Extension of The Best Approximation Operator in Orlicz Spaces

Abstract

Let (Ω,𝒜,𝜇) be a probability space and ℒ⊂𝒜 a sub-𝜎-lattice of the 𝜎-algebra 𝒜. We study an extension of the best 𝜙-approximation operator from an Orlicz space 𝐿^𝜙 to the space 𝐿^𝜙′, where 𝜙^′ denotes the derivative of the convex, but not necessarily a strictly convex function 𝜙. We obtain convergence results when a sequence of 𝜎-algebras ℬ_n converges to ℬ_∞ in a suitable way.

1. Introduction and Notations

Let (Ω,𝒜,𝜇) be a probability space and let ℳ=ℳ(Ω,𝒜,𝜇) be the set of all 𝒜-measurable real-valued functions defined on Ω. Given a 𝐶¹ convex function 𝜙∶[0,∞)→[0,∞) such that 𝜙(0)=0,𝜙(𝑡)>0, when 𝑡>0, let 𝐿^𝜙=𝐿^𝜙(Ω,𝒜,𝜇) be the space of all functions 𝑓∈ℳ such that _Ω𝜙(𝜆|𝑓|)𝑑𝜇<∞, (1.1) for some 𝜆>0. Since we only deal with a Δ₂ function 𝜙, that is, there exists a constant 𝐾>0 such that 𝜙(2𝑥)≤𝐾𝜙(𝑥) for all 𝑥≥0, the space 𝐿^𝜙 can be defined as the space of all function 𝑓∈ℳ, where (1.1) holds for every positive number 𝜆. The space 𝐿^𝜙′ is analogously defined, where 𝜙^′ is the derivative of the function 𝜙. Besides, observe that for a Δ₂ function 𝜙 it holds the next inequality 𝜙(𝑥)≤𝑥𝜙^′(𝑥)≤𝜙(2𝑥)≤𝐾𝜙(𝑥) (1.2) for all 𝑥≥0, and therefore 𝐿^𝜙⊂𝐿^𝜙′. Moreover 𝜑 is a Δ₂ function if and only if 𝜑^′ is a Δ₂ function.

We say, according to [1], that a collection ℒ of sets in 𝒜 is a 𝜎-lattice if it is closed under countable unions and intersections and contains ∅ and Ω. Given a 𝜎-lattice ℒ, we denote by ℒ the 𝜎-lattice of all the complementary sets of ℒ, that is, ℒ={𝐴^𝑐∶𝐴∈ℒ}. Denote by 𝐿^𝜙(ℒ) all ℒ-measurable functions in 𝐿^𝜙.

A set 𝐶⊂𝐿^𝜙 is called 𝜙-closed if and only if 𝑓_𝑛∈𝐶 and 𝑓_𝑛↗𝑓∈𝐿^𝜙 or (𝑓_𝑛↘𝑓∈𝐿^𝜙) then 𝑓∈𝐶. Then 𝐿^𝜙(ℒ) is a 𝜙-closed convex set and a lattice, that is closed for the maximum and minimum of functions.

We will use the notation 𝑓∧𝑔=min(𝑓,𝑔) and 𝑓∨𝑔=max(𝑓,𝑔). A set 𝐶⊂𝐿^𝜙 is called a 𝜎-complete lattice if and only if ⋀_𝑛∈ℕ𝑓_𝑛=inf_𝑛∈ℕ𝑓_𝑛∈𝐶 and ⋁_𝑛∈ℕ𝑓_𝑛=sup_𝑛∈ℕ𝑓_𝑛∈𝐶 for all sequence {𝑓_𝑛}_𝑛∈ℕ⊂𝐶.

It is well known, see [1], that for every 𝑓∈𝐿^𝜙 there exists an element 𝑔∈𝐿^𝜙(ℒ) such that _Ω𝜙(|𝑓−𝑔|)𝑑𝜇=inf_{ℎ∈𝐿𝜙(ℒ)Ω}𝜙(|𝑓−ℎ|)𝑑𝜇. (1.3) Denote by 𝜇^ℒ_𝜙(𝑓) the set of all 𝑔∈𝐿^𝜙(ℒ) satisfying (1.3). Each element of 𝜇^ℒ_𝜙(𝑓) will be called a best 𝜙-approximation of 𝑓 given 𝐿^𝜙(ℒ), and we will refer to the mapping 𝑓→𝜇^ℒ_𝜙(𝑓) defined on 𝐿^𝜑 as the best approximation operator.

It is showed in [1] that for 𝑓∈𝐿^𝜙 the set 𝜇^ℒ_𝜙(𝑓) is a nonvoid 𝜎-complete lattice. Also it was proved that if 𝑓≤𝑔, both in 𝐿^𝜙,𝑓₁∈𝜇^ℒ_𝜙(𝑓) and 𝑔₁∈𝜇^ℒ_𝜙(𝑔) then we have 𝑓₁∧𝑓₂∈𝜇^ℒ_𝜙(𝑓) and 𝑓₁∨𝑓₂∈𝜇^ℒ_𝜙(𝑔). In this case, we say that the multivalued operator 𝜇^ℒ_𝜙(𝑓) is a monotone operator.

The main purpose of this paper is to extend the best approximation operator to the set 𝐿^𝜑′. The case 𝜑(𝑡)=𝑡^𝑝,𝑝>1, was extensively treated in [2] and the best 𝐿¹ approximation operator is extended to all measurable functions in [3]. The extension from 𝐿^𝜑 to 𝐿^𝜑′ is considered in [4] for a 𝐶¹ function 𝜑 which is strictly convex and 𝜑^′(0)=0.

Now in this paper we consider a 𝐶¹ convex function 𝜙,𝜙^′(0)=0, but not necessarily a strictly convex function. Extension of best approximation operator when the approximation classes are the constants is treated in [5–7].

The extension of the best approximation operator is 𝜇^ℒ_𝜙(𝑓); for 𝑓∈𝐿^𝜙′, we will be denoted by ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓). In Theorem 2.12, we prove that ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓)≠∅ for every 𝑓∈𝐿^𝜙′, while in Theorem 2.16 it is proved that it is indeed an extension, that is, ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓)=𝜇^ℒ_𝜙(𝑓) for 𝑓∈𝐿^𝜙. Additional properties are obtained for the set ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓) when the 𝜎-lattice ℒ is a 𝜎-algebra (see Theorem 2.17) and similar results hold when ℒ is the class of monotone functions in 𝐿^𝜙′ (see Theorem 3.2). A martingale-type result is given in Theorem 4.1 which generalizes [8, Theorem 2.8] for the particular family of measures considered in this paper.

2. Extension of the Best 𝜙-Approximation Operator

We begin with some definitions and auxiliary results. The proof of the next two lemmas can be found in [4].

Lemma 2.1. A necessary and sufficient condition for a continuously differentiable and convex function 𝜙 to satisfy the Δ₂ condition is that there exists a constant 𝛼>1 such that 𝑢𝜙^′(𝑢)≤𝛼𝜙(𝑢)∀𝑢≥0. (2.1)

Lemma 2.2. Let 𝑓,𝑔 be in 𝐿^𝜙. Then 𝜙^′(|𝑓|)𝑔 is an integrable function.

According to Brunk and Johansen [8], we set the following definitions.

Definition 2.3. Let 𝜈 be a signed measure on 𝒜 and let ℒ be a 𝜎-lattice contained in 𝒜. Say that 𝑃∈ℒ is a 𝜈-positive set, if for all 𝐷∈ℒ, then 𝜈(𝑃∩𝐷)≥0. A set 𝑁∈ℒ is called 𝜈-negative if for all 𝐶∈ℒ one has 𝜈(𝑁∩𝐶)≤0.

