Real Analysis - Homework solutions
Chris Monico, May 2, 2013
1.1 (a) Rings (resp. σ-rings) are closed under finite (resp. countable) intersections.
(b) If R is a ring (resp. σ-ring) then R is an algebra (resp. σ-algebra) iff X ∈ R.
(c) If R is a (nonempty) σ-ring then {E ⊂ X : E ∈ R or E c ∈ R} is a σ-algebra.
(d) If R is a σ-ring, then {E ⊂ X : E ∩ F ∈ R for all F ∈ R} is a σ-algebra.
Solution:
(a) If R is a ring and E1 , E2 ∈ R, then since R is closed under differences, E1 − (E1 − E2 ) ∈ R. But
E1 − (E1 − E2 ) = E1 − (E1 ∩ E2c ) = E1 ∩ (E1c ∪ E2 ) = E1 ∩ E2 .
It follows inductivelySthat R is closed under finite intersections. Suppose now that R isS a σ-ring and
fn = A − En . Then E
fn ∈ R for all n ≥ 1 and so F = E1 − ∞ E
f
{En } ⊂ R. Let A = Ej and E
j=2 j is also
in R. But




∞
∞
∞
∞
∞
\
\
\
\
\
c
fj  = E1 ∩  (Ac ∪ Ej ) =
E
((E1 ∩ Ac ) ∪ (E1 ∩ Ej )) =
(E1 ∩ Ej ) =
Ej .
F = E1 ∩ 
j=2
j=2
j=2
j=2
j=1
(b) If R is a ring (resp. σ-ring) and X ∈ R, then for all E ∈ R we have E c = X − E ∈ R so R is closed
under complements and therefore an algebra (resp. σ-algebra). On the other hand if R is an algebra (resp.
σ-algebra), then it’s nonempty so there exists E ∈ R and hence X = E ∪ E c ∈ R.
(c) Suppose R is a nonempty σ-ring and M = {E ⊂ X : E ∈ R or E c ∈ R}. Clearly M is closed
under \
complements. Suppose that {En } ⊂ M. By part (a), R is [
closed under countable intersections so
A=
Ejc ∈ R. Since R is closed under countable unions, B =
Ej ∈ R. Now
j≥1
Ejc ∈R
j≥1
Ej ∈R

[
j≥1
Ej



 [
  [




= 
Ej  ∪ 
E
j


 
j≥1
Ejc ∈R
j≥1
Ej ∈R
c c
 \
  [
 


 
= 
Ejc  ∩ 
E
j 

 


j≥1
Ejc ∈R
=


j≥1
Ej ∈R
(A − B)c .
S
Since A, B ∈ R and R is closed under differences, A − B ∈ R, hence Ej = (A − B)c ∈ M. Therefore M
is also closed under countable unions, hence it’s a σ-algebra.
(d) Suppose R is a σ-ring and A = {E ⊂ X : E ∩ F ∈ R for all F ∈ R}. Suppose E ∈ A and let
F ∈ R. Then E ∩ F ∈ R and since R is closed under differences, E c ∩ F = F − E = F − (E ∩S
F ) ∈ R, and
therefore
A
is
closed
under
complements.
If
{E
}
⊂
A
and
F
∈
R
then
E
∩
F
∈
R
and
so
(
En ) ∩ F =
n
n
S
(En ∩ F ) ∈ R for all n, so A is closed under countable unions as well and therefore is a σ-algebra.
1.3 Let M be an infinite σ-algebra. (a) M contains an infinite sequence of (distinct) disjoint sets.
(b) card(M) ≥ c.
Solution:
Note: the word ’distinct’ here is not given as part of the problem, but part (a) becomes trivial without it,
and it is extremely helpful in solving part (b) anyway.
The elements of M are partially ordered by inclusion. By the Hausdorff Maximal Principle, there is
a maximal linearly ordered subset {Eα }α∈A ⊂ M. By way of contradiction, suppose that A is finite, say
A = {1, 2, . . . , n}. Since E1 ⊂ E2 ⊂ · · · ⊂ En , the σ-algebra M({E1 , . . . , En }) is clearly finite so there exists
a set E ∈ M − M({E1 , . . . , En }). We must have E ⊂ En , otherwise E1 ⊂ E2 ⊂ · · · ⊂ En ⊂ En ∪ E would
contradict the maximality of {Eα }. Now since


n
n
[
[
E = E ∩ En = E ∩ E1 ∪
(Ej − Ej−1 ) = (E ∩ E1 ) ∪
(E ∩ (Ej − Ej−1 )),
j=2
j=2
it follows that either E ∩ E1 6∈ M({E1 , . . . , En }) or E ∩ (Ej − Ej−1 ) 6∈ M({E1 , . . . , En }) for some j ≥ 2. In
the first case,
E ∩ E1 ⊂ E1 ⊂ E2 ⊂ · · · ⊂ En ,
would contradict the maximality of {Eα }. In the second case, letting E 0 = Ej−1 ∪ (E ∩ (Ej − Ej−1 )) we have
E1 ⊂ · · · ⊂ Ej−1 ⊂ E 0 ⊂ Ej ⊂ · · · ⊂ En .
But E 0 = Ej−1 would imply that E ∩ (Ej − Ej−1 ) = ∅ ∈ M({E1 , . . . , En }), so E 0 6= Ej−1 . And E 0 = Ej
would imply that E ∩ (Ej − Ej−1 ) = Ej − Ej−1 ∈ M({E1 , . . . , En }), so E 0 6= Ej . Therefore, this contradicts
the maximality of {Eα }, and hence A must be infinite.
So there must exist an infinite increasing sequence {En } of distinct sets in M. Letting F1 = E1 and
Fk+1 = Ek+1 − Ek for all k ≥ 1, it follows that {Fn } is an infinite
of distinct disjoint sets in M.
P∞ sequence
−k
(b) For each x ∈ [0, 1) fix a binary representation x =
b
2
and
let F1 , F2 , . . . be an infinite
k=1 k
sequence of distinct disjoint sets in M. Define a function
ψ : [0, 1)
X
bk 2−k
−→
7−→
M
[
Fk .
k≥1
bk =1
Then ψ is an injection and hence card(M) ≥ card([0, 1)) = c.
1.5 If M = M(E) then M is the union of all σ-algebras generated by F as F ranges over all countable
subsets of E. Hint: show that the latter object is a σ-algebra.
Solution:
[
Let M0 =
M(F). As per the hint, we’ll first show that M0 is a σ-algebra. Suppose {En } ⊂ M0 .
F ⊂E
countable
S
For each n ≥ 1 there is a countable subset Fn ⊂ E such that En ∈ M(Fn ). Then F = Fn is a countable
subset of ESand for all k ≥ 1, Fk ⊂ F implies M(Fn ) ⊂ M(F), so Ek ∈ M(Fk ) ⊂ M(F). Since M(F) is a
σ-algebra, Ek ∈ M(F). And since F is countable, M(F) ⊂ M0 .
Now if E ∈ M0 then there exists a countable subset F ⊂ E such that E ∈ M(F), in which case
E c ∈ M(F) ⊂ M0 . Therefore, M0 is a σ-algebra.
For all F ⊂ E, M(F) ⊂ M(E), so M0 ⊂ M(E). On the other hand if E ∈ E then E is contained in
M({E}) and {E} is countable, so E ∈ M0 and hence E ⊂ M0 , so that M(E) ⊂ M0 .
2
1.6 Complete the proof of Theorem 1.9.
Solution:
It’s routine to verify that µ is a complete measure, so we’ll show only the uniqueness. For this, suppose µ0
is another measure defined on M which extends µ. Then for E ∪ F ∈ M with E ∈ M and F ⊂ N ∈ N , we
have
µ0 E ≤ µ0 (E ∪ F ) ≤ µ0 (E ∪ N )
= µ(E ∪ N ) (since E ∪ N ∈ M )
≤ µE + µN = µE = µ0 E (since E ∈ M).
Therefore µ0 (E ∪ F ) = µE = µ(E ∪ F ), and hence µ0 = µ.
1.9 If (X, M, µ) is a measure space and E, F ∈ M then µE + µF = µ(E ∪ F ) + µ(E ∩ F ).
Solution:
Let E, F ∈ M. Then E − F, E ∩ F , and F − E are disjoint elements of M so
µE + µF
=
(µ(E − F ) + µ(E ∩ F )) + (µ(F − E) + µ(F ∩ E))
=
(µ(E − F ) + µ(E ∩ F ) + µ(F − E)) + µ(F ∩ E)
= µ(E ∪ F ) + µ(E ∩ F ).
1.10 Given a measure space (X, M, µ) and E ∈ M, define µE (A) = µ(A ∩ E) for all A ∈ M. Then
µE is a measure.
Solution:
Certainly µE ∅ = µ(∅ ∩ E) = µ(∅) = 0. Suppose now that {An }n≥1 ⊂ M is a disjoint collection. Then since
{E ∩ An }n≥1 ⊂ M is a disjoint collection it follows that







[
[
[
X
X
µE 
An  = µ E ∩ 
An  = µ 
(E ∩ An ) =
µ(E ∩ An ) =
µE An .
n≥1
n≥1
n≥1
n≥1
n≥1
So µE is countably additive and hence a measure.
1.13 Every σ-finite measure is semifinite.
Solution:
Suppose (X, M, µ) is a measure space and µ is σ-finite. If µ is actually finite then the statement is vacuously
true, so suppose
there exists an E ∈ M with µE = ∞. Since µ is σ-finite, there is a collection {Xn }n≥1 ⊂ M
S
with X = Xn and µXn < ∞ for all n. Then
[
X
∞ = µE = µ(E ∩ X) = µ
(E ∩ Xn ) ≤
µ(E ∩ Xn ),
P
so
µ(E ∩ Xn ) = ∞. In particular, there is a k ≥ 1 for which µ(E ∩ Xk ) > 0. On the other hand,
∞ > µ(Xk ) ≥ µ(E ∩ Xk ) > 0, so E ∩ Xk is a measurable subset of E with positive finite measure. Thus µ
is semifinite.
1.14 If µ is a semifinite measure and µE = ∞ then for every C > 0 there exists F ⊂ E with
C < µF < ∞.
Solution:
Let C = {F ⊂ E : µF < ∞} and α = sup C. By way of contradiction, suppose
that α < ∞. For each
Sn
n ≥ 1 there is an Fn ∈ C such that α ≥ µFn ≥ α − 1/n. Define Gn = k=1 Fk . Then Gn ⊂ E and
3
S
µGn ≥ µFn ≥ α − 1/n for all n. By continuity of measure, µ
G
= limn→∞ µGn and since α ≥ µGn
n
n≥1
for all n ≥ 1, we have
α ≥ lim µGn ≥ lim(α − 1/n) = α,
S
S
so that µ ( Gn ) = α, and hence Gn ∈ C. But
[ [ [ ∞ = µE = µ
Gn + µ E −
Gn = α + µ E −
Gn ,
S
S
so µ (E S
− Gn ) = ∞. Since µ is semifinite
there exists F 0 ⊂ (E − Gn ) with 0 < µF 0 < ∞. Then
S
µ (F 0 ∪ Gn ) = µF 0 + α > α. Since F 0 ∪ Gn ∈ C, this contradicts the fact that α = sup C, and therefore
α = ∞.
1.17 If µ∗ is San outerPmeasure on X and {Aj } is a sequence of disjoint µ∗ -measurable sets then
µ∗ (E ∩ Aj ) = µ∗ (E ∩ Aj ) for all E ⊂ X.
Solution:
Let {Aj } be such a sequence and E ⊂ X. By countable subadditivity,
[ [
X
µ∗ E ∩
Aj
= µ∗
(E ∩ Aj ) ≤
µ∗ (E ∩ Aj ).
Set Bn =
Sn
j=1
Aj . For each n ≥ 2 since An is µ∗ -measurable we have
µ∗ (E ∩ Bn )
= µ∗ (E ∩ Bn ∩ An ) + µ∗ (E ∩ Bn ∩ Acn )
= µ∗ (E ∩ An ) + µ∗ (E ∩ Bn−1 ).
S
Pn
By induction, µ∗ (E∩Bn ) = j=1 µ∗ (E∩An ) for all n ≥ 1. By monotonicity of outer measure, µ∗ E ∩
≥
j≥1 Aj
S
P
P
n
∞
µ∗ (E ∩ Bn ) = j=1 µ∗ (E ∩ An ) for all n ≥ 1, and so µ∗ E ∩
≥ j=1 µ∗ (E ∩ An ).
j≥1 Aj
1.18 Let A ⊂ P(X) be an algebra, µ0 a premeasure on A and µ∗ the induced outer measure.
(a) For all E ⊂ X and > 0 there exists A ∈ Aσ with E ⊂ A and µ∗ A ≤ µ∗ E + .
(b) If µ∗ E < ∞ then E is µ∗ -measurable iff there exists B ∈ Aσδ with E ⊂ B and µ∗ (B − E) = 0.
(c) If µ0 is σ-finite, the restriction µ∗ E < ∞ in (b) is superfluous.
Solution:
S
P
∗
Let E ⊂ X
S and > 0. By definition, there is a sequence {An } ⊂ A so that E ⊂ An and µ0 An ≤ µ E +.
Let A = An . Then A ∈ Aσ and
X
X
µ∗ A ≤
µ∗ An =
µ0 An ≤ µ∗ E + .
For (b), suppose first that E is µ∗ -measurable with µ∗ E < ∞. ByTPart (a), for each n ≥ 1 there
exists Bn ∈ Aσ such that E ⊂ Bn and µ∗ Bn ≤ µ∗ E + 1/n. Set B = Bn . Then E ⊂ B ∈ Aσδ and
µ∗ B ≤ µ∗ Bn ≤ µ∗ E + 1/n for all n ≥ 1, so µ∗ B ≤ µ∗ E. But since E ⊂ B it follows that µ∗ B = µ∗ E. Since
E is µ∗ -measurable,
µ∗ E = µ∗ B
=
µ∗ (B ∩ E) + µ∗ (B ∩ E c )
=
µ∗ E + µ∗ (B − E).
Since µ∗ E < ∞, it follows that µ∗ (B − E) = 0. Conversely, suppose that there exists B ∈ Aσδ with E ⊂ B
and µ∗ (B − E) = 0. Then B is µ∗ measurable since B ∈ Aσδ and B − E is µ∗ -measurable by the proof of
Carathéodory’s Theorem. So (B − E)c is µ∗ -measurable as is
B ∩ (B − E)c = B ∩ (B c ∪ E) = B ∩ E = E.
∗
∗
Finally, for c suppose that µ0 is σ-finite.
S Then so is µ , so there is a disjoint sequence {Xn } of µ ∗
measurable sets with µ Xn < ∞ and X = Xn .
4
Suppose E ⊂ X is µ∗ -measurable and > 0. Then each E ∩ Xn is measurable with finite measure, and
by Part (a) there exists An,k ∈ Aσ such that E ∩ Xn ⊂ An,k and µ∗ An,k < µ∗ (E ∩ Xn ) + k21n . Since E ∩ Xn
is measurable,
µ∗ An,k
=
µ∗ (An,k ∩ (E ∩ Xn )) + µ∗ (An,k ∩ (E ∩ Xn )c )
=
µ∗ (E ∩ Xn ) + µ∗ (An,k − (E ∩ Xn )),
∗
and since µS
(E ∩ Xn ) < ∞, we thus have that µ∗ (An,k − (E ∩ Xn )) = µ∗ An,k − µ∗ (E ∩ Xn ) <
∞
with Bk = n=1 An,k we have Bk ∈ Aσ , E ⊂ Bk , and
!
∞
[
∗
∗
c
∗
c
µ (Bk − E) = µ (Bk ∩ E ) = µ
(An,k ∩ E )
≤
∞
X
n=1
∞
X
1
k2n .
Thus,
∞
X
1
1
= .
µ (An,k − E) ≤
µ (An,k − (E ∩ Xn )) ≤
n
k2
k
n=1
n=1
n=1
∗
∗
T∞
Now set B = k=1 Bk . Then E ⊂ B ∈ Aσδ and µ∗ (B − E) ≤ µ∗ (Bk − E) ≤ k1 for all k ≥ 1, hence
µ∗ (B − E) = 0. The converse follows from our proof of the corresponding implication in Part (b) (which did
not make use of the µ∗ E < ∞ assumption).
1.20 Let µ∗ be an outer measure on X, M∗ the σ-algebra of µ∗ -measurable sets, µ = µ∗ |M∗ , and µ+
the outer measure induced by µ.
(a) If E ⊂ X then µ∗ E ≤ µ+ E with equality iff ∃A ∈ M∗ with E ⊂ A and µ∗ A = µ∗ E.
(b) If µ∗ is induced by a premeasure then µ∗ = µ+ .
(c) If X = {0, 1} there exists an outer measure µ∗ on X such that µ∗ 6= µ+ .
Solution:
S
P
Let ES ⊂ X. Then µ+ E = inf{ µAn : {An } ⊂ M∗ , E ⊂ An }. If {An } ⊂ M∗ is a collection with
E ⊂ An then
[ µ∗ E ≤ µ∗
An
(by monotonicity)
X
≤
µ∗ An
(by countable subadditivity)
X
=
µAn
(since µ = µ∗ |M∗ ).
It follows that µ∗ E ≤ µ+ E.
S∞
Suppose now that µ∗ E = µ+ E. Then for each k ≥ 1 there exists {An,k } ⊂ M∗ with E ⊂ n=1 An,k and


