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Unformatted text preview: nn1
32
For each n we see easily that Þ0 l Pn, f
1 Þ0 u Pn, f
1 0 and 1
n and therefore Þ0 l Pn, f
1 lim nÝ nÝ
lim Þ0 u Pn, f
1 0. 2. Suppose that f is defined on the interval in such a way that whenever x
0, 1 and x has the form 1 for some
n
1
positive integer n we have f x 0 and whenever x belongs to an interval of the form n1 , 1 for some
n
positive integer n we have
f x 1 1 n.
Draw a rough sketch of the graph of this function and explain why it is integrable on the interval 0, 1 . Solution: 0 11 1
65 4 1
2 1
3 1 For each positive integer n we define P n to be the following partition of the interval 0, 1 :
P n 0, 1 , 1 , , 1 , 1 , 1 .
nn1
32
For each n we see easily that Þ0 l Pn, f
1 n1
j1 1
j1 1
j 1 1 j and Þ0 u Pn, f 2
n
1 n1
j1 and so 271 1
j 1
j1 1 1 j Þ0 u Pn, f Þ0 l Pn, f
1 lim nÝ 1 nÝ2 0
lim n and we have shown that f is integrable on 0, 1 .
Incidentally, we have also shown that
n1 lim
Þ0 f n Ý
1 1
j j1 1
j1 1 1 j n nÝ
lim 2
2j 2j 1 j1
n nÝ
lim 1
j 2j 1 j1 In the chapter on infinite series you will learn how to show that the latter limit is 2 2 log 2. 3. Given that f is a bounded function on an interval a, b , prove that the following conditions are equivalent:
a. The function f is integrable on the interval a, b .
b. For every number 0 there exist step functions s and S on the interval a, b such that s Þa S
b c. For every number
such that if S and f S and s. 0 there exist step functions s and S on the interval a, b such that s
E x f a, b Sx sx , we have m E .
To show that condition a implies condition b we assume that condition a holds. In other words, we
assume that f is integrable on a, b . Suppose that 0. Using the first criterion for integrability we
choose a partition P of a, b such that Þ a w P, f
b . We define S u P, f and s l P, f and observe that the functions s and S have the desired
properties.
The proof that condition a implies condition c is very similar. This time, the partition P is chosen
using the second criterion for integrability.
To prove that condition b implies condition a we assume that condition b holds. What we shall
show is that the first criterion for integrability holds. Suppose that 0. Using condition b we
choose step functions s and S on the interval a, b such that s f S and Þa S
b s. Choose a partition P of a, b such that both s and S step within P and observe that since
s l P, f
u P, f
S
we have Þ a w P, f
b Þ a u P, f
b Þa S
b l P, f s. The proof that condition c implies condition a is very similar. This time we show that f is integrable
by showing that the second criterion for integrability holds.
4. Suppose that f is a bounded function on an interval a, b and that for every number 0 there exists an
elementary subset E of a, b such that m E and such that the function f 1 E is Riemann integrable
on a, b . Prove that f must be Riemann integrable on the interval a, b . 272 Solution: In order to show that f is integrabl...
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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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