Analysis

7.3 Kuratowski’s Lemma

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.

Proof: Let C₁ be a 2⁻¹ net of E. Suppose C₁,

,C_m have been chosen such that C_k is a 2^−k net and C_i+1 ⊇ C_i for all i. Then consider E ∖∪. If this set is empty, let C_m+1 = C_m. If it is nonempty, let _i=1^r be a 2−

(m+1)

net for this compact set. Then let C_m+1 = C_m ∪_i=1^r. It follows _m=1^∞ satisfies C_m is a 2^−m net and C_m ⊆ C_m+1.

Let

_k=1m

(1)

equal C₁. Let

{ } A1 ≡ ω : ψ (x1,ω) = max ψ(x1,ω) 1 1 k k

For ω ∈ A₁¹, define s₁

≡ x₁¹. Next let

{ } A12 ≡ ω ∕∈ A11 : ψ (x12,ω) = max ψ (x1k,ω ) k

and let s₁

≡ 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

,s₂,,s_m are simple functions with the property that if m > 1,

for each k + 1 ≤ m, only the second and third assertions holding if m = 1. Letting C_m = _k=1^N, it follows s_m is of the form

∑N sm (ω) = xkXAk (ω),Ai ∩Aj = ∅. (7.6) k=1

(7.6)

meaning that s_m

has value x_k on A_k. Denote by _i=1^n₁ those points of C_m+1 which are contained in B. Letting A_k play the role of Ω in the first step in which s₁ was constructed, for each ω ∈ A₁ let s_m+1 be a simple function which has one of the values y_1i and satisfies

ψ (sm+1(ω),ω) = mi≤anx1 ψ(y1i,ω)

for each ω ∈ A₁. Next let

_i=1^n₂ be those points of C_m+1 different than _i=1^n₁ which are contained in B. Then define s_m+1 on A₂ to have values taken from _i=1^n₂ and

ψ (sm+1(ω),ω) = max ψ(y2i,ω) i≤n2

for each ω ∈ A₂. Continuing this way defines s_m+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

(7.8)

To see this is so, pick ω ∈ Ω. Let

max {ψ(x,ω) : x ∈ Cm+1 } = ψ (yrj,ω) (7.9)

(7.9)

where y_rj ∈

_i=1^n_r ⊆ C_m+1 and out of all the balls B, let the first one which contains y_rj be B. Then by the construction, s_m+1 = y_rj because ψ is at least as large as ψ for all the other y_sj. This and 7.9 verifies 7.8.

From 7.7 it follows s_m

converges uniformly on Ω to a measurable function, f. Then from the construction, ψ ≥ ψ for all m and ω. Now pick ω ∈ Ω and let z be such that ψ = max_x∈Eψ. Letting y_k → z where y_k ∈ C_k, it follows from continuity of ψ in the first argument that

To show ω → ψ is measurable, note that since E is compact, there exists a countable dense subset, D. Then using continuity of ψ in the first argument, which equals a measurable function of ω because D is countable. ■