[
X
1
µAn,k ≤ µ+ E + .
µ∗ 
An,k  ≤
k
n≥1
Then with A =
T
k≥1
S
n≥1
A
, we have A ∈ M∗ , E ⊂ A, and µ∗ A ≤ µ+ E + 1/k for all k, hence
n,k
n≥1
µ∗ A ≤ µ+ E = µ∗ E. But since E ⊂ A we have also that µ∗ E ≤ µ∗ A and so µ∗ A = µ∗ E. Conversely,
suppose A ∈ M∗ with A ⊃ E and µ∗ A = µ∗ E. Then by definition of µ+ and since A ∈ M∗ we have
µ+ E ≤ µA = µ∗ A, so µ+ E = µ∗ A.
For (b), suppose µ∗ is induced by a premeasure on an algebra A ⊂ X and let E ⊂ X. As per the
hint, we invoke Exercise 18aTto conclude that for each n ≥ 1 there exists An ∈ Aσ with An ⊃ E and
µ∗ An ≤ µ∗ E + 1/n. Set A = n≥1 An ∈ Aσδ ⊂ M∗ and it follows that A ⊃ E and µ∗ A ≤ µ∗ E. The reverse
inequality follows by monotonicity, so µ∗ A = µ∗ E, and so by Part (a), µ∗ E = µ+ E.
For (c), let X = {0, 1}. Define µ∗ ∅ = 0, µ∗ {0} = 1, µ∗ {1} = 2, and µ∗ {0, 1} = 2. Then µ∗ is an outer
measure on X. Let M∗ be the σ-algebra of µ∗ -measurable sets. Since
µ∗ (X ∩ {0}) + µ∗ (X ∩ {0}c ) = 3 > µ∗ X,
5
it follows that {0} 6∈ M∗ . Similarly, {1} 6∈ M∗ , so M∗ = {∅, X} and µ = µ∗ |M∗ . Now
nX
[ o
µ+ {0} = inf
µEn : {En } ⊂ M∗ and {0} ⊂
En
= µX = 2.
But µ∗ {0} = 1, so µ∗ 6= µ+ .
1.23 Let A be the collection of finite unions of sets of the form (a, b] ∩ Q, where −∞ ≤ a < b ≤ ∞ (if
b = ∞, then it is understood that (a, b] = (a, ∞)).
(a) A is an algebra on Q.
(b) The σ-algebra generated by A is P(Q).
(c) Define µ0 on A by µ0 ∅ = 0 and µ0 A = ∞ for A 6= ∅. Then µ0 is a premeasure on A and there
is more than one measure on P(Q) whose restriction to A is µ0 .
Solution:
As per the hint for (a), let E = {∅} ∪ {(a, b] ∩ Q : −∞ ≤ a < b ≤ ∞}. Clearly E is closed under intersections
and for ∅ =
6 A ∈ E we have that A = (a, b] ∩ Q for some a < b. The complement of A is taken with respect
to Q (since the claim is that A is a σ-algebra on Q), and so Ac = (−∞, a] ∪ (b, ∞] is a disjoint union of
members of E. Therefore E is an elementary family and so the collection of finite disjoint unions of members
of E is an algebra by Proposition 1.7. But every finite union of sets of the form (a, b] can be written as a
disjoint union of such sets, and hence this algebraTis precisely A.
For (b), suppose that r ∈ Q. Then {r} = n≥1 ((r − 1/n, r] ∩ Q) ∈S M(A). If X ⊂ Q, then X is
countable, say X = {x1 , x2 , . . . }. But {xj } ∈ M(A) for all j ≥ 1, so X = j≥1 {xj } is in M(A). Therefore
M(A) = P(Q).
It’s clear that µ0 is a premeasure on A. The induced outer measure is


X
[ 
µ∗ E = inf
µ0 En : {En } ⊂ A, E ⊂
En


n≥1
∞, if E 6= ∅
=
0,
otherwise.
It follows by Theorem 1.14 that µ = µ∗ is a measure on P(Q) which extends µ0 .
If ν is the counting measure on P(Q), then ν∅ = 0 = µ0 ∅. And for a < b, E = (a, b] ∩ Q has infinite
cardinality, so νE = ∞ = µ0 E, and therefore ν|E = µ0 |E . By finite additivity of ν, it follows that νE =
∞ = µ0 E for all ∅ 6= E ∈ A, so that ν|A = µ0 . But ν{1} = 1 < ∞ = µ{1}, so ν is a different measure on
P(Q) which extends µ0 .
1.25 Complete the proof of Theorem 1.19.
Solution:
The only thing left to do is to show that (a) implies (b) and (c) in the general case, having already shown
it for for sets of finite measure, so let E ∈ Mµ . For j ∈ Z let Ej = E ∩ (j, j + 1]. Then µEj ≤ µ(j, j +
1] = F (j + 1) − F (j) < ∞. Thus there are Gδ sets Vj and null sets Nj so that Ej = Vj − Nj . Since
Ej = Ej ∩ (j, j + 1] = (Vj ∩ (j, j + 1]) − Nj and (j, j + 1] is Gδ , we may further assume that Vj ⊂ (j, j + 1]
and Nj ⊂ (j, j + 1]. With this assumption, it follows that

 

[
[
[
[
E=
Ej =
(Vj ∩ Njc ) = 
Vj  − 
Nj  .
j∈Z
j∈Z
j∈Z
j∈Z
S
A countable union of null sets is null, so we need
T only show that j∈Z Vj is a Gδ set. Clearly, for each j ∈ Z,
the set Vj0 = Vj ∪ (j, j + 1]c is Gδ , and so is j∈Z Vj0 . But for each i ∈ Z we have that
\
\
(i, i + 1] ∩
Vj0 =
((Vj ∪ (j, j + 1]c ) ∩ (i, i + 1]) = Vi .
j∈Z
Therefore
T
j∈Z
Vj0 =
S
j∈Z
j∈Z
Vj is Gδ . The proof that (a) implies (c) is similar.
6
1.26 Prove Proposition 1.20.
Solution:
See Exam 1, Problem #4.
1.27 Prove Proposition 1.22a.
Solution:
This is straightforward - just argue on the ternary expansions.
1.28 Let F be increasing and right-continuous, and let µF be the associated measure. Then µF {a} =
F (a) − F (a−), µF [a, b) = F (b−) − F (a−), µF [a, b] = F (b) − F (a−), and µF (a, b) = F (b−) − F (a).
Solution:
Let a, b ∈ R with a < b. Since F is increasing and F (a) < ∞, the limit F (a−) exists and limn→∞ F (a−1/n) =
F (a−). Since µF (a − 1, a] < ∞, we have by continuity of measure that


