(047104222-6) Bartle - Solutions manual Bartle -The Elements of Integration and Lebesgue Measure (2)

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This is not an official Solutions manual. 
I found this solutions online and 
decided to join them in this pdf in order 
to make my life easier.
This pdf has two parts : 
1- Solutions from University of Oslo
2 - Solutions from University of 
Washington 
Credits are in the beginning of each 
part 
I hope it helps... 
Solutions to The Elements 
of Integration and 
Lebesgue Measure - 
Robert G Bartle
Solutions University 
of Oslo
Credits : Hans Brodersen
Matematisk Institutt 
Universitetet i Oslo 
MAT 3300/4300: Solutions for problems assigned week 35 and 36.
Problem 2.A.
We have that [a, b] ⇢ (a� 1n , b +
1
n ) for each n, and consequently that
[a, b] ⇢
1T
n=1
(a� 1n , b +
1
n ). Assume x < a, and choose k such that x < a�
1
k . Then
x /2 (a� 1k , b+
1
k ), and it follows that x /2
1T
n=1
(a� 1n , b+
1
n ). By a similar argument,
x > b implies that x /2
1T
n=1
(a� 1n , b+
1
n ), and it follows that [a, b] =
1T
n=1
(a� 1n , b+
1
n ).
Since every �-algebra is closed under countable intersections, we will get that
every �-algebra which contains all open intervals (a, b) also also contains all closed
intervals [a, b] (and from this we will get that every closed interval [a, b] is a Borel
set).
For each n we have that [a + 1n , b�
1
n ] ⇢ (a, b), so we get that
1S
n=1
[a + 1n , b�
1
n ] ⇢ (a, b). If x 2 (a, b) and
1
n < min{x� a, b� x}, then
x 2 [a + 1n , b�
1
n ].
From this it follows that (a, b) =
1S
n=1
[a + 1n , b�
1
n ] and this shows that every �-
algebra that contains all closed intervals [a, b] also contains all open intervals (a, b)
(and this shows that the Borel algebra B also is equal the �-algebra generated of
all closed intervals [a, b]).
Problem 2.B.
Let C be the �-algebra generated by all half-open intervals (a, b]. Since
(a, b) =
1T
n=1
(a, b� 1n ], it follows that (a, b) belongs to C and consequently that every
Borel set also belongs to C. On the other hand, we have that (a, b] =
1T
n=1
(a, b+ 1n ).
This shows that (a, b] is a Borel set, and consequently that every set which belongs
to C also is a Borel set. All together we get that C = B.
Since (a,1) =
1S
n=1
(a, a + n) and
(a, b) = (�1, b)\ (a,1) = (C(
1T
n=1
(b� 1n ,1)))\ (a,1), it is easy to see that every
set which belongs to the �-algebra generated by the intervals (a,1) is a Borel set
and vice versa. It follows that B also is generated by the intervals (a,1).
Problem 2.C.
(En) is monotonically increasing since En ⇢ En [ An+1 = En+1. Consider n and
m with m < n. Let x 2 Fn. Then x 2 An and x /2 En�1 and consequently x /2 Ak
when k < n. Especially we have x /2 Am. This implies that x /2 Fm and we get
that Fn \ Fm = ?.
It is obvious that
1S
n=1
En =
1S
n=1
An and that
1S
n=1
Fn ⇢
1S
n=1
An. Let x 2 An.
Let k be the smalest integer such that x 2 Ak. Then x 2 Fk. This shows that
1
2
1S
n=1
An ⇢
1S
n=1
Fn and we therefore get that
1[
n=1
En =
1[
n=1
Fn =
1[
n=1
An.
Problem 2.D.
If x 2 A, then x 2 An for infinitely many integers n, so for each integer m we
can find nm � m such that x 2 Anm . Since nm > m, this means that for all m
x 2
1S
n=m
An. We therefore also get that x 2
1T
m=1
[
1S
n=m
An].
Conversely, let x 2
1T
m=1
[
1S
n=m
An]. For each m we have that x 2
1S
n=m
An, so for each
m, we can consequently find nm � m such that x 2 Anm . Since each nm � m, the
set of all these nm’s must be an infinite set, so we get that x 2 A.
Problem 2.E.
Let x 2 B. If x 2 An for all n, we obviously have that x 2
1S
m=1
[
1T
n=m
An]. Otherwise
we can find finitely many integers n1, .., nk such that x 2 An for all n such that
n 6= nj , j = 1, .., k. We may assume that n1 < n2 · · · < nk. Then x 2 An for all
n > nk, so x 2
1T
n=nk+1
An, and we consequently have that x 2
1S
m=1
[
1T
n=m
An].
Conversely; assume that x 2
1S
m=1
[
1T
n=m
An]. Then there exist m such that x 2
1T
n=m
An. This means that x 2 An for all n � m, so x 2 An for all n except possibly
for n = 1, ..,m� 1. So x 2 B.
Problem 2.F.
Assume (An) is a general sequence of subsets of X. If x belongs to all but a finite
number of the sets An, x also belongs to infinitely many of the sets An. In general,
we therefore always have lim inf An ⇢ lim supAn. Now assumme that (En) is a
monotone increasing sequence. It is obvious that lim supEn is a subset of
1S
n=1
En,
and we therefore only have to prove that
1S
n=1
En ⇢ lim inf En. Let x 2
1S
n=1
En.
Then there is k such that x 2 Ek. Since the sets En are monotone increasing, we
have that x 2 En for all n except possibly for n = 1, .., k � 1. So x belongs to all
but a finite number of the sets En hence x 2 lim inf En and we are done.
Problem 2.H.
In problem 2.F we proved that lim inf En ⇢ lim supEn and the other inclusions are
obvious.
Now let S
1
= {z 2 C ( complex numbers ) : |z| = 1}. Let N be an integer
N > 3 and let An = { z = ei✓ : ✓ 2 [ 2(n�1)⇡N ,
2n�1⇡
N ]}. Now it is clear that for all
m,
1S
n=m
An = S1, so lim sup An = S1. On the other hand since An \An+2 = ?, we
must have
1T
n=m
An = ? for all m so lim inf An = ?.
MAT 3300/4300: Solutions for problems assigned week 36 and 37.
Problem 2.I.
Let X = R and let X be the �-algebra consisting of the sets which either are
countable or have countable compliments. Let f(x) = �1 if x < 0 and f(x) = 1 if
x � 0. Then f is not measurable since {x : f(x) > 0} = {x : x � 0}, and this set
is neither countable nor has countable compliment. On the other hand, we have
that both f2 and |f | are constant (equal 1) and therefore measurable.
Problem 2.K.
Let ↵ 2 R. If ↵ � A, then {x 2 X : fA(x) > ↵} = ?. If ↵ < A, then
{x 2 X : fA(x) > ↵} = X. Finally if �A  ↵ < A, then {x 2 X : fA(x) > ↵} =
{x 2 X : f(x) > ↵} . Since f is measurable all these sets belongs to X, hence fA
is measurable.
Oppgave 2.M.
Proving the statements in this exercise is a straight forward exercise using the
definition of the inverse image and the definitions of the various set-theoretic oper-
ations, and I will omit the details. The purpose of this exercise is to point out that
the inverse image(pull back) commutes with the standard set-theoretic operations,
and it will follow from this that the pullback of a �-algebra is a �-algebra.
Problem 2.O.
Let Y = {E ⇢ Y : f�1(E) 2 X}. Then we have that A ⇢ Y. From exercise 2.N.
it follows that Y is a �-algebra. Let Z be the �-algebra generated by A. Then we
get that Z ⇢ Y, but this means that if F 2 Z, then f�1(F ) 2 X.
Oppgave 2.P.
Let A be the collection of all half-rays (↵,1). By definition, a function f is
measurable if and only if f�1(E) 2 X for all E 2 A. From problem 2.B., we know
that the �-algebra generated by A is equal the Borel-algebra B. If f is measurable
it will then follow from problem 2.O. that f�1(E) 2 X for all E 2 B. Conversely,
if f�1(E) 2 X for all E 2 B, we will get that f�1(E) 2 X for all E 2 A (since
A ⇢ B) so f is measurable.
Oppgave 2.Q.
From the text-book example (c) page 9, we get that if ' is continous, then ' is
Borel-measurable. It therefore follows that E↵ = '�1(↵,1) belongs to B for every
↵. Since f er X-measurable, it follows from 2.P. that f�1(E) 2 X for every E 2 B.
We therefore get that f�1(E↵) = (' � f)�1(↵,1) 2 X, So ' � f is X-measurable.
Oppgave 2.R.
The only property ' needed to satisfy to get the conclusion in problem 2.Q. was
that ' was Borel-measurable (and we only used the continuity of ' to conclude
that). We can thus replace ' of 2.Q. with any Borel-measurable function and
get the same conclusion, namely that � f is X-measurable if f is X-measurable.
Oppgave 3.A.
Since µ is a measure, then �(?) = µ(A\?) = µ(?) = 0, and �(E) = µ(A\E) � 0.
Let (En) be a disjoint sequence of sets in X. Then (A\En) is also a disjoint sequence
of sets in X, and since µ is a measure, we get that �(
1S
n=1
En) = µ(A \ (
1S
n=1
En)) =
µ(
1S
n=1
(A \ En)) =
1P
n=1
µ(A \ En) =
1P
n=1
�(En). This shows that � is a measure.
1
2
Oppgave 3.B.
Since each µj is a measure, we get that µj(?) = 0, and consequently that �(?)=
nP
j=1
ajµj(?) =
nP
j=1
aj0 = 0. Also since each µj is a measure, we get that µj(E) � 0
when E 2 X, hence we get that �(E) =
nP
j=1
ajµj(E) � 0. Finally let (Em) be a
disjoint sequence of sets in X. For each j we have that µj(
1S
m=1
Em) =
1P
m=1
µj(Em),
which gives us
�(
1[
m=1
Em) =
nX
j=1
ajµj(
1[
m=1
Em) =
nX
j=1
aj(
1X
m=1
µj(Em)) =
1X
m=1
(
nX
j=1
ajµj(Em)) =
1X
m=1
(
nX
j=1
�(Em)).
This proves that � is a measure.
Oppgave 3.C.
From the definition it is clear that �(?) = 0 and that �(E) � 0. Let (Em) be a
disjoint sequence of sets in X. Then we have (since each µn is a measure):
�(
1[
m=1
Em) =
1X
n=1
2
�nµn(
1[
m=1
Em) =
1X
n=1
1X
m=1
2
�nµn(Em) =
1X
m=1
1X
n=1
2
�nµn(Em) =
1X
m=1
�(Em).
Proving that � is a measure.
To see that we are allowed to interchange the succession of summation as we
have done above, we use the following: If amn is a double indexed sequence of
non-negative real numbers such that for each n,
1P
m=1
amn < 1 and such that
1P
n=1
1P
m=1
amn < 1, then we have that
1P
n=1
amn < 1 for each m, and moreover we
have that
1P
m=1
1P
n=1
amn =
1P
n=1
1P
m=1
amn. (You can probably find a reference to this
in a su�ciently advanced text-book in real analysis or you can try to prove it your-
self (This can be proven directly with no reference to measure theory, but in fact,
letting X = N and µ equal the counting measure, it is easy to see that it also can be
proven using Monotone Convergence Theorem.)) Putting amn = 2�nµn(Em), we
have that
1P
m=1
amn = 2�nµn(
1S
m=1
Em)  2�nµn(X) = 2�n, and consequently that
1P
n=1
1P
m=1
amn <
1P
n=l
2
�n
= 1 < 1, and this will justify that we can change the suc-
cession of summation. Finally we have that �(X) =
1P
n=1
2
�nµn(X) =
1P
n=1
2
�n
= 1.
3
Let X = N be the natural numbers. Let An = N � {n }. Then An is neither
monoton increasing nor decreasing. We have
1S
n=m
An = N for all m, so lim sup An =
N. On the other hand, if m > 1,
1T
n=m
An = {1, 2, . . . ,m� 1}, so lim inf An = N.
MAT 3300/4300: Solutions for problems assigned week 37 and 38.
Problem 2.V.
In general, countable unions and intersections of sets in a �-algebra are members
of the �-algebra. So, in particular, countable unions of monotone increasing sets
and countable intersections of monotone decreasing sets in the �-algebra are sets
in the �-algebra. Hence any �-algebra is a monotone class. Let A be a nonempty
collection of subsets of a space X. Since there exists �-algebras containing A, there
exists monotone classes containing A. Moreover, it is easily seen that given a family
of monotone classes then their intersection (the class containing all subsets which
belongs to every class in the family) is a monotone class. So the smallest monotone
class containing A will be the intersection of all the monotone classes containing
A.
Problem 3.J.
Recall that lim supEn =
1T
m=1
1S
n=m
En. The sequence Fm =
1S
n=m
En is montone
decreasing, and µ(F1) = µ(
1S
n=1
En) < 1, hence (by Lemma 3.4) we have that
µ(lim sup En) = µ(
1T
m=1
Fm) = lim µ(Fm). Let xm = sup
n�m
µ(En). Then xm is a
decreasing sequence so lim supµ(En) = inf
m
sup
n�m
µ(En) = lim xm. Now µ(Fm) =
µ(
1S
n=1
En) � µ(En) for any n � m. We therefore have that sup
n�m
µ(En) = xm 
µ(Fm), and consequently that
lim supµ(En) = lim xm  lim µ(Fm) = µ(lim sup En).
To see that the assumption µ(
1S
n=1
En) <1 cannot be dropped, let En = [n, n + 1]
and µ equal the Lebesgue measure. Then lim supEn = ? so µ(lim sup En) = 0, but
µ(En) = 1 so lim supµ(En) = 1.
Problem 3.K.
Z is not a �-algebra unless we do have µ(X) = 0 (otherwise X /2 Z) and µ is the
zero-measure.
Let E be in Z and let F 2 X. Since E \ F ⇢ E, we have that 0  µ(E \ F ) 
µ(E) = 0, so µ(E \ F ) = 0 and E \ F 2 Z.
Let (En) be a sequence of sets in Z. Let F1 = E1, and Fn+1 = En+1 �
nS
k=1
Ek.
Then (Fn) is a disjoint sequence with
1S
n=1
Fn =
1S
n=1
En. Since Fn ⇢ En and En 2 Z,
we have that µ(Fn) = 0, and we get that
µ(
1[
n=1
En) = µ(
1[
n=1
Fn) =
1X
n=1
µ(Fn) = 0.
This shows that
1S
n=1
En 2 Z.
1
2
Oppgave 3.L.
Consider a set (E [Z1)�Z2 where E 2 X and Zi ⇢ Z̃i, i = 1, 2 with Z̃i 2 Z (note
that we do not require Zi 2 X). Then we have that (E [ Z1) � Z2 = (E � Z̃2) [
((E \ Z̃2�Z2)[ (Z1�Z2)). Here (E� Z̃2) 2 X and ((E \ Z̃2�Z2)[ (Z1�Z2)) ⇢
(Z̃2 [ Z̃1) 2 Z. Consider E [ Z = E [ Z �? where E 2 X and Z is a subset of a
set in Z. Since ? 2 Z, it is clear that E [ Z 2 X0.
It is clear that X ⇢ X0 so X0 contains ? and X. Again consider E [ Z where
E 2 X and Z is a subset of a set Z̃ in Z. We have that X � (E [ Z) = (X �
(E [ Z̃)) [ (Z̃ � (Z [ E)). Here X � (E [ Z̃) 2 X and Z̃ � (Z [ E) ⇢ Z̃ 2 Z so
X � (E [Z) 2 X0. Consider a sequence En [Zn where En 2 X and Zn is a subset
of a set Z̃n in Z. Then
1S
n=1
(En [Zn) = (
1S
n=1
En)[ (
1S
n=1
Zn), where
1S
n=1
En 2 X and
1S
n=1
Zn ⇢
1S
n=1
Z̃n 2 Z, so
1S
n=1
(En [ Zn) 2 X0. This shows that. X0 is a �-algebra.
Oppgave 3.M.
Assume E [ Z = E0 [ Z 0, where E, E0 are sets in X and Z ⇢ Z̃, Z 0 ⇢ Z̃ 0 with Z̃
and Z̃ 0 belonging to Z. Then E = (E \E0)[ (E�E0) . Now since E�E0 2 X and
we must have that E � E0 ⇢ Z 0 ⇢ Z̃ 0 and µ(Z̃ 0) = 0, we get that µ(E � E0) = 0.
This implies µ(E) = µ(E \ E0). By symmetri, µ(E0) = µ(E \ E0), and alltogether
this implies that µ0 is well-defined.
Writing ? = ? [ ?, we get that µ0(?) = µ(?) = 0. Moreover µ0(E \ Z) =
µ(E) � 0.
Consider a disjoint sequence En[Zn, where each En 2 X and Zn is a subset of a
set Z̃n in Z. It is then clear that En is a disjoint sequence. Moreover, we have that
1S
n=1
(En [ Zn) = (
1S
n=1
En) [ (
1S
n=1
Zn), where
1S
n=1
En 2 X and
1S
n=1
Zn ⇢
1S
n=1
Z̃n 2 Z.
Therefore, we get that
µ0(
1[
n=1
(En [ Zn) = µ0((
1[
n=1
En) [ (
1[
n=1
Zn)) =
µ(
1[
n=1
En) =
1X
n=1
µ(En) =
1X
n=1
µ0(En [ Zn).
This shows that µ0 is a measure on X0.
Finaly if E 2 X, µ0(E) = µ0(E [?) = µ(E), so µ0 agrees with µ on X.
Oppgave 4.C.
It is clear that any value of and ! is either a value of '1 or of '2. Since '1 and
'2 both are simple functions, they have together finitely many values. Thus and
! both have only finitely many values, and since we know that they are measurable
functions (Lemma 2.9), they must be simple functions.
Oppgave 4.H.
In this case
R
fnd� = �([0, n]) = n which tends to +1 when n ! 1 so The
Monotone Convergence Theorem apply.
Oppgave 4.I.R
fnd� =
1
n�([n, +1)) = +1. So limn!1
R
fnd� = +1, and since f = 0 and
therefore
R
fd� = 0, we get that 0 =
R
fd� 6= lim
n!1
R
fnd� = +1.
3
Oppgave 4.K.
Since lim inf fn = f , it follows from Fatou’s lemma that
R
fdµ  lim inf
R
fndµ.
Let ✏ > 0. Since fn ! f uniformly, there exists N such that |fn(x)� f(x)| < ✏ for
all x if n � N . From this we get that fn  f + ✏ when n � N , and we therefore
have that
R
fndµ 
R
fdµ + ✏µ(X) when n � N . It follows that lim sup
R
fndµ R
fdµ + ✏µ(X), and since this is valid for all ✏ and µ(X) < 1, we must have that
lim sup
R
fndµ 
R
fdµ. From all this we get that
Z
fdµ  lim inf
Z
fndµ  lim sup
Z
fndµ 
Z
fdµ.
Which implies that
Z
fdµ = lim inf
Z
fndµ = lim sup
Z
fndµ = lim
Z
fndµ.
MAT 3330/4300: Solutions for problems assigned week 39 and 40.
Exercise 5.O.
Let E = {x :
1P
n=1
|fn(x)| =1}. Since
1P
n=1
|fn| is measurable, it follows that E 2 X.
Let N be a positive integer. It follows from Monotonee Convergence Theorem that
1 >
1X
n=1
Z
|fn|dµ =
Z
(
1X
n=1
|fn|)dµ �
Z
N�Edµ = Nµ(E).
It follows from this inequality that µ(E) = 0. Let f(x) =
1P
n=1
fn(x) for x 2 X � E
and f(x) = 0 for x 2 E. Let gk(x) =
kP
n=1
fn(x) for x 2 X�E and gk(x) = 0 for x 2
E and g(x) =
1P
n=1
|fn(x)| for x 2 X �E and g(x) = 0 for x 2 E. Since g =
1P
n=1
|fn|
µ-almost everywhere it follows from Corollary 4.10 that
R
gdµ =
1P
n=1
R
|fn|dµ <1,
so g is integrable. Since |gk|  g, every gk is also integrabel. From Lebesgue
Dominated Convergence Theorem we get that
R
fdµ =
R
lim gkdµ = lim
R
gkdµ.
On the otherhand we have that gk =
kP
n=1
fn µ-almost everywhere, so
R
gkdµ =
kP
n=1
R
fndµ and therefore we get that
R
fdµ = lim
R
gkdµ =
1P
n=1
R
fndµ. Moreover
it is clear that the series
1P
n=1
fn converges to f on X�E so it converges to f µ-almost
everywhere. (Note that since an integrable function by definition only takes (finite)
real values, we cannot use Lebesgue Dominated Convergence Theorem directely to
the sequence
kP
n=1
fn since this sequence not necessarily is dominated by an integrable
function with finite values. Therefore we need all the above definitions to guarantee
that our functions only have real values.).
Exercise 5.P.
We have that lim |fn| = |f |, so from Fatous Lemma we get thatR
|f |dµ  lim inf
R
|fn|dµ. Since |fn|  |f | + |fn � f |, we have that
R
|fn|dµ R
|f |dµ +
R
|fn � f |dµ. This give us that lim sup
R
|fn|dµ  lim sup(
R
|f |dµ +R
|fn � f |dµ) =
R
|f |dµ (because lim
R
|fn � f |dµ = 0). So, we consequently get
that: Z
|f |dµ  lim inf
Z
|fn|dµ  lim sup
Z
|fn|dµ 
Z
|f |dµ.
Which shows that lim
R
|fn|dµ =
R
|f |dµ.
Exercise 5.Q.
We have that
Z 1
0
etxdx = lim
b!1
Z b
0
etxdx = lim
b!1
�1
t
e�tx
����
b
0
= lim
b!1
[
1
t
� 1
t
e�tb] =
1
t
.
Since s ! es is strictly increasing e�tx  e�ax for t � a > 0 and x � 0. Let
