[Merged by Bors] - feat: the product of Borel spaces is Borel when either of them is second-countable #6689
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…ond-countable (#6689) We have currently that the product of two Borel spaces is Borel when both of them are second-countable. It is in fact sufficient to assume that only one of them is second-countable. We prove this in this PR. Also move the definition of `SecondCountableEither` from `Function.StronglyMeasurable` to `BorelSpace.Basic` to be able to use it in the statement of the above theorem.
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We have currently that the product of two Borel spaces is Borel when both of them are second-countable. It is in fact sufficient to assume that only one of them is second-countable. We prove this in this PR.
Also move the definition of
SecondCountableEitherfromFunction.StronglyMeasurabletoBorelSpace.Basicto be able to use it in the statement of the above theorem.