royden3 - Royden, Real Analysis 3rd ed. Chapter 3 Problem...

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Royden, Real Analysis 3rd ed. Chapter 3 Problem 3-4 Assume E 1 ,E 2 ,... are mutually disjointed, it is easy to check ¯ ¯ ¯ ¯ ¯ n [ i =1 E i ¯ ¯ ¯ ¯ ¯ = n X i =1 | E i | (1) f rst for n =2 and then for n =3 , 4 ,... and so on. We still need to verify ¯ ¯ ¯ ¯ ¯ [ i =1 E i ¯ ¯ ¯ ¯ ¯ = X i =1 | E i | . (2) It is easy to see that (2) holds when the RHS is . Otherwise, all E i = for i larger than some integer, say, n. Then (2) is reduced to (1) and we are done. Problem 3-9 By assumption m A = m ( A E )+ m ( A E c ) for any A. Us ingProb lem3-7andrep lac ing A above by A y, m A = m ( A y ) = m (( A y ) E )+ m (( A y ) E c ) = m ((( A y )+ y ) ( E + y )) + m ((( A y )+ y ) ( E c + y )) = m ( A ( E + y )) + m ( A ( E + y ) c ) . Hence E + y is measurable. Problem 3-11 Consider E n =( n, ) . Then n =1 E n = and mE n = . Problem 3-12 By Proposition 2, it is enough to show m à A [ i =1 E i ! X i =1 m ( A E i ) . By monotonicity and Lemma 9, m à A [ i =1 E i ! m à A n [ i =1 E i ! n X i =1 m ( A E
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Now let n →∞ . Problem 3-13 a. (i), (ii) and (iv) are equivalent. (i) (ii). Applying Proposition 5 to E n = E [ n,n +1] , we can f nd an open set O n such that O n E n and m ( O n ) m ( E n )+ ε 3 · 2 n . Since m E n m [ n,n +1]=1 and both O n ,E n are measurable, m ( O n \ E n )= m O n m E n ε 3 · 2 n , (1) by Lemma 9. Consider O = n = −∞ O n , wh ichisanopense t . Then O n = −∞ E n = E. By (1) and Problem 3-12, m ( O \ E )= m à [ n = −∞ ( O n \ E n ) ! X n = −∞ m ( O n \ E n ) X n = −∞ ε 3 · 2 n = ε . (2) (ii) (iv). Let ε =1 /n in (2) and we have open set O n E and m ( O n \ E ) 1 /n. Let G = n =1 O n . Then G E and 0 m ( G \ E ) m ( O n \ E ) 1 n 0 , as n
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royden3 - Royden, Real Analysis 3rd ed. Chapter 3 Problem...

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