The Packing Measure of the Trajectory of a One-Dimensional Symmetric Cauchy Process

Let 𝑋𝑡={𝑋(𝑡),𝑡≥0} be a one-dimensional symmetric Cauchy process. We prove that, for any measure function, 𝜑,𝜑−𝑝(𝑋[0,𝜏]) is zero or infinite, where 𝜑−𝑝(𝐸) is the 𝜑-packing measure of 𝐸, thus solving a problem posed by Rezakhanlou and Taylor in 1988.


Introduction
Let X t {X t , t ≥ 0} be a strictly stable Levy process taking values in R n n-dimensional Euclidean space of index α ∈ 0, 2 , that is, a Markov process with stationary independent increment whose characteristic function is given by E e i u,X t e −tψ α u .1.1 Here, u and X t are points in R n , u, x is the ordinary inner product in R n , and x 2 x, x .The Levy exponent ψ α u is of the form

1.3
μ dy is an arbitrary finite measure on the unit sphere S n in R n , not supported on a diametrical plane.If in 1.2 μ is the uniform distribution on S n , X t is called the isotropic stable Levy process with index α.In this case, ψ α u λ|u| α for some λ > 0. When α 1, μ must also have the origin as its center of mass, that is, and the resulting process is the symmetric Cauchy process.
If S n yμ dy / 0 for α 1, we have the strictly asymmetric Cauchy process.When α 2, we obtain the standard Brownian motion.
We assume that our process has been defined so that the strong Markov property is valid and all sample paths are right continuous and have left limits everywhere.
It is well known that the sample paths X t of strictly stable Levy processes determine trajectories in R n that are random fractal sets.
We are interested in the range of the processes, that is, the random set R τ generated by X t and defined by The Hausdorff and packing measures serve as useful tools for analyzing fine properties of Levy processes.
The problem of determining the exact Hausdorff measure of the range of those processes for α ∈ 0, 2 has been completely solved.See, for example, 1 .
The study of the exact packing measure of the range of a stochastic process has a more recent history, starting with the work of Taylor and Tricot 2 .
The packing measure of the trajectory was found in 2 by Taylor and Tricot for transient Brownian motion.The corresponding problem for the range of strictly stable processes,α < n, was solved by Taylor 3 .Further results on the asymmetric Cauchy process and subordinators have been established by Rezakhanlou and Taylor 4 and Fristedt and Taylor 5 , respectively.
For the critical cases, α n, the only known result is due to Le Gall and Taylor 6 .They proved that if X t is a planar Brownian motion, α n 2, ϕ − p X 0, t is either zero or infinite for any measure function ϕ.Hence, the packing measure problem of the symmetric Cauchy process on the line remained open.
The main objective of this paper is to show that for α n 1, a similar result to that of planner Brownian motion holds for the packing measure of the range of a one-dimensional symmetric Cauchy process with different criteria on ϕ.

Preliminaries
In this section, we start by recalling the definition and properties of packing measure and packing dimension introduced by Taylor and Tricot 2 .
Let Φ be the class of functions: which are right continuous and monotone increasing with ϕ 0 0 and for which there is a finite constant k > 0 with The inequality 2.2 is a weak smoothness condition usually called a doubling property.A function ϕ in Φ is often called a measure function: where B x i , r i denotes the closure of the open ball B x i , r i which is centered at x and has radius r.
A sequence of closed balls satisfying the condition on the right side of 2.3 is called a ε-packing of E.
ϕ − P is a ϕ-packing premeasure.The ϕ-packing measure on R n , denoted by ϕ − p, is obtained by defining It is proved in 2 that ϕ − p E is a metric outer measure, and hence every Borel set in R n is ϕ − p measurable.We can see that for any This gives a way to determine the upper bound of ϕ − p E .Using the function ϕ s s α , α > 0 gives the fractal index called the packing dimension of E.
In order to calculate the packing measure, we will use the following density theorem of Taylor and Tricot 2 , which we will call Lemma 2.1.
Lemma 2.1.For a given ϕ ∈ Φ, there exists a finite constant k > 0 such that for any Borel measure μ on R n with 0 < μ μ R n < ∞ and any Borel set where is the lower ϕ-density of μ at x.
One then uses the sample path X t to define the random measure known as the occupation measure of the trajectory; |•| denotes the Lebesgue measure.
This gives a Borel measure with μ R u τ, and it is concentrated on X 0, τ and spreads evenly on it. If is the sojourn time of X t in the ball B x, r up to the time τ.Define τ inf{t > 0 : then by a result in 7 about τ one has E 0 τ 1, where E 0 is the associated expectation for the process started at 0. Denote ∞ 0 by 0 .If x 0, one denotes T x, r by T r .In 8 , we exhibited a measure function ϕ satisfying the following criteria.

