1 Introduction
The main purpose of this article is to study the regularity property of the integral of a deterministic function with respect to stochastic measure μ, where μ is defined on Borel subsets of ${[0,1]^{d}}$. In basic statements of the paper, for μ we assume only σ-additivity in probability. We study the regularity with respect to a parameter and with respect to the upper limit of the integral. The case $d=1$ was considered in [20, Section 2.3], but methods of [20] do not work for $d\ge 2$. In our paper, we study the integral using the Fourier–Haar expansion of the integrand.
Sample path behavior of various stochastic processes was studied in many works; we mention only some of them. The conditions for sample boundedness and continuity of α-stable processes were established in [24, Chapter 10]. Hölder continuity of harmonizable operator scaling stable random field is given by Corollary 5.5 [5]. In Theorem 4.1 from [18] the uniform modulus of continuity for some self-similar $S\alpha S$ random field was obtained, while the upper bound of the modulus of continuity of stable random fields is given by Proposition 5.1 and Corollary 5.3 [6]. These results, for example, imply Hölder continuity of mentioned fields.
Conditions of continuity of Gaussian random fields may be found in Theorems 3.3.3, 3.4.1 and 3.4.3 [1]. Hölder continuity of centered Gaussian random fields under certain conditions is proved in [30, Theorem 4.2]. Theorem 2.1 [31] gives an estimate of the modulus of continuity of general random fields on metric space, these results are applied to harmonizable stable random fields.
In our article, we consider stochastic processes which are integrals with respect to a general stochastic measure. The definition of these measures, their properties and examples as well as information about the Fourier–Haar series may be found in Section 2. The continuity and differentiability of sample paths of the parameter-dependent integral are established in Section 3. The continuity of realizations of the integral as a function of the upper limit is studied in Section 4.
2 Preliminaries
2.1 Stochastic measures
In this subsection, we give basic information concerning stochastic measures in a general setting. In statements of Sections 3 and 4, this set function is defined on Borel subsets of ${[0,1]^{d}}$.
Let ${\mathsf{L}_{0}}={\mathsf{L}_{0}}(\Omega ,\mathcal{F},\mathsf{P})$ be the set of all real-valued random variables defined on the complete probability space $(\Omega ,\mathcal{F},\mathsf{P})$. Convergence in ${\mathsf{L}_{0}}$ means the convergence in probability. Let $\mathsf{X}$ be an arbitrary set and $\mathcal{B}$ a σ-algebra of subsets of $\mathsf{X}$.
Definition 1.
A σ-additive mapping $\mu :\hspace{2.5pt}\mathcal{B}\to {\mathsf{L}_{0}}$ is called stochastic measure (SM).
In other words,
We do not assume the moment existence or martingale properties for SM. We can say that μ is an ${\mathsf{L}_{0}}$-valued measure.
We note the following examples of SMs in multidimensional spaces. All measures, if nothing else is mentioned, are considered on Borel σ-algebra of sets in ${[0,1]^{d}}$.
Example 1.
Let μ be a random orthogonal measure – it is defined, for example, in [9, Section 5.3] and [14, Section 2.3] – with a finite structural function. Then μ is an SM in the sense of Definition 1; condition (1) follows from [9, Section 5.3, Theorem 1(d)]. For example, the set function $\mu (A)={\textstyle\int _{{[0,1]^{d}}}}{\mathbf{1}_{A}}(t)\hspace{0.1667em}dW(t)$, where we take multiple Wiener–Itô integral (see its properties, for example, in [17, Section 1.1]), is a random orthogonal measure with the structural function $\mathsf{m}(A)=d!{\lambda _{d}}(A)$, where ${\lambda _{d}}$ is the d-dimensional Lebesgue measure.
Example 2.
Let μ be an independently scattered α-stable random measure for $\alpha \in (0,2]$ with a finite control measure. According to the definition – see [24, Definition 3.3.1] – for disjoint Borel sets $\{{A_{n}}:n\ge 1\}$ the following holds:
Therefore, μ is an SM.
Example 3.
Consider the set function $\mu :\mathcal{B}({[0,1]^{d}})\to {\mathsf{L}_{0}}$ such that
where ${Z_{H}^{q}}$ is the Hermite sheet (see, for example, [7] or [29, Section 4]), $H=({H_{1}},\dots ,{H_{d}})$, $1/2\lt {H_{i}}\lt 1$. Under these conditions, integral in (2) is well defined, and μ is an SM in the sense of Definition 1; the statement (1) can be proved using the representation of μ via Wiener integral (see [29, (4.12)]). Proposition 3 of [7] implies the Hölder continuity of μ.
If $q=1$ then the Hermite sheet coincides with the fractional Brownian sheet.
Theorem 10.1.1 of [15] states the sufficient conditions under which the product of one-dimensional SMs generates an SM.
SMs may be used for study of stochastic dynamical systems (see [3]).
For deterministic measurable functions $f:\mathsf{X}\to \mathbb{R}$, an integral of the form ${\textstyle\int _{\mathsf{X}}}f\hspace{0.1667em}d\mu $ is studied in [15, Chapter 7], [20, Chapter 1]. In particular, every bounded measurable f is integrable w. r. t. any μ. An analogue of the Lebesgue dominated convergence theorem holds for this integral (see [15, Proposition 7.1.1] or [20, Theorem 1.5]).
Below we will use the following statement for SMs defined on arbitrary σ-algebra $\mathcal{B}$.
Lemma 1 (Lemma 3.1 [19]).
Let μ be an SM on $(\mathsf{X},\mathcal{B})$, ${f_{k}}:\hspace{2.5pt}\mathsf{X}\to \mathbb{R}$, $k\ge 1$, be measurable functions such that $\hat{f}(x)={\textstyle\sum _{k=1}^{\infty }}|{f_{k}}(x)|$ is integrable w. r. t. μ. Then
2.2 Besov spaces
We recall the definition of Besov spaces ${B_{p,p}^{\alpha }}({[0,1]^{d}})$ following [8] and [12]. Other approaches to the definition and properties of these spaces are considered in [4], [23, Sections 2.1 and 2.4], [27], [28, Sections 2 and 5]. The equivalence of different definitions is discussed in [8], [16, Section 17.2], [23, Section 2.3].
