1873_solutions

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n and such that m U n  m E n  n 1 . 2 b. For each n define n Vn   Uj, j1 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 n1 Ý  En U , n1 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 n1 it follows from the theorem on unions that U is -null. Some Exercises on Integrability 1. If S is a set of real number...
View Full Document

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.

Ask a homework question - tutors are online