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Unformatted text preview: e on a, b we shall show that f satisfies the second criterion for integrability. Suppose that 0.
Using the given property of f we choose an elementary subset A of a, b such that m A /2 and such
that the function f 1 A is integrable on a, b . Choose a partition P 1 of a, b such that the function A
steps within P. Now, using the fact that f 1 A satisfies the second criterion for integrability we choose
a partition P 2 of a, b such that if we define
B x
a, b
w P, f 1 A x
.
then m B /2. We now define P to be the common refinement of P 1 and P 2 and we express P as
P x 0 , x 1 , , x n .
For each j 1, 2, , n, if the open interval x j 1 , x j is not included in A Þ B then, since the functions f
and f 1 A agree in the interval x j 1 , x j the condition
must hold whenever x w P, f x w P, f 1 A
x j 1 , x j . Since
m AÞB
m A m B x
2
2
we have succeeded in showing that f satisfies the second criterion for integrability.
5.
a. Suppose that f is a nonnegative function defined on an interval a, b and that for every number
set
x
a, b
fx 0, the is finite. Prove that f must be integrable on a, b and that Þ f 0.
a
We shall show that f satisfies the first criterion for integrability. Suppose that 0. Choose a
finite set S such that
fx
ba
whenever x
a, b
S and define P to be the partition of a, b whose points are the numbers a
and b and the members of S arranged in increasing order. Since l P, f is nonnegative and
u P, f never exceeds the value / b a , we have
b Þ a w P, f
b Þ a u P, f
b Þ a u P, f
b l P, f . Since f satisfies the first criterion for integrability, f is integrable on a, b and the same
b
b
from which we deduce that Þ f 0.
argument shows that whenever 0 we have Þ f
a a b. Prove that if f is the ruler function that was introduced in an earlier example then f is an integrable
function on the interval 0, 1 , even though f is discontinuous at every rational number in the interval.
The ruler function obviously has the property described in part a.
6. Given that f is a bounded nonnegative function defined on an interval a, b , prove that the following
conditions are equivalent:
a. The function f is integrable on the interval a, b and Þ f 0.
b a b. For every number 0 there exists an elementary set E such that m E
x
a, b
fx
E. and such that To show that condition a implies condition b we assume that f is integrable on a, b . Suppose that
0. To obtain the desired set E we shall use the same sort of technique as was used in the proof
of Theorem 11.8.4. Choose a step function S f such that 273 Þa S
b 2 and define
E x a, b . Sx We observe that
x
a, b
fx
E.
Now since S is a step function, the set E is elementary and we have
2 Þ a S ÞE S ÞE
b mE from which we deduce that m E .
To show that condition b implies condition a we assume that condition b holds. Once again we
borrow from the proof of proof of Theorem 11.8.4. Using the fact that f is bounded we choose a
number k such that f x k fo...
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 Fall '08
 STAFF
 Math, Calculus

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