Upload
phamhanh
View
248
Download
3
Embed Size (px)
Citation preview
Real Analysis Homework: #1
Yingwei Wang ∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1 Banach space
Question: Let (xn) ⊂ X be a Banach space, and∑
∞
n=1‖xn‖ is convergent. Proof that
∑∞
n=1xn is
convergent in X.
Proof: Suppose∑
∞
n=1‖xn‖ = M < ∞, then ∀ ε > 0,∃N, s.t.
∑∞
n=N ‖xn‖ < ε.
Let sn =∑n
i=1xi, then ∀n > m > N ,
‖sn − sm‖ = ‖n∑
i=m
xi‖ 6
∞∑
i=m
‖xi‖ <∞∑
i=N
‖xi‖ < ε.
Hence, (sn) is a Cauchy sequence and must converge to an element in X.
2 Continuous function
Question: f : (X, τX ) → (Y, τY ) is continuous ⇔ ∀x0 ∈ X and any neighborhood V of f(x0), thereis a neighborhood U of x0 such that f(U) ⊂ V .
Proof: “⇒”: Let x0 ∈ X f(x0) ∈ Y . For each open set V containing f(x0), since f iscontinuous, f−1(V ) which containing x0 is open. Then, there is a neighborhood U of x0 such thatx0 ∈ U ⊂ f−1(V ), that is to say f(U) ⊂ V .
“⇐”: Let V ∈ Y . ∀y ∈ V , choose Vy satisfy y ∈ Vy ⊂ V, Vy ∈ τY . Then V =⋃
Vy.
Let x = f−1(y), then ∀Vy,∃Ux ∈ τX , s.t. x ∈ Ux and f(Ux) ⊂ Vy.
Let U =⋃
Ux, then U ∈ τX and f(U) = V .
Therefore, f : (X, τX) → (Y, τY ) is continuous.
∗
E-mail address: [email protected]; Tel : 765 337 3504
I
Yingwei Wang Real Analysis
3 Closure
3.1
Question: Topological space (X, τ), x ∈ X,E ⊂ X, x ∈ E ⇔ ∀ neighborhoodV of x, E⋂
V 6= ∅.
Proof: Actually, this question is equal to the following one:
(*) Topological space (X, τ), x /∈ E ⇔ There exist an open set V containing x that E⋂
V = ∅.
We just need to prove (*).
“⇒ ” If x /∈ E, then V = X\E is an open set containing x and E⋂
V = ∅.
“⇐” If there exist an open set V containing x that E⋂
V = ∅, then X\V is a closed setcontaining E.
By the definition of closer E, that is the intersection of all closed sets containing E, the setX\V must contain E.
Thus, x /∈ E.
3.2
Question: Metric space (X, d), x ∈ X,E ⊂ X, x ∈ E ⇔ there is a sequence (en)∞n=1in E such
that limn→∞ d(en, x) = 0.
Proof: Similarly as Section 3.1, the above question is equal to the following one:
(**) Metric space (X, d), x /∈ E ⇔ ∀e ∈ E,∃ ε0, s.t. d(e, x) > ε0.
We just need to prove (**).
“⇒ ” Suppose x /∈ E. Since X\E is an open set, then ∃B(x, δ) ⊂ X\E. That is to say∀e ∈ E, d(e, x) > δ.
“⇐” If ∀e ∈ E,∃ ε0, s.t. d(e, x) > ε0, then B(x, ε0
2) is an open set that containing x and
B(x, ε0
2)⋂
E = ∅. So X\B(x, ε0
2) is a closed set containing E.
By the definition of E, E ⊂ X\B(x, ε0
2).
Since x ∈ B(x, ε0
2), x /∈ E.
4 Distance between point and set
If E is a nonempty subset of a metric space X, define the distance from x ∈ X to E by
d(x,E) = infz∈E
d(x, z). (4.1)
4.1
(a) Prove that d(x,E) = 0 ⇔ x ∈ E.
Proof: If infz∈E d(x, z) = 0, then ∀ε, ∃ zε, s.t. d(x, zε) < ε.
II
Yingwei Wang Real Analysis
Choosing ε = 1
n, then we can get a sequence (zn)∞n=1
satisfying limn→∞ d(x, zn) = 0.
According to Section 3.2, x ∈ E.
“⇐” If x ∈ E, by the previously Section 3.2, there is a sequence (zn)∞n=1in E such that
limn→∞ d(zn, x) = 0.
Then d(x,E) = infz∈E d(x, z) = limn→∞ d(zn, x) = 0.
4.2
(b) Prove that x 7→ d(x,E) is a uniformly continuous function on X, by showing that
|d(x,E) − d(y,E)| ≤ d(x, y).
for all x ∈ X, y ∈ X.
Proof: By the definition of (4.1), we can get:
∀x, y ∈ X,∀ε > 0,∃zx ∈ E, s.t. d(x,E) = d(x, zx) + ε, ∃zy ∈ E, s.t. d(y,E) = d(y, zy) + ε.
Without loss of generality, we suppose that d(x,E) ≤ d(y,E). Then when ε is sufficient small,d(y, zy) ≤ d(y, zx).
According to triangle inequality,
d(x, y) ≥ d(y, zx) − d(x, zx) ≥ d(y, zy) − d(x, zx) = d(x,E) − d(y,E).
Besides, in the case that d(x,E) ≥ d(y,E), we can get d(x, y) ≥ d(y,E) − d(x,E). So we canconclude that |d(x,E) − d(y,E)| ≤ d(x, y).
∀x0 ∈ X,∀ε > 0, just choose δ = ε2, then ∀x ∈ B(x0, δ), we have |d(x,E) − d(x0, E)| ≤
d(x, x0) < δ < ε. That is to say x 7→ d(x,E) is a uniformly continuous function on X.
III
Real Analysis Homework: #2
Yingwei Wang ∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1 Banach space
Question: Let C([a, b]) denote the linear space of continuous function f : [a, b] → R. Show thatC[a, b] is a Banach space with respect to the norm
‖f‖ = max{|f(t)| : t ∈ [a, b]}.
Proof: Let (fn)∞n=1 be a Cauchy sequence in C([a, b]). For ∀t ∈ [a, b],∀m,n ∈ N, |fm(t)−fn(t)| ≤‖fn − fm‖, which means (fn(t))∞n=1 is a Cauchy sequence in R and must converge to an elementin R. So we can define a function f : [a, b] → R that
f(t) = limn→∞
fn(t), ∀t ∈ [a, b].
First, we will show that fn → f when n → ∞. ∀ε > 0, there is an N s.t. ∀n,m ≥ N we have‖fn − fm‖ < ε. For each x ∈ [a, b] we have
|fn(x) − f(x)| = limm→∞
|fn(x) − fm(x)|
≤ limm→∞
‖fn − fm‖
≤ ε,
for all n ≥ N . Then for all n ≥ N ,
‖fn − f‖ = maxx∈[a,b]
{|fn(x) − f(x)|} ≤ ε.
Thus fn → f .
∗
E-mail address: [email protected]; Tel : 765 237 7149
I
Yingwei Wang Real Analysis
Second, we will show that f ∈ C([a, b]). Choose a point t0 ∈ [a, b]. Since fn ∈ C([a, b]), for theprevious ε, ∃ δn, s.t. ∀t ∈ (t0 − δ, t0 + δ), |fn(t) − fn(t0)| < ε. For t ∈ (t0 − δ, t0 + δ), n > N wehave
|f(t) − f(t0)| ≤ |f(t) − fn(t)| + |fn(t) − fn(t0)| + |fn(t0) − f(t0)|
< ε + ε + ε
< 3ε.
Thus f ∈ C([a, b]) and C([a, b]) is a Banach space.
2 Banach space
Question: Let l1(N) be the linear space of all the functions f : N → R with the property that|f |1 :=
∑∞n=1 |f(n)| < ∞. Prove that (l1(N), ‖ · ‖1) is a Banach space.
Proof: Let (fi)∞i=1 be a Cauchy sequence in l1(N). For ∀n ∈ N,∀i, j ∈ N, |fi(t) − fj(t)| ≤
‖fi − fj‖, which means (fi)∞i=1(n) is a Cauchy sequence in R and must converge to an element in
R. So we can define a function f : N → R that
f(n) = limi→∞
fi(n), ∀n ∈ N.
