I think this is one of the most remarkable theorems I have seen and this is a good
place to put it since it is completely dependent on metric spaces, continuity and
measurability.
Theorem 7.3.1Let 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)
ismeasurable. Then there exists a measurable function, f having values in E suchthat
ψ(f (ω),ω ) = supψ (x,ω).
x∈E
Furthermore, ω → ψ
(f (ω),ω)
is measurable.
Proof: Let C1 be a 2−1 net of E. Suppose C1,
⋅⋅⋅
,Cm have been chosen such that Ck is a
2−k net and Ci+1⊇ Ci for all i. Then consider E ∖∪
{ ( −(m+1)) }
B x,2 : x ∈ Cm
. If this set is
empty, let Cm+1 = Cm. If it is nonempty, let
{yi}
i=1r be a 2−
(m+1)
net for this compact set.
Then let Cm+1 = Cm∪
{yi}
i=1r. It follows
{Cm}
m=1∞ satisfies Cm is a 2−m net and
Cm⊆ Cm+1.
Let
{ }
x1k
k=1m
(1)
equal C1. Let
{ }
A1 ≡ ω : ψ (x1,ω) = max ψ(x1,ω)
1 1 k k
For ω ∈ A11, define s1
(ω)
≡ x11. Next let
{ }
A12 ≡ ω ∕∈ A11 : ψ (x12,ω) = max ψ (x1k,ω )
k
and let s1
(ω)
≡ x21 on A21. Continue in this way to obtain a simple function, s1 such
that
ψ(s1(ω),ω) = max{ψ (x,ω) : x ∈ C1}
and s1 has values in C1.
Suppose s1
(ω )
,s2
(ω )
,
⋅⋅⋅
,sm
(ω )
are simple functions with the property that if m > 1,
d(sk(ω),sk+1 (ω)) < 2− k,
ψ(s (ω),ω ) = max{ψ (x,ω) : x ∈ C }
k k
sk has values in Ck
for each k + 1 ≤ m, only the second and third assertions holding if m = 1. Letting
Cm =
{xk}
k=1N, it follows sm
(ω )
is of the form
∑N
sm (ω) = xkXAk (ω),Ai ∩Aj = ∅. (7.6)
k=1
(7.6)
meaning that sm
(ω )
has value xk on Ak. Denote by
{y1i}
i=1n1 those points of Cm+1 which
are contained in B
−m
(x1,2 )
. Letting Ak play the role of Ω in the first step in which s1 was
constructed, for each ω ∈ A1 let sm+1
(ω)
be a simple function which has one of the values y1i
and satisfies
ψ (sm+1(ω),ω) = mi≤anx1 ψ(y1i,ω)
for each ω ∈ A1. Next let
{y2i}
i=1n2 be those points of Cm+1 different than
{y1i}
i=1n1
which are contained in B
(x2,2−m )
. Then define sm+1
(ω )
on A2 to have values taken from
{y2i}
i=1n2 and
ψ (sm+1(ω),ω) = max ψ(y2i,ω)
i≤n2
for each ω ∈ A2. Continuing this way defines sm+1 on all of Ω and it satisfies
−m
d(sm (ω ),sm+1 (ω)) < 2 for all ω ∈ Ω (7.7)
(7.7)
It remains to verify
ψ (s (ω),ω) = max {ψ (x,ω ) : x ∈ C }. (7.8)
m+1 m+1