The Riesz “ rising sun ” lemma for arbitrary Borel measures with some applications

The Riesz “rising sun” lemma is proved for arbitrary locally finite Borel measures on the real line. The result is applied to study an attainability problem of the exact constant in a weak (1, 1) type inequality for the corresponding Hardy-Littlewood maximal operator.


Introduction
Let M + be the one-sided Hardy-Littlewood maximal operator on the real line where m stands for the Lebesgue measure.The following equality (1) is well known and sometimes called the Riesz "rising sun lemma" (see [4]) since it can be readily obtained from the following lemma which usually carries this name (see [7], [8]): Lemma.Suppose G is a continuous function on R. Let E be the set of points x such that G(x + h) > G(x) for some h = h x > 0 .If (a, b) is a finite connected component of E , then G(a) = G(b).
The equation ( 1) is an important tool in studying various problems related with maximal functions ant there exist its several proofs (see [3], [8]) which do not depend on a geometric structure of the Lebesgue measure and can be generalized for absolutely continuous measures (see Lemma 1 in [1]).In this note we formulate and prove the lemma in the most general setting for which it remains true.
Throughout the paper let ν be a locally finite signed Borel measure and μ be a positive Borel measure on R. We assume without lose of generality that μ(a, b) > 0 for each open interval.Define the one sided maximal function For a signed measure ν , let ν + and ν − be respectively the positive and the negative part of ν .We assume that ν − (R) < ∞.Theorem 1.If measures ν − and μ are free of atoms, i.e. ν{x} ≥ 0 and μ{x} = 0 for each x ∈ R, and We construct the counterexamples where the equality (3) fails to hold when measures have inadmissable atoms or (2) is not satisfied.Nevertheless the one sided inequality λμ{M μ + ν > λ} ≤ ν{M μ + ν > λ} is always true, i.e. the weak (1,1) type inequality holds for the operator M μ + with exact constant 1 , and we show this fact separately.
Theorem 2. Let ν and μ be as in the introduction.For each λ > 0 , we have We use the above results to answer the question considered below on the attainability of the exact constant in the weak type (1, 1) inequality for the two-sided Hardy-Littlewood maximal operator.Namely, let It is well known that the constant 2 is exact in the following weak type (1, 1) inequality ( 5) for the operator M , i.e., the inequality (5) holds when C = 2 and fails to hold for some f ∈ L(R) and λ > 0 whenever C < 2 .The question arises whether the exact constant 2 can be achieved for some nontrivial integrable function, i.e. if there exist f ∈ L(R) (except f ≡ 0) and λ > 0 such that A similar problem of the attainability of the exact constants is considered in [6].We give a negative answer to the above posed question in the general setting below.
It is well known that C = 2 is also the exact constant in the weak (1,1) type inequality for the Hardy-Littlewood maximal function M μ ν corresponding to the measures ν and μ: and if ν is allowed to have atoms, say ν = δ {0} , then for each λ > 0 (7) For the sake of completeness, we give the proof of a slightly improved version of (6) and show that the equality (7) cannot be achieved (except for the trivial case) if ν + is free of atoms.Proposition 1.For any λ > 0 , we have Proposition 2. Let ν + be a finite measure free of atoms and λ > 0 .If the equality holds, then both sides of (9) are zero.

The proof of the main result
Note that in general {M μ + ν > λ} may not be open, but it is always open from the left, i.e. for each x ∈ {M μ + ν > λ} there exists ε > 0 such that (x − ε, x] ⊂ {M μ + ν > λ} .Hence its representation as a disjoint union of connected components has the form (10) where the angle " " indicates that b n either belongs or does not belong to Then obviously (4) follows, because of (10).We will prove that for each x ∈ (a, b and one can get (11) by passing to the limit in (12) as x tends to a from the right.For each x ∈ (a, b , define (13) (the latter set is not empty).The limiting argument shows that Note that ≥ 0 and the positiveness of λ is justified by (2).If μ(R) < ∞ and λ < 0, then take λ > |λ| and consider the maximal function of measure ν + λ μ with respect to μ, i.e.M μ + (ν + λ μ).It can be readily checked that (ν + λ μ) − does not have atoms and If we now apply the theorem for positive λ + λ , then we get (λ + λ )μ{M μ + ν > λ} = (ν + λ μ){M μ + ν > λ} , which readily implies (3).
Since it follows from the condition (2) that −∞ < a n for each n in the representation (10), we have λμ(a n , b n ≥ ν(a n , b n by virtue of (20).Thus which together with (4) implies (3).

