Covering theorem and applications

1 Vitali's covering theorem for Lebesgue measure 21.1.. Vitali's covering theorem for Lebesgue measure.. Covering theorem has many applications in the study of integrals and limits ofint

HANOI PEDAGOGICAL UNIVERSITY 2

DEPARTMENT OF MATHEMATICS

LUONG THI TUYEN

COVERING THEOREMS AND APPLICATIONS

GRADUATION THESIS

Hanoi, 2019

HANOI PEDAGOGICAL UNIVERSITY 2

DEPARTMENT OF MATHEMATICS

LUONG THI TUYEN

COVERING THEOREMS AND APPLICATIONS

Speciality: Analysis

GRADUATION THESIS

Supervisor: Dr Do Hoang Son

Hanoi, 2019

I assure that the results in this thesis are true and the topic of this thesis is notidentical to other topic I also assure that all the help for this thesis has been acknowl-edge and that the used literature and other auxiliary resources have been completelyreferenced

The authorLuong Thi Tuyen

1 Vitali's covering theorem for Lebesgue measure 21.1 A 5r-covering theorem 21.2 Vitali's covering theorem for Lebesgue measure 4

2 Vitali's covering theorem for Radon measures 72.1 Besicovitch's covering theorem 72.2 Proof of Theorem 2.1 13

Tabel of notation

We introduce here the notation for some basic concepts which are not defined

in the thesis

Z, the set of integers

R, the set of real numbers

Rn, the n- dimensional euclidean space equipped with the inner product x · yand the norm |x|

Sn−1 = {x ∈ Rn : |x| = 1} ,the unit sphere

[a, b], (a, b), [a, b) and (a, b] are the closed, open and half-open

intervals in R

Ln, the Lebesgue measure on Rn

α(n) = Ln{x ∈ Rn: |x| ≤ 1} , the volume of the unit ball

A = ClA, the closure of the set A

χA, the characteristic function of A

Covering theorem has many applications in the study of integrals and limits ofintegrals.The theorem was first discovered and proved by Giuseppe Vitali in 1908 Itstates that if a subset E of is covered by a Vitali's covering, then there exists a family

of spheres is drawn from that covering such that the union of those is covered with E

N, where N is a set of Lebesgue 0 measure Vitali's covering theorem is then proved forthe case Hausdorff measure It also states that Vitali's covering theorem is generallyincorrect for the infinite dimensional case

The purpose of this thesis is to learn about Vitali's covering theorem and some ofits applications.Vitali's covering theorem is an interesting result in the analytical andtheoretical topology

This thesis consists of three chapters:

Chapter 1: We prove Vitali's covering theorem for Lebesgue measure

Chapter 2: We prove covering theorem of Radon measures

Chapter 3: We apply these covering theorems to prove some results about ferentiation of measures

So, B(xk, r(xk)) and B(xl, r(xl))are disjoint We choose the balls B(xi, r(xi))are joint in view of the definition of A1

dis-We need to prove there is only finitely many of B(xi, r(xi))

Indeed, assume that A is finite So, there exists A is finite: x1, , xk+1,

It implies that there exists a subsequence xk∈ A is converge

Similary, we only have finitely many of B(xi, r(xi)) are disjoint, say k2 such that A0

Proceeding in this manner we find the required balls

In the above proof, we have two restrictions

1 Firstly, we assumed that for each x ∈ A, there is only one ball B(x, r(x)) To fix it, wecan select for each centre x a ball B(x, r(x)) ∈ B such that r(x) > 14

15sup {r : B(x, r) ∈ B}and instead of choosing number 3 in (1), we use 8

3

2 Secondly, we assumed that the centres lie in a bounded set To avoid this the proofcan be modified by choosing the new points xi not too far from a fixed point a ∈ A;for example if x and y were possible selections and d(y, a) > 2d(x, a) we would make arule that we cannot pick y