For functions $f\in {\mathsf{L}_{p}}({[0,1]^{d}})$, we consider the value
\[ \| f{\| _{{B_{p,p}^{\alpha }}({[0,1]^{d}})}}=\| f{\| _{{\mathsf{L}_{p}}({[0,1]^{d}})}}+{\Big({\int _{0}^{1}}{({\omega _{p}}(f,r))^{p}}{r^{-\alpha p-1}}\hspace{0.1667em}dr\Big)^{1/p}},\]
where ${\omega _{p}}$ denotes the ${\mathsf{L}_{p}}$-modulus of continuity,
\[\begin{aligned}{}& {\omega _{p}}(f,r)=\underset{|h|\le r}{\sup }{\Big({\int _{{I_{h}}}}|f(x+h)-f(x){|^{p}}\hspace{0.1667em}dx\Big)^{1/p}},\\ {} & \hspace{1em}\text{where}\hspace{1em}{I_{h}}=\{x\in {[0,1]^{d}}:\hspace{2.5pt}x+h\in {[0,1]^{d}}\},\end{aligned}\]
and $|h|$ denotes the Euclidean norm in ${\mathbb{R}^{d}}$. Then
\[ {B_{p,p}^{\alpha }}({[0,1]^{d}})=\{f\in {\mathsf{L}_{p}}({[0,1]^{d}}):\hspace{2.5pt}\| f{\| _{{B_{p,p}^{\alpha }}({[0,1]^{d}})}}\lt +\infty \},\]
and $\| \cdot {\| _{{B_{p,p}^{\alpha }}({[0,1]^{d}})}}$ is a norm in this space.Let μ be an SM defined on the Borel σ-algebra of ${[0,1]^{d}}$. For $x=({x_{1}},{x_{2}},\dots ,{x_{d}})\in {[0,1]^{d}}$ set
In the sequel, we will refer to the following assumption on stochastic measure μ.
Assumption A1.
There exists a real-valued finite measure $\mathsf{m}$ on $(\mathsf{X},\mathcal{B})$ with the following property: if a measurable function $g:\mathsf{X}\to \mathbb{R}$ satisfies ${\textstyle\int _{\mathsf{X}}}{g^{2}}\hspace{0.1667em}d\mathsf{m}\lt +\infty $ then g is integrable w. r. t. μ on $\mathsf{X}$.
For orthogonal measure, we can take its structural measure as $\mathsf{m}$. For an α-stable random measure, Assumption A1 holds for the control measure $\mathsf{m}$ (see (3.4.1) [24]). For an SM from Example 3 we can take measure $\mathsf{m}$ such that
\[ \mathsf{m}(A)={\int _{A}}dv{\int _{{[0,1]^{d}}}}{\prod \limits_{i=1}^{d}}|{u_{i}}-{v_{i}}{|^{2{H_{i}}-2}}du,\]
where u and v are d-dimensional vectors with the components ${u_{i}}$, ${v_{i}}$.We have the following statement concerning the Besov regularity of SMs defined on Borel subsets of ${[0,1]^{d}}$.
2.3 Haar functions
In order to construct a version of the stochastic integral, we use the approximation of the integrand with Fourier–Haar series. In [21] Fourier–Haar series were already used in approximation of the solution to stochastic wave equation when $d=1$.
We follow the definition of one-dimensional Haar functions ${\chi _{n}}(x)$, $x\in [0,1]$, from [13, Section 3]. For $n={2^{j}}+i$, $1\le i\le {2^{j}}$, $j\ge 0$, we write
(3)
\[\begin{aligned}{}& {\Delta _{n}}={\Delta _{j}^{i}}=\big((i-1){2^{-j}},i{2^{-j}}\big),\hspace{1em}{\overline{\Delta }_{n}}=\big[(i-1){2^{-j}},i{2^{-j}}\big],\\ {} & {\Delta _{1}}={\Delta _{0}^{0}}=(0,1),\hspace{1em}{\overline{\Delta }_{1}}=[0,1],\\ {} & {\Delta _{n}^{+}}={\big({\Delta _{j}^{i}}\big)^{+}}=\big((i-1){2^{-j}},(2i-1){2^{-j-1}}\big)={\Delta _{j+1}^{2i-1}},\\ {} & {\Delta _{n}^{-}}={\big({\Delta _{j}^{i}}\big)^{-}}=\big((2i-1){2^{-j-1}},i{2^{-j}}\big)={\Delta _{j+1}^{2i}}.\end{aligned}\]Now let ${\chi _{1}}=1$, and
For $g\in {\mathsf{L}_{1}}([0,1])$, we define the Fourier–Haar coefficients and sums in a standard way:
\[ {\chi _{n}}(x)=\left\{\begin{array}{l@{\hskip10.0pt}l}0,& x\notin {\overline{\Delta }_{n}},\\ {} {2^{j/2}},& x\in {\Delta _{n}^{+}},\\ {} -{2^{j/2}},& x\in {\Delta _{n}^{-}},\end{array}\right.\]
for ${2^{j}}+1\le n\le {2^{j+1}}$. In discontinuity points and at the endpoints of $[0,1]$ we define
(4)
\[\begin{aligned}{}& {\chi _{n}}(x)=\big({\chi _{n}}(x-)+{\chi _{n}}(x+)\big)/2,\hspace{1em}x\in (0,1),\\ {} & {\chi _{n}}(0)={\chi _{n}}(0+),\hspace{1em}{\chi _{n}}(1)={\chi _{n}}(1-).\end{aligned}\]
\[\begin{aligned}{}& {c_{n}}(g)={\int _{[0,1]}}g(x){\chi _{n}}(x)\hspace{0.1667em}dx={2^{j/2}}\Big({\int _{{\Delta _{n}^{+}}}}g(x)\hspace{0.1667em}dx-{\int _{{\Delta _{n}^{-}}}}g(x)\hspace{0.1667em}dx\Big),\\ {} & {S_{N}}(g,x)={\sum \limits_{n=1}^{N}}{c_{n}}(g){\chi _{n}}(x).\end{aligned}\]
We will approximate g taking the Fourier–Haar sums for values $N={2^{k}}$, $k\in {\mathbb{Z}_{+}}$.By (III.8), (III.8′) of [13], for $x\ne i{2^{-k}}$, $0\le i\le {2^{k}}$, it holds
\[ {S_{{2^{k}}}}(g,x)={\sum \limits_{i=1}^{{2^{k}}}}{2^{k}}{\mathbf{1}_{{\Delta _{k}^{i}}}}(x){\int _{{\Delta _{k}^{i}}}}g(t)\hspace{0.1667em}dt,\]
and
\[\begin{aligned}{}& {S_{{2^{k}}}}\Big(g,\frac{i}{{2^{k}}}\Big)={2^{k-1}}\Big({\int _{{\Delta _{k}^{i}}}}g(t)\hspace{0.1667em}dt+{\int _{{\Delta _{k}^{i+1}}}}g(t)\hspace{0.1667em}dt\Big),\hspace{1em}i=1,2,\dots {2^{k}}-1,\\ {} & {S_{{2^{k}}}}(g,0)={2^{k}}{\int _{{\Delta _{k}^{1}}}}g(t)\hspace{0.1667em}dt,\hspace{1em}{S_{{2^{k}}}}(g,1)={2^{k}}{\int _{{\Delta _{k}^{{2^{k}}}}}}g(t)\hspace{0.1667em}dt.\end{aligned}\]
For integrable functions $f:{[0,1]^{d}}\to \mathbb{R}$ we will use approximation by multivariate Fourier–Haar sums.