Definition 2.4. Let {𝜈_𝑎}_𝑎∈ℝ be a family of measures on 𝒜, and let ℒ be a 𝜎-lattice contained in 𝒜. An ℒ-measurable function 𝑔 is called a Lebesgue-Radon-Nikodym function (LRN function) for {𝜈_𝑎} given ℒ if and only if the set {𝑔>𝑎} is 𝜈_𝑎-positive for all 𝑎∈ℝ and the set {𝑔<𝑎} is 𝜈_𝑎-negative for all 𝑎∈ℝ.

Remark 2.5. We note that in Definition 2.4 it is sufficient to impose the conditions for all 𝑎 in a dense set in ℝ, see [8, page 588].

For 𝑓∈𝐿^𝜙,𝑔∈𝐿^𝜙(ℒ), and 𝑎∈ℝ, we define the following measures on 𝒜: 𝜇_𝑔(𝐴)=_𝐴𝜙^′(𝑓−𝑔)𝑑𝜇,𝜇_𝑎(𝐴)=_𝐴𝜙^′(𝑓−𝑎)𝑑𝜇, (2.2) where 𝜙^′(𝑥)=𝜙^′(|𝑥|)sign(𝑥). Note that when 𝑓∈𝐿^𝜙′ and 𝑔∈𝐿^𝜙′(ℒ), the measure 𝜇_𝑔 and 𝜇_𝑎 are well defined.

The next theorem is a characterization of 𝜇^ℒ_𝜙(𝑓), see [9, Theorem 3.2].

Theorem 2.6. Let 𝑓∈𝐿^𝜙,ℒ⊂𝒜 be a 𝜎-lattice and 𝑔∈𝐿^𝜙(ℒ). Then the following statement are equivalent.
(1)𝑔∈𝜇^ℒ_𝜙(𝑓).(2)(a) a set {𝑔>𝑎} is 𝜇_𝑔-positive for all 𝑎∈ℝ; and(b) a set {𝑔<𝑎} is 𝜇_𝑔-negative for all 𝑎∈ℝ.(3)𝑔 is an LRN function for the family {𝜇_𝑎}_𝑎∈ℝ given ℒ.

Now we extend the operator 𝜇^ℒ_𝜙(⋅) to the space 𝐿^𝜙′.

Definition 2.7. Let ℒ be a 𝜎-lattice and let 𝑓∈𝐿^𝜙′. Then 𝑔 is an extended best 𝜙-approximation if and only if 𝑔∈𝐿^𝜙′(ℒ) and
(i)the set {𝑔>𝑎} is 𝜇_𝑔-positive for all 𝑎∈ℝ;(ii)the set {𝑔<𝑎} is 𝜇_𝑔-negative for all 𝑎∈ℝ. For 𝑓∈𝐿^𝜙′ we denote by ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓) the set of all extended best 𝜙-approximation functions.

Remark 2.8. Let 𝑓∈𝐿^𝜙′ and let 𝑔 be a function in 𝐿^𝜙′(ℒ) such that the set {𝑔>𝑎} is 𝜇_𝑔-positive and the set {𝑔<𝑎} is 𝜇_𝑔-negative for all 𝑎∈ℝ. Then we have the following.
(i)For all ℎ∈𝐿^∞(ℒ) and ℎ≥0, _{𝑔>𝑎}𝜙^′(𝑓−𝑔)ℎ𝑑𝜇≥0. (2.3) For all ℎ∈𝐿^∞(ℒ) and ℎ≥0, _{𝑔<𝑎}𝜙^′(𝑓−𝑔)ℎ𝑑𝜇≤0. (2.4)(ii)𝜙^′(𝑓−𝑔)𝑑𝜇=0. (2.5)

Proof. We prove inequality (2.3). Since the set {𝑔>𝑎} is 𝜇_𝑔-positive, that is, for each 𝐷∈ℒ and 𝑎∈ℝ, we have _{{𝑔>𝑎}∩𝐷}𝜙^′(𝑓−𝑔)𝑑𝜇≥0. (2.6) For ℎ=∑^𝑁_𝑘=1𝑐_𝑘𝒳_𝐷𝑘, where 𝐷_𝑘∈ℒ and 𝑐_𝑘≥0, 𝑘=1,…,𝑁, then by (2.6), we have _{𝑔>𝑎}𝜙^′(𝑓−𝑔)ℎ𝑑𝜇=^𝑁_𝑘=1𝑐_{𝑘{𝑔>𝑎}∩𝐷𝑘}𝜙^′(𝑓−𝑔)𝑑𝜇≥0. (2.7) All nonnegative ℎ∈𝐿^∞(ℒ) can be obtained as a limit of functions of the above type. The proof of inequality (2.4) is similar.
The equality (2.5) is obtained using Lebesgue's theorem when 𝑎→∞ with ℎ=1 in (2.4) and if 𝑎→−∞ consider in (2.3) also ℎ=1.

As a reference, we note that (2) is equivalent to (3) in Theorem 2.6 for 𝑓∈𝐿^𝜙′,𝑔∈𝐿^𝜙′. We have the next remark.

Remark 2.9. For 𝑓∈𝐿^𝜙′ and 𝑔∈𝐿^𝜙′(ℒ), the following statements are equivalent:
(1)𝑔∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓);(2)𝑔 is an LRN function for the family {𝜇_𝑎}_𝑎∈ℝ given ℒ.

The next lemma is a particular case of [8, Theorem 1.8].

Lemma 2.10. Let 𝑓∈𝐿^𝜙′ and 𝑔∈𝐿^𝜙′(ℒ). Then the following statements are equivalent.
(1)𝑔 is an LRN function for the family {𝜇_𝑎}_𝑎∈ℝ given ℒ.(2)There exists a countable set 𝐷 such that {𝑔≤𝑎} is 𝜇_𝑎-negative for all 𝑎∈𝐷^𝑐 and the set {𝑔≥𝑎} is 𝜇_𝑎-positive for all 𝑎∈𝐷^𝑐.

We need the following auxiliary result.

Lemma 2.11. Let 𝑓_𝑛∈𝐿^𝜙′ be a sequence of functions such that 𝑓_𝑛↗𝑓(𝑓_𝑛↘𝑓), where 𝑓∈𝐿^𝜙′. Let 𝑔_𝑛∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓_𝑛) be such that 𝑔_𝑛↗𝑔(𝑔_𝑛↘𝑔). Then 𝑔∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓).