We can now easily derive a Vitali-type covering theorem for the Lebesgue sure Ln

1.2 Vitali's covering theorem for Lebesgue

mea-sure

Theorem 1.2 Let A ⊆ Rn Let B be the family of closed balls in Rn satisfying:

∀x ∈ A : inf {d(B) : x ∈ B ∈ B} = 0

Then there exists {Bi}∞

i=1∈ B such thati)Bi1 ∩ Bi2 = ∅ ∀i1 6= i2

Moreover, for every ε > 0, we can choose Bi such that

Applying a 5r-covering theorem, ∃Bi ∈ B, i = 1, , k,

such that Bi 1 ∩ Bi2 = ∅ ∀i1 6= i2; Bi ⊂ V

7n − 1

6n

Now A1 is bounded set and choose V1 such that

B = {B((x, y), y) : x ∈ R, 0 < y < ∞} covers A = {(x, 0) : x ∈ R} in the sense of theorem 1.2 but for any countable subcol-lection B1, B2, we have

Then there exists {Bi}∞

i=1 such that:

the-2.1 Besicovitch's covering theorem

We shall begin with a simple lemma from plane geometry Instead of the ing elementary geometric considerations one can also easily deduce it from the cosineformula for the angle of a triangle in terms of the side-lengths

follow-Lemma 2.1 Assume that a, b ∈ R2 satisfy two following conditions:

i, 0 < |a| < |a − b|

ii, 0 < |b| < |a − b|

≥ 1

Proof Let a(x1, y1) and b(x2, y2)

From the hypothesis, we have

(x1− x2)2+ (y1+ y2)2 > x21+ y12(x1− x2)2+ (y1+ y2)2 > x22+ y22

=p(cosθ1− cosθ2)2+ (sin θ1− sinθ2)2

=p2 − 2(cosθ1cosθ2+ sinθ1sinθ2)

Proof For each i = 1, , k, we suppose that ai 6= 0 and

ai ∈ B(a/ j, rj) ⇒ |ai− aj| > rj (2)From (1) and (2), we have

|ai| < ri < |ai− aj| fori 6= j

Applying lemma 2.1 with a = ai and b = bj for i 6= j in the two- dimensional plane,

we have

≥ 1 for i 6= j (∗)Since the unit sphere Sn−1 is compact there is an integer N(n) with the followingproperty: if y1, , yk ∈ Sn−1 with |yi− yj| ≥ 1for i 6= j, then k ≤ N(n) By (*), N(n)

is what we want

Theorem 2.2 (Besicovitch's covering theorem) Assume that A is a bounded subset of

Rn B is a family of closed balls such that each point of A is the centre of some ball of

where P (n) is an integer depending only on n

(2) There exists family B1, , BQ(n)⊂ B covering A such that

ii, B ∩ B0 = ∅ for B, B0 ∈ Bi with B 6= B0

where Q(n) is an integer depending only on n

Proof (1) For each x ∈ A, pick one ball B(x, r(x)) ∈ B As A is bounded, we mayassume that

M1 = sup

x∈A

r(x) < ∞

x1 ∈ A with r(x1) ≥ M1

2and then inductively

k2 < ,a decreasing sequence of positive numbers Miwith 2Mi+1≤ Mi, and a sequence

of balls Bi = B(xi, r(xi))inB with the following properties Let

To establish also the second statement of (1), suppose that a point x belongs to p balls

p ≤ P (n) = 16nN (n)with N(n) as in Lemma 2.2

Using (5) and Lemma 2.2 we see that the indices mi can belong to at most N(n)different blocks Ij, that is,

card {j : Ij∩ {mi : i = 1, 2, , p} 6= ∅} ≤ N (n)

Consequently it suffices to show that

card (j : Ij ∩ {mi : i = 1, 2, , p}) ≤ 16n for j = 1, 2, (6)Fix j and write

Ij ∩ {mi : i = 1, 2, , p} = {l1, , lq}

By (3) and (4) the balls B(xl i,1

4r(xli)), i = 1, , q,are disjoint and they are contained

4r(xli))



≤ Ln(B(x, 2Mj))

= α(n)(2Mj)n.and so q ≤ 16n as disired This proves (6), and thus also (1)

(2) Let B1, B2, be the balls found in (1) Letting Bi = B(xi, ri), there are for each

 > 0 only finitely many balls Bi with ri ≥  because of (1) and the boundedless of A.Thus we may assume r1 ≥ r2 ≥

Let B1,1 = B1 and then inductively if B1,1, B1,j have been chosen, B1,j+1 = Bk where

k is the smallest integer with

We then have to show that m ≤ 4nP (n).

