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 n1
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 .
n1 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 nonoverlapping 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
j1 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.
 Fall '08
 STAFF
 Math, Calculus

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