SOLUTIONS OF MIDTERM EXAMS
MINGFENG ZHAO
March 09, 2011
1. (Problem 9, in Page 49) Let f : [0, 1] → [0, 1] be the Cantor function in Page 38 and Page 39, and
let g(x) = f (x) + x. Show that
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, then m(g(C)) = 1.
c. By Exercise 29 of Chapter 1, in Page 39, g(C) contains a Lebesgue non measurable set A. Let
B = g −1 (A). Then B is Lebesgue measurable but not Borel.
d. There exists a Lebesgue measurable function F and a continuous function G on R such that
F ◦ G is not Lebesgue measurable.
Proof. a. Recall the definition of f . For all x ∈ C, we know
(
C=
x ∈ [0, 1] : x =
∞
X
ai
i=1
2i
)
, where ai ∈ {0, 1} for all i ≥ 1 .
And [0, 1]\C is open, then we can assume [0, 1]\C =
S∞
k=1 (Ak , Bk ),
where Ak , Bk ∈ C, and
(Ai , Bi ) ∩ (Aj , Bj ) = φ if i 6= j. So we can define


P∞
P∞ ai



if x = i=1 a2ii ∈ C
i=1 2i ,
f (x) =



 f (A ), if x ∈ (A , B ) ⊂ [0, 1]\C
k
k
k
Then f |C is increasing strictly. So f only has jump discontinuity points in [0, 1]. But we know
f (C) = [0, 1], and f (0) = 0, f (1) = 1, so f is continuous. Then we know that f is onto, increasing,
and continuous on [0, 1].
When we define g(x) = f (x) + x, then g(x) is continuous. Since g(0) = f (0) + 0 = 0, and
g(1) = f (1) + 1 = 2. So g : [0, 1] −→ [0, 2] is onto. On the other hand, if g(x1 ) = g(x2 ), x1 > x2 , that
1
2
MINGFENG ZHAO
is, f (x1 ) + x1 = f (x2 ) + x2 . So x1 − x2 = f (x2 ) − f (x1 ). Since f is increasing, then f (x2 ) − f (x1 ) ≤ 0,
but x1 − x2 > 0, contradiction. Therefore, we have g is increasing strictly. So we conclude that g is a
bijection from [0, 1] to [0, 2], and g is increasing strictly on [0, 1].
Now we show that h = g −1 is also continuous on [0, 2]. For any xn = g −1 (yn ), and yn → y, as
n → ∞. For any limit point x of {xn }∞
n=1 , then by the continuity of g, we get g(x) = y. If we have
0
two differential limit points x and x0 of {xn }∞
n=1 , without loss of generality, we assume x > x , then
we have g(x) = g(x0 ), contradicts with g is increasing strictly on [0, 1]. Therefore, {xn }∞
n=1 just has
only one limit point, which implies that xn → x for some x ∈ [0, 1], and g(x) = y, that is g −1 (y) = x.
Therefore, g −1 is continuous.
b. Since [0, 1]\C =
S∞
k=1 (Ak , Bk ),
then we have
m(g(C))
=
=
2 − m(g([0, 1]\C))
2−m
∞
[
!
(Ak , Bk )
k=1
=
2−m g
∞
[
!!
(Ak , Bk )
k=1
=
2−
∞
X
m(g(Ak , Bk ))
k=1
For any k ≥ 1, we know when x ∈ (Ak , Bk ), g(x) = f (x) + x = f (Ak ) + x, so g((Ak , Bk )) ⊂
(f (Ak ) + Ak , f (Bk ) + Bk ). So m(g(Ak , Bk )) = Bk − Ak = m((Ak , Bk )). Hence
m(g(C))
=
2−
∞
X
m((Ak , Bk ))
k=1
=
2 − m([0, 1]\C)
=
2−1
=
1.
SOLUTIONS OF MIDTERM EXAMS
3
c. Since g −1 (A) = B ⊂ C, then m∗ (B) = m(C) = 0. Since m is complete, so B is Lebesgue
measurable.
