1873_solutions

# 2p 2 2p 1 2p and so 2p 2 2p 1 1 2p 2p and

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Unformatted text preview: n and such that m U n  m E n  n 1 . 2 b. For each n define n Vn   Uj, j1 observe that Vn  Un Þ U1 E1 Þ U2 and then deduce that m V n  for each n. E2 Þ  Þ Un 1 En 1 9. Prove that if  is an increasing function and A is -nul and if  0 then there exists an expanding sequence U n of open elementary sets whose union includes A and for which var , U n  for every n. The solution to this exercise is almost identical to that of Exercise 8. 10. Prove that if  is an increasing function and if H is a closed bounded -null set then for every  0 there exists an elementary set E such that H E and var , E  . The solution to this exercise is almost a carbon copy of the proof of the theorem on closed bounded sets of measure zero. 11. Prove that if  is an increasing function and x is a real number then the set x is -null if and only if  is continuous at x. A direct proof of this assertion is very simple but it also follows at once from Exercise 10. 12. Prove that if  is an increasing function and U is an interval then U is -null if and only if  is constant on U. In the event that an increasing function  is constant on an interval U, the set U is clearly -null. In the event that an increasing function  is not constant on an interval U, then  fails to be constant on a closed bounded interval H that is included in U, and it follows from Exercise 10 that the subset H fails to be -null. 13. Prove that if  is an increasing function and A is a countable set then A is -null if and only if  is continuous on A. This assertion follows at once from Exercise 11 and the theorem on unions. 14. Prove that if  is the Cantor function and C is the Cantor set then, although C has measure zero, it is not -null. As we know, the Cantor set C has measure zero. Now suppose that  is the Cantor function. To obtain a contradiction, suppose that the Cantor set C is  null, and, using Exercise 10, choose an elementary set E that includes C such that var , E  1. Choose an open elementary set U such that E U and var , U  1. We now define the elementary sets E n as in the discussion of the Cantor set. Since Ý  En U n1 Ý  En U , n1 it follows from the the Cantor intersection theorem that for some positive integer we have E n 398 U. Since var , E n  1 we have contradicted the fact that var , U  1. 15. Prove that if  is an increasing function and U is an open set that is not -null then U has a closed bounded subset that is not -null. We shall show that if every closed bounded subset of a given open set U is -null then U is -null. Suppose that U is open and that every closed bounded subset of U is -null. Suppose that  0. For each positive integer n we define 1 Hn  x R R U x n, n n where  R U stands for the distance function of the set R U. Since each set H n is closed and bounded, it is -null. Therefore, since U Ý  Hn n1 it follows from the theorem on unions that U is -null. Some Exercises on Integrability 1. If S is a set of real number...
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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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