Proof. We will prove the result just for the increasing case, the proof for the decreasing case follows the same pattern. The function 𝑔=lim_𝑛→∞𝑔_𝑛 is obviously ℒ-measurable function. Now, we prove that 𝑔∈𝐿^𝜙′, and it satisfies (i) and (ii) of Definition 2.7. We have that 𝑓_𝑛−𝑔_𝑛≤𝑓−𝑔_𝑛≤𝑓−𝑔₁. (2.8) Using (2.5), _Ω𝜙^′(𝑓_𝑛−𝑔_𝑛)𝑑𝜇=0foreach𝑛∈ℕ. (2.9) According to (2.8) and (2.9), we have, since 𝜙^′(𝑥) is an increasing function, _Ω𝜙^′(𝑓−𝑔_𝑛)𝑑𝜇≥0. (2.10) Since 𝜙^′ is a continuous function and 𝜙^′(0)=0, we have by (2.8) and Lebesgue's theorem _Ω(𝜙^′(𝑓−𝑔)∨0)𝑑𝜇≤_Ω(𝜙^′(𝑓−𝑔₁)∨0)𝑑𝜇. (2.11) Now by (2.10) it holds _Ω(−𝜙^′(𝑓−𝑔_𝑛)∨0)𝑑𝜇≤_Ω(𝜙^′(𝑓−𝑔_𝑛)∨0)𝑑𝜇. (2.12) Using Fatou in (2.12), we have _Ω(−𝜙^′(𝑓−𝑔)∨0)𝑑𝜇≤_Ω(𝜙^′(𝑓−𝑔)∨0)𝑑𝜇. (2.13) Therefore, using (2.11) and (2.12), we get 𝑔∈𝐿^𝜙′(Ω) and ∫_Ω𝜙^′(𝑓−𝑔)𝑑𝜇≥0.
Let 𝑎∈ℝ,𝐷∈ℒ, and 𝐶∈ℒ we know for each 𝑛_{{𝑔𝑛>𝑎}∩𝐷}𝜙^′(𝑓_𝑛−𝑔_𝑛)𝑑𝜇≥0,_{{𝑔𝑛<𝑎}∩𝐶}𝜙^′(𝑓_𝑛−𝑔_𝑛)𝑑𝜇≤0. (2.14) Since {𝑔_𝑛}_𝑛∈ℕ is an increasing sequence, we get ⋃_𝑛∈ℕ{𝑔_𝑛>𝑎}={𝑔>𝑎}, and by (2.14), we have _{{𝑔>𝑎}∩𝐷}𝜙^′(𝑓−𝑔)𝑑𝜇≥0. (2.15)Hence, the set {𝑔>𝑎} is 𝜇_𝑔-positive for all 𝑎∈ℝ.
Now, for 𝑎∈ℝ and 𝑘,𝑛∈ℕ, we define 𝐵_𝑛={𝑔_𝑛<𝑎} and 𝐵_𝑛,𝑘={𝑔_𝑛<𝑎−1/𝑘}. We have that, for 𝑛→∞,𝐵_𝑛,𝑘↘𝐴_𝑘 for some 𝒜-measurable set such that 1{𝑔<𝑎−𝑘}⊂𝐴_𝑘1⊂{𝑔≤𝑎−𝑘}. (2.16) We observe that 𝒳_𝐴𝑘→𝒳_{𝑔<𝑎},a.e. Then taking limit as 𝑘→∞,𝑛→∞ and using Lebesgue's theorem, we obtain _{{𝑔<𝑎}∩𝐶}𝜙^′(𝑓−𝑔)𝑑𝜇=lim_𝑘→∞𝜙^′(𝑓−𝑔)𝒳_{𝐴𝑘∩𝐶}𝑑𝜇=lim_𝑘→∞lim_{𝑛→∞𝐵𝑛,𝑘∩𝐶}𝜙^′(𝑓−𝑔)𝑑𝜇≤0. (2.17)

Theorem 2.12. Let ℒ be a 𝜎-lattice and 𝑓∈𝐿^𝜙′, then ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓)≠∅.

Proof. For 𝑓∈𝐿^𝜙′ we can define the following sequences. For each 𝑚∈ℕ, let 𝑓_𝑚=𝑓∨(−𝑚) and when 𝑚→∞ we have 𝑓_𝑚↘𝑓. Set 𝑓^𝑛_𝑚=(𝑓∨(−𝑚))∧𝑛 for all 𝑛,𝑚 in ℕ, then we have 𝑓_𝑚𝑛↗𝑓_𝑚, when 𝑛→∞. Since for each 𝑚,𝑛∈ℕ we have 𝑓^𝑛_𝑚∈𝐿^𝜙, there exist 𝑔^𝑛_𝑚∈𝜇^ℒ_𝜙(𝑓^𝑛_𝑚). As 𝜇^ℒ_𝜙(⋅) is a monotone operator over 𝐿^𝜙 we can take a new sequence that we call again 𝑔^𝑛_𝑚∈𝜇^ℒ_𝜙(𝑓^𝑛_𝑚), such that 𝑔^𝑛_𝑚≤𝑔_𝑚^𝑛+1 for all 𝑛∈ℕ.
Since 𝑓^𝑛_𝑚≥𝑓^𝑛_𝑚+1 and using again that 𝜇^ℒ_𝜙(⋅) is a monotone operator, we have ^0𝑥0𝑔^{02𝑑𝑐𝑛}_𝑚+1≤𝑔^{0𝑥002𝑑𝑐𝑛}_𝑚, where ^0𝑥0𝑔^{02𝑑𝑐𝑛}_𝑚∈𝜇^ℒ_𝜙(𝑓^𝑛_𝑚) is the sequence defined by ^0𝑥0𝑔^{02𝑑𝑐𝑛}₁=𝑔^𝑛₁ and ^0𝑥0𝑔^{02𝑑𝑐𝑛}_𝑚+1=𝑔^{0𝑥002𝑑𝑐𝑛}_𝑚∧𝑔^𝑛_𝑚+1. Furthermore, it is easy to check that ^0𝑥0𝑔^{02𝑑𝑐𝑛}_𝑚≤𝑔^{0𝑥002𝑑𝑐}_𝑚^𝑛+1.
Then, we have that for each 𝑚∈ℕ that 𝑓^𝑛_𝑚↗𝑓_𝑚, when 𝑛→∞, and since 𝜇^ℒ_𝜙(𝑓^𝑛_𝑚)⊂𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓^𝑛_𝑚) we have ^0𝑥0𝑔^{02𝑑𝑐𝑛}_𝑚∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓^𝑛_𝑚) and if we define 𝑔_𝑚=lim𝑔_𝑛→∞^{0𝑥002𝑑𝑐𝑛}_𝑚 by Lemma 2.11 we obtain 𝑔_𝑚∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓_𝑚) and 𝑔_𝑚≥𝑔_𝑚+1 for all 𝑚∈ℕ. If we take 𝑚→∞, we have 𝑓_𝑚↘𝑓 and by Lemma 2.11 we get 𝑔∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓), where 𝑔=lim_𝑚→∞𝑔_𝑚.

To see that the extended best 𝜙-approximation is an extension of the best 𝜙-approximation operator, we must prove ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓)=𝜇^ℒ_𝜙(𝑓) for every 𝑓∈𝐿^𝜙. First, we need to prove the following lemmas.

Lemma 2.13. Let 𝜙 be a 𝐶¹ convex function and assume that it satisfies the Δ₂-condition. Then 𝐾𝜙(𝑎)+2𝑎𝜙^′(𝑥−𝑎)≤𝐾+22𝜙(𝑥), (2.18) for 𝑎,𝑥≥0, where 𝐾 is the constant for the Δ₂ condition.

