NEW CHARACTERIZATIONS OF SOME Lp-SPACES

For a complete measure space (X,Σ,μ), we give conditions which force Lp(X,μ), for 1 ≤ p < ∞, to be isometrically isomorphic to p(Γ) for some index set Γ which depends only on (X,μ). Also, we give some new characterizations which yield the inclusion Lp(X,μ)⊂ Lq(X,μ) for 0<p < q.


Introduction.
Suppose X is a nonempty set, Σ is σ -algebra of subsets of X, µ a positive measure on Σ.For each positive number p, let L p (X, µ) denote the space of all real valued Σ-measurable functions f on X such that x | f | p dµ < ∞, and L ∞ (X, µ) denote the space of all essentially bounded, real valued Σ-measurable functions on X.In [2,3,5] some characterizations of the positive measure µ on (X, Σ) for which L p (X, µ) ⊆ L q (X, µ), 0 < p < q, are given.The purpose of this note is to give some new characterizations of such measure µ which yield the inclusion L p (X, µ) ⊆ L q (X, µ) for 0 < p < q.Our proofs are more transparent, direct, and work even if the measure µ is not σ -finite.Further we show that in a situation when L p (X, µ) ⊆ L q (X, µ) for some pair p, q with 0 < p < q, then L p (X, µ), for 1 ≤ p < ∞, is isometrically isomorphic to p (Γ ) for some index set Γ which depends only on the measure space (X, Σ,µ).

Preliminaries.
Throughout the following (X, Σ,µ) is a positive measure space.We assume that the measure µ is complete.For the sake of simplicity, we write L p (µ) for L p (X, µ) and ) is said to be atomic if every measurable set of positive measure contains an atom.For more information on measurable spaces and related topics we refer to [1,2,4].We collect some interesting and useful properties of atomic and nonatomic sets in the following proposition.Proposition 2.1.Let (X, Σ,µ) be a complete measure space.(a) If {A n } is a sequence of distinct atoms, then there exists a sequence {B n } of disjoint atoms such that for each n, B n ⊆ A n and From part (a) of the proposition, there exists a sequence {B n } of disjoint atoms such that B n ⊆ A n for each n and ∪A n = ∪B n .Obviously The following lemmas are quite useful in the proof of the main result.Lemma 2.2.Let (X, Σ,µ) be a complete measure space.(a) If {B n } is a sequence of measurable sets of positive measure and µ(B n ) → 0 as n → ∞, then there exists a sequence {C n } of disjoint measurable sets of positive measure such that µ(C n ) → 0 as n → ∞.
(b) If {E n } is a sequence of disjoint measurable sets of positive measure such that µ(E n ) → 0 as n → ∞, then for any positive number m > 1 there exists a subsequence Proof.(a) Without loss of generality, we may assume that µ(B n ) < 1 for each n.If for some positive integer k, B k is nonatomic, by using an argument similar to that of Proposition 2.1(c), we can construct a sequence C n of disjoint measurable sets of positive measure such that µ(C n ) → 0 as n → ∞.Suppose that B k is atomic for each positive integer k, let A 1 be an atom contained in B 1 .Since µ(B n ) → 0 as n → ∞, µ(A 1 ∩ B k ) can be positive only for finitely many k > 1.Let n 1 be the smallest positive integer such that µ(A 1 ∩ B n 1 ) = 0. Now choose an atom A 2 contained in B n 1 .Obviously A 2 is indistinguishable from A 1 .Also, µ(A 2 ∩ B k ) can be positive for at most finitely many k greater than n 1 .Let n 2 be the smallest positive integer greater than n 1 such that µ(A 2 ∩ B n 2 ) = 0. Now choose an atom A 3 contained in B n 2 .Clearly A 3 is indistinguishable from A 1 and A 2 .Continuing in this fashion, we get a sequence {A k } of atoms which are indistinguishable and . This completes the proof of part (a).
(b) Let {E n } be a sequence of measurable sets of positive measure such that µ(E n ) → 0 as n → ∞.Without loss of generality, we may assume that {µ(E n )} is a strictly decreasing sequence.Let m > 1.Let k 0 > 2 be a positive integer such that 1/2 < (k Continuing inductively in this way, we can choose strictly increasing sequences of positive integers {k i } and . This completes the proof of part (b).Lemma 2.3.If L p (µ) ⊆ L q (µ) for 0 < p < q, then there does not exist a disjoint sequence {E n } of measurable sets of positive measure such that µ(E n ) → 0 as n → ∞.
Proof.Suppose there exists a disjoint sequence {E n } of measurable sets of positive measure such that µ Clearly m > 1.By Lemma 2.2(b), there exists a subsequence {E n i } of {E n } and a strictly increasing sequence of positive integers On the other hand, (2.4) Thus f ∈ L p (µ) but f ∉ L q (µ).This completes the proof of the lemma.

