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\topmatter
\title 
On the Poisson Limit Theorems of Sinai and Major 
\endtitle
\author Nariyuki Minami \endauthor
\affil Institute of Mathematics, Unversity of Tsukuba \\
Tsukuba, Ibaraki 305-8571, Japan \\
E-mail: minami\@ sakura.cc.tsukuba.ac.jp
\endaffil

\abstract
Let  $f(\varphi)$  be a positive continuous function on  
$0\leq\varphi\leq\Theta$ , where  $\Theta\leq2\pi$ , and let  $\xi$  
be the number of two-dimensional lattice points in the domain  
$\Pi_R(f)$  between the curves  $r=(R+c_1/R)f(\varphi)$  and  
$r=(R+c_2/R)f(\varphi)$ , where  $c_1<c_2$  are fixed. Randomizing 
the function  $f$  according to a probability law  $\bold P$ , and 
the parameter  
$R$  according to the uniform distribution  $\mu_L$  on the interval  
$[a_1L,a_2L]$ , Sinai showed that the distribution of  $\xi$  under  
$\bold P\times\mu_L$  converges to a mixture of the 
Poisson distributions as  $L\to\infty$ . 
Later Major showed that for $\bold P$-almost all  $f$ , 
the distribution of  $\xi$  
under  $\mu_L$  converges to a Poisson distribution as  $L\to\infty$ . 
In this note, we shall 
give shorter and  more transparent proofs to these interesting theorems, 
at the same time extending the 
class of  $\bold P$  and strengthening the statement of Sinai.\par
\endabstract
\endtopmatter

\document

\head {\bf \S1 Introduction} \endhead
In connection to a question in the theory of quantum chaos, Sinai 
proposed ([S1]) and solved ([S2]) the following problem. \par
Given a planar curve which is described in the polar coordinate as
$$r=f(\varphi)\ ,\ 0\leq\varphi\leq\Theta\ , \tag 1.1$$
with  $\Theta\in(0,2\pi]$  and  $f(\cdot)>0$  continous, consider the 
domain  $\Pi_R(f)$  defined by
$$\align
\Pi_R(f)=\{x\in\bold R^2 &\vert\ 0\leq\varphi(x)\leq\Theta\ ,\tag1.2\\
& (R+c_1/R)f(\varphi(x))<\vert x\vert<(R+c_2/R)f(\varphi(x))\}\ .
\endalign$$
Here  $c_1<c_2$  are real constants and  $R>0$  is a parameter which 
we shall let tend to infinity. Moreover  $\vert x\vert=r=\sqrt{x_1^2
+x_2^2}$  is the distance of  $x$  from the origin and  $\varphi(x)$  
is the angle between the vector  $x=(x_1,x_2)$  and the positive real 
axis.
 If we set
$$\lambda(f)=(c_2-c_1)\int_0^{\Theta}f(\varphi)^2d\varphi\ ,\tag 1.3$$
then as  $R\to\infty$ , the area of  $\Pi_R(f)$  is asymptotically given by  
$\lambda(f)(1+O(R^{-2}))$ . Thus the domain  $\Pi_R(f)$  becomes thinner 
and thinner but its area is asymptotically constant. Now consider the 
quantity
$$\xi=\xi(R;f)=\sharp\{\Pi_R(f)\cap\bold Z^2\}\ ,\tag 1.4$$
which is the number of the lattice points in  $\Pi_R(f)$ . Then the problem, 
which is of physical significance, is the following:\par
Suppose the parameter  $R>0$  is randomly distributed on the interval  
$I_L=(a_1L,a_2L)$ , $0<a_1<a_2$ , according to the uniform distribution  
$\mu_L$ . What is the limiting probability distribution of  $\xi(\cdot;f)$  
under  $\mu_L$  as  $L\to\infty$ ? Numerical studies suggest that for 
typical smooth curves  $f$  which are obtained from Hamiltonians of 
integrable dynamical systems, the limiting distribution is Poissonian. 