Proof. We consider two cases. First, we assume 0<𝑥≤𝑎. Since 𝜙 is Δ₂-convex function, we have that 𝜙(𝑎)=𝜙(𝑎−𝑥+𝑥)≤(𝐾/2)(𝜙(𝑎−𝑥)+𝜙(𝑥)). Using 𝑥𝜙^′(𝑥)≥𝜙(𝑥) for all 𝑥, we get 𝜙(𝑥)+𝑎𝜙^′(𝑎−𝑥)=𝜙(𝑥)+(𝑎−𝑥+𝑥)𝜙^′(𝑎−𝑥)≥𝜙(𝑥)+𝜙(𝑎−𝑥)+𝑥𝜙^′≥2(𝑎−𝑥)𝐾𝜙(𝑎)+𝑥𝜙^′2(𝑎−𝑥)≥𝐾𝜙(𝑎). (2.19) Then we obtain 𝐾𝜙(𝑎)−2𝑎𝜙^′𝐾(𝑎−𝑥)≤2𝜙(𝑥). (2.20) For 0≤𝑎<𝑥, we have 𝐾𝜙(𝑎)+𝑎2𝜙^′𝐾(𝑥−𝑎)≤𝜙(𝑎)+2^𝑥_𝑥−𝑎𝜙^′𝐾(𝑡)𝑑𝑡≤𝜙(𝑥)+2𝜙(𝑥)=𝐾+22𝜙(𝑥). (2.21)

Lemma 2.14. Let 𝑓∈𝐿^𝜙 and 𝑔∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓), then _{𝑔>0}𝜙^′(𝑓−𝑔)𝑔𝑑𝜇≥0. (2.22)

Proof. Since {𝑔>𝑎} is 𝜇_𝑔-positive for all 𝑎∈ℝ, then for all 𝐷∈ℒ, we have that _{{𝑔>𝑎}∩𝐷}𝜙^′(𝑓−𝑔)𝑑𝜇≥0. (2.23) In particular, it holds that for all 𝑎∈ℝ, _{𝑔>𝑎}𝜙^′(𝑓−𝑔)𝑑𝜇≥0, (2.24) that is, _{{𝑔>𝑎}∩{𝑓>𝑔}}𝜙^′(|𝑓−𝑔|)𝑑𝜇≥_{{𝑔>𝑎}∩{𝑓≤𝑔}}𝜙^′(|𝑓−𝑔|)𝑑𝜇. (2.25) Now, we have ^∞_{0{𝑔>𝑎}∩{𝑓>𝑔}}𝜙^′(|𝑓−𝑔|)𝑑𝜇𝑑𝑎≥^∞_{0{𝑔>𝑎}∩{𝑓≤𝑔}}𝜙^′(|𝑓−𝑔|)𝑑𝜇𝑑𝑎. (2.26) By the Fubini's theorem, we get _{𝑔>0}^𝑔₀𝜙^′(|𝑓−𝑔|)𝒳_{𝑓>𝑔}𝑑𝑎𝑑𝜇≥_{𝑔>0}^𝑔₀𝜙^′(|𝑓−𝑔|)𝒳_{{𝑓≤𝑔}}𝑑𝑎𝑑𝜇. (2.27) Thus _{𝑔>0}𝜙^′(|𝑓−𝑔|)𝒳_{𝑓>𝑔}𝑔𝑑𝜇≥_{𝑔>0}𝜙^′(|𝑓−𝑔|)𝒳_{{𝑓≤𝑔}}𝑔𝑑𝜇. (2.28) To see that inequality (2.22) is equivalent to (2.28) we will prove that 𝜙^′(𝑓−𝑔)𝒳_{𝑓>𝑔}𝑔∈𝐿¹({𝑔>0}). In fact _{𝑔>0}𝜙^′(𝑓−𝑔)𝒳_{𝑓>𝑔}𝑔𝑑𝜇≤_{𝑔>0}𝜙^′(𝑓−𝑔)𝒳_{𝑓>𝑔}𝑓𝑑𝜇≤_{𝑔>0}𝜙^′(𝑓)𝑓. (2.29)Since 𝑓∈𝐿^𝜙 by Lemma 2.2, the last integral is finite.

The following properties of the set ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓) can be easily proved.

Proposition 2.15. Let 𝑓∈𝐿^𝜙′, then
(1)−𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(−𝑓)=𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓),(2)^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓+𝑡)=𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓)+𝑡 for all ℎ∈ℝ.

Now we prove that the operator ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓) is in fact an extension of the operator 𝜇^ℒ_𝜙(𝑓).

Theorem 2.16. Let 𝑓∈𝐿^𝜙, then ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓)=𝜇^ℒ_𝜙(𝑓).

Proof. For 𝑓∈𝐿^𝜙, we will prove only that ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓)⊂𝜇^ℒ_𝜙(𝑓). The other inclusion follows from Theorem 2.6. Let 𝑔∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓) and again using Theorem 2.6 it remains to prove that 𝑔∈𝐿^𝜙. Recall that 𝜙(0)=0, then _Ω𝜙(|𝑔|)𝑑𝜇=_{𝑔>0}𝜙(𝑔)𝑑𝜇+_{𝑔<0}𝜙(−𝑔)𝑑𝜇. (2.30) By Lemma 2.13 we obtain the following inequality: _{𝑔>0}𝐾𝜙(𝑔)𝑑𝜇+2_{𝑔>0}𝜙^′(|𝑓|−𝑔)𝑔𝑑𝜇≤𝐾+22_{𝑔>0}𝜙(|𝑓|)𝑑𝜇. (2.31) Applying Lemma 2.14 and taking into account that 𝜙^′(𝑥) is an increasing function, we get 0≤_{𝑔>0}𝜙^′(𝑓−𝑔)𝑔𝑑𝜇≤_{𝑔>0}𝜙^′(|𝑓|−𝑔)𝑔𝑑𝜇. (2.32) Thus using (2.32) in (2.31), we have _{𝑔>0}𝜙(𝑔)𝑑𝜇≤𝐾+22_Ω𝜙(|𝑓|)𝑑𝜇. (2.33) For the set {𝑔<0}, again by Lemma 2.13, we obtain _{𝑔<0}𝐾𝜙(−𝑔)𝑑𝜇+2_{𝑔<0}𝜙^′(|𝑓|+𝑔)(−𝑔)𝑑𝜇≤𝐾+22_{𝑔<0}𝜙(|𝑓|)𝑑𝜇. (2.34) Since −𝑓+𝑔≤|𝑓|+𝑔, we have 𝜙^′(−𝑓+𝑔)≤𝜙^′(|𝑓|+𝑔). Thus _{𝑔<0}𝜙^′(−𝑓+𝑔)(−𝑔)𝑑𝜇≤_{𝑔<0}𝜙^′(|𝑓|+𝑔)(−𝑔)𝑑𝜇. (2.35) By (1) in Proposition 2.15, −𝑔∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(−𝑓), and by Lemma 2.14, we have that _{𝑔<0}𝜙^′(−𝑓+𝑔)(−𝑔)𝑑𝜇≥0. (2.36) Therefore, ∫_{𝑔<0}𝜙(−𝑔)𝑑𝜇≤((𝐾+2)/2)∫𝜙(|𝑓|)𝑑𝜇. By (2.33), we have _Ω𝜙(|𝑔|)𝑑𝜇≤𝐾+22_Ω𝜙(|𝑓|)𝑑𝜇, (2.37) and therefore 𝑔∈𝐿^𝜙.

Now, if consider a 𝜎-subalgebra ℬ instead of a 𝜎-lattice ℒ, the extended best 𝜙-approximation operator has the following properties.

Theorem 2.17. Let 𝑓,𝑓₁,and𝑓₂ be in 𝐿^𝜙′, if ℬ is a sub-𝜎-algebra of the 𝜎-algebra 𝒜, then the following hold.
(1)The set-valued function ^0𝑥0𝜇^{02𝑑𝑐ℬ}_𝜙(𝑓) is a monotone operator.(2)The set ^0𝑥0𝜇^{02𝑑𝑐ℬ}_𝜙(𝑓) is a 𝜎-complete lattice, and there exist 𝑈_ℬ,𝑉_ℬ∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓) such that 𝑈_ℬ≤𝑔≤𝑉_ℬ a.e. for every 𝑔∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓).

