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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 onscreen 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.
 Fall '08
 STAFF
 Math, Calculus

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