To give a rigorous proof to this conjecture for a concretely given  $f$  
remains as a difficult open question. What Sinai did was to randomize 
the function  $f$  according to some probability law  $\bold P$  
on some
 function space. Then he showed that under the probability measure  
$\bold P\times\mu_L$ , the distribution of  $\xi(\cdot;\cdot)$  
converges, as  
$L\to\infty$ , to a mixture of Poisson distributions each of which 
having the parameter  
$\lambda(f)$ . He also showed that there is a sequence  
$\bar{L}_n\to\infty$  such that for  $\bold P$-almost all  $f$ , the law of  
$\xi(\cdot;f)$  under  $\mu_{\bar{L}_n}$  converges as  $n\to\infty$  to 
the Poisson distribution with parameter  $\lambda(f)$ . Later Major [M] 
extended Sinai's idea and succeeded in getting rid of subsequences. 
Namely by a more elaborate analysis, he proved that as  $L\to\infty$ , the 
distribution of  $\xi(\cdot;f)$  under  $\mu_L$  converges to the Poisson 
distribution with parameter  $\lambda(f)$ , for  $\bold P$-almost all  $f$ .
\par
Under the conditions on  $\bold P$  posed by Sinai and Major (and which we 
shall assume also in this note), the sample functions  $f(\varphi)$  
is not smooth in  $\varphi$  (see Remark 2 of [M]), so that the direct 
connection of their results to the problem of quantum chaos is lost. 
But still the theorems of Sinai and Major are interesting from the viewpoint 
of elementary probability theory, since they provide us a new limit theorem 
which gives the Poisson distribution. The present author has the opinion 
that these theorems deserve to be made more popular among probabilists 
who are not necessarily interested in such topics as quantum chaos. 
The author also believes that in the future study of quantum chaos too, 
one will 
need to prove variants of these theorems under various different conditions, 
e.g. for higher dimensional lattices. For these 
purposes, however, the proofs of these theorems should be made as transparent 
and readable as possible. In fact, Sinai's paper [S2] contains many 
unnecessary arguments and is obscure at some crucial points. 
(For example, the interrelation among his conditions I) to VI) are not clear 
enough.)
There is no 
such ambiguity in Major's paper [M], but some of his arguments are also 
dispensable.
\par
In this paper, we will make a more direct attack to the theorems of Sinai and 
Major, to improve them in the following way:
\par
\roster
\item The distribution under  $\bold P$  of  $\xi(R;\cdot)$  converges, as  
$R\to\infty$ , to a mixture of Poisson distribution. This result has 
a much simpler proof than Sinai's one and implies Sinai's original 
assertion as a trivial corollary.

\item We shall give a slightly different formulation of Major's theorem, 
which turns out to be equivalent to the original one but technically more 
tractable and 
more closely related to the problem in quantum chaos raised by Berry 
and Tabor [BT] . (See also [Mi].)

\item The theorems are extended to the case where  $f(\cdot)$  are not 
Lipschitz continuous. Major's proof cannot cover this case because he 
makes explicit use of the sample Lipschitz continuity. But in fact, we 
need to use the samplewise Lipschitz continuity
 only in combination with the upper bound of the probability 
density, so that these two conditions ((1.7) and (1.8) in 
(II-2) below) can entirely be replaced 
by a single condition (II-1). As we shall see below, the number  
$\sigma\in(0,1)$  in (II-1) plays the roll of  $\tau-1$  in (II-2), 
$\tau$ appearing in (1.8), where 
the number  $1$  comes from the Lipschitz continuity of  $f(\cdot)$ .

\item We can slightly loosen the technical condition on the partial 
derivative of the probability density as in (III). (Compare this with (2b) of 
Property A of [M].) Moreover even if we do not assume this condition, 
we still have a partial result, Thorem 2.

\item We prove the limit theorems by directly proving the convergence of 
factorial moments of all order as in (2.2) and (3.5). These assertions are 
actually stronger than those of Sinai and Major because they proved the 
convergence of moments not for the random variables $\xi(R;f)$ themselves 
but for their modifications. As discussed by the present author in [Mi], 
the convergence of $k$-th moments may have applications in a problem of 
quantum chaos even if it is proven only for finite numbers of $k$'s, so 
that the Poisson limit theorem is not available.
