myweb.ttu.edu/bban
lev.ban@ttu.edu
Real Analysis
Byeong Ho Ban
Mathematics and Statistics
Texas Tech University
Chapter 2. Integration
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.
Proof. .
(→)
Suppose that f is (M, BR ) measurable. Note that {−∞} , {∞} ∈ BR since {∞}∩R = {−∞}∩R = ∅ ∈ BR .
Thus, by definition of (M, BR ) measurable, f −1 ({−∞}) ∈ M, f −1 ({∞}) ∈ M.
Also observe the below
f −1 (B) ∩ Y = f −1 (B) ∩ f −1 (R) = f −1 (B ∩ R) ∈ M
∀B ∈ BR
since B ∩ R ∈ BR and f is M measurable. Thus, f is measurable on Y .
(←)
Suppose that f −1 ({−∞}), f −1 ({∞}) ∈ M, and f is measurable on Y, and let B ∈ BR . Observe the below.
f −1 (B) = f −1 ((B ∩ R) ∪ (B ∩ {∞}) ∪ (B ∩ {−∞}))
= f −1 (B ∩ R) ∪ f −1 (B ∩ {−∞}) ∪ f −1 (B ∩ {∞})
= (f −1 (B) ∩ f −1 (R)) ∪ f −1 (B ∩ {−∞}) ∪ f −1 (B ∩ {∞}) ∈ M
since f is measurable on Y and B ∩ {∞} and B ∩ {−∞} are ∅ or {∞} and {−∞} respectively.
Thus, f is measurable.
2.
Proof.
3. If {fn } is a sequence of measurable functions on X, then {x : lim fn (x)exists} is a measurable set.
Proof. .
Observe the below.
{x : lim fn (x)exists} =
x : lim sup fn = lim inf fn
n→∞
=
n→∞
x : lim sup fn − lim inf fn = 0
n→∞
n→∞
Since lim supn→∞ fn − lim inf n→∞ fn is a measurable function and the set above is the inverse image of
{0} ∈ BR , {x : lim fn (x)exists} is measurable.
1
2
4. If f : X → R and f −1 ((r, ∞]) ∈ M for each r ∈ Q, then f is measurable.
Proof. .
Let a ∈ R.
Note that ∃ {rn }n∈N ⊂ Q such that rn > a ∀n ∈ N and limn→∞ rn = a. Thus, observe the below.
[
(a, ∞] =
(rn , ∞]
n∈N
Since f −1 ((rn , ∞]) ∈ M ∀n ∈ N and M is σ-algebra,
!
f −1 ((a, ∞]) = f −1
[
(rn , ∞]
n∈N
=
[
f −1 ((rn , ∞]) ∈ M
n∈N
Since a is arbitrary real number, f is measurable.
5.
Proof.
6.
Proof.
7.
Proof.
8. If f : R → R is monotone, then f is Borel measurable.
Proof. .
Note that f is (BR , BR )-measurable and BR = M(E), where E = {(a, b) : a < b and a, b ∈ R}. Thus, it is
enough to prove that f −1 ((a, b)) ∈ BR ∀a, b ∈ R.
Let (a, b) ∈ BR and f be increasing function.
Claim f −1 ((a, b)) is an interval.
Proof. .
If f −1 (a, b) = ∅ or it has only one element, then trivially f −1 (a, b) ∈ BR .
Let α, β ∈ f −1 ((a, b)) and let γ ∈ (α, β). Then observe the below.
a < f (α) ≤ f (γ) ≤ f (β) < b
since f is increasing. Thus, γ ∈ (α, β) ⊂ f −1 ((a, b)). Therefore, f −1 ((a, b)) is an interval.
Since any kind of interval is in BR , f −1 ((a, b)) ∈ BR . Thus, f is Borel measurable.
If f is decreasing, −f is increasing so it is Borel measurable by the argument above, so −(−f ) = f is also
Borel measurable.
3
H
9. Let f : [0, 1] → [0, 1] be the Cantor function ( 1.5), and let g(x)=f(x)+x.
a. g is a bijection from [0, 1] → [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 29 of Chapter 1, g(C) contains a Lebesgue nonmeasurable set A. Let B = g −1 (A). Then B
is Lebesgue measurable but not Borel.
d. There exist a Lebesgue measurable function F and a continuous function G on R such that F◦G is not
Lebesgue measurable.
Proof. .
Solution for a
Proof. Recall that f is increasing and onto [0,1]. Since x is strictly increasing, g is strictly increasing, so
g is injective. Also, since x is onto [0,1], it is clear that g is onto [0,2]. Thus, g is bijective.
Let a, b ∈ [0, 1] where a ≤ b. And, recalling that f is continuous, observe the below.
h−1 ((a, b)) = g((a, b)) = (g(a), g(b)) ⊂ [0, 2]
so h−1 ((a, b)) is an open interval contained in [0,2].