Theorem 2.2. Suppose ϕ rh r , where h r is a monotone nondecreasing function and
where X t is a one-dimensional symmetric Cauchy process.
For any t 0 ≥ 0, X t t 0 − X t 0 is also a symmetric Cauchy process on the line since the finite-dimensional distribution of X t t 0 − X t 0 is independent of t 0 ; see, for example, 1 for the strong Markov property of Cauchy processes.
The following corollary is then immediate.
Corollary 2.3.Let X t , t ≥ 0, be a one-dimensional symmetric Cauchy process.Then, for any t 0 ≥ 0 with probability one,

2.14
where ϕ is as defined in Theorem 2.2.
One will also need an estimate for the small ball probability of the sojourn time T r , taken from 8, Theorem 3.1 .

Lemma 2.4. Suppose ϕ r rh r
, where h r is a monotone increasing function.For T defined in 2.12 , then for any fixed constant c 1 and a k ρ −k , ρ > 1,

2.15
In the next section, we will use the above results and some known techniques to calculate the packing measure of the trajectory of the one-dimensional symmetric Cauchy process.

The measure of the trajectory
In this section, we proceed to the main result.Theorem 3.1.Let X t {X t : t ≥ 0} be a one-dimensional symmetric Cauchy process.If ϕ r rh r , where h is a nondecreasing function, then with probability one, where ϕ − p X 0, τ is the ϕ-packing measure of X 0, τ .
Proof.In order to apply the density Lemma 2.1, we have to calculate But by Corollary 2.3, for each fixed t 0 ∈ 0, τ with probability one, lim inf Then a Fubini argument gives Using an application of the inequality of the density Lemma 2.1, we have and thus ϕ − pX 0, t ∞ with probability one if 0 h s /s ln 1/s ∞.In order to prove the upper bound, we use density Lemma 2.1 in the other direction, as well as a "bad-point" argument similar to that in 3 .
For each point x ∈ R, let V k x denote a semidyadic interval with length 2 −k whose complement is at distance 2 −k−2 from a dyadic interval of length 2 −k−2 which contains x. Let We use the intervals in Γ E to replace the balls B x, r in 2.3 with length 2 −k replacing 2r diamB x, r .
This gives a new premeasure ϕ − P xx E comparable to ϕ − P as follows.There exist positive finite constants k 1 , k 2 such that, for all Borel sets E ⊂ R, where be the set of "good" points.A Fubini argument tells us that |G| τ < ∞ a.s.; then using the density lemma in the other direction, we have where is the set of "bad" points.
For t ∈ G j , by monotonicity, we have for a positive constant c, μ B X t , 2 −k ≤ cjϕ 2 −k , 3.14 for infinitely many k.
For fixed j, we can only get a contribution to ϕ−P xx X G j from semidyadic intervals of length 2 −k if the dyadic interval of length 2 −k−2 is entered by X t at time t ≤ τ and the restarted process leaves the interval of length 2 −k−2 in less than jϕ 2 −k .
Thus, if S k is a semidyadic interval of length 2 −k , then S k is bad if X s enters inside dyadic interval of length 2 −k−2 but spends less than jϕ 2 −k in S k ; otherwise it is "good".Any t ∈ G j will be in infinitely many such bad S k .
The probability that S k is bad given that it is entered is at most Thus, using 3.16 , we have sufficiently small.It follows that ϕ − p xx X G j 0 a.s., and from 3.8 , ϕ − p xx X G j 0 a.s.So ϕ − p xx X ∞ j 1 G j ≤ ϕ − p xx X G j 0. By 3.12 , G∪ ∞ j 1 G j 0, τ , and therefore ϕ−pX 0, τ 0 if 0 h s /s ln 1/s ds < ∞.This completes the proof.
log u, y .

. 15 by 3 ∞ k k 0
Lemma 2.4.Let N k τ be the number of intervals of length 2 −k that are entered by the time τ, and let B k τ denote the number of those that are bad; thenEB k τ ≤ EN k τ h 2 −k k .3.16Leaving out the nonoverlapping requirement, we have, for a positive constant c 3 ,Eϕ − p xx X G j ≤ c EB k τ ϕ 2 −k .Now,by 1, Lemma 4.1 , EN k s ≤ c 2 2 m , for a positive constant c 2 .