\
µF {a} = µF 
(a − 1/n, a] = lim µF (a − 1/n, a] = lim (F (a) − F (a − 1/n)) = F (a) − F (a−).
n→∞
n≥1
n→∞
The remaining properties are easily shown using this; for example,
µF [a, b) = µF (a, b] + µF {a} − µF (b) = F (b) − F (a) + F (a) − F (a−) − F (b) + F (b−) = F (b−) − F (a−).
1.29 Let E be a Lebesgue measurable set.
(a) If E ⊂ N where N is the nonmeasurable set from §1.1, then mE = 0.
(b) If mE > 0 then E contains a nonmeasurable set.
Solution:
Suppose E ⊂ N is measurable. In the notation of §1.1, let
Er = {x + r : x ∈ E ∩ [0, 1 − r)} ∪ {x + r − 1 : x ∈ E ∩ [1 − r, 1).
Then for all r ∈ R = Q ∩ [0, 1), since Er ⊂ Nr and the collection {Nr }r∈R is a disjoint collection, it follows
that {Er }r∈R is a disjoint collection. By the translation invariance and finite additivity of Lebesgue measure,
we have that each Er is measurable and mEr = mE for all r ∈ R. So
!
[
X
X
1≥m
Er =
mEr =
mE,
r∈R
r∈R
r∈R
and therefore mE = 0.
For (b), suppose that E ⊂ [0, 1] is a subset with the property that every subset of E is measurable. Then
for each r ∈ R, the set E ∩ Nr is measurable. By the translation invariance and finite additivity of Lebesgue
measure, the set E1−r ∩ N is therefore measurable, and hence must have measure zero by Part (a). Since
the collection {Nr }r∈R is disjoint we have that
!
[
X
X
mE = m
(E ∩ Nr ) =
m(E ∩ Nr ) =
m(E1−r ∩ N ) = 0.
r∈R
r∈R
r∈R
7
1.30 If E ∈ L and mE > 0 then for every α < 1 there is an open interval I such that m(E ∩ I) > αmI.
Solution:
It’s clear if α ≤ 0, so assume that 0 < α < 1. Suppose first that E ∈ L with
mE < ∞. By way of
. Then there is
contradiction suppose that m(E ∩ I) ≤ αmI for every open interval I. Let ∈ 0, (1−α)mE
α
S
P
a disjoint collection {Ik }k≥1 of open intervals such that E ⊂ Ik and
mI)k ≤ mE + . It follows that
[ [
mE = m E ∩
Ik
= m
(E ∩ Ik )
X
=
m(E ∩ Ik )
X
≤
αmIk
X
= α
mIk ≤ α(mE + ) < αmE + (1 − α)mE = mE,
a contradiction.
Now if mE = ∞ then there is a k ∈ Z such that E 0 = E ∩ (k, k + 1] has positive measure. So for each
α < 1 there is an open interval I such that m(E 0 ∩ I) > αmI, and since m(E ∩ I) ≥ m(E 0 ∩ I), we’re done.
1.31 If E ∈ L and mE > 0 then the set E − E = {x − y : x, y ∈ E} contains an interval centered at
zero.
Solution:
As per the hint, we invoke Problem #30 and let I = (x0 − α, x0 + α) be an interval such that m(E ∩ I) >
(3/4)mI. Let 0 ≤ δ < α and suppose BWOC that δ 6∈ E − E. Then x − y 6= δ for all x, y ∈ E. Define
E1
= E ∩ (x0 − α, x0 ]
E2
= E ∩ (x0 , x0 + α).
Then for all x ∈ E1 , we have x + δ ∈ I and x + δ 6∈ E, so x + δ ∈ I − E. Therefore E1 + δ ⊂ I − E. Similarly,
we have that E2 − δ ⊂ I − E. Then E1 + δ ⊂ I − E implies
mE1 = m(E1 + δ) ≤ m(I − E) = mI − m(I ∩ E).
Similarly,
mE2 = m(E2 − δ) ≤ m(I − E) = mI − m(I ∩ E).
It follows that m(E∩I) = mE1 +mE2 ≤ 2mI−2m(I∩E), and so m(E∩I) ≤ (2/3)mI < (4/3)(2/3)m(E∩I) =
(8/9)m(E ∩ I), a contradiction. Therefore δ ∈ E − E and so [0, α) ⊂ E − E. But z ∈ E − E iff −z ∈ E − E,
so in fact, (−α, α) ⊂ E − E.
2.1 Let f : X −→ R and Y = f −1 (R). Then f is measurable iff f −1 ({−∞}) ∈ M, f −1 ({∞}) ∈ M
and f is measurable on Y .
Solution:
Suppose f is measurable. Then {∞}, {−∞} ∈ BR so f −1 ({−∞}), f −1 ({∞}) ∈ M. Let B ∈ BR . Then
f −1 (B) ∈ M. Since R ∈ BR , Y = f −1 (R) ∈ M hence f −1 (B) ∩ Y ∈ M so f is measurable on Y .
For the converse, let B ∈ BR . Set B1 = B∩R, B2 = B∩{−∞, ∞}. Since f is measurable on Y , f −1 (B1 ) =
−1
f (B1 ) ∩ Y ∈ M and by hypothesis, f −1 (B2 ) ∈ M, so f −1 (B) = f −1 (B1 ∪ B2 ) = f −1 (B1 ) ∪ f −1 (B2 ) ∈ M.
8
2.3 If {fn } is a sequence of [complex-valued] measurable functions on X, then {x ∈ X
lim fn (x) exists } is a measurable set.
Solution:
For n, N, m ≥ 1 let
En,N,m
=
{x ∈ X : |(fn − fN )(x)| < 1/m}
=
(fn − fN )−1 (B1/m (0)),
:
where B1/m (0) is the open ball of radius 1/m centered at zero. Since fn − fN is measurable, En,N,m is
measurable for all n, N, m ≥ 1, and hence
\ [ \
En,N,m ,
E=
m≥1 N ≥1 n≥N
is measurable. Furthermore, x ∈ E iff ∀m ≥ 1 ∃N ≥ 1 such that |fn (x) − fN (x)| < 1/m for all n ≥ N , so
E = {x ∈ X : lim fn (x) exists }.
2.4 If f : X −→ R and f −1 ((r, ∞]) ∈ M for each r ∈ Q then f is measurable.
Solution:
For all a ∈ R,
[
(a, ∞] =
(r, ∞],
r∈Q
r>a
so the σ-algebra generated by {(r, ∞] : r ∈ Q} contains {(a, ∞] : a ∈ R}, and hence it contains BR .
Therefore f is measurable by Proposition 2.1.
2.5 If X = A ∪ B where A, B ∈ M, a function f on X is measurable iff f is measurable on both A
and B.
Solution:
Suppose f is measurable and B is a Borel set. Then f −1 (E) ∈ M and hence both f −1 (E) ∩ A ∈ M and
f (E) ∩ B ∈ M, so f is measurable on A and B.
Conversely, suppose f is measurable on A and B and E is a Borel set. Then f −1 (E) ∩ A ∈ M and
−1
f (E) ∩ B ∈ M so
f −1 (E) = f −1 (E) ∩ X = f −1 (E) ∩ A ∪ f −1 (E) ∩ B ∈ M,
−1
so f is measurable.
2.8 If f : R −→ R is monotone then f is Borel measurable.
Solution:
Without loss of generality, assume f is increasing. Suppose (a, b) is a finite open interval in R. We claim
that f −1 (a, b) is an open interval. For this, suppose that x1 , x2 ∈ f −1 (a, b) and x1 < x2 . Let x ∈ (x1 , x2 ).
Then since f is increasing,
a < f (x1 ) ≤ f (z) ≤ f (x2 ) < b,
so that z ∈ f −1 (a, b) proving the claim. But the collection of finite open intervals in R generates BR , so f is
Borel measurable by Proposition 2.1.
9
2.9 Let f : [0, 1] −→ [0, 1] be the Cantor-Lebesgue function from §1.5 and let g(x) = f (x) + x.
a. g is a bijection from [0, 1] to [0, 2] and h = g −1 is continuous from [0, 2] to [0, 1].
b. If C is the Cantor set, m(g(C)) = 1.
c. By Exercise 1.29, g(C) contains a Lebesgue nonmeasurable set A. Let B = g −1 (A). Then B
is Lebesgue measurable but not Borel.
Solution:
(a) Since f is increasing and a(x) = x is strictly increasing, it follows that g is strictly increasing, hence
it is one-to-one. Since f is continuous, so is g, and so it’s clearly onto, hence bijective. Since g is strictly
increasing and continuous, for 0 ≤ a < b ≤ 1 we have
h−1 ((a, b)) = g((a, b)) = (g(a), g(b)).
It follows that for every open O ⊂ [0, 1], h−1 (O) is open in [0, 2], so h is continuous.
S
1] so that C = [0, 1] − k≥1 Ik and
P (b) There is a collection {I
Sk } of disjoint closed intervals
S in [0, S
mIk = 1. Then [0, 1] − C = Ik , and so g([0, 1] − C) = g ( Ik ) = g(Ik ). Since g is one-to-one, {g(Ik )}
is a disjoint collection so
X
2 = m(g([0, 1])) = m(g(C)) +
m(g(Ik )).
(0.1)
If Ik = [ak , bk ] then f is constant on Ik by construction, so g(x) = f (x) + x = f (ak ) + x for all x ∈ Ik , and
thus g(Ik ) = Ik + f (ak ). Therefore m(g(Ik )) = mIk and we have from (0.1) that
X
2 = m(g(C)) +
mIk = m(g(C)) + 1,
hence m(g(C)) = 1.
(c) Since B = g −1 (A) ⊂ g −1 (g(C)) = C and mC = 0, it follows that B is Lebesgue measurable. Since h :
[0, 2] −→ [0, 1] is continuous, it is measurable. If B ⊂ [0, 1] were Borel then h−1 (B) = g(B) = g(g −1 (A)) = A
would be measurable, so B cannot be Borel.
R
2.12 Prove Prop. 2.20: If f ∈ L+ and f < ∞ then {x : f (x) = ∞} is a null set and {x : f (x) > 0}
is a σ-finite set.
Solution:
Let E = {x : f (x) = ∞}. Then E is measurable and for all n ≥ 1, ϕn = nχE is a simple function with
0 ≤ ϕn ≤ f , so
Z
Z
f≥
ϕn = nµE.
R
R
Thus, µE ≤ n1 f for all n ≥ 1. Since 0 ≤ f < ∞ it follows that µE = 0.
S
Now let F = {x : f (x) > 0} and Fn = {x : f (x) > 1/n}. Then F = n≥1 Fn . For each n,
R
R
ψn = (1/n)χF is a simple function with 0 ≤ ψn ≤ f , so f ≥ ψn = n1 µFn , from which it follows that
n
R
µFn ≤ n f < ∞ for all n, so F indeed is a σ-finite set.
R
R
R
R
2.13 Suppose {fn } ⊂ L+ , fn −→ f pointwise, and
f = lim
fn < ∞. Then E f = lim E fn for all
R
R
E ∈ M. However, this need not be true if f = lim fn = ∞.
Solution:
Let A ∈ M and note that f χA ≤ f so
Z
R
A
f≤
R
f < ∞. By Theorem 2.15 we have that
Z
f=
Z
(f χA + f χAc ) =
10
Z
f+
A
f.
Ac
Since fn χA −→ f χA, we have from Fatou’s Lemma that
Z
Z
f =
(lim inf fn χA)
A
Z
≤ lim inf
fn
A
Z
Z
fn −
fn
= lim inf
c
Z
Z A
=
f − lim sup
fn .
Ac
All terms above are finite; from this and another application of Fatou’s Lemma,
Z
Z
Z
Z
Z
lim sup
fn ≤ f −
f=
f ≤ lim inf
fn .
Ac
R
Ac
A
R
Ac
R
R
Therefore, lim Ac fn exists and equals Ac f , and Rhence lim E fnR = E f for all E ∈ M.
To see that this need not be true when lim fRn = ∞ = R f , considerR (X, M) = (R, L)Rand fn =
nχ(0,1/n] + 1χ[1,∞). Then fn −→ χ[1,∞) on R and lim fn = ∞ = f . But lim [0,1] fn = 1 6= 0 = [0,1] f .
R
+
+
2.14 If
R f ∈ L R let λE = E f dµ for E ∈ M. Then λ is a measure on M and for every g ∈ L ,
g dλ = f g dµ. Hint: first suppose that g is simple.
Solution:
R
R
R
First note that λ∅ = ∅ f dµ = f χ∅ dµ = 0 dµ = 0. Suppose that {En } is a disjoint sequence in M. Then
Z
Z
[ En
λ
=
f dµ = f χS E dµ
S
n
En
Z X
XZ
(by Thm. 2.15)
=
f χE dµ =
f χE dµ
n
n
Z
X
X
=
f dµ =
λEn .
En
Therefore λ is a measure.
Suppose that E ∈ M. Then
Z
Z
χE dλ =
Z
dλ = λE =
E
Z
f dµ =
E
f χE dµ.
R
R
It follows from linearity (for nonnegative functions and nonnegative constants) that ϕ dλ = f ϕ dµ for
every nonnegative simple function ϕ. Suppose now that g ∈ L+ . By Thm. 2.10, there is a sequence {ϕn } of
increasing, nonnegative simple functions converging pointwise to g. By the Monotone Convergence Theorem
we have
Z
Z
Z
g dλ = lim ϕn dλ = lim f ϕn dµ.
But ϕn ≤ ϕn+1 and f ≥ 0 implies that f ϕn ≤ f ϕn+1 , so {f ϕn } is an
R increasingR sequence of nonnegative
functions tending to f g, so again applying the MCT we find that lim f ϕn dµ = f g dµ.
R
R
R
2.15 If {fn } ⊂ L+ , fn decreases pointwise to f , and f1 < ∞, then f = lim fn .
Solution:
R
R
Since
fRn is a decreasing
sequence converging
to f , we have Rf ≤ f1 and consequently
f ≤ f1 < ∞. Now
R
R
R
R
R
R f + (f1 − fR) = fR1 , and since f < ∞, it follows that (f1 − f ) = f1 − f . It similarly follows that
(f1 − fn ) = f1 − fn . Since {f1 − fn } is an increasing sequence of nonnegative functions tending to
f1 − f , it follows from the MCT that
Z
Z
Z
Z
Z
Z
f1 − f = (f1 − f ) = lim (f1 − fn ) = f1 − lim fn .
11
R
+
2.16 If
R f ∈ LR and f < ∞ then for every > 0 there exists E ∈ M such that µE < ∞ and
f − .
f>
E
Solution:
R
R R
Let > 0. By definition there is a simple function
ϕ with 0 ≤ ϕ ≤ f and ϕ >
f − . Since ϕ < ∞,
Pn
the support of ϕ is finite; that is, if ϕ = j=1 aj χE is the canonical representation, then aj = 0 for at
j
R
P
most one value of j, say a1 = 0. Then ∞ > ϕ =
aj µEj ≥ min{a2 , . . . , an }(µE2 + . . . µEn ), and so
E = E2 ∪ · · · ∪ En has finite measure. Then
Z Z
Z
Z
f≥
f − .
ϕ= ϕ>
E
E
2.17 Assume Fatou’s Lemma and deduce the MCT from it.
Solution:
Suppose that {fn } is an increasing sequence of nonnegative functions tending to f . By Fatou’s Lemma,
Z
Z
Z
f = (lim inf fn ) ≤ lim inf fn .
R
R
R
R
On the other hand, since fn ≤ f for all n ≥ 1, fn ≤ f for all n ≥ 1 and so lim sup fn ≤ f . Thus,
Z
Z
Z
lim sup fn ≤ f ≤ lim inf fn ,
so lim
R
fn exists and equals
R
f.
2.18 Fatou’s Lemma remains valid if the hypothesis that fn ∈ L+ is replaced by the hypothesis that
fn is measurable and fn ≥ −g where g ∈ L+ ∩ L1 .
Solution:
Suppose {fn } is a measurable sequence, g ∈ L+ ∩ L1 and fn ≥ −g for all n ≥ 1. Set fen = fn + g. Then
fen ≥ 0 so by Fatou’s Lemma
Z
Z
Z
Z
Z
Z
g + lim inf fn = lim inf fen ≤ lim inf fen = g + lim inf fn .
Since g ∈ L1 ,
R
g is finite so it may be subtracted from both sides to obtain
R
lim inf fn ≤ lim inf
R
fn .
2.19 Suppose {fn } ⊂ L1 (µ) and fn −→ fR uniformly.
R
(a) If µX < ∞ then f ∈ L1 (µ) and fn −→ f .
(b) If µX = ∞ the conclusions of (a) can fail.
Solution:
Suppose {fn } ⊂ L1 , fn −→ f uniformly, and µX < ∞. Then there exists N such that |fn (x) − f (x)| < 1
for all n ≥ N and all x ∈ X. Therefore
Z
Z
Z
Z
Z
Z
|f | ≤ |f − fN | + |fN | ≤ 1 + |fN | = µX + |fN | < ∞,
R
R
so f ∈ L1 (µ). RSet g =R 1 + |f |. Then |g| = µX + |f | < ∞, so g ∈ L1 (µ) and |fn | ≤ g for all n ≥ N . So
by LDCT, lim fn = f .
For (b), consider the measure space (R, L, m). For an example where the first conclusion fails, consider
fn (x) = (1/x)χ[1,n)(x) for n ≥ 2. Each fn is certainly in L1 (m) and fn (x) −→ (1/x)χ[1,∞)(x) uniformly on
R, but (1/x)χ[1,∞)(x) 6∈ L1 (m).
For an example where the first conclusion holds but the second fails, consider fn = (1/n)χ[0,n). Then
R
R
1
1
Rfn ∈ L (m) for all n and fn −→ 0 uniformly and 0 ∈ L (m). But fn = 1 for all n so fn −→ 1 while
0 = 0.
12
R
R
R
R
2.20 If fn , gn , f, g ∈ L1 , fn → f , gn → g a.e., |fn | ≤ gn , and gn → g then fn → f .
Solution:
As in the proof of Theorem 2.24, begin by taking real and imaginary parts, so that it suffices to consider
the case where all functions are real-valued. Then |fn | ≤ gn implies gn + fn ≥ 0 and gn − fn ≥ 0. Since
lim(fn + gn ) = f + g, applying Fatou’s Lemma in each of these cases 1 we have
Z
Z
Z
Z
Z
Z
g + f = lim(gn + fn ) ≤ lim inf (gn + fn ) = g + lim inf fn ,
and
Z
Z
g−
Since both
so lim
R
R
g and
R
Z
Z
lim(gn − fn ) ≤ lim inf
f=
Z
(gn − fn ) =
Z
g − lim sup
fn .
f are finite, it follows that
Z
Z
Z
lim sup fn ≤ f ≤ lim inf fn ,
fn exists and equals
R
f.
2.22 Let µ be the counting measure on N. Interpret Fatou’s Lemma, MCT, and LDCT as statements
about infinite series.
Solution:
First note that every function f : N −→ C is measurable in this case since each singleton {n} with n ∈ N
is measurable and every subset of N is a countable union of singletons. Furthermore, since µ{n} = 1 for all
n ∈ N,
Z
Z
∞
X
|f (n)|.
ϕ dµ =
|f | dµ = sup
0≤ϕ≤|f |
1
n=1
P
Thus, f ∈ L iff
f (n) is absolutely convergent.
P∞
Fatou’s Lemma in this case translates to: If {ak,n } is a doubly-indexed nonnegative series with n=1 ak,n
absolutely convergent for every k ≥ 1, then
∞
X
(lim inf ak,n ) ≤ lim inf
n=1
k
k
∞
X
ak,n .
n=1
The Monotone Convergence Theorem
says: If {ak,n } is a doubly-indexed nonnegative series for which
P∞
ak+1,n ≥ ak,n for all k, n ≥ 1 and n=1 ak,n is absolutely convergent for every k ≥ 1 then
∞
X
n=1
lim ak,n = lim
k→∞
k→∞
∞
X
ak,n .
n=1
Finally, the Lebesgue Dominated Convergence
Theorem yields the result that: If {ak,n } is a doublyP∞
indexed series of complex
numbers
with
a
absolutely
convergent for every k ≥ 1 and there exists a
n=1 k,n
P
sequence {bn } with
bn absolutely convergent and |ak,n | ≤ bn for all k, n ≥ 1, then limk→∞ ak,n exists for
each n and
∞
∞
X
X
lim ak,n = lim
ak,n .
n=1
k→∞
k→∞
n=1
1 Note: It’s not generally true that lim inf(a + b ) = lim inf a + lim inf b ; produce a counterexample to convince yourself
k
k
k
k
of this fact. However, we are using in this case the result that if lim ak = a then lim inf(ak + bk ) = a + lim inf bk . If you’re
unfamiliar with this, you should prove it.
13
2.25 Let f (x) = x−1/2 for 0 <P
x < 1 and 0 otherwise. Let Q = {r1 , r2 , . . . } be an enumeration of the
rationals and set g(x) = f (x − rn )/2n .
(a) g ∈ L1 (m) and, in particular, g < ∞ a.e..
(b) g is discontinuous everywhere, unbounded on every interval, and remains so after any modification on a Lebesgue null set.
(c) g 2 < ∞ a.e. but g 2 is not integrable on any interval.
Solution:
Since f is measurable, so is each f (x − rn )/2n , and hence g is measurable as well. Notice that for each n ≥ 1,
√ 1
, if rn < x < 1 + rn ,
x−rn
f (x − rn ) =
0, otherwise.
So for each n ≥ 1,
Z
1
f (x − rn )
dx = n
2n
2
Z
1+rn
√
rn
dx
1
= n−1 .
2
x − rn
By Theorem 2.15,
Z
Z X
∞
X Z f (x − rn )
X
f (x − rn )
1
|g(x)| dx =
dx =
dx =
= 2,
n−1
2n
2n
2
n=1
so indeed g ∈ L1 (m). By Prop. 2.20, g is finite a.e.
(b) Let x0 ∈ R and suppose first that g(x0 ) is finite. Let > 0 and 0 < δ < 1. By the density of Q in R
there exists rn ∈ Q such that x0 < rn < x0 + δ. Let x0 ∈ (rn , x0 + δ) such that
1
√
> + g(x0 ).
2n x0 − rn
Then
g(x0 ) ≥
1
f (x0 − rn )
= n√ 0
> + g(x0 ),
n
2
2 x − rn
and |x0 − x0 | < δ.
Since 0 < δ < 1 was otherwise arbitrary, g is discontinuous at x0 . On the other hand, if g(x0 ) = ∞ and δ > 0
then there is an x0 ∈ (x0 , x0 + δ) for which g(x0 ) is finite since g is finite a.e.. Therefore g is discontinuous
on R.
Let I ⊂ R be a (nondegenerate) interval and rN ∈ Q a rational
point in the interior of I. Let ∈ (0, 1)
2
1
such that (rN , rN + ) ⊂ I, and let α > 2N . Then for all x ∈ rN , rN + 2N1 α
we have
g(x) =
∞
X
f (x − rN )
1
1
f (x − rn )
≥
= N√
> N 1 = α.
n
N
2
2
2
x − rN
2
2N α
n=1
2
Thus, for all sufficiently large α, m g −1 ((α, ∞)) ≥ 2N1 α > 0. This shows that g is unbounded on I and
remains so after any modification on a Lebesgue null set.
(c) If g is finite on E then so is g 2 . Therefore g 2 < ∞ a.e. Again, let I ⊂ R be an interval and rN ∈ Q
be a point in the interior of I. Let ∈ (0, 1) so that (rN , rN + ) ⊂ I. Then
Z
I
g2 =
2
Z X
f (x − rn )
I
2n
Z dx ≥
I
f (x − rN )
2N
2
Z
rN +
dx ≥
rN
so g 2 is not integrable over I.
14
dx
1
= N
N
4 (x − rN )
4
Z
rN +
rN
dx
= ∞,
x − rN
Rx
2.26 If f ∈ L1 (m) and F (x) = −∞ f (t) dt then F is continuous on R.
Solution:
Let x0 ∈ R and let {xn } be a sequence converging to x0 . Set fn = f χ(−∞,x ]. Then the sequence {fn }
n
converges pointwise to either f χ(−∞,x0 ) or f χ(−∞,x0 ]. In any case |fn | ≤ |f | and |f | ∈ L1 (m), so by the LDCT
Z
0
x0
F (x ) =
Z
f (t) dt = lim
n→∞
−∞
Z
xn
f = lim F (xn ).
fn = lim
n→∞
n→∞
−∞
Since {xn } was an arbitrary sequence converging to x0 , limx→x0 F (x) = F (x0 ), so F is continuous at x0 , and
hence continuous on R.
2.27 Let P
fn (x) R= ae−nax − be−nbx where 0 < a < b.
∞
∞
(a) n=1 0 |fn (x)| dx = ∞.
P∞ R ∞
(b) n=1 0 fn (x) dx = 0.
R ∞ P∞
P∞
(c) n=1 fn ∈ L1 ([0, ∞), m) and 0
n=1 fn (x) dx = log(b/a).
Solution:
For (a), we have
∞ Z ∞
X
n=1
|fn | ≥
0
∞ Z
X
∞
ae−nax − be−nbx dx
≥
1
an
n=1
=
∞ Z
X
n=1
∞
X
n=1
=
∞
(ae−nax − be−nbx ) dx
1
an
1
1
−
ne neb/a
∞
1
1 X1
− b/a
= ∞.
e e
n
n=1
For (b), notice that for every n ≥ 1,
−nax
Z ∞
Z ∞
e−nbx
−e
−nax
−nbx
+
fn =
(ae
− be
) dx =
n
n
0
0
P∞ R
so n=1 fn = 0.
P∞
Finally for (c), let f = n=1 fn . Then for each x > 0 we have
f
=
=
=
∞
X
ae−nax − be−nbx
n=1
∞
X
a
a
n=1
∞ X
n=1
=
e−nax − b
1
eax
n
∞
X
∞
= 0,
0
e−nbx
n=1
∞ X
−b
n=1
1
ebx
(since both series conv. absolutely)
n
a
b
−
eax − 1 ebx − 1
Note that for x > 0 and 0 < a < b we have
eax − 1
ax2
a2 x3
bx2
b2 x3
ebx − 1
=x+
+
+ ··· ≤ x +
+
+ ··· =
,
a
2!
3!
2!
3!
b
and so f (x) =
a
eax −1
−
b
ebx −1
≥ 0. Therefore,
a
b
gn (x) =
−
χ[1/n,∞)(x) ≥ 0,
eax − 1 ebx − 1
15
and lim gn = f , so by the Monotone Convergence Theorem,
Z
Z
Z
1 − e−b/n
1 − e−b/n
gn = lim log
|f | = f = lim
=
log
lim
= log(b/a) < ∞.
n→∞
n→∞
n→∞ 1 − e−a/n
1 − e−a/n
2.28 Compute the
R ∞following limits and justify the calculations.
a. limn→∞ 0 (1 + (x/n))−n sin(x/n) dx.
R1
b. limn→∞ 0 (1 + nx2 )(1 + x2 )−n dx.
R∞
c. limn→∞ 0 n sin(x/n)[x(1 + x2 )]−1 dx.
R∞
d. limn→∞ a n(1 + n2 x2 )−1 dx.
Solution:
2
2
. Since
a. Notice that for x ≥ 0, (1 + nx )n ≥ 1 + n1 nx + n2 nx2 = 1 + x + n(n−1)x
2n2
n ≥ 2, it follows that (1 + nx )n ≥ 1 + x + x2 /4 for n ≥ 2 and all x ≥ 0. Therefore
so by the LDCT,
Z
lim
n→∞
∞
0
1+
x −n
sin(x/n) dx =
n
∞
Z
sin(x/n)
1
≤
x n
1+ n
1+x+
and
x2
0
4
∞
Z
lim
n→∞
0
1+
dx
1+x+
x2
4
n(n−1)
2n2
=
1
2
−
1
2n
≥
1
4
for
< ∞,
x −n
sin(x/n) dx =
n
Z
∞
0 dx = 0.
0
b. With fn = (1 + nx2 )(1 + x2 )−n we have for n ≥ 2 that
1 + nx2
≤ 1,
1 + nx2 + n2 x4
|fn | ≤
and 1 ∈ L1 ([0, 1], m). Since lim fn = χ{0}, it follows from the LDCT that
Z
1
lim
n→∞
c. Let fn =
n sin(x/n)
x(1+x2 ) .
Z
fn dx =
0
0
1
χ{0} dx = 0.
Since | sin α| ≤ |α| for all α ∈ R, we have that | sin(x/n)| ≤ |x/n| and so
|fn | ≤
and g ∈ L1 ([0, ∞), m). Furthermore, fn →
Z
lim
n · nx
1
=
= g(x),
x(1 + x2 )
1 + x2
x
x(1+x2 )
∞
Z
=
∞
fn dm =
0
0
1
1+x2 ,
so by LDCT (with dominating function g),
dx
= tan−1 x
1 + x2
∞
0
=
π
.
2
d. Here we simply compute the limit directly:
Z
lim
n→∞
a
∞
n
dx = lim tan−1 (nx)
2
n→∞
1 + (nx)
∞
a
= lim
n→∞

0,

π/2,
− tan−1 (an) =

2
π,
π
if a > 0,
if a = 0,
if a < 0.
How does this accord with the various convergence theorems? Let
n
0, if x 6= 0,
f (x) = lim
=
∞, if x = 0.
n→∞ 1 + (nx)2
R∞
Since a f = 0 for all a, it is possible that the convergence theorems may apply for a > 0, but the hypotheses
of MCT and LDCT cannot be satisfied for a ≤ 0. Indeed, {fn } is not monotone on any interval containing
zero since {fn (0)} is an increasing sequence while for all x 6= 0, {fn (x)} is eventually decreasing.
16
If a > 0 then |fn | ≤ 1/x2 on [a, ∞), so LDCT may be applied in this case, but there is no function
g ∈ L1 ([a, ∞)) that dominates {fn } on [a, ∞) if a ≤ 0.
Fatou’s Lemma, however, does apply and it supplies us with the conclusion that
Z
Z ∞
f ≤ lim inf fn ,
0=
a
which is rather weak since the sequence {fn } is nonnegative.
R∞
R∞
2.29 Show that 0 xn e−x dx = n! by differentiating the equation 0 e−tx dx = 1/t. Similarly show
√
p
R ∞ 2n −x2
R∞
2
π
that −∞ x e
dx = (2n)!
by differentiating the equation −∞ e−tx dx = π/t.
4n n!
Solution:
R∞
We first show by induction that for all n ≥ 1 and t ∈ [1, 2], n!/tn+1 = 0 xn e−tx dx. Evaluation at t = 1
then produces the desired result.
−tx
exists for t ∈ [1, 2]. With g(x) = xe−x we have g ∈ L1 ([0, ∞))
Let f (x, t) = e−tx . Then ∂f
∂t = −xe
and
∂f
∂t
≤ g for all (x, t) ∈ [0, ∞) × [1, 2]. So, by Theorem 2.27,
−1
∂
=
2
t
∂t
∞
Z
∞
Z
f (x, t) dx =
0
0
∂f
dx = −
∂t
Z
∞
xe−tx dx.
0
R∞
xn e−tx dx. Then
Z ∞
∂
n!
∂
(n + 1)!
=
−
xn e−tx dx.
=
−
tn+2
∂t tn+1
∂t 0
Suppose now that n ≥ 1 and n!/tn+1 =
0
For all t ∈ [1, 2],
∂ n −tx x e
= xn+1 e−tx ≤ xn+1 e−x ∈ L1 ([0, ∞)),
∂t
so we may again apply Theorem 2.27 to conclude that
Z ∞
Z ∞
(n + 1)!
∂ n −tx =
−
x
e
dx
=
xn+1 e−tx dx.
tn+2
∂t
0
0
The second part of the problem is similar.
2.31 Derive the following:
R∞
√
2
2
a. For a > 0, −∞ e−x cos ax dx = πe−a /4 .
R1
P∞
b. For a > −1, 0 xa (1 − x)−1 log x dx = − k=1 (a + k)−2 .
R ∞ a−1 x
P∞
(e − 1)−1 dx = Γ(a)ζ(a), where ζ(a) = n=1 n−a .
c. For a > 1, 0 x
R ∞ −ax −1
d. For a > 1, 0 e
x sin x dx = arctan(a−1 ).
R ∞ −ax
P∞
e. For a > 1, 0 e
J0 (x) dx = (a2 + 1)−1/2 , where J0 (x) = n=0 (−1)n x2n 4−n (n!)−2 is the
Bessel function of order 0.
Solution:
a. Using the power series expansion of cos ax we have
e
−x2
cos ax =
∞
X
2
e−x
n=0
(ax)2n
(−1)n .
(2n)!
Furthermore,
∞
X
n=0
e−x
2
∞
∞
2n
k
2 X (ax)
2 X (ax)
2
2
(ax)2n
(−1)n = e−x
≤ e−x
= e−x eax = eax−x .
(2n)!
(2n)!
k!
n=0
k=0
17
2
Since eax−x ∈ L1 (R) it follows from Theorem 2.25 and the result of Exercise 2.29 that
Z ∞ X
Z
∞
∞
2n
X
(−1)n a2n ∞ 2n −x2
−x2 (ax)
n
x e
dx
e
(−1) dx =
(2n)!
(2n)!
−∞ n=0
−∞
n=0
√
∞
X
(−1)n a2n (2n)! π
=
·
(2n)!
4n n!
n=0
n
∞ √ X
√
2
−a2
1
π
=
= πe−a /4 .
4
n!
n=0
P∞
a
log x
b. Using the formula for the geometric series, we have for 0 < x < 1 that x 1−x
= n=0 xn+a log x =
P∞ n+a
− n=0 x
log(1/x). Since each function xn+a log(1/x) is nonnegative on (0, 1), we have by Theorem 2.15
that
Z 1X
Z 1 a
∞
x log x
dx = −
xn+a log(1/x) dx
1−x
0
0 n=0
∞ Z 1
X
= −
xn+a log(1/x) dx.
(0.2)
n=0
0
For j ≥ 1, let gj = xn+a log(1/x)χ[1/j,1]. Then {gj } is an increasing sequence of functions, so by MCT and
integration by parts,
Z 1
Z
Z
n+a
x
log(1/x) dx =
lim gj dm(x) = lim
gj dm(x)
(0,1] j→∞
0
j→∞
(0,1]
1
Z
=
1/j
=
=
xn+a log(1/x) dx
lim
j→∞
lim
j→∞
xn+a+1
−xn+a+1 log x
+
n+a+1
(n + a + 1)2
1
(n + a + 1)2
1
.
1/j
( a > −1 was used here)
Note: this type of argument shows that the usual technique for dealing with such ‘improper’ integrals
works, provided the integrand is nonnegative for almost all values sufficiently close to the limit of integration;
henceforth, I will assume this rather than detail out the steps as I’ve done here.
Combining with (0.2), we have
Z 1 a
∞
∞
X
X
x log x
1
1
dx = −
=
−
.
2
1
−
x
(n
+
a
+
1)
(a
+
k)2
0
n=0
k=1
c. Using the formula
α
1−α
2
3
= α + α + α + . . . which is valid for |α| < 1, we have
xa−1
= xa−1
ex − 1
e−x
1 − e−x
=
∞
X
xa−1 e−kx ,
for x > 0.
k=1
In this last expression, the summands are nonnegative, so
Z ∞ a−1
∞ Z ∞
X
x
dx
=
xa−1 e−kx dx =
ex − 1
0
0
k=0
=
=
∞ Z
X
k=1
∞
X
k=1
∞
X
k=1
18
∞
u a−1
k
0
1
ka
Z
e−u
du
k
∞
ua−1 e−u du
0
Γ(a)
= Γ(a)ζ(a).
ka
d. Using the power series expansion for sin x we have
x−1 e−ax sin x =
∞
X
(−1)k x2k
k=0
Letting fk =
k
(−1) x
(2k+1)!
∞ Z
X
k=0
2k
(2k + 1)!
e−ax .
e−ax , it follows that
∞
|fk | =
0
∞
X
k=0
1
(2k + 1)!
Z
∞
X
∞
2k −ax
x e
dx
=
0
k=0
∞
X
=
k=0
∞
X
=
k=0
So by Theorem 2.25,
Z
∞
0
∞
X
!
fk
k=0
=
∞ Z
X
fk
=
k=0
∞
X
=
k=0
∞
X
=
k=0
e. Let
fn =
∞
Z
t2k e−t dt
0
Γ(2k + 1)
(2k + 1)!a2k+1
1
< ∞,
(2k + 1)a2k+1
for a > 1.
Z ∞
∞
X
(−1)k
x2k e−ax dx
(2k + 1)! 0
∞
0
k=0
1
(2k + 1)!a2k+1
(−1)k Γ(2k + 1)
(2k + 1)!a2k+1
(−1)k
= arctan(a−1 ).
(2k + 1)a2k+1
(−1)n x2n −ax
e
.
4n (n!)2
Then
∞ Z
X
n=0
0
∞
∞
X
1
|fn | =
n
4 (n!)2
n=0
Z
∞
2n −ax
x e
dx =
0
=
=
If you are familiar with the well-known bound
inductively, or sloppily by:
2n
n
∞
X
1
n
4 (n!)2 a2n+1
n=0
∞
X
n=0
∞
X
Z
∞
t2n e−t dt
0
Γ(2n + 1)
4n (n!)2 a2n+1
(2n)!
.
n (n!)2 a2n+1
4
n=0
≤ 22n you could use it here. Otherwise, deduce it
3
1
...
≤ 1.
4
2
P∞ R ∞
(The right-hand side here cannot be taken as 1/2 if n = 0). It thus follows that n=0 0 |fn | < ∞. So by
Theorem 2.25 we have that
Z ∞ X X
Z ∞
∞
∞ Z ∞
X
(−1)n
fn =
fn =
x2n e−ax dx
n (n!)2
4
0
0
0
n=0
n=0
(2n)!
(2n)(2n − 1)(2n − 2)(2n − 3) . . . (2)(1)
=
=
n
2
4 (n!)
(2n)(2n)(2n − 2)(2n − 2) . . . (2)(2)
2n − 1
2n
2n − 3
2n − 2
=
∞
X
(−1)n Γ(2n + 1)
4n (n!)2 a2n+1
n=0
=
√
1
.
a2 + 1
(For the last equality, you may either look it up or derive it).
19
2.45 If (Xj , Mj ) is a measurable space for j = 1, 2, 3 then ⊗31 Mj = M1 ⊗ M2 ⊗ M3 . Moreover, if µj
is a σ-finite measure on (Xj , Mj ) then µ1 × µ2 × µ3 = (µ1 × µ2 ) × µ3 .
Solution:
By Prop. 1.3, M1 ⊗ M2 is generated by E1 = {E1 × E2 : Ej ∈ Mj }. Since E3 = M3 generates M3 , we
have by Prop. 1.4 that (M1 ⊗ M2 ) ⊗ M3 is generated by
E
= {A × B : A ∈ E1 , B ∈ E3 }
= {(E1 × E2 ) × E3 : Ej ∈ Mj }.
Under the natural identification where we take (X1 × X2 ) × X3 = X1 × X2 × X3 , we thus have
E = {E1 × E2 × E3 : Ej ∈ Mj },
and this set generates ⊗31 Mj by Proposition 1.3.
Suppose now that µ1 , µ2 , µ3 are σ-finite. Then for all E1 ∈ M1 , E2 ∈ M2 , E3 ∈ M3 we have
(µ1 × µ2 ) × µ3 ((E1 × E2 ) × E3 )
=
(µ1 × µ2 )(E1 × E2 )µ3 (E3 )
= µ1 (E1 )µ2 (E2 )µ3 (E3 )
=
(µ1 × µ2 × µ3 )(E1 × E2 × E3 ).
It follows by countable additivity that ((µ1 × µ2 ) × µ3 )(A) = (µ1 × µ2 × µ3 )(A) for every set A which is a
finite disjoint union of measurable cuboids (rectangular parallelpipeds). The collection A of all such sets A
is an algebra and the σ-algebra generated by A is ⊗31 Mj ; since the measures (µ1 × µ2 ) × µ3 and µ1 × µ2 × µ3
are σ-finite and agree on A, they are equal by the uniqueness assertion in Theorem 1.14.
2.46 Let X = Y = [0, 1], M = N = B[0,1] , let µ be the Lebesgue measure
RR and ν the
RR counting
measure. If D = {(x, x) : x ∈ [0, 1]} is the diagonal in X × Y , then
χD dµ dν,
χD dν dµ
R
and χD d(µ × ν) are all unequal.
Solution:
R
R
For all y ∈ [0, 1] we have χD(x, y) dµ(x) = {y} 1 dµ(x) = 0, so
ZZ
ZZ
χD dµ dν =
For all x ∈ [0, 1],
R
χD(x, y) dν(y) =
R
{x}
ZZ
Z
χD(x, y) dµ(x) dν(y) =
1 dν(y) = ν{x} = 1, so
ZZ
χD dν dµ =
0 dν(y) = 0.
Z
χD(x, y) dν(y) dµ(x) =
1 dµ(x) = 1.
[0,1]
R
Finally, χD d(µ × ν) = (µ × ν)(D) which we will now show is ∞, by arguing on the outer
S measure of
D. SupposeSthat {An × Bn } is a countable
collection
of
measurable
rectangles
with
D
⊂
(An × Bn ).
S
Then D = (D ∩ An × Bn ), and so (An ∩ Bn ) ⊃ [0, 1]. It follows that there is an N ≥ 1 for which
m(AN ∩ BN ) > 0. In particular, mAN > 0 and BN is infinite so that νBN = ∞. It follows that the outer
measure of D is infinite and since D is measurable, (µ × ν)(D) = ∞.
2.49 Prove Theorem 2.39 by using Thm. 2.37 and Prop. 2.12 together with the following lemmas.
a. If E ∈ M × N and (µ × ν)(E) = 0 then ν(Ex ) = µ(E y ) = 0 for almost all x, y.
b.
and f = 0 λ-a.e., then fx and f y are integrable for almost all x, y and
R If f is RL-measurable
y
fx dν = f dµ = 0 for almost all x, y.
Solution:
a. One may prove the stronger result for E ∈ M ⊗ N with (µ × ν)(E) = 0 trivially by invoking Theorem
2.36; I think this is what was intended, since it’s the first thing I’m going to do in Part (b) anyway.
20
b. Let E = {(x, y) : f (x, y) 6= 0}. Then λE = 0 and since λ is the completion of µ × ν, there is a set
e ∈ M ⊗ N such that E
e ⊃ E and (µ × ν)(E)
e = 0. By Theorem 2.36,
E
Z
Z
e
e
e y dν(y).
0 = (µ × ν)(E) = ν Ex dµ(x) = µE
ex = 0 a.e. [µ] and µE
e y = 0 a.e. [ν]. Since Ex ⊂ E
ex it follows by the completeness of ν
Therefore ν E
that νEx = 0 a.e. [µ]. Similarly, from the completeness of µ it follows that µE y = 0 a.e. [ν]. Since
{y : fx (y) 6= 0} = Ex and νEx = 0 a.e. [µ], it follows that for almost all x, fx = 0 a.e.. By RProp. 2.11,
we therefore have that for almost all x, fx is measurable, |fx | = 0Ra.e. so fx is integrable, and fx dν = 0.
Similarly, we conclude that for almost all y, f y is integrable, and |f y | dµ = 0.
Thm. 2.39: Assume the hypotheses of the Theorem and suppose that f is L-measurable and f ≥ 0. By
Prop. 2.12, there is an M ⊗ N -measurable function fe so that fe = f a.e. [λ]. We may further assume that
fe ≥ 0 (by possibly modifying fe on a µ × ν-null set). Set g = f − fe. Then g = 0 a.e. [λ], so by Part (b),
gx , g y are integrable, and in particular measurable, for almost all x, y. fex , fey are measurable by Prop. 2.34,
y
and so
for almost all x, y respectively. Furthermore,
byRTheorem 2.37, the functions
R fx , f are measurable
R
R
e
x 7→ fx dν and y 7→ fey dµ are measurable. By Part (b), we have gx dν = g y dµ = 0 for almost all x, y
resp., so the functions
Z
Z
Z
x 7→
fx dν = (fex + gx ) dν = fex dν,
for almost all x
Z
Z
Z
y 7→
f y dµ = (fey + g y ) dµ = fey dµ,
for almost all y
are measurable for almost all x, y respectively. By Theorem 2.37,
ZZ
ZZ
ZZ
Z
f dν dµ =
(fex + gx ) dν dµ =
fex dν dµ = fed(µ × ν),
RR
R
R
R
e
and similarly,
f dµ dν = fed(µ × ν), so it remains only
R to show that R f d(µ × ν) = f dλ.
If E is M ⊗ N measurable then (µ × ν)(E) = λ(E), so χE d(µ × ν) = χE dλ. By linearity, we have the
same result for M ⊗ N-measurable simple functions. Let {ϕn } be a sequence of simple functions increasing
to fe. Using the Monotone Convergence Theorem twice, we deduce that
Z
Z
Z
Z
fed(µ × ν) = lim ϕn d(µ × ν) = lim ϕn dλ = fedλ,
R
R
and since f = fe a.e. [λ], it follows that fedλ = f dλ.
Tonelli’s Theorem now follows from Fubini’s as in the proof of Theorem 2.37.
3.8 ν µ iff |ν| µ iff ν + µ and ν − µ.
Solution:
Suppose ν µ and µE = 0. Let X = P ∪ N be a Hahn decomposition w.r.t. ν, so that ν + E = ν(E ∩ P )
and ν − E = −ν(E ∩ N ). Then
µE = 0 ⇒ µ(E ∩ P ) = 0 ⇒ ν(E ∩ P ) = 0 ⇒ ν + E = 0,
and
µE = 0 ⇒ µ(E ∩ N ) = 0 ⇒ ν(E ∩ N ) = 0 ⇒ ν − E = 0,
so |ν|E = ν + E + ν − E = 0, and thus |ν| µ.
Now if |ν| µ and µE = 0 then ν + E + ν − E = 0, and so ν + E = ν − E = 0 since ν + , ν − are positive
measures. Therefore ν + µ and ν − µ.
Finally, if ν + µ and ν − µ and µE = 0 then ν + E = 0 and ν − E = 0 so νE = ν + E − ν − E = 0, and
thus ν µ.
21
3.9 Suppose {νj } is
Pa sequence of positive measures. If νj ⊥ µ for all j then (
for all j then ( νj ) µ.
Solution:
P
νj ) ⊥ µ. If νj µ
Suppose first that νTj ⊥ µ for all P
j. Then for each j there isSa measurable Aj so that νj ≡ 0 on Aj and
µ ≡ 0 on Acj . Set A = Aj . Then ( νj ) ≡ 0 on A and Ac = Acj , so for every measurable E ⊂ Ac ,
µE = µ(E ∩ Ac ) ≤
X
µ(E ∩ Acj ) = 0.
second part, suppose νP
j µ for all j, and E has µE = 0. Then νj E = 0 for all j, so
P For the P
( νj )E =
νj E = 0, and hence ( νj ) µ.
3.10 Theorem 3.5 may fail when ν is not finite.
Solution:
As
R per the hint, consider dν(x) = dx/x and dµ(x) = dx on (0, 1). We have ν µ since µE = 0 ⇒ νE =
(1/x) dx = 0.
E
On the other hand, suppose = 1. Then for all δ > 0, µ((0, δ/2)) < δ and
Z
ν((0, δ/2)) =
(0,δ/2)
dx
=
x
δ/2
Z
0
dx
= ∞ > 1 = ,
x
so the conclusion of Theorem 3.5 does not hold in this case.
3.11 (a) Every finite subset of L1 (µ) is uniformly integrable.
(b) If {fn } ⊂ L1 (µ) converges in the L1 metric to f ∈ L1 (µ), then {fn } is uniformly integrable.
Solution:
For (a), suppose {f1 , . . . , fn } ⊂ L1 (µ) and > 0. By Corollary 3.6, for each 1 ≤ j ≤ n there is a δj > 0 s.t.
Z
fj dµ < .
µE < δj =⇒
E
Then δ = min{δ1 , . . . , δn } answers the -challenge for uniform integrability.
(b) Suppose {fn } ⊂ L1 (µ) and fn −→ f ∈ L1 (µ) in the L1 metric. Let > 0. Then there is a δ1 > 0 so
that
Z
µE < δ1 =⇒
f dµ < /2.
E
R
Since fn → f in L1 , there is an N so that |fn − f | dµ < /2 for all n ≥ N . Thus, for E with µE < δ1 and
n ≥ N,
Z
Z
Z
Z
Z
fn dµ =
f dµ + (fn − f ) dµ ≤
f dµ +
|fn − f | dµ < /2 + /2.
E
E
E
E
E
By the first part, there is a δ2 > 0 responding to the -challenge for uniform integrability of {f1 , . . . , fN −1 },
so δ = min{δ1 , δ2 } does the job for every fn .
3.16 Suppose µ, ν are σ-finite measures on (X, M) with ν µ and let λ = µ + ν. If f = dν/ dλ, then
0 ≤ f < 1 µ-a.e. and dν/ dµ = f /(1 − f ).
Solution:
dν
dλ
Since λ µ, we have by Prop 3.9 that dµ
= f dµ
.
22
Since
dλ
dµ
=1+
dν
dµ
=1+f
dλ
dµ
and f,
dλ
dµ
can be taken to be finite everywhere, it follows that
(1 − f )
dλ
= 1.
dµ
(0.3)
Let E = {x : f (x) ≥ 1}. Then
Z
Z
Z
Z
dλ
µE =
dµ =
1 dµ =
(1 − f ) dλ ≤
(1 − f )
0 dλ = 0.
dµ
E
E
E
E
But since µ is a positive measure, it follows that µE = 0. Therefore, 1 − f > 0 µ-a.e. and from (0.3) we
f
1
dν
dλ
dλ
= 1−f
, and so dµ
= f dµ
= 1−f
.
deduce that dµ
Extra: Suppose µ is P
a σ-finite measure on (X, M) and {En } a sequence of measurable sets. Define ν on
dν
M by νE = µ(E ∩ En ). Find the Radon-Nikodym derivative dµ
.
Solution:
Clearly ν is σ-finite and ν µ, so the Radon-Nikodym derivative in question exists. Furthermore, for every
E ∈ M,
X
νE =
µ(E ∩ En )
Z X
XZ
XZ
χE dµ,
χE dµ =
=
χE∩E dµ =
n
so that
dν
dµ
=
P
E
n
n
χE .
n
3.25 If E is a Borel set in Rn define DE (x) = limr→0 m(E ∩ B(r, x))/mB(r, x), if the limit exists.
a. Show that DE (x) = 1 a.e. on E and DE (x) = 0 a.e. on E c .
b. Find examples of E and x s.t. DE (x) is a given number α ∈ (0, 1) and an example for which
DE (x) does not exist.
Solution:
a. This is a direct application of Thm. 3.22: let E ∈ BRn and define νA = m(A ∩ E) for all Borel sets A
dν
(i.e., ν = m|E ). Then ν m and dm
= χE. Since the family {B(r, x)}r>0 shrinks nicely to x, it follows by
Theorem 3.22 that DE (x) = limr→0 νB(r, x)/mB(r, x) = χE(x) for m-almost all x ∈ Rn .
b. Let α ∈ (0, 1) and E = {(t sin θ, t cos θ) : t > 0, 0 ≤ θ ≤ 2πα} ⊂ R2 , and x = (0, 0). For each r > 0,
m(E ∩ B(r, x)) = 2πr2 α, so DE (x) = α.
For an example in which the limit does not exist, consider the subset E of R given by
[ 1
1
1 1
1 1
E=
,
=
,
∪
,
∪ ...,
22n+1 22n
8 4
32 16
n≥1
1
and x = 0. Consider the sequence rk = 2−k . Then mB(rk , 0) = 2k−1
. Now if k is even, say k = 2k 0 , then
X 1
X 1
1
1
1
4
m(E ∩ B(rk , 0)) =
m 2n+1 , 2n =
=
=
,
0 ·
2n+1
k
2
2
2
2·4
3
3 · 2k−1
0
2n≥k
n≥k
and so m(E ∩ B(rk , 0))/mB(rk , 0) = 1/3 when k is even. Now for odd k, say k = 2k 0 + 1 we have
X 1
X
1
1
1
4
1
m(E ∩ B(rk , 0)) =
m 2n+1 , 2n =
=
· =
,
2n+1
k0 +1 3
2
2
2
2
·
4
3
·
2k
0
2n≥k
n≥k +1
and so m(E ∩ B(rk , 0))/mB(rk , 0) = 1/6 when k is odd. Therefore, limk→∞ m(E ∩ B(rk , 0))/mB(rk , 0) does
not exist and hence DE (0) is undefined.
23
3.26 If λ and µ are positive, mutually singular Borel measures on Rn and λ + µ is regular then so are
λ and µ.
Solution:
For compact K ⊂ Rn , (λ + µ)K < ∞, and so λK < ∞, µK < ∞ since λ and µ are positive.
Let A ∈ BRn so that µA = λAc = 0. Then for all E ∈ BRn ,
λE = λ(E ∩ A)
=
(λ + µ)(E ∩ A)
=
inf{(λ + µ)U : U is open and U ⊃ E ∩ A}
≥ inf{λU : U is open and U ⊃ E ∩ A}
≥ λ(E ∩ A) = λE.
So we have λE = inf{λU : U is open and U ⊃ E ∩ A}, Below, we will show that for each > 0 there is an
open set O ⊃ Ac such that λO < . With this, it follows that for each open U ⊃ E ∩ A, the set O ∪ U is
open, O ∪ U ⊃ E and λ(O ∪ U ) ≤ λU + so that inf{λG : G is open and G ⊃ E} ≤ λE by above, while
the reverse inequality follows from monotonicity. The corresponding property for µ follows by interchanging
µ and λ in this argument, so showing the existence of such O will complete the proof.
Let > 0. Since λ + µ is regular and Rn is a countable
S union of compact sets, λ + µ is σ-finite. So there
is a countable disjoint collection {Xj }j≥1 with Ac = Xj and (λ + µ)Xj < ∞. For each j ≥ 1, let Oj be
an open set with Oj ⊃ Xj such that
(λ + µ)Oj ≤ (λ + µ)Xj +
and set O =
S
,
2j
Oj . Then O is open, O ⊃ Ac and
λOj = λ(Oj ∩ A) = (λ + µ)(Oj ∩ A) = (λ + µ)Oj − (λ + µ)(Oj ∩ Ac ) ≤ (λ + µ)Xj +
Therefore λO ≤
P
− µXj = j .
j
2
2
λOj < .
3.30 Construct an increasing function on R whose set of discontinuities is Q.
Solution:
One method is to enumerate the rationals Q = {x1 , x2 . . . } and construct a Borel measure µ for which
µE = µ(E ∩ Q) for all Borel sets E and µ{xj } = 2−j . Then apply Theorem 1.16 to get an increasing F with
µF = µ.
Another method is to construct an increasing function f : [0, ∞) −→ [0, ∞) which is discontinuous on Q;
the result then follows by extending f to an odd function via f (−x) = −f (x) for x > 0. To that end, let
X
f (x) =
0≤ p
q ≤x
where the sum is taken over all rational
0 ≤ f (x) ≤
p
q
1
,
q3
with gcd(p, q) = 1. Notice first that for x > 0,
∞
X
X
q=1 0≤p≤qx
∞
∞
X
X
1
bqxc
1
xπ 2
=
≤
x
=
,
q3
q3
q2
6
q=1
q=1
so the sum converges. Clearly f is increasing. If x =
|f (x) − f (x − )| =
a
b
X
0≤ p
q ≤x
so f is discontinuous at x.
24
is a positive rational, then for all x > > 0,
1
−
q3
X
0≤ p
q ≤x−
1
1
≥ 3,
q3
b
P∞
Suppose x is irrational and > 0. Choose N ∈ N so that q=N q12 < . For each b = 1, 2, . . . , N let nb ∈ N
for which x − nbb = δb is minimal. Since x is irrational, each δb is positive. Let δ = min{1, δ1 , δ2 , . . . , δN }.
Then x −
p
q
< δ ⇒ q > N , so for 0 < y < δ,
0 ≤ f (x) − f (x − δ) =
X
x−δ< p
q ≤x
∞
X
1
<
q3
∞
∞
∞
X
X
X
1
1 + qδ
1
1
≤
=
+
δ
< + δ ≤ 2.
q3
q3
q3
q2
X
q=N qx−qδ<p≤qx
q=N
q=N
q=N
Similarly, 0 ≤ f (x + δ) − f (x) ≤ 2, so f is continuous at x.
Rb
3.33 If F is increasing on R then F (b) − F (a) ≥ a F 0 (t) dt.
Solution:
Let −∞ < a < b < ∞ and assume WOLG that F is NBV by properly redefining it outside the interval [a, b].
Let µF be the Borel measure associated with F and
λ ⊥ m, ρ m,
µF = λ + ρ,
its Lebesgue-Radon-Nikodym decomposition. Both λ and ρ are regular measures, so by Theorem 1.16 there
exist right-continuous increasing functions G, H such that
R x λ = µG , ρ =R xµH . Both G, H ∈ N BV , so by
Proposition 3.30, G0 = 0 a.e. and for x ∈ [a, b], H(x) = −∞ H 0 (t) dt = a H 0 (t) dt. But µF = µG + µH =
µG+H , so F = G + H a.e. and since G0 and H 0 exist a.e., it follows that F 0 = G0 + H 0 = H 0 a.e., so
Z b
Z b
0
F (b) − F (a) = µF (a, b] = µG (a, b] + µH (a, b] ≥ µH (a, b] =
H (t) dt =
F 0 (t) dt.
a
a
3.37 Suppose F : R −→ C. There is a constant M s.t. |F (x) − F (y)| ≤ M |x − y| for all x, y ∈ R iff F
is AC and |F 0 | ≤ M a.e..
Solution:
Suppose first that such a constant M > 0 exists. Let > 0 and set δ = /M . Suppose {(ai , bi )}N
i=1 is a
PN
disjoint collection of open intervals for which i=1 (bi − ai ) < δ. Then
N
X
|F (bi ) − F (ai )| ≤ M
N
X
(bi − ai ) < M δ = ,
i=1
i=1
so F is absolutely continuous. Furthermore, for all h > 0,
|F (x + h) − F (x)|
|F (x + h) − F (x)|
=
≤ M,
h
(x + h) − x
so |F 0 (x)| ≤ M for all x ∈ R.
Conversely, suppose F is AC and |F 0 | ≤ M a.e.. Let x, y ∈ R with X < y. Then
Z y
Z y
Z y
0
0
|F (y) − F (x)| =
F (t) dt ≤
|F (t)| dt ≤
M dt = M |y − x|.
x
x
x
P
3.39 If {Fj } is a sequence of nonnegative
Fj (x) < ∞
P 0 increasing functions on [a, b] so that F (x) =
0
for all x ∈ [a, b], then F (x) = Fj (x) for almost all x ∈ [a, b].
Solution:
As per the hint, it suffices to assume (by properly extending the Fj ’s and F to R) that the Fj ’s and F are
NBV. Let µF and µFj be the associated Borel measures. These are positive σ-finite Borel measures, so they
admit Lebesgue-Radon-Nikodym decompositions:
dµF = dλ + f dm,
and dµFj = dλj + fj dm,
25
P
where all of these measures are positive and λ ⊥ m and λj ⊥ m. Since µF and
µFj agree on h-intervals,
they are equal, and so
X
X
X
dµF =
dµFj =
dλj +
fj dm .
P
P
Furthermore, since ( dλj ) ⊥ m and ( fj dm) m by Exercise
3.8, it follows
P
Pfrom the uniqueness
assertion
in
the
Lebesgue-Radon-Nikodym
Theorem
that
dλ
=
dλ
and
f
dm
=
fj dm; in particular,
j
P
f=
fj m-almost everywhere. Thus, with Er = (x, x + r] we have by Theorem 3.22 for m-almost all x,
F 0 (x) = lim
r→0
But for almost all x, f (x) =
µF Er
F (x + r) − F (x)
= lim
= f (x).
r→0 mEr
r
P
fj (x) and another application of Theorem 3.22 yields for almost all x that
X
X
X
µF Er
f (x) =
fj (x) =
lim j
=
Fj0 (x).
r→0 mEr
4.3 Every metric space is normal.
Solution:
Suppose (X, ρ) is a metric space. If x, y ∈ X with x 6= y then r = ρ(x, y) > 0 and B(r/2, x) is an open set
containing x but not y, so X is T1 .
Now suppose A, B are disjoint closed subsets of X. Let U = {x ∈ X : ρ(x, A) < ρ(x, B)} and
V = {x ∈ X : ρ(x, B) < ρ(x, A)}. We claim that U is open. If x ∈ U then r = ρ(x, B) − ρ(x, A) > 0 and
B(r/2, x) ⊂ U since ρ(z, x) < r/2 implies
ρ(z, A) ≤ ρ(z, x) + ρ(x, A) < r/2 + ρ(x, A) =
ρ(x, B) − ρ(x, A)
ρ(x, B) + ρ(x, A)
+ ρ(x, A) =
< ρ(x, B).
2
2
The same argument applied to V shows that V is open. By construction, we have that U ∩ V = ∅ and
A ⊂ U , B ⊂ V , so indeed the space is normal.
4.5 Every separable metric space is second countable.
Solution:
Suppose (X, ρ) is separable and {xn } is a countable dense subset. Let
= {B(1/m, xn ) : m, n ∈ N}.
Certainly
is a countable collection of open sets, and we claim it is a base for the metric space topology
on X. To show this, we will appeal to Proposition 4.2.
Let U ⊂ X be an open set. For each y ∈ U there is an ry > 0 so that B(ry , y) ⊂ U
and there
exists ny
B
B
so that ρ(xny , y) < ry /2. Let my ∈ N such that
z ∈ B m1y , xny we have that
1
my
<
ry
2
− ρ(xny , y). Then xny ∈ B
1
my , xny
ρ(y, z) ≤ ρ(y, xny ) + ρ(xny , z)
ry
1
<
+
< ry − ρ(xny , y) < ry ,
2
my
S
so z ∈ B(ry , y) ⊂ U and hence B m1y , xny ⊂ U . Therefore, U = y∈U B m1y , xny , so
metric space topology on X by Prop. 4.2.
26
and for all
B is a base for the
Spr.2012 prelim # 8: Suppose f is continuous and real-valued on [a, b] and
that f ≡ 0 on [a, b].
Solution:
Rb
a
xn f (x) dx = 0 for all n = 0, 1, . . . . Show
Let > 0. Since polynomial functions are dense in C([a, b], R), there is a polynomial p(x) such that
Rb
kf − pku < . From the hypothesis, it follows by linearity that a p(x)f (x) dx = 0, so
b
Z
Z
b
Z
f (x)2 dx,
a
a
a
b
(p(x) − f (x))f (x) dx +
p(x)f (x) dx =
0=
and so
Z
b
2
Z
Z
b
Z
a
a
b
|f (x)| dx.
|f (x) − p(x)||f (x)| dx ≤ (f (x) − p(x))f (x) dx ≤
f (x) dx =
a
b
a
Rb
Rb
Since > 0 is arbitrary and f is fixed and a |f (x)| dx < ∞, it follows that a f (x)2 dx = 0, so f = 0 a.e.
on [a, b]. But since f is continuous, this implies f ≡ 0 on [a, b].
5.1 If X is a normed vector space over K (= R or C) then addition and scalar multiplication are
continuous from X × X and K × X to X. Moreover, the norm is continuous from X to [0, ∞); in
fact, kxk − kyk ≤ kx − yk.
Solution:
Let A : X × X −→ X be given by A(x1 , x2 ) = x1 + x2 . Then A is a linear map from the NVS X × X to
the NVS X. Furthermore, for all (x1 , x2 ) ∈ X × X,
kA(x1 , x2 )k = kx1 + x2 k ≤ kx1 k + kx2 k ≤ 2 max{kx1 k, kx2 k} = 2k(x1 , x2 )k,
so A is bounded, hence continuous, by Prop. 5.2.
Since K is a NVS over itself, K × X is a NVS and M : K × X −→ X given by M (λ, x) = λx is a linear
map. We will show that M is continuous at (0, 0), and result then follows from Prop. 5.2. Let > 0, and
δ = min{, 1}. Then for all (λ, x) ∈ K × X with k(λ, x)k < δ we have max{|λ|, kxk} < δ ≤ and δ 2 ≤ δ, so
kM (λ, x)k = kλxk = |λ|kxk < δ 2 ≤ δ ≤ ,
so M is continuous at (0, 0).
Finally we show that k · k is continuous. Let > 0, set δ = and suppose x, y ∈ X such that kx − yk < δ.
Then kxk = kx − y + yk ≤ kx − yk + kyk, so kxk − kyk ≤ kx − yk. Also, kyk = ky − x + xk ≤ ky − xk + kxk,
so kyk − kxk ≤ ky − xk = kx − yk, and therefore kxk − kyk ≤ kx − yk < , so k · k is uniformly continuous
on X.
5.2 L(X, Y ) is a vector space and the function k · k defined by (5.3) is a norm on it.
Solution:
The space Y X of all functions from X to Y is a vector space, so it suffices to show that L(X, Y ) is nonempty,
closed under addition and closed under scalar multiplication (that is, that L(X, Y ) is a subspace of Y X ).
Certainly the function Z(x) = 0 is in L(X, Y ) so it’s nonempty. If T1 , T2 ∈ L(X, Y ) then there are positive
constants C1 , C2 such that kT1 xk ≤ C1 kxk and kT2 xk ≤ C2 kxk for all x ∈ X. Then T1 + T2 is linear and for
all x ∈ X,
k(T1 + T2 )xk = kT1 x + T2 xk ≤ kT1 xk + kT2 xk ≤ (C1 + C2 )kxk,
so T1 + T2 is bounded, hence T1 + T2 ∈ L(X, Y ). If α ∈ K then for all x ∈ X we have
k(αT1 )xk = kα(T1 x)k = |α|kT1 xk ≤ (|α|C1 )kxk,
27
so αT ∈ L(X, Y ). Therefore L(X, Y ) is a vector space.
By definition, kT k ≥ 0 for all T ∈ L(X, Y ) and kT k = 0 iff kT xk ≤ 0kxk = 0 for all x ∈ X, so kT k = 0
iff kT k = 0. Suppose T1 , T2 ∈ L(X, Y ). Then for all x ∈ X,
k(T1 + T2 )xk ≤ kT1 xk + kT2 xk ≤ kT1 kkxk + kT2 kkxk = (kT1 k + kT2 k)kxk,
so kT1 + T2 k ≤ kT1 k + kT2 k. Finally, if α ∈ K then
kαT1 k
=
sup{k(αT1 )xk : kxk = 1}
=
sup{|α|kT1 xk : kxk = 1}
= |α| sup{kT1 xk : kxk = 1}
= |α|kT1 k,
so k · k is a norm on L(X, Y ).
5.3 Complete the proof of Prop. 5.4.
Solution:
We have that {Tn } is a Cauchy sequence in L(X, Y ) and T x = lim Tn x for all x ∈ X. We need to show that
T ∈ L(X, Y ) and kTn − T k → 0. If x1 , x2 ∈ X and α1 , α2 ∈ K then
T (α1 x1 + α2 x2 ) = lim Tn (α1 x1 + α2 x2 )
n→∞
=
=
lim (α1 Tn x1 + α2 Tn x2 )
n→∞
α1 lim Tn x1 + α2 lim Tn x2 ,
n→∞
n→∞
since addition and scalar multiplication on X are continuous by Problem 1. But the last expression equals
α1 T x1 + α2 T x2 , so T is linear.
Since {Tn } is Cauchy in L(X, Y ) and kTm − Tn k ≥ kTm k − kTn k , it follows that {kTn k} is Cauchy in
R, hence convergent. From continuity of the norm on X we have that
kT xk = k lim Tn xk = lim kTn xk ≤ kxk lim kTn k,
n→∞
n→∞
n→∞
so T is bounded, and hence T ∈ L(X, Y ).
Let > 0 and N ∈ N such that kTn − Tm k < for all m, n ≥ N . Then for all x ∈ X with kxk = 1 we
have that for all n ≥ N ,
k(Tn − T )xk = kTn x − T xk
=
=
lim kTn x − Tm xk
m→∞
lim k(Tn − Tm )xk ≤ kxk lim kTn − Tm k ≤ .
m→∞
m→∞
Therefore, kTn − T k ≤ for all n ≥ N , and so kTn − T k → 0. In particular, since kTn k − kT k ≤ kTn − T k,
we also have that kTn k → kT k.
5.6 Suppose
that XPis a finite dimensional vector space. Let e1 , . . . , en be a basis for X and define
Pn
n
k 1 aj ej k1 = 1 |aj |.
a. k · k1 is a norm on X. P
n
b. The map (a1 , . . . , an ) 7→ 1 aj ej is continuous from K n with the usual Euclidean topology to
X with the topology defined by k · k1 .
c. {x ∈ X : kxk1 } is compact in the topology defined by k · k1 .
d. All norms on X are equivalent.
Solution:
P
a. It’s clear that for all x ∈ X, kxk1 ≥ 0 and kxk1 = 0 iff x = 0. If x =
aj ej and α ∈ K then
X
X
X
kαxk1 =
αaj ej =
|αaj | = |α|
|aj | = αkxk1 .
1
28
Finally, if x =
P
aj ej and y =
P
bj ej then
X
X
kx + yk1 =
|aj + bj | ≤
(|aj | + |bj |) = kxk1 + kyk1 ,
so k · k1 is a norm on X.
1/2
b. If we endow the vector space K n with the usual norm, k(a1 , . . . , an )k = |a1 |2 + · · · + |an |2
, then
n
n
K n is a NVS and
this
norm
induces
the
Euclidean
toplogy
on
K
.
The
function
L
:
K
−→
X
given
by
P
L(a1 , . . . , an ) = aj ej is linear, so by Prop. 5.2 is suffices to show that L is continuous at 0. Let > 0 and
set δ = /n. Let a ∈ K n with kak < δ. Then
δ > kak = |a1 |2 + · · · + |an |2
1/2
⇒ δ 2 > |a1 |2 + · · · + |an |2 ⇒ δ 2 > |aj |2 for all 1 ≤ j ≤ n,
Pn
so /n > |aj | for all j and hence kL(a)kP
1 =
1 |aj | < , so L is continuous at zero, and hence continuous.
n
c. Let F = {(a1 , . . . , an ) ∈ K n :
|aj | = 1}. Then F is a bounded
P subset of KP and we will show
c
c
that it is closed by showing
that F is open. If b ∈ F then either
|bj | < 1 or
|bj | > 1. In the
P
first case, let = 1 −P |bj | > 0. P
Then for all c ∈ K n with kc − bk < /n we have |cj | P
< |bj | + /n for
all j and so kck1 =
|cj | P
< +
|bj | = 1. Therefore, B(/n, b) ⊂ F c . Similarly, if
|bj | > 1 then
B(0 /n, b) ⊂ F c where 0 =
|bj | − 1. So F c is open and hence F is closed. Since K n = Rn or K n = Cn
and F ⊂ K n is closed and bounded, F is compact. Since the function L in part (b) is continuous, it follows
that L(F ) = {x ∈ X : kxk1 = 1} is compact in X.
d. Let k · k be an arbitrary norm on X. Let C2 = max{ke1 k, . . . , ken k} and x ∈ X. Then
kxk = kx1 e1 + · · · + xn en k ≤ |x1 |ke1 k + · · · + |xn |ken k ≤ |x1 |C2 + · · · + |xn |C2 = C2 kxk1 .
We claim now that k · k is continuous with respect to the topology induced by k · k1 . Let x ∈ X and
> 0. With δ = /C2 we have that for all y ∈ X, if ky − xk1 < δ then ky − xk ≤ C2 ky − xk1 < , so k · k is
continuous at x, and hence continuous on X (in fact, uniformly continuous).
By part (c), the set B = {x ∈ X : kxk1 = 1} is compact in the k · k1 topology, and since k · k is
continuous with respect to the k · k1 topology, it has a minimum on B, say C1 . Since 0 6∈ B, we have C1 > 0
and so for all 0 6= x ∈ X,
x
≥ C1 ,
so
kxk ≥ C1 kxk1 .
kxk1
Therefore, k · k is equivalent to k · k1 .
5.7 Let X be a Banach space.
a. If T ∈ L(X,
P∞ X) and kI − T k < 1 where I is the identity operator, then T is invertible; in fact,
the series 0 (I − T )n converges in L(X, X) to T −1 .
b. If T ∈ L(X, X) is invertible and kS − T k < kT −1 k−1 , then S is invertible. Thus the set of
invertible operators is open in L(X, X).
Solution:
2
Notice that for all A ∈ L(X, X) and all x ∈ X, kA2 xk = kA(Ax)k ≤ kAkkAxk ≤ kAk
kxk, so kA2 k ≤
P∞
2
n
n
n
kAk
,
and
inductively,
kA
k
≤
kAk
for
all
n
≥
1.
Since
kI
−
T
k
<
1,
it
follows
that
0 k(I − T ) k ≤
P∞
P∞
n
n
kI
−
T
k
<
∞,
so
the
series
(I
−
T
)
is
absolutely
convergent,
and
hence
convergent
by
Theorem
0
0
5.1 since L(X, X) is a Banach space by Prop 5.4.
PN
For N ≥ 0 let SN = n=0 (I − T )n . Then
T SN = (I − (I − T ))SN = SN −
N
X
(I − T )n+1 = I − (I − T )N +1 .
n=0
Since T is continuous,
T ( lim SN ) = lim T SN = I − lim (I − T )N +1 .
N →∞
N →∞
N →∞
29
But
lim (I − T )N +1 = lim k(I − T )N +1 k ≤ lim kI − T kN +1 = 0,
N →∞
N →∞
N +1
so limN →∞ (I − T )
N →∞
= 0. Therefore
I = T ( lim SN ) = T
N →∞
∞
X
(I − T )n ,
(0.4)
n=0
P∞
P∞
and so, if T is bijective, then n=0 (I − T )n = T −1 . We have already that n=0 (I − T )n ∈ L(X, X), so it
remains only to show that T is bijective.
Suppose BWOC that x ∈ ker T and x 6= 0. Then
kxk = k(I − T )xk ≤ kI − T kkxk < kxk,
a contradiction, so ker T = {0} and T is therefore one-to-one. It follows immediately from (0.4) that T is
surjective.
b. kST −1 − Ik = k(S − T )T −1 k ≤ kS − T kkT −1 k < 1, so by part (a), ST −1 is invertible, say A =
(ST −1 )−1 . Then S(T −1 A) = (ST −1 )A = I, and since T −1 A ∈ L(X, X), S is invertible.
5.13 Suppose k · k is a seminorm on X and let M = {x ∈ X : kxk = 0}. Then M is a subspace of X
and the map x + M 7→ kxk is a norm on X/M.
Solution:
M is clearly nonempty since 0 ∈ M. Suppose x1 , x2 ∈ M. Then kx1 + x2 k ≤ kx1 k + kx2 k = 0, so
kx1 + x2 k = 0 and hence x1 + x2 ∈ M. Additionally, if α ∈ K then kαx1 k = |α|kx1 k = 0, so αx1 ∈ M and
therefore M is a subspace of X.
Denote the map in question by kx + Mk = kxk. We first need to see that this map is well-defined on
X/M. Suppose x1 + M = x2 + M. Then x1 − x2 ∈ M, so kx1 − x2 k = 0. It follows that
kx1 k
= kx1 − x2 + x2 k ≤ kx1 − x2 k + kx2 k = kx2 k,
kx2 k
= kx2 − x1 + x1 k ≤ kx2 − x1 k + kx1 k = kx1 k,
so kx1 k = kx2 k and therefore the map is well-defined.
Finally, to see that this is a norm on X/M, suppose x1 + M, x2 + M ∈ X/M. Then
k(x1 + M) + (x2 + M)k = k(x1 + x2 ) + Mk = kx1 + x2 k ≤ kx1 k + kx2 k = kx1 + Mk + kx2 + Mk.
If α ∈ K then kα(x1 + M)k = kαx1 + Mk = kαx1 k = |α|kx1 k = |α|kx1 + Mk. Finally, if kx + Mk = 0 then
kxk = 0, so x ∈ M and hence x + M = 0 + M, so k · k is a norm on x/M.
5.17 A linear functional f on a NVS X is bounded iff f −1 ({0}) is closed.
Solution:
Suppose f is bounded. Since {0} is closed in K and f is continuous, f −1 ({0}) is closed in X.
(A hint is given to use Exercise 12(b) for the converse, but I don’t immediately see how to use it; so here
is a direct proof instead.)
Suppose that f −1 ({0}) is closed. If f −1 ({0}) = X, then f = 0 and hence bounded. So assume there is
an x0 ∈ X such that f (x0 ) 6= 0. By scaling, we may assume that f (x0 ) = 1, and by linearity we have that
f −1 ({1}) = x0 + f −1 ({0}), so that f −1 ({1}) is closed. BWOC assume that f is not bounded. Then for
xn
each n ∈ N there exists xn ∈ X such that kxn k = 1 and |f (xn )| > n. Set yn = f (x
so that f (yn ) = 1 and
n)
−1
kyn k < 1/n. Since {yn } is contained in the closed subset f ({1}) and yn → 0, it follows that 0 ∈ f −1 ({1}),
a contradiction. Therefore f is bounded.
30
5.19(a) Let X be an infinite dimensional vector space. There is a sequence {xj } in X such that kxj k = 1
for all j and kxj − xk k ≥ 1/2 for j 6= k.
Solution:
Let x1 be an arbitrary element of X with kx1 k = 1. We inductively construct xn as follows: suppose that
x1 , . . . , xn−1 have been constructed which satisfy the conclusions. Let M = Span {x1 , . . . , xn−1 }. By Ex.
18b, M is closed. Since X is infinite dimensional, M is a proper subspace. So, by Ex. 12b with = 1/2,
there is an element xn ∈ X such that kxn k = 1 and kx + Mk ≥ 1/2. In particular, since −xi ∈ M for
1 ≤ i < n, it follows that
kxn − xi k ≥ inf{kx + yk : y ∈ M} = kxn + Mk ≥ 1/2,
for all 1 ≤ i < n.
5.22 Suppose X, Y are NVS and T ∈ L(X, Y ).
a. Define T † : Y ∗ −→ X ∗ by T † f = f ◦ T . Then T † ∈ L(Y ∗ , X ∗ ) and kT † k = kT k.
b. Applying the construction in (a) twice, one obtains T †† ∈ L(X ∗∗ , Y ∗∗ ). If X and Y are
b Yb in X ∗∗ , Y ∗∗ , then T †† |X = T .
identified with their natural images X,
†
c. T is injective iff the range of T is dense in Y .
d. If the range of T † is dense in X ∗ then T is injective; the converse is true if X is reflexive.
Solution:
a. It’s straightforward to verify that T † is linear. To see that it’s bounded, let g ∈ Y ∗ . Then for all x ∈ X
we have
k(T †g)(x)k = k(g ◦ T )(x)k = kg(T (x)k ≤ kgkkT (x)k ≤ kgkkT kkxk,
so kT † gk ≤ kgkkT k, and hence kT † k ≤ kT k, so T † is bounded.
Claim: kT † k ≥ kT xk for all x ∈ X with kxk = 1. If T x = 0 the claim holds trivially, so assume x0 ∈ X
with kx0 k = 1 and T x0 6= 0. Set y0 = T x0 . By Thm 5.8b, there exists g0 ∈ Y ∗ such that kg0 k = 1 and
g0 (y0 ) = ky0 k. Then
kT † k
=
sup{kT † gk : kgk = 1}
≥
kT † g0 k = kg0 ◦ T k
=
sup{k(g0 ◦ T )(x)k : kxk = 1}
≥
kg0 (T (x0 ))k = kg0 (y0 )k = ky0 k = kT x0 k.
This proves the claim and it follows that kT † k ≥ sup{kT xk : kxk = 1} = kT k, hence kT † k = kT k.
b. Let x ∈ X. Then T †† x
b=x
b ◦ T † . For all g ∈ Y ∗ we have that
(b
x ◦ T † )(g) = x
b(T † g) = (T † g)(x) = g(T x).
b and Y with Yb .
Therefore T †† = x
b ◦ T † = Tcx, so T †† |X = T , after identification of X with X
∗
†
c. Suppose first that T (X) is dense in Y . Suppose g ∈ Y and T g = 0. Then for all x ∈ X,
0 = (T † g)(x) = g(T (x)). Since g is continuous and vanishes on the dense subset T (X) of Y , it is identically
zero, hence T † is injective.
Conversely, suppose T † is injective. BWOC, suppose that T (X) is not dense in Y . Then M = T (X)
is a closed proper subspace of Y by Ex. 5.5. Let y0 ∈ Y − M, By Thm 5.8a, there exists g ∈ Y ∗ such
that g|M = 0 and g(y0 ) 6= 0. Then for all x ∈ X, (T † g)(x) = g(T (x)) = 0, so T † g = 0 ∈ X ∗ . But since
g 6= 0 ∈ Y ∗ , it follows that T † is not injective, a contradiction.
d. If the range of T † is dense in X ∗ , then by part (c), T †† is injective. Under the (bijective) identification
in part (b), T †† |X = T , so T is injective.
b = X ∗∗ so T †† = T after identification of X with
Suppose X is reflexive and T is injective. Then X
∗∗
††
b
b
X = X and Y with Y . Since T is injective, by part (c), the range of T † is dense in X ∗ .
5.27 There exist meager subsets of R whose complements have Lebesgue measure zero.
Solution:
For each n ≥ 1, there is an open subset On ⊂ R such that Q ⊂ On and mOn < 1/n. Set Fn = Onc . Then
31
each Fn is closed and Fn ∩ Q = ∅, so Fno = ∅, hence Fn is nowhere dense. Let X =
countable union of nowhere dense sets, hence meager, and