\endroster

Now let us formulate our conditions and theorems. We suppose that our 
probability measure  $\bold P$  is defined on the function space
$$
\Cal X=\Cal X_{b_1,b_2}=\{f\vert f\ \text{is continuous on}\ 
[0,\Theta]\ 
\text{and}\ b_1\leq f(\varphi)\leq b_2\}
\tag 1.5$$
with  $0<b_1<b_2$ , and we assume that  $\bold P$  satisfies the following 
two 
conditions (I) and (II).\par
\bigskip
\definition{Condition (I)} For each  $k\geq1$ , and  $(\varphi_1,\ldots,
\varphi_k)$  ($\varphi_i\ne\varphi_j$) 
in  $[0,\Theta]^k$ , the distribution of  
$(f(\varphi_1),\ldots,f(\varphi_k))$  has a density  
$$p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)$$
which is  $C^1$  on  
$$[b_1,b_2]^k\times\{(\varphi_1,\ldots,\varphi_k)\in[0,\Theta]^k\ \vert\ 
\varphi_i\ne\varphi_j\ (i\ne j)\}\ .$$
We note that if  $\pi$  is a permutation of  $\{1,\ldots,k\}$ , 
then
$$p_k(y_{\pi(1)},\ldots,y_{\pi(k)}\vert\varphi_{\pi(1)},\ldots,
\varphi_{\pi(k)})=p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)\ .$$
\enddefinition
\definition{Condition (II)} Either of the conditions (II-1) and (II-2) below 
holds:
\bigskip
\roster
\item"(II-1)" There exists a $\sigma\in(0,1)$ such that the inequality
$$p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)
\ \leq\ B_k\prod_{j=2}^k(\varphi_j-\varphi_{j-1})^{-\sigma}\tag1.6$$
holds with some constant $B_k$ for all $k\geq2$  and  
$0\leq\varphi_1<\cdots<\varphi_k\leq\Theta$ .
\item"(II-2)" There is a constant  $b_3>0$  such that for  $\bold P$-almost 
all  $f\in\Cal X$ , 
$$Y(f)\equiv\sup_{0\leq\varphi_1<\varphi_2\leq\Theta}
\frac{\vert f(\varphi_2)-f(\varphi_1)\vert}{\varphi_2-\varphi_1}\ \leq\ b_3\ .
\tag1.7$$
Moreover there exists a  $\tau\in(1,2)$  such that the inequality
$$p_k(y_1,\ldots,y_k\vert\varphi_1\ldots,\varphi_k)\ \leq\ 
C_k\prod_{j=2}^k(\varphi_j-\varphi_{j-1})^{-\tau} \tag 1.8$$
holds with some constant $C_k$ for all $k\geq2$  and  
$0\leq\varphi_1<\cdots<\varphi_k\leq\Theta$ .
\endroster
\enddefinition
\bigskip
Under these conditions, we can prove Theorems 1 and 2 below:
\bigskip
\proclaim{Theorem 1} As  $R\to\infty$ , the distribution of  $\xi(R;\cdot)$
  under the probability measure  $\bold P$  converges to the mixture of 
Poisson 
distributions each of which having the 
parameter  $\lambda(f)$ , where $\lambda(f)$  was defined in (1.3). 
Namely we have
$$\lim_{R\to\infty}\bold P(\xi(R;\cdot)=k)=\bold E_{\bold P}
\left[e^{-\lambda(f)}\frac{\lambda(f)^k}{k!}\right]\ ,\ k=0,1,\ldots. 
\tag 1.9$$
\endproclaim
\bigskip
As an immediate corollary, we obtain Sinai's theorem ([S2]).