Let G ∈ [0, 1], then, since any open set in R is a countable disjoint union of open intervals{In },
!
h−1 (G) = h−1
[
In
n∈N
=
[
h−1 (In )
n∈N
Since countable union of open intervals is open, h−1 (G) is open, to h is continuous from [0,2] to [0,1].
Solution for b
Proof. Observe that
2 = m([0, 2]) = m(g([0, 1])) = m (g(([0, 1] \ C) ∪ C)
= m(g([0, 1] \ C)) + m(g(C))
!!
[
=m g
I
+ m(g(C))
I∈V
where
1 n−1 ai
2
n−1 ai
V=
Σi=1 i + n , Σi=1 i + n : ai ∈ {0, 2} , n ∈ N \ {0}
3
3
3
3
Observe that
m (g(I)) = m(I)
∀I ∈ V
since
m(g((a, b))) = g(b) − g(a)(∵ g : continuous)
= (f (b) + b) − (f (a) + a)
= b − a(∵ f is constant on [0,1] \ C)
4
Since intervals in V are disjoints and g is injection,
!!
[
m g
I
=m
I∈V
!
[
g(I)
I∈V
= ΣI∈V m(g(I))
= ΣI∈V m(I)
2n−1
3n
1/3
=
=1
1 − 2/3
= Σ∞
n=1
Therefore,
m(g(C)) = 1
Solution for c
Proof. Observe that
B = g −1 (A) ⊂ g −1 (g(C)) = C
Since m is complete and m(C) = 0, B is Lebesgue measurable.
If B is Borel measurable, since g −1 is continuous so is measurable,
(g −1 )−1 (B) = g(B) = A ∈ BR
it is contradiction.
Solution for d
Proof. Let’s set like below.
G=h
F = XB
where B is from c Clearly, F is Lebesgue measurable since characteristic functions are L-measurable. Also,
G is continuous for a.
However, F ◦ G is not L-measurable since
(F ◦ G)−1 ({1}) = (XB ◦ h)−1 ({1})
= h−1 ◦ XB−1 ({1})
= h−1 (B)
= A 6∈ BR
10.
Proof.
11.
Proof.
5
12. Prove Proposition 2.20.(See Proposition 0.20, where a special case is proved.)
Proposition 2.20.
R
If f ∈ L+ and f < ∞, then {x : f (x) = ∞} is a null set and {x : f (x) > 0} is σ-finite.
Proof. .
R
R
R
Let E =R{x : f (x) = ∞}. Note that E = f −1 (∞) ∈ M. Observe that E f ≤ f . If µ(E) > 0, E f = ∞,
so ∞ ≤ f < ∞. It is contradiction.
S
Let F = {x : f (x) > 0}, and note that F = n∈N Fn where Fn = {x : f (x) > n1 }. Note that n1 χFn < f , so
Z
Z
1
1
µ(Fn ) =
χF < f < ∞ ∀n ∈ N
n
n n
Thus,
Z
µ(Fn ) < n
f < ∞ ∀n ∈ N
Therefore, F is σ-finite
R
R
R
R
13. Suppose {fn } ⊂ L+ , fn → f pointwise,
and
f
=
lim
f
<
∞.
Then,
f
=
lim
f for all
n
E
E n
R
R
E ∈ M. However, this need not be true if f = lim fn = ∞.
Proof. .
Let E ∈ M. By Fatou’s lemma, considering the sequence {fn χE } ⊂ L+ where fn χE → f χE ,
Z
Z
Z
Z
Z
f=
lim fn χE = lim inf fn χE ≤ lim inf fn χE = lim inf
fn
n→∞
E
n→∞
n→∞
n→∞
Note that, by Thm 2.15
Z
Z
Z
Z
Z
Z
f = (f χE + f χE c ) = f χE + f χE c =
f+
E
f <∞
Ec
E
and, similarly,
Z
Z
Z
fn =
fn +
Thus,
Z
Z
f ≤ lim inf
Ec
n→∞
Z
Z
Z
fn −
fn = lim inf (
Ec
n→∞
∀n ∈ N
fn
Ec
E
Z
fn − lim sup
fn ) = lim
n→∞
E
Z
fn =
n→
Z
fn ≤
lim sup
n→∞
Since lim inf n→∞
R
E
Thus, limn→∞
E
f−
fn ≤ lim supn→∞
Z
Z
f ≤ lim inf
f=
Ec
E
E
n→∞
fn
E
R
fn ,
Z
Z
Z
lim inf
fn =
f = lim sup fn
E
n→∞
R
Z
E
fn exists and it is equal to
E
R
E
n→∞
E
f.
If M = BR and fn = nχ[0, 1 ] + χ[ 1 ,∞) , then fn → f = χ(0,∞) . And observe that
n
n
Z
Z
lim
fn = ∞ = f
n→∞
f − lim sup
→
E
Arranging terms,
Z
Z
fn
E
6
However,
Z
1
lim
fn = lim (2 − ) = 2 6= 1 =
n→∞ [0,1]
n→∞
n
Z
Z
χ(0,∞) =
[0,1]
f
[0,1]
R
+
+
14.