X = [0,1), B the Borel sets on X and let � be the Lebesque measure on B.
1
2
Let 1 > a > 0, and let g(x) = e� ax2 . From exercise 4.M and the calculation
above, we have that
R
gd� = 2a , so g is integrable. Let gn(x) = x
ne�ax. Since
lim
x!1
gn(x)
g(x) = limx!1x
ng(x) = 0(this is a well known limit), gn is also integrable
for each n. For t 2 [a, 1] let fn(x, t) = xne�tx. Since |fn(x, a)| = gn(x) and
|@fn@t (x, t)| = fn+1(x, t)  gn+1(x), the hypothesis of Corrolary 5.9 is satisfied. We
consequently get that Fn(t) =
R
fn(x, t)d�(x) is di↵erentiable and
dFn
dt
=
Z
@fn
@t
(x, t)d�(x) =
Z
�xn+1e�txd�(x) =
Z 1
0
�xn+1e�txdx,
where we have used the conclusion of 5.9 and exercise 4.M. Putting n = 0, 1, 2, . . . ,
we get from all this:
F0(t) =
Z 1
0
e�txdx =
1
t
dF0
dt
=
Z 1
0
�xe�txdx = �F1(t) =
d
dt
(
1
t
) = � 1
t2
dF1
dt
=
Z 1
0
�x2e�txdx = �F2(t) =
d
dt
(
1
t2
) = � 2
t3
dF2
dt
=
Z 1
0
�x3e�txdx = �F3(t) =
d
dt
(
2
t3
) = �3 · 2
t4
. . .
After n-iterations we get that n!tn+1 =
R1
0 x
ne�txdx. Letting t = 1 we get the
formula
n! =
Z 1
0
xne�xdx.
Exercise 5.R.
Firstly, assume that t0 6= t1 and let t 6= t0 and also let t 6= t1. Then we have that
|f(x, t)|  |f(x, t0)| + |f(x, t)� f(x, t0)|
 |f(x, t0)| + |f(x, t)� f(x, t1)| + |f(x, t1)� f(x, t0)|
 |f(x, t0)| + |t� t1|g(x) + |t1 � t0|g(x).
Since |f(x, t0)| + |t� t1|g(x) + |t1 � t0|g(x) obviously is integrable, we have that
x ! f(x, t) is integrable for every such t. In the same manner, for t = t1 we have
that |f(x, t1)|  |f(x, t0)| + |t1 � t0|g(x). If t0 = t1 then |f(x, t)|  |f(x, t0)| +
|t� t0|g(x). So for all t the function x! f(x, t) is bounded by integrable functions
and is therefore integrable. Let F (t) =
R
f(x, t)dµ(x). Let (sn) be a sequence such
that sn ! t1 and assume that sn 6= t1. Then F (sn)�F (t1)sn�t1 =
R f(sn)�f(t1)
sn�t1 dµ(x), and
since | f(sn)�f(t1)sn�t1 |  g(x) Lebesgue Dominated Convergence Theorem will imply
that
lim
n!1
F (sn)� F (t1)
sn � t1
=
Z
lim
n!1
f(sn)� f(t1)
sn � t1
dµ(x) =
Z
@f
@t
(x, t1)dµ(x).
Here, the right hand side is independent of the choice of sequence (sn) so F (t) is
di↵erentiable at t1, and we consequently get that dFdt (t1) =
R @f
@t (x, t1)dµ(x).
3
Exercise 5.S.
Since the improper Riemann integral
1R
�1
|f(x, t)|dt exists for each x, the integral
1R
�1
f(x, t)dt will also exist, and we have that
1R
�1
f(x, t)dt = lim
n!1
nR
�n
f(x, t)dt.
Put gn(x) =
nR
�n
f(x, t)dt. Since every Riemann sum for the integral
nR
�n
f(x, t)dt only
is a linear combination of measurable functions x ! f(x, t), such Riemann sums
become measurable functions on X, and since gn is a limit of such Riemann sums,
each gn is a measurable function. Then, it also follows that lim gn(x) =
1R
�1
f(x, t)dt
is a measurable function on X. Moreover it is clear thar |gn(x)|  h(x) for all n.
Lebesgue Dominated Convergence Theorem therefore implies that
lim
Z
gn(x)dµ(x) =
Z
[
1Z
�1
f(x, t)dt]dµ(x)
that is
lim
n!1
Z
[
nZ
�n
f(x, t)dt]dµ(x) =
Z
[
1Z
�1
f(x, t)dt]dµ(x).(*)
(⇤) shows that x !
1R
�1
f(x, t)dt is integrable. Letting [a, b] = [�n, n] the ass-
sumptions of Corollary 5.10 are satisfied, so from the conclusion of 5.10 we get
that Z n
�n
[
Z
f(x, t)dµ(x)]dt =
Z
[
Z n
�n
f(x, t)dt]dµ(x).
letting n!1 and using (⇤), we get that
lim
Z n
�n
[
Z
f(x, t)dµ(x)]dt =
Z
[
1Z
�1
f(x, t)dt]dµ(x).
If f(x, t) was non-negativthen this will show that the improper Riemann integralR1
�1[
R
f(x, t)dµ(x)]dt exsits and we have that
Z 1
�1
[
Z
f(x, t)dµ(x)]dt =
Z
[
1Z
�1
f(x, t)dt]dµ(x).
The improper Riemann integral is however defined by letting the lower and upper
limit in the integral approach �1 and 1 independently, so dealing with functions
of both positive and negative values, we actually must show that both the limitsR 0
�1[
R
f(x, t)dµ(x)]dt and
R1
0 [
R
f(x, t)dµ(x)]dt exist. In our case this is however
obvious, because we can in the above arguments replace f(x, t) by |f(x, t)| and
prove the conclusion for this function, and the integrals
R 0
�1[
R
f(x, t)dµ(x)]dt and
4
R1
0 [
R
f(x, t)dµ(x)]dt must then exist since they are dominated of similar integrals
of |f(x, t)| which then must be finite.
Problem 5.T.
For a measurable function f we define fn by
fn(x) = f(x), hvis |f(x)|  n
= n, if |f(x)| > n
= �n, if |f(x)| < �n.
We have |fn|  |f | and lim fn = f ,so if f is integrable then |f | is also integrable,
and we will get from Lebesgue Dominated Convergence Theorem that
R
fdµ =
lim
R
fndµ. Moreover we have that lim |fn| = |f |, so from Fatous Lemma we get
that Z
|f |dµ  lim inf
Z
|fn|dµ  sup
Z
|fn|dµ.
So if sup
R
|fn|dµ <1, e |f |and therefore f is integrable.
Problem 6.A.
. If f 2 C[0, 1 and ↵ 2 R, then |f(o)| � 0 and |(↵f)(0)| = |↵f(0)| = |↵||f(0)|, and
if f, g 2 C[0, 1] then |(f + g)(0)| = |f(0) + g(0)|  |f(o)| + |g(0)| and from all this
we get that N0 is a semi-norm.
Problem 6.B.
Let f 2 C[0, 1]. Since |f | � 0, we get that N1(f) =
1R
0
|f |dx � 0. If f(x) = 0 for
all x then it is clear that N1(f) = 0. If f 2 C[0, 1] og and there exists x 2 [0, 1]
such that f(x) 6= 0, then there exists an interval I ⇢ [0, 1] with length l(I) > 0
shuch that x 2 I and |f(t)| > |f(x)|2 for t 2 I. Since |f | �
|f(x)|
2 �I , we will get
that N1(f) �
1R
0
|f(x)|
2 �Idx =
|f(x)|
2 l(I) 6= 0. Let ↵ 2 R. since the Riemannintegral
is linear we get that N1(↵f) = |↵|N1(f). Let f, g 2 C[0, 1]. Then we have that
|f + g|  |f |+|g|. Integrating this inequality we get that N1(f+g)  N1(f)+N1(g).
Alltogether this shows that N1 is a norm on C[0, 1].
Let m > n. Then we have that fn = fm on [0, (1� 1n )/2] [ [
1
2 , 1], so
N1(fn � fm) =
Z 1
0
|fn � fm|dx =
Z 1
2
(1� 1n )/2
|fn � fm|dx 
Z 1
2
(1� 1n )/2
2dx =
1
n
.
Since 1n ! 0 when n!1, it follows that (fn) is a Cauchy sequence.
Assume now that there exists f 2 C such that lim fn = f in the norm N1, that is
lim N1(f � fn) = 0. Let m be a positive integer. Then
0 = lim
n!1
Z 1
0
|f � fn|dx � lim
n!1
Z (1� 1m )/2
0
|f � fn|dx + lim
n!1
Z 1
1
2
|f � fn|dx =
lim
n!1
Z (1� 1m )/2
0
|f |dx + lim
n!1
Z 1
1
2
|f � 1|dx =
Z (1� 1m )/2
0
|f |dx +
Z 1
1
2
|f � 1|dx.
5
This shows that for every m we must have f(x) = 0 on [0, (1� 1m )/2] and f(x) = 1
on [ 12 , 1]. So f becomes equal 0 on [0,
1
2 ) and equal 1 on [
1
2 , 1], and such a function
f is not continous. So we have got a contradiction.
Problem 6.C.
From property (i) of a norm it follows that d(u, v) = N(u� v) � 0. From property
(ii) it follows that d(u, v) = N(u � v) = 0 , u � v = 0 , u = v. From property
(iv) we get that if u, v and w are vectors in V , then
d(u, v) = N(u�v) = N((u�w)+(w�v))  N(u�w)+N(w�v) = d(u, w)+d(w,v).
This shows that d is a metric.
MAT 3330/4300: Solutions for problems assigned week 40 and 41.
Oppgave 6.D.
Let p > 1. First we show that the sequence (t + s)p � tp � sp for t, s � 0. This is
obvious when t = 0. Assume that t > 0. Let h(s) = (t+s)p�tp�sp. Then we have
that h(0) = 0, and h0(s) = p(t + s)p�1 � psp�1 > 0 since t + s > s. So h is strictly
increasing and h(s) � h(0) = 0 for s � 0. Let ✏ > 0. Since (f+)p 2 L1, there exists
a simple function 1 such that 0  1  (f+)p and 0 
R
((f+)p � 1)dµ < ✏
p
2 .
We have 1 =
nP
i=1
ai�Ei where the Ei’s are disjoint and ai � 0. Let �1 =
nP
i=1
a
1
p
i �Ei .
Then �p1 = 1, and we have that 0  �1  f+. From the inequality above it follows
that
(f+)p � 1 = (f+)p � �p1 = ((f+ � �1) + �1)p � �
p
1 � (f+ � �1)p � 0.
So we get that
✏p
2 >
R
((f+)p � 1)dµ �
R
(f+ � �1)pdµ.
in a similar manner, we can find a simple function �2 such that 0  �2  f� and
such that
✏p
2 >
R
(f� � �2)pdµ. Let � = �1 � �2. Then we have that |f � �|p =
(f+ � �1)p + (f� � �2)p, so from the integral inequality above it follows that
✏p >
Z
|f � �)|pdµ = ||f � �||pp,
hence ||f � �||p < ✏ (so the simple functions are dense in i Lp).
Now let f 2 L1. We may assume that there exists a K > 0 such that |f | < K
(every µ-equivalence class in L1 has such representative). Let ✏ > 0 and let n be
an integer such that
2K
n < ✏. Consider the partition �K = t0 < t1 < · · · < tn = K
where ti+1 � ti = 2Kn . Let Ei = {x : ti < f(x)  ti+1}. Let � =
n�1P
i=0
ti�Ei . Then
we get that 0 < f(x)� �(x)  ti+1 � ti = 2Kn < ✏, and � is a simple function with
||f � �||1 < ✏. So the answer of the last question in the problem is yes!
Oppgave 6.H.
Let En = {x 2 X. |f(x)| � 1n}. Then
R
|f |p dµ � 1np µ(En). So f 2 Lp implies that
µ(En) < +1. So, since E = [En, E is �-finite.
Oppgave 6.F.
We have that
1 >
Z
|f |pdµ �
Z
|f�En |pdµ �
Z
|n�En |pdµ = npµ(En).
Since the sequence npµ(En) is bounded by ||f ||pp, but np !1, we must have that
µ(En)! 0 when n!1.
Oppgave 6.H.
It is well known that
1P
n=1
1
n2 < 1 so �(X) =
1P
n=1
1
n2 < 1. Let �n = f
p�[1,..,n].
Then �n is a monotone sequence and lim�n = fp. It follows from The Monotone
Convergence Theorem that
Z
fpd� = lim
Z
�nd� = lim
nX
k=1
k
p
2
k2
=
1X
n=1
n(
p
2�2).
1
2
We know that the series above is convergent if and only if
p
2 �2 < �1 consequently
if and only if p < 2. So we have that f 2 Lp if and only if 1  p < 2.
Oppgave 6.J.
Here we assume that f has only real values (otherwise the conclusion of the exer-
cise is false, since the measurable function f(x) = +1 satisfies the criterion of the
problem for any p (because En = ? for any n) but f is obviously not in Lp for any
p). Then we have that X =
1S
n=1
En, and this is a disjoint union. Let gk =
kP
n=1
n�En
and let g =
1P
n=1
n�En . Then we have g = lim gk, so this g is a measurable func-
tion. The Monotone Convergence Theorem then gives us that
R
gdµ =
1P
n=1
nµ(En).
Moreover g > |f |, so f 2 L1 if
R
gdµ =
1P
n=1
nµ(En) <1.
Define h =
1P
n=1
(n� 1)�En . The Monotone Convergence Theorem again gives that
R
hdµ =
1P
1=n
(n� 1)µ(En). Moreover we have that |f | � h, so if f 2 L1 we get that
1 >
R
|f |dµ �
R
hdµ =
1P
1=n
(n � 1)µ(En). Since
1P
n=1
µ(En) = µ(X) < 1, we get
that
1P
n=1
nµ(En) =
1P
n=1
(n� 1)µ(En) +
1P
n=1
µ(En) <1.
With g and gk defined as above, we get that gp =
1P
n=1
np�En and that g
p
k =
kP
n=1
np�En . Moreover g
p
k is monotonically increasing and we have that g
p
= lim gpk.
The Monotone Convergence Theorem then gives us that
R
gpdµ = lim
R
gpkdµ =
1P
n=1
npµ(En). So if
1P
n=1
npµ(En) < 1, we get that g 2 Lp, and since gp � |f |p we
consequently get that f 2 Lp.
Define h as above. With a same sort reasoning as above we get that
R
hpdµ =
1P
n=1
(n�1)pµ(En). Assume that f 2 Lp. Since |f | � h we get that h 2 Lp
that is
1P
n=1
(n � 1)pµ(En) < 1. From the limit comparison test it now follows that
1P
n=1
npµ(En) <1.
Oppgave 6.K.
If f 2 Lp, then it follows from 6.J. that
1P
n=1
npµ(En) < 1. Since npµ(En) �
nrµ(En), we also get that
1P
n=1
nrµ(En) < 1. So again applying 6.J. we get that
f 2 Lr. Since f 2 Lp, we have that |f |r 2 L p
r
. Let q = pp�r . Then
1
q +
1
p
r
= 1.
Moreover, since µ(X) <1, g 2 Lq where g ⌘ 1. Hölder’s inequality gives that
|||f |rg||1 = |||f |r||1 = ||f ||rr  |||f |r|| pr ||g||q = ||f ||
r
p(µ(X))
1
q .
3
This gives
||f ||r  ||f ||p(µ(X))
1
rq = ||f ||p(µ(X))s.
(Since
1
rq =
(p�r)
pr = (1/r)� (1/p) = s). Especially if µ(X) = 1, then ||f ||r  ||f ||p.
Oppgave 6.L.
Given f then
R
|f |pdµ =
1P
n=1
|f(n)p|p. So f 2 Lp if
1P
n=1
|f(n)|p < 1. If f 2 Lp,
then |f(n)|p ! 0, and therefore f(n) ! 0. So for big n we must have f(n) < 1.
If s � p, we consequently have that |f(n)|p � |f(n)|s when n is big. From the
comparison test for series it follows that
1P
n=1
|f(n)|s < 1 if
1P
n=1
|f(n)|p < 1, and
it follows that f 2 Ls if f 2 Lp. Let us prove that ||f ||s  ||f ||p. Let a, b � 0. we
will first show that (ap + bp)
1
p � (as + bs) 1s . We may assume that a, b > 0. Then
we have that:
(ap + bp)
1
p � (as + bs) 1s , 1 + (a
b
)
p � (1 + (a
b
)
s
)
p
s � 0.
Let h(t) = 1+tp�(1+ts)
p
s . We have that h(0) = 0 and h0(t) = ptp�1(1�(1+ 1ts )
p�s
s ).
Let t > 0. Since 1 + 1ts > 1 and
p�s
s < 0,we get that (1 +
1
ts )
p�s
s < 1 and we
therefore get that h0(t) > 0. So since h is strictly increasing on [0,1) and h(t) � 0.
So h(ab ) = 1 + (
a
b )
p � (1 + (ab )
s
)
p
s > 0 and therefore (ap + bp)
1
p � (as + bs) 1s for
all a, b � 0. Let (ai) be a sequence with ai � 0. It follows from induction and the
inequality above that
(
n+1X
i=1
api )
1
p = (((
nX
i=1
api )
1
p )
p
+ apn+1)
1
p � (((
nX
i=1
api )
1
p )
s
+ asn+1)
1
s �
(((
nX
i=1
asi )
1
s )
s
+ asn+1)
1
s = (
n+1X
i=1
asi )
1
s ,
so for all finite n we get that (
nP
i=1
api )
1
p � (
nP
i=1
asi )
1
s . Let n!1 then
(
1X
i=1
api )
1
p � (
1X
i=1
asi )
1
s , or
||f ||p = (
1X
i=1
|f(i)|p)
1
p � (
1X
i=1
|f(i)|s) 1s = ||f ||s.
Oppgave 6.M.
An extension of problem 4M will show that if g � 0 is continous on (0,+1), g is �-
integrable on (0,+1) f if and only if the improper Riemann integral
R +1
0 g(x) dx
exists. This is equivalent with the following; that there exixts � > 0 and M > 0
such that the two improper integrals
R �
0 g(x) dx and
R +1
M g(x) dx both exist. Let
r > 0, s > 0. Then I claim that lim
x!0+
xr(log x)s = 0. It is obviously enough to
prove this for any integer s, and using L’Hopitals rule we find that
lim
x!0+
xr(log x)s = lim
x!0+
(log x)s
1
xr
= lim
x!0+
�s
r
xr(log x)s�1,
4
and the claim will follow by induction on s.
Let p > 0 and r > 0. From the limit we have considered follows that there exists
a � = �(r, p) such that xr < 1(2| log x|)p when x 2 (0, �). Clearly this � can also be
chosen so small that (2| log x|)p > (1 + | log x|)p when x 2 (0, �). All this will imply
that f(x)p > x
r
x
p
2
on x 2 (0, �). Now we know that g(x) = x
r
x
p
2
is not integrable on
(0, �) when p2 � r � 1, hence when p � 2 + 2r. Therefore for any small r > 0 and
any p � 2 + 2r, f is not in Lp. So if p > 2 we get that f /2 Lp (choosing r so small
that p � 2 + 2r also holds).
From above we find that
lim
x!+1
(log x)s
xr
= lim
t!0+
tr(� log t)s = 0,
for any s, r > 0. So we can find M = M(r, p) > 0 such that when x 2 [M,+1),
x�r < 1(2 log x)p . This M can also be chosen such that 1+log x < 2 log x on [M,+1)
So, for x 2 [M,+1) we have that (f(x))p > 1
x
p
2 (2 log x)p
> 1
x
p
2 +r
. g(x) = 1
x
p
2 +r
is
not integrable on [M,+1) when p2 + r  1, that is p  2 � 2r.Therefore for any
small r > 0 and any p  2 � 2r, f is not in Lp. So if p < 2 we get that f /2 Lp
(choosing r so small that p  2� 2r also holds).
Now we will prove that f 2 L2. It is easy to see that it is enough to prove there
exists � > 0 and M > 0 such that the two improper integrals
R �
0
1
x(log x)2 dx andR +1
M
1
x(log x)2 dx both exist.
Let G(x)= � log | log x|log x and g(x) =
1
x(log x)2 (log | log x|� 1). Then G
0
(x) = g(x) and
there exists � > 0 and M > 0 such that g(x) > 1x(log x)2 on (0, �) [ [M,+1). It is
easy to prove that lim
x!0+
G(x) = lim
x!+1
G(x) = 0 and from this it will follow that
the integrals
R �
0 g(x)dx and
R +1
M g(x)dx exist. From this we get that
R �
0
1
x(log x)2 dx
and
R +1
M
1
x(log x)2 dx exist and we consequently get that f 2 L2.
MAT 3330/4300: Solutions for problems assigned week 42 and 43.
Problem 7.B.
For each x there is N such that x /2 [ 1n ,
2
n ] when n � N , so fn(x) = 0 when n � N .
From this we see that the sequence converges pointwise towards f ⌘ 0. However
we have that
||fn � f ||p = (
Z
|fn|pd�)
1
p = (
Z
np�[ 1n , 2n ]d�)
1
p = n
p�1
p .
So ||fn � f ||p ! 1 if p = 1 and ||fn � f ||p ! 1 if p > 1 hence (fn) does not
converge to f in Lp.
Oppgave 7.H.
Since Lp = Lp(X,X, µ) is a metric space (with a metric induced by the Lp-norm)
limits of sequences are unique. Therefore f and g must be representatives of the
same µ-class of functions hence f = g µ-almost everywhere.
Oppgave 7.I.
Let fn = �Enand assume that fn ! f in Lp. For each m, let Im = (�1,� 1m ) [
( 1m , 1�
1
m ) [ (1 +
1
m ,+1). Let Fm = f
�1(Em). Since fn only has values 0 and 1,
it is clear that x 2 Fn ) |f(x)� fn(x)| � 1m . This implies that
||f � fn||pp �
Z
Fm
|f � fn|dµ �
1
mp
µ(Fm).
Since ||f � fn||pp ! 0 when n! +1, we must have that µ(Fm) = 0 for each m. Let
F = [mFm. Then µ(F ) = 0 and X � F = f�1({0, 1}) So we must have that f is
equal �f�1(1) outside F hence equal a characteristic function µ- almost everywhere.
Problem 8.C.
Let P, N be a Hahn-decomposition for �. Suppose that M is a null set. Then, since
P\M ⇢M , P\M ⇢M , we get that �+(M) = �(P\M) = ��(M) = ��(N\M) =
0 hence |�|(M) = �+(M)+��(M) = 0. Now assume that |�|(M) = 0. Let E ⇢M
be measurable. Since |�| is a measure we get that |�|(E) = 0. Then we must have
�+(E) = ��(E) = 0, and we get that �(E) = �+(E) � ��(E) = 0. This proves
that M is a nullset.
Problem 8.E.
Assume µ3(E) = 0. Since µ2 << µ3, we have that µ2(E) = 0. Since µ1 << µ2, we
have that µ1(E) = 0. This proves that µ1 << µ3.
Let µ2 be the Lebesgue measure on B. Let µ1 be the measure on B given by
µ1(E) = 0 for all E 2 B. Then µ1 << µ2, but if E = (a, b) where a < b, then
µ2(E) = b� a 6= 0 whilst µ1(E) = 0, so we cannot have µ2 << µ1.
Oppgave 8.K.
Since Nn is negative for � � nµ, we have that (� � nµ)(Nn)  0. So �(Nn) 
nµ(Nn) < 1 (since µ is finite). So N =
1S
n=1
Nn is a countable union of sets with
finite �-measure and therefore �-finite with respect to �.
Let E ⇢ P . Then E ⇢ Pn for all n, and since Pn is positive for ��nµ, we get that
(� � nµ)(E) � 0. So �(E) � nµ(E) for all n. If µ(E) > 0 and we let n ! 1 we