By the result of part a., we know that for any open set U ⊂ [0, 1], we have g(U ) is open in [0, 2].
So g maps Borel set in [0, 1] to Borel set in [0, 2]. If B is Borel measurable, then g(B) = A is Borel
measurable, contradiction with A is not m-measurable.
Therefore, we get B is Lebesgue measurable but not Borel.
d. We let A, B be the sets in the part c, and A ⊂ (0, 2). We define F (x) = 1B (x), which is Lebesgue
measurable, and
G(x)
=




0,
if x < 0





g −1 (x), if 0 ≤ x ≤ 2







 1,
if x > 2
Then G is continuous.
Claim: F ◦ G is not Borel measurable.
In fact, for any Borel set U in R, and 1 ∈ U , then
(F ◦ G)−1 (U ) = G−1 ◦ F −1 (U ) = G−1 (B) = A
Which is not Borel measurable. Therefore, F ◦ G is not Borel measurable.
2. (Problem 16, in Page 52) If f ∈ L+ , and
such that µ(E) < ∞, and
Proof. Since f ∈ L+ , and
R
E
f dµ >
R
X
R
X
R
X
f dµ < ∞, then for every > 0, there exists E ∈ M
f dµ − .
f dµ < ∞, by the Problem 14, in page 52, we know that the function
ν : M −→ R given by
Z
f dµ for all F ∈ M
ν(F ) =
F
4
MINGFENG ZHAO
is a finite measure on X. For any n ≥ 1, let
x ∈ X : f (x) >
En =
1
n
and E0 = {x ∈ X : f (x) > 0}. Then we have En ⊂ En+1 for all n ≥ 1, and
∞
[
E0 =
En
n=1
For each n ≥ 1, we have
Z
µ(En )
=
dµ
En
Z
dµ
=
{x∈X:
Z
1
f (x)> n
}
f (x)
≤
=
≤
1
n
dµ Since
{x∈X: f (x)> n1 }
Z
n
f (x)dµ
{x∈X: f (x)> n1 }
Z
n
f (x)dµ
f (x)
1
n
≥ 1, and the monotonicity of integral
X
<
∞
By the continuity from below of the measure ν, we know that
lim ν(En ) = ν(E0 ).
n→∞
On the other hand, we know that
Z
ν(En )
=
f dµ
En
Z
ν(E0 )
=
f dµ
E0
Z
f dµ since f (x) = 0 if x ∈
/ E0
=
X
Hence we get
Z
lim
n→∞
Z
fd =
En
f dµ,
X
SOLUTIONS OF MIDTERM EXAMS
5
which implies that for every > 0, there exists N ≥ 1, whenever n ≥ N , we have µ(En ) < ∞, and
R
En
f dµ >
R
X
f dµ − .
3. (Problem 20, in Page 59) (A generalized Dominated Convergence Theorem) If fn , gn , f, g ∈
L1 , fn −→ f , and gn −→ g a.e, as n → ∞, and |fn | ≤ gn , and
R
fn −→
R
R
gn −→
R
g, as n → ∞. Then
f , as n → ∞.
Proof. Since |fn | ≤ gn , then gn + fn ≥ 0, and gn − fn ≥ 0. By the Fatou’s Lemma 2.18 in Page 52,
we have
Z
Z
lim inf (gn + fn ) ≤
lim inf
lim inf (gn − fn ) ≤
lim inf
(gn − fn )
Z
Z
n→∞
(gn + fn )
n→∞
Z
Z
n→∞
n→∞
Since fn −→ f , gn −→ g, as n → ∞, then we have
Z
(g + f ) ≤
lim inf
Z
Z
(g − f ) ≤
Since
R
gn −→
R
gn +
n→∞
lim inf
n→∞
fn
Z
gn −
fn
g, as n → ∞, then we have
Z
Z
g+
Z
Z
f
≤
Z
g−
Z
g + lim inf
fn
n→∞
Z
f
≤
Z
g − lim sup
fn
n→∞
Since
R
g < ∞, then we get
Z
lim sup
n→∞
Therefore, we get limn→∞
R
fn =
R
Z
fn ≤
Z
f ≤ lim inf
n→∞
fn
f.
