Math 516
Professor Lieberman
April 24, 2009
HOMEWORK #7 SOLUTIONS
Chapter 11
29. Let A, B be a Hahn decomposition for ν, and let E be a measurable set. Then we
have
νE = ν(E ∩ A) + ν(E ∩ B).
Because A is a positive set, we have ν(E ∩ A) ≥ 0, so νE ≥ ν(E ∩ B). But the
definition of ν − says that ν − E = ν(E ∩ B), and hence νE ≥ ν − E. Similarly, B is a
negative set, so ν(E ∩ B) ≤ 0, and ν + E = ν(E ∩ A), so νE ≤ ν + E.
33. (a) Since X is σ-finite, there is a sequence (Xi ) of disjoint measurable sets with
µXi < ∞ for all i and X = ∪Xi . Then, for each i, the finite-measure version of
the Radon-Nikodym theorem gives a nonnegative integrable function fi on Xi
such that
Z
νE =
fi dµ
E
for any measurable subset E of Xi . We now define f on X by f (x) = fi (x) if
x ∈ Xi . If E is a measurable subset of X, we set Ei = E ∩ Xi , so
Z
νEi =
fi dµ,
Ei
and E = ∪Ei . Since ν is a measure, it follows that
Z
X
XZ
XZ
νE =
νEi =
fi dµ =
f dµ =
f dµ.
Ei
Ei
E
Hence f is the desired function.
(b) Suppose f and g are two nonnegative functions on a σ-finite measure space
(X, B, µ) such that
Z
Z
f dµ =
E
g dµ
E
for all measurable sets E. Then there
S is a sequence (Xi ) of disjoint measurable
sets of finite measure with X =
Xi . Fix numbers N > M > 0 and let
E = {x : f (x) > N, g(x) < M } and Ei = E ∩ Xi . Then
Z
Z
f dµ ≥ N µEi ,
g dµ ≤ M µEi .
Ei
Ei
Since µEi < ∞ and N > M , it follows that µEi = 0 for all i, and hence µE = 0.
We now note that
[
{x : f (x) > g(x)} =
{x : f (x) > N, g(x) < M },
N >M >0 are in Q
1
2
so µ({x : f (x) > g(x)} = 0. Similarly, µ({x : f (x) < g(x)} = 0, so f = g a.e.
h i
dν
. First, suppose f is a nonnegative,
34. (a) For simplicity of notation, we set g = dµ
P
simple function: f =
ai χEi . Then
Z
X
X Z
f dν =
ai ν(Ei ) =
ai
g dµ
Ei
Z X
X Z
ai χEi g dµ
=
ai χEi g dµ =
Z · ¸
dν
= f
dµ.
dµ
If f is a nonnegative, measurable function, then it is the limit of an increasing
sequence (fn ) of nonnegative simple functions, hence
Z
Z
f dν = lim fn dν
by the Monotone Convergence Theorem. In addition, (fn g) is an increasing
sequence of measurable functions which converges to f g, so another application
of the Monotone Convergence Theorem says that
Z
Z
lim fn g dµ = f g dµ.
R
R
Since fn dν = fn g dµ, we conclude that
Z
Z
f dν = f g dµ
in this case, as well.h i
i
(b) Now we write gi = dν
for i = 1, 2. Then, for any measurable set E, we have
dµ
Z
Z
(ν1 + ν2 )(E) = ν1 (E) + ν2 (E) =
g1 dµ +
g2 dµ
E
E
Z
= (g1 + g2 ) dµ.
E
h
i
Since the Radon-Nikodym derivative d(ν1dµ+ν2 ) is unique, it follows that
·
¸ ·
¸ ·
¸
d(ν1 + ν2 )
dν1
dν2
=
+
.
dµ
dµ
dµ
(c) Now we set
·
¸
· ¸
dν
dµ
f=
, g=
.
dµ
dλ
Let E be a measurable set. From the definition of the Radon-Nikodym derivative, we have that
Z
ν(E) =
f dµ,
E
3
and part (a) implies that
Z
Z
f dµ =
f g dλ.
