1873_solutions

1873_solutions

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Unformatted text preview: a A  Þa B Þa A B b b  m A m B b mA B. 2. Prove that if E is an elementary set and m E  0 then E must be finite. Choose an interval a, b such that the function  E is zero outside a, b and a partition P  x 0 , x 1 , , x n of a, b such that  E steps within P. Since  E is nonnegative and its sum over P is zero, the constant value of  E in each interval x j 1 x j must be zero. Therefore  E can be nonzero only at points of P, in other words, every member of E is a point of P and we conclude that the set E is finite. 3. Explain why the set of all rational numbers in the interval 0, 1 is not elementary. The desired result will be clear when we have proved the stronger assertion that is made in Exercise 4. 4. Prove that if E is an elementary subset of 0, 1 and if every rational number in the interval belongs to E then the set 0, 1 E must be finite. We assume that E is an elementary subset of 0, 1 and that every rational number in 0, 1 belongs to E. Choose a partition P  x 0 , x 1 , , x n of a, b such that  E steps within P. For each j, since the interval x j 1 , x j contains some rational numbers, the constant value of  E in x j 1 , x j must be 1. Since the numbers in 0, 1 that do not belong to E have to be points of P, the set 0, 1 E must be finite. 5. Give an example of a set A of numbers such that if E is any elementary subset of A we have m E  0 and if E is any elementary set that includes A we have m E 1. Hint: Take another look at Exercise 4. The set 0, 1 Q has the desired properties. 6. Given that E is an elementary set that is not closed and that F is a closed elementary subset of E, prove that m E F  0. 265 Solution: Choose a lower bound a and an upper bound b of the set E. Since the set E F is elementary, if we want to show that m E F  0 then, from Exercise 2, all we have to show is that the set E F cannot be finite. The fact that E F is not finite follows from the fact that finite sets are always closed, that F is closed and that the set E, which isn’t closed is the union of the two sets F and E F. 7. Given that f is a step function, that E is an elementary set and that f x  0 whenever x R E, prove that Ý ÞE f  Þ Ý f. Solution: The desired equality follows at once from the definitions and the fact that f  f E . 8. Given that f and g are step functions, that E is an elementary set and that f x that g x whenever x E, prove ÞE f ÞE g. Choose an interval a, b outside of which both of the functions f and g are zero. Since f E follows from the nonnegativity property of integrals of step functions that g E , it ÞE f  Þ a f E Þ a g E  ÞE g b b 9. Given that f is a nonnegative step function, that A and B are elementary sets and that A B, prove that ÞA f ÞB f. The desired inequality follows at once from the fact that f A f B . We choose an interval a, b that includes the set B and use the nonnegativity property to obtain ÞA f  Þ a f A Þ a f B  ÞB f b b 10. Given that f is a step function and that E i...
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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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