\bigskip
\proclaim{Corllary} As  $L\to\infty$ , the distribution of  
$\xi(\cdot;\cdot)$  under the probability measure  $\mu_L\times\bold P$  
converges to the mixture of Poisson distribution, each of which having the 
parameter  
$\lambda(f)$ , namely
$$\lim_{L\to\infty}(\mu_L\times\bold P)(\xi=k)=
\bold E_{\bold P}\left[e^{-\lambda(f)}\frac{\lambda(f)^k}{k!}\right]\ ,
\ k=0,1,\ldots. \tag 1.10$$
\endproclaim
\bigskip
Sinai proved this assertion under conditions which are close to our conditions 
(I) and (II-2) with  
$\tau=1$ . Under the same conditions, he also showed the following theorem, 
which we shall prove, simultaneously with Theorem 3,
 under wider conditions (I) and (II).\par
\bigskip
\proclaim{Theorem 2} There is a sequence  $\bar{L}_n\to\infty$  
such that for  $\bold P$-almost all  $f\in\Cal X$ , the distribution of  
$\xi(\cdot;f)$  under  $\mu_{\bar{L}_n}$  converges to the Poisson 
distribution with parameter  $\lambda(f)$ , namely we have 
$$\lim_{n\to\infty}\mu_{\bar{L}_n}(\xi(\cdot;f)=k)=
e^{-\lambda(f)}\frac{\lambda(f)^k}{k!}\ ,\ k=0,1,\ldots, \tag 1.11$$
for  $\bold P$-almost all  $f\in\Cal X$ .
\endproclaim
\bigskip
If we assume, in addition to Conditions (I) and (II), the following 
technical condition, we can get rid of subsequences.\par
\bigskip
\definition{Condition (III)} For some constants  $A_k>0$ , 
$\nu_k\in(0,d/2)$ , where  $k\geq2$  and  $d=1-\sigma$ or $=2-\tau$  
according to which of (II-1) and (II-2)  $\bold P$ satisfies, we have
$$\vert\partial_i p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)\vert
\ \leq\ A_k\exp(\beta^{-\nu_k}) \tag 1.12$$
whenever  $k\geq2$ , $b_1\leq y_j\leq b_2$ , 
$\vert\varphi_i-\varphi_j\vert\geq\beta>0$ , $i,j=1,\ldots,k$ . Here  
$\partial_i$  denotes either  $\partial/\partial y_i$  or  
$\partial/\partial\varphi_i$ .
\enddefinition
\bigskip
\proclaim{Theorem 3} If we assume Condition (III) in addition to Conditions 
(I) and (II), then for  $\bold P$-almost all  $f\in\Cal X$ , the 
distribution of 
$\xi(\cdot;f)$  under  $\mu_L$  converges to the Poisson distribution 
with parameter  $\lambda(f)$  as  $L\to\infty$ , namely we have
$$\lim_{L\to\infty}\mu_L(\xi(\cdot;f)=k)=e^{-\lambda(f)}
\frac{\lambda(f)^k}{k!}\ ,\ k=0,1\ldots, \tag 1.13$$
for  $\bold P$-almost all  $f\in\Cal X$ .
\endproclaim
\bigskip
This was proved by Major under Conditions (I), (II-2) and a slightly more 
restrictive condition than (III).\par

\bigskip
Before closing this introduction, we shall give an example of  $\bold P$  
satisfying our conditions (I), (II-1) and (III). (An example which 
satisfies (II-2) instead of (II-1) was discussed by Sinai [S2] and 
Major [M].)\par
Let  $\{X_t;t\ge0\}$  be the reflecting Brownian motion on the interval  
$S=[b_1,b_2]$  with fixed  $X_0=b$ , and let  $\bold P$  be its 
probability law. Then if we set  
$f(\varphi)=X_{1+\varphi}$ , the law of  $\{f(\varphi);
0\le\varphi\le\Theta\}$  under  $\bold P$  satisfies our requirements.\par
In fact, the infinitesimal generator of  $\{X_t\}$  is the Neumann Laplacian  
$\frac1{2}\Delta$  on  the interval $S$ . Let  
$\cdots<-\lambda_n<\cdots<-\lambda_1
<-\lambda_0=0$  be its eigenvalues and  $\psi_n(x)$ , $n\ge0$  be the 
corresponding 
normalized eigenfunctions. The transition density  $P_t(x,y)$  of  
$\{X_t\}$  has the eigenfunction expansion:
$$P_t(x,y)=\sum_{n\ge0}e^{-\lambda_n t}\psi_n(x)\psi_n(y)\ ,\tag1.16$$
and if  $0\le\varphi_1<\cdots<\varphi_k\le\Theta$ , the joint distribution 
of  $(f(\varphi_1),\ldots,f(\varphi_k))$  under  $\bold P$  has the density
$$\align
&\ p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k) \tag1.17\\
&=P_{1+\varphi_1}(b,y_1)P_{\varphi_2-\varphi_1}(y_1,y_2)\cdots
P_{\varphi_k-\varphi_{k-1}}(y_{k-1},y_k)\ .