The multivariate Haar functions are defined with equalities
\[ {\chi _{{n_{1}},\dots ,{n_{d}}}^{(d)}}(x)={\chi _{{n_{1}}}}({x_{1}}){\chi _{{n_{2}}}}({x_{2}})\dots {\chi _{{n_{d}}}}({x_{d}}),\]
where ${\chi _{{n_{s}}}}$ are one-dimensional Haar functions, $x=({x_{1}},\dots ,{x_{d}})\in {[0,1]^{d}}$. Consider the Fourier–Haar coefficients
\[ {c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}(f)={\int _{{[0,1]^{d}}}}f(x){\chi _{{n_{1}},\dots ,{n_{d}}}^{(d)}}(x)\hspace{0.1667em}dx,\]
and sums
\[ {S_{{2^{k}}}^{(d)}}(f,x)=\sum \limits_{{n_{1}},\dots ,{n_{d}}\in \{1,2,3,\dots {2^{k}}\}}{c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}(f){\chi _{{n_{1}},\dots ,{n_{d}}}^{(d)}}(x).\]
In the sequel, we denote by ${\mathbb{N}_{k}}$ the set of elements $({n_{1}},\dots ,{n_{d}})$ such that $1\le {n_{s}}\le {2^{k}}$ for all $s,1\le s\le d$. In other words, ${\mathbb{N}_{k}}={\{1,2,3,\dots ,{2^{k}}-1,{2^{k}}\}^{d}}$.The following statement on the uniform convergence of multivariate Fourier–Haar sums for continuous functions follows from formula (14) in [22, Lemma 2] for $p=\infty $. For $d=1$ this fact may be found, for example, in [13, Theorem III.2]. A similar statement for multivariate periodic functions was proved in [2].
3 Parameter dependent integral
In this section, we study the properties of random function
where Z is a metric space. Note that a parameter dependent stochastic integral w.r.t. Brownian sheet was considered in [25, 26].
If f is continuous in x then Lemma 2 and analogue of the dominated convergence theorem ([20, Theorem 1.5]) imply that for each fixed z
is the version of $\eta (z)$.
\[ {\int _{{[0,1]^{d}}}}f(z,x)\hspace{0.1667em}d\mu (x)=\text{p-}\hspace{-0.1667em}\hspace{-0.1667em}\hspace{-0.1667em}\underset{k\to \infty }{\lim }{\int _{{[0,1]^{d}}}}{S_{{2^{k}}}^{(d)}}(f,x)\hspace{0.1667em}d\mu (x),\]
where p-lim is considered as the limit in probability. Therefore,
(6)
\[ \tilde{\eta }(z)={\int _{{[0,1]^{d}}}}{S_{1}^{(d)}}(f,x)\hspace{0.1667em}d\mu (x)+{\sum \limits_{k=1}^{\infty }}{\int _{{[0,1]^{d}}}}({S_{{2^{k}}}^{(d)}}(f,x)-{S_{{2^{k-1}}}^{(d)}}(f,x))\hspace{0.1667em}d\mu (x)\]In (6) we have integrals of simple functions, and each integral is equal to a respective linear combination of values of μ. We fix the same version of μ for all these values.
We will use notation ${e_{i}}=(0,\dots ,0,1,0,\dots ,0)$, where 1 stays on i-th position, ${e_{i}}\in {\mathbb{R}^{d}}$.
Lemma 3.
Assume that there exists $l\ge 0$ such that function $f(z,x):Z\times {[0,1]^{d}}\to \mathbb{R}$ is continuously differentiable l times on ${[0,1]^{d}}$ for each z, and
for some constant ${C_{f}}\gt 0$, which is independent of z, and all $0\le r\le l$, ${s_{1}},\dots ,{s_{r}}\subset \{1,\dots ,d\}$. Moreover, let l-th derivatives be Hölder continuous with an exponent $\alpha \gt 0$, i. e.
(7)
\[ \Big|\frac{{\partial ^{r}}f(z,x)}{\partial {x_{{s_{1}}}}\dots \partial {x_{{s_{r}}}}}\Big|\le {C_{f}},\hspace{1em}x\in {[0,1]^{d}},\]
If $l+\alpha \gt d/2$, then the version (6) satisfies the inequality
where ${C_{\mu }^{(d)}}(\omega )$ denotes a random constant that depends only of dimension d and SM μ, ${C_{\mu }^{(d)}}(\omega )\lt \infty $ a.s.
Moreover, series (6) converges absolutely and uniformly, i. e.
Proof.
We have that
For each ${n_{s}}\ge 2$, take ${j_{s}}$ such that ${2^{{j_{s}}}}+1\le {n_{s}}\le {2^{{j_{s}}+1}}$, ${j_{s}}\ge 0$. Denote by $\{{j^{(k)}},1\le k\le d\}$ the permutation of $\{{j_{k}},1\le k\le d\}$ such that
If ${n_{{s_{1}}}}=\cdots ={n_{{s_{i}}}}=1$, we set ${j^{(1)}}=\cdots ={j^{(i)}}=0$. We rewrite the expression for ${c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}(f)$ in the following way:
In equality (*) we opened the brackets in the product
(11)
\[\begin{aligned}{}& {c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}(f)={\int _{{[0,1]^{d}}}}f(z,t){\chi _{{n_{1}},\dots ,{n_{d}}}^{(d)}}(t)\hspace{0.1667em}dt={\int _{{[0,1]^{d}}}}f(z,t)\prod \limits_{1\le s\le d}{\chi _{{n_{s}}}}({t_{s}})\hspace{0.1667em}dt\\ {} & \hspace{1em}={2^{{\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}{j_{s}}/2}}{\int _{{[0,1]^{d}}}}f(z,t)\prod \limits_{1\le s\le d,{n_{s}}\ge 2}\Big({\mathbf{1}_{{\Delta _{{n_{s}}}^{+}}}}({t_{s}})-{\mathbf{1}_{{\Delta _{{n_{s}}}^{-}}}}({t_{s}})\Big)\hspace{0.1667em}dt\\ {} & \hspace{1em}={2^{{\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}{j_{s}}/2}}{\int _{{[0,1]^{d}}}}f(z,t)\prod \limits_{1\le s\le d,{n_{s}}\ge 2}\Big({\mathbf{1}_{{\Delta _{{n_{s}}}^{+}}}}({t_{s}})-{\mathbf{1}_{{\Delta _{{n_{s}}}^{+}}}}({t_{s}}-{2^{-{j_{s}}-1}})\Big)\hspace{0.1667em}dt\\ {} & \hspace{1em}\stackrel{(\ast )}{=}{2^{{\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}{j_{s}}/2}}\\ {} & \hspace{2em}\times {\int _{{[0,1]^{d}}}}\sum \limits_{{\varepsilon _{s}}\in \{0,1\},{n_{s}}\ge 2}{(-1)^{{\varepsilon _{1}}+\cdots +{\varepsilon _{d}}}}f(z,{t_{1}}+{\varepsilon _{1}}{2^{-{j_{1}}-1}},\dots ,{t_{d}}+{\varepsilon _{d}}{2^{-{j_{d}}-1}})\\ {} & \hspace{2em}\times \prod \limits_{1\le s\le d,{n_{s}}\ge 2}{\mathbf{1}_{{\Delta _{{n_{s}}}^{+}}}}({t_{s}})\hspace{0.1667em}dt={2^{{\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}{j_{s}}/2}}\\ {} & \hspace{2em}\times {\int _{{\textstyle\prod _{1\le s\le d,{n_{s}}\ge 2}}{\Delta _{{n_{s}}}^{+}}}}\sum \limits_{{\varepsilon _{s}}\in \{0,1\},{n_{s}}\ge 2}{(-1)^{{\varepsilon _{1}}+\cdots +{\varepsilon _{d}}}}\\ {} & \hspace{2em}\times f(z,{t_{1}}+{\varepsilon _{1}}{2^{-{j_{1}}-1}},\dots ,{t_{d}}+{\varepsilon _{d}}{2^{-{j_{d}}-1}})\hspace{0.1667em}dt.\end{aligned}\]