Proof. To prove (1), recall that this set-valued operator is monotone if 𝑓₁≤𝑓₂; then if 𝑔₁∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓₁) and 𝑔₂∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓₂), we have that 𝑔₁∧𝑔₂∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓₁) and 𝑔₁∨𝑔₂∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓₂). Since 𝐿^𝜙(ℬ) is a lattice, we know 𝑔₁∧𝑔₂∈𝐿^𝜙(ℬ) and 𝑔₁∨𝑔₂∈𝐿^𝜙(ℬ). We will prove first that 𝑔₁∧𝑔₂∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓₁). Set 𝜇^𝑓𝑖_𝑎(𝐴)=∫_𝐴𝜙^′(𝑓_𝑖−𝑎)𝑑𝜇, where 𝑎∈ℝ and 𝑖=1,2. We will see that 𝑔₁∧𝑔₂ is an LRN function for the family of measures {𝜇^𝑓1_𝑎}_𝑎∈ℝ given ℬ. First, we will see that for each 𝑎∈ℝ and for all 𝐵∈ℬ, we have 𝜇^𝑓1_𝑎({𝑔₁∧𝑔₂>𝑎}∩𝐵)≥0. (2.38) Since {𝑔₁∧𝑔₂>𝑎}∩𝐵={𝑔₁>𝑎}∩{𝑔₂>𝑎}∩𝐵 and {𝑔₂>𝑎}∩𝐵∈ℬ and using that 𝑔₁ is an LRN function of the family {𝜇^𝑓1_𝑎}_𝑎∈ℝ, we obtain that for all 𝐵∈ℬ_{{𝑔1∧𝑔2>𝑎}∩𝐵}𝜙^′(𝑓₁−𝑎)𝑑𝜇=_{{𝑔1>𝑎}∩{𝑔2>𝑎}∩𝐵}𝜙^′(𝑓₁−𝑎)𝑑𝜇≥0. (2.39) Now, we see that {𝑔₁∧𝑔₂<𝑎} is 𝜇^𝑓1_𝑎-negative for all 𝑎∈ℝ. For 𝐵∈ℬ, we have {𝑔₁∧𝑔₂<𝑎}∩𝐵=({𝑔₁<𝑎}∩𝐵)∪({𝑔₁≥𝑎}∩{𝑔₂<𝑎}∩𝐵). (2.40) Using 𝑓₁≤𝑓₂ and that 𝜙^′(⋅) is a nondecreasing function, we obtain _{{𝑔1∧𝑔2<𝑎}∩𝐵}𝜙^′(𝑓₁−𝑎)𝑑𝜇=_{{𝑔1<𝑎}∩𝐵}𝜙^′(𝑓₁−𝑎)𝑑𝜇+_{{𝑔1≥𝑎}∩{𝑔2<𝑎}∩𝐵}𝜙^′(𝑓₁−𝑎)𝑑𝜇≤_{{𝑔1<𝑎}∩𝐵}𝜙^′(𝑓₁−𝑎)𝑑𝜇+_{{𝑔1≥𝑎}∩{𝑔2<𝑎}∩𝐵}𝜙^′(𝑓₂−𝑎)𝑑𝜇≤0. (2.41) Thus _{{𝑔1∧𝑔2<𝑎}∩𝐵}𝜙^′(𝑓₁−𝑎)𝑑𝜇≤0. (2.42) By (2.39) and (2.42), we have 𝑔₁∧𝑔₂∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓₁).
Now we show 𝑔₁∨𝑔₂∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓₂). Since 𝑓₁≤𝑓₂ and {𝑔₁>𝑎} is a 𝜇^𝑓1_𝑎-positive for all 𝑎∈ℝ and for all 𝐵∈ℬ, we have _{{𝑔1∨𝑔2>𝑎}∩𝐵}𝜙^′(𝑓₂−𝑎)𝑑𝜇=_{{𝑔2>𝑎}∩𝐵}𝜙^′(𝑓₂−𝑎)𝑑𝜇+_{{𝑔1>𝑎}∩{𝑔2≤𝑎}∩𝐵}𝜙^′(𝑓₂−𝑎)𝑑𝜇≥_{{𝑔2>𝑎}∩𝐵}𝜙^′(𝑓₂−𝑎)𝑑𝜇+_{{𝑔1>𝑎}∩{𝑔2≤𝑎}∩𝐵}𝜙^′(𝑓₁−𝑎)𝑑𝜇≥0. (2.43) Since _{{𝑔1∨𝑔2<𝑎}∩𝐵}𝜙^′(𝑓₂−𝑎)𝑑𝜇=_{{𝑔1<𝑎}∩{𝑔2<𝑎}∩𝐵}𝜙^′(𝑓₂−𝑎)𝑑𝜇≤0, (2.44) the inequalities (2.43) and (2.44) prove that 𝑔₁∨𝑔₂∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓₂).
As the statement (1) proves in particular that ^0𝑥0𝜇^{02𝑑𝑐ℬ}_𝜙(𝑓) is a lattice, we will see that the set is a 𝜎-complete lattice. Given a sequence {𝑔_𝑛}_𝑛∈ℕ in ^0𝑥0𝜇^{02𝑑𝑐ℬ}_𝜙(𝑓), we have that ⋁^𝑛₁𝑔_𝑘∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓); then, from Lemma 2.11 we obtain that ⋁_𝑛∈ℕ𝑔_𝑛=lim_𝑛→∞⋁^𝑛₁𝑔_𝑛∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓). The proof ⋀_𝑛∈ℕ𝑔_𝑛∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓) is similar.
By [10, Proposition II.4.1] there exists a sequence 𝑔_𝑛∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓) such that inf𝑔_𝑛≤𝑔≤sup𝑔_𝑛, for every 𝑔∈𝜇^{0𝑥002𝑑𝑐ℬ}_𝜙(𝑓). Set 𝑈_ℬ=inf𝑔_𝑛 and 𝑉_ℬ=sup𝑔_𝑛, then 𝑈_ℬ and 𝑉_ℬ are in ^0𝑥0𝜇^{02𝑑𝑐ℬ}_𝜙(𝑓) since this set is a 𝜎-complete lattice.

3. Extended Best 𝜙-Approximation with Nondecreasing Functions

When the approximation class is the monotone functions defined on [0,1] we can obtain similar results as those of Theorem 2.17. Now Ω=[0,1],𝜇 is Lebesgue measure on the measurable sets, and ℒ={(𝑎,1),[𝑎,1),∅,ℝ}_𝑎∈ℝ. Therefore, 𝐿^𝜙′(ℒ) is the set of nondecreasing functions in 𝐿^𝜙′[0,1].

Remark 3.1. Let 𝑔 be a nondecreasing function on [0,1]. Given 𝑎∈ℝ, the set {𝑔<𝑎} is one of the intervals [0,𝛼_𝑎) or [0,𝛼_𝑎] and similarly the set {𝑔>𝑎} is (𝛽_𝑎,1] or [𝛽_𝑎,1] with 0≤𝛼_𝑎≤𝛽_𝑎≤1. Then 𝐻_𝑔={𝑎∈ℝ∶𝛼_𝑎=𝛽_𝑎} is a dense set in ℝ. In fact, the complement set of 𝐻_𝑔 is a countable set.