Counterexamples
The following example shows that in general ν − cannot have an atom in Theorem 1.
Observe that (22) implies that (23) We say that a finite system of covering intervals {I j } n j=1 is minimal if ∪ n j=1 I j ⊃ K and K ⊂ ∪ n l =j=1 I j for each l = 1, 2, . . ., n.Since from every system of covering intervals one can select a minimal subsystem, we can assume that {I j } n j=1 is minimal.Let I j = (a j , b j ), j = 1, 2, . . ., n.It can be observed that a i = a j and b i = b j whenever i = j , since the system is minimal.We can assume that a 1 < a 2 < . . .< a n .The minimality of the system also implies that (a j , b j ) ∩ (a j+2 , b j+2 ) = ∅ for each j ≤ n − 2, since otherwise either (a j+1 , b j+1 ) can be excluded from the system (kept K covered) if b j+1 < b j+2 or (a j+2 , b j+2 ) can be excluded from the system if b j+1 > b j+2 .Thus we have the two open sets consisting of disjoint intervals By developing further the idea of the proof of Proposition 1, one can prove the following covering lemmas, which leads to the Lebesgue differentiation theorem for arbitrary measures.Although it is essential in the following two lemmas that we deal with the one-dimensional case, nevertheless we emphasize that the measure μ is arbitrary and may not satisfy the doubling condition.

Lemma 1.
Let {[a j , b j ]} j∈J be a system of closed intervals that covers a measurable set A ⊂ R with μ(A) < ∞.Then, for each ε > 0 , there exists Proof.Clearly, one can associate to each point So, if we denote by Δ x the interval with length 1  2 |Δ x | and with the same left or right endpoint x, then the system of intervals {Δ x } x∈Q will be disjoint.Thus Q is numerable.
The system of open intervals {(a j , b j )} j∈J covers A\Q .Take an arbitrary ε and a compact set K ⊂ (A\Q) such that (24) μ(K) > μ(A\Q) − ε.
We can cover K by a finite subsystem of intervals {(a j , b j )} j∈J1 , i.e.J 1 is a finite subset of J and and we can take another finite subsystem of closed intervals {[a j , b j ]} j∈J2 which covers A ∩ Q up to a set of μ-measure less than ε , It follows from ( 24),( 25) and ( 26) that Thus, we have found a finite system of closed intervals {[a j , b j ]} j∈J3 which covers A up to a set of μ-measure 2ε .
The rest of the proof repeats the proof of Proposition 1. Assume ∪ j∈J3 [a j , b j ] = A J and let {[a j , b j ]} n j=1 be a minimal cover of A J (with exactly the same definition as in the proof of Proposition 1).We can assume without loss of generality that a 1 < a 2 < . . .< a n .Then

consist of disjoint closed intervals and
either for i = 1 or 2 and the lemma follows.
Using Lemma 2, one can prove the following lemma exactly in the same way as Theorem 2.8 is proved in [5].

Lemma 2.
Let A ⊂ R be a measurable set and {[a j , b j ]} j∈J be a system of closed intervals such that for each x ∈ A and ε > 0 there exist j where dνc dμ is the Radon-Nikodym derivative.Using Lemma 2, one can get the Lebesgue differentiation theorem for the pair of measures ν and μ in a standard way (see, e.g., the proof of Theorem 2.12 in [5]).