\[ \prod \limits_{1\le s\le d,{n_{s}}\ge 2}\Big({\mathbf{1}_{{\Delta _{{n_{s}}}^{+}}}}({t_{s}})-{\mathbf{1}_{{\Delta _{{n_{s}}}^{+}}}}({t_{s}}-{2^{-{j_{s}}-1}})\Big)\]
and used the change of the variables ${t_{s}}-{2^{-{j_{s}}-1}}\to {t_{s}}$ in ${\mathbf{1}_{{\Delta _{{n_{s}}}^{+}}}}({t_{s}}-{2^{-{j_{s}}-1}})$.The last sum in (11) can be represented as the sum of at most ${2^{d-1}}$ summands of the form
\[ f(z,{t^{\ast }})-f(z,{t^{\ast }}+{2^{-{j^{(d)}}-1}}{e_{{s_{d}}}})={\int _{0}^{{2^{-{j^{(d)}}-1}}}}\frac{\partial f(z,{t^{\ast }}+{h_{{s_{d}}}})}{\partial {t_{{s_{d}}}}}d{h_{{s_{d}}}},\hspace{1em}{j^{(d)}}={j_{{s_{d}}}}.\]
Now we repeat the same thoughts $l-1$ times and use the properties of a function f, which leads to the inequality
Further, we have the estimate
Here we denoted by A the set of indexes i, for which ${n_{s}}=1$. It is left to prove that for each fixed $\beta \gt 0$, ${P_{2}}\le {C_{\mu }^{(d)}}(\omega )$. As follows from Lemma 1, it is sufficient to show that
where by ${C^{(d)}}$ we denote positive constant that depends only on d. The inequality (14) holds true, as is proved below.
\[\begin{aligned}{}& {\sum \limits_{k=1}^{\infty }}\underset{z\in Z}{\sup }\Big|{\int _{{[0,1]^{d}}}}({S_{{2^{k}}}^{(d)}}(f,x)-{S_{{2^{k-1}}}^{(d)}}(f,x))\hspace{0.1667em}d\mu (x)\Big|\\ {} & \hspace{1em}\le {\sum \limits_{k=1}^{\infty }}\underset{z\in Z}{\sup }\Big|\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}}}{c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}(f){\int _{{[0,1]^{d}}}}{\chi _{{n_{1}},\dots ,{n_{d}}}^{(d)}}(x)\hspace{0.1667em}d\mu (x)\Big|\\ {} & \hspace{1em}\stackrel{\text{(12)}}{\le }{2^{d}}{C_{f}}{\sum \limits_{k=1}^{\infty }}\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}}}\hspace{-24.0pt}{2^{{\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}-{j_{s}}/2}}{2^{-({j^{(d)}}+\cdots +{j^{(d-l+1)}}+\alpha {j^{(d-l)}})}}\\ {} & \hspace{2em}\times \Big|{\int _{{[0,1]^{d}}}}{\chi _{{n_{1}},\dots ,{n_{d}}}^{(d)}}(x)\hspace{0.1667em}d\mu (x)\Big|\\ {} & \hspace{1em}\le {2^{d}}{C_{f}}{\bigg({\sum \limits_{k=1}^{\infty }}\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}}}{2^{\beta {\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}{j_{s}}}}{2^{-2({j^{(d)}}+\cdots +{j^{(d-l+1)}}+\alpha {j^{(d-l)}})}}\bigg)^{1/2}}\\ {} & \hspace{2em}\times {\bigg({\sum \limits_{k=1}^{\infty }}\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}}}{2^{(-\beta -1){\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}{j_{s}}}}{\Big({\int _{{[0,1]^{d}}}}{\chi _{{n_{1}},\dots ,{n_{d}}}^{(d)}}(x)\hspace{0.1667em}d\mu (x)\Big)^{2}}\bigg)^{1/2}}\\ {} & \hspace{1em}:={2^{d}}{C_{f}}{P_{1}^{1/2}}{P_{2}^{1/2}}.\end{aligned}\]
For a sufficiently small $\beta \gt 0$ we have the following estimate for ${P_{1}}$:
(13)
\[\begin{aligned}{}{P_{1}}& =\underset{k\ge 1}{\sup }\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}}{2^{\beta {\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}{j_{s}}}}{2^{-2({j^{(d)}}+\cdots +{j^{(d-l+1)}}+\alpha {j^{(d-l)}})}}\\ {} & =\underset{k\ge 1}{\sup }\sum \limits_{A\subset \{1,\dots ,d\}}\sum \limits_{\substack{0\le {j_{i}}\le k,\\ {} i\in A}}{2^{(\beta +1){\textstyle\sum _{i\in A}}{j_{i}}-2({j^{(d)}}+\cdots +{j^{(d-l+1)}}+\alpha {j^{(d-l)}})}}\\ {} & \le {2^{d}}\underset{k\ge 1}{\sup }\sum \limits_{0\le {j_{i}}\le k}{2^{(\beta +1-2(l+\alpha )/d){\textstyle\textstyle\sum _{i=1}^{d}}{j_{i}}}}={2^{d}}C.\end{aligned}\](14)
\[ {\tilde{P}_{2}}={\sum \limits_{k=1}^{\infty }}\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}}}{2^{(-\beta /2-1/2){\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}{j_{s}}}}|{\chi _{{n_{1}},\dots ,{n_{d}}}^{(d)}}(x)|\le {C^{(d)}},\]
\[\begin{aligned}{}{\tilde{P}_{2}}& =\underset{k\ge 1}{\sup }\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}}{2^{(-\beta /2-1/2){\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}{j_{s}}}}|{\chi _{{n_{1}}}}({x_{1}})\dots {\chi _{{n_{d}}}}({x_{d}})|\\ {} & \le \underset{k\ge 1}{\sup }\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}}{2^{(-\beta /2){\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}{j_{s}}}}\\ {} & \hspace{1em}\times {\prod \limits_{s=1}^{d}}\Big({\mathbf{1}_{{\Delta _{{n_{s}}}}}}({x_{s}})+\frac{1}{2}{\mathbf{1}_{\{(i-1){2^{-{j_{s}}}}\}}}({x_{s}})+\frac{1}{2}{\mathbf{1}_{\{i{2^{-{j_{s}}}}\}}}({x_{s}})\Big)\\ {} & \stackrel{(\ast )}{\le }\underset{k\ge 1}{\sup }\sum \limits_{A\subset \{1,\dots ,d\}}\sum \limits_{\substack{0\le {j_{s}}\le k,\\ {} s\in A}}{2^{(-\beta /2){\textstyle\sum _{s\in A}}{j_{s}}}}\le {2^{d}}\underset{k\ge 1}{\sup }\sum \limits_{0\le {j_{s}}\le k}{2^{(-\beta /2){\textstyle\textstyle\sum _{s=1}^{d}}{j_{s}}}}={C^{(d)}}.\end{aligned}\]
Here in inequality (*) we used that for each set ${j_{1}},\dots ,{j_{d}}$, for each x, ${x_{s}}\ne i{2^{-{j_{s}}}}$, exists exactly one set $({n_{1}},\dots ,{n_{d}})$ such that ${\chi _{{n_{1}},\dots ,{n^{d}}}^{(d)}}(x)\ne 0$ and ${2^{{j_{s}}}}+1\le {n_{s}}\le {2^{{j_{s}}+1}}$. If we have coordinates ${x_{s}}=i{2^{-{j_{s}}}}$, we take into account that ${\mathbf{1}_{\{i{2^{-{j_{s}}}}\}}}({x_{s}})$ has the coefficient $1/2$, as follows from (4). Now the statement of the lemma is a consequence of (13) and (14). □Remark 1.