Note that each 𝐶∈ℒ is of the form (𝑐,1] or [𝑐,1] and 𝐷∈ℒ is 𝐷=[0,𝑑) or [0,𝑑]. Thus _{{𝑔>𝑎}∩𝐷}𝜙^′(𝑓−𝑎)𝑑𝜇=^𝑑_𝛽𝑎𝜙^′(𝑓−𝑎)𝑑𝜇,_{{𝑔<𝑎}∩𝐶}𝜙^′(𝑓−𝑎)𝑑𝜇=^𝛼𝑎_𝑐𝜙^′(𝑓−𝑎)𝑑𝜇. (★)

Theorem 3.2. Let 𝐿^𝜙(ℒ) be the class of the 𝜙^′-integrable nondecreasing functions in [0,1]. Then the following hold.
(1)The set mapping ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓) is a monotone operator.(2)For every 𝑓∈𝐿^𝜙′, the set ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓) is a 𝜎-complete lattice.

Proof. First we prove (1), that is given 𝑓₁, 𝑓₂ in 𝐿^𝜙′ with 𝑓₁≤𝑓₂, for each 𝑔_𝑖∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓_𝑖),𝑖=1,2 we will see that 𝑔₁∧𝑔₂∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓₁) and 𝑔₁∨𝑔₂∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓₂). Let 𝐻 be 𝐻_𝑔1∩𝐻_𝑔2, where 𝐻_𝑔𝑖 is the set given in Remark 3.1. Recall that 𝑔₁∧𝑔₂∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓₁) if and only if for each 𝑎∈𝐻 and 𝑐,𝑑∈ℝ_{{𝑔1∧𝑔2>𝑎}∩(0,𝑑)}𝜙^′(𝑓₁−𝑎)𝑑𝑥≥0,_{{𝑔1∧𝑔2<𝑎}∩(𝑐,1)}𝜙^′(𝑓₁−𝑎)𝑑𝑥≤0. (3.1) Also 𝑔₁∨𝑔₂∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓₂) if and only if for each 𝑎∈𝐻, and 𝑐,𝑑∈ℝ, _{{𝑔1∨𝑔2>𝑎}∩(0,𝑑)}𝜙^′(𝑓₂−𝑎)𝑑𝑥≥0,_{{𝑔1∨𝑔2<𝑎}∩(𝑐,1)}𝜙^′(𝑓₂−𝑎)𝑑𝑥≤0. (3.2) First we prove (3.1). Now we see that _{{𝑔1∧𝑔2>𝑎}∩(0,𝑑)}𝜙^′(𝑓₁−𝑎)𝑑𝑥≥0, (3.3) with {𝑔₁∧𝑔₂>𝑎}=(𝛽^𝑎₁,1]∩(𝛽^𝑎₂,1]. Since ∫_{(𝛽𝑎1,𝑑)}𝜙^′(𝑓₁−𝑎)𝑑𝑥≥0 to prove (3.3), we have to see that _{(𝛽𝑎2,𝑑)}𝜙^′(𝑓₁−𝑎)𝑑𝑥≥0, (3.4) where 𝛽^𝑎₁≤𝛽^𝑎₂. Indeed by (★), we get 0≤_𝛽𝑎1𝑎^𝑑𝜙^′(𝑓₁−𝑎)𝑑𝑥=^𝛽𝑎2_𝛽𝑎1𝜙^′(𝑓₁−𝑎)𝑑𝑥+^𝑑_𝛽𝑎2𝜙^′(𝑓₁−𝑎)𝑑𝑥. (3.5) Since 𝜙^′(⋅) is a nondecreasing function, we have 0≤^𝛽𝑎2_𝛽𝑎1𝜙^′(𝑓₂−𝑎)𝑑𝑥+^𝑑_𝛽𝑎2𝜙^′(𝑓₁−𝑎)𝑑𝑥. (3.6) As ∫^𝛽𝑎2_𝛽𝑎1𝜙^′(𝑓₂−𝑎)𝑑𝑥≤0 (𝛽^𝑎₂=𝛼^𝑎₂), we have ∫^𝑑_𝛽𝑎2𝜙^′(𝑓₁−𝑎)𝑑𝑥≥0, that is (3.4).
Now we will prove that _{{𝑔1∧𝑔2<𝑎}∩(𝑐,1)}𝜙^′(𝑓₁−𝑎)𝑑𝑥≤0. (3.7) In fact _{{𝑔1∧𝑔2<𝑎}∩(𝑐,1)}𝜙^′(𝑓₁−𝑎)𝑑𝑥=_{{𝑔1<𝑎}∩(𝑐,1)}𝜙^′(𝑓₁−𝑎)𝑑𝑥+_{{𝑔1≥𝑎}∩{𝑔2<𝑎}∩(𝑐,1)}𝜙^′(𝑓₁−𝑎)𝑑𝑥≤_{{𝑔1<𝑎}∩(𝑐,1)}𝜙^′(𝑓₁−𝑎)𝑑𝑥+_{{𝑔2<𝑎}∩[{𝑔1≥𝑎}∩(𝑐,1)]}𝜙^′(𝑓₂−𝑎)𝑑𝑥. (3.8) The last two integrals in (3.8) are less or equal than zero, so (3.7) holds. A similar argument shows that 𝑔₁∨𝑔₂∈𝜇^{0𝑥002𝑑𝑐ℒ}_𝜙(𝑓₂). Therefore, the extended best 𝜙-approximation operator is a monotone operator. By (1), we have that ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓) is a lattice, just setting 𝑓=𝑓₁=𝑓₂. Now by Lemma 2.11 we obtain that ^0𝑥0𝜇^{02𝑑𝑐ℒ}_𝜙(𝑓) is a 𝜎-complete lattice.

4. A Limit Theorem for Extended Best 𝜙-Approximations

Given a sequence {ℬ_𝑛}_𝑛∈ℕ of 𝜎-algebras contained in the 𝜎-algebra 𝒜, we consider two cases, ℬ_𝑛⊂ℬ_𝑛+1 for all 𝑛∈ℕ and we set ℬ_∞ for the 𝜎-algebra generated by ⋃_𝑛ℬ_𝑛, and if ℬ_𝑛⊇ℬ_𝑛+1 for all 𝑛∈ℕ, we set ℬ_∞=⋂_𝑛∈ℕℬ_𝑛.

The next result is a particular case of [8, Theorem 2.8] when 𝜙 is a strictly convex function. This assumption on the function 𝜙 assures that the family of measures {𝜇_𝑎}_𝑎∈ℝ decreases at zero as required by Brunk and Johansen in that theorem.

Theorem 4.1. Let {ℬ_𝑛}_𝑛∈ℕ be an increasing or decreasing sequence of 𝜎-algebras in 𝒜, and let ℬ_∞ be the limit of the sequence. If 𝑓∈𝐿^𝜙′, then we have for all 𝑔_𝑛∈𝜇^{0𝑥002𝑑𝑐ℬ𝑛}_𝜙(𝑓),𝑛∈ℕ, that 𝑙𝑖𝑚𝑖𝑛𝑓_𝑛→∞𝑔_𝑛,𝑎𝑛𝑑𝑙𝑖𝑚𝑠𝑢𝑝_𝑛→∞𝑔_𝑛 are in ^0𝑥0𝜇^{02𝑑𝑐ℬ∞}_𝜙(𝑓).