\endalign$$
Now it is elementary to calculate  $\lambda_n$  and  $\psi_n(x)$  
explicitly:
$$\lambda_n=\frac{\pi^2 n^2}{2(b_2-b_1)^2}
\quad;\ \psi_n(x)=\sqrt{\frac2{b_2-b_1}}\cos\frac{n\pi}{b_2-b_1}(x-b_1)\ ,\ 
n\ge0\ .\tag1.18$$
>From this, it is clear that  $P_t(x,y)$  is   $C^1$  on  
$[0,\infty)\times[b_1,b_2]^2$  and hence from (1.17), our  
$p_k(\cdot\vert\cdot)$  is  $C^1$  in all 
of its variables $y_1,\ldots,y_k$ and $\varphi_1,\ldots,\varphi_k$ 
as required in Condition (I).\par

It is clear from (1.16), (1.17) and (1.18) that there is a constant  
$B'_k$  such that
$$p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)\leq 
B'_k\prod_{j=2}^k\left(\sum_{n\ge0}e^{-\lambda_n(\varphi_j-\varphi_{j-1})}
\right)\ .\tag1.19$$
On the other hand, if we let
$$N(\lambda)=\sum_{\lambda_n\le\lambda}1=\left[\frac{\sqrt{2\lambda}}
{\pi}(b_2-b_1)\right]\ ,
\tag1.20$$
then
$$\sum_{n\ge0}e^{-\lambda_n t}=\int_{0-}^{\infty}e^{-\lambda t}dN(\lambda) 
\tag1.21$$
and since
$$N(\lambda)\sim\frac{\sqrt{2}(b_2-b_1)}{\pi}\sqrt{\lambda} \tag1.22$$
as  $\lambda\to+\infty$ , we can use the Abelian theorem to obtain
$$\sum_{n\ge0}e^{-\lambda_n t}\sim\text{const.}t^{-1/2} \tag1.23$$
as  $t\downarrow0$ , where the constant can be explicitly given.
 Hence by (1.19), there is another constant  $B_k>0$  such that
$$p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)\leq 
B_k\prod_{j=2}^k(\varphi_j-\varphi_{j-1})^{-1/2}\ ,\tag1.24$$
so that Condition (II-1) is valid with  $\sigma=1/2$ .\par
As is obvious from (1.17), in order to verify Condition (III), it suffices 
to obtain an upper bound of  $\vert(\partial/\partial y)P_t(x,y)\vert$  
and  $\vert(\partial/\partial t)P_t(x,y)\vert$ . Differentiating (1.16) 
term by term, we obtain
$$\left\vert\frac{\partial}{\partial y}P_t(x,y)\right\vert
\leq\text{const.}\int_{0-}^{\infty}e^{-\lambda t}dN_1(\lambda) \tag1.25$$
and
$$\left\vert\frac{\partial}{\partial t}P_t(x,y)\right\vert
\leq\text{const.}\int_{0-}^{\infty}e^{-\lambda t}dN_2(\lambda)\ ,\tag1.26$$
where we have set