First, we will show that fi → f when i → ∞. ∀ε > 0, there is an I s.t. ∀i, j ≥ I we have‖fi − fj‖ < ε. For each n ∈ N we have
|fi(n) − f(n)| = limj→∞
|fi(x) − fj(x)|
≤ limj→∞
‖fi − fj‖
≤ ε,
for all i ≥ I. Then for all i ≥ N ,
‖fi − f‖ = maxn∈N
{|fi(n) − f(n)|} ≤ ε.
Thus fi → f .
Second, we will show that f ∈ l1(N).
‖f‖1 =
∞∑
n=1
|f(n)|
=∞∑
n=1
| limi→∞
fi(n)|
≤ limi→∞
∞∑
n=1
|fi(n)| < ∞.
II
Yingwei Wang Real Analysis
Then f ∈ l1(N) and (l1(N), ‖ · ‖1) is a Banach space.
3 Application of Baire Theorem
Question: Let f : R → R be a smooth function (i.e. C∞). Suppose that for each t ∈ R there isnt ∈ N such that f (nt)(t) = 0. Prove that there is an interval I of positive length such that therestriction of f to I is a polynomial.
Proof: Define Tn := {t ∈ R : f (n)(t) = 0}. It is easy to verify that for each n the restrictionof f to Tn is a polynomial with at most (n + 1) degree. Then we just need to show that ∃n0, s.t.(Tn0
)◦ 6= ∅.
First, we claim that ∀n ≥ 1, Tn is a closed set. Consider the set T cn = R\Tn: ∀t ∈ T c
n, f (n) 6= 0.Since f ∈ C∞, f (n) is continuous, we know that ∃ δ, s.t. ∀x ∈ (t − δ, t + δ), f (n)(x) 6= 0. That isto say (t − δ, t + δ) ⊂ T c
n. Then R\Tn is an open set while Tn is a closed set.
Second, we claim that∞⋃
n=1Tn = R. ∀t ∈ R, ∃ nt ∈ N, s.t. f (nt)(t) = 0, which means t ∈ Tnt
.
Thus,∞⋃
n=1Tn = R.
According to the Baire Category Theorem, ∃n0, s.t. (Tn0)◦ 6= ∅. That is to say ∃ interval I ⊂
(Tn)◦, s.t. the restriction of f to I is a polynomial.
4 Outer measure
Question: Let m∗(A) denote the outer measure of A ⊂ R. Show that if B ⊂ R and m∗(B) = 0,then m∗(A ∪ B) = m∗(A).
Proof: On one hand,A ⊂ (A ∪ B) ⇒ m∗(A) ≤ m∗(A ∪ B).
On the other hand,
m∗(A ∪ B) ≤ m∗(A) + m∗(B) = m∗(A) + 0 = m∗(A).
Thus,m∗(A ∪ B) = m∗(A).
5 Continuity of function
Question: Let (xn)∞n be an enumeration of Q. Define f : R → R by f(x) =∑ 1
2n . where thesummation is extended over all n such that xn < x. Prove that f is discontinuous at each rationalnumber and that f is continuous at each irrational number.
III
Yingwei Wang Real Analysis
Proof: Define Ax = {i ∈ N : xi < x & xi ∈ {xn}∞n = Q}. Then f(x) =
∑
n∈Ax
12n , ∀x ∈ R.
5.1 Rational points ⇒ Discontinuity
Fix xr ∈ Q, r ∈ N. Let xn ∈ Q and xn > xr. It is obviously that r ∈ (Axn\ Axr
). Then
|f(xn) − f(xr)| =
∣
∣
∣
∣
∣
∣
∑
p∈Axn
1
2p−
∑
q∈Axr
1
2q
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
∑
i∈(Axn\Axr )
1
2i
∣
∣
∣
∣
∣
∣
>1
2r.
which means limx∈Q, x→x+
r
f(x) 6= f(xr). That is to say f is discontinuous at each rational number.
5.2 Irrational points ⇒ Continuity
Since∞∑
n=1
12n = 1, we can know that ∀ε > 0, ∃N ∈ N, s.t.
∞∑
n=N+1
12n < ε. Let the set Sε =
{x1, x2, · · · , xN}. We can rearrange the elements of the set Sε such that x1 < x2 < · · · < xN .
∀α ∈ (R \ Q), we can choose δ > 0 as the following way:
δ =
12(x1 − α) if α < x1,12(min{α − xi, xi+1 − α}) if xi < α < xi+1,12(α − xN ) if α > xN .
Then (Ax+δ \ Ax−δ) ∩ ASε= ∅. That is to say ∀n ∈ (Ax+δ \ Ax−δ), n ≥ N + 1.
So ∀x ∈ (x− δ, x+ δ), |f(x)− f(α)| ≤ |f(x+ δ)− f(x− δ)| =∑
n∈(Ax+δ\Ax−δ)
12n <
∞∑
n=N+1
12n < ε,
which means f is continuous at the point α ∈ (R \ Q).
IV
Real Analysis Homework: #3
Yingwei Wang ∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1 Measure inequality
Question: Let (X,A, µ) be a measure space. Let (Ak)∞
k=1be a sequence of sets in A. Prove that
µ
(
∞⋃
n=1
∞⋂
k=n
Ak
)
≤ limk→∞
inf µ(Ak).
Proof: Let Bn =∞⋂
k=n
Ak, then Bn = Bn+1 ∩ An, Bn ⊂ Bn+1. So
µ(
m⋃
n=1
Bn) = µ(Bm) =
∞⋂
k=m
Ak ⊂ Ak, ∀k ≥ m
⇒ µ
(
m⋃
n=1
Bn
)
≤ µ(Ak), ∀k ≥ m
⇒ µ
(
∞⋃
n=1
Bn
)
≤ limk→∞
inf µ(Ak).
2 Example of measure
Question: Let µ be a measure on (R,B), where B are the Borel sets, such that µ([0, 1)) = 1 andµ(x + B) = µ(B) for all x ∈ R and B ∈ B. Prove that
(1) µ([0, 1/n)) = 1/n for all integers n ≥ 1 and
(2) µ([a, b)) = b − a for all real numbers a < b.
∗
E-mail address: [email protected]; Tel : 765 237 7149
I
Yingwei Wang Real Analysis
Proof: (1) By the assumption that µ(x + B) = µ(B), we can know that µ([0, 1/n)) =µ([1/n, 2/n)) = · · · = µ([(n − 1)/n, 1)). Then
1 = µ([0, 1]) =
n∑
k=1
µ([(k − 1)/n, k/n)) = nµ([0, 1/n))
⇒ µ([0, 1/n)) = 1/n.
(2) Since for any a < b, µ([a, b)) = µ([0, b − a)), we only need to consider the cases thata = 0, b > 0.
If b ∈ Q, then b = pq
where p, q ∈ N, q 6= 0.
µ([0, b)) = µ([0,p
q)) =
p−1∑
k=0
µ([k
q,k + 1
q)) = p µ([0,
1
q)) =
p
q= b.
If b ∈ R\Q, then ∃ (bn)∞n=1 ⊂ Q s.t. limn→∞
bn = b. Then
µ([0, b)) = limn→∞
µ([0, bn)) = limn→∞
bn = b.
3 Lebesgue outer measure
Question: Let m∗ be the Lebegue outer measure on R. For any two sets A,B ⊂ R, prove theinequality:
m∗(A) + m∗(B) ≤ 2m∗(A△B) + 2m∗(A ∩ B). (3.1)
Proof: For the set A, we have
m∗(A) ≤ m∗(A\B) + m∗(A ∩ B) (since A = (A\B) ∪ (A ∩ B))
≤ m∗(A△B) + m∗(A ∩ B). (since (A\B) ⊂ (A△B)) (3.2)
Similarly, for the set B, we have
m∗(B) ≤ m∗(A△B) + m∗(A ∩ B). (3.3)
By (3.2)-(3.3), we can get (3.1).
4 Zero measure set
Question: Let m denote the Lebesgue measure on R. Let A ⊂ R be a Lebesgue measurable set.Suppose that m(A ∩ [a, b]) < (b − a)/2 for all a < b real numbers. Show that m(A) = 0.
II
Yingwei Wang Real Analysis
Proof: On one hand, by the definition of Lebesgue measure,
m(A) = inf{∞∑
n=1
l(In) : In ⊂ R are open intervals, A ⊂∞⋃
n=1
In},
we know that ∀ε > 0, ∃ {In}∞
n=1, s.t.
m(A) ≥∞∑
n=1
l(In) − ε,
where A ⊂∞⋃
n=1
In and Ii ∩ Ij = ∅, i 6= j.
We can choose ε = 1
4m(A) in the above inequality, then we can get
m(A) ≥4
5
∞∑
n=1
l(In). (4.1)
On the other hand, by the assumption, m(A ∩ In) < 1
2l(In),∀n ∈ N+. So we have
m(A) =
∞∑
n=1
m(A ∩ In) ≤1
2
∞∑
n=1
l(In). (4.2)
From (4.1)-(4.2) we can conclude that m(A) = 0.
5 Measure function
Question: Let m denote the Lebesgue measure on R. Let A ⊂ R be a Lebegue measurableset with m(A) < ∞. Show that the function f : R → [0,∞), f(x) = m(A ∩ (−∞, x)) iscontinuous. Deduce that for every β ∈ [0,m(A)] there is a Lebesgue measurable set B ⊂ A suchthat m(B) = β.
Proof:
5.1 The continuity of the auxiliary function g
Since m(A) < ∞, by the Proposition 15 on Page 63 in Royden’s book, given ∀ε > 0, there is afinite union U of open intervals such that
m∗(U△A) < ε. (5.1)
III
Yingwei Wang Real Analysis
Suppose U =n⋃
i=1
(ai, bi), and a = a1, b = bn, then U ⊂ [a, b].
On the interval [a, b], we can define the function g such that
g(x) = m(A ∩ [a, x)), ∀x ∈ [a, b].
Then for ∀∆x > 0,
A ∩ [a, x + ∆x) = (A ∩ [a, x))⋃
(A ∩ [x, x + ∆x))
⇒ g(x + ∆x) ≤ g(x) + ∆x
⇒ g(x + ∆x) − g(x) ≤ ∆x
For ∆x < 0, we can get the similar result:
g(x + ∆x) − g(x) ≥ −∆x
Then we have|g(x + ∆x) − g(x)| ≤ |∆x|
which means g ∈ C([a, b]).
5.2 The continuity of f
For ∀x ∈ [a, b], choose ∆x < ε, then
|g(x) − f(x)| ≤ m(U△A) < ε
⇒ |f(x + ∆x) − f(x)| ≤ |f(x + ∆x) − g(x + ∆x)| + |g(x + ∆x) − g(x)| + |g(x) − f(x)| < 3ε.
which means f ∈ C([a, b]).
For x ≤ a, |f(x)| ≤ m(U△A) < ε, so f ∈ C((−∞, a]). Similarly, we have f ∈ C([b,∞)).Thus, f ∈ C(R).
5.3 The Intermediate Value Theorem
Since g(a) = 0, g(b) = m(A), by the Intermediate Value Theorem of continuous functions, for∀β ∈ [0,m(A)] there is an x0 ∈ [a, b] such that g(x0) = β. Then we can choose B = A ∩ [a, x0].
IV
Real Analysis Homework: #4
Yingwei Wang ∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1 Measurable function
Question: Let (X,A, µ) be a measure space. Let (fn)∞n=1 be a sequence of measurablefunctions: fn : X → R. Show that the set of points x where the limit lim
n→∞fn exists
(finite or infinite) is a measurable set.
Proof: Let Ak = {x ∈ X : |fn(x) − fm(x)| < 1k , ∀m,n > k}, A = lim
k→∞Ak, where
m,n, k ∈ N. It is easy to know that each Ak is measurable set since fn is a measurablefunction. We have this observation: Ak+1 ⊂ Ak ⇒ A ⊂ Ak,∀k. Thus, A is a measurableset.
Let B = {x ∈ R : limn→∞
fn(x) exists}. I will show that A = B.
On one hand, let x ∈ A, then for ∀k ∈ N, |fn(x) − fm(x)| < 1k , ∀m,n > k, which
mean fn(x) is a Cauchy sequence in R. So limn→∞
fn(x) exists and then x ∈ B.
On the other hand, let x ∈ B, then ∀ε > 0, ∃N ∈ N s.t. |fn(x)−fm(x)| < ε, ∀m,n >
N , which means x ∈ AN . Then if ε → 0, N → ∞, x ∈ A.
Now we can conclude that B is a measurable set.
2 Differential function
Question: Let f : R → R be a differential function. Prove that f ′ is Lebesgue measurable.
∗
E-mail address: [email protected]; Tel : 765 237 7149
I
Yingwei Wang Real Analysis
Proof: Define a sequence of functions: gn(x) = f(x+1/n)−f(x)1/n , x ∈ R, n ∈ N. Then
f is measurable
⇒ gn(x) is measurable ,∀n,
⇒ f ′(x) = limn→∞
gn is measurable.
3 The sets with measure zero
Question: Let f : R → R be defined by f(x) = x5 + sin x. Suppose that a set A ⊂ R hasLebesgue measure zero. Show that f(A) has Lebesgue measure zero.
Proof: I want to prove that for any function f ∈ C1(R), the image of the set withLebesgue measure zero has also Lebesgue measure zero.
Let AN = [−N,N ] ∩ A, then A =∞⋃
N=1AN , m(A) = 0 ⇒ m(AN ) = 0,∀N ∈ N. We
will focus the problem on each AN .
By the definition of Lebesgue measure, for ∀ε > 0, there exists a sequence of intervals(In)∞n=1, Ii ∩ Ij = ∅, i 6= j and AN ⊂ ∪In, s.t.
∑
n
l(In) < ε, since m(AN ) = 0. (3.1)
Let In = (an, bn), In = [an, bn], then m(In) = m(In). Then we have
m(f(AN )) <∑
n
m(f(In)) =∑
n
m(f(In)). (3.2)
Since f is continuous, we can know that
m(f(In)) = maxx∈In
f(x) − minx∈In
f(x).
Since In is a closed interval and f ∈ C1(In), we can find x1, x2 ∈ In s.t. f(x1) =maxx∈In
f(x), f(x2) = minx∈In
f(x). Without loss of generality, we can assume that x1 ≤ x2.
Then
m(f(In)) = f(x1) − f(x2)
≤ |f ′(ξ)|(x2 − x1), ξ ∈ [x1, x2],
≤ maxx∈In
f ′(x) l(In)
≤ maxx∈In
f ′(x) ε.
II
Yingwei Wang Real Analysis
Since maxx∈Inf ′(x) < ∞ and let ε → 0, we can get m(f(In)) = 0.
By (3.2) we have m(f(AN )) = 0. So
m(f(A)) = m(f(∪AN )) ≤∑
N
m(f(AN )) = 0.
4 Measurable function
Question: Let f : R → R be a Lebesgue measurable function. Define g : R → R byg(x) =
∑∞n=1
f(x)n
n!(n+3) . Prove that g is Lebesgue measurable.
Proof: First, we claim that ∀a ∈ R, the series sn =∑n
k=1ak
k!(k+3) is convergent.
∀a ∈ R,∃N ∈ N, s.t. |a| < N . Then for ∀k > 2N , we have∣
∣
∣
∣
ak
k!(k + 3)
∣
∣
∣
∣
≤|a|N
1 · · ·N·
|a|N
(N + 1) · · · (2N)·
|a|k−2N
(2N + 1) · · · k·
1
k + 3.
Let M = |a|N
1···N , we can choose k > M , s.t. |a|N
1···N · 1k+3 < 1. Since |a|N
(N+1)···(2N) < 1,
|a|k−2N
(2N+1)···k <(
12
)k−2N, we have
∣
∣
∣
∣
ak
k!(k + 3)
∣
∣
∣
∣
<
(
1
2
)k−2N
.
Since for fixed N ,∑∞
k=1
(
12
)k−2Nis convergent, we have for fixed a,
∑∞k=1
ak
k!(k+3) isalso convergent.
Second, let fn =∑n
k=1f(x)n
n!(n+3) which is measurable and
g(x) = limn→∞
fn(x), for ∀x ∈ R.
