hw5solutions

# hw5solutions - Math 318 HW#5 Solutions 1 Show that the sets...

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Math 318 HW #5 Solutions 1. Show that the sets E x defined as part of showing that m * is not countably additive (cf. Example 7.7) form equivalence classes. Use this to conclude that for any x, y [0 , 1], either E x = E y or E x E y = . Proof. We want to show that the relation x y when y E x is an equivalence relation. The relation is certainly reflexive, as x E x . Also, if x y , then by definition y E x , meaning that y - x = r for some r Q . But then x - y = - r Q , so x E y , and hence y x . For transitivity, suppose x y and y z , meaning that y E x and z E y . In turn, this means that y - x = r 1 and z - y = r 2 for some r 1 , r 2 Q . But then z - x = ( z - y ) + ( y - z ) = r 2 + r 1 Q , so z E x , which means x z . Therefore, we’ve shown that is an equivalence relation. Hence, for any x, y [0 , 1], either x y or x y . If x y , then for any z E x , x z , so transitivity implies y z , which means z E y , so we have that E x E y . The reverse containment is given by a similar argument, so E x = E y . On the other hand, if x y , then no element of E x can be an element of E y (and vice versa ), so we see that E x E y = . 2. Exercise 9.8. (a) Show that the two expressions for μ in Example 6.5 are equivalent. Proof. Suppose, first, that A n i =1 ( a i , b i ). Let f be a step function which is equal to 1 on [ a i , b i ] for each i , and zero everywhere else. Then f χ A and 1 0 f = n i =1 ( b i - a i ) . Therefore, we see that n i =1 ( b i - a i ) : A n i =1 ( a i , b i ) 1 0 f : f a step function and f χ A on [0 , 1] . Hence, inf 1 0 f : f a step function and f χ A on [0 , 1] inf n i =1 ( b i - a i ) : A n i =1 ( a i , b i ) . 1

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On the other hand, suppose f is a step function on [0 , 1] such that f χ A . Let [ a 1 , b 1 ] , . . . , [ a n , b n ] be the intervals on which f 1. Then A n i =1 [ a i , b i ] n i =1 ( a i - 1 /m, b i + 1 /m ) for any positive integer m . Therefore, x m := n i =1 ( b
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