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Papageorgiou We prove that the convergence of a sequence of functions in the spaceL0of measurable functions, with respect to the topology of convergence in measure, implies the conver-ge

Volume 2007, Article ID 63439, 17 pages

doi:10.1155/2007/63439

Research Article

Rearrangement and Convergence in Spaces of

Measurable Functions

D Caponetti, A Trombetta, and G Trombetta

Received 3 November 2006; Accepted 25 February 2007

Recommended by Nikolaos S Papageorgiou

We prove that the convergence of a sequence of functions in the spaceL0of measurable functions, with respect to the topology of convergence in measure, implies the conver-genceμ-almost everywhere (μ denotes the Lebesgue measure) of the sequence of

rear-rangements We obtain nonexpansivity of rearrangement on the spaceL ∞, and also on Orlicz spacesL Nwith respect to a finitely additive extended real-valued set function In the spaceL ∞and in the spaceEΦ, of finite elements of an Orlicz spaceLΦof aσ-additive

set function, we introduce some parameters which estimate the Hausdorff measure of noncompactness We obtain some relations involving these parameters when passing from a bounded set ofL ∞, orLΦ, to the set of rearrangements

Copyright © 2007 D Caponetti et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

The notion of rearrangement of a real-valuedμ-measurable function was introduced by

Hardy et al in [1] It has been studied by many authors and leads to interesting results

in Lebesgue spaces and, more generally, in Orlicz spaces (see, e.g., [2–5]) The spaceL0

is a space of real-valued measurable functions, defined on a nonempty setΩ, in which

we can give a natural generalization of the topology of convergence in measure using a group pseudonorm which depends on a submeasure defined on the power setᏼ(Ω) of

Ω (see [6,7] and the references given there) In the second section of this note we study rearrangements of functions of the space L0 The rearrangements belong to the space

T0([0, +∞)) of all real-valued totallyμ-measurable functions defined on [0, + ∞) We ex-tend to this setting some convergence results (see, e.g., [3,5]) Precisely, we prove that the convergence in the spaceL0implies the convergenceμ-almost everywhere of

rearrange-ments Moreover, by the convergence inL of a nondecreasing sequence of nonnegative

functions, we obtain the convergence in measure of the corresponding nondecreasing se-quence of rearrangements In the third section we introduce, in a natural manner, the spaceL ∞as the closure of the subspace of all simple functions ofL0 with respect to the essentially supremum norm The spaceL ∞ so defined is contained inL0, and we prove nonexpansivity of rearrangement on this space In the last section we obtain nonexpan-sivity of rearrangement on Orlicz spacesL Nof a finitely additive extended real-valued set function

We recall (see [8]) that for a bounded subsetY of a normed space (X,  ·  ) the Haus-dorff measure of noncompactness γ X(Y ) of Y is defined by

γ X(Y ) =inf

ε > 0 : there is a finite subset F of X such that Y ⊆ ∪ f ∈ F B X(f , ε)

, (1.1) whereB X(f , ε) = { g ∈ X :  f − g  ≤ ε } In sections3and4we introduce, respectively, a parameterω L ∞inL ∞and a parameterω EΦin the spaceEΦof finite elements of a classical Orlicz spaceLΦof aσ-additive set function By means of these parameters, we derive an

exact formula inL ∞and an estimate inEΦfor the Hausdorff measure of noncompact-ness Then as a consequence of nonexpansivity of rearrangement we obtain inequalities involving such parameters, when passing from a set of functions inL ∞, or inLΦ, to the set of rearrangements We denote byN,Q, andRthe set of all natural, rational, and real numbers, respectively.,

LetΩ be a nonempty set andR Ωthe set of all real-valued functions onΩ with its natural Riesz space structure LetᏭ be an algebra in the power set ᏼ(Ω) of Ω and let η : ᏼ(Ω) →

[0, +∞] be a submeasure (i.e., a monotone, subadditive function withη( ∅)=0) Then

 f 0=inf

a > 0 : η

| f | > a

< a

where{| f | > a } = { x ∈Ω :| f (x) | > a }and where inf∅ =+∞defines a group pseudo-norm onR Ω (i.e.,00=0, f 0=  − f 0and f + g 0≤  f 0+ g 0for all f , g ∈

R Ω) We denote by

S(Ω,Ꮽ)=

 n



i =1

a i χ A i:n ∈ N,a i ∈ R,A i ∈Ꮽ



(2.2)

the space of all real-valuedᏭ-simple functions on Ω; hereby χ Adenotes the characteris-tic function ofA defined on Ω By L0:= L0(Ω,Ꮽ,η) we denote the closure of the space

S(Ω,Ꮽ) in (R Ω, · 0)

For each function f ∈ RΩ, set| f | ∞ =supΩ| f |and denote byB(Ω,Ꮽ) the closure of the spaceS(Ω,Ꮽ) in (RΩ,| · | ∞) As f 0≤ | f | ∞, we haveB(Ω,Ꮽ)⊆ L0 If forM ∈ᏼ(Ω)

we setη(M) =0 ifM =∅andη(M) =+∞ifM , then (L0, · 0)=(B(Ω,Ꮽ),| · | ∞) We point out that the spaceB(Ω,ᏼ(Ω)) coincides with the space of all real-valued bounded functions defined onΩ, and clearly B(Ω,Ꮽ) ⊆ B(Ω,ᏼ(Ω))

Throughout this note, given a finitely additive set functionν : Ꮽ →[0, +∞], we denote

byν ∗:ᏼ(Ω)→[0, +∞] the submeasure defined byν ∗(E) =inf{ ν(A) : A ∈ Ꮽ and E ⊆

A } Moreover, whenever Ω is a Lebesgue measurable subset ofRn, we denote byμ the

Lebesgue measure on theσ-algebra of all Lebesgue measurable subsets ofΩ, we write

μ-a.e for μ-almost everywhere.

Example 2.1 (see [9, Chapter III]) LetΩ be a Lebesgue measurable subset ofRn,Ꮽ the

σ-algebra of all Lebesgue measurable subsets of Ω and η = μ ∗ If η( Ω) < + ∞, thenL0

coincides with the spaceM( Ω) of all real-valued μ-measurable functions defined on Ω If η(Ω)=+∞, thenL0coincides with the spaceT0(Ω) of all real-valued totally μ-measurable functions defined onΩ

The following definitions are adapted from [10, Chapter 4]

Definition 2.2.

(i) A subsetA of Ω is said to be an η-null set if η(A) =0

(ii) A function f ∈ RΩ is said to be anη-null function if η( {| f | > a })=0 for every

a > 0.

(iii) Two functionsf , g ∈ RΩare said to be equal η-almost everywhere, and is used the

notationf = g η-a.e if f − g is an η-null function.

(iv) A function f ∈ RΩis said to be dominated η-almost everywhere by a function g,

and is used the notationf ≤ g η-a.e if there exists an η-null function h ∈ RΩsuch that f ≤ g + h.

Observe that a function f ∈ RΩis anη-null function if and only if  f 0=0

The distribution function η f of a function f ∈ L0is defined by

η f(λ) = η

| f | > λ

Observe thatη f = η | f |andη f may assume the value +∞ In the next proposition, we state some elementary properties of the distribution functionη f (see [2, Chapter 2])

Proposition 2.3 Let f , g ∈ L0and a 0 Then the distribution function η f of f is non-negative and decreasing Moreover,

(i)η a f(λ) = η f(λ/ | a | ) for each λ ≥ 0,

(ii)η f +g(λ1+λ2)≤ η f(λ1) +η g(λ2) for each λ1,λ2≥ 0.

Proposition 2.4 Let f , g ∈ L0 If  f − g 0= 0 then η f = η g μ-a.e.

Proof Let f , g ∈ L0 andh ∈ L0 be anη-null function such that g = f + h Let I and J

denote the intervals { λ ≥0 :η f(λ) =+∞}and { λ ≥0 :η g(λ) =+∞}, respectively We start by proving thatμ(I) = μ(J) Assume μ(I) μ(J) and μ(I) < μ(J) Then I ⊂ J and μ(J \ I) > 0 Denoted by int(J \ I) the interior of the interval J \ I, we have η g(λ) =+∞and

η f(λ) < + ∞for eachλ ∈int(J \ I) Fix λ1∈int(J \ I) and λ2> 0 such that λ1+λ2∈int(J \ I).

By property (ii) ofProposition 2.3, we have

+∞ = η g

λ1+λ2



= η f +h

λ1+λ2



≤ η f

λ1



+η h

λ2



= η f

λ1



< + ∞, (2.4)

that is a contradiction Setλ =supI =supJ and let λ0∈[λ, + ∞) be a point of continuity

of both the functionsη f andη g By property (ii) ofProposition 2.3, it follows that

η f

λ0 

=lim

n η g − h

λ0+1

n ≤ η g

λ0 

+ lim

n η h

1

n = η g

λ0 

Similarly, we findη g(λ0)≤ η f(λ0) Henceη f = η g μ-a.e. 

Proposition 2.5 Let f , g ∈ L0 If | f | ≤ | g | η-a.e., then η f ≤ η g μ-a.e.

Proof Let h ∈ L0be anη-null function such that | f | ≤ | g |+h Then η | f | ≤ η | g |+hand, by

Proposition 2.4,η | g | = η | g |+ h μ-a.e Hence η | f | ≤ η | g | μ-a.e., which gives the assert. 

Observe that, when (Ω,Ꮽ,ν) is a totally σ-finite measure space and η= ν ∗, the distri-bution functionη f of f ∈ L0is right continuous (see [2]) In our setting this is not true anymore, as the following example shows

Example 2.6 (see [9, Chapter III, page 103]) LetΩ=[0, 1) and letᏭ be the algebra of all finite unions of right-open intervals contained inΩ Denote again by μ the Lebesgue

measureμ restricted to Ꮽ Let η = μ ∗ Consider the function f : [0, 1) → Rdefined as

f (x) =0, ifx ∈[0, 1)\ Q, and as f (x) =1/q, if x = p/q ∈[0, 1)∩ Qin lowest terms Then

 f 0=0 and so f is an η-null function but f is not null μ-a.e since η( {| f | > 0 })=1 Moreover,η f(λ) =0 ifλ > 0 and η f(0)=1 Thenη f is not right continuous in 0 Throughout, without loss of generality, we will assume that the distribution function

η f of a function f ∈ L0is right continuous, which together withProposition 2.4yields

η f = η gwheneverf , g ∈ L0and f − g 0=0

The decreasing rearrangement f ∗of a function f ∈ L0is defined by

f ∗(t) =inf

λ ≥0 :η f(λ) ≤ t

Clearly, by the above assumption onη f, ∗ = g ∗iff , g ∈ L0with f − g 0=0

Proposition 2.7 Let f ∈ L0 If f ∗(t) =+∞ , then t = 0.

Proof Assume that f ∗(t) =+∞ Thenη f(λ) > t for all λ ≥0 Since f 0< + ∞, for some

λ ≥0 we haveη f(λ) < + ∞ Hence, asη f is decreasing, there exists finite limλ →+∞ η f(λ) =

l ≥0 The thesis follows by proving thatl =0 Assumel > 0 and choose a function s ∈

S(Ω,Ꮽ) such that f − s 0≤ l/2.

Fixλ > l + maxΩ| s |and putA = {| f | > λ }, thenη(A) = η f(λ) ≥ l and

f (x) − s(x) ≥  f (x) − s(x) ≥ l (2.7) for eachx ∈ A So that  f − s 0≥ l So we obtain l ≤  f − s 0≤ l/2: a contradiction. 

The following proposition contains some properties of rearrangements of functions

ofL0 The proofs of (i)–(iv) (except some slight modifications) are identical to that of [2] for rearrangements of functions of a Banach function space, and we omit them

Proposition 2.8 Let f , g ∈ L0and a ∈ R Then f ∗ is nonnegative, decreasing, and right continuous Moreover,

(i) (a f ) ∗ = | a | f ∗ ;

(ii) f ∗(η f(λ)) ≤ λ, (η f(λ) < + ∞ ) and η f(f ∗(t)) ≤ t, ( f ∗(t) < + ∞ );

(iii) (f + g) ∗(t1+t2)≤ f ∗(t1) +g ∗(t2) for each t1,t2≥ 0;

(iv) if | f | ≤ | g | η-a.e., then f ∗ ≤ g ∗ μ-a.e.

Proof Clearly f ∗is nonnegative and decreasing We prove that f ∗is right continuous Fixt0≥0 and assume that limt → t+

0 f ∗(t) = a < f ∗(t0)< + ∞ Chooseb ∈(a, f ∗(t0)) Ob-serve that, sinceb < f ∗(t0), we have thatη f(b) > t0by the definition of f ∗ Moreover, since limt → t+

0 f ∗(t) = a, there exists t1> 0 such that t0< t1< η f(b) and f ∗(t1)< b From

the definition of f ∗we obtain thatη f(b) ≤ t1 It follows thatt1< η f(b) ≤ t1 which is a contradiction Then limt → t+

0 f ∗(t) = f ∗(t0)

To complete the proof, suppose that f ∗(0)=+∞, and assume that limt →0+f ∗(t) =

a < + ∞ Choose b > a Then η f(b) > 0 and since lim t →0+f ∗(t) = a we have that there

existst2> 0 such that t2< η f(b) and f ∗(t2)< b From the definition of f ∗we obtain that

η f(b) ≤ t2 It follows thatt2< η f(b) ≤ t2which is contradiction Hence limt →0+f ∗(t2)=

Now we show that the rearrangement of a function ofL0 is a function of the space

T0([0, +∞)) of all real-valued totallyμ-measurable functions defined on [0, + ∞), intro-duced in [9, Chapter III, Definition 10] (see alsoExample 2.1) InT0([0, +∞)), we write

| · |0instead of · 0

Theorem 2.9 Let f ∈ L0 Then

(i) f and f ∗ are equimeasurable, that is, η f(λ) = μ f ∗(λ) for all λ ≥ 0;

(ii) f ∗ ∈ T0([0, +∞ )) and | f ∗ |0=  f 0.

Proof (i) Fixed λ ≥0 such that η f(λ) < + ∞, by the first inequality of property (ii) of

Proposition 2.8, we have that f ∗(η f(λ)) ≤ λ Moreover, since f ∗is decreasing, we have

f ∗(t) ≤ λ for each t such that η f(λ) < t It follows that μ f ∗(λ) =sup{ f ∗ > λ } ≤ η f(λ) It

remains to prove thatη f(λ) ≤ μ f ∗(λ) Suppose that f ∗(0)=+∞ Thenμ f ∗(λ) =sup{ f ∗ >

λ } for allλ ≥0 Assume that there existsλ0≥0 such that η f(λ0)> μ f ∗(λ0) Fixedt ∈

(μ f ∗(λ0),η f(λ0)), we have that f ∗(t) ≤ λ0sincet > μ f ∗(λ0)=sup{ f ∗ > λ0} On the other hand, since t < η f(λ0), by the definition of f ∗, we obtain f ∗(t) > λ0 which is a con-tradiction The same proof breaks down if f ∗(0)< + ∞and λ < f ∗(0) If f ∗(0)< + ∞

and λ ≥ f ∗(0) then μ f ∗(λ) =0 Moreover, by the second part of the property (ii) of

Proposition 2.8, it follows thatη f(f ∗(0))=0 and thenη f(λ) =0 for allλ ≥ f ∗(0) This completes the proof

The next theorem states two well-known convergence results (see, e.g., [5, Lemma 1.1] and [3, Lemma 2], resp.)