will get that �(E) =1. If µ(E) = 0, we get, since � << µ, that �(E) = 0.
1
2
Problem 8.N.
Let � =
nP
i=1
ai�Ei be a simpel function with ai � 0. Then
Z
�d� =
nX
i=1
ai�(Ei) =
nX
i=1
ai
Z
Ei
fdµ =
nX
i=1
ai
Z
�Eifdµ =
Z nX
i=1
ai�Eifdµ =
Z
�fdµ.
Let (�n) be a sequence of monotonically increasing, non-negative simple functions
which is pointwise convergent to g (by Lemma 2.11, we can find such sequence).
Then (�nf) will be a sequence of monotonically increasing, non-negative functions
which is pointwise convergent to gf , so using The Monotone Convergence Theorem
twice we get from above that:
Z
gd� = lim
Z
�nd� = lim
Z
�nfdµ =
Z
gfdµ.
Problem 8.O.
Let E be X-measurable, using the conclusion of 8.N, we get that
⌫(E) =
Z
E
d⌫
d�
d� =
Z
�E
d⌫
d�
d� =
Z
�E
d⌫
d�
d�
dµ
dµ =
Z
E
d⌫
d�
d�
dµ
dµ.
Since any two Radon-Nikodym derivatives of ⌫ with respect to µ are equal µ-
almost everywhere, then given any three d⌫d� ,
d�
dµ and
d⌫
dµ , we can conclude that that
d⌫
dµ =
d⌫
d�
d�
dµ µ-almost everywhere.
We have
(�1 + �2)(E) = �1(E) + �2(E) =
Z
E
d�1
dµ
dµ +
Z
E
d�2
dµ
dµ =
Z
E
(
d�1
dµ
+
d�2
dµ
) dµ.
This shows that for any choice of Radon-Nikodym derivatives we get d(�1+�2)dµ =
d�1
dµ +
d�2
dµ µ-almost everywhere.
Problem 8.P.
Let d�dµ and
dµ
d� be two choices of Radon-Nikodym derivatives. Since we can take
d�
d� = 1, we get from 8.O that
d�
dµ
dµ
d� = 1 almost everywhere. This implies that
dµ
d� 6= 0 almost everywhere and that we can write
d�
dµ =
1
dµ
d�
almost everywhere.
MAT 3330/4300: Solutions for problems assigned week 43 and 44..
Problem 9.F.
Let E ⇢
mS
k=1
Ink . Since l(In) = 1, then l⇤(E) 
mP
k=1
l(Ink) = m < +1. Let En =
(n+1�2�|n|, n+1], and let E =
1S
n=0
En. Then l⇤(E\In) = l⇤(En) = 2�|n| 6= 0, and
l⇤(E) =
1P
n=�1
l⇤(En) =
1P
n=�1
2�|n| = 3 < +1. Since each In is a Borel set and
consequently Lebesgue measurable, it follows that E \ In is Lebesgue measurable
if E is Lebesgue measurable.
If E \ In is Lebesgue measurable then E =
1S
n=�1
E \ In is Lebesgue measurable
since the collection of the Lebesgue measurable sets is a �-algebra.
Oppgave 9.G.
If l⇤(A) = +1, we can take G✏ = R. Let us assume that l⇤(A) < +1. Let F be
the algebra generated by the half-open intervals. From the definition of the outer
measures there exists a sequence (En) of sets in F such that l⇤(A) 
1P
n=1
l(En) <
l⇤(A) + ✏2 and A ⇢
1S
n=1
En. Since l⇤(A)  l⇤(
1S
n=1
En) 
1P
n=1
l(En)  +1 we
may assume that the En’s are disjoint of type En = (an, bn]. Then we get that
l⇤(A) 
1P
n=1
(bn � an) < l⇤(A) + ✏2 . Put G✏ =
1S
n=1
(an, bn + ✏2n+1 ). Then A ⇢ G✏, G✏
is open and
l⇤(A)  l⇤(G✏) 
1X
n=1
l(an, bn +
✏
2n+1
) =
1X
n=1
(bn � an) +
1X
n=1
✏
2n+1
< l⇤(A) +
✏
2
+
✏
2
= l⇤(A) + ✏.
Oppgave 9.H.
Since a set consisting of a single point is Lebesgue measurable with measure equal
0, we get that B � {n + 1} is Lebesgue measurable and l⇤(B) = l⇤(B � {n + 1}).
We may therefore assume that B ⇢ (n, n + 1). Let A = (n, n + 1) � B. By 9.G
there is an open set G✏ such that A ⇢ G✏ and l⇤(A)  l⇤(G✏)  l⇤(A) + ✏. We
may assume that G✏ ⇢ (n, n + 1). Let K✏ = [n + ✏2 , n + 1�
✏
2 ] � G✏. Then K✏ is
compact and K✏ ⇢ B. Now we have
1� ✏ = l⇤([n + ✏
2
, n + 1� ✏
2
]) = l⇤(K✏) + l⇤(G✏ \ [n +
✏
2
, n + 1� ✏
2
])
 l⇤(K✏) + l⇤(G✏)  l⇤(K✏) + l⇤(A) + ✏.
We get that l⇤(B) + l⇤(A) = 1  l⇤(K✏) + l⇤(A) + 2✏, which gives us l⇤(B) 
l⇤(K✏) + 2✏ and the claim in the exercise will follow from this.
Oppgave 9.I.
The first claim (about open sets) follows directly from the conclusion of 9.G. As
for the second let us first prove the claim when A ⇢ (�n, n]. Let ✏ > 0. For
1
2
k = �n, . . . n � 1, let Ak = A \ (k, k + 1]. Using the conclusion of 9.H, we can
find a compact set Kk ⇢ Ak such that l⇤(Kk)  l⇤(Ak)  l⇤(Kk) + ✏2n . Put
K =
n�1S
k=�n
Kk. Then K is compact with K ⇢ A. Looking more closely into the
solution of 9.H., we see that the compact sets constructed there actually were Borel
sets (beeing the di↵erence of a closed interval and an open set). Therefore K and
each Kk are Lebesgue measurable, and since our union is disjoint (and the Lebesgue
measure is equal the outer measure on the Lebesgue measurable sets), we get that
l⇤(K) =
n�1P
k=�n
l⇤(Kk). So we get that
l⇤(K)  l⇤(A) 
n�1X
k=�n
l⇤(Ak) 
n�1X
k=�n
(l⇤(Kk) +
✏
2n
) = l⇤(K) + ✏.
The second claim of the problem in the case when A is bounded will directly from
this.
Now let us assume that A is not bounded. Let ✏ > 0. Let this time An = A \
(�n, n + 1]. It follows from above that we can find a compact set Kn ⇢ An such
that l⇤(Kn)  l⇤(An)  l⇤(Kn) + ✏n . Now since An is an increasing sequence of
Lebesgue measurable set whose union is equal A, we get that lim l⇤(An) = l⇤(A).
On the other hand the above inequality implies that lim l⇤(Kn) = lim l⇤(An) and
the conclusion for an arbitrary A will follow from this.
Oppgave 9.J.
Let ✏ > 0. Using 9.G and the fact that any open set can be written as a disjoint
union of open intervals, we get that A ⇢
1S
n=1
In where (In) is a disjoint sequence
of open intervals and �(A) 
1P
n=1
l(In)  �(A) + ✏2 . Since �(A) +
✏
2 < +1 the
series
1P
n=1
l(In) is convergent, so there isa number m such that
1P
n=m+1
l(In) < ✏2 .
Put G =
mS
n=1
In. From this we easily get that ||�A � �G||1 = �(A M G) = �(A �
G) + �(G� A)  ✏. It is easy to see that �G can be approximated by a continous
piecewise linear non-negative map f such that ||�G � f ||1 < ✏. Using this and
Minkowski ’s Inequality we get the second inequality of the problem (here I leave
out further details).
Oppgave 9.K.
Sketch of the proof. First assume that A ⇢ (n, n + 1]. By 9.G. let Gm be open
such that A ⇢ Gm and �(A)  �(Gm)  �(A) + 1m . Put B =
1T
m=1
Gm. Then this
B has the desired properties. For general A, put An = A\ (n, n + 1], choose Borel
set Bn satisfying the conclusion with respect to An and modify Bn by intersecting
with (n, n + 1]. Then the conclusion still holds. Now put B =
1T
n=1
Bn. It is easy to
see that this set satisfy the desired conclusion.
To prove that every Lebesgue measurable set is the union of a Borelset and a
set of Lebesgue measure zero, we can argue in a similar manner, but instead of
3
approximating from above by open set we use 9.H and approximate from below by
compact set which by the arguments of 9.H. also could be taken to be Borel-sets.
Oppgave 9.L.
Sketch of proof: It is easy to see that it follows from 9.J. that the conclusion holds
for non-negative simple function and then for arbitrary non-negative integrable
functions (by the definition of the integral and Minkowsi’s inequality. Then split-
ing f in a positive and negative part, and approximating these with contionous
functions separately, we obtain the conclusion.
Solutions 
University of 
Washington 
Credits : Mary 
Radcliffe 
Math 426: Homework 1 Solutions
Mary Radcli↵e
due 9 April 2014
In Bartle:
2B. Show that the Borel algebra B is also generated by the collection of all
half-open intervals (a, b] = {x 2 R | a < x  b}. Also show that B is
generated by the collection of all half-rays {x 2 R | x > a}.
Proof. Let S1 = {(a, b) | a < b}, and S2 = {(a, b] | a < b}. By definition,
�(S1) = B. To show that �(S2) = B, it su�ces to show that S1 ⇢ �(S2)
and S2 ⇢ �(S1), as then, since �(A) is the smallest �-field containing A,
the result follows immediately.
Note that if a < b, then (a, b) = [n2Z+(a, b� 1n ], and (a, b] = \n2Z+(a, b+
1
n ). Therefore, (a, b) 2 �(S2) and (a, b] 2 �(S1), completing the proof.
For the second property, let S3 = {(a,1) | a 2 R}. Again, it su�ces to
show that S3 ⇢ �(S1) and S1 ⇢ �(S3).
Let a < b. Then (a, b) = (a,1)\
�
\n2Z+(b� 1n ,1)
�
2 �(S3). Moreover,
(a,1) = [n2Z+(a, a+ n) 2 �(S1), completing the proof.
2K. Show directly that if f is measurable and A > 0, then the truncation fA
defined by
fA(x) =
8
<
:
f(x) if |f(x)|  A
A if f(x) > A
�A if f(x) < �A
is measurable.
Proof. It su�ces to prove that for all ↵ 2 R, we have f�1A ((↵,1)) is mea-
surable. Note that if ↵ > A, then f�1A ((↵,1)) = ;, which is measurable.
If ↵ = A, then f�1A (↵) = f
�1([A,1)), which is measurable by Prop 2.4. If
�A  ↵ < A, then f�1A ((↵,1)) = f�1((↵,1)), which is measurable since
f is. Finally, if ↵ < �A, then f�1A ((↵,1)) = X, which is measurable.
Therefore, fA is measurable.
3E. Let X be an uncountable set and let F be the family of all subsets of X.
Define µ on E by requiring that µ(E) = 0 if E is countable, and µ(E) = 1
if E is uncountable. Show that µ is a measure on F .
Proof. Clearly, µ(;) = 0 and µ(E) � 0 for all E 2 F . Let {En} ⇢ F be
a sequence of pairwise disjoint subsets of F . If all En are countable, then
µ(En) = 0 for all n. Moreover, as the countable union of countably many
subsets is countable, we have also that µ([En) = 0. Thus, µ([En) =P
µ(En).
If not, there exists some k such that Ek is uncountable. Then [En is
uncountable also, so µ([En) = 1 = µ(Ek) =
P
µ(En).
1
Therefore, µ is countably additive, and is thus a measure on F .
3F. Let X = Z+ and let F be the family of all subsets of X. If E is finite, let
µ(E) = 0, if E is infinite, let µ(E) = 1. Is µ a measure on F?
Solution. No, µ is not a measure on X. Let En = {n}. Then En is finite
for all n, so µ(En) = 0. However, [En = Z+, so µ([En) = 1. Thus, µ is
not countably additive and thus is not a measure.
3H. Show that Lemma 3.4(b) may fail if the finiteness condition µ(F1) < 1
is dropped.
Proof. Let Fn = R\[�n, n], so that µ(Fn) = 1 for all n, and Fn � Fn+1.
Then \Fn = ;, so µ(\Fn) = 0. However, limµ(Fn) = 1, and thus the
result fails for {Fn}.
3T. Show that the Lebesgue measure of the Cantor set C is zero.
Proof. Let E0 = [0, 1], E1 = E0\( 13 ,
2
3 ), E2 = E1\(
1
9 ,
2
9 )\(
7
9 ,
8
9 ), etc., so
that C = \En. Then by Prop 3.4(b), since En � En+1 and �(E0) = 1 <
1, we have �(C) = lim�(En).
Note that by definition, �((a, b)) = b � a. Moreover, by Prop 3.4(b), we
have [a, b] = \n2Z+(a� 1n , b+
1
n ), and thus �([a, b]) = lim�((a�
1
n , b+
1
n )) =
lim(b�a+ 2n ) = b�a. Therefore, we have �(En) = �(En�1)�
1
3�(En�1) =
2
3�(En�1). Thus, �(En) =
�
2
3
�n
, so �(C) = lim
�
2
3
�n
= 0.
3U. By varying the construction of the Cantor set, obtain a set of positive
Lebesgue measure that contains no nonvoid open interval.
Proof. The construction here is called a “Fat Cantor set,” although you
certainly may have come up with something else. We begin as with the
construction of the Cantor set: Let E0 = [0, 1]. Take E1 to be E0 with the
middle third interval deleted, as before: E1 = E0\( 13 ,
2
3 ). Now, to form
E2, from each interval of E1, delete an open segment, the sum of whose
lengths is 1/6. There are many ways to do this, but, for example, we could
take E1\( 324 ,
5
24 )\(
19
24 ,
21
24 ), so that the total measure of the removed sets is
1
6 . Similarly, construct Ek from Ek�1 by choosing an interval from each
connected set in Ek�1, the total measure of which will be
1
3⇤2k . Then take
E = \Ek. By Lemma 3.4(b) we have that �(E) = 1�
P1
k=0
1
3⇤2k =
1
3 > 0,
but as we delete from every interval at every step, the resulting intersection
contains no nonvoid open intervals.
3V. Suppose that E is a subset of a set N 2 F with µ(N) = 0, but E /2 F .
Then the sequence {fn}, fn = 0 converges µ-a.e. to �E . Hence the
almost-everywhere limit of a sequence of measurable functions may not
be measurable.
Proof. Note that limn!1 fn(x) = 0 for all x 2 X. Moreover, �E(x) = 0
for all x /2 N . Thus, fn ! �E for all x /2 N , so {x | fn(x) 6! �E(x)} has
measure 0 under µ. Therefore, fn converges µ-a.e. to �E .
Additional Exercises:
1. Let (X,F) be a measurable space.
2
(a) Let ⇤ = {� | � is a charge on F}. Prove that ⇤ is a vector space
over R.
Proof. As ⇤ is a subset of all real-valued functions on F , and the
set of all real-valued functions is itself a vector space, we need only
verify the closure conditions required for a subset of a vector space
to be a vector subspace.
First, clearly the 0-function is in ⇤. Thus, we need only show that if
�1,�2 2 ⇤ and c1, c2 2 R, then c1�1 + c2�2 2 ⇤.
Let � = c1�1+c2�2. Clearly �(;) = 0. Moreover, if {En} are disjoint
in F , then
�([En) = c1�1([En) + c2�2([En)
= c1
X
�1(En) + c2
X
�2(En)
=
X
(c1�1(En) + c2�2(En)
=
X
�(En),
where we take the convention that 1�1 = 0. Thus, � is countably
additive, and therefore � is a charge on F . Thus, ⇤ is a vector
space.
(b) Let M = {µ | µ is a measure on F}. Is M a vector space over R?
Explain.
Proof. No, M is not a vector space. Since measures cannot take on
negative values, �µ /2 M for all µ 2 M . However, M is related
to a vector space in the sense that any finite linear combination of
measures is a measure, provided that the constants used are positive
(such a space is called a convex cone).
Extra stu↵:
2D. Let {An} be a sequence of subsets of a set X. If A consists of all x 2 X
which belong to infinitely many of the sets An, show that
A =
1\
m=1
1[
n=m
An.
The set A is often called the limit superior of {An} and is denoted by
lim supAn.
Proof. Let G =
1\
m=1
1[
n=m
An.
Suppose x 2 A. Then for all m, there exists nm > m such that x 2 Anm .
Thus x 2
S1
n=m An, and thus x 2 G. Therefore, A✓ G.
On the other hand, suppose x 2 G. Then for all m, x 2
S1
n=m An, so for
all m there exists nm > m such that x 2 Anm . Therefore, x is in infinitely
many of the sets An, so x 2 A. Therefore, G ✓ A.
Thus G = A as desired.
2E. Let {An} be a sequence of subsets of a set X. If B consists of all x 2 X
which belong to all but a finite number of the sets An, show that
B =
1[
m=1
1\
n=m
An.
3
The set B is often called the limit inferior of {An} and is denoted by
lim inf An.
Proof. Let G =
1[
m=1
1\
n=m
An.
Suppose x 2 B. Then there exists N such that x 2 An for all n � N .
Thus, x 2
T1
n=N An, so x 2 G. Thus, B ✓ G.
On the other hand, suppose x 2 G. Then there exists N such that x 2T1
n=N An, so x 2 An for all n � N , and thus there are at most N � 1 sets
An for which x /2 An. Therefore, x 2 B, so G ✓ B.
Thus, G = B as desired.
3I. Let (X,F , µ) be a measure space and let {En} ⇢ F . Show that
µ(lim inf En)  lim inf µ(En).
Proof. Let Fn = \m�nEm, so that Fn = Fn+1 \ En, and thus Fn ⇢
Fn+1. Moreover, lim inf En = [Fn. Then by Prop 3.4(a), we have
µ(lim inf En) = limµ(Fn). As Fn ⇢ En for all n, we have that µ(Fn) 
µ(En), and thus limµ(Fn)  lim inf µ(En), where we have traded the
limit for the limit inferior, since we cannot be assured that limµ(En)
exists. Therefore, µ(lim inf En)  lim inf µ(En), as desired.
3J. Let (X,F , µ) be a measure space and let {En} ⇢ F . Show that
lim supµ(En)  µ(lim supEn)
when µ([En) < 1. Show that this inequality may fail if µ([En) = 1.
Proof. As in the previous problem, takeGn = [m�nEm, so that lim supEn =
\Gn. Note that Gn = Gn+1 [ En, so Gn+1 ✓ Gn, and by hypothesis,
µ(G1) = µ([En) < 1. Therefore, by Prop 3.4(b), we have µ(lim supEn) =
µ(\Gn) = limµ(Gn). Moreover, as En ✓ Gn, we have that µ(Gn) �
µ(En), so limµ(Gn) � lim supµ(En), where again we have traded the
limit for the limit superior, in case limµ(En) does not exist. Therefore,
µ(lim supEn) � lim supµ(En), as desired.
Note that if µ([En) = 1, we find ourselves in a similar situation as
with problem 3H. Indeed, the same counterexample applies: take En =
R\[�n, n] for all n. Then Fn = En for all n, and �(Fn) = �(En) = 1 for
all n. On the other hand, lim supEn = \Fn = ;, so �(lim supEn) = 0 6�
lim supµ(En) = 1.
4
The set B is often called the limit inferior of {An} and is denoted by
lim inf An.
Proof. Let G =
1[
m=1
1\
n=m
An.
Suppose x 2 B. Then there exists N such that x 2 An for all n � N .
Thus, x 2
T1
n=N An, so x 2 G. Thus, B ✓ G.
On the other hand, suppose x 2 G. Then there exists N such that x 2T1
n=N An, so x 2 An for all n � N , and thus there are at most N � 1 sets
An for which x /2 An. Therefore, x 2 B, so G ✓ B.
Thus, G = B as desired.
3I. Let (X,F , µ) be a measure space and let {En} ⇢ F . Show that
µ(lim inf En)  lim inf µ(En).
Proof. Let Fn = \m�nEm, so that Fn = Fn+1 \ En, and thus Fn ⇢
Fn+1. Moreover, lim inf En = [Fn. Then by Prop 3.4(a), we have
µ(lim inf En) = limµ(Fn). As Fn ⇢ En for all n, we have that µ(Fn) 
µ(En), and thus limµ(Fn)  lim inf µ(En), where we have traded the
limit for the limit inferior, since we cannot be assured that limµ(En)
exists. Therefore, µ(lim inf En)  lim inf µ(En), as desired.
3J. Let (X,F , µ) be a measure space and let {En} ⇢ F . Show that
lim supµ(En)  µ(lim supEn)
when µ([En) < 1. Show that this inequality may fail if µ([En) = 1.
Proof. As in the previous problem, takeGn = [m�nEm, so that lim supEn =
\Gn. Note that Gn = Gn+1 [ En, so Gn+1 ✓ Gn, and by hypothesis,
µ(G1) = µ([En) < 1. Therefore, by Prop 3.4(b), we have µ(lim supEn) =
µ(\Gn) = limµ(Gn). Moreover, as En ✓ Gn, we have that µ(Gn) �
µ(En), so limµ(Gn) � lim supµ(En), where again we have traded the
limit for the limit superior, in case limµ(En) does not exist. Therefore,
µ(lim supEn) � lim supµ(En), as desired.
Note that if µ([En) = 1, we find ourselves in a similar situation as
with problem 3H. Indeed, the same counterexample applies: take En =
R\[�n, n] for all n. Then Fn = En for all n, and �(Fn) = �(En) = 1 for
all n. On the other hand, lim supEn = \Fn = ;, so �(lim supEn) = 0 6�
lim supµ(En) = 1.
4
 