6
MINGFENG ZHAO
4. (Problem 21, in Page 59) Suppose fn , f ∈ L1 , and fn −→ f a.e, as n → ∞. Then
R
as n → ∞ iff
Proof. (=⇒) If
|fn | −→
R
R
R
|fn − f | −→ 0
|f |, as n → ∞.
|fn − f | −→ 0 as n → ∞, then we have
Z
Z
|fn | − |f |
Z
= (|fn | − |f |)
Z
≤
||fn | − |f ||
Z
≤
|fn − f |
−→ 0,
So we get
(⇐=) If
R
R
|fn | −→
|fn | −→
R
as n → ∞.
|f |, as n → ∞.
|f |, as n → ∞. Consider hn = |fn − f |. Since fn , f ∈ L1 , and fn → f a.e,
R
as n → ∞, then hn ∈ L1 , and hn → 0 a.e, as n → ∞. Since |hn | ≤ |fn | + |f |, so |fn | + |f | ∈ L1 ,
and
R
R
(|fn | + |f |) −→
R
R
2|f | = 2 |f |, as n → ∞. By the result of Problem 20, in Page 58, we have
hn −→ 0, as n → ∞. Therefore, we have
R
|fn − f | −→ 0 as n → ∞.
1
5. (Problem 25, in Page 59) Let f (x) = x− 2 if 0 < x < 1, and f (x) = 0 otherwise. Let {rk }∞
k=1 be an
enumeration of the rationals, and set g(x) =
P∞
n=1
2−n f (x − rn ). Then
a. g ∈ L1 (m), and in particular g < ∞ a.e.
b. g is discontinuous at every point and unbounded on every interval, and it remains so after
any modification on a Lebesgue null set.
c. g 2 < ∞ a.e, but g 2 is not integrable on any interval.
Proof. a. Let fn (x) = 2−n f (x − rn ) ≥ 0, then g(x) =
R
R
fn (x)dx =
R rn +1
rn
Z
g(x)dx
R
=
2−n f (x − rn )dx =
Z X
∞
R n=1
fn (x)dx
R1
0
1
P∞
n=1
fn (x). For any n ≥ 1, we have
2−n x− 2 dx = 21−n . Hence we get
SOLUTIONS OF MIDTERM EXAMS
=
∞ Z
X
=
By the Monotone Convergence Theorem 2.14 in Page 50
R
n=1
∞
X
fn (x)dx
7
21−n
n=1
=
1
Therefore, we get g ∈ L1 (m), and in particular g < ∞ a.e.
c. Since g < ∞, a.e, then g 2 < ∞, a.e. If g 2 is integrable on some interval I, without loss of generality,
we assume (a, b) ⊂ I, and g 2 ∈ L1 ((a, b)), where −∞ < a < b < ∞. Take any rationals rn ∈ (a, b),
since (a, b) is open, then there exists some δ > 0 such that rn + x ∈ (a, b) for all x < δ. Then we have
Z
rn +δ
(g(x))2 dx
rn +δ
Z
(fn (x))2 dx
≥
rn
rn
rn +δ
Z
=
(2−n f (x − rn ))2 dx
rn
yδ
Z
(2−n f (y))2 dy
=
Let y = x − rn
0
δ
Z
2−2n y −1 dy
=
0
= ∞.
Which contradicts with
R rn +δ
rn
(g(x))2 dx ≤
R
(a,b)
g 2 dx < ∞. Therefore, we have g 2 is not integrable
on any interval.
b. Claim I: g is unbounded on every interval.
If not, we assume g is bounded on an interval I. Then we have g 2 is bounded on an interval I,
in particular, we get g 2 is integrable on I, which contradicts with the result of part c. Therefore, we
have g is unbounded on every interval.
Claim II: g is discontinuous at every point.
If not, we assume g is continuous at x = x0 , then by the continuity, we get there exists > 0 such
that g is bounded on (x0 − , x0 + ), which contradicts with Claim I. Therefore, g is discontinuous at
every point.
