2.2 there exists x
(ω)
such that ω → x
(ω)
is measurable and a
sequence
{ }
bk(ω)
_{k(ω)
=1}^{∞},lim_{k(ω)
→∞}b_{k(ω)
}
(ω )
= x
(ω)
. This x
(ω)
is also a fixed
point.
Consider this last claim. x
(ω)
= ∑_{i=0}^{n}t_{i}
(ω)
x_{i} and after applying f
(⋅,ω)
, the result
is ∑_{i=0}^{n}s_{i}
(ω)
x_{i}. Then b_{k(ω)
}∈ σ_{k}
(ω)
where σ_{k}
(ω)
is a simplex having vertices
{ k k }
y0 (ω),⋅⋅⋅,y n(ω)
and the value of
[ k k ]
y 0 (ω),⋅⋅⋅,yn (ω )
is P_{n}. Re ordering these if
necessary, we can assume that the label for y_{i}^{k}
(ω)
= p_{i} which implies that, as noted
above,
si(ω)-≤ 1, si(ω) ≤ ti(ω)
ti(ω)
the i^{th} coordinate of f
( )
yki (ω),ω
with respect to the original vertices of S decreases and
each i is represented for i =
{0,1,⋅⋅⋅,n}
. Thus
yki (ω) → x(ω)
and so the i^{th} coordinate of y_{i}^{k}
(ω)
,t_{i}^{k}
(ω)
must converge to t_{i}
(ω)
. Hence if the i^{th}
coordinate of f
(yk(ω),ω)
i
is denoted by s_{i}^{k}
(ω)
,
ski (ω) ≤ tki (ω)
By continuity of f, it follows that s_{i}^{k}
(ω)
→ s_{i}
(ω)
. Thus the above inequality is preserved
on taking k →∞ and so
0 ≤ si(ω) ≤ ti(ω)
this for each i. But these s_{i} add to 1 as do the t_{i} and so in fact, s_{i}
(ω )
= t_{i}
(ω)
for each i
and so f
(x (ω ),ω)
= x
(ω )
. This proves the following theorem which gives the existence
of a measurable fixed point.
Theorem 47.2.3Let S be a simplex
[x0,⋅⋅⋅,xn]
such that
{xi − x0}
_{i=1}^{n}areindependent. Also let f
(⋅,ω)
: S → S be continuous for each ω and ω → f
(x,ω)
ismeasurable, meaning inverse images of sets open in S are in ℱ where
(Ω,ℱ )
is ameasurable space. Then there exists x
(ω)
∈ S such that ω → x
(ω)
is measurableand f
(x(ω),ω )
= x
(ω)
.
Corollary 47.2.4Let K be a closed convex bounded subset of ℝ^{n}. Let f
(⋅,ω)
:
K → K be continuous for each ω and ω → f
(x,ω)
is measurable, meaning inverseimages of sets open in K are in ℱ where
(Ω,ℱ )
is a measurable space. Then thereexists x
(ω)
∈ K such that ω → x
(ω)
is measurable and f
(x (ω ),ω)
= x
(ω )
.
Proof: Let S be a large simplex containing K and let P be the projection map onto
K. Consider g
(x,ω)
≡ f
(P (x),ω)
. Then g satisfies the necessary conditions for
Theorem 47.2.3 and so there exists x
(ω)
∈ S such that ω → x
(ω)
is measurable and
g
(x (ω),ω)
= x
(ω)
. But this says x
(ω)
∈ K and so g
(x(ω) ,ω )
= f
(x(ω),ω )
.
■
Much shorter proof
The above gives a proof of a measurable fixed point as part of a proof of the Brouwer
fixed point theorem directly but it is a lot easier if you simply begin with the
existence of the Brouwer fixed point and show it is measurable. We sent the
above to be considered for publication and the referee pointed this out. I totally
missed it because I had forgotten about the Kuratowski selection theorem. The
functions f : Ω × E → ℝ in which f
(⋅,ω)
is continuous are called Caratheodory
functions.
Kuratowski [?] which is presented next. It is Theorem 9.1.11 in this collection.
Theorem 47.2.5Let E be a compact metric space and let
(Ω,ℱ )
be a measure space.Suppose ψ : E × Ω → ℝ has the property that x → ψ
(x,ω)
is continuous and ω → ψ
(x,ω)
is measurable. Then there exists a measurable function, f having values in E suchthat
ψ(f (ω ),ω) = sup ψ(x,ω).
x∈E
Furthermore, ω → ψ
(f (ω),ω)
is measurable.
Theorem 47.2.6Let K be a closed convex bounded subset of ℝ^{n}. Let f
(⋅,ω)
:
K → K be continuous for each ω and ω → f
(x,ω)
is measurable, meaning inverseimages of sets open in K are in ℱ where
(Ω,ℱ )
is a measurable space. Then thereexists x
(ω)
∈ K such that ω → x
(ω)
is measurable and f
(x (ω ),ω)
= x
(ω )
.
Proof:Simply consider E = K and ψ
(x,ω )
≡−
|x − f (x,ω)|
. It has a maximum
x
(ω )
for each ω thanks to continuity of f
(⋅,ω)
. Thanks to the Brouwer fixed point
theorem, this x
(ω)
must be a fixed point. By the above Kuratowski theorem, one of these
x