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Anahi Jensen

Anahi Jensen

Answered question

2022-05-26

Let S , T , I be non-empty sets, for every i I let A i be a sigma-algebra on T, and let f : S T. Is it true that f 1 ( i I A i ) = i I f 1 ( A i )?

Remarks
1. It is known that f 1 ( i I B i ) = i I f 1 ( B i ) if B i T for every i I. But this result is not the same as what needs to be proved, since i I B i T, whereas i I A i P T.
2. I was able to show that f 1 ( i I A i ) i I f 1 ( A i ), so the question reduces to whether the converse containment is valid.

Answer & Explanation

zepplinkid7yk

zepplinkid7yk

Beginner2022-05-27Added 11 answers

Not true. Let Ω = [ 0 , 1 ] and let A 1 = { , Ω , A , A c } and A 2 = { , Ω , B , B c } where A = [ 0 , 1 2 ) , B = [ 0 , 1 3 ). Then the left side is { , Ω }. I leave it to you to find an f for which f 1 ( A ) = f 1 ( B ) ( , Ω) so the right side contains f 1 ( A ).

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