8
MINGFENG ZHAO
We assume that h(x) = g(x) for a.e x ∈ R. Then we have h is unbounded on every interval, and
h2 is not integrable on any interval.
Claim III: h is discontinuous at every point.
If not, we assume h is continuous at x = x0 , then by the continuity, we get there exists > 0 such
that h is bounded on (x0 − , x0 + ), which contradicts with h2 is not integrable on any interval..
Therefore, h is discontinuous at every point.
6. (Problem 28, in Page 60) Compute the following limits and justify the calculations.
a. limn→∞
R∞
d. limn→∞
R∞
0
a
(1 + (x/n))−n sin(x/n)dx.
n(1 + n2 x2 )−1 dx, where a ∈ R.
Proof. a. Since for all x 6= 0, we have
1
n
1 + nx
=
≤
1
1+n·
x
n
1+n·
x
n
+
n(n−1)
2
·
x 2
n
·
x 2
n
1
+
n(n−1)
2
1
=
1+x+
n(n−1) x2
2
n2
≤
1
1+x+
n2 x 2
2 n2
=
1
1+x+
x2
2
=
2
2 + 2x + x2
≤
2
1 + x2
Since x ≥ 0
Also sine nx ≤ 1, then we get
1
x 2
n sin ≤
x
1+ n
n 1 + x2
+ ··· +
x n
n
Since x ≥ 0
SOLUTIONS OF MIDTERM EXAMS
2
1+x2 ,
For
we know
∞
Z
0
2
dx
1 + x2
r
Z
=
=
=
lim
r→∞
0
2
dx
1 + x2
By the Monotone Convergence Theorem
lim arctan r
r→∞
π
< ∞.
2
And for all x > 0, we have
1
n
1 + nx
lim
n→∞
=
lim h
n→∞
1
1+
x
n
nx ix
1
ex
=
And
lim sin
n→∞
x
= 0.
n
Hence we have
lim
n→∞
x
1
sin = 0.
x n
n
1+ n
By the Dominated Convergence Theorem, we get
∞
Z
0
x
1
sin dx = 0.
x n
n
1+ n
b. Case I: a = 0.
For
R∞
0
Z
0
n
1+n2 x2 dx,
∞
we have
n
dx
1 + n2 x2
Z
=
lim
r→∞
0
Z
=
=
=
r
n
lim
r→∞
0
r
n
dx
1 + n2 x2
1
dy
1 + y2
lim arctan r
r→∞
π
.
2
By the Monotone Convergence Theorem
Let y = nx
9
10
MINGFENG ZHAO
Therefore, we have
Z
lim
n→∞
0
∞
n
π
dx = .
1 + n2 x2
2
Case II: a > 0.
Since
n
1 + n2 x2
≤
n
n2 x2
=
1
nx2
≤
1
.
x2
And
Z
∞
a
1
dx
x2
Z
r
1
dx By the Monotone Convergence Theorem
r→∞ a x2
1 1
= lim
−
r→∞
a r
=
lim
=
1
.
a
And also for all x ≥ a, we have
lim
n→∞
n
1 + n2 x2
=
0.
By the Dominated Convergence Theorem, we have
∞
Z
a
n
dx = 0.
1 + n2 x2
Case III: a < 0.
Since
Z
a
∞
n
dx
1 + n2 x2
Z
0
=
a
Z
=
a
0
n
dx +
1 + n2 x2
Z
0
π
n
dx +
2
2
1+n x
2
∞
n
dx
1 + n2 x2
By the result of Case I
SOLUTIONS OF MIDTERM EXAMS
For
R0
n
dx,
a 1+n2 x2
11
we know
Z
a
0
Z
n
dx
1 + n2 x2
0
=
an
1
dy
1 + y2
Let y = nx
= − arctan an.
So
0
n
π
dx = .
1 + n2 x2
2
∞
n
dx = π.
1 + n2 x2
Z
lim
n→∞
a
Therefore, we have
Z
lim
n→∞
a
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu