Math 432 - Real Analysis II
Solutions to Homework due February 22
Let f be a bounded function defined on [a, b] and let S ⊂ [a, b]. Define
M (f, S) = sup{f (x) | x ∈ S} and m(f, S) = inf{f (x) | x ∈ S}.
Given a partition
P = {a = t0 < t1 < t2 < . . . < tn = b},
define the Upper Darboux Sum of f with respect to P to be
U (f, P ) =
n
X
M (f, [tk−1 , tk ]) · (tk − tk−1 )
k=1
and the Lower Darboux Sum of f with respect to P to be
L(f, P ) =
n
X
m(f, [tk−1 , tk ]) · (tk − tk−1 ).
k=1
Question 1. For the interval [0, 1], consider its partition into n equally sized subintervals. In other words,
k
n−1
1 2
,1 .
P = 0, , , . . . , , . . . ,
n n
n
n
(a) With the partition P mentioned above and f (x) = x, compute U (f, P ) and L(f, P ). Your answer should
be in terms of n.
(b) With the partition P mentioned above and f (x) = x2 , compute U (f, P ) and L(f, P ). Your answer
should be in terms of n.
Solution 1.
(a) First, we compute M (f, P ) and m(f, P ). Since f (x)
= x is an increasing, continuous function, it will
k−1 k
attain its max at the right endpoint of [tk−1 , tk ] =
,
. Thus,
n
n
M (f, P ) =
k
.
n
Similarly, the minimum will be attained at its left endpoint, and so
m(f, P ) =
k−1
.
n
Also, note that tk−1 − tk = 1/n. Computing the upper Darboux sum, we get
U (f, P ) =
n
n
X
k1
1 X
n(n + 1)
= 2
k=
.
nn
n
2n2
k=1
k=1
Similarly, the Lower Darboux sum is given by
L(f, P ) =
n
X
(k − 1) 1
k=1
n
n
=
n
n−1
1 X
1 X
(n − 1)(n)
k
−
1
=
j=
.
n2
n2 j=0
2n2
k=1
1
(b) We compute M (f, P ) and m(f, P ). Since f (x) = x2 is an increasing, continuous function on [0, 1], it
will attain its max and the right endpoint of every sub-interval. Thus,
M (f, P ) =
k2
.
n2
Similarly, the min will occur at the left endpoint and
m(f, P ) =
(k − 1)2
.
n2
Computing the upper Darboux sum, we get that
M (f, P ) =
n
n
X
k2 1
1 X
n(n + 1)(2n + 1)
.
=
k=
n2 n
n3
6n3
k=1
k=1
For the Lower Darboux sum, we get
L(f, P ) =
n
n−1
X
(k − 1)2 1
1 X
(n − 1)(n)(2n − 1)
.
=
j=
n2
n
n3 j=1
6n3
k=1
Question 2. Consider the function
f (x) =
0 if x ∈ Q
.
1 if x 6∈ Q
(a) For any interval [t, t0 ] ⊂ R with t < t0 , compute M (f, [t, t0 ]) and m(f, [t, t0 ]).
(b) Consider the interval [a, b] for any a < b. Let P = {a = t0 < t1 < t2 < . . . < tn = b} be any partition of
[a, b]. Use (a) to compute U (f, P ) and L(f, P ).
Solution 2.
(a) Inside any interval [t, t0 ] there exists infinitely many rational and irrational numbers. Thus, for our f (x),
the max and min are attained at the irrational and rational values, respectively. Thus,
M (f, [t, t0 ]) = 1 and m(f, [t, t0 ]) = 0.
(b) Since m(f, [t, t0 ]) = 0, then for any partition of [a, b], we get a sum of zeroes and thus
L(f, P ) = 0.
For the Upper Darboux sum, since every M (f, [t, t0 ]) = 1, we end up with a sum of the length of the
intermediate subintervals and thus
U (f, P ) = b − a.
We defined the upper and lower Darboux integrals of f on [a, b] in the following way, respectively:
U (f ) = inf{U (f, P ) | P is a partition of [a, b]}
L(f ) = sup{L(f, P ) | P is a partition of [a, b]}.
We say that f is (Darboux) integrable over [a, b] if L(f ) = U (f ).
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Question 3. We will use the above computations to prove that the functions in Question 1 are integrable
and the function in 2 is not.
(a) Let Pn be the partition of n equal subintervals of [0, 1] described in Question 1. Show that for any of
these partitions Pn , the following is true:
1
≤ U (x, Pn )
2
L(x, Pn ) ≤
.
(b) As in (a), show that
1
≤ U (x2 , Pn ).
3
L(x2 , Pn ) ≤
(c) For each Pn , compute
U (x, Pn ) − L(x, Pn ) and U (x2 , Pn ) − L(x2 , Pn ).
Be sure to simplify as much as possible.
(d) Use your results from (c) and a theorem from class to show that both x and x2 are integrable on [a, b]
and that
Z 1
Z 1
1
1
and
x2 dx = .
x dx =
2
3
0
0
(e) Use your computations from Question 2 to show that this f (x) is non-integrable on any interval [a, b]
with a < b.
Solution 3.
(a) Previously, we computed that L(x, Pn ) =
n(n−1)
2n2 .
Note that
n(n − 1)
n2 − n 1
1
=
· ≤ .
2
2n
n2
2
2
Similarly, we compute that U (x, Pn ) =
n(n+1)
2n2 .
Thus,
n2 + n 1
1
n(n + 1)
=
≥ .
2n2
n2 2
2
Thus, L(x, Pn ) ≤
1
2
≤ U (x, Pn ).
(b) For f (x) = x2 , we compute that L(x2 , Pn ) = (n−1)(n)(2n−1)
. Since n − 1 < n and 2n − 1 < 2n, we have
6n3
that
(n − 1)(n)(2n − 1)
2n3
1
≤
= .
6n3
6n3
3
Furthermore, we computed U (x2 , Pn ) =
n(n+1)(2n+1)
.
6n2
Since n < n + 1 and 2n < 2n + 1, we get that
n(n + 1)(2n + 1)
2n3
1
≥
= .
3
3
6n
6n
3
Thus, we have that L(x2 , Pn ) ≤
1
3
≤ U (x2 , Pn ).
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(c) Computing the difference, we get that
U (x, Pn ) − L(x, Pn ) =
2n
1
n(n + 1) n(n − 1)
−
= 2 = .
2n2
2n2
2n
n
For the next function, we get that
U (x2 , Pn ) − L(x2 , Pn ) =
n(n + 1)(2n + 1) (n − 1)(n)(2n − 1)
−
=
6n3
6n3
1
6n2
= .
6n3
n
(d) Since U (x, Pn ) − L(x, Pn ) = n1 and we can choose an n large enough so that n1 < ε for any ε > 0, we get
that U (x, Pn ) − L(x, Pn ) < ε. Thus, f (x) = x is integrable. Notice that L(x, Pn ) is arbitrarily close to
1/2 and always strictly less than 1/2. Thus, L(f ) ≥ 1/2. Similarly, since U (x, Pn ) ≥ 1/2 and arbitrarily
R1
close to it, then U (f ) ≤ 1/2. Since f is integrable, 1/2 ≤ U (f ) = L(f ) ≤ 1/2. Thus, 0 x dx = 21 .
R1 2
Similar arguments give you that 0 x dx = 13 .
(e) We proved in Question 2 that for any interval [a, b] and for any partition P , U (f, P ) = b − a and
L(f, P ) = 0. Thus, since this is true for any partition, U (f ) = b − a and L(f ) = 0. Since L(f ) 6= U (f ),
f is not integrable over [a, b].
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