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Measure Theory/Measures on Polish spaces

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Theorem (von Neumann's big theorem on algebra maps):

Let Ω,Ψ be Polish spaces that are part of the outer regular measure spaces (Ω,ℱ,μ) and (Ψ,𝒢,λ). Suppose that ϕ:ℱ→𝒢 is an algebra map such that for all D∈𝒢, there exists an A∈ℱ such that λ(ϕ(A)ΔD)=0. Then there exists Ω′⊆Ω and Ψ′⊆Ψ such that μ(Ω∖Ω′)=λ(Ψ∖Ψ′)=0 and a bijective function f:Ω′→Ψ′ such that

  1. for A∈ℱ, λ(f(A∩Ω′)Δϕ(A))=0 and
  2. for D∈𝒢, μ(f−1(D∩Ψ′)δA)=0, where A∈ℱ is any set such that λ(ϕ(A)ΔD)=0.

Proof: First, we write

Ω=⨆n∈ℕKn∪N1,

where N1 is a nullset and each Kj is closed and of diameter ≤1. Then write Jj′:=ϕ(Kj) and Jj:=Jj′‾. Then we write

Ψ=⨆n1,n2∈ℕJn1,n2∪M2,

where Jn1,n2⊆Jn1 is closed and each Jn1,n2 has diameter ≤1/2. Then set Kn1′,n2 to be a set such that λ(ϕ(Kn1′,n2)ΔJn1,n2)=0 and Kn1,n2 its closure.


Continuing this ping-pong game, we obtain nested sequences Kn1,…,nk and Jn1,…,nk dependent on different numbers of indices. Moreover, if we set

N:=⋃k∈ℕNk and M:=⋃k∈ℕMk,

then μ(N)=λ(M)=0, so that for any given number of indices, the corresponding sequence of nested sets K resp. J covers almost all of Ω resp. Ψ, dependent on whether the number of indices is even or odd. Define Ω′:=Ω∖N and Ψ′:=Ψ∖M. Then define a function f:Ω′→Ψ′ as follows: If x∈Ω′, then for any given number of indices k there exists one and only one Kn1,…,nk that contains x. Moreover, ◻