Due to the fact that the balls Bi cover A, we can find i with x ∈ Bi Then for each

k = 1, , m, Bi ∈ B/ k, which means by the construction of Bk that Bi ∩ Bk,ik 6= ∅ forsome ik for which ri ≤ rk,ik, ri and rk,i k being the radii of Bi and Bk,i k, respectively.Hence, there are balls B0

k of radius ri

2 contained in (2Bi) ∩ Bk,ik for all k = 1, , m.Since each point of Rn is contained in at most P (n) balls Bk,i k, k = 1, , m,this is alsotrue for the smaller balls B0

2.2 Proof of Theorem 2.1

Suppose that µ(A) > 0 Firstly, we will assume A is bounded

µis a Radon measure on Rn,then

µ(A) = inf {µ(U ) : A ⊂ U, U is open } for A ⊂ Rn

It implies that there is an open set U such that A ⊂ U and

µ(U ) ≤



1 + 14Q(n)

µ(A)where Q(n) is as in Besicovitch's covering theorem

As a result of that theorem, we can find B1, , BQ(n) ∈ B such that {Bi}Q(n)i=1 ⊂B; Bi 1 ∩ Bi 2 = ∅ ∀i1 6= i2 and

i of Bi, we haveµ(A) ≤ 2Q(n).µ[B0i⇒ µ[Bi0≥ 1

2Q(n).µ(A)Put A1 = A \S B0

2Q(n)

µ(A)

=



1 − 14Q(n)

µ(A)

= u.µ(A)

As above, we have

µ(A2) ≤



1 − 14Q(n)

µ(A1)

≤



1 − 14Q(n)

2

µ(A)

= u2µ(A)where A2 = A1\S B00

i, Bi00 is finite subfamily of B0

i.Proceeding in this manner, we get

D(µ, λ, x) = lim inf

r↓0

µ(B(x, r))λ(B(x, r))are called respectively the upper and lower derivatives of µ with respect to λ at a point

con-µ(A) = 0 for any A ⊂ Rn such that λ(A) = 0

The absolute continuity of µ with respect to λ is denoted by µ  λ

In the otherword, the property µ  λ is equavalent to following statement: forany ε > 0, there is a δ > 0 such that

µ(A) < ε for every A ⊂ Rn with λ(A) < δ

The following lemma plays the key role for the proof of our main result (Theorem 3.1)

Lemma 3.1 Let A be a subset of Rn and µ, λ be Radon measures on Rn Then, foreach 0 < t < ∞

i)If D(µ, λ, x) ≤ t for all x ∈ A, then µ(A) ≤ tλ(A),

ii) If D(µ, λ, x) ≥ t for all x ∈ A, then µ(A) ≥ tλ(A)

Proof (i)λ is Radon measure So, we can find an open set U such that

λ(A) = inf {λ(U ) : A ⊂ U }

It implies that ∃δ > 0 : λ(U) ≤ λ(A) + δ

Appling Theorem 2.1, ∃Bi ⊂ U, Bi1 ∩ Bi2 = ∅ for i1 6= i2 and satisfy

(ii)We have

D(µ, λ, x) ≥ t ⇔ lim inf

r↓0

µ(B(x, r))λB(x, r) ≥ t

⇔ lim sup

r↓0

λ(B(x, r))µB(x, r) ≤ 1

t

⇔ D(λ, µ, x) ≤ 1

t.Then, by the part (i), we have

λ(A) ≤ 1

tµ(A).