Assume that constants in inequalities (7) and (8) depend not only on f, but also on z. Then the series (6) converges a.s. for each fixed $z\in Z$ and the following analogue of (9):
\[ {\sum \limits_{k=1}^{\infty }}\Big|{\int _{{[0,1]^{d}}}}({S_{{2^{k}}}^{(d)}}(f,x)-{S_{{2^{k-1}}}^{(d)}}(f,x))\hspace{0.1667em}d\mu (x)\Big|\le {C_{f,z}}{C_{\mu }^{(d)}}(\omega ).\]
This statement is proved similarly to Lemma 3. We just refer to ${C_{f,z}}$ everywhere instead of ${C_{f}}$.The following statement gives the conditions of the continuity with respect to parameter $z\in Z$.
Theorem 2.
Let Z be a metric space, and the function $f(z,x):Z\times {[0,1]^{d}}\to \mathbb{R}$ be such that all paths $f(\cdot ,x)$ are continuous on Z for each x, $f(z,\cdot )$ is continuously differentiable l times on ${[0,1]^{d}}$ for fixed $l\ge 0$ and all $z\in Z$, while
\[ \Big|\frac{{\partial ^{r}}f(z,x)}{\partial {x_{{s_{1}}}}\dots \partial {x_{{s_{r}}}}}\Big|\le {C_{f}},\hspace{1em}x\in {[0,1]^{d}},\]
for some constant ${C_{f}}\gt 0$ (independent of z) and all $0\le r\le l$, ${s_{1}},\dots ,{s_{r}}\subset \{1,\dots ,d\}$. Let l-th derivatives be Hölder continuous with the exponent α and
Proof.
From (11) it follows that
\[\begin{aligned}{}{c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}(f)& ={2^{{\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}{j_{s}}/2}}{\int _{{\textstyle\prod _{1\le s\le d,{n_{s}}\ge 2}}{\Delta _{{n_{s}}}^{+}}}}\sum \limits_{{\varepsilon _{s}}\in \{0,1\},{n_{s}}\ge 2}{(-1)^{{\varepsilon _{1}}+\cdots +{\varepsilon _{d}}}}\\ {} & \hspace{1em}\times f(z,{t_{1}}+{\varepsilon _{1}}{2^{-{j_{1}}-1}},\dots ,{t_{d}}+{\varepsilon _{d}}{2^{-{j_{d}}-1}})\hspace{0.1667em}dt,\end{aligned}\]
and coefficients ${c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}$ are continuous functions of variable z. Now we can see from representation (10) that ${\textstyle\int _{{[0,1]^{d}}}}{S_{{2^{k}}}^{(d)}}(f,x)\hspace{0.1667em}d\mu (x)$ is continuous in variable z for all $k\ge 1$, $\omega \in \Omega $.Finally we refer to Lemma 3 and obtain that ${\textstyle\int _{{[0,1]^{d}}}}{S_{{2^{k}}}^{(d)}}(f,x)\hspace{0.1667em}d\mu (x)$ converges to $\tilde{\eta }(z)$ uniformly a.s. as $k\to \infty $, that implies the statement of our theorem. □
Theorem 3.
Let the function $f(z,x):Z\times {[0,1]^{d}}\to \mathbb{R}$, $Z=[a,b]$, be continuous differentiable $l+1$ times on $Z\times {[0,1]^{d}}$ while
\[ \Big|\frac{{\partial ^{r+1}}f(z,x)}{\partial z\partial {x_{{s_{1}}}}\dots \partial {x_{{s_{r}}}}}\Big|\le {C_{f}},\hspace{1em}x\in {[0,1]^{d}},\]
for all $0\le r\le l$, ${s_{1}},\dots ,{s_{r}}\subset \{1,\dots ,d\}$. Moreover, let $l+1$-th derivatives be Hölder continuous with the exponent $\alpha \gt 0$, i. e.
\[ \Big|\frac{{\partial ^{l+1}}f(z,{x^{(1)}})}{\partial z\partial {x_{{s_{1}}}}\dots \partial {x_{{s_{l}}}}}-\frac{{\partial ^{l+1}}f(z,{x^{(2)}})}{\partial z\partial {x_{{s_{1}}}}\dots \partial {x_{{s_{l}}}}}\Big|\le {C_{f}}|{x^{(1)}}-{x^{(2)}}{|^{\alpha }}.\]
If, in addition, $l+\alpha \gt d/2$, then paths of a random function $\tilde{\eta }$, which is defined by (6), have bounded derivatives on Z,
Proof.