R If f R∈ L , let λ(E) = E f dµ for E ∈ M. Then λ is a measure on M, and for any g ∈ L ,
gdλ = f gdµ. (First suppose that g is simple.)
Proof. .
R
It is trivial that λ(∅) = ∅ f dµ = 0.
S
Let {En }n∈N ⊂ M be a sequence of disjoint sets and let E = n∈N En .
Z
Z
Z
λ(E) =
f dµ = f χE dµ = Σn∈N f χEn dµ
E
Z
= Σn∈N f χEn dµ
Z
= Σn∈N
f dµ
En
= Σn∈N λ(En )
As we can observe, λ has countably additivity. So it is measure.
Let g ∈ L+ be a simple function where the standard form is below.
φ = Σnk=1 ak χEk
And note
Z that
Z
n
n
φdλ = Σk=1 ak λ(Ek ) = Σk=1 ak
f dµ =
Σnk=1 ak
Z
Z
f χEk dµ =
f Σnk=1 ak χEk dµ
Z
=
f φdµ
Ek
Now, let g ∈ L+ be any positive measurable function. Then there exists a sequence of simple functions
{φn }n∈N which increases to g. Then, by monotone convergence theorem, since {f φn } increases to f g
Z
Z
Z
Z
gdλ = lim φn dλ = lim
f φn dµ = f gdµ
n→
n→∞
15. If {fn } ⊂ L+ , fn decreases pointwise to f , and
R
f1 < ∞, then
Proof. .
Since {fn } is decreasing sequence, we know that
Z
Z
fn ≤ f1 < ∞ ∀n ∈ N
Also, note the below.
Z
Z
fn +
Since
R
fn < ∞ and
R
Z
(f1 − fn ) =
f1
f <∞
Z
Z
Z
(f1 − fn ) = f1 − fn
Z
Z
Z
(f1 − f ) = f1 − f
R
R
f = lim fn
7
Now, let’s consider the sequence {f1 −fn }n∈N ⊂ L+ , and note that the sequence is increasing and convergent
to f1 − f . Thus, by monotone convergence theorem,
Z
Z
Z
Z
Z
Z
f1 − f = (f1 − f ) = lim
(f1 − fn ) = f1 − lim
fn
n→∞
n→∞
R
As a conclusion, since f1 < ∞
Z
Z
lim
fn = f
n→∞
16. If f ∈ L+ and
R
f < ∞, for every > 0 there exists E ∈ M such that µ(E) < ∞ and
R
E
R
f > ( f ) − .
Proof. .
Let > 0 be given. Then we can choose φ ∈ L+ which is a simple function such that 0 ≤ φ ≤ f and
Z
Z
φ> f −
Since φ is simple function, let
φ = Σnk=1 ak χEk + 0χE0
be the standard form where E0 = f −1 ({0})
Since Ek ∈ M and we have
Z
min (a1 , . . . , an )(µ(E1 ) + · · · + µ(En )) ≤ φ ≤
S
S
, we know that µ( nk=1 Ek ) = Σnk=1 µ(Ek ) < ∞ and E = nk=1 Ek ∈ M.
Thus,
Z
Z
Z
Z
f
φ=
f −< φ=
E
Z
f <∞
E
17.
Proof.
18.
Proof.
19.
Proof.
20. (A generalized Dominated Convergence Theorem)
R
R
R
R
If fn , gn , f , g ∈ L1 , fn → f and gn → g a.e., |fn | ≤ gn , and gn → g, then fn → f .
(Rework the proof of the dominated convergence theorem)
8
Proof. .
Note that −gn ≤ fn ≤ gn so gn + fn ≥ 0 and gn − fn ≥ 0 ∀n ∈ N. Then by Fatou’s lemma,
Z
Z
Z
Z
Z
Z
g + f = lim inf (gn + fn ) ≤ lim inf (gn + fn ) = g + lim inf fn
n→∞
n→∞
n→∞
Z
Z
Z
Z
Z
Z
g − f = lim inf (gn − fn ) ≤ lim inf (gn − fn ) = g − lim sup fn
n→∞
Thus, since
R
n→∞
n→∞
g < ∞, we have
Z
lim sup
Z
fn ≤
f ≤ lim inf
n→∞
n→∞
Therefore, limn→∞
R
fn exists, and it is equal to
R
Z
fn
f.
21.
Proof.
22. Let µ be counting measure on N. Interpreting Fatou’s lemma and the monotone and dominated convergence theorems as statement about infinite series.
Proof. .
Observe the below.