Hence, we know that g(x) is also measurable.
5 Measurable function
Question: Let f : R → R be a Lebesgue measurable function. Suppose that A ⊂ R is aBorel set. Show that the set {x ∈ A : f(x) > x} is Lebesgue measurable.
Proof: Define a function g : R → R by g(x) = f(x) − x, then g(x) is measurable (bythe Theorem 6 in the page 259 of Royden’s book). So the set B = {x ∈ R : g(x) > 0} ismeasurable. Then, {x ∈ A : f(x) > x} = A ∩ B is measurable.
III
Real Analysis Homework: #5
Yingwei Wang ∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
Note: In this paper, {f(x) satisfies some property} = {x : f(x) satisfies some property}
1 Integrable function
1.1 a
Question: Show that if f is integrable then the set {f(x) 6= 0} is of σ-finite measure.
Proof: On one hand,
{f(x) 6= 0} = {|f(x)| > 0} =
∞⋃
n=1
{|f(x)| ≥1
n}. (1.1)
On the other hand,
∞ >
∫
|f |dµ ≥
∫
{|f(x)|≥ 1
n}|f |dµ ≥
1
nµ{|f(x)| ≥
1
n}, ∀n ∈ N, (1.2)
which means
µ{|f(x)| ≥1
n} < ∞,∀n ∈ N. (1.3)
From (1.1) and (1.3), we can know that the set {f(x) 6= 0} is of σ-finite measure.
1.2 b
Question: Show that if f is integrable, f ≥ 0, then f = lim ϕn for some increasing sequenceof simple functions each of which vanishes outside a set of finite measure.
Proof: By proposition 7 on Page 260 in Royden’s book, since the set {f(x) 6= 0} is ofσ-finite measure, we can find a sequence (ϕn) of simple functions defined on {f(x) 6= 0} withϕn+1 ≥ ϕn such that f = lim ϕn and each ϕn vanishes outside a set of finite measure.
Then we can define ϕn(x) = 0 on the set {f(x) = 0}, and get the conclusion.
∗
E-mail address: [email protected]; Tel : 765 237 7149
I
Yingwei Wang Real Analysis
1.3 c
Question: Show that if f is integrable with respect to µ, then given ǫ > 0 there is a simplefunction ϕ such that
∫
|f − ϕ|dµ < ǫ.
Proof: By the assumption, f+ and f− are nonnegative integrable functions. By (b), thereare increasing sequence (φn) and (ϕm) such that f+ = lim φn and f− = lim ϕm. By theMonotone Convergence Theorem, we have
∫
f+dµ = lim
∫
φn,∫
f−dµ = lim
∫
ϕm.
So given ǫ > 0, there are φN and ϕM such that∫
f+dµ −
∫
φNdµ <ǫ
2,
∫
f−dµ −
∫
ϕMdµ <ǫ
2.
Let ϕ = φN − ϕM . Then ϕ is also a simple function and satisfies∫
|f − ϕ|dµ
≤
∫
|f+ − φN |dµ +
∫
|f− − ϕM |dµ
=
(∫
f+dµ −
∫
φNdµ
)
+
(∫
f−dµ −
∫
ϕMdµ
)
< ǫ.
2 Measurable set
Question: Let f : R → R be defined by f(x) = x3 − 2x2 + x − 1. Show that if A ⊂ R isLebesgue measurable, then f(A) ⊂ R is Lebesgue measurable.
Proof: Since f(x) is a polynomial, then it is measurable and there exists a sequence ofstep functions (ϕn) s.t. ϕn(x) → f(x),∀x ∈ R.
By assumption, A ⊂ R is Lebesgue measurable, which means ∀E ⊂ R,
m(E) = m(E ∩ A) + m(E ∩ Ac).
II
Yingwei Wang Real Analysis
For each step function ϕn(x), if A is measurable, then ϕn(A) is just a set of points so itis measurable. Thus f(A) = lim
n→∞ϕn(A) is measurable.
3 Limits
Question: Let (X,A, µ) be a measure space with µ(X) < ∞. Let f : X → (−1, 1) be ameasurable function. Consider the sequence
an =
∫
X(1 + f + · · · + fn)dµ.
Show that either (an) converges to a finite number or otherwise limn→∞
an = ∞. (In other words
(an) cannot have two distinct limit points.)
Proof: Since ∀x ∈ X, |f(x)| < 1, the series∑∞
n=0 |f(x)|n is convergent for ∀x ∈ X. Itfollows that all of the functions g(x) =
∑∞n=0(f(x))n and h0(x) =
∑∞n=0(f(x))2n, h1(x) =
∑∞n=0(−f(x))2n+1 are absolutely convergent.
Let A = {f ≥ 0}, B = {f < 0}, then X = A ∪ B.
a+n =
∫
A
n∑
j=0
(f(x))jdµ
a−n =
∫
B
n∑
j=0
(f(x))jdµ =
{
∫
B
∑kj=0 f2jdµ −
∫
B
∑kj=0(−f)2j+1dµ if n = 2k + 1,
∫
B
∑kj=0 f2jdµ −
∫
B
∑kj=0(−f)2j−1dµ if n = 2k.
Hence,
limn→∞
an = limn→∞
(a+n + a−n ) =
∫
Ag(x)dµ +
∫
Ah0(x)dµ −
∫
Bh1(x)dµ.
which means limn→∞
an can not have two distinct limit points.
4 Convergence
Question: Show that the following sequence is convergent in R.
an = n
∫ 1
1/n
cos(x + 1/n) − cos x
x3/2dx.
Proof: Consider the function
fn(x) =cos(x + 1/n) − cos x
1/n
X[1/n,1](x)
x3/2, x ∈ [0, 1],
III
Yingwei Wang Real Analysis
where XA(x) is the characteristic function.
It is obvious that
limn→∞
fn = −sin x
x3/2, x ∈ [0, 1].
Let f(x) = − sinxx3/2
, then |f(x)| ≤ xx3/2
= 1x1/2
since sin x ≤ x, x ∈ [0, 1].
Let g(x) = 1x1/2
, then ∃N ∈ N, s.t. ∀n > N, |fn(x)| < g(x). Then by Lebesgue dominatedconvergence theorem, we have
limn→∞
an = limn→∞
∫ 1
0fn(x)dx =
∫ 1
0lim
n→∞fn(x)dx =
∫ 1
0−
sin x
x3/2dx <
∫ 1
0
1
x1/2dx = 2.
So an is convergent.
5 Infinitive sum
Question: Let (X,A, µ) be a complete measure space. Let (fn)∞n=1 be a sequence of A-measurable functions, fn : X → R. Suppose that
∞∑
n=1
∫
X|fn|dµ < ∞. (5.1)
Show that the series∑∞
n=1 fn(x) is absolutely convergent µ-almost everywhere on X.
Proof: First, claim that
∫
X
(
∞∑
n=1
|fn|
)
dµ =∞∑
n=1
∫
X|fn|dµ. (5.2)
Let Fn =n∑
i=1|fi|, F∞ =
∞∑
i=1|fi|, ∀x ∈ X. Then the statement (5.2) becomes
∫
XF∞dµ = lim
n→∞
n∑
i=1
∫
X|fi|dµ = lim
n→∞
∫
XFndµ. (5.3)
Since Fn(x) ≤ Fn+1(x) and Fn → F∞, by the Monotone Convergence Theorem, we canget (5.3).
Second, by the assumption (5.1) and the claim (5.2), we can get
∫
X
(
∞∑
n=1
|fn|
)
dµ =∞∑
n=1
∫
X|fn|dµ < ∞, (5.4)
which means the measure µ ({∑∞
n=1 |fn| = ∞}) = 0. That is to say the series∑∞
n=1 fn(x) isabsolutely convergent µ-almost everywhere on X.
IV
Real Analysis Homework: #6
Yingwei Wang ∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1 Integral function
Question: Let f : R → R be a Lebesgue integrable function. Prove that g(x) is alsoLebesgue integrable.
Proof: First, if f(x) is a non-negative simple function, then
f(x) =
n∑
k=1
ckχEi(x), ∀x ∈ R,
g(x) =
n∑
k=1
ckχEi+{1}(x), ∀x ∈ R.
Hence, g(x) is also a non-negative simple function and Lebesgue integrable.