Hence µ(A) ≥ tλ(A)

Theorem 3.1 Assume that µ and λ are Radon measures on Rn Then, the followingstatements are fulfilled:

i)The derivative D(µ, λ, x) always exists for λ almost all

holds provided that µ  λ

iii)µ  λ if and only if D(µ, λ, x) < ∞ for µ almost all x ∈ Rn

Proof For each 0 < r < ∞, 0 < s < t < ∞, let

As,t,r =x ∈ B(r) : D(µ, λ, x) ≤ s ≤ t ≤ D(µ, λ, x) and

At,r =x ∈ B(r) : D(µ, λ, x) ≥ t

As a result of lemma 3.1, we have

tλ(As,r,t) ≤ µ(As,t,r) ≤ sλ(As,t,r) < ∞and

uλ(Au,r) ≤ µ(Au,r) ≤ µ(B(r)) < ∞These inequalities yeild λ(As,t,r) = 0 since s < t, and

λ(\

u>0

Au,r) = lim

u→∞λ(Au,r) = 0

But the complement of the set {x : ∃D(µ, λ, x) < ∞} is the union of the sets As,t,r andT

Corollary 3.1 Assume that λ is a Radon measure on Rn.

i, If A is a λ-measure set on Rn then the following limit holds

1 for λ almost all x ∈ A

0 for λ almost all x ∈ Rn\ A

Z

B(x,r)

f dλ = f (x)

for λ almost ally x ∈ Rn

Proof We will prove (ii) first

Assume that f ≥ 0 Define the Radon measure µ by µ(A) = R

To prove (i), we substitue f = χ(A)

Definition 3.3 Assume that µ and λ are Radon measures on Rn If there exist a set

A ⊂ Rn such that

λ(A) = 0 = µ(Rn\ A)then µ and λ are called mutually singular

Theorem 3.2 Let µ and λ be finite Radon measures on Rn Then there exists a Borelfunction f and a Radon measure ν such that λ and ν are mutually singular and

µ1xA and ν = µx(Rn\ A)

Then it is easy to see that

µ = µ1+ ν,

and λ and ν are mutually singular by theorem 3.1.

Moreover, Lemma 3.1 gives µ1  λ has desired representation by theorem 3.1 with

f dλ + ν(B) for Borel sets B ⊂ Rn

The last statement is obvious

Corollary 3.2 ( The Radon- Nikodym theorem for finite Radon measures) Assumethat µ and λ are finite Radon measures on Rn If µ  λ (i.e, µ is absolutely continuouswith respect to λ), then there exists a Borel function f : Rn→ [0, ∞) such that for anyBorel set A ⊆ Rn, we have

hdλ, then f = h λ − almost everywhere

The function f is called the Radon-Nikodym derivative and is denoted by dµ

dλ

Corollary 3.3 (Lebegue decomposition theorem for finite Radon measures)

Assume that µ and λ are two finite Radon measures on Rn Then there exists two

σ-finite measures µs and µa such that

In this thesis, we have presented systematically the following results:

(1) The 5r- covering theorem

(2) The Vitali's covering theorems for Lebesgue measure and for Radon measures.(3) The Besicovitch's covering theorem

(4) Some results about differentiation of measures

[A] References in Vietnamese

[1] Ho ng Töy, H m thüc v  gi£i t½ch h m , NXB DHQGHN (2003)

(1) The 5r- covering theorem

(2) The Vitali''s covering theorems for Lebesgue measure and for Radon measures.(3) The Besicovitch''s covering theorem

(4) Some results... 3.2 Let µ and λ be finite Radon measures on Rn Then there exists a Borelfunction f and a Radon measure ν such that λ and ν are mutually singular and

µ1xA and ν =... class="page_container" data-page="25">

and λ and ν are mutually singular by theorem 3.1.

Moreover, Lemma 3.1 gives µ1  λ has desired representation by theorem 3.1 with

f dλ