Lemma 3 implies that
\[\begin{array}{r}\displaystyle {\int _{{[0,1]^{d}}}}\frac{\partial f(z,x)}{\partial z}\hspace{0.1667em}d\mu (x)={\int _{{[0,1]^{d}}}}{S_{1}^{(d)}}\Big(\frac{\partial f}{\partial z},x\Big)\hspace{0.1667em}d\mu (x)\\ {} \displaystyle +{\sum \limits_{k=1}^{\infty }}{\int _{{[0,1]^{d}}}}\Big({S_{{2^{k}}}^{(d)}}\Big(\frac{\partial f}{\partial z},x\Big)-{S_{{2^{k-1}}}^{(d)}}\Big(\frac{\partial f}{\partial z},x\Big)\Big)\hspace{0.1667em}d\mu (x),\end{array}\]
where the series on the right-hand side converges uniformly. According to equalities
\[\begin{aligned}{}& {\int _{{[0,1]^{d}}}}{S_{{2^{k}}}^{(d)}}\Big(\frac{\partial f}{\partial z},x\Big)\hspace{0.1667em}d\mu (x)\\ {} & \hspace{1em}=\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}}{c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}\Big(\frac{\partial f}{\partial z}\Big){\int _{{[0,1]^{d}}}}{\chi _{{n_{1}},\dots ,{n_{d}}}^{(d)}}(x)\hspace{0.1667em}d\mu (x)\\ {} & \hspace{1em}\stackrel{\text{(11)}}{=}\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}}\frac{\partial {c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}(f)}{\partial z}{\int _{{[0,1]^{d}}}}{\chi _{{n_{1}},\dots ,{n_{d}}}^{(d)}}(x)\hspace{0.1667em}d\mu (x)\\ {} & \hspace{1em}=\frac{d}{dz}{\int _{{[0,1]^{d}}}}{S_{{2^{k}}}^{(d)}}(f,x)\hspace{0.1667em}d\mu (x)\end{aligned}\]
and Remark 1, we can differentiate the series in (6) and obtain that
\[\begin{aligned}{}\frac{d\tilde{\eta }(z)}{dz}& =\frac{d}{dz}{\int _{{[0,1]^{d}}}}{S_{1}^{(d)}}(f,x)\hspace{0.1667em}d\mu (x)\\ {} & \hspace{1em}+{\sum \limits_{k=1}^{\infty }}\frac{d}{dz}{\int _{{[0,1]^{d}}}}\big({S_{{2^{k}}}^{(d)}}(f,x)-{S_{{2^{k-1}}}^{(d)}}(f,x)\big)\hspace{0.1667em}d\mu (x)\\ {} & ={\int _{{[0,1]^{d}}}}\frac{\partial f(z,x)}{\partial z}\hspace{0.1667em}d\mu (x),\end{aligned}\]
which finishes the proof. □4 Integral as a function of upper limit
For a continuous function $f(x):{[0,1]^{d}}\to \mathbb{R}$ and $y=({y_{1}},\dots ,{y_{d}})\in {[0,1]^{d}}$ we consider the random function
Lemma 2 implies that
is the version of $\xi (y)$.
(16)
\[\begin{aligned}{}\tilde{\xi }(y)& ={\int _{{\textstyle\textstyle\prod _{s=1}^{d}}[0,{y_{s}}]}}{S_{1}^{(d)}}(f,x)\hspace{0.1667em}d\mu (x)\\ {} & \hspace{1em}+{\sum \limits_{k=1}^{\infty }}{\int _{{\textstyle\textstyle\prod _{s=1}^{d}}[0,{y_{s}}]}}({S_{{2^{k}}}^{(d)}}(f,x)-{S_{{2^{k-1}}}^{(d)}}(f,x))\hspace{0.1667em}d\mu (x)\end{aligned}\]We refer to the following assumptions on SM μ.
Assumption A2.
Random function $\mu (x)=\mu \Big({\textstyle\prod _{i=1}^{d}}[0,{x_{i}}]\Big)$, $x\in {[0,1]^{d}}$, has continuous paths.
In the following condition on the uniform modulus of continuity we take the continuous version of μ.
Assumption A3.
If $d\ge 2$ then ${\textstyle\sum _{k=1}^{\infty }}{k^{d-2}}\omega (\mu ,{2^{-k}})\lt \infty $ a.s.
It is easy to see that Asumption A3 holds for $\mu (x)$ with Hölder continuous paths. It also holds if, for example, $\omega (\mu ,\tau )\le C|\ln \tau {|^{-\epsilon }}$, $\epsilon \gt d-1$.
Example 4.
Denote
Here ${\delta _{x,\bar{x}}}=\| {B^{q}}(x)-{B^{q}}(\bar{x}){\| _{{L^{2}}(\Omega )}}\le Cq(|x-\bar{x}|)$, D is a diameter of $[t,T]$. From (17) it follows that for all $x,\bar{x}\in [t,T]$,
\[ q(\tau )=|\ln \tau {|^{-\gamma }},\hspace{1em}\gamma \gt d-1/2,\hspace{2em}\mathcal{K}(z)=\sqrt{\frac{d{q^{2}}(z)}{dz}}.\]
We introduce the stochastic process
\[ {B^{q}}(x)={\int _{{\textstyle\textstyle\prod _{s=1}^{d}}[0,{x_{s}}]}}{\prod \limits_{s=1}^{d}}\mathcal{K}({x_{s}}-{y_{s}})dW(y),\hspace{1em}x=({x_{1}},\dots ,{x_{d}}),\]
where W is a d-dimensional Wiener process. Theorem 3.1 in [11] implies the existence of rectangle $[t,T]={\textstyle\prod _{i=1}^{d}}[{t_{i}},{T_{i}}]\subset {[0,1]^{d}}$ such that
(17)
\[ \underset{\varepsilon \to 0+}{\lim }\underset{\substack{x,\bar{x}\in [t,T]\\ {} {\delta _{x,\bar{x}}}\le \varepsilon }}{\sup }\frac{|{B^{q}}(x)-{B^{q}}(\bar{x})|}{{\delta _{x,\bar{x}}}\sqrt{\ln \Big(\frac{D}{{q^{-1}}({\delta _{x,\bar{x}}})}\Big)}}=C\hspace{1em}\text{a.s.}\]
\[\begin{aligned}{}|{B^{q}}(x)-{B^{q}}(\bar{x})|& \le C{\delta _{x,\bar{x}}}\sqrt{\ln D+{\delta _{x,\bar{x}}^{-1/\gamma }}}\\ {} & \stackrel{\gamma \gt 1/2}{\le }C|\ln |x-\bar{x}|{|^{-\gamma }}\sqrt{\ln D+|\ln |x-\bar{x}||}\\ {} & \le C|\ln |x-\bar{x}|{|^{1/2-\gamma }}.\end{aligned}\]
Therefore, ${\omega _{[t,T]}}({B^{q}},\tau )\le C|\ln \tau {|^{1/2-\gamma }}$.Now we are ready to formulate the main result of the section.
Theorem 4.
Proof.
