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Unformatted text preview: 2n respectively in
2j 1
2j
each interval 1 2n , 1 2n of the partition.
We see that
Þ 1 Sn
1 as n 2n s n d 0J , 0
j1 2
2n 2
n 1
n 0 Ý. Now for each n we have
2n Þ 1 S n d 1J , 0
1 and the latter expression approaches 3 as n 2j
2n
j1
1 2nn 1 1
n Ý. 7. Prove that the integral Þ 0 xd
1 exists and evaluate it.
We define x x whenever x
Pn x 0, 1 . For each positive integer n we define
0 2 , 1 2 , 2 2 , , n 2 ,
n
n
n
n
j
n we define s n and S n to be the step functions that take the value
partition P n and that takes the constant values
j1
n 2 , j
n 2 2 j1
n and j
n 2 2 at each point j
n 2 of the respectively on each interval of the partition. For each n we see that Þ0 Sn
1 n s n d
j1 j
n 2 and the latter expression approaches 0 as n
each n we see that 306 j 1
n 2 j
n j 1
n 1
n Ý. Therefore the required integral exists. Now for Þ 0 S n d n 1 j1
1
3 and the latter expression approaches 2 j
n j j
n 1
n 2
2n 32n 1
6n Ý. Therefore as n Þ 0 xd
1 x 1.
3 Some Exercises on RiemannStieltjes Integrability
1. Suppose that
1
n 1 if x has the form fx for some positive integer n
. 0 otherwise
Prove that f is Riemann integrable on the interval 0, 1 and that Þ f 0.
1 0 Solution: 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 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 Riemann 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
j 307 1
j1 1 1 j and
n1 Þ0 u Pn, f
1 2
n 1
j1 1
j j1 1 1 j and so 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
lim
Þ0 f n Ý
1 n1 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 is an increasing function f is a bounded function on an interval a, b , prove that the following
conditions are equivalent:
a. The function f is integrable with respect to 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 a, b S and f S and s d . 0 there exist step functions s and S on the interval a, b such that s
E x f Sx sx , we have var , E .
d. For eve...
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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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