Let Γ be an arbitrary index set and (S,S) a measure space. I want to study the product space SΓ. For that we define
Xγ:SΓ→S,ω=(ωγ)γ∈Γ↦Xγ(ω):=ωγ
where Xγ is the coordinate map on the γ-th coordinate. Moreover, we define
SΓ=σ(Xγ,γ∈Γ):=σ({{Xγ∈Aγ};γ∈Γ,Aγ∈S})
this should be the smallest σ-algebra, such that all Xγ's are measurable. Now my question, why is the following set a generator of this σ-algebra?
M:={ω∈SΓ;ωj∈Aj,j∈J}=∏j∈JAj×∏k∈JcS
for all J⊂Γ which are finite and Aj∈S. Why is this true? It would be appreciated if someone could give me a proof or post a reference.
As I saw the statement, I had to think about the product topology. This can be described in a similar way (using a basis). However, in topology one has often that a property is true just for finite intersections or similar things. In measure theory this is not the case. So I do not see where the finiteness come in.
Thank you for your help.
cheers
math
Answer
Similar as in product topology we care about finite intersections, in measure theory we care about countable operations. But note one thing: let Γ be infinite and J⊂Γ be some infinite countable set. Let us numerate: j1,j2,…,jn,⋯∈J.
If SΓ is the product σ-algebra, and Aj∈S then any set of the form
Bn=n∏k=1Ajn×∏k∈Γ∖{j1,j2,…,jn}S∈SΓ
for any finite n, hence
∏k∈JAj×∏k∈JcS=∞⋂n=1Bn∈SΓ.
So it means that there is no need to ask the product σ-algebra to be generated by a basis of products where there are possibly countably many sets which are not equal to S.
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