Exercise 8.1.8

Finish the argument.

Answers

Consider an arbitrary 𝜖 > 0 . We have that with some gauge δ 1 ( x ) , we have | R ( f , P ) A 1 | < 𝜖 2 for all P which are δ 1 ( x ) -fine. Similarly we have δ 2 ( x ) where | R ( f , P ) A 2 | < 𝜖 2 . Let δ ( x ) = min { δ 1 ( x ) , δ 2 ( x ) } and note that δ is also a gauge, and that any partition P which is δ ( x ) -fine is also δ 1 ( x ) -fine and δ 2 ( x ) -fine. Find a partition P which is δ ( x ) -fine.

We then have | R ( f , P ) A 1 | < 𝜖 2 and | A 2 R ( f , P ) | < 𝜖 2 , so | A 2 A 1 | < 𝜖 , for arbitrary 𝜖 > 0 . Therefore A 1 = A 2 .

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2022-01-27 00:00
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