$$N_1(\lambda)=\sum_{n;\lambda_n\le\lambda}n \quad;\ 
N_2(\lambda)=\sum_{n;\lambda_n\le\lambda}n^2\ .\tag1.27$$
Since as  $\lambda\to\infty$  we have
$$N_1(\lambda)\sim\text{const.}\lambda\quad;\ 
N_2(\lambda)\sim\text{const.}\lambda^{3/2}\ ,\tag1.28$$
we can again use the Abelian theorem to obtain
$$\left\vert\frac{\partial}{\partial y}P_t(x,y)\right\vert
=O(t^{-1}) \tag1.29$$
and
$$\left\vert\frac{\partial}{\partial t}P_t(x,y)\right\vert
=O(t^{-3/2}) \tag1.30$$
as  $t\downarrow0$ . Hence we can conclude that for some constant  
$A_k>0$ ,
$$\vert\partial_i p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)\vert
\leq A_k\beta^{-3/2} \tag1.31$$
whenever  $b_1\le y_j\le b_2$ , $\vert\varphi_i-\varphi_j\vert\ge\beta>0$ ,
 $i,j=1,\ldots,k$ , which is much stronger than (1.12).\par

\head {\bf \S2 Proof of Theorem 1} \endhead
For  $k\geq1$ , define
$$\xi_k=\xi_k(R;f)=\pmatrix \xi \\ k \endpmatrix 1_{\{\xi\geq k\}}
=\sum_{n=k}^{\infty}\frac{n!}{k!(n-k)!}1_{\{\xi=n\}}\ .\tag 2.1$$
As in [S2], for the proof of Theorem 1, it suffices to prove
$$\lim_{R\to\infty}\bold E_{\bold P}[\xi_k(R;\cdot)]
=\frac1{k!}\bold E_{\bold P}[\lambda(f)^k] \tag 2.2$$
for each  $k\geq1$ . On the other hand, by a simple combinatorial 
argument, we can write
$$\xi_k=\sum_{\sharp A=k}1_{\{A\subset\Pi_R(f)\}}\ ,\tag 2.3$$
where the summation ranges over all  $k$-point subsets  $A$  of  
$\bold Z^2$ .
\par
If  $m_1\ ,m_2\in\bold Z^2$  are distinct but  $\varphi(m_1)=\varphi(m_2)$ , 
then we must have  $\vert m_1-m_2\vert\geq1$ . Hence by the definition 
of  $\Pi_R(f)$ , if $R>0$ is large,  then $m_1\ ,m_2\in\Pi_R(f)$ with 
$m_1\ne m_2$ is possible only if  $\varphi(m_1)\ne\varphi(m_2)$ . Noting  
$b_1\leq f\leq b_2$ , we can therefore rewrite (2.3) as 
$$\xi_k=\sum_{(m_1,\ldots,m_k)\in\Cal Z_k(R)}1_{\{m_1,\ldots,m_k\in\Pi_R(f)\}}
\tag2.4$$
when  $R>0$  is large, where we have set
$$\align
\Cal Z_k(R)=\{(m_1,\ldots,m_k)\in(\bold Z^2)^k\vert\ &
\frac{b_1}2 R\leq\vert m_j\vert\leq2b_2R\ , j=1,\ldots,k \tag 2.5\\
&\text{and}\ 0\leq\varphi(m_1)<\cdots<\varphi(m_k)\leq\Theta\}\ .
\endalign$$
Now it is easy to see that  $m\in\Pi_R(f)$  is equivalent to
$$f(\varphi(m))\in I_m^R\equiv\left(\frac{\vert m\vert}{R+c_2/R},
\frac{\vert m\vert}{R+c_1/R}\right)\ ,\tag2.6$$
and hence
$$\bold E_{\bold P}[\xi_k(R;\cdot)]=\sum_{\Cal Z_k(R)}\bold P
(f(\varphi(m_j))\in I_{m_j}^R\ ,\ j=1,\ldots,k)\ .\tag 2.7$$
Take a  $\beta>0$  and define
$$\Cal Z_k(\beta;R)=\{(m_1,\ldots,m_k)\in\Cal Z_k(R)\vert\ 
\varphi(m_j)-\varphi(m_{j-1})\geq\beta\ ,\ j=2,\ldots,k\}\ ;\tag 2.8$$
and
$$\Cal Z'_k(\beta;R)=\Cal Z_k(R)\setminus\Cal Z_k(\beta;R) \tag 2.9$$
so that
$$\sum_{\Cal Z_k(R)}=\sum_{\Cal Z_k(\beta;R)}+\sum_{\Cal Z'_k(\beta;R)}
\ .\tag 2.10$$
Let us first estimate the sum over  $\Cal Z_k(\beta;R)$ . By the definition 
(2.6) of  $I_m^R$ , we have as  $R\to\infty$ ,
$$\text{the center of}\ I_m^R=\frac{\vert m\vert}R(1+O(R^{-2})) \tag2.11$$
and
$$\text{the length of}\ I_m^R=\frac{(c_2-c_1)\vert m\vert}{R^3}
(1+O(R^{-2}))\ .\tag2.12$$
In particular, if  $(m_1,\ldots,m_k)\in\Cal Z_k(R)$ , the length of  
$I_m^R$  is of  $O(R^{-2})$  for  $j=1,\ldots,k$ .\par
Define for  $\beta>0$
$$\align
\Psi_k(\beta)=&\sup\Sb (y_1,\ldots,y_k)\\ \in[b_1,b_2]^k\endSb
\sup\Sb (\varphi_1,\ldots,\varphi_k)\in[0,\Theta]^k\\ 
\varphi_j-\varphi_{j-1}\ge\beta\\ j=2,\ldots,k\endSb
\max_{1\le i\le k} \tag2.13\\
&\max\left\{\vert\frac{\partial}{\partial y_i}
p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)\vert\ ,\ 
\vert\frac{\partial}{\partial \varphi_i}
p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)\vert\right\}
\endalign$$
and
$$\psi_k(\beta)=\sup\Sb (y_1,\ldots,y_k)\\ \in[b_1,b_2]^k\endSb
\sup\Sb (\varphi_1,\ldots,\varphi_k)\in[0,\Theta]^k\\ 
\varphi_j-\varphi_{j-1}\ge\beta\\ j=2,\ldots,k\endSb
p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)\ . \tag2.14$$
Then by Condition (I), and (2.11), (2.12) and the remark following it, 
we will have
$$\align
&\ \bold P(f(\varphi(m_j))\in I_{m_j}^R\ ,\ j=1,\ldots,k) \tag2.15\\
&=\int_{I_{m_1}^R}dy_1\cdots\int_{I_{m_k}^R}dy_k\ 
p_k(y_1,\ldots,y_k\vert\varphi(m_1),\ldots,\varphi(m_k))\\
&=p_k(\frac{\vert m_1\vert}R,\ldots,\frac{\vert m_k\vert}R
\vert\varphi(m_1),\ldots,\varphi(m_k))\prod_{j=1}^k
\frac{(c_2-c_1)\vert m_j\vert}{R^3}\\
&\quad +O((\psi_k(\beta)+\Psi_k(\beta))R^{-2k-2})\ .
\endalign$$
Hence noting  $\sharp\Cal Z_k(R)=O(R^{2k})$ , we get
$$\align
&\ \sum_{\Cal Z_k(\beta;R)}\bold P(f(\varphi(m_j))\in I_{m_j}^R\ ,
\ j=1,\ldots,k)
\tag2.16\\
&=(c_2-c_1)^k\sum_{\Cal Z_k(\beta;R)}
p_k(\frac{\vert m_1\vert}R,\ldots,\frac{\vert m_k\vert}R
\vert\varphi(m_1),\ldots,\varphi(m_k))\left(\prod_{j=1}^k
\frac{\vert m_j\vert}{R}\right)\left(\frac1{R^2}\right)^k\\
&\quad +O((\psi_k(\beta)+\Psi_k(\beta))R^{-2})\ .
\endalign$$
Now let us consider a function on  $(\bold R^2)^k$  defined by
$$R(x_1,\ldots,x_k)=p_k(\vert x_1\vert,\ldots,\vert x_k\vert\ \vert\ 
\varphi(x_1),\ldots,\varphi(x_k))\left(\prod_{j=1}^k\vert x_j\vert\right)\ .