Main results.
For the sake of clarity, we first start with a definition.For any nonempty set Γ , and p > 0, we define p (Γ ) to be the set of all extended real valued functions f on Γ such that f is nonzero only on a countable subset of Γ and When p ≥ 1, p (Γ ) becomes a Banach space under the norm defined by f p (Γ ) = ( α | f (α) | p ) 1/p .Now, we are ready to state the main result.Theorem 3.1.Let (X, Σ,µ) be a complete measure space.The following six conditions are equivalent: (1) L p (µ) ⊂ L q (µ) for some pair of real numbers p and q with 0 < p < q.
(5) There is no sequence {B n } in Σ such that µ(B n ) > 0 for each n and µ(B n ) → 0 as n → ∞.
Moreover, these statements imply that: for each positive number p ≥ 1, L p (µ) is isomerically isomorphic to p (Γ ) for some index set Γ which depends only on (X, Σ,µ).
(1) ⇒( 2): suppose that L p ⊂ L q for some pair p, q with 0 < p < q.We claim L p ⊂ L ∞ .Suppose there is an f in L p which is not essentially bounded.Then there exists a strictly increasing sequence {n k } of positive integers such that for each k ≥ 1, the set (2) ⇒(3): suppose that L p (µ) ⊂ L ∞ (µ) for some p > 0. Let r be any positive real number.We show L r (µ) (3) ⇒( 4): suppose that L p ⊂ L ∞ for all p > 0. Let g ∈ L p .Write A = {x : |g(x)| > 1}.If A is a nonatomic set of positive measure, by Proposition 2.1(c), A contains a disjoint sequence {E n } of measurable subsets of A of positive measure such that µ(E n ) → 0 as n → ∞.As is noted in the proof of Lemma 2.3, we can construct a function f in L p which is not in L ∞ .Hence A contains an atom.Since the measure of A is finite, in view of Proposition 2.1(a), A cannot contain infinitely many atoms.Therefore, A can be written as a finite disjoint union of atoms.Suppose that A = ∪ n i=1 θ i , where θ i 's are disjoint atoms.By Proposition 2.1(d), g is constant on each θ i , Let g θ i be the value of g on θ i .Then for any q > p, Hence L p ⊂ L q for q > p.
contain infinitely many atoms.Therefore, B can be written as finite disjoint union of atoms.Since g is constant on each atom, it follows that g ∈ L ∞ .
Finally, we show that for p ≥ 1, one of the statements (1) through (6) (and hence all of them) imply statement (7).Let (X, Σ,µ) be a measure space such that L p (µ) ⊆ L q (µ) for some 1 ≤ p < q.Let {θ i } i∈Γ be the collection of all atoms in X where Γ is some index set.Let f ∈ L p (µ) be an arbitrary nonzero element of f .By Proposition 2.1(d) f is constant almost everywhere on any atom.We denote the value of f on an atom θ lies in the support of f by f θ .Since the support of f is σ -finite, and by statement (5) of the theorem any measurable set of finite measure is disjoint union of finitely many atoms, the support of f can be written as countable union of atoms.Let {θ n (f )} be the set of all atoms that forms the support of f .We define F : L p (µ) → p (Γ ) by for any nonzero f in L p (µ).The function F is well defined since any two functions that are equal in L p (µ) are equal almost everywhere and thus share the same support.
If A is a nonatomic set of positive measure, then there exists a sequence {E n } of disjoint measurable subsets of A of positive measure such that µ(E n ) → 0 as n → ∞.(d)If f ∈ L p (µ) and A is an atom in Σ, then f is constant almost everywhere (a.e.) on A.
atoms, and A is an atom contained in ∪A n , then there exists a unique m such that A is indistinguishable from A m .(c)