For version (16), we have that
Here ${S_{{2^{k}}}^{(d)}}(f,x)$ is a simple function. By our assumption, SM μ has continuous paths. Therefore, for each k, the random function of variable y
(18)
\[ \tilde{\xi }(y)=\underset{k\to \infty }{\lim }{\int _{{\textstyle\textstyle\prod _{s=1}^{d}}[0,{y_{s}}]}}{S_{{2^{k}}}^{(d)}}(f,x)\hspace{0.1667em}d\mu (x).\]
\[ {\int _{{\textstyle\textstyle\prod _{s=1}^{d}}[0,{y_{s}}]}}{S_{{2^{k}}}^{(d)}}(f,x)\hspace{0.1667em}d\mu (x)\]
has continuous paths. We will prove that the convergence in (18) is uniform in $y\in {[0,1]^{d}}$ a.s., and this will imply the continuity of $\tilde{\xi }$.We aim to show that
\[ {\sum \limits_{k=1}^{\infty }}\underset{y\in {[0,1]^{d}}}{\sup }\Big|{\int _{{\textstyle\textstyle\prod _{s=1}^{d}}[0,{y_{s}}]}}({S_{{2^{k}}}^{(d)}}(f,x)-{S_{{2^{k-1}}}^{(d)}}(f,x))\hspace{0.1667em}d\mu (x)\Big|\le {\tilde{C}_{f,\mu }^{(d)}}(\omega )\]
for some random constant ${\tilde{C}_{f,\mu }^{(d)}}(\omega )\lt \infty $ a.s. that may depend on d, f, μ.Recall that paths of $\mu ({x_{1}},\dots ,{x_{d}})$ are continuous, therefore for any cut of the set ${\textstyle\prod _{s=1}^{d}}[{y_{s1}},{y_{s2}}]\subset {[0,1]^{d}}$ we have
Thus, we obtain
\[\begin{aligned}{}& {\int _{{\textstyle\textstyle\prod _{s=1}^{d}}[0,{y_{s}}]}}({S_{{2^{k}}}^{(d)}}(f,x)-{S_{{2^{k-1}}}^{(d)}}(f,x))\hspace{0.1667em}d\mu (x)\\ {} & \hspace{1em}=\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}}}{c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}(f){\int _{{\textstyle\textstyle\prod _{s=1}^{d}}[0,{y_{s}}]}}{\chi _{{n_{1}},\dots ,{n_{d}}}^{(d)}}(x)\hspace{0.1667em}d\mu (x)\\ {} & \hspace{1em}\stackrel{\text{(19)}}{=}\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}}}{c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}(f){2^{({j_{1}}+\cdots +{j_{d}})/2}}\\ {} & \hspace{2em}\times {\int _{{\textstyle\textstyle\prod _{s=1}^{d}}[0,{y_{s}}]}}\prod \limits_{1\le s\le d,{n_{s}}\ge 2}\Big({\mathbf{1}_{{\Delta _{{n_{s}}}^{+}}}}({x_{s}})-{\mathbf{1}_{{\Delta _{{n_{s}}}^{-}}}}({x_{s}})\Big)\hspace{0.1667em}d\mu (x)\\ {} & \hspace{1em}=\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}}}{c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}(f){2^{({j_{1}}+\cdots +{j_{d}})/2}}\\ {} & \hspace{2em}\times \sum \limits_{{\varepsilon _{s}}\in \{+,-\}}\mu \Big(\prod \limits_{1\le s\le d}{\Delta _{{n_{s}}}^{{\varepsilon _{s}}}}\cap \prod \limits_{1\le s\le d}[0,{y_{s}}]\Big)\\ {} & \hspace{1em}:={A_{k1}}(y)+{A_{k2}}(y).\end{aligned}\]
Here ${A_{k1}}(y)$ is the sum of terms with $({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}}$, ${\varepsilon _{s}}\in \{+,-\}$ such that
\[ {\prod \limits_{s=1}^{d}}{\Delta _{{n_{s}}}^{{\varepsilon _{s}}}}\subset {\prod \limits_{s=1}^{d}}[0,{y_{s}}]\]
(here for ${n_{s}}=1$ we take only ${\varepsilon _{s}}=+$ and ${\Delta _{{n_{s}}}^{{\varepsilon _{s}}}}=(0,1)$). Taking into account the definition of ${\Delta _{{n_{s}}}^{+}}$ and ${\Delta _{{n_{s}}}^{-}}$ in (3), we get that in ${A_{k1}}(y)$
\[ {\prod \limits_{s=1}^{d}}{\Delta _{{n_{s}}}^{{\varepsilon _{s}}}}\cap {\prod \limits_{s=1}^{d}}[0,{y_{s}}]={\prod \limits_{s=1}^{d}}{\Delta _{{n_{s}}}^{{\varepsilon _{s}}}}={\prod \limits_{s=1}^{d}}{\Delta _{{m_{s}}}}\]
for some $({m_{1}},\dots ,{m_{d}})\in {\mathbb{N}_{k+1}}\setminus {\mathbb{N}_{k}}$. Therefore,
\[\begin{aligned}{}& {\sum \limits_{k=1}^{\infty }}\underset{y\in {[0,1]^{d}}}{\sup }|{A_{k1}}(y)|\\ {} & \hspace{1em}\le {\sum \limits_{k=1}^{\infty }}\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}}}|{c_{{n_{1}},\dots ,{n_{d}}}^{(d)}}(f)|{2^{({j_{1}}+\cdots +{j_{d}})/2}}\sum \limits_{{\varepsilon _{s}}\in \{+,-\}}\Big|\mu \Big(\prod \limits_{1\le s\le d}{\Delta _{{n_{s}}}^{{\varepsilon _{s}}}}\Big)\Big|\\ {} & \hspace{1em}\stackrel{\text{(12)}}{\le }C{\sum \limits_{k=1}^{\infty }}\sum \limits_{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}}}{2^{-({j_{1}}+\cdots +{j_{d}})}}\sum \limits_{{\varepsilon _{s}}\in \{+,-\}}\Big|\mu \Big(\prod \limits_{1\le s\le d}{\Delta _{{n_{s}}}^{{\varepsilon _{s}}}}\Big)\Big|\\ {} & \hspace{1em}\le C{\sum \limits_{k=1}^{\infty }}\sum \limits_{({m_{1}},\dots ,{m_{d}})\in {\mathbb{N}_{k+1}}\setminus {\mathbb{N}_{k}}}{2^{d}}{2^{-({j^{\prime }_{1}}+\cdots +{j^{\prime }_{d}})}}\Big|\mu \Big(\prod \limits_{1\le s\le d}{\Delta _{{m_{s}}}}\Big)\Big|.\end{aligned}\]
Here ${j^{\prime }_{s}}$ are taken such that ${2^{{j^{\prime }_{s}}}}+1\le {m_{s}}\le {2^{{j^{\prime }_{s}}+1}}$, and ${j^{\prime }_{s}}={j_{s}}+1$ for respective ${m_{s}}$. From estimates in (13), (14) it follows that
\[ {\sum \limits_{k=1}^{\infty }}\underset{y\in {[0,1]^{d}}}{\sup }|{A_{k1}}(y)|\le {\tilde{C}_{\mu }^{(d)}}(\omega )\]
for some random constant ${\tilde{C}_{\mu }^{(d)}}(\omega )$ that depends only of d and μ. It is easy to see that all estimates in (13), (14) remain valid if we change ${\chi _{{n_{1}},\dots ,{n_{d}}}^{(d)}}(x)$ to
Further, we estimate ${A_{k2}}(y)$, i. e. the sum of terms with $({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}}$ such that ${y_{s}}\in {\Delta _{{n_{s}}}}$ for some s.
