1873_solutions

# From the theorem on unions that the set 0 1 q does

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Unformatted text preview: s then the boundary of S is defined to be the set S R S. Prove that if S is a subset of an interval a, b then the function  S is Riemann integrable on a, b if and only if the boundary of S has measure zero. The function  S fails to be continuous at a number x if and only if x belongs to the boundary of S. 2. If S is a set of real numbers then the boundary of S is defined to be the set S R S. Prove that if S is a subset of an interval a, b then the function  S is Riemann integrable on a, b if and only if the boundary of S has nineteenth century measure zero. The boundary of a set is closed and bounded. Therefore Exercises 1 and 2 are saying exactly the same thing. 3. Suppose that E n is a contracting sequence of closed elementary subsets of 0, 1 whose intersection C does not contain any interval of positive length. Prove that the function  C is Riemann integrable on 0, 1 if and only if the set C has measure zero. The fact that C does not contain any interval of positive length tells us that 0, 1 0, 1 C. Therefore, since C is closed, the boundary of C must be C itself. The desired result therefore follows at once from Exercise 1. 4. Suppose that 0   1/3. The -Cantor set C  is defined as follows: We begin by defining E 0   0, 1 . For every nonnegative integer n, if the elementary set E n has already been defined then we obtain E n1  from E n  by removing from each of its component intervals the centrally located open interval of length  1/3 n . Then we define C  Ý  En  . n1 a. Show that   1/3 then the -Cantor set is just the “usual” Cantor set C that was defined earlier. Solution: This assertion was proved in our earlier discussion of the Cantor set. b. Prove that the -Cantor set contains no interval of positive length. 399 Solution: Since each set E n  is the union of 2 n closed intervals each of length less than C contains no interval of positive length follows from the discussion of the Cantor set that was given earlier. 1/2 n , that c. Evaluate the integral Þ0 E 1 n and deduce that the -Cantor set has measure zero if and only if   1/3. Because of this fact, it is traditional to call the Cantor set C 1/3 the thin Cantor set and to call the sets C  fat Cantor sets when 0    1/3. Solution: For each n, the set E n  is obtained by removing 2 n non-overlapping open intervals of length  1/3 from the set E n 1  . Thus E n  is obtained from E n 1  by removing an elementary set with measure  2/3 n 1 . Therefore, if n is any positive integer, the set E n  is obtained from the interval 0, 1 by removing an elementary set with measure n1 n j1 j1 2 3  3 1 2 3 n and we conclude that n 3  3 2 3 0 as n Ý that the set C  has In the event that   1/3, it follows from the fact that m E n  measure zero. Suppose now that   1/3. Since the set C  is closed and bounded, in order to show that C  fails to have measure zero, all we have to do is show that whenever an elementary set E includes C  , we have m E 1 3. To...
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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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