Wednesday, 13 March 2019

measure theory - Generator of the sigma-algebra on a product space




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:SΓS,ω=(ωγ)γΓXγ(ω):=ωγ



where Xγ is the coordinate map on the γ-th coordinate. Moreover, we define



SΓ=σ(Xγ,γΓ):=σ({{XγAγ};γΓ,AγS})



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:={ωSΓ;ωjAj,jJ}=jJAj×kJcS



for all JΓ which are finite and AjS. 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 JΓ be some infinite countable set. Let us numerate: j1,j2,,jn,J.



If SΓ is the product σ-algebra, and AjS then any set of the form
Bn=nk=1Ajn×kΓ{j1,j2,,jn}SSΓ
for any finite n, hence
kJAj×kJcS=n=1BnSΓ.



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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