Theorem 2.10 Let Ω be a Lebesgue measurable subset ofRn , and let { f n } be a sequence of elements of the space T0(Ω) of all real-valued totally μ-measurable functions defined on Ω

(i) If { f n } converges in measure to f , then f n ∗(t) converges to f ∗(t) in each point t of continuity of f ∗

(ii) If { f n } is a nondecreasing sequence of nonnegative functions convergent to f μ-a.e, then f n ∗ is a nondecreasing sequence convergent to f ∗ pointwise.

The remainder of this section will be devoted to extend these convergence results to the general setting of the spaceL0 We need the following lemma

Lemma 2.11 Let f n, ∈ L0(n =1, 2, .) be such that  f n − f 0→ 0 Then η f n(λ) → η f(λ) for each point λ of continuity of η f Moreover, if lim λ → λ+

0η f(λ) =+∞ then lim n →+∞ η f n(λ0)=

+∞

Proof Let λ > 0 be a point of continuity of η f and assumeη f n(λ)η f(λ) Then there are

ε0> 0 and a subsequence (η f nk) of (η f n) such that| η f nk(λ) − η f(λ) | > ε0for eachk ∈ N Put

I1=k ∈ N:η f nk(λ) > η f(λ) + ε0



, I2=k ∈ N:η f nk(λ) < η f(λ) − ε0



EitherI1orI2is infinite Leth > 0 such that

η f(λ − h) < η f(λ) + ε0

2, η f(λ + h) > η f(λ) − ε0

SupposeI1is infinite and letk ∈ I1 Consider the sets

A λ − h =x ∈Ω : f (x) > λ − h

,

A n k,λ =x ∈Ω : f n

k(x) > λ

Then η(A λ − h)= η f(λ − h) and η(A n k,λ)= η f n k(λ) We have that η f nk(λ) − η f(λ − h) >

ε0/2 Moreover,

η

A n k,λ \ A λ − h



≥ η

A n k,λ



− η

A λ − h



> ε0

Letx ∈ A n k,λ \ A λ − h Then| f (x) | ≤ λ − h and | f n k(x) | > λ Therefore | f n k(x) | − | f (x) | > h.

Hence

η

x ∈Ω : f n

k(x) − f (x) > h

≥ η

x ∈Ω : f n

k(x) − f (x) > h

≥ η

A n k,λ \ A λ − h

> ε0

2,

(2.12)

and this is a contradiction since f n − f 0→0 The proof is similar in the case the setI2

is infinite The second part of the proposition follows analogously 

Theorem 2.12 Let f n, ∈ L0(n =1, 2, .) be such that  f n − f 0→ 0 Then f n ∗(t) → f ∗(t) for each point t of continuity of f ∗ Moreover, if lim t →0+f ∗(t) =+∞ then lim n →+∞ f n ∗(0)=

+∞

Proof Let t0> 0 be a point of continuity of f ∗and assume f n ∗(t0)f ∗(t0) Then there areε0> 0 and a subsequence ( f n ∗) of (f n ∗) such that| f n ∗(t0)− f ∗(t0)| > ε0for eachk ∈ N

I1=k ∈ N: n ∗ k

t0



> f ∗

t0



+ε0



, I2=k ∈ N: f n ∗ k

t0



< f ∗

t0



− ε0



. (2.13) EitherI1orI2is infinite Leth > 0 such that

f ∗

t0− h

< f ∗

t0



+ε0

∗

t0+h

> f ∗

t0



− ε0

SupposeI1is infinite Fixk ∈ I1,t ∈[t0− h, t0] andσ ∈[f ∗(t0) +ε0/2, f ∗(t0) +ε0] Then

f ∗(t) ≤ f ∗

t0



+ε0

2 ≤ σ,

f n ∗ k(t) > f ∗

t0



Henceη f(σ) ≤ t0− h < t0andη f k(σ) ≥ t0 This shows thatη f n(σ)η f(σ) for all k ∈ I1

andσ ∈[f ∗(t0) +ε0/2, f ∗(t0) +ε0] which byLemma 2.11is a contradiction The second

Lemma 2.13 Let f n, ∈ L0(n =1, 2, .) be such that { f n } is a nondecreasing sequence of nonnegative functions and  f n − f 0→ 0 Then | η f n − η f |0→ 0.