Second, if f(x) is a non-negative measurable function, then there exists a sequenceof non-negative simple monotonic increasing functions {ϕk(x)}∞k=1 s.t.
limk→∞
ϕk(x) = f(x), ∀x ∈ R.
It is obvious that {ϕk(x − 1)}∞k=1 is also a monotonic increasing sequence and
limk→∞
ϕk(x − 1) = f(x − 1), ∀x ∈ R.
∗
E-mail address: [email protected]; Tel : 765 237 7149
I
Yingwei Wang Real Analysis
Hence,
∫
R
g(x) dm
=
∫
R
f(x − 1) dm
= limk→∞
∫
R
ϕk(x − 1) dm
= limk→∞
∫
R
ϕk(x) dm
=
∫
R
f(x)dm < ∞.
So we have g(x) is Lebesgue integrable on R.
Finally, if f(x) is an arbitrary function on R, we can rewrite f as f = f+ − f−,where both f+ and f− are non-negative integrable function, and then get the sameconclusion about g(x).
2 Zero function
Question: Let f : [0, 1] → R be a Lebesgue measurable function. Suppose that forany function g : [0, 1] → R, fg is Lebesgue integrable and
∫
[0,1] fg dm = 0. Prove thatf = 0 almost everywhere with respect to the Lebesgue measure.
Proof: Suppose that there exists a set A ⊂ [0, 1], m(A) > 0, s.t. ∀x ∈ A, f(x) 6= 0.Without loss of generality, we can assume that f(x) > 0,∀x ∈ A.
∀ε > 0, we can find a sequence of intervals In = (a, b) ⊂ [0, 1] such that A ⊂ ∪In
and m(A\ ∪ In) < ε. Choose g(x) as follows:
g(x) =
{
−(x − an)(x − bn), x ∈ [an, bn],0, x ∈ [0, 1]\ ∪ In.
It is easy to verify that g(x) is continuous and satisfies: g(x) > 0, x ∈ ∪In, andg(x) = 0, x ∈ [0, 1]\ ∪ In. Hence, fg ≥ 0, x ∈ [0, 1] and fg > 0, x ∈ A. So we have
∫
[0,1]fg dm =
∫
∪In
fgdm =
∫
A
fgdm +
∫
∪In\Afgdm,
II
Yingwei Wang Real Analysis
On one hand,∫
Afgdm > 0. On the other hand, since m(A\ ∪ In) < ε, we can
choose ε such that∣
∣
∣
∫
∪In\Afgdm
∣
∣
∣<
∫
Afgdm. It follows that
∫
[0,1] fg dm > 0, which
contradicts that∫
[0,1] fg dm = 0,∀g ∈ C1([0, 1]).
Thus, f = 0, a.e. x ∈ [0, 1].
Note: Another method is by the statement that “Any measurable function can beapproximated by a sequence of continuous functions”.
3 σ-finite
Question: Let (X,A), µ be a measure space and let f : X → R be A-measurable.Suppose that µ is σ-finite. Suppose also that there is c ≥ 0 such that for all E ∈ Awith µ(E) < ∞ one has |
∫
Ef dµ| ≤ c. Show that f is Lebesgue integrable on X.
Proof: Since µ is σ-finite, we can choose a sequence of sets (Xn) such that X = ∪Xn,X1 ⊂ X2 · · · ⊂ Xn ⊂ · · · ⊂ X, lim
n→∞Xn = X and
µ(Xn) < ∞,∀n ∈ N.
Define fn(x) = f(x)χXn,∀x ∈ X, then
limn→∞
fn(x) = f(x) and |fn(x)| ≤ |f(x)|, ∀x ∈ X.
Since µ(Xn), by assumption,
∣
∣
∣
∣
∫
X
fn dµ
∣
∣
∣
∣
=
∣
∣
∣
∣
∫
Xn
fdµ
∣
∣
∣
∣
≤ c, ∀n ∈ N.
By Lebesgue dominated control theorem, we have
∫
X
f dµ = limn→∞
∫
X
fn dµ ≤ limn→∞
∣
∣
∣
∣
∫
X
fn dµ
∣
∣
∣
∣
≤ c,
which implies f is Lebesgue integrable on X.
III
Yingwei Wang Real Analysis
4 Limits
Question: Let (X,A), µ be a measure space. Let (fn)∞n=1 be a sequence of A-measurablefunctions, fn : X → [0,+∞]. Show that
∫
X
lim infn→∞
fndµ ≤ lim infn→∞
∫
X
fndµ.
Proof: Let gn(x) = infk≥n
fn(x), f(x) = lim infn→∞
fn(x), ∀x ∈ X. Then gn are nonneg-
ative and gn increases to f , limn→∞
gn(x) = f(x). Therefore,
∫
X
gndµ ≤ infk≥n
∫
X
fkdµ.
If we take the limit as n → ∞, on the left side we obtain∫
Xfdµ by the monotone
convergence theorem, while on the right side we obtain lim infn→∞
∫
Xfndµ.
IV
Real Analysis Homework: #7
Yingwei Wang ∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1 Limits
Question: Compute the following limits and justify your calculation.
(i) limn→∞
∫∞0
(
1 + xn
)−nsin x
ndx.
Solution: Let fn(x) =(
1 + xn
)−nsin x
n , then |fn(x)| ≤ e−x,∀x ∈ [0,∞). Besides, limn→∞
fn(x) =
0. Then we have
limn→∞
∫ ∞
0fndx = 0.
(ii) limn→∞
∫ n0
(
1 − xn
)nex/2dx.
Solution: Let fn(x) =(
1 − xn
)nex/2χ[0,n], f(x) = e−x/2 then lim
n→∞fn(x) = f(x).
Let gn = 2f − fn, then gn monotonic increasingly converges to f as n → ∞. Besides, whenn is sufficient large, gn ≥ 0. By Monotonic Convergence Theorem, we can see that
limn→∞
gn(x) = f(x).
So we have
limn→∞
∫ n
0
(
1 −x
n
)nex/2dx = lim
n→∞
∫ ∞
0fndx =
∫ ∞
0fdx = 2.
(ii) limn→∞
∫ n0
(
1 + xn
)ne−2xdx.
Solution: Let fn(x) =(
1 + xn
)ne−2xχ[0,n], f(x) = e−x then lim
n→∞fn(x) = f(x) and fn(x) ≤
f(x),∀x ∈ [0,∞).
So we have
limn→∞
∫ n
0
(
1 +x
n
)ne−2xdx = lim
n→∞
∫ ∞
0fndx =
∫ ∞
0fdx = 1.
∗
E-mail address: [email protected]; Tel : 765 237 7149
I
Yingwei Wang Real Analysis
2 Limits
Question: Let (X,A, µ) be measure space and let f : X → [0,+∞) be A-measurable. Supposethat 0 < c =
∫
X fdµ < ∞. Show that
limn→∞
∫
Xn ln
(
1 +
(
f
n
)α)
dµ =
+∞, if 0 < α < 1,c, if α = 1,0, if 1 < α < ∞.
Proof: Let
fn(x) = n ln
(
1 +
(
f
n
)α)
= ln
(
1 +
(
f
n
)α)n
= ln
(
(
1 +fα
nα
)nα)n1−α
= n1−α · ln
(
(
1 +fα
nα
)nα)
If 0 < α < 1, thenlim
n→∞fn(x) = +∞ · fα = +∞.
By Fatou’s Lemma, we have
limn→∞
∫
Xfn(x)dµ ≥ lim
n→∞
∫
Xfn(x)dµ ≥
∫
Xlim
n→∞fn(x)dµ = +∞.
If α = 1, thenlim
n→∞fn(x) = f(x).
Besides, ∀x ∈ X, fn(x) ≤ f(x). By Lebesgue Dominated Control Theorem, we have
limn→∞
∫
Xfn(x)dµ =
∫
Xf(x)dµ = c.
If 1 < α < ∞, thenlim
n→∞fn(x) = 0 · fα = 0.
By Monotonic Convergence Theorem, we have
limn→∞
∫
Xfn(x)dµ =
∫
X0 dµ = 0.