Z
f dµ =
Σ∞
n=1
Z
∞
f dµ = Σ∞
n=1 f (n)µ({n}) = Σn=1 f (n)
{n}
N
Here, we can say f (n) = an and fk (n) = akn
Fatou0 s lemma
If {akn }k,n∈N is any sequence of nonnegative real numbers, then
∞
Σ∞
n=1 lim inf aka ≤ lim inf Σn=1 akn
k→∞
n→∞
Monotone Convergence Theorem
If {akn }k,n∈N is any sequence of nonnegative real numbers such that akn ≤ a(k+1)n ∀k ∈ N and limk→∞ akn =
an , then
lim Σn∈N akn = Σn∈N lim akn
k→∞
k→∞
Dominated Convergence Theorem
If {akn }k,n∈N is any sequence of real numbers such that Σn∈N akn is absolutely convergent ∀k, (a) akn → an
∀n and (b)there exists a nonnegative sequence {bn }n∈N such that |akn | ≤ bn ∀k. Then Σn∈N an is absolutely
convergent and Σn∈N an = limk→∞ Σn∈N akn .
23. Given a bounded function f : [a, b] → R, let
H(x) = lim sup f (y)
δ→0 |y−x|≤δ
h(x) = lim inf f (y)
δ→0 |y−x|≤δ
Prove Theorem 2.28b by establishing the following lemmas:
a. H(x) = h(x) iff f is continuous at x.
b. In the notation of the proof of Theorem 2.28a, H = G a.e. and h = g a.e. Hence H and h are Lebesgue
R
R
b
measurable, and [a,b] Hdm = I a (f ) and [a,b] hdm = I ba
9
Proof. .
Proof for a
If H(x) = h(x), h(x) = H(x) = limy→x f (y) exists. Since h(x) ≤ f (x) ≤ H(x) is true, we have
limy→x f (y) = f (x). Thus, f is continuous at x.
Conversely, if f is continuous at x, limy→x f (y) exists, so limy→x f (y) = H(x) = h(x).
Proof for b.
Let {Pn }n∈N be a sequence of partitions of [a, b] such that min (Pn ) = a, max (Pn ) = b ∀n ∈ N, and
Pn ⊂ Pn+1 ∀n ∈ N.
Note that
n
mnk χ(tk−1 ,tk ]
gPn = ΣJk=1
lim gPn = g
n
Mkn χ(tk−1 ,tk ]
GPn = ΣJk=1
lim GPn = G
n→∞
n→∞
,where Jn is the numbers of partition points in Pn .
Assume that x ∈ [a, b] is a point such that x 6∈ Pn ∀n ∈ N. If n is a natural number, ∃k ∈ {1, 2, . . . , Jn }
such that x ∈ (tk−1 , tk ] where tk−1 , tk ∈ Pn .
Let δ > 0 be arbitrary real number such that [x − δ, x + δ] ⊂ (tk−1 , tk ] where (tk−1 , tk ] is a partition of Pn .
Then we can always choose a partition
Ps = {a = s0 < s1 < · · · < sJl = b}
such that x ∈ (sk−1 , sk ] and
(sk−1 , sk ] ⊂ [x − δ, x + δ] ⊂ (tk−1 , tk ]
where s > n. Thus,
Mks = GPs (x) ≤ sup f (y) ≤ GPn (x) = Mkn
|y−x|≤δ
mnk
= gPn (x) ≤ inf f (y) ≤ gPs (x) = msk
|x−y|<δ
Since s → ∞ and δ → 0 if n → ∞,
H(x) = lim sup f (y) = G(x) and h(x) = lim inf f (y) = g(x)
δ→0 |y−x|<δ
δ→0 |x−y|<δ
Since {tk }k∈N is countable, it is Lebesgue measure 0. So G = H and h = g a.e.
Additionally, Note that f is Riemann integrable iff G = g iff H = h a.e. iff f is continuous a.e.. Thus, f
is Riemann integrable iff {x ∈ [a, b] : f is discontinuous at x} has Lebesgue measure 0.
24.
Proof.
25.
Proof.
26.
Proof.
10
27.
Proof.
28. Compute
R ∞ the following limits and justify the calculations:
a. limn→∞ 0 (1 + nx )−n sin( nx )dx
R1
b. limn→∞ 0 (1 + nx2 )(1 + x2 )−n dx
R ∞ n sin( x )
c. limn→∞ 0 x(1+xn2 ) dx
R∞
d. limn→∞ a 1+nn2 x2 dx
(The answer depends on whether a > 0, a = 0, or a < 0. How does this accord with the various convergence
theorems?)
Proof. .
Solution of a
n
Note that, since x ≥ 0, 1 + nx ≥ 1 + x +
n(n−1)x2
2n2
sin( nx )
1
≤
x n ≤
(1 + n )
(1 + nx )n
x2
4
x2
4
∀n ∈ N \ {1}. Observe the below.
Z ∞
1
and
dx = π < ∞
x2
+1
0
4
≥1+
1
+1
Thus, by DCT,
Z
lim
n→∞
0
∞
sin( nx )
dx =
(1 + nx )n
∞
Z
0
sin( nx )
lim
dx =
n→∞ (1 + x )n
n
Z
0
∞
0
dx = 0
ex
Solution of b
Note that, since x ≥ 0, (1 + x2 )n ≥ 1 + nx2 ∀n ∈ N. So observe the below.
Z 1
1 + nx2
1 + nx2
1dx = 1 < ∞
≤
= 1 and
(1 + x2 )n
1 + nx2
0
Therefore, by DCT,
Z
lim
n→∞
0
1
1 + nx2
dx =
(1 + x2 )n
Z
0
1
1 + nx2
lim
dx
n→∞ (1 + x2 )n
Note that, if x = 0
1 + nx2
1
lim
= lim n = 1
n→∞ (1 + x2 )n
n→∞ 1
If x 6= 0, by L’hospital’s rule,
1 + nx2
2nx
1
lim
= lim
= lim
= 0.
n→∞ (1 + x2 )n
n→∞ 2xn(1 + x2 )n−1
n→∞ (1 + x2 )n−1
Therefore,
Z
lim
n→∞
0
1
1 + nx2
dx = 0
(1 + x2 )n
11
Solution of c
Note that, since x ≥ 0
Z ∞
n sin( nx )
n nx
1
1
π
≤
=
∀n ∈ N and
dx = < ∞
2
2
2
2
x(1 + x )
x(1 + x )
1+x
1+x
2
0
Thus, by DCT,
Z ∞
Z ∞
Z ∞
n sin( nx )
n sin( nx )
1
π
lim
lim
dx =
=
dx =
2
2
2
n→∞ 0
n→∞ x(1 + x )
x(1 + x )
1+x
2
0
0
Solution of d
Let consider the change of variable ’t = nx’.
Z ∞
Z ∞
Z ∞
χ[an,∞) dt
n
dt
dx =
=
2
2
2
2
1+n x
a
na 1 + t
−∞ 1 + t
Observe that
Z
χ[an,∞)
1
1
≤
∀n
∈
N
and
dt < ∞
2
1 + t2
1 + t2
R 1+t
We will divide this proof into three cases according to the value of a.
Observe that, by DCT,
Z
∞
lim
n→∞
−∞
χ[an,∞] dt
=
1 + t2
χ[an,∞]
dt
2
R n→∞ 1 + t
Z
lim
(1) When a > 0
Z
∞
lim
n→∞
−∞
χ[an,∞] dt
=
1 + t2
χ[an,∞]
lim
dt =
2
R n→∞ 1 + t
Z
Z
{∞}
1
dt = 0
1 + t2
(2) When ’a = 0’,
Z
∞
lim
n→∞
−∞
χ[an,∞] dt
=
1 + t2
χ[an,∞]
lim
dt =
2
R n→∞ 1 + t
Z
χ[0,∞]
lim
dt =
2
R n→∞ 1 + t
Z
∞
Z
0
dt
π
=
2
1+t
2
(3) When ’a < 0’
Z
∞
lim
n→∞
−∞
χ[an,∞] dt
=
1 + t2
χ[an,∞]
lim
dt =
2
R n→∞ 1 + t
Z
Z
R
χ[−∞,∞]
dt =
1 + t2
How does this accord with the various convergence theorem?
Z
R
1
dt = π
1 + t2
29.
Proof.
30.
Proof.
12
31.
Proof.
32.
Proof.
33. If fn ≥ 0 and fn → f in measure, then
R
f ≤ lim inf
R
fn .
Proof. .
R
R
. Note that lim
inf
f
is
a
limit
point
of
a
set
f
:
n
∈
N
,
so
there
exists
a
subsequence,
fnj , of
n
n
R
R
{fn } such that fnj → lim inf fn .
Also, observe that any
subsequence
of {fn } converges to f in measure, thus fnj → f in measure. So there
exists a subsequence, fnji , of fnj such that fnji → f a.e.
Now, by Fatou’s lemma, observe the below.
Z
Z
Z
Z
Z
Z
Z
fnj = lim inf fn
f=
lim fnji = lim inf fnji ≤ lim inf fnji ≤ lim inf fnj = lim
i→∞
i→∞
Therefore,
R
f ≤ lim inf
R
i→∞
j→∞
j→∞
fn .
34.R Suppose R|fn | ≤ g ∈ L1 and fn → f in measure.
a. f = lim fn
b. fn → f in L1
Proof. .
Solution for a.
Note that g − fn , g + fn ∈ L+ and that g − fn → g − f and g + fn → g + f in measure. Then by Homework
question 1(Ex.33),
Z
Z
Z
Z
Z
Z
g − f = g − f ≤ lim inf (g − fn ) = g − lim sup fn
Z
Z
Z
Z
Z
Z
g + f = g + f ≤ lim inf (g + fn ) = g + lim inf fn
R
R
R
R
R
Since g < ∞,
we
have
lim
sup
f
≤
f
≤
lim
inf
f
,
which
means
that
lim
fn exists and that
n
n
n→∞
R
it is equal to
f
.
R
R
Therefore, f = lim fn .
Solution for b.
Note that |f − fn | ≤ |f | + g ∈ L1 , and |f − fn | → 0 in measure. Then by the result of a,
Z
Z
lim |f − fn | = 0 = 0
Thus, fn → f in L1 .
35. fn → f in measure iff for every > 0 there exists N ∈ N such that µ({x : |fn (x) − f (x)| ≥ }) < for
n ≥ N.
13
Proof. .
( =⇒ ).
Let > 0 be given. Then by definition,
µ({x : |fn (x) − f (x)| ≥ }) → 0
as n → ∞
By definition of convergent sequence,
∃N ∈ N such that
µ({x : |fn (x) − f (x)| ≥ }) < ∀n ≥ N
(⇐=)
Suppose that for every > 0 there exists N ∈ N such that µ({x : |fn (x) − f (x)| ≥ }) < for n ≥ N .
Let δ > 0 be given. Then note that ∀ ∈ (0, δ), ∃N ∈ N such that
µ({x : |fn (x) − f (x)| ≥ δ}) ≤ µ({x : |fn (x) − f (x)| ≥ }) < since
{x : |fn (x) − f (x)| ≥ δ} ⊂ {x : |fn (x) − f (x)| ≥ } .
Thus,
lim µ({x : |fn (x) − f (x)| ≥ δ}) = 0
n→∞
It means, fn → f in measure.
36.
Proof.
37.
Proof.
38.
Proof.
39. If fn → f almost uniformly, then fn → f a.e. and in measure.
Proof. .
Suppose that fn → f almost uniformly, then ∀ > 0 ∃ E ⊂ X such that µ(E) < and fn → f uniformly
on E c .
Claim 1. fn → f a.e.
( =⇒ )
By given condition, for any k ∈ N, we can choose Ek ⊂ X such that
µ(Ek ) <
1
k
and
fn → f uniformly on Ekc
14
nT
o
T
j
Let E = k∈N Ek . And note that
E
is decreasing sequence and µ(E1 ) < 1.
k
k=1
Thus by continuity from above,
j
\
1
=0
j→∞
j→∞
j→∞ j
k=1
S
Now, observe that fn → f uniformly on Ekc ∀k ∈ N and that E c = k∈N Ekc .
If x ∈ E c , ∃k ∈ N such that x ∈ Ek . Then fn (x) → f (x) as n → ∞.
It means that fn → f on E c . Therefore, fn → f a.e.
µ(E) = lim µ(
Ek ) ≤ lim µ(Ej ) = lim
Claim 2. fn → f in measure.
( =⇒ )
Let > 0 be given and let En = {x : |fn (x) − f (x)| ≥ }.
Then, there exists E ⊂ X such that µ(E) < and fn → f uniformly on E c .
So, there exists N ∈ N, such that
|fn (x) − f (x)| < ∀n ≥ N
It means the below.
{x : |fn (x) − f (x)| ≥ } ⊂ E ∀n ≥ N
By monotonicity,
µ({x : |fn (x) − f (x)| ≥ }) ≤ µ(E) < ∀n ≥ N
Therefore, by the result of HW question 3(Ex.35), fn → f in measure.
40.
Proof.
41.
Proof.
42.
Proof.
43.
Proof.
44.
Proof.
N
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
Proof.
Claim 1.
N3
1
Mj = (M1 ⊗ M2 ) ⊗ M3
15
Proof. .
Let E = {A1 × A2 × A3 : Aj ∈ Mj , j = 1, 2, 3}, E(12)3 = {A × A3 : A ∈ M1 ⊗ M2 , A3 ∈ M3 } and E(12) =
{A1 × A2 : Aj ∈ Mj , j = 1, 2}. Then note the below.
3
O
Mj = M(E)
j=1
(M1 ⊗ M2 ) ⊗ M3 = M(E(12)3 )
M1 ⊗ M2 = E(12)
N
Note that E ⊂ E(12)3 , thus, clearly, 3j=1 Mj ⊂ (M1 ⊗ M2 ) ⊗ M3 .
n
o
N3
On the other hands, let R = A ∈ M1 ⊗ M2 : A × E ∈ j=1 Mj , ∀E ∈ M3 . By the construction,
R ⊂ M1 ⊗ M2 . And it is clear that E(1,2) ⊂ R. We will show that R is a σ-algebra. Let A ∈ R and
observe the below.
c
c
c
A × E = (A × E) \ (X × E ) ∈
3
O
Mj
∀E ∈ M3
where X = X1 × X2 . Thus, Ac ∈ R.
Also, If {An }∞
n=1 ⊂ R, observe the below.
!
3
∞
∞
O
[
[
(An × E) ∈
Mj
An × E =
∀E ∈ M3
j=1
n=1
n=1
j=1
N3
since
j=1 Mj is a σ-algebra. Therefore, R is a σ-algebra containing E(12) . Thus, M1 ⊗ M2 = R. It
N
N
implies that E(12)3 ⊂ 3j=1 Mj , so (M1 ⊗ M2 ) ⊗ M3 ⊂ 3j=1 Mj .
N
Therefore, 31 Mj = (M1 ⊗ M2 ) ⊗ M3 .
Claim 2. µ1 × µ2 × µ3 = (µ1 × µ2 ) × µ3
Proof. .