E
E
Hence
Z
ν(E) =
f g dλ
E
for any measurable set E. Again the uniqueness of the Radon-Nikodym derivative implies that
· ¸
dν
fg =
.
dλ
(d) Since
· ¸
dµ
,
1=
dµ
we infer from part (c) that
· ¸· ¸
dµ dν
1=
.
dν
dµ
and hence
· ¸ · ¸−1
dν
dµ
=
.
dµ
dν
39. First, we have ν ¿ µ because, if µ(E) = 0, then E = ∅, so ν(E) = 0. For any point
x ∈ X, we have that ν({x}) = 0 while, for any function f , we have
Z
f dµ = f (x).
{x}
Hence any function f which satisfies
Z
ν(E) =
f dµ
E
for all E ∈ B must be identically zero. But this choice for f only works if ν(E) = 0
for all E ∈ B. Since [0, 1] ∈ B and ν([0, 1]) = 1, it follows that there is no such
function f .
48. First, if E ∈ B, then for every horizontal or vertical line L, L∩E or L\E is countable,
so L \ Ẽ = L ∩ E or L ∩ Ẽ = L \ E must be countable. Hence Ẽ ∈ B. Next, if
(Ei ) is a sequence in B, let L be a horizontal or vertical line and set E = ∪Ei . If
L ∩ Ei is countable for all i, then L ∩ E is a countable union of countable sets so it is
countable. On the other hand, if L \ Ei is countable for some i, then L \ E ⊂ L \ Ei ,
so L \ E is countable. Hence E ∈ B. Therefore B is a σ-algebra.
To see that µ is a measure, we just have to check countable additivity. Let (Ei ) be
a countable collection of disjoint subsets of X and set E = ∪Ei . Next, write Ki for
the collection of horizontal or vertical lines L such that L \ Ei is countable, and write
K for the collection of horizontal or vertical lines L such that L \ E is countable. I
claim that K = ∪Ki and that Ki ∩ Kj = ∅ if i 6= j. Since L \ E = ∩(L \ Ei ), it
4
follows that, if L \ E is countable, then L \ Ei is countable for some i, so K ⊂ ∪Ki .
Conversely, if L\Ei is countable for some i, then L\E ⊂ L\Ei , so L\E is countable.
Hence ∪Ki ⊂ K, so K = ∪Ki . Next, let L ∈ Ki . Then L \ Ei is countable. If L \ Ej
is countable (for this same line L), then L \ (Ei ∩ Ej ) = (L \ Ei ) ∪ (L \ Ej ) is also
countable, so L ∩ (Ei ∩ Ej ) is non-empty. Since Ei ∩ Ej = ∅ for i 6= j, it follows that
i = j in this case. Hence L ∈ Ki for some i implies that L ∈
/ Kj for i 6= j. Hence, the
number of elements in K is the sum of the number of elements in each Ki . Hence µ
is a measure. A similar argument (but just using horizontal lines) shows that ν is a
measure.
Next, if f ∈ L1 (µ), then
¯ Z
¯Z
Z
¯
¯
¯ f dν ¯ ≤ |f | dν ≤ |f | dµ,
¯
¯
so F is a bounded linear functional on L1 (µ).
R Now, suppose that1 there is a locally measurable function g such that F (f ) =
f g dµ for all f ∈ L (µ). Let H be a horizontal line. Then, for any vertical line L,
L \ H is uncountable. In addition, if H 0 is a horizontal line H 0 \ H is equal to H 0
and hence uncountable if H 0 6= H while H 0 \ H = ∅ and hence countable if H 0 = H.
Therefore µH = 1. We now set Z = {hx, yi : g(x, y) = 0}. Then, for any horizontal
line H, Z ∩ H is measurable, so either H ∩ Z or H \ Z must be countable. If H \ Z
were countable, then we would have 1 = ν(H ∩ Z) because ν(H ∩ Z) ≤ µH = 1 and
H is a horizontal line such that H \ (H ∩ Z) = H \ Z is countable. On the other
hand,
Z
ν(H ∩ Z) =
g dµ = 0
H∩Z
because g = 0 on H ∩ Z. It follows that H ∩ Z is countable for any horizontal line
H.
For a vertical line V , we have µV = 1 just as before, so Z ∩ V is measurable and
hence either V ∩ Z or V \ Z must be countable. Now 0 = ν(V \ Z) because, for any
vertical line H \ (V \ Z) is uncountable, and
Z
ν(V \ Z) =
g dµ
V \Z
which implies that mu(V \ Z) = 0 because g 6= 0 on V \ Z. Hence V \ Z is countable
for any vertical line V .