\
\
mX c = m 
Fnc  = m 
On  = 0.
n≥1
S
n≥1
Fn . Then X is a
n≥1
Extra prob. If X, Y are NVS’s and T ∈ L(X, Y ) then Γ(T ) is closed in X × Y .
Solution:
We will show that the complement is open. Let (x0 , y0 ) ∈ Γ(T )c . Then r = kT x0 − y0 k > 0. Let δ 0 > 0
such that kxk < δ 0 implies kT xk < r/3 and set δ = min{δ 0 , r/3}. Then for all (x, y) ∈ X × Y such that
k(x, y) − (x0 , y0 )k < δ we have kx − x0 k < δ and ky − y0 k < δ, so
kT x − yk
= kT (x − x0 ) + T x0 − (y − y0 ) − y0 k
≥
kT x0 − y0 k − kT (x − x0 )k − ky − y0 k
= r − kT (x − x0 )k − ky − y0 k
≥ r − r/3 − r/3 = r/3.
Therefore B(δ, (x0 , y0 )) ⊂ Γ(T )c and so Γ(T )c is open, hence Γ(T ) is closed.
5.32 Let k · k1 , k · k2 be norms on a vector space X such that k · k1 ≤ k · k2 . If X is complete with
respect to both norms then the norms are equivalent.
Solution:
Let X1 = (X, k · k1 ), X2 = (X, k · k2 ) and define T : X2 −→ X1 by T (x) = x. Clearly T is linear and for all
x ∈ X, kT xk1 = kxk1 ≤ kxk2 , so kT k ≤ 1. Thus, T ∈ L(X2 , X1 ) and T is bijective, so by Corollary 5.11,
T is an isomorphism, and hence T −1 ∈ L(X1 , X2 ). In particular, T −1 is bounded, so there is a C > 0 such
that for all x ∈ X, kxk2 = kT −1 xk2 ≤ Ckxk1 , and therefore the norms are equivalent.
5.54 For every nonempty set A, `2 (A) is complete.
Solution:
Notice first that for f ∈ `2 (A),
kf k2 = hf, f i =
X
f (a)f (a) =
a∈A
X
|f (a)|2 .
a∈A
Suppose {fn }n≥1 is a Cauchy sequence in `2 (A). Then for each > 0 there is an N ∈ N such that for all
m, n ≥ N
X
2 > kfm − fn k2 =
|fm (a) − fn (a)|2 ,
a∈A
and in particular, for each a0 ∈ A we have > |fm (a0 ) − fn (a0 )| for all m, n ≥ N . Therefore {fn (a0 )} is a
Cauchy sequence in C for each a0 ∈ A, so we define f (a) = limn→∞ fn (a) for all a ∈ A.
Claim: f ∈ `2 (A). Since {fn } is Cauchy and kfm k − kfn k ≤ kfm − fn k for all m, n ≥ 1, it follows that
kfn k is Cauchy in R, hence convergent. By continuity of the norm, kf k = limn→∞ kfn k < ∞, so f ∈ `2 (A).
Furthermore,
lim kf − fn k = k lim (f − fn )k = 0,
n→∞
n→∞
2
so fn → f in ` (A).
32
5.55 Let H be a Hilbert space.
a. (The polarization identity) For all x, y ∈ H,
hx, yi =
1
kx + yk2 − kx − yk2 + ikx + iyk2 − ikx − iyk2 .
4
b. If H 0 is another Hilbert space, a linear map from H to H 0 is unitary iff it is isometric and
surjective.
Solution:
For (a), fix x, y ∈ H and let α denote the RHS of the given equation. Then
α
=
1
(hx + y, x + yi − hx − y, x − yi + ihx + iyi − ihx − iy, x − iyi)
4
1
i
(2hx, yi + 2hy, xi) + (2hx, iyi + 2hiy, xi)
4
4
i
1
hx, yi + hx, yi +
hx, iyi + hx, iyi
2
2
1
i
(2 Re hx, yi) + (2 Re hx, iyi)
2
2
Re hx, yi + i Re (−ihx, yi)
=
Re hx, yi + i Im hx, yi = hx, yi.
=
=
=
=
For (b), suppose that U : H −→ H 0 is unitary. Then U is surjective by definition and for all x ∈ H,
kU xk2 = hU x, U xi = hx, xi = kxk2 , so U is an isometry.
Conversely, suppose U : H −→ H 0 is a surjective isometry. Then U is one-to-one since U x = 0 ⇔ kU xk =
0 ⇔ kxk = 0 ⇔ x = 0. So U is bijective and therefore has a linear inverse. For each y ∈ H 0 there is a unique
x ∈ H such that U x = y, so that kyk = kU xk = kxk, and so kU −1 yk = kxk = kyk, so U −1 is a bounded
linear map. Finally, to see that U is unitary it remains only to show that it preserves the inner product. For
all x, y ∈ H we have by (a) that
1
kU x + U yk2 − kU x − U yk2 + ikU x + iU yk2 − ikU x − iU yk2
4
1
kU (x + y)k2 − kU (x − y)k2 + ikU (x + iy)k2 − ikU (x − iy)k2
=
4
1
=
kx + yk2 − kx − yk2 + ikx + iyk2 − ikx − iyk2
4
= hx, yi.
hU x, U yi =
6.2 a. If f and g are measurable functions on X then kf gk1 ≤ kf k1 kgk∞ . If f ∈ L1 and g ∈ L∞ ,
kf gk1 = kf k1 kgk∞ iff |g(x)| = kgk∞ a.e. on the set where f (x) 6= 0.
b. k · k∞ is a norm on L∞ .
c. kfn − f k∞ → 0 iff there exists E ∈ M such that µE c = 0 and fn → f uniformly on E.
d. L∞ is a Banach space.
e. The simple functions are dense in L∞ .
Solution:
R
R
a. Since |g| ≤ kgk∞ a.e., kf gk1 = |f ||g| ≤ |f |kgk∞ = kf k1 kgk∞ . Let E = {x ∈ X : f (x) 6= 0}.
It’s clear that if |g(x)|
R = kgk∞
R a.e. on E then kf gk1 = kf k1 kgk∞ , so we will show the converse. Suppose
f ∈ L1 , g ∈ L∞ and |f g| = |f |kgk∞ . Then
Z
Z
Z
0 = (|f |kgk∞ − |f g|) = |f |(kgk∞ − |g|) =
|f |(kgk∞ − |g|),
E
since kgk∞ − |g| is finite on X. Furthermore, since kgk∞ − |g| ≥ 0 a.e., the integrand is nonegative a.e., so
|f |(kgk∞ − |g|) = 0 a.e. on E, and hence kgk∞ = |g| a.e. on E since f 6= 0 on E.
33
b. The only norm property which is not immediately obvious is the triangle inequality, so suppose
f, g ∈ L∞ . Then |f | ≤ kf k∞ a.e. and |g| ≤ kgk∞ a.e. imply |f |+|g| ≤ kf k∞ +kgk∞ a.e., but |f +g| ≤ |f |+|g|,
so |f + g| ≤ kf k∞ + kgk∞ a.e., hence kf + gk∞ ≤ kf k∞ + kgk∞ .
c. Suppose kfn − f k∞ → 0. Then for all k ∈ N there is an Nk so that kfn − f k∞ ≤ 1/k for all n ≥ Nk .
So, for
Ek ∈ M such that µEkc = 0 and |fn − f | ≤ 1/k on Ek for all n ≥ Nk . Set
T∞each k ∈ N therec is an S
∞
E = k=1 Ek . Then µE = µ ( k=1 Ekc ) = 0.
We will now show that fn → f uniformly on E. Let > 0. Then there is a k ∈ N such that 1/k ≤ and
for all n ≥ Nk we have |fn − f | ≤ 1/k on Ek for all n ≥ Nk . Since 1/k ≤ and E ⊂ Ek , it follows that
|fn − f | ≤ on E for all n ≥ Nk . The converse is clear.
d. The only thing that remains to show is that L∞ is complete. Suppose that {fn } is a Cauchy sequence
in L∞ . For each k ∈ N there exists Nk such that kfm − fn k∞ < 1/k T
for all m, n ≥ Nk . Let Ek ∈ M so
∞
that |fm − fn | < 1/k on Ek for all m, n ≥ Nk and µEkc = 0. Set E = k=1 Ek . Then µE c = 0 and for all
x ∈ E, {fn (x)} is Cauchy, so let f (x) = limn→∞ fn (x)χE(x). Then fn → f uniformly on E, so by part (c),
kfn − f k∞ → 0. In particular, there is an N so that kfN − f k∞ < 1 and hence kf k∞ ≤ kfN k + 1 < ∞, so
f ∈ L∞ and L∞ is therefore complete.
e. Let f ∈ L∞ and > 0. It suffices to show that there is a simple function ϕ ∈ L∞ with kf − ϕk∞ < .
Suppose first that f is real-valued. Let n ∈ N such that 1/n < and let M = kf k∞ . For each integer
j ∈ [−M n, M n] let
j+1
j
≤ f (x) ≤
,
Ej = x ∈ X :
n
n
PmN
and set ϕ = j=−M n (j/n)χE . Then ϕ ∈ L∞ is simple and for all x with |f (x)| ≤ M there is a unique j
j
such that x ∈ Ej and so |f (x) − ϕ(x)| ≤ 1/n < . Thus, |f − ϕ| < a.e., so kf − ϕk∞ < .
For the general case, let ϕ1 , ϕ2 be simple functions for which k Re (f )−ϕ1 k∞ < /2 and k Im (f )−ϕ2 k∞ <
/2. Then ϕ = ϕ1 + iϕ2 is a simple function in L∞ and kf − ϕk∞ = k Re (f ) − ϕ1 + i( Im (f ) − ϕ2 )k∞ ≤
k Re (f ) − ϕ1 k∞ + ki( Im (f ) − ϕ2 )k∞ < .
6.5 Suppose 0 < p < q < ∞. Then Lp 6⊂ Lq iff X contains sets of arbitrarily small positive measure.
Lq 6⊂ Lp iff X contains sets of arbitrarily large finite measure. What about the case q = ∞?
Solution:
Suppose X contains sets of arbitrarily small positive measure. Let F1 ⊂ X be a measurable set with
0 < µF1 < S1/2. Inductively, let Fn+1 be a measurable set with 0 < µFn+1 < (1/4)µFn . For n ≥ 1 set
∞
En = Fn − k=n+1 Fk . Then for i < j,
Ei ∩ Ej = Fi ∩
∞
\
k=i+1
!
Fkc