N
We already know from class that µ1 × µ2 × µ3 is a measure on 31 Mj . Note, for any Aj ∈ Mj (j = 1, 2, 3)
the below.
((µ1 × µ2 ) × µ3 )(A1 × A2 × A3 ) = (µ1 × µ2 )(A1 × A2 )µ3 (A3 )
= µ1 (A1 )µ2 (A2 )µ3 (A3 ) = (µ1 × µ2 × µ3 )(A1 × A2 × A3 )
Since (µ1 × µ2 × µ3 ) = (µ1 × µ2 ) × µ3 on A, and since (µ1 × µ2 × µ3 ) is a premeasure on an algebra A
which is a collection of a finite disjoint unions of elements
of A, (µ1 × µ2 ) × µ3 is a premeasure on A and
N3
(µ1 × µ2 ) × µ3 is a measure on (M1 ⊗ M2 ) ⊗ M3 = 1 Mj . Since each µj are σ-finite, so is (µ1 × µ2 ) × µ3
and µ1 × µ2 × µ3 , µ1 × µ2 × µ3 = (µ1 × µ2 ) × µ3 by uniqueness of measure.
46. Let X = Y = [0, 1], M = N = B[0,1] , µ=Lebesgue
measure,
RR
R R and ν= counting
R measure. If D =
{(x, x) : x ∈ [0, 1]} is the
χD dµdν,
χD dνdµ, and χD d(µ × ν) are all
R diagonal in X × Y , then
unequal. (To compute χD d(µ × ν), go back to the definition of µ × ν.)
16
Proof. Observe the following calculations.
Z Z
Z Z
χD dµdν =
χyD dµ
Z Z
dν =
χyD dµ
Z
dν =
0dν = 0
{y}
Z Z
Z Z
χD dνdµ =
χDx dν dµ =
Z
1dµ = µ([0, 1]) = 1
{x}
As for the last integration,first note the below.
Z Z
χD d(µ × ν) = inf Σn∈N µ(An )ν(Bn )| {An × Bn }n∈N is any covering of D
Let {An × Bn }n∈N be a covering of D. And observe that [0, 1] ⊂
least one n ∈ N, such that
S
n∈N
(An ∩ Bn ). Thus, there exist at
µ(An ∩ Bn ) > 0.
And observe the below.
0 < µ(An ∩ Bn ) ≤ µ(An )
∞ = ν([0, 1]) ≤ ν(Bn )
Therefore,
Z Z
χD d(µ × ν) = inf Σn∈N µ(An )ν(Bn )| {An × Bn }n∈N is any covering of D = inf {∞} = ∞
Thus, in this case, we cannot change the order of integration. And that is because ν is not σ-finite.
47.
Proof.
48. Let X = Y = N, M = N = P(N), µ = ν =counting measure.
Define f (m, n) =R R1 if m = n,
R
fR (m,
f dµdν and
R n) = −1 if m = n + 1, and f (m, n) = 0 otherwise. Then |f |d(µ × ν) = ∞, and
f dνdµ exists and are unequal.
17
Proof. Observe the following calculations. (L = {(n, n) : n ∈ N})
Z
Z
∞ = Σn∈N | − 1| =
|f |d(µ) ≤ |f |d(µ × ν)
L
Z Z
Z Z
f dµdν =
Z
Z
1dν +
=
{1}
Z Z
Z
f (m)dµ(m) dν(n) =
n
Z
Z
f (m)dµ(m) dν(n) +
n
{1}
Z
n
f (m)dµ(m) dν(n)
A\{1}
0dν = 1
A\{1}
Z Z
f dνdµ =
Z
Z
fm (n)dν(n) dµ(m)
Z
Z
=
fm (n)dν(n) +
fm (n)dν(n) +
fm (n)dν(n) dµ(m)
N\{1}
{m}
{m−1}
N\{m,m−1}
Z
Z Z
+
f1 (n)dν(n) +
f1 (n)dν(n) dµ(m)
{1}
{1}
N\{1}
Z
Z
=
(1 − 1 + 0)dµ +
(1 + 0)dµ
N\{1}
{1}
Z
Z
=
0dµ +
1dµ
N\{1}
{1}
=0+0=0
49.
Proof.
50.
Proof.
51.
Proof.
52. The Fubini-Tonelli theorem is valid when (X, M, µ) is an arbitrary measure space and Y is a countable
set, N = P(Y ), and ν is counting measure on Y. (Cf. Theorem 2.15 and 2.25)
Proof. .
Since Y is countable, let Y = {yn }n∈N . We need to show the below.
Z
Z
f d(µ × ν) = f (x, yn )dµ(x)
X×{yn }
for any f ∈ L+ (X × Y ) and for any f ∈ L1 (µ × ν) and for all n ∈ N.
Firstly, let’s consider a characteristic function χE where E ∈ M ⊗ N .
18
Observe the below.
Z
Z
yn
yn
χE d(µ × ν) = (µ × ν)(E ∩ (X × {yn })) = µ(E )ν({yn }) = µ(E ) = χE (x, yn )dµ
X×{yn }
If yn 6∈ E, χE (x, yn ) = 0 and (µ × ν)(E ∩ (X × {yn })) = 0, so both sides are 0.