We now use a definition that was very convenient in Chapter 12. For each x ∈ A,
we write Zx = {hx, bi : b ∈ B} and, for each y ∈ B, we write Z y = {ha, yi : a ∈ A}.
Then we consider the two possibilities on the comparison of the sizes of A and B. If
A has fewer elements than B, we observe that
\
Z=
Zy,
y∈B
Since V \ Z is countable for each vertical line and Z y = V ∩ Z for some vertical line
V , it follows that Z y is non-empty for each y ∈ B and hence Z has no fewer elements
5
than B. On the other hand,
Z=
\
Zx ,
x∈A
and Zx = Z ∩ H for the vertical line H = {hx, bi : b ∈ B} so Z has no more elements
than A. But we can’t have the three conditions
(i) A has fewer elements than B,
(ii) Z has no fewer elements than B,
(iii) Z has no more elements than A.
Therefore, there is no function g in this case.
If A has more elements than B, we set W = (A × B) \ Z, and define Wx and W y
as before (with W in place of Z). A similar argument shows that W has no more
elements than B but no fewer elements than A. Again, there is no function g in this
case.
Chapter 12
2. First, we let E1 and E2 be disjoint measurable sets, we let A be any set and we set
B = A ∩ (E1 ∪ E2 ). Then B ∩ E1 = A ∩ E1 and B ∩ E2 = A \ E1 . Because E1 is
measurable, we conclude that
µ∗ (A ∩ (E1 ∪ E2 )) = µ∗ B = µ∗ (B ∩ E1 ) + µ∗ (B \ E2 )
= µ∗ (A ∩ E1 ) + µ∗ (A ∩ E2 ).
Now let (Ei ) be a sequence of disjoint measurable sets, set E =
n
[
Gn =
Ei .
i=1
A simple induction argument shows that
n
X
∗
µ (A ∩ Gn ) =
µ∗ (A ∩ Ei ).
i=1
By monotonicity, we conclude that
∗
µ (A ∩ E) ≥
n
X
µ∗ (A ∩ Ei )
i=1
for any positive integer n and hence
µ∗ (A ∩ E) ≥
X
µ∗ (A ∩ Ei ).
On the other hand, countable subadditivity implies that
X
µ∗ (A ∩ E) ≤
µ∗ (A ∩ Ei ).
Combining these two inequalities gives
X
µ∗ (A ∩ E) =
µ∗ (A ∩ Ei ).
S
Ei and define
6
4. (a) Let A ∈ A and suppose A can be written as a finite union in two ways: A =
Sn
Sk
i=1 Ci and A =
j=1 Dj with Ci and Dj finite disjoint collections of sets in C .
S
Then for each i, we have Ci = kj=1 (Ci ∩ Dj ) and Ci ∩ Dj ∈ C , so
µCi =
k
X
µ(Ci ∩ Dj ).
j=1
Hence
n
X
µCi =
i=1
n X
k
X
µ(Ci ∩ Dj ).
i=1 j=1
Similarly,
k
X
µDj =
i=1
k X
n
X
µ(Dj ∩ Ci ).
j=1 i=1
Because the sums are all finite, we have
n X
k
X
µ(Ci ∩ Dj ) =
i=1 j=1
and hence
k X
n
X
µ(Dj ∩ Ci ),
j=1 i=1
n
X
µCi =
i=1
k
X
µDj .
i=1
(b) Let (Ai ) be a countable disjoint collection of sets in A and suppose A = ∪Ai is in
A . Then
S there is a finite disjoint collection of sets {B1 , . . . , Bn } in C such that
A = ni=1 Bi . In addition, for each positive integer i, there is a finite collection
of sets {B1i ,S
. . . , Bk(i)i } in C such that Ai = ∪Bji . We now
Pset Bijk = Bji ∩ Bk .
Then Bk = i,j Bijk , so property (ii) implies that µBk ≤ i,j µBijk . From part
(a), we know that
X
X
Bijk
µA =
µBk ≤
i,j,k
and hence
µA ≤
X
µAi .
