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hw7solutions

hw7solutions - Math 318 HW#7 Solutions 1 Exercise 20.12 Let...

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Math 318 HW #7 Solutions 1. Exercise 20.12. Let C be the Cantor set. Let D [0 , 1] be a nowhere dense measurable set with m ( D ) > 0. Then there is a non-measurable set B D . At each stage of the construction of C and of D , a certain finite number of open intervals of [0 , 1] are deleted (put into [0 , 1] \ C or [0 , 1] \ D ). Let g map the intervals put into [0 , 1] \ D at the n th stage linearly onto the intervals put into [0 , 1] \ C at the n th stage, for n = 1 , 2 , . . . . Thus g is monotone and defined at every element of [0 , 1] \ D , mapping onto [0 , 1] \ C . (a) Using the fact that g is increasing and that [0 , 1] \ D is dense in [0 , 1], prove that for every x 0 D , lim x x - 0 g ( x ) = sup { g ( x ) : x [0 , 1] \ D and x < x 0 } and lim x x + 0 g ( x ) = inf { g ( x ) : x [0 , 1] \ D and x > x 0 } . Proof. Since [0 , 1] \ D is dense, we know that every x 0 D is a limit point of [0 , 1] \ D , so lim x x - 0 g ( x ) makes sense. Now, { g ( x ) : x [0 , 1] \ D and x < x 0 } is nonempty (unless x 0 = 0, in which case lim x x - 0 g ( x ) is vacuous) and, since g is increasing, bounded above by g ( x 1 ) for any x 1 > x 0 such that x 1 [0 , 1] \ D . Therefore, sup { g ( x ) : x [0 , 1] \ D and x < x 0 } = M exists. But then, for any > 0, there exists x [0 , 1] \ D with x < x 0 such that g ( x ) > M - . Letting δ = x 0 - x , we know that x < x < x 0 implies that g ( x ) g ( x ) M , and so g ( x ) - M g ( x ) - M 0 . Therefore, since g ( x ) - M > - , the above implies that - < g ( x ) - M 0 , so we can conclude that | x - x 0 | < δ and x < x 0 implies that | g ( x ) - M | < . Since the choice of > 0 was arbitrary, we conclude that lim x x - 0 g ( x ) = M = sup { g ( x ) : x [0 , 1] \ D and x < x 0 } , as desired. An equivalent argument shows that lim x x + 0 g ( x ) = inf { g ( x ) : x [0 , 1] \ D and x > x 0 } . (b) Use (a) to show that for x 0 D , lim x x - 0 g ( x ) lim x x + 0 g ( x ). 1

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Proof. Since g is monotone increasing, every element of the set { g ( x ) : x [0 , 1] \ D and x > x 0 } is an upper bound on the set { g ( x ) : x [0 , 1] \ D and x < x 0 } . Therefore, by defi- nition of the supremum, lim x x - 0 g ( x ) = sup { g ( x ) : x [0 , 1] \ D and x < x 0 } is less than or equal to every element of { g ( x ) : x [0 , 1] \ D and x > x 0 } . In other words, lim x x - 0 g ( x ) is a lower bound on the set { g ( x ) : x [0 , 1] \ D and x > x 0 } , and so, by the definition of the infimum, is less than or equal to lim x x - 0 g ( x ) inf { g ( x ) : x [0 , 1] \ D and x > x 0 } = lim x x + 0 g ( x ) , as desired.
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hw7solutions - Math 318 HW#7 Solutions 1 Exercise 20.12 Let...

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