Proof. Define 𝑔=limsup_𝑛→∞𝑔_𝑛 and 𝑔=liminf_𝑛→∞𝑔_𝑛, then we only prove that 𝑔,𝑔∈𝜇^{0𝑥002𝑑𝑐ℬ∞}_𝜙(𝑓), when {ℬ_𝑛}_𝑛∈ℕ is an increasing sequence of 𝜎-algebras, the proof for the decreasing case is similar.
First, we prove for each 𝑓∈𝐿^𝜙′ that the set {𝑔≥𝑎} is 𝜇_𝑎-positive for all 𝑎∈ℝ. Let 𝐵∈ℬ_𝑚, and for 𝐻_𝑛=⋃_𝑘≥𝑛{𝑔_𝑘>𝑎−𝜖_𝑛}, where 𝜖_𝑛 decreases to zero, we have that {𝑔≥𝑎}=⋂_𝑛≥1𝐻_𝑛, and for all 𝑛∈ℕ,𝐻_𝑛+1⊂𝐻_𝑛. Now for each 𝑛∈ℕ, we define the following disjoint sets. For 𝑝≥𝑛, set 𝐻_𝑛,𝑛={𝑔_𝑛>𝑎−𝜖_𝑛},…,𝐻_𝑛,𝑝={𝑔_𝑝>𝑎−𝜖_𝑛}∩{𝑔_𝑝−1≤𝑎−𝜖_𝑛}∩⋯∩{𝑔_𝑛≤𝑎−𝜖_𝑛}. Thus, 𝐻_𝑛=⋃_𝑝≥𝑛𝐻_𝑛,𝑝, and then _{𝐻𝑛∩𝐵}𝜙^′(𝑓−(𝑎−𝜖_𝑛))𝑑𝜇=^∞_{𝑝=𝑛𝐻𝑛,𝑝∩𝐵}𝜙^′(𝑓−(𝑎−𝜖_𝑛))𝑑𝜇. (4.1) As ℬ_𝑛⊂ℬ_𝑛+1, and for 𝑝≥𝑛, we have that {𝑔_𝑝−1≤𝑎−𝜖_𝑛}∩⋯∩{𝑔_𝑛≤𝑎−𝜖_𝑛}∈ℬ_𝑝. As 𝐵∈ℬ_𝑚 then 𝐵∈ℬ_𝑝 and ∫_{𝐻𝑛,𝑝∩𝐵}𝜙^′(𝑓−(𝑎−𝜖_𝑛))𝑑𝜇≥0 for 𝑚≤𝑛≤𝑝. Thus ∫_{𝐻𝑛∩𝐵}𝜙^′(𝑓−(𝑎−𝜖_𝑛))𝑑𝜇≥0, and by Lebesgue's theorem, we get _{{𝑔≥𝑎}∩𝐵}𝜙^′(𝑓−𝑎)𝑑𝜇=lim_{𝑛→∞𝐻𝑛∩𝐵}𝜙^′(𝑓−(𝑎−𝜖_𝑛))𝑑𝜇≥0 (4.2) for all 𝐵∈⋃_𝑛∈ℕℬ_𝑛. Now we have (4.2) for all 𝐵∈ℬ_∞. In fact, the set 𝐷={𝐵∈𝒜∶∫_{{𝑔≥𝑎}∩𝐵}𝜙^′(𝑓−𝑎)𝑑𝜇≥0} is a monotone class, that is, the set 𝐷 is closed for increasing and decreasing sequences of sets. As ⋃_𝑛∈ℕℬ_𝑛⊂𝐷 and this union is an algebra of sets, the monotone class generated by it is ℬ_∞, that is, ℬ_∞⊂𝐷.
Now let us prove that the set {𝑔≤𝑎} is 𝜇_𝑎-negative for all 𝑎∈ℝ. As {𝑔≤𝑎}=⋂_𝑛∈ℕ{𝑔<𝑎+1/𝑛} and _{{𝑔≤𝑎}∩𝐵}𝜙^′(𝑓−𝑎)𝑑𝜇=lim_{𝑛→∞{𝑔<𝑎+1/𝑛}∩𝐵}𝜙^′(𝑓−𝑎)𝑑𝜇, (4.3) then we have to prove that for all 𝑎∈ℝ, the set {𝑔<𝑎} is 𝜇_𝑎-negative. Since {𝑔<𝑎}=⋃_𝑛≥1⋂_𝑘≥𝑛{𝑔_𝑘<𝑎−𝜖_𝑛}=⋃_𝑛≥1𝐻_𝑛, where 𝐻_𝑛=⋂_𝑘≥𝑛{𝑔_𝑘<𝑎−𝜖_𝑛} and 𝜖_𝑛↘0, we have for all 𝑛∈ℕ that 𝐻_𝑛⊂𝐻_𝑛+1. Then _{{𝑔<𝑎}∩𝐵}𝜙^′(𝑓−𝑎)𝑑𝜇=lim_{𝑛→∞𝐻𝑛∩𝐵}𝜙^′(𝑓−(𝑎−𝜖_𝑛))𝑑𝜇, (4.4) for a fixed set 𝐵∈ℬ_𝑚.
Set 𝐺_𝑛=⋂_{𝑘≥𝑛+1}{𝑔_𝑘<𝑎−𝜖_𝑛} and note that 𝐻_𝑛=𝐺_𝑛∩{𝑔_𝑛<𝑎−𝜖_𝑛}. Then for 𝑚≤𝑛, we have _{{𝑔𝑛<𝑎−𝜖𝑛}∩𝐵∩𝐺𝑛}𝜙^′(𝑓−(𝑎−𝜖_𝑛))𝑑𝜇+_{{𝑔𝑛<𝑎−𝜖𝑛}∩𝐵∩𝐺𝑐𝑛}𝜙^′(𝑓−(𝑎−𝜖_𝑛=))𝑑𝜇_{{𝑔𝑛<𝑎−𝜖𝑛}∩𝐵}𝜙^′(𝑓−(𝑎−𝜖_𝑛))𝑑𝜇≤0. (4.5) Now, we prove the following inequality: _{{𝑔𝑛<𝑎−𝜖𝑛}∩𝐵∩𝐺𝑐𝑛}𝜙^′(𝑓−(𝑎−𝜖_𝑛))𝑑𝜇≥0. (4.6) We can see that {𝑔_𝑛<𝑎−𝜖_𝑛}∩𝐵∩𝐺^𝑐_𝑛=_{𝑘≥𝑛+1}𝐴_𝑘, (4.7)where 𝐴_𝑘 are the following disjoint sets 𝐴_𝑛+1={𝑔_𝑛+1≥𝑎−𝜖_𝑛}∩𝐵∩{𝑔_𝑛<𝑎−𝜖_𝑛⋮𝐴},_𝑘={𝑔_𝑘≥𝑎−𝜖_𝑛}∩^𝑘−1_𝑖=𝑛+1{𝑔_𝑖<𝑎−𝜖_𝑛}∩{𝑔_𝑛<𝑎−𝜖_𝑛}∩𝐵. (4.8) Then _{{𝑔𝑛<𝑎−𝜖𝑛}∩𝐵∩𝐺𝑐𝑛}𝜙^′(𝑓−(𝑎−𝜖_𝑛))𝑑𝜇=_{𝑘≥𝑛+1𝐴𝑘}𝜙^′(𝑓−(𝑎−𝜖_𝑛))𝑑𝜇. (4.9) Since 𝐴_𝑘={𝑔_𝑘≥𝑎−𝜖_𝑛}∩𝐵_𝑘, where 𝐵_𝑘∈ℬ_𝑘, we have (4.6). Therefore by (4.5), we have _{𝐻𝑛∩𝐵}𝜙^′(𝑓−(𝑎+𝜖_𝑛))𝑑𝜇≤0, (4.10) for all 𝐵∈⋃_𝑛∈ℕℬ_𝑛. Thus by (4.10), (4.4), and (4.3), we get _{{𝑔≤𝑎}∩𝐵}𝜙^′(𝑓−𝑎)𝑑𝜇≤0 (4.11) for all 𝐵∈⋃_𝑛∈ℕℬ_𝑛. Therefore, the result is satisfied for all 𝐵∈ℬ_∞. Thus 𝑔∈𝜇^{0𝑥002𝑑𝑐ℬ∞}_𝜙(𝑓).