II
Yingwei Wang Real Analysis
3 Limits
Question: Let (X,A, µ) be measure space. Let (fn)∞n=1 be sequence of A-measurable func-tions, fn : X → [0,+∞). Suppose that limn→∞ fn(x) = f(x) for all x ∈ X and thatlimn→∞
∫
X fn(x)dµ =∫
X f(x)dµ < ∞. Show that limn→∞
∫
A fn(x)dµ =∫
A f(x)dµ for ev-ery set A ∈ A.
Proof: For any set A ∈ A, define gn(x) = fn(x)χA, hn(x) = fn(x)χX\A, the we have
limn→∞
fn(x) = f(x),
limn→∞
gn(x) = f(x)χA,
limn→∞
hn(x) = f(x)χX\A,
gn(x) + hn(x) = fn(x), ∀n and ∀x.
On one hand, by Fatou’s Lemma, we have∫
XfχAdµ ≤ lim
n→∞
∫
Xgndµ, (3.1)
∫
XfχX\Adµ ≤ lim
n→∞
∫
Xhndµ, (3.2)
⇒
∫
Xfdµ ≤ lim
n→∞
∫
Xgndµ + lim
n→∞
∫
Xhndµ. (3.3)
On the other hand,
limn→∞
∫
Xgndµ + lim
n→∞
∫
Xhndµ = lim
n→∞
(∫
Xgndµ +
∫
Xhndµ
)
≤ limn→∞
∫
Xfndµ =
∫
Xfdµ. (3.4)
By (3.3) and (3.4), we can see that
limn→∞
∫
Xgndµ + lim
n→∞
∫
Xhndµ =
∫
Xfdµ. (3.5)
Then by (3.1), (3.2) and (3.5), we can see that∫
XfχAdµ = lim
n→∞
∫
Xgndµ, (3.6)
∫
XfχX\Adµ = lim
n→∞
∫
Xhndµ, (3.7)
So we can get
limn→∞
∫
Xfn(x)χAdµ =
∫
Xf(x)χAdµ
⇒ limn→∞
∫
Afn(x)dµ =
∫
Af(x)dµ.
III
Yingwei Wang Real Analysis
4 The absolute continuity of integral
Question: Let f : R → [0,∞] be a Lebesgue integrable function. Show that for each α > 0there is β > 0 such that if B ⊂ R is Lebesgue measurable and m(B) < β, then
∫
B fdm < α.
Proof: Since f is nonnegative and Lebesgue integrable, ∀α > 0,∃ a nonnegative simplefunction ϕ(x), such that
∫
R
f − ϕdm =
∫
R
fdm −
∫
R
ϕdm <α
2.
Let ϕ(x) < M , then we can choose B ⊂ R s.t. m(B) < α2M , then
∫
Bfdm ≤
∫
Bf − ϕdm +
∫
Bϕdm
≤
∫
R
f − ϕdm + M · m(B)
<α
2+ M ·
α
2M= α.
IV
Real Analysis Homework: #8
Yingwei Wang ∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
Note: In this paper, {f(x) satisfies some property} = {x : f(x) satisfies some property}
1 Integral function
Question: Let f : (0, 1) → R be a Lebesgue integrable function. Suppose that∫(0,x)
fdm for all x ∈ (0, 1). (1.1)
Show that f(x) = 0 a.e.
Proof: First I want to claim that∫ 10 fdm = 0. Let fn = fχ(0,1−1/n), then |fχ(0,1/n)(x)| ≤
|f(x)|, ∀n ∈ N and x ∈ (0, 1) By Lebesgue dominated convergence theorem,∫ 1
0fdm = lim
n→∞
∫(0,1)
fndm = limn→∞
∫(0,1−1/n)
fdm = 0.
Second, let A = {f(x) > 0}. Suppose m(A) > 0, then ∃ closed set B ⊂ A, m(B) > 0.Let C = [0, 1]\B, then C is an open set, i.e. C = ∪(an, bn).∫
Bfdm +
∫C
fdm =
∫(0,1)
fdm = 0,
⇒
∫C
fdm 6= 0, (since
∫B
fdm > 0)
⇒
∫∪(an,bn)
fdm 6= 0,
⇒ ∃ (an, bn), s.t.
∫(an,bn)
fdm 6= 0,
⇒
∫(0,bn)
fdm −
∫(0,an)
fdm 6= 0,
∗
E-mail address: [email protected]; Tel : 765 237 7149
I
Yingwei Wang Real Analysis
which contradicts with (1.1).
So m(A) = 0. Similarly, we can prove that m({f(x) < 0}) = 0. That is to say f(x) = 0a.e.
2 Convergence of L1 norm and pointwise
Question: Let (fn) be a convergent sequence in L1(X,A, µ) with limit f . Show that (fn)has a subsequence (fnk
) such that fnk→ f for almost x ∈ X.
Proof: Since (fn) be a convergent sequence in L1(X,A, µ), then (fn) is a Cauchy se-quence in L1(X,A, µ), which means
∫X|fn − fm|dµ → 0, as n,m → ∞.
⇒ ∀ε > 0, ∃N , s.t. ∀n,m > N ,
m({|fn − fm| 6= 0}) < ε.
⇒ ∀i,∃ki ∈ N s.t. ∀n,m > ki,
m({|fn − fm| ≥1
2i}) <
1
2i.
We can assume that k1 < k2 < · · · < ki < ki+1 < · · · , and define
Ei = {|fki− fki+1
| ≥1
2i},
then m(Ei) < 12i .
Let
E =
∞⋂j=1
∞⋃i=j
Ei,
then m(E) = 0.
∀x ∈ X\E, ∃j s.t.
x ∈ X
∞⋃i=j
Ei.
Then ∀i ≥ j,
|fki+1(x) − fki(x)| <
1
2i,
II
Yingwei Wang Real Analysis
which means∑
|fki+1(x) − fki(x)| is convergent, so fk1
+∑
(fki+1(x) − fki(x)) is also
convergent.
Denote limki→∞
fki(x) = g(x),∀x ∈ X\E, then |fnk
− g| → 0, as n → ∞, ∀x ∈ X\E. So
∫X|fnk
− g|dµ =
∫X\E
|fnk− g|dµ → 0,
⇒ ‖fnk− g‖L1 → 0.
Since ‖fnk− f‖L1 → 0, by the completeness of L1(X,A, µ), f = g for almost all x ∈ X.
So we have a subsequence (fki) which is convergent to f for almost all x ∈ X.
3 Integral functions and continuous functions
Question: Let f : [0, 1] → R be a Lebesgue integrable function. Show that there is asequence of continuous functions fn : [0, 1] → R which converges pointwise to f almosteverywhere.
Proof: First, I want to introduce the Lusin’s Theorem (Problem 3.31 on Page 74 inRoyden’s book):
Theorem 3.1 (Lusin). Let f be a measurable real-valued function on an interval [a, b].Then given δ, there is a continuous function ϕ on [a, b] such that m({f 6= ϕ}) < δ.
By this theorem, for f ∈ L1([0, 1]), ∀n ∈ N, ∃ a sequence of continuous functionsϕn ∈ C([0, 1]) s.t.
m({ϕn 6= f}) <1
2n.
Then we have∫X|ϕn − f |dm =
∫{ϕn 6=f}
|ϕn − f |dm → 0, as n → ∞,
since m({δn 6= f}) → 0 as n → ∞.
It implies that ϕn converges to f in the L1 norm. The conclusion of the last problemtells us that (ϕn) has a subsequence (ϕnk
) such that ϕnk→ f for almost x ∈ X.
III
Real Analysis Homework: #9
Yingwei Wang ∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1 Absolutely continuous
Question: Let f, g : [a, b] → R be two absolutely continuous functions. Prove or disprovethat h(x) = ef(x)g(x) is absolutely continuous on [a, b].
Proof: First, I want to show that if both f and g are absolutely continuous, then fg isalso absolutely continuous.
There exists M > 0 such that |f(x)| ≤ M and |g(x)| ≤ M for ∀x ∈ [a, b]. Given ε > 0,there exits δ > 0 such that
n∑i=1
|f(xi) − f(yi)| <ε
2M, and
n∑i=1
|g(xi) − g(yi)| <ε
2M,
for any finite collection {(xi, yi)} of disjoint intervals in [a, b] with∑n
i=1 |xi − yi| < δ.
