MATH 104, SUMMER 2006, HOMEWORK 12 SOLUTION
BENJAMIN JOHNSON
Due August 16
Assignment:
Section 33: 33.2, 33.4, 33.12
Section 34: 34.2, 34.5, 34.6
Section 33
33.2 Let S be a non-empty bounded subset of R. For fixed c > 0, let cS = {cs : s ∈ S }. Show that
sup(cS ) = c · sup(S ) and inf(cS ) = c · inf(S ).
Proof. Let x ∈ cS . Then x = c · s for some s ∈ S. Since inf S is a lower bound for S , and sup S is
an upper bound for S , we have inf S ≤ s ≤ sup S . Hence c · inf S ≤ x ≤ c · sup S .
Let m, M ∈ R. Assume that m is a lower bound for cS and that M is an upper bound for cS .
Then (∀x ∈ cS )(m ≤ x ≤ M). Let s ∈ S. Then m ≤ cs ≤ M, so mc ≤ s ≤ Mc . Thus mc is a lower
bound for S , and Mc is an upper bound for S . Since inf S is the greatest lower bound for S and
sup S is the least upper bound for S , we must have mc ≤ inf S and sup S ≤ Mc . Hence m ≤ c · inf S
and c · sup S ≤ M.
We’ve shown that c · inf S satisfies the definition for inf cS , and that c · sup S satisfies the
definition for sup cS . Since suprema are unique, c · inf S = inf cS and c · sup(S ) = sup(cS ).
33.4 Give an example of a function f on [0, 1] that is not integrable for which | f | is integrable.



1 if x ∈ [0, 1] ∩ Q
Define f (x) = 

−1 if x ∈ [0, 1] ∩ (R r Q)
It is easy to show that L01 ( f ) = −1 and U01 ( f ) = 1, so f is not integrable over [0, 1]. But | f | is the
R1
constant function 1, which is integrable over [0, 1], with 0 | f | = 1.
33.12 Let f be the function
described in Exercise 17.14.



0 if x ∈ R r Q
]
[I.e., f (x) = 
1


if
x
∈
Q
(and
x
is
reduced
to
lowest
terms)
the denominator of x
(a) Show that f is not piecewise continuous or piecewise monotonic on any interval [a, b].
Proof. Let a < b ∈ R.
To say that f is not piecewise monotone on [a, b] is to say precisely, ...
(∀P : P is a size-n partition of [a, b])(∃k ∈ N : k < n)( f is not monotone on [tk , tk+1 ]).
Let P be any partition of [a, b]. Consider [t0 , t1 ]. There exist two distinct irrational numbers
x < y in [t0 , t1 ], and a rational number c between them. We have f (x) = f (y) = 0, and
f (c) > 0. Since x < c and f (x) < f (c), f is not decreasing. Since c < y and f (c) > f (y), f is
not increasing. So f is not monotone on [a, t1 ].
Date: August 16, 2006.
1
2
BENJAMIN JOHNSON
Analogously, to say that f is not piecewise continuous on [a, b] is to say precisely, ...
(∀P : P is a size-n partition of [a, b])(∃k ∈ N : k < n)( f is not continuous on [tk , tk+1 ]).
Let P be any partition of [a, b]. Consider [t0 , t1 ]. There exists at least one rational number
c ∈ [t0 , t1 ]. Exercise 17.4 (for which I distributed a solution) showed that f is discontinuous
at each rational point of R. So f is not continuous at c and hence not continuous on [t0 , t1 ].
Rb
(b) Show f is integrable on every interval [a, b] and that a f = 0.
Proof. Let a <b ∈ R. Let > 0. Fix n to be an integer with n > b−a
. Then 1n < b−a
.
1


 f (x) if f (x) ≤ n
Define g(x) = 
.

0 if f (x) > 1
n
Observe that 0 ≤ g(x) ≤ n1 for every x ∈ [a, b] (since f – hence g – is non-negative, and by the
definition of g). Also observe that on any interval of the form [tk , tk+1 ], there exists at least one
irrational number c; and for this c we have g(c) = f (c) = 0. So inf{g(x) : x ∈ [tk , tk+1 ]} = 0.
Let P be any partition of [a, b].
Then
n−1
X
b
inf{g(x) : x ∈ [tk , tk+1 ]} · (tk+1 − tk )
La (g, P) =
k=0
=
n−1
X
0 · (tk+1 − tk )
k=0
=0
and
Uab (g, P)
=
n−1
X
sup{g(x) : x ∈ [tk , tk+1 ]} · (tk+1 − tk )
k=0
≤
n−1
X
1
k=0
n
(tk+1 − tk )
1
= (tn − t0 )
n
1
= (b − a)
n
<
(b − a)
b−a
=
So U(g, P) − L(g, P) < . This shows g is integrable.
R b The calculation involving L(g, P) was
b
exact, and implies that La (g) = 0. So we must have a g = 0.
Now the solution to Exercise 17.4 (HW 6) proves that the set {x ∈ [a, b] : f (x) < n1 } is finite.
So g differs from f at only finitely
many
Rb
R b points of [a, b]. By Exercise 32.7 (HW 11) we have
that f is also integrable and a f = a g = 0.
Section 34
34.2 Calculate
MATH 104, SUMMER 2006, HOMEWORK 12 SOLUTION
(a) lim x→0
3
R
1 x t2
e dt
x 0
Rx 2
t
Let F(x) =
e dt. Then
Z
1 x t2
F(x)
lim
e dt = lim
x→0 x 0
x→0
x
Z 0
F(x) − F(0)
2
= lim
(since F(0) =
et dt = 0)
x→0
x−0
0
0
= F (0) (definition of the derivative)
0
2
= e0 (by FTC)
=1
(b) limh→0
R
1 3+h t2
e dt
h 3
An alternative definiton for the derivative of a function at a that you might remember from
f (a)
Stewart’s Calculus book, is f 0 (a) = limh→0 f (a+h)−
. The fact that this new limit agrees with
h
our definiton for f 0 (a) is given as an exercise 28.14. (It follows easily from L’Hospital’s rule,
0
= f 0 (a)).
since limh→0 f (a + h) − f (a) = 0, limh→0 h = 0, and limh→0 f (a+h)
1
Rx 2
Let F(x) = 0 et dt. Then
R 3+h 2
R3 2
Z
et dt − 0 et dt
1 3+h t2
0
lim
e dt = lim
h→0 h 3
h→0
h
F(3 + h) − F(3)
= lim
h→0
h
= F 0 (3)
2
= e3
= e9
34.5 Let f be continuous on R and define
F(x) =
x+1
Z
f (t)dt for x ∈ R.
x−1
Show that F is differentiable and compute F 0 .
Rx
Define G(x) = 0 f (t)dt. Define k(x) = x − 1. Define h(x) = x + 1. Then F = (G ◦ h) − (G ◦ k).
Since G, h, and k are all differentiable on R, so is F, and by the various derivative rules (including
the chain rule), F 0 (x) = G0 (h(x))h0 (x) − G0 (k(x))k0 (x) = f (h(x)) − f (k(x)) = f (x + 1) − f (x − 1).
34.6 Let f be continuous on R and define
G(x) =
sin x
Z
f (t)dt for x ∈ R.
0
Show that G is differentiable and compute G0 .
Rx
Define F(x) = 0 f (t)dt. Define h(x) = sin x. Then G = (F ◦h). Since F, and h are differentiable
on R, so is G, and by the chain rule, G0 (x) = F 0 (h(x))h0 (x) = f (sin x) · cos x.