We have {𝑔≤𝑎}=⋂_𝑛≥1𝐻_𝑛, where 𝐻_𝑛=⋃_𝑘≥𝑛{𝑔_𝑘<𝑎+𝜖_𝑛} and 𝜖_𝑛↘0, then 𝐻_𝑛+1⊂𝐻_𝑛 for all 𝑛∈ℕ. Since 𝑓∈𝐿^𝜙′ we have for all 𝐵∈ℬ_∞ that _{{𝑔≤𝑎}∩𝐵}𝜙^′(𝑓−𝑎)𝑑𝜇=lim_{𝑛→∞𝐻𝑛∩𝐵}𝜙^′(𝑓−(𝑎+𝜖_𝑛))𝑑𝜇. (4.12) For 𝑝>𝑛 define the following disjoint sets 𝐻_𝑛,𝑛={𝑔_𝑛<𝑎+𝜖_𝑛} and 𝐻_𝑛,𝑝={𝑔_𝑝<𝑎+𝜖_𝑛}∩{𝑔_𝑝−1≥𝑎+𝜖_𝑛}∩⋯∩{𝑔_𝑛≥𝑎+𝜖_𝑛}. Then for 𝐵∈ℬ_𝑚 we have _{𝐻𝑛∩𝐵}𝜙^′(𝑓−(𝑎+𝜖_𝑛))𝑑𝜇=_{𝑝≥𝑛𝐻𝑛,𝑝∩𝐵}𝜙^′(𝑓−(𝑎+𝜖_𝑛))𝑑𝜇. (4.13) Now if 𝑚≤𝑛≤𝑝,𝐻_𝑛,𝑝∩𝐵={𝑔_𝑘<𝑎+𝜖_𝑛}∩𝐵^⋆, where 𝐵^⋆∈ℬ_𝑝. Thus _{𝐻𝑛,𝑝∩𝐵}𝜙^′(𝑓−(𝑎+𝜖_𝑛))𝑑𝜇≤0. (4.14) Then by (4.12) and (4.14), we have for all 𝐵∈⋃_𝑛∈ℕℬ_𝑛 that _{{𝑔≤𝑎}∩𝐵}𝜙^′(𝑓−𝑎)𝑑𝜇≤0. (4.15) Therefore, we have (4.15) for all 𝐵∈ℬ_∞.
Let us see now that _{{𝑔≥𝑎}∩𝐵}𝜙^′(𝑓−𝑎)𝑑𝜇≥0 (4.16) for all 𝐵∈⋃_𝑛∈ℕℬ_𝑛.
As {𝑔≥𝑎}=⋂_𝑛∈ℕ{𝑔>𝑎−1/𝑛}, we have to prove that for all 𝑎∈ℝ, the set {𝑔>𝑎} is 𝜇_𝑎-positive. We have that {𝑔>𝑎}=⋃_𝑛≥1⋂_𝑘≥𝑛{𝑔_𝑘≥𝑎+𝜖_𝑛}=⋃_𝑛≥1𝐻_𝑛, where 𝐻_𝑛 is the increasing sequence 𝐻_𝑛=⋂_𝑘≥𝑛{𝑔_𝑘≥𝑎+𝜖_𝑛} and 𝜖_𝑛↘0. Then we have _{{𝑔>𝑎}∩𝐵}𝜙^′(𝑓−𝑎)𝑑𝜇=lim_{𝑛→∞𝐻𝑛∩𝐵}𝜙^′(𝑓−(𝑎+𝜖_𝑛))𝑑𝜇. (4.17) Set 𝐺_𝑛=⋂_{𝑘≥𝑛+1}{𝑔_𝑘≥𝑎+𝜖_𝑛} and note that 𝐻_𝑛=𝐺_𝑛∩{𝑔_𝑛≥𝑎+𝜖_𝑛}. Then for 𝐵∈ℬ_𝑚,𝑚≤𝑛, we have 0≤_{{𝑔𝑛≥𝑎+𝜖𝑛}∩𝐵}𝜙^′(𝑓−(𝑎+𝜖_𝑛=))𝑑𝜇_{{𝑔𝑛≥𝑎+𝜖𝑛}∩𝐵∩𝐺𝑛}𝜙^′(𝑓−(𝑎+𝜖_𝑛))𝑑𝜇+_{{𝑔𝑛≥𝑎+𝜖𝑛}∩𝐵∩𝐺𝑐𝑛}𝜙^′(𝑓−(𝑎+𝜖_𝑛))𝑑𝜇. (4.18) Now, we prove _{{𝑔𝑛≥𝑎+𝜖𝑛}∩𝐵∩𝐺𝑐𝑛}𝜙^′(𝑓−(𝑎+𝜖_𝑛))𝑑𝜇≤0. (4.19) We can see that {𝑔_𝑛≥𝑎+𝜖_𝑛}∩𝐵∩𝐺^𝑐_𝑛=_{𝑘≥𝑛+1}𝐴_𝑘, (4.20) where 𝐴_𝑘 are the following disjoint sets: 𝐴_𝑛+1={𝑔_𝑛+1<𝑎+𝜖_𝑛}∩𝐵∩{𝑔_𝑛≥𝑎+𝜖_𝑛⋮𝐴},_𝑘={𝑔_𝑘<𝑎+𝜖_𝑛}∩^𝑘−1_𝑖=𝑛+1{𝑔_𝑖≥𝑎+𝜖_𝑛}∩{𝑔_𝑛≥𝑎+𝜖_𝑛}∩𝐵. (4.21)Then _{{𝑔𝑛≥𝑎+𝜖𝑛}∩𝐵∩𝐺𝑐𝑛}𝜙^′(𝑓−(𝑎+𝜖_𝑛))𝑑𝜇=_{𝑘≥𝑛+1𝐴𝑘}𝜙^′(𝑓−(𝑎+𝜖_𝑛))𝑑𝜇. (4.22) Since 𝐴_𝑘={𝑔_𝑘<𝑎+𝜖_𝑛}∩𝐵_𝑘, where 𝐵_𝑘∈ℬ_𝑘, we have (4.19). Therefore by (4.18), we have _{𝐻𝑛∩𝐵}𝜙^′(𝑓−(𝑎+𝜖_𝑛))𝑑𝜇≥0 (4.23) for all 𝐵∈⋃_𝑛∈ℕℬ_𝑛. Thus by (4.23) and (4.17), we get _{{𝑔≥𝑎}∩𝐵}𝜙^′(𝑓−𝑎)𝑑𝜇≥0 (4.24) for all 𝐵∈⋃_𝑛∈ℕℬ_𝑛. Therefore, the result is satisfied for all 𝐵∈ℬ_∞. Thus 𝑔∈𝜇^{0𝑥002𝑑𝑐ℬ∞}_𝜙(𝑓).

Acknowledgment

This work was supported by CONICET and UNSL grants.