Then
n∑i=1
|f(xi)g(xi) − f(yi)g(yi)|
≤
n∑i=1
|f(xi)||g(xi) − g(yi)| +
n∑i=1
|g(yi)||f(xi) − f(yi)|
<ε
2+
ε
2= ε.
Second, I want to show that if q : [a, b] → [a′, b′] is absolutely continuous on [a, b] andp : [a′, b′] → R is Lipschitz continuous on [a′, b′], then p(q(x)) is absolutely continuous on[a, b].
∗
E-mail address: [email protected]; Tel : 765 237 7149
I
Yingwei Wang Real Analysis
There exists N > 0 such that for ∀x, y ∈ [a′, b′],
|p(x) − p(y)| ≤ N |x − y|.
Given ε, there exits δ > 0 such that
n∑i=1
|q(xi) − q(yi)| <ε
N
for any finite collection {(xi, yi)} of disjoint intervals in [a, b] with∑n
i=1 |xi − yi| < δ.
Thenn∑
i=1
|p(q(xi)) − p(q(yi))|
≤ N
n∑i=1
|q(xi) − q(yi)|
< ε.
Finally, since ex is Lipschitz continuous in any finite interval [a′, b′], so h = efg isabsolutely continuous if both f and g are absolutely continuous.
2 Bounded variation
Question: Let f ∈ BV [a, b]. Show that T baf ≥
∫ ba |f ′(x)|dx.
Proof: By Jordan Theorem, f(x) can be written as the difference of two monotoneincreasing functions on [a, b],
f(x) = g(x) − h(x),
where
g(x) =1
2(T x
a f + f(x)),
h(x) =1
2(T x
a f − f(x)).
It is easy to verify that both g and h are increasing, so g′ and h′ are nonnegative.
By Lebesgue Theorem, ∫ b
ag′(x)dx ≤ g(b) − g(a),
∫ b
ah′(x)dx ≤ h(b) − h(a).
II
Yingwei Wang Real Analysis
Since f ′ = g′ − h′, |f ′| ≤ g′ + h′. Then we have
∫ b
a|f ′(x)|dx ≤
∫ b
ag′(x) + h(x)dx ≤ g(b) − g(a) + h(b) − h(a) = T b
af.
3 Bounded variation
Question: Let f ∈ BV [a, b]. Show that f has at most countably many points of disconti-nuity.
Proof: First, I want to show that ∀x0 ∈ [a, b], both limx→x+
0
f(x) and limx→x−
0
f(x)exist.
By Jordan Theorem, f = g − h where g and h are monotone increasing functions on[a, b]. Let G = supx∈[a,x0) g(x) and H = supx∈[a,x0) h(x). It is obvious that A,B < ∞.
Given ε > 0, there exists δ > 0 such that
G −ε
2< g(x0 − δ) ≤ G,
H −ε
2< h(x0 − δ) ≤ H.
Then for ∀x ∈ (x0 − δ, x0), we have
G −ε
2< g(x) ≤ G ⇒ 0 ≤ G − g(x) <
ε
2,
H −ε
2< h(x) ≤ H ⇒ 0 ≤ H − h(x) <
ε
2.
It follows that|G − H − f(x)| ≤ (G − g(x)) + (H − h(x)) < ε,
for ∀x ∈ (x0 − δ, x0). So limx→x+
0
f(x) = G−H. Similarly we can know that limx→x−
0
f(x)
also exists.
Second, I want to show that the set of discontinuities of f is at most countably.
Let E = {x ∈ [a, b] : f(x+) 6= f(x−)}, E1 = {x ∈ [a, b] : g(x+) > g(x−)}, E2 = {x ∈[a, b] : h(x+) > h(x−)}. Since g and h are monotone increasing and f = g − h, we haveE = E1 ∪ E2.
For ∀x ∈ E1, ∃rx ∈ Q such that g(x−) < rx < g(x+). Besides, if x1 < x2, theng(x1+) ≤ g(x2−) so rx1
6= rx2. Thus x → rx is a bijection between E1 and a subset of Q,
which means E1 is at most countably.
Similarly, E2 is at most countably and further E = E1 ∪ E2 is also at most countably.
III
Yingwei Wang Real Analysis
4 L1 space
Question: Let (fn) be a sequence of nonnegative functions in L[0, 1] such that∫ 10 fn(t)dt
and∫ 11/n dt ≤ 1/n for all n ≥ 1. If g(x) = supn fn(x) show that g /∈ L1[0, 1].
Proof: Suppose that g ∈ L1[0, 1], I want to get a contradiction.
By the absolutely continuity of the integral, for ε = 13 , there exists δ such that for any
measurable subset E ⊂ [0, 1], if m(E) < δ, then
∫E
gdm <1
3. (4.1)
Choosing an integer n satisfying n ≥ 3 and 1/n < δ, then since∫ 10 fn(t)dt and
∫ 11/n dt ≤
1/n, we have ∫ 1/n
0fndt = 1 −
∫ 1
1/nfndt ≥ 1 − 1/n,
for all n > 3.
By the definition of g, we have
∫ 1/n
0gdt ≥
∫ 1/n
0fndt ≥ 1 − 1/n ≥ 1 − 1/3 ≥
2
3. (4.2)
But by (4.1), we have ∫ 1/n
0gdt <
1
3. (4.3)
It is obvious that (4.2) contradicts with (4.3).
IV
Real Analysis Homework: #10
Yingwei Wang ∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1 Convergence series
Question: Let (an)∞n=1 be sequence of non-negative numbers such that
∞∑
n=1
an < ∞. (1.1)
Let r1, r2, . . . be an enumeration of the rational numbers. Show that the series
∞∑
n=1
a2n
|x − rn|(1.2)
converges for almost all x ∈ R.
Proof: I found two methods to prove this. Hopefully, both of them are correct.
1.1 Method one: integral
Consider the new series:∞∑
n=1
an√
|x − rn|(1.3)
If (1.3) converges, then (1.2) is sure to converge.
If |x − rn| ≥ 1 for some n, then
an√
|x − rn|≤ an. (1.4)
∗
E-mail address: [email protected]; Tel : 765 237 7149
I
Yingwei Wang Real Analysis
So the convergence of the series is obvious.
If |x − rn| < 1 for some n, then rn − 1 < x < rn + 1. Consider the integral
∫ rn+1
rn−1
an√
|x − rn|dx
= an
(∫ rn
rn−1
1√rn − x
dx +
∫ rn+1
rn
1√x − rn
dx
)
= an
(∫
1
0
1√tdt +
∫
1
0
1√tdt
)
= 4an, (1.5)
which means∑∞
n=1an√|x−rn|
is convergent almost everywhere for x ∈ (rn − 1, rn + 1).
In sum, (1.3) converges for x ∈ R, a.e. and so does (1.2).
1.2 Method two: measure
Consider the setEε
n = {x : |x − rn| < εan}, (1.6)
then the measure of this set ism(Eε
n) = 2εan. (1.7)
For ∀x ∈ R\ ∪ Eεn, we have
an
|x − rn|≤ 1
ε
⇒ a2n
|x − rn|≤ an
ε
⇒∞
∑
n=1
a2n
|x − rn|≤ 1
ε
∞∑
n=1
an. (1.8)
As ε → 0, m(∪Eεn) → 0. Hence,
∑∞n=1
a2n
|x−rn|converges for almost all x ∈ R.
II
Yingwei Wang Real Analysis
2 Bounded variation functions
Question: Find all functions f ∈ BV [0, 1] with the property that
f(x) + (T x0 f)1/2 = 1,∀x ∈ [0, 1]. (2.1)
Solution: Let x = 0 in Eqn.(2.1), we can get
f(0) = 1.
Since F (x) = (T x0 f)1/2 is an increasing function, f(x) should be a decreasing function.
Hence, F (x) = (T x0 f)1/2 = (f(0) − f(x))1/2. So
f(x) + (f(0) − f(x))1/2 = 1,
⇒ (f(0) − f(x)) = (1 − f(x))2,
⇒ (1 − f(x)) = (1 − f(x))2,
⇒ f(x) = 1, ∀x ∈ [0, 1].
3 Absolutely continuous function
Question: Let f : [0, 1] → [0, 1], f(x) =√
x. Show that f ∈ AC[0, 1].