Proof Assume by contradiction | η f n − η f |00 Sinceη f n ≤ η f n+1 ≤ η f, we findε0> 0,

σ0> 0 and n ∈ Nsuch that

μ

λ ≥0 :η f(λ) − η f n(λ) > ε0



> σ0 (2.16) for alln ∈ Nwithn ≥ n Set B n = { λ ≥0 :η f(λ) − η f n(λ) > ε0}, then∩ n ≥ n B nis nonempty, and forλ0∈ ∩ n ≥ n B nwe have

sup

n ≥ n

η f n



λ0



≤ η f



λ0



Then we chooseh > 0 such that

η f

λ1



− η f n

λ2



≥ ε0

for allλ1,λ2∈[λ0,λ0+h] and all n ≥ n In particular, we have

η f



λ0+h

− η f n



λ0



≥ ε0

Then using the same notations and considerations similar to that ofLemma 2.11, we find



x ∈ Ω : f (x) − f n(x) > h

⊇ A λ0 +h \ A n,λ0,

η

A λ0 +h \ A n,λ0



≥ η f

λ0+h

− η f n



λ0 

≥ ε0

2

(2.20)

Theorem 2.14 Let f n, ∈ L0(n =1, 2, .) be such that { f n } is a nondecreasing sequence

of nonnegative functions and  f n − f 0→ 0 Then | f n ∗ − f ∗ |0→ 0.

Proof The proof, usingLemma 2.13, is analogous to the proof ofTheorem 2.12 

We remark that if{ f n }is a sequence of elements of the spaceT0(Ω),Theorem 2.14

yields (ii) ofTheorem 2.10

We introduce the notion of essentially boundedness, following [10] For f ∈ RΩ, set

 f  ∞ = inf

A ⊆ Ω, η(A) =0

sup

then ·  ∞defines a group pseudonorm onR Ω, for each submeasureη onᏼ(Ω)

We recall that, ifν is a finitely additive extended real-valued set function on an

alge-braᏭ⊆ ᏼ(Ω) and η = ν ∗, the spaceL∞(Ω,Ꮽ,ν) of all real-valued essentially bounded functions introduced in [10] is defined by

L∞(Ω,Ꮽ,ν)=f ∈ RΩ: f  ∞ < + ∞. (3.2)

In our setting it is natural to define a space L ∞(Ω,Ꮽ,η) of all real-valued essentially

bounded functions as follows

Definition 3.1 The space L ∞:= L ∞(Ω,Ꮽ,η) is the closure of the space S(Ω,Ꮽ) in (R Ω,

 ·  ∞)

Let f ∈ RΩ Since f 0≤  f  ∞, we haveL ∞ ⊆ L0 Moreover, f 0=0 if and only

if f  ∞ =0 In the remainder part of this note we will identify functions f , g ∈ RΩfor which f − g 0=0 Then (L0, · 0) and (L ∞, ·  ∞) become anF-normed space (in the

sense of [11]) and a normed space, respectively

Proposition 3.2 Let ν be a finitely additive extended real-valued set function on an algebra

Ꮽ in ᏼ(Ω) and η = ν ∗ Then the spaceL∞(Ω,Ꮽ,ν) coincides with the space L∞(Ω,ᏼ(Ω),η)

Proof Given f ∈ L ∞(Ω,ᏼ(Ω),η), find a simple function s∈ S(Ω,ᏼ(Ω)) such that f −

s  ∞ < + ∞ From f  ∞ ≤  f − s  ∞+ s  ∞, we get f ∈L∞(Ω,Ꮽ,ν) On the other hand,

if f ∈L∞(Ω,Ꮽ,ν) then there exists A⊆ Ω such that η(A) =0 and such that supΩ\A | f | <

+∞ Consider the real functiong on Ω defined by g = f onΩ\ A and by g =0 onA Of

courseg ∈L∞(Ω,Ꮽ,ν) and  f − g  ∞ =0 Moreover,g ∈ B(Ω,ᏼ(Ω))⊆ L ∞(Ω,ᏼ(Ω),η).

