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Unformatted text preview: Math 245A Homework 4 Brett Hemenway November 9, 2005 Chapter 2 9. Let f : [0 , 1] [0 , 1] be the Cantor function. Then f is a continuous, increasing function from [0 , 1] onto [0 , 1]. Let g ( x ) = f ( x ) + x . a. g ( x ) is continuous and increasing because it is the sum of two continuous increasing functions. g takes [0 , 1] onto [0 , 2] because g (0) = 0, g (1) = 2, and g is continuous and increasing. If x < y , then f ( x ) f ( y ), so g ( x ) = f ( x ) + x < f ( y ) + y = g ( y ), so g is onetoone. Hence g is a bijection from [0 , 1] to [0 , 2]. Because g is continuous and bijective, we have that h = g 1 is a continuous function from [0 , 2] to [0 , 1]. b. Let C be the Cantor set. Notice that [0 , 1] \ C is a union of open intervals [0 , 1] \ C = S n =1 ( a n , b n ), and that f is constant on each of these intervals. The image of an interval ( a i , b i ) under g is then just the interval ( f ( a i ) + a i , f ( a i ) + b i ). So we have m (( a i , b i )) = m ( g (( a i , b i )). If i 6 = j , then the intervals g (( a i , b i )) = g (( a j , b j )) are disjoint because the intervals ( a i , b i ), ( a j , b j ) are disjoint and g is injective. So we have that m ( g ([0 , 1] \ C )) = m ([0 , 1] \ C ) = 1 Since m ( g ([0 , 1]) = m ([0 , 2]) = 2, we have that m ( g ( C )) = 1. c. Since m ( g ( C )) > 0 it contains a Lebesgue nonmeasurable subset by exercise 29, in Chapter 1. Let A g ( C ) be a Lebesgue nonmeasurable set. Let B = g 1 ( A ). 1 Then B C , so it is Lebesgue measurable because C is a null set, and the Lebesgue measure is complete. If B were Borel mea surable, then corollary 2.2 would give us that ( g 1 ) 1 ( B ) = A is Borel measurable, but this is false, since A is not Lebesgue measurable. So we conclude that B is not Borel measurable.is not Borel measurable....
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.
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