∩ Fj ∩ 
∞
\

Fkc 
k=j+1
⊂
∞
\
!
Fkc
∩ Fj = ∅ (since )i + 1 ≤ j.
k=i+1
P∞
Furthermore µEn ≥ µFn − k=n+1 µFk ≥ µFn − (1/2)µFn > 0 and µEn ≤ µFn < 2−n . Therefore {En } is
a disjoint sequence
0 < µEn < 2−n .
P∞ with
1
Set ϕ = n=1 (nµEn )1/q χE . Since the En ’s are disjoint, this sum converges everywhere on X. Finally,
n
we have
Z
∞
∞
X
X
µEn
1
q
q
kϕkq = |ϕ| =
=
= ∞,
nµE
n
n
n=1
n=1
so ϕ 6∈ Lq . But
kϕkpp
Z
=
∞
∞
∞
X
X
X
µEn
(µEn )1−p/q
1−p/q
|ϕ| =
=
<
(µEn )
<
2−n(1−p/q) < ∞,
(nµEn )p/q
np/q
n=1
n=1
n=1
n=1
p
∞
X
since 21−p/q > 1. Therefore ϕ ∈ Lp and so Lp 6⊂ Lq .
34
Conversely, suppose Lp 6⊂ Lq and let f ∈ Lp with f 6∈ Lq . For each positive integer n let En = {x ∈ X :
|f (x)| > n}. By Prop. 6.10, if f ∈ L∞ then f ∈ Lq , so we must have kf k∞ = ∞ and therefore µEn > 0 for
all n ≥ 1. On the other hand, for each n ≥ 1,
Z
Z
Z
p
p
p
kf kp = |f | ≥
|f | ≥
np = np µEn ,
En
En
kf kp
p
np
so limn→∞ µEn ≤ limn→∞
= 0, so µEn → 0 as n → ∞.
For the second question, suppose first that X contains sets of arbitrarily
Pn large finite measure. Let
F1 ⊂ X with 1 ≤SµF1 < ∞. Inductively, let Fn+1 ⊂ X with ∞ > µFn+1 > k=1 µFk + P
1. Set E1 = F1 and
n
n
En+1P= Fn+1 − k=1 Fk for n ≥ 1. The sequence {En } is disjoint and µEn ≥ µFn+1 − k=1 µFk > 1. Let
∞
1
f = n=1 (nµEn )1/p χE . Then
n
Z
∞
X
µEn
kf kpp = |f |p =
= ∞,
nµE
n
n=1
so f 6∈ Lp . But
kf kqq =
Z
|f |q =
∞
∞
X
X
(µEn )1−q/p
1
µEn
=
<
< ∞,
q/p
q/p
q/p
(nµE
)
n
n
n
n=1
n=1
n=1
∞
X
so f ∈ Lq and hence Lq 6⊂ Lp .
Conversely,
suppose Lq 6⊂ Lp and let f ∈ Lq − Lp . Let En = {x ∈ X : |f (x)| > 1/n}. Since
R
q
q
n
kf kq = |f | > µE
all n ≥ 1, it follows that µEn ≤ nq kf kqq < ∞ for all n. Furthermore, the sets En are
nq for S
increasing, so with E = n≥1 En we have µE = limn→∞ µEn . But E = {x ∈ X : f (x) 6= 0}, so if µE < ∞
then by Prop. 6.12 applied to the measure space (E, M|E , µ|E ), we’d have kf kp < ∞, a contradiction. Thus
limn→∞ µEn = ∞. (Alternatively, apply Hölder as in the proof of Theorem 6.12 to the function f χE.)
What about the case q = ∞? The first assertion continues to hold with q = ∞. Suppose X contains
−n
sets of arbitrarily
small positive measure. As P
above, let {EP
.
n } be a disjoint sequence with 0 < µEn < 2
P
∞ np
p
∞
p
p
p
n µEn < n=1 2n < ∞, so f ∈ L and hence L 6⊂ L∞ .
Set f =
nχE . Then f 6∈ L but kf kp =
n
Conversely, if Lp 6⊂ L∞ then let f ∈ Lp − L∞ . Set En = {x ∈ X : |f (x)| ≥ n}. Since f 6∈ L∞ we have
R
kf kp
µEn > 0 for all n and kf kpp ≥ En |f |p ≥ np µEn so µEn ≤ npp → 0 as n → ∞, so X contains sets of
arbitrarily small positive measure.
For the second statement, the “only if” fails with q = ∞. Let X = {x0 }, M = {∅, X} and µ∅ = 0,
µX = ∞. The constant function f = 1 is in L∞ but not Lp , so L∞ 6⊂ Lp bu X does not contain sets
of arbitrarily large finite measure. The converse, however, does still hold with q = ∞. If X contains sets
of arbitrarily
large positive measure, let {En } be a disjoint sequence
withS1 ≤ µEn < ∞ for all n. Let
R
P∞
f = n=1 χE . Then |f | ≤ 1 on X so f ∈ L∞ , but kf kpp = |f |p = µ ( En ) = ∞, so f 6∈ Lp , hence
n
L∞ 6⊂ Lp .
6.6 Suppose 0 < p0 < p1 ≤ ∞. Find examples of functions f on (0, ∞) (with Lebesgue measure) such
that f ∈ Lp iff (a) p0 < p < p1 , (b) p0 ≤ p ≤ p1 , (c) p = p0 .
Solution:
Assume first that p1 < ∞. For (a), set f = x−1/p1 χ(0,1) + x−1/p0 χ[1,∞). Then
Z
p
Z
|f | =
0
1
dx
xp/p1
Z
+
1
∞
dx
.
xp/p0
Both integrals are nonnegative, so kf kp < ∞ iff both integrals are finite. The first is finite iff p < p1 and the
second is finite iff p > p0 , hence f ∈ Lp iff p0 < p < p1 .
For (b), set
1
1
f = 1/p
χ
+
χ
.
x 1 | log x|2/p1 (0,1/2) x1/p0 | log x|2/p0 [2,∞)
R
It is a calculus exercise to verify that |f |p < ∞ iff p0 ≤ p ≤ p1 . For (c), take p1 = p0 in the function
defined f defined in (b).
1
If p1 = ∞, for (a) take f = | log(x−1)|
χ[1,∞). For (b) take f = x1/p0 | log
χ
.
x1/p0
x|2/p0 [2,∞)
35
6.7 If f ∈ Lp ∩ L∞ for some p < ∞ so that f ∈ Lq for all q > p, then kf k∞ = limq→∞ kf kq .
Solution:
It’s clear if kf k∞ = 0, so assume kf k∞ > 0. Let 0 < < kf k∞ , and set A = {x ∈ X : |f (x)| > kf k∞ − }.
Then f ∈ Lp implies that µA < ∞, and by construction, µA > 0. For all q ≥ p,
1/q Z
1/q
Z
q
q
≥
|f |
≥ (kf k∞ − )(µA)1/q ,
|f |
kf kq =
A
so lim inf q→∞ kf kq ≥ (kf k∞ − ) lim inf q→∞ (µA)1/q = kf k∞ − . Since this holds for all > 0, it follows
that lim inf q→∞ kf kq ≥ kf k∞ .
For the reverse inequality, we appeal to Prop. 6.10 with r = ∞ to conclude that for all q > p, kf kq ≤
p/q
1−p/q
p/q
1−p/q
kf kp kf k∞ . Therefore, lim supq→∞ kf kq ≤ lim supq→∞ kf kp kf k∞
= kf k∞ .
6.10 Suppose 1 ≤ p < ∞. If fn , f ∈ Lp and fn → f a.e., then kfn − f kp → 0 iff kfn kp → kf kp .
Solution:
Suppose first that kfnRkp → kf kp . Then |fnR− f |p ≤ (|fn | + |f |)p ≤ (2 max{|f |, |fn |})p ≤ 2p (|f |p + |fn |p ).
By assumption, limn→∞ 2p (|f |p + |fn |p ) = (lim 2p (|f |p + |fn |p ) < ∞, so by the Generalized LDCT it
follows that
Z
Z
lim kf − fn kpp = lim |f − fn |p = lim |f − fn |p = 0.
The converse is easy: kf − fn kp → 0 implies that fn → f in Lp . Since the norm k · kp is continuous on Lp ,
it follows that limn→∞ kfn kp = k limn→∞ fn kp = kf kp .
6.12 If p 6= 2 the Lp norm does not arise from an inner product on Lp except in the trivial cases when
dim Lp ≤ 1.
Solution:
Let 1 ≤ p ≤ ∞, and suppose that (X, M, µ) is a measure space and dim Lp > 1. If every E ∈ M had
µE ∈ {0, ∞} then the dimension of Lp would be zero, so there must exist a set E1 ∈ M with 0 < µE1 < ∞.
Furthermore, if every E ∈ M with 0 < µE < ∞ satisfied µ(E4E1 ) = 0, then every function in Lp would
equal a constant multiple of χE a.e., which would give dim Lp = 1. Thus, there is a set E2 ∈ M with
1
0 < µE2 < ∞ and E1 ∩ E2 = ∅.
Suppose now that p < ∞. Set f1 = χE /(µE1 )1/p and f2 = χE /(µE2 )1/p . Then kf1 kp = kf2 kp = 1.
1
2
Suppose that the norm on Lp does arise from an inner product. Then the parallelogram law holds, so that
kf1 + f2 k2p + kf1 − f2 k2p = 2(kf1 k2p + kf2 k2p ) = 4.
Computing the LHS we have
kf1 +
f2 k2p
+ kf1 −
f2 k2p
Z
χE
χE
p 2/p
Z
χE
χE
p 2/p
+
+
−
(µE1 )1/p
(µE2 )1/p
(µE1 )1/p
(µE2 )1/p
Z
Z χ
2/p
2/p
χE
χE
χE
E1
2
1
2
=
+
+
−
(since E1 ∩ E2 = ∅)
µE1
µE2
µE1
µE2
=
=
2
1
1
2
2
22/p + 22/p = 21+ p .
2
So we must have 21+ p = 4, so that 1 + p2 = 2 and hence p = 2.
On the other hand, if p = ∞, set f1 = χE and f2 = χE . Then the parallelogram law is violated since
1
2
kf1 + f2 k2∞ + kf1 − f2 k2∞ = 2 6= 4 = 2(kf1 k2∞ + kf2 k2∞ ).
So the inner product on L∞ also does not arise from an inner product.
36
6.13 Lp (R, m) is separable for 1 ≤ p < ∞. However, L∞ (R, m) is not separable.
Solution:
Pn
Let 1 ≤ p < ∞, and S = { i=1 ri χ(c ,d ) : n ∈ N, ri , ci , di ∈ Q}. Clearly S is countable and we will
i i
show that S is dense in Lp (R, m).
Pn
Let f ∈ Lp (R, m) and > 0. By Prop. 6.7 there is a simple function ϕ = i=1 ai χE with mEi < ∞
i
1
for 1 ≤ i ≤ n such that kf − ϕkp < /3. Let M = max{|ai − aj | : 1 ≤ i < j ≤
p L 1, so by
R n}. Then ϕ ∈
Theorem 2.26 there is a step function ψ with finite measure support such that |ϕ − ψ| < 3 M p−1 . It
follows that
1/p
1/p
kϕ − ψk1 kϕ − ψk1−1/p
≤ kϕ − ψk1 M 1−1/p < /3.
∞
But ϕ − ψ ∈ L1 ∩ L∞ so by Prop. 6.10 it follows that
1/p
kϕ − ψkp ≤ kϕ − ψk1 kϕ − ψk1−1/p
< /3.
∞
kri χ(c
i ,di )
to do it).
Pn
0
0
0
i=1 ri χ(c0 ,d0 ) with the intervals (ci , di ) disjoint. For each 1 ≤ i ≤ n let ri , ci , di ∈ Q
i i
0
− ri χ(c0 ,d0 )kp < /(3n) (it’s clear that they exist, but the argument is straightforward if
i i P
Pn
n
Then s = i=1 ri χ(c ,d ) ∈ S and ks − ψkp ≤ i=1 kri χ(c ,d ) − ri0 χ(c0 ,d0 )kp < /3 and so
i i
i i
Suppose ψ =
i
such that
you want
i
kf − skp ≤ kf − ϕkp + kϕ − ψkp + kψ − skp < .
Therefore S is dense in Lp (R, m).
Let {fn } be a countable sequence in L∞ (R, m). For each n ≥ 1, let
−1, if fn ≥ 0 a.e. on [n, n + 1),
an =
1, otherwise.
P∞
Define g(x) = n=1 an χ[n,n+1). By construction, kgk∞ = 1 so g ∈ L∞ (R, m) and kg − fn k∞ ≥ 1 for all n,
so {fn } is not dense in L∞ (R, m). Therefore no countable sequence in L∞ (R, m) is dense and so L∞ (R, m)
is not separable.
37