Then there exists a sequence (s n) in S(Ω,ᏼ(Ω)) such that | g − s n | ∞ →0 Since  f −

s n  ∞ ≤  f − g  ∞+ g − s n  ∞ = | g − s n | ∞, we have that f ∈ L ∞(Ω,ᏼ(Ω),η) 

We write brieflyB([0, + ∞)) instead ofB(Ω,Ꮽ), when Ω=[0, +∞),Ꮽ is the σ-algebra

of all Lebesgue measurable subsets ofΩ and η = μ ∗ The next proposition establishes that the rearrangement of a function ofL ∞is a function ofB([0, + ∞))

Proposition 3.3 Let f ∈ L ∞ Then f ∗ ∈ B([0, + ∞ )) and | f ∗ | ∞ = f ∗(0)=  f  ∞ Proof Let ε > 0 Then there is A ⊆ Ω such that η(A) =0 and supΩ\A | f | <  f  ∞+ε.

Hence{| f | >  f  ∞+ε } ⊆ A, so that η( {| f | >  f  ∞+ε })=0

Therefore| f ∗ | ∞ = f ∗(0)≤  f  ∞+ε so that | f ∗ | ∞ ≤  f  ∞ Now we have to prove that f  ∞ ≤ | f ∗ | ∞ Assume| f ∗ | ∞ < c <  f  ∞ Then for eachA ⊆ Ω such that η(A) =0

we have supΩ\A | f | > c and η f(c) = η( {| f | > c })> 0 For t ∈[0,η f(c)), by the definition

of the function f ∗, we obtainf ∗(t) ≥ c > | f ∗ | ∞ = f ∗(0) which is a contradiction, since

Our next aim is to prove nonexpansivity of rearrangement onL ∞ We need the follow-ing two lemmas

Lemma 3.4 Let s1,s2∈ S( Ω,Ꮽ) Then | s ∗1− s ∗2| ∞ ≤  s1− s2 ∞

Proof Let s1,s2∈ S(Ω,Ꮽ) and put s1− s2 ∞ = ε Let { A1, , A n }be a finite partition of

Ω in Ꮽ such that s1=n

i =1 a i χ A iands2=n

i =1 b i χ A i Set

s = n



i =1

min a i , b i χ A

whereη(A) =0 and| s1(x) − s2(x) | ≤ ε for all x ∈Ω\ A It suffices to prove that

s(x) ≤ s1(x) ≤ s ε(x), s(x) ≤ s2(x) ≤ s ε(x), (3.4) for allx ∈Ω\ A, where s ε = | s |+ε In fact, from this and from property (iv) ofProposition 2.8, it follows that

s ∗ ≤ s ∗1 ≤ s ∗ ε μ-a.e., s ∗ ≤ s ∗2 ≤ s ∗ ε μ-a.e., (3.5) and thus| s ∗1 − s ∗2| ∞ ≤ | s ∗ ε − s ∗ | ∞ = ε Fix x ∈Ω\ A and let i ∈ {1, , n }such thatx ∈

A i \ A Now, if s(x) = | a i |we have

s(x) = s1(x) ≤ a i +ε = s ε(x). (3.6)

Ifs(x) = | b i |, since s1− s2 ∞ = ε implies 0 ≤ | a i | − | b i | ≤ | a i − b i | ≤ ε, we have

s(x) ≤ a i = s1(x) ≤ b i +ε = s ε(x). (3.7)

Analogously we obtains(x) ≤ | s2(x) | ≤ s ε(x) for x ∈Ω\ A, and the lemma follows. 

