hw5solutions

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

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Unformatted text preview: 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 6∼ 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 6∼ 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 ⊂ S 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 Z 1 f = n X i =1 ( b i- a i ) . Therefore, we see that n X i =1 ( b i- a i ) : A ⊂ n [ i =1 ( a i ,b i ) ⊆ Z 1 f : f a step function and f ≥ χ A on [0 , 1] . Hence, inf Z 1 f : f a step function and f ≥ χ A on [0 , 1] ≤ inf n X i =1 ( b i- a i ) : A ⊂ n [ i =1 ( a i ,b i ) ....
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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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hw5solutions - Math 318 HW #5 Solutions 1. Show that the...

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