Secondly, let’s consider a positive simple function φ = ΣN
i=1 an χEi .
Observe the below.
Z
Z
φd(µ × ν) =
ΣN
i=1 ai χEn d(µ × ν)
X×{yn }
X×{yn }
Z
N
= Σi=1
ai χEi d(µ × ν)
X×{yn }
Z
N
= Σi=1 ai
χEi d(µ × ν)
X×{yn }
Z
N
= Σi=1 ai χEi (x, yn )dµ
Z
= ΣN
i=1 ai χEi (x, yn )dµ
Z
= φ(x, yn )dµ
Thirdly, let’s consider positive measurable function f ∈ L+ . Note that there exists increasing sequence of
simple functions {φj }j∈N , such that limn→∞ φn = f
Due to monotone convergence theorem, observe the below.
Z
Z
f d(µ × ν) = lim
φj d(µ × ν)
j→∞ X×{y }
X×{yn }
n
Z
= lim
φj (x, yn )dµ
j→∞
Z
= f (x, yn )dµ
R
Now, let’s verify Tonelli’s theorem. We know
that g(x) = fx dν = Σn∈N f (x, yn ) ∈ L+ (X) since each
R
f (x, yn ) ∈ L+ . Also, we have that h(yn ) = f yn dµ ∈ L+ (Y ). Also, observe the below.
Z
Z
Z
Z Z
f d(µ × ν) = Σn∈N
f d(µ × ν) = Σn∈N f (x, yn )dµ =
f dµ dν
X×{yn }
Z
Z
Z
Z
Z Z
f d(µ × ν) = Σn∈N
f d(µ × ν) = Σn∈N f (x, yn )dµ = Σn∈N f (x, yn )dµ =
f dν dµ
X×{yn }
Thus, Tonelli’s theorem is valid.
Fourthly, let’s consider a function f ∈ L1 (µ × ν).
Observe the below.
Z
Z
Z
+
f d(µ × ν) =
f d(µ × ν) −
f − d(µ × ν)
X×{yn }
X×{yn }
X×{yn }
Z
Z
+
= f (x, yn )dµ − f − (x, yn )dµ
Z
= f (x, yn )dµ
19
Now, let’s see if Fubini’s theorem is valid. Observe the below.
Z
Z
Z
|f |d(µ × ν) = Σn∈N |f (x, yn )|dµ = Σn∈N |f (x, yn )|dµ < ∞
Therefore, f (x, yn ) ∈ L1 (µ) ∀n ∈ N and f (x, yn ) ∈ L1 (ν) a.e x ∈ X. Also, observe that
Z
g(x) = fx dν = Σn∈N f (x, yn ) ∈ L1 (µ)
Z
h(yn ) = f yn dµ ∈ L1 (ν)
Z
Z
Z
Z Z
f d(µ × ν) = Σn∈N
f d(µ × ν) = Σn∈N f (x, yn )dµ =
f dµ dν
X×{yn }
Z
Z
f d(µ × ν) = Σn∈N
Z
f d(µ × ν) = Σn∈N
Z Z
Z
f (x, yn )dµ =
Σn∈N f (x, yn )dµ =
f dν dµ
X×{yn }
Thus, Fubini’s theorem is valid.
53.
Proof.
54. How much of Theorem 2.44 remains valid if T is not invertible?
Proof. Recall the Theorem 2.44 below.
Suppose T ∈ GL(n, R)
a. If f is Lebesgue measurable function on Rn , so is f ◦ T . If f ≥ 0 or f ∈ L1 (m), then
Z
Z
f (x)dx = |detT | f ◦ T (x)dx
b. If E ∈ Ln , then T (E) ∈ Ln and m(T (E)) = |detT |m(E)
Suppose that T is not invertible.
Since T is still continuous mapping, f ◦ T is Lebesgue measurable.
R
However, since |detT | = 0 and f dx can be non zero, the second statement of (a) does not hold.
Let E ∈ Ln , and note that, since T is not invertible, T (E) is a subset of a subspace of Rn spanned
by the column vectors of the matrix T . And with proper transformation, M (∈ GL(n, R)), we have the
condition below.
M ◦ T (E) ⊂ Rn−1 × {0}
Since m(Rn−1 × {0}) = 0 and Lebesgue measure is complete, M ◦ T (E) ∈ Ln .
And by Thm 2.44, M −1 ◦ M ◦ T (E) = T (E) ∈ Ln
Also,
0 = m(M ◦ T (E)) = |detM |m(T (E)) =⇒ m(T (E)) = 0 (∵ |detM | =
6 0)
And since |detT | = 0,
m(T (E)) = 0 = |detT |m(E)
55.
Proof.
20
56.
Proof.
57.
Proof.
58.
Proof.
59.
Proof.
60.
Proof.
61.
Proof.
62.
Proof.
63.
Proof.
64.
Proof.