Comptes Rendus Mathématique

. Let ρ be a symmetric measure of Lebesgue type, i.e.,


Introduction
Let µ be a Borel probability measure on R n .We call µ a spectral measure if there exists a countable discrete set Λ ⊂ R n such that E (Λ) := {e 2πi 〈λ,x〉 : λ ∈ Λ} forms an orthogonal basis for the Hilbert space L 2 (µ), and the Λ is called the spectrum of µ.The research about spectral measures was initiated by Fuglede [7], whose famous conjecture claimed that Ω is a spectral set on R n if and only if Ω is a translational tile on R n .Although the conjecture is disproved in dimension three or higher by Tao et al. [9,13,15], it is still an open problem in dimension 1 and 2. In 1998, Jorgensen and Pedersen [8] found the first singular and non-atomic spectral measure (1/4-Cantor measure), and the research about spectral measures was developed toward the field of fractal by this discovery.For the details and recent advances, one can refer to [2][3][4] and so on.
Besides spectral measures, people are also interested in looking for measures that admit exponential frames (also called Fourier frames) and exponential Riesz bases [1,5,6,14].In 2018, Lev [12] studied the addition of two measures supported respectively on two orthogonal subspaces embedded in the ambient space R n and showed that these measures admit Fourier frames.Recently, Lai, Liu and Prince [10] studied the Riesz bases and orthogonal bases for these measures.In this paper, we continue the line of research into the spectrality of the addition of measures supported on two orthogonal subspaces and extend some results of Lai, Liu and Prince [10].
Recall that a Borel measure µ on R is continuous if µ({x}) = 0 for all x ∈ R. Let µ and ν be two continuous Borel probability measures on R. We embed them into the x and y axes in R 2 respectively.The additive space for µ and ν is the space L 2 (ρ), where ρ is the measure and δ 0 is the Dirac measure at 0. We will refer to the compact support of µ and ν as the component spaces of the measure ρ.If µ = ν, we say that ρ is symmetric.If 0 ∉ supp(µ) ∩ supp(ν), we call ρ non-overlapping (Here, supp(µ) denotes the compact support of µ).If µ, ν are Lebesgue measures supported on intervals of length one, we call ρ the additive space of Lebesgue type, the additive space is defined as L 2 (ρ).Recently, Lai, Liu and Prince [10] proved the following partial results.
(1) If t = 0, then ρ is spectral and has a unique spectrum up to translations. ( , where a > 1 is a positive integer, then ρ is not spectral.
The authors leave many questions in [10].In this paper, we obtain a sufficient and necessary condition for ρ to be spectral under the assumption that t ∈ Q \ {− where Λ 0 is the spectrum of the Lebesgue measure supported on The proof depends on the analysis of the so called Orthogonality Equation.We firstly prove that the spectrum of ρ is contained in a straight line if t ∈ Q\{− 1 2 } (Proposition 8).Then we give an interesting lemma about the Orthogonality Equation (Lemma 7) and extend a result about lower Beurling density (Lemma 6).At last, we prove that ρ is not spectral if t ∈ Q \ ( 12 Z) by reduction to absurdity.Furthermore, we construct the spectrum of ρ under the condition that t ∈ 1 2 Z.This paper is organized as follows.Section 1 is an introduction and we state our main result.Section 2 presents some preliminary results.Section 3 is devoted to prove Theorem 1.We conclude in Section 4 with some remarks and a conjecture.

Preliminaries
In this section, we introduce some preliminary definitions and basic results which are used in our proof.
Let ρ be a symmetric measure of Lebesgue type, i.e., where Note also that for e a,b , the x projection is e a , and the y projection is e b .If Λ ⊂ R 2 and (a, b) ∈ Λ, then the x projection of (a, b) is a and the y projection is b.Λ x is the set and similarly for Λ y .We will use these observations frequently in the paper.
Let E (Λ) be an orthogonal set of exponential functions with exponent in Λ for L 2 (ρ) and (a, b), (c, d ) are any two distinct points of Λ.
The above equation is also called Orthogonality Equation in [10].As the right side is real, if This fact will be used many times in this paper.
In the following, we give some useful results proved in [10].

Definition 2 ([10, Definition 7]
).Let Λ be the set of exponents for a set of exponential functions on an additive space.The multiplicity of Λ is the largest number of points on any vertical or horizontal line through Λ, if such a maximum exists.We say that Λ has bounded multiplicity in this case.
Similarly, the multiplicity of u ∈ Λ x (or v ∈ Λ y is the number of points on a vertical (or horizontal) line through u (or v), if this number is finite.

Theorem 4 ([10, Theorem 4.2]). Let E (Λ) be an orthonormal basis for an additive space. Then
(i) Λ has multiplicity one; (ii) Suppose that µ and ν are Lebesgue measures supported on intervals of length one.Then Λ cannot be a subset of Z 2 .
For t ∈ Q \ {− 1 2 }, by the following lemma, we can assume that there are infinite integer points in Λ if E (Λ) is infinite.Lemma 5. Let ρ be a symmetric measure of Lebesgue type, where the component measure µ is the Lebesgue measure supported on [t , t + 1] and t the lemma follows from taking (u, v) = (0, 0).In the following, we consider the case # As the right side is real and λ 1 , λ 2 ∉ Z, we obtain (λ 1 − λ 2 ) 1 , λ 2 ), (λ 1 , λ 2 ) be any two distinct points of Λ \ Z 2 .Write by the Orthogonality Equation ( 2), we have sin π λ sin π λ Note that {k j } ∞ j =1 is a sequence of integers.Without loss of generality, we assume that k j = k i (mod p) for any j ̸ = i .It follows that sin π λ Hence λ This together with k j 2 − k j 1 ∈ pZ implies that λ ∈ 1 2 Z for all j 2 > j 1 .Clearly, there exists a subsequence {n k } such that λ In the same manner we can see that there exists a subsequence {n k } such that λ ), the lemma is proved.

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Recall the Beurling density of countable sets.Let Λ be a countable set in R n .For r > 0, the lower Beurling density corresponding to r (or r -Beurling density) of Λ is defined by the formula Proof.Without loss of generality, we assume that α, β > 0. It follows that

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With the help of above preparations, we can prove our main result.
According to Proposition 8, there exists α ̸ = 0, 1 such that with gcd(p, q) = 1.Without loss of generality, we only consider the case p, q > 0.
We divide the following proof into four steps.
When q = 2, then p ≥ 3 is odd.By the expression of ( 9), (10), without loss of generality, we assume that The last equality follows from r is odd and a = 2 p .Hence Λ is not a spectrum of ρ.Then we have proved that Λ can't be the spectrum of ρ if t ∈ Q \ ( 1 2 Z).
Completion of the proof of Theorem 1.In the final step, we construct a spectrum of ρ under the condition that t ∈ 1 2 Z, i.e., q = 1.Combining with the expression of Λ in ( 9), ( 10), (11), we know if p is odd, then there exists an odd r ∈