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b b m A m B b mA B. 2. Prove that if E is an elementary set and m E 0 then E must be finite.
Choose an interval a, b such that the function E is zero outside a, b and a partition
P x 0 , x 1 , , x n
of a, b such that E steps within P. Since E is nonnegative and its sum over P is zero, the
constant value of E in each interval x j 1 x j must be zero. Therefore E can be nonzero only at
points of P, in other words, every member of E is a point of P and we conclude that the set E is
finite.
3. Explain why the set of all rational numbers in the interval 0, 1 is not elementary.
The desired result will be clear when we have proved the stronger assertion that is made in
Exercise 4.
4. Prove that if E is an elementary subset of 0, 1 and if every rational number in the interval belongs to E then
the set 0, 1
E must be finite.
We assume that E is an elementary subset of 0, 1 and that every rational number in 0, 1 belongs
to E. Choose a partition
P x 0 , x 1 , , x n
of a, b such that E steps within P. For each j, since the interval x j 1 , x j contains some rational
numbers, the constant value of E in x j 1 , x j must be 1. Since the numbers in 0, 1 that do not
belong to E have to be points of P, the set 0, 1
E must be finite.
5. Give an example of a set A of numbers such that if E is any elementary subset of A we have m E 0 and if
E is any elementary set that includes A we have m E
1. Hint: Take another look at Exercise 4.
The set 0, 1 Q has the desired properties. 6. Given that E is an elementary set that is not closed and that F is a closed elementary subset of E, prove that
m E F 0. 265 Solution: Choose a lower bound a and an upper bound b of the set E. Since the set E F is
elementary, if we want to show that m E F 0 then, from Exercise 2, all we have to show is that the set
E F cannot be finite. The fact that E F is not finite follows from the fact that finite sets are always
closed, that F is closed and that the set E, which isn’t closed is the union of the two sets F and E F. 7. Given that f is a step function, that E is an elementary set and that f x 0 whenever x R E, prove that Ý ÞE f Þ Ý f. Solution: The desired equality follows at once from the definitions and the fact that
f f E .
8. Given that f and g are step functions, that E is an elementary set and that f x
that g x whenever x E, prove ÞE f ÞE g.
Choose an interval a, b outside of which both of the functions f and g are zero. Since f E
follows from the nonnegativity property of integrals of step functions that g E , it ÞE f Þ a f E Þ a g E ÞE g
b b 9. Given that f is a nonnegative step function, that A and B are elementary sets and that A B, prove that ÞA f ÞB f.
The desired inequality follows at once from the fact that f A f B . We choose an interval a, b that
includes the set B and use the nonnegativity property to obtain ÞA f Þ a f A Þ a f B ÞB f
b b 10. Given that f is a step function and that E i...
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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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