Math 426: Homework 2 Solutions
Mary Radcli↵e
due 16 April 2014
In Bartle:
9B. Show that the family G of all finite unions of sets of the form (a, b), (�1, b),
(a,1), (�1,1) is not an algebra of sets in R.
Solution. Consider E = (�1, 0) [ (1,1) 2 G. Note that R\E = [0, 1].
Since [0, 1] is closed, it cannot be written as the finite union of open sets.
Therefore, R\E /2 G, and thus G is not an algebra.
9E. Prove that if E is a countable subset of R, then it has Lebesgue measure
0.
Proof. Note that by definition, �((a, b]) = b�a. Moreover, as we have seen,
�([a, b]) = b � a. Thus, �(a) = �([a, b]\(a, b]) = �([a, b]) � �((a, b]) = 0,
so singletons have measure 0. By countable additivity, as every countable
set can be written as the countable union of singletons, it must also have
measure 0.
9S. Let g be a monotone increasing and right continuous function on R to R.
If µg is as defined on page 105, prove that µg is a measure on F (and can
hence be extended to a measure on F⇤ using the Carthéodory Extension
Theorem).
Proof. Clearly, µg(;) = 0, and µg((a, b]) � 0 for all a  b since g is a
monotonic function. Thus, it su�ces to check that µg is additive; that
is, if E 2 F can be written as an (at most) countable union E = [En of
pairwise disjoint sets in F , then µg(E) =
P
µg(En).
As with `, it su�ces to verify this property for generating sets of F , of
which there are four types: (a, b], (�1, b], (a,1), and (�1,1).
Case 1: E = (a, b] = [1n=1En, where En = (an, bn], and the En are
disjoint. Let (a1, b1], . . . , (ak, bk] be any finite subcollection of the En
(renumbered as 1 through k for convenience of notation), such that a 
a1 < b1  a2 < b2  . . . bk  b. Then
Pk
n=1 µg(En) =
P1
n=1(g(bn) �
g(an))  g(b)� g(a) = µg((a, b]). As this holds for any finite collection of
the En, we must have that
P1
n=1 µg(En)  µg((a, b]).
For the opposite inequality, choose ✏ > 0, and let {✏j} be a sequence of
positive numbers such that
P
✏j < ✏. For each j, let cj be such that g(bj+
cj)� g(bj) < ✏j ; this is possible since g is right continuous. Then consider
Ij = (aj , cj) � (aj , bj ]. Note that for any � > 0, we have [a+ �, b] ⇢ [Ij ,
and thus {Ij} has a finite subcover of [a+ �, b]. Let us take this subcover
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to be I1, I2, . . . , Ik, and order these such that a1 < a + �, bk + ck > b,
and if 2  j  k, we have aj < bj�1 + cj�1 (note: we may remove any
redundant or unnecessary sets to do this ordering). Then we have
g(b)� g(a+ �)  g(bk + ck)� g(a1)

