- #1
MWG@berlin
- 15
- 0
Dear mates,
I run into a difficulty on the captioned area and am looking forward for enlightenments.
Consider two measurable spaces (X,A) and (Y,B). The standard way to generate the product sigma algebra (XxY,AxB) is to consider the smallest
sigma algebra that contains:
Ring/Algebra- S:{E in XxY; E=finite union of AixBi pairwise disjoint,Ai in A Bi in B}
The question is if we allow 'countably infinite union to replace finite union in the above Ring/Algebra, I conjecture (my presumptious inspection with the properties) that this set is already 'a sigma algebra' hence contains the smallest sigma algebra generated as in the literature. (Since S contains {E1xE2inXxY; E1 in A, E2 in B})
Consequence (BIG PROBLEM): As a result, we inevitably arrive at a very useful Theorem (which cannot be right):
let E be an element in the product sigma algebra (generated as in the literature), it follows E=countably union of AixBi such that Ai in A, Bi in B
Remarks:
Note that with the Theorem in the consequence, many related proofs can be further simplified. That is why I think I am missing something. I believe that when I release the restriction of "finite" union, it "still" cannot possibly be a sigma algebra.
Is S really a sigma algebra if the finite union is replaced by countably instead?
I run into a difficulty on the captioned area and am looking forward for enlightenments.
Consider two measurable spaces (X,A) and (Y,B). The standard way to generate the product sigma algebra (XxY,AxB) is to consider the smallest
sigma algebra that contains:
Ring/Algebra- S:{E in XxY; E=finite union of AixBi pairwise disjoint,Ai in A Bi in B}
The question is if we allow 'countably infinite union to replace finite union in the above Ring/Algebra, I conjecture (my presumptious inspection with the properties) that this set is already 'a sigma algebra' hence contains the smallest sigma algebra generated as in the literature. (Since S contains {E1xE2inXxY; E1 in A, E2 in B})
Consequence (BIG PROBLEM): As a result, we inevitably arrive at a very useful Theorem (which cannot be right):
let E be an element in the product sigma algebra (generated as in the literature), it follows E=countably union of AixBi such that Ai in A, Bi in B
Remarks:
Note that with the Theorem in the consequence, many related proofs can be further simplified. That is why I think I am missing something. I believe that when I release the restriction of "finite" union, it "still" cannot possibly be a sigma algebra.
Is S really a sigma algebra if the finite union is replaced by countably instead?
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