8. (a) If µ+ E = ∞, then the inequality is automatically true, so we may assume that
µ+ E is finite. Now let ε P
> 0 be given, and suppose that E ⊂ ∪Ai , where each
+
Ai is in A and µ E ≥
µ̄Ai − ε. Now set A = ∪Ai . Then µ∗ Ai = µ̄Ai so
subadditivity implies that
X
X
µ∗ A ≤
µ∗ Ai =
µ̄Ai .
by monotonicity, we know that µ∗ E ≤ µ∗ A, and hence
µ∗ E ≤ µ+ E + ε.
Since this inequality is true for any ε > 0, it follows that µ∗ E ≤ µ+ E.
7
(b) Suppose first that E is a set such that µ+ E = µ∗ E.If µ∗ E = ∞, then X is a
µ∗ -measurable set with E ⊂ X and µ∗ X = µ∗ E. If µ∗ E < ∞, then, for any
(n)
positive integer n, there is a sequence (Ai ) of µ∗ -measurable sets such that
S (n)
E ⊂ i Ai and
X (n)
1
1
µ̄Ai ≤ µ+ E + = µ∗ E + .
n
n
i
S (n)
P ∗ (n)
Setting A(n) = i Ai , we see that T
µ∗ A(n) ≤
µ Ai by subadditivity, so
1
∗ (n)
∗
(n)
µ A ≤ µ E + n . We now set A = A and use monotonicity to conclude
that A is µ∗ -measurable with µ∗ A ≤ µ∗ E + n1 for any positive integer n and
hence µ∗ A ≤ µ∗ E. Since E ⊂ A, it follows from monotonicity that µ∗ A ≥ µ∗ E
so µ∗ A = µ∗ E.
Now suppose that E is a set and that there is a µ∗ -measurable set A with E ⊂ A
and µ∗ A = µ∗ E. Then µ+ E ≤ µ+ A = µ∗ A = µ∗ E. Using the inequality from
part (a), we conclude that µ+ E = µ∗ E.
(c) Suppose first that µ+ E = µ∗ E for all sets E. Then part (b) gives a set A ⊃ E
such that µ∗ A = µ∗ E, so µ∗ A < µ∗ E + ε for any ε > 0. (In this case, we can
use the same set A for any ε. Hence µ∗ is regular.
Conversely, suppose µ∗ is regular and fix a set E. Since µ∗ is regular, for each
positive integer n, there is a µ∗ -measurable set An such that E ⊂ An and
µ∗ An ≤ µ∗ E + n1 . If we set A = ∩An , then E ⊂ A and µ∗ A ≤ µ∗ E + n1 for any
n. Since µ∗ A ≥ µ∗ E, we conclude that µ∗ A = µ∗ E, so µ+ E = µ∗ E by part (b).
(d) If µ∗ is regular, then parts (b) and (c) imply that µ+ E = µ∗ E for any set E.
Hence µ∗ is induced by µ̄.
Conversely, if µ∗ is induced by a measure on an algebra, then we must have that
µ+ E = µ∗ E for all sets E. Hence µ∗ is regular by part (c).
(e) Write the set X = {a, b} and define µ∗ by µ∗ ∅ = 0, µ∗ {a} = 1, µ∗ {b} = 2,
and µ∗ X = 2.5. It’s easy to check that µ∗ is an outer measure and that {b}
is measurable while {a} is not. In addition µ∗ is not regular because {a} is
a subset E of X but there is no measurable subset A of X with E ⊂ A and
µ∗ A < µ∗ E + 1.
P
19. Fubini’s Theorem says that if i,j |xij | < ∞ (in the sense of problem 2.22), then the
following sums are all finite:
∞
∞
∞ X
∞
∞ X
∞
X
X
X
X
|xij |,
|xij |,
|xij |,
|xij |.
i=1
In addition,
i=j
∞ X
∞
X
j=1 i=1
j=1 i=1
xij =
X
i,j
xij =
i=1 j=1
∞ X
∞
X
xij .
i=1 j=1
Tonelli’s Theorem says that if all the numbers xij are nonnegative, then
∞ X
∞
∞ X
∞
X
X
X
xij .
xij =
xij =
j=1 i=1
i,j
i=1 j=1
8
Parts (i), (i)0 , (ii), and (ii)0 of Tonelli’s Theorem are true for any nonnegative function
xij because all functions are measurable in this case.