hw7solutions

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

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Unformatted text preview: 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 D , lim x x- g ( x ) = sup { g ( x ) : x [0 , 1] \ D and x < x } and lim x x + g ( x ) = inf { g ( x ) : x [0 , 1] \ D and x > x } . Proof. Since [0 , 1] \ D is dense, we know that every x D is a limit point of [0 , 1] \ D , so lim x x- g ( x ) makes sense. Now, { g ( x ) : x [0 , 1] \ D and x < x } is nonempty (unless x = 0, in which case lim x x- g ( x ) is vacuous) and, since g is increasing, bounded above by g ( x 1 ) for any x 1 > x such that x 1 [0 , 1] \ D . Therefore, sup { g ( x ) : x [0 , 1] \ D and x < x } = M exists. But then, for any > 0, there exists x [0 , 1] \ D with x < x such that g ( x ) > M- . Letting = x- x , we know that x < x < x implies that g ( x ) g ( x ) M , and so g ( x )- M g ( x )- M . Therefore, since g ( x )- M >- , the above implies that- < g ( x )- M , so we can conclude that | x- x | < and x < x implies that | g ( x )- M | < . Since the choice of > 0 was arbitrary, we conclude that lim x x- g ( x ) = M = sup { g ( x ) : x [0 , 1] \ D and x < x } , as desired. An equivalent argument shows that lim x x + g ( x ) = inf { g ( x ) : x [0 , 1] \ D and x > x } . (b) Use (a) to show that for x D , lim x x- g ( x ) lim x x + g ( x ). 1 Proof. Since g is monotone increasing, every element of the set { g ( x ) : x [0 , 1] \ D and x > x } is an upper bound on the set { g ( x ) : x [0 , 1] \ D and x < x } . Therefore, by defi- nition of the supremum, lim x x- g ( x ) = sup { g ( x ) : x [0 , 1] \ D and x < x } is less than or equal to every element of { g ( x ) : x [0 , 1] \ D and x > x } . In other words, lim x x- g ( x ) is a lower bound on the set...
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This note was uploaded on 08/20/2011 for the course MATH 318 taught by Professor Staff during the Spring '08 term at Haverford.

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

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