We get
\[\begin{aligned}{}|{A_{k2}}(y)|& \le \sum \limits_{\substack{({n_{1}},\dots ,{n_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}},\\ {} \exists {y_{s}}\in {\Delta _{{n_{s}}}^{{\varepsilon _{s}}}}}}{2^{-{\textstyle\sum _{1\le s\le d,{n_{s}}\ge 2}}{j_{s}}}}\Big|\mu \Big(\prod \limits_{1\le s\le d}({({\Delta _{{n_{s}}}})^{{\varepsilon _{s}}}}\cap [0,{y_{s}}])\Big)\Big|\\ {} & ={2^{d}}\sum \limits_{\substack{({m_{1}},\dots ,{m_{d}})\in {\mathbb{N}_{k}}\setminus {\mathbb{N}_{k-1}},\\ {} \exists {y_{s}}\in {\Delta _{{m_{s}}}}}}{2^{-{\textstyle\sum _{1\le s\le d,{m_{s}}\ge 3}}{j^{\prime }_{s}}}}\Big|\mu \Big(\prod \limits_{1\le s\le d}({\Delta _{{m_{s}}}}\cap [0,{y_{s}}])\Big)\Big|\\ {} & \le {2^{d}}\omega (\mu ,{2^{-k-1}})\sum \limits_{\substack{({m_{1}},\dots ,{m_{d}})\in {\mathbb{N}_{k+1}}\setminus {\mathbb{N}_{k}},\\ {} \exists {y_{s}}\in {\Delta _{{m_{s}}}}}}{2^{-{\textstyle\sum _{1\le s\le d,{m_{s}}\ge 3}}{j^{\prime }_{s}}}}=:{D_{k+1}}.\end{aligned}\]
Now we check the convergence of the series ${\textstyle\sum _{k=1}^{\infty }}{D_{k}}$ with the help of the sum
\[ \sum \limits_{\substack{({m_{1}},\dots ,{m_{d}})\in {\mathbb{N}_{k}},\\ {} \exists {y_{s}}\in {\Delta _{{m_{s}}}}}}{2^{-{\textstyle\sum _{1\le s\le d,{m_{s}}\ge 3}}{j^{\prime }_{s}}}}:={T_{k}}.\]
Notice that for each fixed set $({j^{\prime }_{1}},\dots ,{j^{\prime }_{d}})$ there exist at most
\[ {2^{{\textstyle\sum _{1\le s\le d,{m_{s}}\ge 3}}{j^{\prime }_{s}}}}-\prod \limits_{1\le s\le d,{m_{s}}\ge 3}\big({2^{{j^{\prime }_{s}}}}-1\big)\]
sets $({m_{1}}\dots ,{m_{d}})$, for which $\exists {y_{s}}\in {\Delta _{{m_{s}}}}$. Denoting by A the set of indices s, which satisfy the equality ${n_{s}}=1$, we obtain that
\[\begin{aligned}{}\underset{y}{\sup }{T_{k}}& \le \sum \limits_{A\subset \{1,\dots ,d\}}\sum \limits_{\substack{1\le {j^{\prime }_{i}}\le k,\\ {} i\in A}}\bigg(1-\prod \limits_{i\in A}\big(1-{2^{-{j^{\prime }_{i}}}}\big)\bigg)\\ {} & =\sum \limits_{A\subset \{1,\dots ,d\}}\sum \limits_{\substack{1\le {j^{\prime }_{i}}\le k,\\ {} i\in A}}\sum \limits_{\substack{B\subset A,\\ {} B\ne \varnothing }}{(-1)^{|B|+1}}{2^{-{\textstyle\sum _{i\in B}}{j^{\prime }_{i}}}}\\ {} & =\sum \limits_{A\subset \{1,\dots ,d\}}\sum \limits_{\substack{B\subset A,\\ {} B\ne \varnothing }}{(-1)^{|B|+1}}\sum \limits_{\substack{1\le {j^{\prime }_{i}}\le k,\\ {} i\in A}}{2^{-{\textstyle\sum _{i\in B}}{j^{\prime }_{i}}}}\\ {} & \stackrel{(\ast )}{=}{\sum \limits_{u=1}^{d}}{\sum \limits_{v=1}^{u}}{(-1)^{v+1}}\left(\hspace{-3.0pt}\begin{array}{c}d\\ {} u\end{array}\hspace{-3.0pt}\right)\left(\hspace{-3.0pt}\begin{array}{c}u\\ {} v\end{array}\hspace{-3.0pt}\right)\sum \limits_{\substack{1\le {j^{\prime }_{i}}\le k,\\ {} 1\le i\le u}}{2^{-{\textstyle\sum _{1\le i\le v}}{j^{\prime }_{i}}}}\\ {} & ={\sum \limits_{u=1}^{d}}{\sum \limits_{v=1}^{u}}{(-1)^{v+1}}\left(\hspace{-3.0pt}\begin{array}{c}d\\ {} u\end{array}\hspace{-3.0pt}\right)\left(\hspace{-3.0pt}\begin{array}{c}u\\ {} v\end{array}\hspace{-3.0pt}\right){(1-{2^{-k}})^{v}}{k^{u-v}}\\ {} & ={\sum \limits_{u=1}^{d}}\left(\hspace{-3.0pt}\begin{array}{c}d\\ {} u\end{array}\hspace{-3.0pt}\right)({k^{u}}-{(k-1+{2^{-k}})^{u}})\\ {} & ={(k+1)^{d}}-{(k+{2^{-k}})^{d}}.\end{aligned}\]
Here in $(\ast )$ we used the fact that sum ${(-1)^{|B|+1}}{\textstyle\sum _{\substack{1\le {j^{\prime }_{i}}\le k,\\ {} i\in A}}}{2^{-{\textstyle\sum _{i\in B}}{j^{\prime }_{i}}}}$ depends only on $|A|$ and $|B|$. Therefore,
\[\begin{aligned}{}{D_{k}}& \le {2^{d}}\omega (\mu ,{2^{-k}})\big({(k+1)^{d}}-{(k+{2^{-k}})^{d}}-{k^{d}}+(k-1+{2^{-k+1}})\big)\\ {} & ={2^{d}}\omega (\mu ,{2^{-k}})\big({(k+1)^{d}}-2{k^{d}}+{(k-1)^{d}}+O({2^{-\beta k}})\big),\end{aligned}\]
where $0\lt \beta \lt 1$. Applying Assumption A3 we get the statement of the theorem. □If $\mu (x)$ have a Hölder continuous paths with exponent $\gamma \gt 0$, then the statement of Theorem 4 follows from Theorem 16 [10]. Moreover, Theorem 16 [10] states that ξ has the version that is Hölder continuous with the same exponent $\gamma \gt 0$. In [10], integral is considered in the Young sense, and its value coincides with the value of our integral with respect to Hölder continuous μ.