Proof: Since√
x = 1
2
∫ t0
t−1/2dt, by the Absolutely continuity of the integral (Propo-sition 4.14 in Page 88 of Royden’s book), it is easy to know that f(x) =
√x ∈ AC[0, 1].
4 Non-absolutely continuous function
Question: Construct f : [0, 1] → [0, 1], f ∈ AC[0, 1] such that√
f /∈ AC[0, 1].
Solution: Let f(x) be as following
f(x) =
{
x2 sin2( 1
x), 0 < x ≤ 1,0, x = 0.
(4.1)
It is easy to know that f(x) ∈ AC[0, 1] since f ′(x) = 2x sin2(1/x)−2 sin(1/x) cos(1/x)is bounded in [0, 1].
III
Yingwei Wang Real Analysis
However, for√
f ,√
f(x) =
{
x sin( 1
x), 0 < x ≤ 1,0, x = 0,
(4.2)
we choose the partition
0 <1
nπ + π/2<
1
nπ<
1
(n − 1)π + π/2< · · · <
1
π + π/2<
1
π<
1
π/2< 1.
Then
n∑
k=0
|√
f(xk + 1) −√
f(xk)| = | sin 1 − 2
π| +
n∑
k=0
1
kπ + π/2+
1
nπ + π/2.
So T 10
√f = ∞ and
√f /∈ BV [0, 1] and further
√f /∈ AC[0, 1].
IV
Real Analysis Homework: #11
Yingwei Wang ∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1 Open sets
Question: Show that every open subset U of R can be written uniquely as the union of acountable family of mutually disjoint open intervals.
Proof: Let U ∈ R be open. For x1 ∈ U , consider the interval I1 = (a1, b1) where
a1 = inf{t ∈ R : (t, x) ⊂ U},
b1 = sup{t ∈ R : (x, t) ⊂ U}.
Then for x2 = U\I1, consider the interval I2 = (a2, b2) where
a2 = inf{t ∈ R : (t, x) ⊂ U\I1},
b2 = sup{t ∈ R : (x, t) ⊂ U\I1}.
In this way, we get a sequence of disjoint open intervals {In}∞
n=1 such that ∪∞
n=1In = U .
2 Measurable sets
Question: Let f : R → R be a continuous functions with the property that m∗(f(A)) = 0whenever m∗(A) = 0, A ⊂ R. Show that if A ⊂ R is Lebesgue measurable, then f(A) isLebesgue measurable.
Proof: That A is measurable means ∀ε > 0, ∃ disjoint open intervals {In}∞
n=1 s.t.A ⊂ ∪In and m(∪In − A) < ε.
Since m∗(A) = inf{∑
l(In) : A ⊂ ∪In}, we have m∗(∪In) → m∗(A) and m∗(∪In−A) →0 as ε → 0.
By the assumption, m∗(f(∪In − A)) → 0, so m(∪f(In) − f(A)) < ε, for ∀ε > 0.
Since ∪f(In) is an open set and f(A) ⊂ ∪f(In), we know that f(A) is measurable.
∗
E-mail address: [email protected]; Tel : 765 237 7149
I
Yingwei Wang Real Analysis
3 Absolutely continuity
Question: Let f : R → R ∈ AC[a, b] for all a < b, a, b ∈ R. Show that if A ⊂ R is Lebesguemeasurable, then f(A) is Lebesgue measurable.
Proof: It is suffices to show that f ∈ AC[a, b] for all a < b, a, b ∈ R implies that f hasthe property that m(f(E)) = 0 whenever m(E) = 0, A ⊂ R.
That f is absolutely continuous means ∀ε > 0, ∃ δ such that for disjoint intervals{Ii = (xi, yi)}
ni=1 satisfying
n∑
i=1
(yi − xi) < δ,
thenn
∑
i=1
|f(yi) − f(xi)| < ε.
Suppose E ⊂ R and m(E) = 0. Let G = ∪(xi, yi) ⊃ E. Choose ci, di ∈ [xi, yi] such that
f([xi, yi]) = [f(ci), f(di)],
thenm(f(E)) ≤ m(f(G)) ≤
∑
|f(di) − f(ci)| < ε.
So m(f(E)) = 0.
By the conclusion of the Problem 2, we can know that if E ⊂ R is Lebesgue measurable,then f(E) is Lebesgue measurable.
4 Lebesgue measure
Question: Let f ∈ AC[a, b] for all a < b where a, b ∈ R. Suppose that f is strictlyincreasing. Show that if A is Lebesgue measurable then
m(f(A)) =
∫
A
f ′(t)dt. (4.1)
Proof: Case 1: A is an open interval A = (a0, b0).
On one hand, since f is strictly increasing, m(f(A)) = f(b0)−f(a0); on the other hand,∫
Af ′(t)dt =
∫ b0a0
f ′(t)dt = f(b0) − f(a0). Thus, (4.1) holds.
Case 2: If A is an open set, then A =⊔
In where Ii ∩ Ij = ∅, i 6= j, In = (an, bn).
II
Yingwei Wang Real Analysis
On one hand,
m(f(A)) =
∞∑
n=1
(f(bn) − f(an)) = f(b∞) − f(a1).
On the other hand,
∫
A
f ′(t)dt =
∞∑
n=1
∫ bn
an
f ′(t)dt =
∞∑
n=1
(f(bn) − f(an)) = f(b∞) − f(a1).
It implies the (4.1).
Case 3: If A is a Gσ-set, it suffices to show that∫
A
f ′(x)dx ≤ m∗(f(A)). (4.2)
Choose open intervals In such that f(A) ⊂ ∪In, then E ⊂ ∪Jn, where Jn = f−1(In).
Choose sequences {α(n)k } and {β
(n)k } such that
limk→∞
α(n)k = inf
x∈Jn
{x},
limk→∞
β(n)k = sup
x∈Jn
{x}.
Then we get∫
Jn
f ′(x)dx = limk→∞
∫ β(n)k
α(n)k
f ′(x)dx ≤ m(In).
As a sequence,∫
A
f ′(x)dx ≤
∫
∪Jn
f ′(x)dx ≤∑
∫
Jn
f ′(x)dx ≤ m(In).
By the definition of outer measure, we can get (4.2).
5 Signed measure
Question: Let µ = µ+ −µ− be the Jordan decomposition of a signed measure µ defined ona measurable space (X,A). Define the total variation of µ to be the measure |µ| = µ++µ−.Show that for each A ∈ A,
|µ|(A) = sup
{
n∑
i=1
|µ(Ai)| : (Ai)ni is a partition of A with Ai ∈ A, n ≥ 1
}
. (5.1)
III
Yingwei Wang Real Analysis
Proof: Let E,F be a Hahn decomposition for µ. That is to say X = E ∪F , E ∩F = ∅,and E is positive while F is negative respect to µ.
On one hand, we know that
µ+A = µ(A ∩ E),
µ−A = −µ(A ∩ F ),
⇒ |µ|(A) = µ(A ∩ E) − µ(A ∩ F ). (5.2)
On the other hand,
∫
A
(χA∩E − χA∩F ) dµ =
∫
A∩E
1dµ −
∫
A∩F
1dµ = µ(A ∩ E) − µ(A ∩ F ). (5.3)
By Eqn.(5.2) and Eqn.(5.3), we know that
|µ|(A) =
∫
A
(χA∩E − χA∩F ) dµ
=
n∑
i=1
∫
Ai
(χAi∩E − χAi∩F ) dµ,
where (Ai)ni=1 is a partition of A with Ai ∈ A.
Since for each Ai,∫
Ai(χAi∩E − χAi∩F ) dµ ≥ |µ(Ai)|, so we have
|µ|(A) =
n∑
i=1
∫
Ai
(χAi∩E − χAi∩F ) dµ ≥ sup
{
n∑
i=1
|µ(Ai)|
}
, (5.4)
for any partition (Ai)ni=1.
Besides, if we choose A1 = A ∩ E and A2 = A ∩ F , then
|µ|(A) = µ(A1) − µ(A2) = |µ(A1)| + |µ(A2)|,
which means
sup
{
n∑
i=1
|µ(Ai)|
}
≥ |µ|(A). (5.5)
By Eqn.(5.4) and (5.5), we can get the conclusion (5.1).
IV