kX
j=1
(g(bj + cj)� g(aj))

kX
j=1
(g(bj) + ✏j � g(aj))

1X
j=1
(g(bj)� g(aj)) + ✏.
As this holds for all � > 0 and g is right continuous, we can take � ! 0
to obtain that g(b)� g(a) 
P1
j=1(g(bj)� g(aj)) + ✏. Taking ✏ ! 0 then
yields the result.
Case 2: E = (a,1) = [1n=1En, where En = (an, bn] and the En are
pairwise disjoint. (Note: here we allow bn 2 R). Let L = limx!1 g(x) 2
R. Suppose bn = 1 for some n; wolog, let b1 = 1. Then [1n=2En =
(a, a1], which by the above argument has µg([1n=2En) =
P1
n=2 µg(En) =
g(a1)� g(a). Moreover, by definition, µg(E1) = L� g(a1). Thus,
µg(E) = L� g(a) = µg(E1) +
1X
n=2
µg(En) =
1X
n=1
µg(En).
If no bn = 1, then choose M, � > 0. Note that [a+�, a+M ] ⇢ (a,1), and
thus we may construct an open cover of [a+�, a+M ] as in Case 1. This will
yield that g(a+M)�g(a+�) 
P1
n=1 µg(En)+✏ for all ✏ > 0, and thus by
taking first � and then ✏ to 0, we have that g(a+M)�g(a) 
P1
n=1 µg(En)
for all M > 0. Taking M ! 1 yields that µg((a,1)) 
P1
n=1 µg(En).
On the other hand, if we arrange E1, E2, . . . , Ek such that a  a1 < b1 
a2 < b2  · · ·  bk, then we have
Pk
n=1 µg(E1) =
Pk
n=1(g(bn)� g(an)) 
g(bk) � g(a)  L � g(a), so for any finite sum,
Pk
n=1 µg(En)  µg(E).
Thus, takingk ! 1 yields the result.
For Cases 3 and 4, proceed similarly as with Case 2. We omit the details
here.
Notation: Throughout, � refers to Lebesgue measure, and we assume the
Axiom of Choice, forever and ever.
1. Prove Theorem 3 in the additional notes: Let E 2 L, x 2 R. Define the
following sets:
(a) E + x = {y + x | y 2 E}
(b) xE = {xy | y 2 E}
Then E + x, xE 2 L, and
(a) �(E + x) = �(E)
(b) �(xE) = |x|�(E)
Proof. (a) Note that it is immediately clear that for any interval I, we
have µ⇤(I + x) = µ⇤(I). Therefore, by definition, we also have that
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for any A ⇢ R, µ⇤(A + x) = µ⇤(A). Thus, it remains only to show
that if E is measurable, then E+x is also measurable. Note, however,
that if A ⇢ R, since E 2 L, we have
µ⇤(A) = µ⇤(A� x) = µ⇤((A� x) \ E) + µ⇤((A� x) \ Ec)
= µ⇤(A \ (E + x)) + µ⇤(A \ (Ec + x))
= µ⇤(A \ (E + x)) + µ⇤(A \ (E + x)c),
and thus, E + x is measurable.
(b) Again, it is immediately clear by definition that for any interval I,
we have µ⇤(xI) = |x|µ⇤(I). Thus, it is clear that µ⇤(xA) = |x|µ⇤(A)
for any A 2 L. Therefore, it remains only to show that if E is
measurable, then xE is also measurable. Note that if x = 0, this is
obvious, so we assume that x 6= 0.
Let ✏ > 0, and assume E 2 L. Then there exists an open set U
with E ⇢ U and �(U\E) < ✏|x| . Note that x(U\E) = (xU\xE), and
that xU is clearly open. Moreover, as E is assumed to be Lebesgue,
we have that U\E is Lebesgue and thus for any ↵, �(↵(U\E)) =
|↵|�(U\E). Thus we have xE ⇢ xU and �(xU\xE) = �(x(U\E)) =
|x|�(U\E) < ✏. But then xE satisfies condition (2) of Theorem 4,
and therefore xE 2 L.
2. Complete the proof of Theorem 4 in the additional notes.
Proof. It remains only to show that the third condition is equivalent to
the first two.
Suppose E 2 L. Then Ec 2 L, and thus by condition (2), for all ✏ there
exists an open set U such that Ec ⇢ U and �(U\(Ec)) < ✏. Let C = U c.
Then as Ec ⇢ U , we have C ⇢ E. Moreover, E\C = E \ Cc = E \ U =
U\(Ec), so �(E\C) < ✏. Therefore, condition (3) is true.
Now, let us assume that (3) holds. We wish to show that E is measurable.
Note that it is clear, by the same argument as above, that for all ✏ there
exists an open set U such that Ec ⇢ U and �(U\Ec) < ✏. But this implies
that Ec 2 L, and thus E 2 L.
3. Prove the following theorem:
Let A be an algebra, µ a measure on �(A), and B 2 �(A) with
µ(B) < 1. For any ✏ > 0, there exists A 2 A with µ(A�B) < ✏,
where A�B is the symmetric di↵erence as defined in Theorem
5 Version 2.
Use this theorem to prove Theorem 5 Version 2 from the additional notes.
Proof. Note that as �(A) is a �-field, it is also a field. Let µ⇤ be the
extension of µ to �(A)⇤ from Carthéodory’s Extension Theorem, so that
µ = µ⇤ on �(A). Note also that by countable additivity and the defini-
tion of outer measure, µ⇤ ⌘ (µ|A)⇤, where (µ|A)⇤ is the extension of the
measure µ|A on the field �.
Thus if µ(B) < 1, we have µ(B) = µ⇤(B) = (µ|A)⇤(B) = inf
P
µ(En),
where the inf is over all collections {En} ⇢ A with B ⇢ [En. Take a
collection En with µ⇤(B) <
P
µ(En) +
✏
2 . Note that this implies that
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µ(([En)\B) < ✏2 . Now, since µ(B) < 1, the sum converges, so we may
choose N su�ciently large that
�����
NX
n=1
µ(En)�
1X
n=1
µ(En)
����� <
✏
2
.
Let A = [Nn=1En, and E = [1n=1En. Note that the previous inequality
implies that µ(E\A) < ✏2 . Now, if x 2 B\A, then x 2 E and x /2 A,
so x 2 E\A. If x 2 A\B, then x 2 E\B. Therefore, we have that
µ(A�B)  µ(E\A) + µ(E\B) < ✏. Moreover, as A is a finite union of
sets in A, we also have that A 2 A, yielding the result.
To prove Thm 5, v. 2, we proceed as follows. Let E 2 L, ✏ > 0. and by
Thm 4, let Un be an open set such that E ⇢ Un and �(Un\E) < ✏2n . Then
by the above result, we may choose An 2 F so that �(An�Un) < ✏2·2n .
Put A = \An. Then A 2 B. Moreover, we have
�(A�E) = �(A\E) + �(E\A)
= �((\An)\E) + �(E\(\An))
= �(\(An\E)) + �([(E\An))
 �(A1\E) +
X
�(E\An)
 ✏
2
+
X
�(Un\An)
 ✏
2
+
X ✏
2 · 2n = ✏,
and the result is verified.
4. Prove that every Lebesgue measurable function is almost equal to a Borel
measurable function.
Proof. We assume, wolog, that f � 0, as if not we may prove the result
for both f+ and f� and simply subtract the two functions.
Let f : R ! R be a nonnegative Lebesgue measurable function. For
each q 2 Q+, let Lq = f�1((q,1). Since f is Lebesgue measurable,
Lq 2 L, so there exists some Borel set Bq and some null set Nq such that
Lq = Bq [ Nq. Let N = [q2Q+Nq. Then since each Nq has �(Nq) = 0
and Q+ is countable, by countable additivity we have �(N) = 0. Define
g(x) = f(x) if x /2 N and g(x) = 0 if x 2 N . Clearly, g = f almost
everywhere.
Now, note that for any q 2 Q+, we have
g�1((q,1)) = {x | g(x) > q}
= {x | x 2 Bp 8p < f(x) and f(x) > q}
=
[
↵>q
 
\
p<↵
Bp
!
=
[
r>q
r2Q
 
\
p<r
Bp
!
as this is a countable union/intersection of Borel sets, and the last line
follows by density of Q. Moreover, for any arbitrary ↵ � 0, we have
g�1((↵,1)) =
[
q>↵
q2Q
g�1((q,1)) 2 B,
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and thus g�1((↵,1)) 2 B for all ↵ 2 R+. Also, g�1((↵,1)) = R for any
↵ < 0, so g is therefore Borel measurable.
5. (a) Let E be the nonmeasurable set defined in Section 3 of the notes.
Prove that if F ⇢ E is Lebesgue measurable, then �(F ) = 0.
Proof. Suppose F ⇢ E is measurable. Then, as in the proof seen
in Section 3, we have that [q2[�1,1]\Q(F + q) ⇢ [q2[�1,1]\Q(E +
q) ⇢ [�1, 2]. By countable additivity, if �(F + q) = ↵ > 0, then
�([q2[�1,1]\Q(F + q)) = 1, which is a contradiction to the mono-
tonicity of �. Thus, �(F ) = 0.
(b) Let A be a Lebesgue measurable set, with �(A) > 0. Prove that
A contains a nonmeasurable subset. (Hint: reduce to the case that
A ⇢ [0, 1]. Then consider [(A \ E + q) as in the example done in
class/notes.)
Solution. Let An = [�n, n] \ A. Since �(A) > 0, there exists some
n such that �(An) > 0. Let Bn =
1
nAn, so that Bn ⇢ [0, 1] and
�(Bn) > 0. Note that if F ⇢ Bn is nonmeasurable, then by Theorem
3, nF ⇢ An ⇢ A is also nonmeasurable. Thus, it su�ces to consider
sets of positive Lebesgue measure in [0, 1]
Suppose A ⇢ [0, 1] is Lebesgue measurable with �(A) > 0. Let
Aq = A \ (E + q), where E is as in Section 3 in the notes. Then
clearly A = [q2[�1,1]A \ (E + q), and A0 ⇢ A. Moreover, if A0 is
measurable, then Aq is also measurable with �(Aq) = �(A0) for all q.
But then �(A) = �([q2[�1,1]A \ (E + q)). If A0 is measurable with
positive measure, this is a contradiction, since �(A) < 1. If A0 is
measurable with measure 0, this is a contradiction, since �(A) > 0.
Thus, A0 is nonmeasurable.
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Powered by TCPDF (www.tcpdf.org)
 
Math 426: Homework 3
Mary Radcli↵e
due 23 April 2014
In Bartle:
4A. If the simple function ' 2 M+ has the representation ' =
Pm
k=1 bk�Fk ,
where bk 2 R and Fk 2 F , prove that
R
' dµ =
Pm
k=1 bkµ(Fk).
Proof. For 1  k  m, let k = bk�Fk + 0�X\Fk . Then k is in standard
form, so
R
 k dµ = bkµ(Fk). Moreover, ' =
Pm
k=1 k, and thus since
integration is linear, we have
R
' dµ =
PR
 k dµ =
Pm
k=1 bkµ(Fk).
4B. The sum, scalar multiple, and product of simple functions are simple func-
tions.
Proof. Let ' =
Pm
k=1 ak�Ek and =
Pn
j=1 bk�Fk be two simple func-
tions, and let c 2 R. Note that c' =
Pm
k=1(cak)�Ek is clearly simple.
Now, as the only possible values of ' are a1, a2, . . . , am and the only pos-
sible values of are b1, b2, . . . , bn, we have that the only possible values of
'+ are a1+b1, a1+b2, . . . , a1+bn, a2+b1, . .. , am+bn and the only pos-
sible values of ' are a1b1, a1b2, . . . , a1bn, a2b1, . . . , ambn, and thus both
' + and ' take only finitely many values. By definition, then, both
'+ and ' are simple functions.
4G. Let X = Z+, F be the set of all subsets of X, and µ be counting measure.
If f is a nonnegative function on X, then f 2 M+(X,F) and
R
f dµ =P1
n=1 f(n).
Proof. Let fk = f�[k], where [k] = {1, 2, . . . , k}. Then fk is a simple
function, as it can take at most k values, and fk can be written in the rep-
resentation fk =
Pk
i=1 f(i)�{i}, so
R
fk =
Pk
i=1 f(i)µ({i}) =
Pk
i=1 f(i).
Moreover, fk  fk+1 for all k, and thus by the MCT, we have
R
f dµ =
lim
R
fk dµ = lim
Pk
i=1 f(i) =
P1
i=1 f(i).
4K. If (X,F , µ) is a finite measure space, and {fn} is a real-valued sequence
in M+(X,F) which converges uniformly to a function f , then f 2 M+
and
R
f dµ = lim
n!1
Z
fn dµ.
Proof. By Cor. 2.10, f 2 M , and as fn � 0 for all n, we also have f � 0,
and thus f 2 M+. Let ✏ > 0, and let N be su�ciently large that if n � N ,
we have |fn(x) � f(x)| < ✏ for all x 2 X. Choose n � N , and let ' be a
simple function with '  fn. Define as
 (x) =
⇢
0 if '(x) < ✏
'(x)� ✏ if '(x) � ✏ .
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As ' takes only finitely many values, we have also that takes only
finitely many values, and if ' =
Pn
i=1 ai�Ei +
Pm
j=1 bj�Fj , where ai � ✏
for all i and bj < ✏ for all j, then =
Pn
i=1(ai � ✏)�Ei . Moreover, as
|fn(x)� f(x)| < ✏ for all x, we have < f . Therefore,
Z
f dµ �
Z
 dµ
=
nX
i=1
(ai � ✏)�Ei
=
nX
i=1
ai�Ei +
mX
j=1
bj�Fj �
mX
j=1
bj�Fj �
nX
i=1
✏�Ei
�
Z
' dµ�
mX
j=1
✏�Fj �
nX
i=1
✏�Ei
=
Z
' dµ� ✏µ(X)
4L. Let X be a finite closed interval [a, b] in R, and let F be the collection of
Borel sets in X, and � be Lebesgue measure on F . If f is a nonnegative
continuous function on X, show that
R
f d� =
R b
a f(x)dx.
Proof. Recall that we can define
R b
a f(x)dx = limn!1
P2n�1
k=0 mn,k
b�a
2n ,
where
mn,k = min
a+(b�a) k2n xa+(b�a)
k+1
2n
f(x).
That is, we take the limit of the step function defined on a partition that
divides [a, b] into 2n equal intervals. As f is continuous, all mn,k are finite,
and this limit is guaranteed to exist and equal the integral1 of f .
Define 'n to be the simple function which takes value mn,k on [a + (b �
a) k2n , a+ (b� a)
k+1
2n ] and 0 elsewhere. Note that the 'n are an increasing
sequence of nonnegative functions with limit f . Therefore, by the MCT,R
f d� = limn!1
R
'n d� = limn!1
P2n�1
k=0 mn,k
b�a
2n =
R b
a f dx.
4M. Let X = [0,1), F the set of Borel subsets of X, and � be Lebesgue
measure on F . If f is a nonnegative continuous function on X, show that
R
f d� = lim
b!1
Z b
0
f(x)dx.
Proof. Take {an} to be any nonnegative sequence that increases to1. De-
fine fn = f�[0,an]. Then the fn are an increasing sequence of nonnegative
functions with limit f , and thus by the MCT, we have
Z
f d� = lim
n!1
Z
fn d� = lim
n!1
Z an
0
f dx,
where we are able to pass from the Lebesgue integral to the Riemann
integral on the bounded interval [0, an] by the previous problem. As
this holds for any sequence tending to 1, we must have that
R
f d� =
limb!1
R b
0 f dx.
1
If you took 425 with me last quarter, we did not define the Riemann integral this way.
However, this interpretation can be proven to be equivalent to ours. You should check that you
can prove it, using the It Doesn’t Matter Lemma (6.7) from Baby Rudin and an appropriately
defined refinement of an arbitrary partition.
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4S. If f 2 M+(X,F) and
R
f dµ < 1, then for any ✏ > 0 there exists a set
E 2 F such that µ(E) < 1 and
R
f dµ =
R
E f dµ+ ✏.
Proof. Let ✏ > 0. For n 2 Z+, define En = {x 2 X | f(x) � 1n}. Note
that as
R
f dµ �
R
En
f dµ � 1nµ(En), we have that µ(En) is finite for
all n. Moreover, if fn = f�En then as the En are increasing sets, the fn
are increasing positive functions with limit f , and therefore by the MVT,R
fndµ !
R
f dµ. Therefore, 9N such that
��R fNdµ�
R
f dµ
�� < ✏.
Therefore, taking E = EN yields the result.
4T. Suppose that {fn} ⇢ M+(X,F), that {fn} converges to f , and that
R
f dµ = lim
n!1
Z
fn dµ < 1. Prove that
R
E f dµ = limn!1
Z
E
fn dµ for all
E 2 F .
Proof. Let us suppose, to the contrary, that there exists some E withR
E f dµ 6= limn!1
R
E fn dµ. Note that by Fatou’s Lemma, this implies
that
R
E f dµ < lim inf
R
E fn dµ. Then we have the following.
Z
f dµ =
Z
E
f dµ+
Z
X\E
f dµ
< lim inf
Z
E
fn dµ+ lim inf
Z
X\E
fn dµ
 lim inf
 Z
E
fn dµ+
Z
X\E
fn dµ
!
= lim
Z
X
fn dµ =
Z
f dµ,
where strict inequality follows in line 2 by Fatou’s Lemma and because
both lim inf
R
E fn dµ and lim inf
R
X\E fn dµ are finite. This is a clear
contradiction, and thus we must have that
R
E f dµ = lim
R
E fn dµ for all
E 2 F .
Also, complete the following:
1. (a) Show, by example, that if {fn} ⇢ M+ is a sequence of functions with
fn(x) � fn+1(x) (i.e., a decreasing sequence) with fn ! f , that it
may not be true that
R
fdµ = lim
R
fndµ (and thus there is no MCT
for decreasing sequences).
Proof. Define fn = �[n,1]. Then fn is decreasing, and fn ! 0 = f .
However,
R
f d� = 0 6= 1 = lim
R
fn d�.
(b) Let (X,F , µ) be a finite measure space. Prove that if {fn} ⇢ M+(X,F)
is a decreasing sequence of uniformly bounded functions with fn ! f ,
then
R
fdµ = lim
R
fndµ.
Proof. Let M be a uniform bound for fn. Define gn = M � fn.
Then gn is an increasing sequence with limit M � f , and thus by the
MCT,
R
(M � f)dµ = lim
R
(M � fn)dµ. Note that
R
Mdµ is finite
since µ is finite, so the result follows by subtracting
R
Mdµ from both
sides.
(c) Show that if {fn} ⇢ M+ is a decreasing sequence of functions with
fn ! f and
R
fndµ < 1 for some n, then lim
R
fndµ =
R
fdµ.
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Proof. Put gk = fn � fn+k�1 for all k � 1. Then gk is an increasing
sequence of functions, with gk ! fn�f . Thus, by the MCT, we haveR
(fn � f)dµ = limk!1
R
(fn � fn+k�1)dµ, and the result follows by
subtracting the (finite) integral
R
fn dµ from both sides.
2. Prove that if X is a finite set under counting measure µ, then for every
sequence of function fn on X with fn ! f , we have
R
fndµ !
R
fdµ.
Proof. WOLOG, we assume that X = {1, 2, . . . , N}. Then
R
fn dµ =PN
k=1 fn(k) and
R
f dµ =
PN
k=1 f(k). Taking a limit, we have lim
R
fn dµ =
lim
n!1
NX
k=1
fn(k) =
NX
k=1
lim
n!1
fn(k) =
NX
k=1
f(k), as desired.
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Powered by TCPDF (www.tcpdf.org)
 
Math 426: Homework 4
Mary Radcli↵e
due 2 May 2014
In Bartle:
5C. If f 2 L(X,F , µ) and g is a F-measurable real-valued function such that
f(x) = g(x) almost everywhere, then g 2 L(X,F , µ) and
R
f dµ =
R
g dµ.
Proof. Let h(x) = |f(x)� g(x)|. Then h 2 M+(X,F), and h = 0 almost
everywhere. Therefore, h is integrable and
R
h dµ = 0. But this implies
that f�g is integrable, and thus g is integrable. Moreover,
��R (f � g) dµ
�� R
h dµ = 0, and thus
R
(f � g) dµ = 0. The result immediately follows by
linearity of integration.
5E. If f 2 L and g is a bounded, measurable function, then fg 2 L.
Proof. Let M be such that |g|  M . Note that by linearity of integration,
we have that Mf 2 L, and |fg|  |Mf | for all x. Then by Corollary 5.4,
fg is integrable.
5I. If f is a complex-valued function on X such that Re f and Im f belong to
L(X,F , µ), we say that f is integrable and define
R
f dµ =
R
Re f dµ +
i
R
Im f dµ. Let f be a complex-valued measurable function. Showthat
f is integrable if and only if |f | is integrable, in which case
��R f dµ
�� R
|f | dµ.
Proof. Let us write f = f1 + if2, so that f1 = Re f and f2 = Im f . We
have that f is measurable if and only if f1 and f2 are measurable, and
|f | =
�
(f1)2 + (f2)2
�1/2
. Thus, f is measurable implies that |f | is also
measurable.
Notice, if a, b 2 R, then
p
a2 + b2 =
p
(a+ b)2 � 2ab 
p
(a+ b)2 =
|a + b|  |a| + |b|. Therefore, if f is integrable, then |f | is a measurable
function such that |f |  |f1| + |f2|, and thus by Corollary 5.4, |f | is
integrable.
Conversely, if |f | is integrable, note that |f1| < |f | and |f2| < |f |, so we
have f1 and f2 are real-valued integrable functions by Corollary 5.4. But
then by definition f is also integrable.
For the inequality, write
R
f dµ = rei✓, so that
��R f dµ
�� = r. Let g(x) =
e�i✓f(x), so that g(x) = g1(x) + ig2(x), and
R
g dµ =
R
e�i✓f(x) dµ =
e�i✓
R
f(x) dµ = r. But then
R
g2 dµ = 0, so g2 = 0 almost everywhere.
Moreover, |f | = |e�i✓||f | = |g| = |g1| almost everywhere.
Therefore, we have
����
Z
f dµ
���� = r =
Z
g1 dµ =
����
Z
g1 dµ
���� 
Z
|g1| dµ =
Z
|f | dµ,
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since g1 is real valued and we may use Theorem 5.3.
5P. Let fn 2 L(X,F , µ), and suppose that {fn} converges to a function f .
Show that if lim
R
|fn � f | dµ = 0, then
R
|f | dµ = lim
R
|fn| dµ.
Proof. Note that as |fn � f | is integrable for n su�ciently large, we have
fn � f is integrable, and thus by linearity f is integrable. Moreover, for
all n, we have |fn|  |fn�f |+ |f |, so |fn|� |f |  |fn�f |. Integrating and
taking limits on both sides, we have lim
R
|fn| dµ�
R
|f | dµ  lim
R
|fn �
f | dµ = 0, and thus lim
R
|fn| dµ =
R
|f | dµ
5Q. If t > 0, then
R1
0 e
�txdx = 1t . Moreover, if t � a > 0, then e
�tx  e�ax.
Use this and Exercise 4M to justify di↵erentiating under the integral sign
and to obtain the formula
R1
0 x
ne�xdx = n!.
Proof. Now,
Z 1
0
e�txdx = lim
b!1
Z b
0
e�txdx
= lim
b!1
�1
t
e�tx
����
b
x=0
= lim
b!1
✓
�1
t
e�tb +
1
t
◆
=
1
t
,
where the integral is obtained by the fundamental theorem of calculus.
Let f(x, t) = e�tx for t 2 [ 12 ,
3
2 ]. Then
@f
@t (x, t) = �xe
�tx, so |@f@t (x, t)| 
xe�
x
2 for all t. Let g(x) = xe�
x
2 . We claim that g is integrable on [0,1).
This can be seen in several ways. Note that g is measurable. Moreover,
note that if x is su�ciently large (in fact, x > 8 ln 4 is su�cient), we
have that x < ex/4. Define h(x) = g(x) if x  8 ln 4 and h(x) = e� x4 if
x > 8 ln 4. Then g  h, and h is clearly integrable, and thus by Corollary
5.4, g is also integrable. But then we have, by Theorem 5.9, that
� 1
t2
=
d
dt
✓
1
t
◆
=
d
dt
✓Z 1
0
e�txdx
◆
=
d
dt
 Z
[0,1)
e�txd�(x)
!
=
Z
[0,1)
@
@t
�
e�tx
�
d�(x)
=
Z
[0,1)
�xe�txd�(x)
= �
Z 1
0
xe�txd�(x).
Taking t = 1 and multiplying by �1 yields the result for n = 1.
Now, proceed by induction on n. Suppose it is known that
R1
0 x
n�1e�txdx =
(n�1)!
tn . Let fn(x, t) = x
n�1e�tx for t 2 [ 12 ,
3
2 ]. As above, we have
@fn
@t (x, t) = �x
ne�tx, so |@fn@t (x, t)|  x
ne�
x
2 . Again, for x su�ciently
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large, we have that xn < ex/4, so as above, we have that xne�
x
2 is inte-
grable on [0,1). Then as above, we have
� n!
tn+1
=
d
dt
✓
(n� 1)!
tn
◆
=
d
dt
✓Z 1
0
xn�1e�txdx
◆
=
Z
[0,1)
@
@t
(xn�1e�tx)d�(x)
= �
Z 1
0
xne�txdx.
Taking t = 1 and multiplying by �1 yields the result for n.
5R. Suppose that f is defined on X ⇥ [a, b] to R and that the function x 7!
f(x, t) is F-measurable for each t 2 [a, b]. Suppose that for some t0, t1 2
[a, b], the function x 7! f(x, t0) is integrable on X, that @f@t (x, t1) exists,
and that there exists an integrable function g onX such that
��� f(x,t)�f(x,t1)t�t1
��� 
g(x) for x 2 X and t 2 [a, b], t 6= t1. Then

d
dt
Z
f(x, t) dµ(x)
�
t=t1
=
Z
@f
@t
(x, t1) dµ(x).
Proof. Choose tn to be any sequence with tn ! t1, tn 6= t1 (we may
start at n = 2 to avoid confusion). Put hn(x) =
f(x,tn)�f(x,t1)
tn�t1 , so by
hypothesis |hn(x)|  g(x) for all x, and hn(x) ! @f@t (x, t1). By the
dominated convergence theorem, then, we have limn!1
R
X hn(x)dµ(x) =R
X
@f
@t (x, t1)dµ(x). On the other hand,
lim
n!1
Z
X
hn(x)dµ(x) = lim
n!1
Z
f(x, tn)� f(x, t1)
tn � t1
dµ(x)
= lim
n!1
1
tn � t1
✓Z
f(x, tn)dµ(x)�
Z
f(x, t1)dµ(x)
◆
=

d
dt
Z
f(x, t)dµ(x)
�
t=t1
,
as desired.
5T. Let f be a F-measurable function on X to R. For n 2 Z+, let {fn}
be the sequence of truncates of f . If f is integrable with respect to µ,
then
R
f dµ = lim
R
fn dµ. Conversely, if sup
R
|fn| dµ < 1, then f is
integrable.
Proof. Note that for all n, we have |fn|  f , and thus if f is integrable,
the DCT implies that
R
fn dµ !
R
f dµ.
For the converse, note that |fn| is a monotonically increasing sequence
of nonnegative functions with limit |f |, and thus the MCT implies thatR
|fn| dµ !
R
|f | dµ. Thus, by hypothesis,
R
|f | dµ < 1, and therefore
|f | and thus f are integrable.
4O. Fatou’s Lemma has an extension to a case where the fn take on negative
values. Let h 2 M+(X,F), and suppose that
R
h dµ < 1. If {fn} is a
sequence in M(X,F) and �h  fn then
R
lim inf fn dµ  lim inf
R
fn dµ.
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Proof. Note that fn + h � 0. Thus we have
Z
lim inf fn dµ+
Z
h dµ =
Z
(lim inf fn + lim inf h) dµ
=
Z
(lim inf(fn + h)) dµ
 lim inf
Z
(fn + h) dµ
= lim inf
✓Z
fn dµ+
Z
h dµ
◆
= lim inf
Z
fn dµ+
Z
h dµ.
Subtracting
R
h dµ yields the result.
Also, complete the following:
1. Compute the following limits. Justify each computational step using con-
vergence theorems and/or calculus.
(a) limn!1
R1
0
n
1+n2x2 dx.
(b) limn!1
R1
0
�
1 + xn
��n
log(2 + cos( xn ))dx
(c) limn!1
R n
�n f
�
1 + xn2
�
g(x)dx, where g : R ! R is (Lebesgue) inte-
grable and f : R ! R is bounded, measurable, and continuous at
1.
Solution. (a) We consider this as two integrals, over [0, 1] and [1,1).
Note that on [0, 1], we have
lim
n!1
Z 1
0
n
1 + n2x2
dx = lim
n!1
[arctan(nx)]10
= lim
n!1
arctan(n) =
⇡
2
.
For the other part, notice that n1+n2x2 
n
n2x2 
1
nx2 
1
x2 , and
moreover,
R1
1
1
x2 = 1 < 1, so
1
x2 2 L([1,1),L,�). Therefore, by
the dominated convergence theorem, we have
lim
n!1
Z 1
1
n
1 + n2x2
dx = lim
n!1
Z
[1,1)
n
1 + n2x2
d�(x)
=
Z
[1,1)
lim
n!1
✓
n
1 + n2x2
◆
d�(x)
=
Z
[1,1)
0d�(x) = 0.
Therefore, we obtain limn!1
R1
0
n
1+n2x2 dx =
⇡
2 .
(b) Note that (1 + xn )
�n  e�x, and as cos( xn )  1, we have log(2 +
cos( xn ))  log 3. Thus,
�
1 + xn
��n
log(2+cos( xn ))  (log 3)e
�x, which
is clearly integrable. Therefore, by the dominated convergence theo-
rem, we have
lim
n!1
Z 1
0
⇣
1 +
x
n
⌘�n
log
⇣
2 + cos
⇣x
n
⌘⌘
dx =
Z 1
0
lim
n!1
✓⇣
1 +
x
n
⌘�n
log
⇣
2 + cos
⇣x
n
⌘⌘◆
dx
=
Z 1
0
e�x log(3)dx
= log(3)
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(c) Let us rewrite the integral as
lim
n!1
Z n
�n
f
⇣
1 +
x
n2
⌘
g(x)dx = lim
n!1
Z
R
f
⇣
1 +
x
n2
⌘
�[�n,n]g(x) d�(x).
LetM be such that |f(x)| < M for all x. Then
��f
�
1 + xn2
�
�[�n,n]g(x)
�� 
M |g(x)|, and since g 2 L, we also have |g| 2 L and thus M |g| 2 L.
Therefore, by the dominated convergence theorem, we have
lim
n!1
Z n
�n
f
⇣
1 +
x
n2
⌘
g(x)dx = lim
n!1
Z
R
f
⇣
1 +
x
n2
⌘
�[�n,n]g(x) d�(x)
=
Z
R
lim
n!1
⇣
f
⇣
1 +
x
n2
⌘
�[�n,n]g(x)
⌘
d�(x)
=
Z
R
f(1)�Rg(x)d�(x)
= f(1)
Z
R
g(x)d�(x),
where the limit exists because f is continuous at 1.
2. Let f(t)=
R t
0 e
�x2dx. We will use DCT to evaluate limt!1 f(t), even
without being able to evaluate the indefinite integral.
(a) Put h(t) = f(t)2 and g(t) =
R 1
0
e�t
2(1+x2)
1+x2 dx. Show that h
0(t) =
�g0(t).
(b) Show that h(t) + g(t) = ⇡4 for all t.
(c) Use the previous parts to conclude that lim
t!1
f(t) =
Z 1
0
e�x
2
dx =
p
⇡
2
.
Solution. (a) Note that h0(t) = 2f(t)f 0(t) = 2e�t
2
f(t), by the funda-
mental theorem of calculus.
For g0(t), let f(x, t) = e
�t2(1+x2)
1+x2 , and note that
@
@t
h
e�t
2(1+x2)
1+x2
i
=
�2te�t2(1+x2) = �2te�t2e�t2x2 . Put ⌘(t) = 2te�t2 . Note that ⌘0(t) =
(2 � 4t2)e�t2 , and thus by the first derivative test, ⌘ has a global
maximum of
p
2/e at t =
p
1/2, and a global minimum of �
p
2/e
at t = �
p
12. Therefore,
����2te�t
2
��� 
p
2/e =: C for all t. Thus,
���@f@t
���  Ce�t
2x2  C, which is clearly integrable on [0, 1]. Therefore,
by Theorem 5.9,
d
dt
"Z 1
0
e�t
2(1+x2)
1 + x2
dx
#
=
Z 1
0
@
@t
 
e�t
2(1+x2)
1 + x2
!
dx
=
Z 1
0
�2te�t
2(1+x2)dx
= �2te�t
2
Z 1
0
e�t
2x2dx
= �2e�t
2
Z t
0
e�u
2
du using the substitution u = tx
= �2e�t
2
f(t) = �h0(t),
as desired.
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(b) As (h + g)0 = 0 for all t, we have that h + g is constant. Note that
h(0) = 0. Moreover, g(0) =
R 1
0
1
1+x2 dx = arctan(1) =
⇡
4 . Therefore,
(h+ g)(0) = ⇡4 , and as h+ g is constant, the result follows.
(c) Note that |f(x, t)|  11+x2 for all x, t, and thus by the dominated
convergence theorem, we have
lim
t!1
g(t) =
Z 1
0
lim
t!1
e�t
2(1+x2)
1 + x2
dx
=
Z 1
0
0 dx = 0.
But then limt!1 h(t) =
⇡
4 , and thus limt!1 f(t) =
p
⇡/4 =
p
⇡
2 .
Extras:
5A. If f 2 L(X,F , µ) and a > 0, show that the set {x 2 X | |f(x)| > a}
has finite measure. In addition, the set {x 2 X | f(x) 6= 0} has �-finite
measure.
Proof. As |f | 2 L, we have that
R
|f | dµ = A < 1. Let Ea = {x 2
X | |f(x)| > a}. Then we have |f | = |f |�Ea + |f |�Eca � a�Ea + |f |�Eca ,
and thus
A =
Z
|f | dµ � aµ(Ea) +
Z
Eca
|f | dµ � aµ(Ea).
Therefore, µ(Ea)  A/a.
Moreover, {x 2 X | f(x) 6= 0} = [n2Z+En, and thus the set has �-finite
measure.
5B. If f is a F-measurable real-valued function and if f(x) = 0 for µ-almost
all x 2 X, then f 2 L(X,F , µ) and
R
f dµ = 0.
Proof. Note that |f | 2 M+ is µ-almost 0, and thus
R
|f | dµ = 0. This
implies that |f | 2 L, and thus f 2 L. Moreover,
��R f dµ
�� 
R
|f | dµ = 0,
and thus
R
f dµ = 0.
5D. If f 2 L(X,F , µ) and ✏ > 0, then there exists a measurable simple function
' such that
R
|f � '| dµ < ✏.
Proof. As f+ and f� are M+, we have measurable simple functions '+
and '� such that '+  f+, '�  f�, and
R
(f+ � '+) dµ < ✏2 ,
R
(f� �
'�) dµ < ✏2 . Let ' = '
+ � '�. Then we have ' is simple, and
Z
|f�'| dµ =
Z
|f+�'+�(f��'�)| dµ 
Z �
|f+ � '+|+ |(f� � '�)|
�
dµ < ✏.
5F. If f belongs to L, it does not follow that f2 belongs to L.
Proof. Let f = 1p
x
on [0, 1]. Then we have
R
[0,1] f d� =
R 1
0
1p
x
dx =
2
p
x|10 = 2. However, f2 =
1
x does not have a finite integral on [0, 1], so
f2 /2 L.
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5G. Suppose that f 2 L(X,F , µ), and that its indefinite integral is �(E) =R
E f dµ for E 2 F . Show that �(E) � 0 for all E 2 F if and only if
f(x) � 0 for almost all x 2 X. Moreover, �(E) = 0 for all E if and only
if f(x) = 0 for almost all x 2 X.
Proof. Write f = f+ � f�. Suppose �(E) >� 0 for all E 2 F . Let
✏ > 0, and put E✏ = {x 2 X | f(x) < �✏}. As f is measurable, E✏ 2 F .
Moreover,
R
E f dµ  �✏µ(E✏) � 0, and thus µ(E✏) = 0. But then {x 2
X | f(x) < 0} = [n2Z+E1/n has measure 0, and therefore f � 0 µ-almost
everywhere.
On the other hand, if f � 0 µ-almost everywhere, then for all E 2 F ,
f�E � 0 µ-almost everywhere, and thus by 5B,
R
f�E dµ � 0 for all
E 2 F .
For the second piece, we repeat the above argument with E✏ = {x 2
X | |f(x)| > ✏}.
5H. Suppose that f1, f2 2 L(X,F , µ), and let �1,�2 be their indefinite inte-
grals. Show that �1(E) = �2(E) for all E 2 F if and only if f1(x) = f2(x)
for almost all x 2 X.
Proof. Note that �1(E) = �2(E) for all E if and only if (�1 � �2)(E) = 0
for all E, if and only if f1 � f2 = 0 µ-almost everywhere (by problem
G).
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Powered by TCPDF (www.tcpdf.org)
 
Math 426: Homework 5 Solutions
Mary Radcli↵e
due 14 May 2014
In Bartle:
6H. Let X = Z+, and let � be the measure on X which has measure 1n2 at the
point n. Show that �(X) < 1. Let f be defined on X by f(n) =
p
n.
Show that f 2 Lp if and only if 1  p < 2.
Solution. We have �(X) =
P1
n=1 �(n) =
⇡2
6 , which is finite. Moreover,
Z
|f |p d� =
1X
n=1
|f(n)|p�(n)
=
1X
n=1
np/2n�2
=
1X
n=1
np/2�2,
which converges if and only if p/2� 2 < �1, if and only if p < 2. But as
p is assumed to be at least 1, since otherwise we do not have a norm, we
have that the integral is finite if and only if 1  p < 2.
6I. Modify the previous exercise to obtain a function on a finite measure space
which belongs to Lp if and only if 1  p < p0.
Solution. Let X = Z+, and let � be the measure on X with �(n) = n�p0 .
Note that as p0 > 1, we have that �(X) =
P1
n=1 n
�p0 < 1.
Take f(n) = n1�1/p0 . Then we have
Z
|f |p d� =
1X
n=1
|f(n)|p�(n)
=
1X
n=1
np�p/p0n�p0
=
1X
n=1
np((p0�1)/p0)�p0 ,
which converges if and only if p((p0 � 1)/p0) � p0 < �1, if and only if
p < (p0 � 1)(p0/(p0 � 1)) = p0, as desired.
6N. Let (X,F , µ) be a measure space, and let f belong to both Lp1 and Lp2 ,
with 1  p1 < p2 < 1. Prove that f 2 Lp for any value of p such that
p1  p  p2.
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Solution. Define E1 = {x 2 X | |f(x)| < 1} and E2 = {{x 2 X | |f(x)| �
1}. Note that if x 2 E1, then |f(x)|p  |f(x)|p1 , and if x 2 E2, then
|f(x)|p  |f(x)|p2 . Therefore we have
Z
|f |p dµ =
Z
E1
|f |p dµ+
Z
E2
|f |p dµ

Z
E1
|f |p1 dµ+
Z
E2
|f |p2 dµ

Z
|f |p1 dµ+
Z
|f |p2 dµ < 1
by hypothesis. Therefore, f 2 Lp.
6P. Let f 2 Lp(X,F , µ), 1  p < 1, and let ✏ > 0. Show that there exists
a set E✏ 2 F with µ(E✏) < 1 such that if F 2 F and F \ E✏ = ;, then
kf�F kp < ✏.
Solution. For n 2 Z+, define Gn = {x 2 X | |f(x)|p > 1n}. Then |f(x)|
p �
1
n�Gn , and thus 1 >
R
|f |p dµ �
R
1
n�Gn dµ =
1
nµ(Gn), so µ(Gn) < 1
for all n.
Now, |f |p�Gn is an increasing sequence of functions with limit |f |p, and
thus by the MCT, we have that
R
|f |p�Gn dµ !
R
|f |p dµ. Let N 2 Z+
such that
��R |f |p�GN dµ�
R
|f |p dµ
�� < ✏p. Then as
R
|f |pdµ =
R
|f |p�GN dµ+R
|f |p�X\GN dµ, we have that
R
|f |p�X\GN dµ < ✏p. Let GN = E✏.
Now, let F ⇢ F with F \ E✏ = ;, so F ⇢ X\GN . Then
kf�F kpp =
Z
|f |p�F dµ 
Z
|f |p�X\GN dµ < ✏
p,
and the desired result follows.
6Q. Let fn 2 Lp(X,F , µ), 1  p < 1, and let �n be defined for E 2 F by
�n(E) =
�R
E |fn|
p dµ
�1/p
. Show that |�n(E) � �m(E)|  kfn � fmkp.
Hence, if {fn} is a Cauchy sequence in Lp, then lim�n(E) exists for each
E 2 F .
Solution. Note that �n(E) = kfn�Ekp. Moreover, by the triangle inequal-
ity, we have that for any two functions f and g, kfkp  kf�gkp+kgkp, and
thus kfkp � kgkp  kf � gkp. Wolog, assume that kfn�Ekp > kfm�Ekp.
Then
|�n(E)� �m(E)| = |kfn�Ekp � kfm�Ekp|
= kfn�Ekp � kfm�Ekp
 k(fn � fm)�Ekp
 kfn � fmkp,
as desired.
6R. Let fn,�n be as in Exercise 6Q. If {fn} is a Cauchy sequence and ✏ > 0,
then there exists a set E✏ 2 F with µ(E✏) < 1 such that if F 2 F and
F \ E✏ = ;, then �n(F ) < ✏ for all n 2 Z+.
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Solution.By problem 6P, such a set exists for each fn; for all ✏ > 0, let
E(n)✏ be such that if F \ E(n)✏ = ;, then kfn�E(n)✏ kp < ✏.
Now, fix ✏ > 0. Let N be su�ciently large that if n,m � N , then kfn �
fmkp < ✏2 . Put E✏ =
N[
k=1
E(k)✏/2.
Let F 2 F with F \ E✏ = ;. If n  N , then F ⇢ Ec✏/2, so kfn�F kp =
�n(F ) < ✏/2. If n > N , then |�n(F ) � �N (F )|  kfn � fNkp < ✏2 , and
thus �n(F ) <
✏
2 + �N (F ) < ✏.
6T. If f 2 L1(X,F , µ), then |f(x)|  kfk1 for almost all x. Moreover,
if A < kfk1, then there exists a set E 2 F with µ(E) > 0 such that
|f(x)| > A for all x 2 E.
Solution. For all ✏ > 0, there exists a set N✏ such that µ(N✏) = 0 and
S(N✏) < kfk1 + ✏. Take N = [1k=1N1/k. Then µ(N) = 0 by countable
additivity, and if x /2 N , then x /2 N1/k for all k. Thus, |f(x)|  S(N1/k)
for all k, and thus |f(x)| < kfk1 + 1/k for all k, so |f(x)|  kfk1 for all
x /2 N . Therefore, |f(x)|  kfk1 for almost all x.
Suppose there exists some A < kfk1 such that there is no E 2 F with
µ(E) > 0 and |f(x)| > A for all x 2 E. Let N = {x 2 X | |f(x)| >
A}. Then by hypothesis, µ(N) = 0, and thus kfk1  S(N)  A, a
contradiction. Thus, the result holds.
Additional Exercises:
1. Let L(Rn,Rm) denote the space of linear transformations from Rn to Rm.
Recall that in 425, we defined the norm of a transformation as follows:
kAk = sup
kxk2=1
kAxk2 (see Baby Rudin, page 208). We proved that this
satisfies the properties of a norm in Theorem 9.7 (again, page 208.) This
is usually called the Euclidean norm, the `2 norm, or the Frobenius norm
for matrices, and is more properly denoted by kAk2.
(a) Let V be a vector space with norm k · k, and let L(V ) be the space
of linear transformations from V to itself. Define N : L(V ) ! R by
N(A) = sup
kvk=1
kAvk. Show that N defines a norm on V . (Usually
we denote this as N(A) = |||A|||). Thus we can define a p-norm for
operators as well.
(b) Show that if |||A||| is defined from a vector norm on V as in part (a),
then we have the inequality |||AB|||  |||A||||||B||| for all A,B 2 L(V ).
This property is usually called submultiplicativity.
(c) Let A be an n ⇥ n real matrix. Define |||A|||p (sometimes we write
just kAkp) to be the matrix norm derived from the Lp norm on Rn
as in part (a). This is usually called the p-norm on linear operators.
Determine an expression for |||A|||1 and |||A|||1 in terms of only A.
Solution. (a) Clearly, N(A) � 0 for all A 2 L(V ), and N(0) = 0. Now,
suppose that A 6= 0. Then there exists some x 2 V such that Ax 6= 0,
so sup
kvk=1
kAvk � kAxk/kxk > 0. Thus, N(A) = 0 if and only if A = 0.
Next, if ↵ 2 R, we have N(↵A) = sup
kvk=1
k↵Avk = sup
kvk=1
|↵|kAvk =
|↵|N(A), since k · k is a norm on V .
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Finally, let A,B 2 L(V ). Then we have
N(A+B) = sup
kvk=1
k(A+B)vk
 sup
kvk=1
(kAvk+ kBvk)
 sup
kvk=1
kAvk+ sup
kvk=1
kBvk = N(A) +N(B).
Therefore, N is a norm on L(V ).
(b) Consider |||AB||| = sup
kvk=1
k(AB)vk. Let y = Bv. Then we have
|||AB||| = sup
kvk=1
k(AB)vk
= sup
kvk=1
kAykkykkyk
= sup
kvk=1
kAyk
kyk kBvk
 sup
y2V
kAyk
kyk supkvk=1
kBvk
= sup
kyk=1
kAyk sup
kvk=1
kBvk = |||A||||||B|||.
(c) We first consider |||A|||1. Note that if v 2 Rn has kvk1 = 1, then we
have
P
|vi| = 1. Thus, if A1, A2, . . . , An are the columns of A, we
have kAvk = kv1A1+ · · ·+ vnAnk 
P
|vi|kAik1  max1in kAik1.
Moreover, if i attains the maximum, taking v = ei yields that kAeik =
max1in kAik1. Therefore, the 1-norm of A is the maximal 1-norm
of the columns of A.
On the other hand, if kvk1 = 1, then max |vi| = 1. Moreover, if
a1, a2, . . . , an are the rows of A, then Av = [a1 · v, a2 · v, . . . , an · v]T ,
so kAvk1 = max1in |ai · v|. If S =
Pn
j=1 |aij | is the maximal row
sum, then, it is clear that kAvk1  S. Moreover, taking v to be a
vector of all ±1 such that aijvj > 0 for all j, we have kAvk1 = S.
Therefore, |||A|||1 is the maximal 1-norm of the rows of A.
2. Let (X,F , µ) be a measure space.
(a) Prove that the simple functions with finite support are dense in
Lp(X) if 1  p < 1. That is, for all f 2 Lp and ✏ > 0, there
exists a simple function � with finite support and kf � �kp < ✏.
(b) Prove that the simple functions with finite support are dense in L1
if X is of finite measure. Give an example of an L1 function that
cannot be approximated by simple functions with finite support if
µ(X) = 1.
Solution. (a) Let f 2 Lp, ✏ > 0. Let En = {x 2 X | |f(x)| � 1/n}. Note
that as |f |p�En increases to |f |p, by the MCT there exists n such
that
R
|f |p dµ�
R
|f |p�En dµ <
�
✏
2
�p
. Let g = |f |�En , so g has finite
support and g 2 Lp.
By Theorem 2.11, there exists a simple function � such that |g��| <
✏
2(µ(En))1/p
for all x, and � = 0 whenever g = 0, so � has finite
support. Then
R
|g � �|p dµ 
R
En
⇣
✏
2(µ(En))1/p
⌘p
=
�
✏
2
�p
.
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Therefore, we have
kf � �kp =
✓Z
|f � �|p dµ
◆1/p
=
 Z
En
|f � �|p dµ+
Z
Ecn
Z
|f � �|p dµ
!1/p

✓Z
En
|g � �|p dµ+
⇣ ✏
2
⌘p◆1/p

⇣
2
⇣ ✏
2
⌘p⌘1/p
= 21/p
✏
2
< ✏,
as desired.
(b) The first part is immediate by applying Theorem 2.11 to f . For the
second part, if f(x) = 1, where f 2 L1(R), it is clear that if g is
any finitely supported function, then kf � gk1 � 1, so f cannot be
approximated by finitely supported simple functions.
3. Consider the real space under Lebesgue measure, (R,B,�). Recall that
C(R) is the family of continuous functions on R. Define Cc(R) to be the
family of continuous, compactly supported functions on R.
(a) Show that for 1  p < 1, Cc(R) is dense in Lp(R). (Hint: Since
we know the simple functions are dense by problem 2, it su�ces
to approximate simple functions by compactly supported continuous
functions.)
(b) Does this result hold in L1? Explain.
Solution. (a) First, we claim that if �(E) < 1, then for all ✏ > 0 there
exists a continuous function f✏ such that kf � �Ekp < ✏.
Let En = E \ [�n, n]. As En ⇢ En+1 and [En = E, we have that
�(En) ! �(E). Choose N su�ciently large that �(En) � �(E) ��
✏
4
�p
.
Let U be an open set such that En ⇢ U and �(U\En) 
�
✏
4
�p
. By
possibly intersecting with the open set (�n� �, n+ �), we can ensure
that U is contained in a compact set. Note, moreover, that as U
is open U can be written as an (at most) countable union of open
intervals. Write U = [Ik, where Ik = (ak, bk).
Define fk as follows. Fix some �k <
✏p(p+1)
2kp+2 . Then put
fk(x) =
8
>><
>>:
1 ak  x  bk
1� 1�k (x� bk) bk < x  bk + �k
1 + 1�k (x� ak) ak � �k  x < ak
0 else
.
Then we have fk is supported on (ak � �k, bk + �k), fk is continuous,
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and
k�Ik � fkkpp =
Z
|�Ik�fk |p d�
=
Z ak
ak��k
����1 +
1
�k
(x� ak)
����
p
dx+
Z bk
ak
0pdx+
Z bk+�k
bk
����1�
1
�k
(x� bk)
����
p
dx
=
Z 1
0
�ku
pdu+
Z 0
1
��kupdu
= 2
�k
p+ 1
< 2
✏p(p+1)
2kp+2
p+ 1
=
⇣ ✏
2k+1
⌘p
Let f =
P
k fk. Then f is continuous, and we have
kf � �Ukp = k
X
k
(f � �Ik)kp

X
k
kf � �Ikkp
<
X
k
✏
2k+1
= ✏/2
Therefore, we have
k�E � fkp  k�E � �Enkp + k�En � �Ukp + k�U � fkp
 µ(E\En)1/p + µ(U\En)1/p +
✏
2
 ✏
4
+
✏
4
+
✏
2
< ✏.
Now, let � be an arbitrary finitely supported simple function. Then
we may write � =
Pn
k=1 ak�Ek , where �(Ek) < 1 and ak 6= 0 for all
k. Then put fk to be a continuous function such that k�Ek � fkkp 
✏
|ak| . Taking f =
P
akfk yields a continuous function satisfying the
desired property by the triangle inequality.
(b) No. We have the same problem with compact (aka finite) support
here as in 2b.
Extras:
6C. Let N be a norm on a linear space V and let d be defined for u, v 2 V by
d(u, v) = N(u� v). Show that d is a metric onV .
Solution. Clearly d(u, v) � 0 for all u, v, and d(u, v) = 0 i↵ u = v. Also,
d(v, u) = N(v�u) = |�1|N(u�v) = N(u�v) for all u, v, and for all u, v, w,
we have d(u,w) = N(u�w) = N(u� v + v �w)  N(u� v) +N(v �w)
by the triangle inequality. Thus, d is a metric.
6D. Whoops, this is a part of additional exercise 2.
6E. If f 2 Lp, 1  p < 1, and if E = {x 2 X | |f(x)| 6= 0}, then E is �-finite.
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Solution. Define En = {x 2 X | |f(x)|p � 1n}. Notice then that [En = E,
and as f 2 Lp, we have 1 >
R
|f |p dµ �
R
En
|f |p dµ � 1nµ(En), and thus
µ(En) < 1 for all n. Therefore, E is �-finite.
6F. If f 2 Lp and if En = {x 2 X | |f(x)| � n}, then µ(En) ! 0 as n ! 1.
Solution. Note that En ⇢ En+1, so µ(En)  µ(En+1). By the same logic
as problem 6E, it is clear that µ(En) is finite, and thus {µ(En)} is a
bounded monotonic sequence in R, and it thus has a limit. Suppose the
limit is not 0, so that µ(En) ! C > 0. Then for all n, we have
R
|f |p dµ �R
En
|f |p dµ � Cnp, and thus the integral is infinite, a contradiction.
6K. If (X,F , µ) is a finite measure space and f 2 Lp, then f 2 Lr for 1  r  p.
Apply Hölder’s Inequality to |f |r in Lp/r and g = 1 to obtain the inequality
kfkr  kfkpµ(X)s, where s = 1/r � 1/p.
Solution. Let E1 = {x 2 X | |f(x)| < 1}. Then if x 2 Ec1, |f(x)|r 
|f(x)|p. Therefore,
Z
|f(x)|r dµ 
Z
E1
|f(x)|r dµ+
Z
Ec1
|f(x)|r

Z
E1
1 dµ+
Z
Ec1
|f(x)|p dµ
< µ(E1) + kfkpp < 1.
Thus, f 2 Lr.
Now, since |f |r 2 Lp/r and 1 2 Lp/(p�r), which are conjugate indices,
Hölder’s Inequality yields
k|f |rk1  k|f |rkp/rk1kp/(p�r).
Note that for any x, y, we have k|f |xky = kfkxxy, and thus we may rewrite
the above inequality as
kfkrr  kfkrpµ(X)(p�r)/p.
Taking 1/r power on both sides yields the given inequality.
6L. Suppose that X = Z+ and µ is the counting measure on Z+. If f 2 Lp,
then f 2 Ls for 1  p  s < 1, and kfks  kfkp.
Solution. If f 2 Lp, then we have
P
|f(n)|p < 1. Then by comparison
test, for n su�ciently large |f(n)| < 1, and we have |f(n)|s < |f(n)|p, and
thus f 2 Ls and kfks =
P
|f(n)|s 
P
|f(n)|p = kfkp.
6U. If f 2 Lp, 1  p  1, and g 2 L1, then the product fg 2 Lp and
kfgkp  kfkpkgk1.
Solution. Let M = kgk1. If there exists N 2 F with µ(N) = 0 and
g(x) > M on N , replace g with a function h that takes value 0 on N
and agrees with g elsewhere, so g = h a.e. and h is bounded. ThenR
|fg|p dµ =
R
|fh|p dµ 
R
Mp|f |p dµ. The result follows.
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Powered by TCPDF (www.tcpdf.org)
 
Math 426: Homework 6
Mary Radcli↵e
due 21 May 2014
In Bartle:
7G. If a sequence {fn} converges in measure to a function f , then every sub-
sequence of {fn} converges in measure to f . More generally, if {fn} is
Cauchy in measure, then every subsequence of {fn} is Cauchy in measure.
Proof. Suppose {fn} converges in measure to f . Let {fnk} be a sub-
sequence of fn. Fix ↵, ✏, and let N be such that if n > N , then
µ({x 2 X | |f(x)� fn(x)| > ↵}) < ✏. Fix K such that nK > N . Then for
all k > K, nk > N , and thus µ({x 2 X | |f(x) � fnk(x)| > ↵}) < ✏, and
thus fnk converges in measure to f .
The proof for Cauchy in measure is similar.
7I. If {fn} is a sequence of characteristic functions of sets in F , and {fn}
converges to f in Lp, show that f is (almost everywhere equal to) the
characteristic function of a set in F .
Proof. Note that fn has a subsequence fnk that converges almost every-
where to f . Since fnk are measurable, they have a measurable limit al-
most everywhere, and thus f is almost everywhere equal to a measurable
function. But any {0, 1}-valued measurable function is the characteristic
function of a set in F , yielding the result.
7Q. Let (X,F , µ) be a finite measure space. If f is a F-measurable function,
define r(f) =
R |f |
1+|f | dµ. Show that fn
µ�! f if and only if r(fn � f) �! 0.
Proof. Suppose that fn
µ�! f . Fix ↵, and define En = {x 2 X | |fn(x) �
f(x)| � ↵}. Note that if x /2 En, then |fn(x)�f(x)|1+|fn(x)�f(x)| < ↵. Thus we have
r(fn � f) =
Z |fn(x)� f(x)|
1 + |fn(x)� f(x)|
dµ
=
Z
En
|fn(x)� f(x)|
1 + |fn(x)� f(x)|
dµ+
Z
Ecn
|fn(x)� f(x)|
1 + |fn(x)� f(x)|
dµ

Z
En
1 dµ+
Z
Ecn
↵ dµ
= µ(En) + ↵(µ(X)� µ(En))
= (1� ↵)µ(En) + ↵µ(X),
and the inequality holds for all ↵, n. Now, for given ✏, choose ↵ < ✏2µ(X) ,
and N such that for n � N , we have µ(En) < ✏2(1�↵) . Then r(fn � f) < ✏
for all n � N . Therefore, r(fn � f) �! 0.
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On the other hand, suppose r(fn�f) �! 0. Note that the function g(x) =
x
1+x is strictly increasing on (0,1) (you can examine the derivative, for
example), and thus we have, for ↵, En as above,
r(fn � f) =
Z |fn(x)� f(x)|
1 + |fn(x)� f(x)|
dµ
�
Z
En
|fn(x)� f(x)|
1 + |fn(x)� f(x)|
dµ
�
Z
En
↵
1 + ↵
dµ
=
↵
1 + ↵
µ(En).
As r(fn � f) �! 0, we thus have that µ(En) �! 0 for any ↵, and therefore
fn
µ�! f .
7R. If the sequence {fn} of measurable functions converges almost everywhere
to a measurable function f , and � : R �! R is continuous, then the se-
quence ��fn converges almost everywhere to ��f . Conversely, if � is not
continuous at every point, then there exists a sequence {fn} that converges
almost everywhere to f but � � fn does not converge almost everywhere
to � � f .
Proof. For the first part, let x 2 X such that fn(x) �! f(x). Then since
� is continuous, we have lim
n�!1�(fn(x)) = �( limn�!1 fn(x)) = �(f(x)), and
therefore ��fn converges everywhere that fn does. Thus ��fn
a.e.��! ��f .
On the other hand, suppose � has a discontinuity at a. Let {an} be a
sequence with an �! a but �(an) 6�! �(a). Let fn(x) = an, the constant
function. Then fn converges everywhere to f(x) = a, but � � fn(x) =
�(an) 6�! �(a) = � � f(x) at every point x.
7V. Let (X,F , µ) be an arbitrary measure space and let � : R �! R be contin-
uous and satisfy
(⇤⇤) 9K � 0 such that |�(t)|  K|t| 8t 2 R.
If f 2 Lp, then � � f 2 Lp. Conversely, if � does not satisfy (⇤⇤), then
there exists a measure space and a function f 2 Lp such that � � f is not
in Lp.
Proof. For the first part, suppose that f 2 Lp and � satisfies (⇤⇤). Then
we have
Z
|�(f(x))|p dµ(x) 
Z
|Kf(x)|p dµ(x)
= Kp
Z
|f |p dµ < 1,
so � � f 2 Lp.
On the other hand, suppose � does not satisfy (⇤⇤). For each n 2 Z+,
let tn be a point such that �(tn) > n|tn|. Clearly we may choose these
points to be distinct, as � is continuous. Moreover, we may choose these
so that {tn | n 2 Z+} does not contain any accumulation points. Thus we
may define �1, �2, . . . inductively so that B�k(tk) is disjoint from B�j (tj)
for all j 6= k, and �(x) > n|x| for all x 2 B�n(tn) (this is possible due to
continuity of �).
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Define a measure µk on R by µk(E) = 12�k �(E \ B�k(tk)). That is, µk
is the measure on B�k(tk) obtained by scaling Lebesgue measure by
1
2�k
,
so that µk(B�k(tk)) = 1. As positive scaling of a measure produces a
measure, this is in fact a measure on R. Moreover, as the sum of measures
is also a measure, we can produce a measure on R by µ(E) =
1X
k=1
µk(E).
Let f(x) = 1
n
1
p
+1
2
if x 2 B�n(tn) and 0 elsewhere. Then we have
Z
|f |p dµ =
1X
n=1
✓
1
n
1
p+
1
2
◆p
µ(B�n(tn))
=
1X
n=1
1
n1+
p
2
< 1,
so f 2 Lp(R,B, µ). However,
Z
|� � f |p dµ �
1X
n=1
✓
n
1
n
1
p+
1
2
◆p
µ(B�n(tn))
=
1X
n=1
1
n1�
p
2
= 1,
so � � f 62 Lp(R,B, µ).
7W. If fn
Lp��! f on an arbitrary measure space, and � is continuous and satisfies
(⇤⇤), then � � fn
Lp��! � � f . Conversely, if � does not satisfy (⇤⇤), then
there exists a measurespace and a sequence fn which converges in Lp to
f but � � fn does not converge in Lp to � � f .
Proof. We here use Vitali’s Convergence Theorem. As fn
µ�! f , it is clear
that � � fn
µ�! � � f . Moreover, by the argument in 7V, for any g 2 Lp,
k� � gkp  Kkgkp.
For each ✏, let E✏ be chosen so that µ(E✏) < 1 and
R
Ec✏
|fn|p dµ < ✏p for
all n. Let F✏ = E ✏K . Then we have
R
F c✏
|� � fn|p dµ < Kp
R
Ec✏
K
|fn|p dµ <
Kp(✏/K)p = ✏p.
For each ✏, let �(✏) be such that if E 2 F and µ(E) < �(✏), thenR
E |fn|
p dµ < ✏p for all n. Let ⌘(✏) = �( ✏K ). Then as above, if E 2 F with
µ(E) < ⌘(✏), then
R
E |� � fn|
p dµ < Kp(✏/K)p = ✏p for all n.
Thus, by Vitali’s Convergence Theorem, the result holds.
For the second part, let µ, f be as in Problem 7V. Define fn(x) =
1
n
1
p
+1
2
if x 2 B�n(tn) and 0 elsewhere. Let gn = f1 + f2 + · · ·+ fn. Then by the
argument above, fn
Lp��! f , but � � f 62 Lp and thus � � fn 6
Lp��! � � f .
Additional Exercises:
In the following exercises, I refer to the “five modes of convergence” as almost
everywhere, uniformly almost everywhere, almost uniformly, in measure, and in
Lp. When I say “prove such and so in each of the five modes” it must be done
for all of them.
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1. Let fn, gn be sequences of measurable functions. Prove that convergence
is linear in each of the five modes of convergence. That is to say, if fn �! f
and gn �! g in one of the five modes, then fn + gn �! f + g in that same
mode.
• Almost Everywhere: Suppose fn �! f and gn �! g almost ev-
erywhere. Let N1, N2 be null sets such that fn �! f on X\N1 and
gn �! g on X\N2. Let x 2 X\(N1 [ N2). Then fn(x) �! f(x) and
gn(x) �! g(x), and thus fn(x) + gn(x) �! f(x) + g(x) (linearity of
convergence for sequences of numbers). Therefore, fn + gn �! f + g
on X\(N1 [ N2). Moreover, µ(N1 [ N2)  µ(N1) + µ(N2) = 0, so
fn + gn �! f + g a.e..
• Uniformly Almost Everywhere: Let N1 be a null set such that
for all ✏ > 0, there exists M✏ 2 Z+ such that if n � M , then |fn(x)�
f(x)| < ✏ for all x 2 X\N1, and likewise N2 a null set such that for all
✏ > 0 there exists P✏ 2 Z+ such that if n � P✏, then |gn(x)�g(x)| < ✏
for all x 2 X\N2.
Fix ✏ > 0, and let N✏ = max{M✏/2, P✏/2}. Then for all x 2 X\(N1 [
N2), we have |fn(x)+gn(x)�f(x)�g(x)|  |fn(x)�f(x)|+ |gn(x)�
g(x)| < ✏, and thus fn + gn �! f + g u.a.e..
• Almost Uniformly: For each � > 0, let E�, F� be measurable sets
such that fn converges uniformly on X\E� and gn converges uni-
formly on X\F�, and µ(E�), µ(F�) < �. Define G� = E�/2 [ F�/2.
Then µ(G�) < �, and as both fn, gn converge uniformly on X\G�,
so too does there sum (by the same argument as above). Thus,
fn + gn �! f + g a.u..
• In Measure: Fix ↵ > 0. Note that if |fn(x)+gn(x)�f(x)�g(x)| >
↵, since |fn(x)+gn(x)�f(x)�g(x)|  |fn(x)�f(x)|+ |gn(x)�g(x)|,
we must have at least one of |fn(x)� f(x)| or |gn(x)� g(x)| is itself
greater than ↵/2. Therefore, we have {x 2 X ||fn(x)+gn(x)�f(x)�
g(x)| > ↵} ✓ {x 2 X |fn(x)�f(x)| > ↵/2}[{x 2 X |gn(x)�g(x)| >
↵/2}. Since both fn, gn converge in measure, we thus obtain that
µ({x 2 X ||fn(x)+gn(x)�f(x)�g(x)| > ↵}) �! 0, so fn+gn �! f+g
in measure.
• In Lp: Note that kfn + gn � f � gkp  kfn � fkp + kgn � gkp �! 0,
so fn + gn �! f + g in Lp.
2. (Squeeze test.) If fn �! 0 in one of the five modes, and |gn|  |fn| almost
everywhere for all n, then gn �! 0 in the same mode.
• Almost Everywhere: Let N1 be a null set such that fn(x) �! 0 for
all x /2 N1. Let N2 be a null set such that |gn(x)|  |fn(x)| for all
x /2 N2 and n 2 Z+. Let x /2 (N1 [N2). Then |gn(x)|  |fn(x)| �! 0,
so by the squeeze test for numbers, gn(x) �! 0. Thus gn �! 0 for all
x /2 N1 [N2, a null set.
• Uniformly Almost Everywhere: Let N1 be a null set such that
for all ✏ > 0 there exists N✏ so that for all n � N✏, |fn(x)| < ✏ for
all x /2 N1. Let N2 be as in part (a). Then for all x /2 N1 [N2, and
n � N✏, we have |gn(x)|  |fn(x)| < ✏. so gn �! 0 u.a.e..
• Almost Uniformly: Let E✏ be a set with µ(E✏) < ✏, and fn �! 0
uniformly on X\E✏. Then likewise gn �! 0 uniformly on X\E✏ (by
the same argument as above), so gn �! 0 a.u..
• In Measure: LetN2 be as in part (a). Note that {x 2 X\N2 | |gn(x)| >
↵} ✓ {x 2 X\N2 | |fn(x)| > ↵}, and thus µ({x 2 X | |gn(x)| > ↵}) 
µ(N2) + µ({x 2 X\N2 | |fn(x)| > ↵}) �! 0, so gn
µ�! 0.
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• In Lp: Note that
R
|gn|p dµ 
R
|fn|p dµ �! 0, and thus kgnkp �! 0
and gn
Lp��! 0.
3. (Lusin’s Theorem) Let f 2 L1([a, b]) under Lebesgue measure, and let
� > 0. Prove that there exists a Lebesgue measurable set S ⇢ [a, b] such
that �(S) < � and f is continuous on [a, b]\S.
Proof. Let fn be a sequence of continuous functions on [a, b] such that
fn
L1��! f . Then there exists a subsequence gk of fn that converges almost
everywhere to f . But then by Egoro↵’s Theorem, gk converges almost
uniformly to f , so for all � > 0 there exists a set S ⇢ [a, b] such that
�(S) < � and gk �! f uniformly on [a, b]\S. But the uniform limit of
continuous functions is always continuous, and thus f is continuous on
[a, b]\S.
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Powered by TCPDF (www.tcpdf.org)
 
Name:
Math 426/576: Midterm Solutions
2 May 2014
Turn o↵ and put away your cell phone.
No notes or books are permitted during this exam.
No calculators or any other devices are permitted during this exam.
Read each question carefully, answer each question completely, and show all of your work.
Write solutions clearly and legibly; credit will not be given for illegible solutions.
If any question is not clear, ask for clarification.
The following notation will be used throughout, unless otherwise specified.
• (X,F , µ) is a measure space.
• � denotes Lebesgue measure on (R,L), where L denotes the �-field of Lebesgue mea-
surable functions.
• All functions take values in R.
• L = L(X,F , µ) denotes the set of Lebesgue integrable functions on X.
# Points Score
1 12
2 11
3 12
4 10
5 10
⌃ 55
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1. Complete each of the following definitions. (3 points each)
(a) Let (X,F) be a measurable space. A measure on F is
Solution: Look this up.
Comments: A measure takes values in R, not just in R. I didn’t take o↵ points
for this.
(b) Let (X,F , µ) be a measure space, and f, g : X ! Y be functions. Then f = g
µ-almost everywhere if
Solution: Look this up.
Comments: Many people put {x 2 X | f(x) 6= g(x)} has measure 0. This is nearly
true, except that this set may not be measurable, so in reality the condition is
that this set is contained in a set of measure 0.
(c) Let f : P(X) ! R be a function. Then f is countably subadditive if
Solution: Look this up.
(d) A measure space (X,F , µ) is said to be complete if
Solution: Look this up.
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2. (a) (6 points) State and prove Fatou’s Lemma.
(b) (5 points) Does there exist a version of Fatou’s Lemma in which lim inf is replaced
with lim sup? If so, state and prove it. If not, explain where/why the proof breaks
down and provide a counterexample.
(a) Solution: Fatou’s Lemma: Let fn 2 M+ for all n 2 Z+. Then
R
lim inf fn dµ 
lim inf
R
fn dµ.
Proof: For m 2 Z+, put gm = inf{fm, fm+1, . . . }. Then gm is an increasing
sequence with lim gm = limm!1 infn�m fn = lim inf fn. Thus, by the monotone
convergence theorem, lim
R
gm dµ =
R
lim gm dµ =
R
lim inf fn dµ.
On the other hand, gm  fn for all n � m, and thus
R
gm dµ 
R
fn dµ for
all n � m, andthus
R
gm dµ  infn�m
R
fn dµ. Therefore, limm!1
R
gm dµ 
limm!1 infn�m
R
fn dµ = lim inf
R
fn dµ.
Thus,
R
lim inf fn dµ = lim
R
gm dµ  lim inf
R
fn dµ.
Comments: A lot of people assumed in the statement of the theorem that fn !
f . This is not part of Fatou’s Lemma. More egregious, some assumed that fn
increased to f , which makes the theorem irrelevant since we can just apply MCT
to the sequence
R
fn dµ directly.
Also, a handful of people missed nonnegativity of the fn here. This is critical.
(b) Solution: Mostly, the answer should be no. The generalization you probably
would want is
R
lim sup fn dµ � lim sup
R
fn dµ. This is false, because in the
proof you would get some gm that are decreasing to lim sup fn, and we don’t
have a decreasing MCT. A counterexample would be our favorite functions, fn =
n�[1,1/n]. You can check that these fail the inequality above.
Some people had some interesting variants in which this can be made to be true.
For example, you can make it be true if you demand that all the fn are uniformly
bounded by a Lebesgue integrable function. Then you can do a trick similar to
what we do in the proof of the DCT. Some people tried to say this if the fn
are simply uniformly bounded, but this doesn’t work (for the same reason as the
decreasing MCT doesn’t work in this case unless your measure is finite.)
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3. Determine if each of the following is true or false. If true, provide a short proof. If
false, provide an explicit counterexample. (4 points each)
(a) If F1 � F2 � F3 � . . . is a decreasing sequence in F , then µ(\Fn) = limµ(Fn).
Solution: False!
Favorite counterexample: Fn = [n,1). Then �(Fn) = 1 for all n but \Fn = ;,
so the limit fails.
In order that this is true, you must have µ(Fn) < 1 for some n.
Comments: Please be careful of arithmetic with potentially infinite numbers.
(b) If f : R ! R is di↵erentiable, then f 0 is Lebesgue measurable.
Solution: True!
Let fn(t) =
f(t+ 1n )�f(t)
1/n . Note that since f is di↵erentiable, it is therefore con-
tinuous and therefore measurable. Thus, fn is a measurable function for all n.
Moreover, fn ! f 0 (pointwise), and thus f 0 can be obtained as the limit of mea-
surable functions, and it is therefore measurable.
Comments: A lot of people said that since f is di↵erentiable, then f 0 is integrable,
which is NOT true. Recall our definition of integrability requires that the integral
be finite, so even the simplest examples (like f(x) = x2 or f(x) = x) fails this
test.
Some people declared that derivatives are always continuous. No. No no no.
A few people almost got the right answer, using something like “let tn ! t, and
define fn(t) =
f(tn)�f(t)
tn�t , then fn ! f
0 and is measurable.” This is close to correct,
except that you need to be careful that your fn converge to f 0 for ALL choices of
t. This is not quite right, since you seem to have selected a special t at the outset.
(c) If f, g 2 L then fg 2 L.
Solution: False!
A common counterexample was f = 1px on [0, 1], and g = f . You can verify that
this is integrable but fg = f 2 = 1x on [0, 1] is not.
Another nice counterexample is a function f taking value n on a sequence of
disjoint intervals of length 1n3 .
Comments: Many people claimed that if f, g are nonnegative and
R
f dµ,
R
g dµ
are both finite, then
R
fg dµ is also finite. This is simply false, as the above
examples show. Intuitively, think if these were sums. Just because
P
an andP
bn are finite, that doesn’t mean that
P
anbn is.
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4. (10 points) For t > 0, determine
R 1
0
xt�1
log x dx as a function of only t. Prove that your
answer is correct with calculus and/or convergence theorems.
Solution: Let F (t) =
R 1
0
xt�1
log x dx. First, consider
@
@t
✓
xt � 1
log x
◆
=
xt log x
log x
= xt  1
for all x 2 [0, 1]. Since 1 is integrable on [0, 1], we may apply Theorem 3 from the
formula sheet to obtain
F 0(t) =
Z 1
0
@
@t
✓
xt � 1
log x
◆
dx =
Z 1
0
xtdx =
1
t+ 1
.
The fundamental theorem of calculus thus implies that F (t) = log(1 + t) + C, where
C is a constant.
To determine C, we shall take a limit as t ! 0+. Note that the right hand side has
limt!0+(log(1 + t) + C) = C. For the left hand side, note that if a < 0, we have
|ea � 1|  |a| (check a Taylor expansion for ea), and thus
���xt�1log x
��� =
��� et log x�1t log x
��� t  t, and
thus we have
0  lim
t!0+
F (t) = lim
t!0+
Z 1
0
xt � 1
log x
dx  lim
t!0+
Z 1
0
t dx = 0.
(Note: I have dropped the absolute value bars since x
t�1
log x is positive on x 2 [0, 1] for
all t > 0.)
Therefore, C = 0, and thus F (t) = log(1 + t).
Comments: This problem was pretty rough. Please verify you remember how to inte-
grate and di↵erentiate.
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5. (10 points) Suppose A 2 L. Prove that �(A) = sup{�(K) | K ⇢ A, K is compact}.
Solution: Note that if K ⇢ A, �(K)  �(A), so we have �(A) � sup{�(K) | K ⇢
A, K is compact} automatically. For the opposite inequality, we consider two cases,
according to if �(A) < 1 or �(A) = 1.
First, suppose �(A) < 1. For each n 2 Z+, put An = A \ [�n, n]. Then An 2 L, and
An ⇢ An+1, and [An = A. Thus we have that limn!1 �(An) = �(A).
Let ✏ > 0, and choose N 2 Z+ such that for all n � N , �(An) � �(A) � ✏2 . Recall
that there exists a closed set C ⇢ An such that �(An\C) < ✏2 , since An is Lebesgue
measurable. Note that C ⇢ [�n, n], so C is closed and bounded, and is thus compact.
Moreover, C ⇢ A and �(C) � �(A)�✏. Therefore, for all ✏ > 0, there exists a compact
set K with �(K) � �(A)� ✏, and thus sup{�(K) | K ⇢ A, K is compact} � �(A).
For the second case, suppose that �(A) = 1. For n 2 Z+, put Bn = {x 2 A | n� 1 
|x| < n}, so that the Bn are disjoint and [Bn = A, and each Bn is measurable.
As [Bn = A, we have �(A) =
P
�(Bn). Let ✏ > 0, M 2 R, and choose N 2 Z+
such that 2N �
PN
n=1 �(Bn) > M + ✏. Choose a closed set C ⇢ [
N
n=1Bn such that
�
��
[Nn=1Bn
�
\C
�
< ✏. Then C ⇢ [�N,N ], so C is compact, and �(C) � M . Therefore,
we have that for all M 2 R, there exists a compact set K ⇢ A such that �(K) � M ,
and therefore, sup{�(K) | K ⇢ A, K is compact} = 1 = �(A).
Comments: A lot of people identified that the right theorem to use here was that any
Lebesgue measurable set can be approximated by closed sets, so that we already have
�(A) = sup{C ⇢ A | C is closed}. However, from there, a lot of very interesting things
happened.
• It is NOT true that these sets are automatically bounded, just closed.
• It is NOT true that every Borel set is a countable union of intervals (closed or
open). You must have intersections as well. An example to see this is true: the
Cantor set is not a countable union of intervals.
• It is NOT true that if �(A) < 1, then A is bounded.
https://www.coursehero.com/file/9770582/Midterm-Exam-Spring-2014-Solutions/
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