1873_solutions

The interval 0 2 then since the irrationality of 2

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Unformatted text preview: s an elementary set, prove that ÞE f ÞE |f |. Hint: Use the fact that |f | E f E |f | E . 11. Given that A and B are elementary sets, prove that ÞA  B  ÞB  A  m A B. Solution: Choose a lower bound a and an upper bound b of the set A Þ B. We see that ÞA  B  Þ a  B  A  Þ a  A  B  ÞB  A . b b The fact that these expressions are equal to m A B follows at once from the fact that AB  A B. The on-screen version of this book contains a special group of exercises that are designed to be done as a special project. These exercises require you to have read some of the chapter on infinite series. To reach this group of exercises, click on the icon . Additional Exercises on Elementary Sets and Infinite 266 Series 1. Given that H is a closed elementary set and U n is a sequence of open elementary sets and that Ý H  Un, n1 use this earlier exercise to deduce that, for some positive integer N we have N mH m Un n1 and deduce that Ý m Un . mH n1 Choose a positive integer N such that N H  Un. n1 We have Ý N n1 mH m N n1  Un m Un . m Un n1 2. Given that E is an elementary set and that U n is a sequence of open elementary sets and that Ý E  Un, n1 prove that Ý m Un . mE n1 Hint: Make use of the theorem on approximation by open sets and closed sets. Given any closed subset H of E we know from Exercise 1 that Ý m Un . mH Since Ý m n1 n1 U n is an upper bound of the set mH H is elementary and closed and H E and since m E is the least upper bound of this set we have Ý m Un . mE n1 3. Given that A n is a sequence of elementary sets and that  0 and that the series m A n is convergent, and given that for each positive integer n the set U n is an open elementary set that includes A n and satisfies the inequality m Un  m An  n , 2 prove that Ý Ý m Un  n1 m An  . n1 The desired result follows at once from the fact that 267 Ý . 2n n1 4. Given that E is an elementary set and that A n is a sequence of elementary sets and that Ý  An, E n1 prove that Ý m An . mE n1 To obtain a contradiction, assume that Ý mE  m An . n1 Choose  0 such that Ý Ý m An  mE   n1 Ý m An  n1 n1 2n . For each n, choose an open elementary set U n that includes A n such that m Un  m An  n . 2 We see that Ý Ý Ý m An  mE  n1 n1 2n  m Un n1 which, in view of Exercise 2, is impossible because Ý  Un. E n1 5. Suppose that E is an elementary set and that A n is a sequence of elementary sets with the property that whenever i and j are positive integers and i j we have Ai Aj  . Suppose that E is an elementary set and that E Ý  An. n1 Prove that Ý mE  m An . n1 Given any postiive integer N we see that N N mE m  An n1  m An . n1 Therefore Ý N mE m An lim NÝ n1 and the desired result therefore follows from Exercise 4. 268  m An , n1 Some Exercises on the Riemann Integral 1. Prove that the integral Þ 1 3x 2 dx 4 exists and has the value 63. Solution: This exercise will become obsolete when we reach the fundamen...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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