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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,
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...
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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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