Lemma 3.5 Let f ∈ L ∞ Then for each ε > 0 there exists a function s ∈ S( Ω,Ꮽ) such that

 f − s  ∞ ≤ ε/2 and | f ∗ − s ∗ | ∞ ≤ ε.

Proof Fix ε > 0 Then similar to [10, page 101] (seeTheorem 3.10), we have that there is

a finite partition{ A1, , A n }ofΩ in Ꮽ and A ⊆ Ω with η(A) =0 such that

sup

x,y ∈ A i \ A

f (x) − f (y) ≤ ε (3.8) for eachi ∈ {1, , n } Set

λ i = inf

x ∈ A i \ A

f (x) , Λi = sup

x ∈ A \ A

f (x) , a i = λ i+Λi

for eachi ∈ {1, , n } Define the simple function

s = n



i =1

Then for eachi ∈ {1, , n }and for eachx ∈ A i \ A we have | f (x) − s(x) | ≤ ε/2 Hence

 f − s  ∞ ≤ ε/2 Now consider the simple function ϕ defined by

ϕ(x) =

⎧

⎪

⎪

⎨

⎪

⎪

⎩

a i+ε 2

, ifx ∈ A i,a i < − ε

2,

0, ifx ∈ A i,− ε

2 ≤ a i ≤ ε

2,

a i − ε

2

, ifx ∈ A i,a i > ε

2.

(3.11)

Then a direct computation shows that

ϕ(x) ≤ f (x) ≤ ϕ ε(x), ϕ(x) ≤ s(x) ≤ ϕ ε(x), (3.12) for allx ∈Ω\ A, where ϕ ε = | ϕ |+ε Put h(x) =(max| a i |)χ A(x) and k(x) = | f (x) | χ A(x).

Thenϕ ≤ | f |+h and | f | ≤ ϕ ε+k As h and k are both η-null functions, from the

prop-erty (iv) ofProposition 2.8it follows thatϕ ∗ ≤ f ∗ ≤ ϕ ∗ ε μ-a.e., and analogously ϕ ∗ ≤

s ∗ ≤ ϕ ∗ ε μ-a.e., hence | f ∗ − s ∗ | ∞ ≤ | ϕ ∗ ε − ϕ ∗ | ∞ = ε. 

Theorem 3.6 Let f , g ∈ L ∞ Then | f ∗ − g ∗ | ∞ ≤  f − g  ∞

Proof Let ε > 0 ByLemma 3.5we can finds, u ∈ S(Ω,Ꮽ) such that

 f − s  ∞ ≤ ε

4,  g − u  ∞ ≤ ε

4,

f ∗ − s ∗

∞ ≤ ε

2, g ∗ − u ∗

∞ ≤ ε

2.

(3.13)

We have that

 s − u  ∞ ≤  f − s  ∞+ f − g  ∞+ g − u  ∞ ≤  f − g  ∞+ε

Then the last inequality andLemma 3.4imply| s ∗ − u ∗ | ∞ ≤  f − g  ∞+ε/2.

Consequently we have

f ∗ − g ∗

∞ ≤ f ∗ − s ∗

∞+ s ∗ − u ∗

∞+ g ∗ − u ∗

∞ ≤  f − g  ∞+ε, (3.15)

Remark 3.7 We observe thatTheorem 3.6does not hold in every spaceL0 In fact, let

L0= M([0, 1]) (seeExample 2.1) and set

s n =

n−1

i =0

(n − i)χ[i/n,(i+1)/n), t n =

n−1

i =1

(n − i)χ[i/n,(i+1)/n), (3.16)

for eachi ∈ {1, , n } Define the simple function

s... (seeTheorem 3.10), we have that there is

a finite partition{ A1, , A n }of< i>Ω in Ꮽ and A ⊆ Ω with η(A) =0... ∈Ω\ A, where s ε = | s |+ε In fact, from this and from property (iv) of< /i>Proposition 2.8, it follows that

s ∗ ≤