\tag2.17$$
Then the sum on the right hand side of (2.16) is the approximation by 
Riemann sum of the integral
$$(c_2-c_1)^k\idotsint_{D_k(\beta)}dx_1\cdots dx_k\ R(x_1,\ldots,x_k)\ ,
\tag 2.18$$
where the domain  $D_k(\beta)$  of integration is given by
$$\align
D_k(\beta)=\{(x_1,\ldots,x_k)\in\bold (\bold R^2)^k\vert
&\frac{b_1}2\le\vert x_j\vert\le2b_2\ ,\ j=1,\ldots,k\ , \tag2.19\\
&0\le\varphi(x_j)\le\Theta\ ,\\
&\varphi(x_j)-\varphi(x_{j-1})\ge\beta\ ,\ j=2,\ldots,k\ \}\ .
\endalign$$
It is easy to see that
$$\max_{1\le j\le k}\left\Vert\frac{\partial R}{\partial x_j}\right\Vert
\leq\text{const.} (\psi_k(\beta)+\Psi_k(\beta)) \tag 2.20$$
holds on $D_k(\beta)$ . 
This gives the accuracy of the above Riemann sum 
approximation, and we obtain, after replacing the domain of integration  
$D_k(\beta)$  by
$$\align
D_k=\{(x_1,\ldots,x_k)\in(\bold R^2)^k\vert
&\frac{b_1}2\le\vert x_j\vert\le2b_2\ ,\ j=1,\ldots,k\ , \tag2.21\\
&0\le\varphi(x_{j-1})<\varphi(x_j)\le\Theta\ ,\ j=2,\ldots,k\}\ ,
\endalign$$
and tranforming the variable into the polar coordinate,
$$\align
&\ \sum_{\Cal Z_k(\beta;R)}\bold P(f(\varphi(m_j))\in I_m^R\ ,\ j=1,\ldots,k)
\tag 2.22\\
&=(c_2-c_1)^k\idotsint_{0\le\varphi_1<\cdots<\varphi_k\le\Theta}
d\varphi_1\cdots d\varphi_k\idotsint_{b_1\le y_j\le b_2\ ,\ j=1,\ldots,k}
dy_1\cdots dy_k\times\\
&\quad\times\left(\prod_{j=1}^k y_j^2\right)
p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)\\
&\quad +O((\psi_k(\beta)+\Psi_k(\beta))R^{-1}+\beta)\\
&=\frac1{k!}\bold E_{\bold P}[\lambda(f)^k]+
O((\psi_k(\beta)+\Psi_k(\beta))R^{-1}+\beta)\ .
\endalign$$
In order to complete the proof of (2.2), it remains to prove
$$\lim_{\beta\downarrow0}\varlimsup_{R\uparrow\infty}
\sum_{\Cal Z'_k(\beta;R)}\bold P(f(\varphi(m_j))\in I_{m_j}^R\ ,\ j=1,\ldots,k)
=0\ .\tag2.23$$
Suppose first that  $\bold P$  satisfies (II-1). By the remark which 
follows (2-12), 
we see that
$$\bold P(f(\varphi(m_j))\in I_{m_j}^R\ ,\ j=1,\ldots,k)
\le \text{const.}R^{-2k}
\prod_{j=2}^k(\varphi(m_j)-\varphi(m_{j-1}))^{-\sigma}\ .\tag2.24$$
Hence Lemma 1 below gives
$$\varlimsup_{R\uparrow\infty}
\sum_{\Cal Z'_k(\beta;R)}\bold P(f(\varphi(m_j))\in I_m^R\ ,\ j=1,\ldots,k)
=O(\beta^{1-\sigma})\tag2.25$$
for small  $\beta>0$ , and (2.23) follows immediately from this.
\bigskip
\proclaim{Lemma 1} For large  $R>0$  and  for small  $\beta>0$ , we have 
the estimate
$$\max_{m\in\Cal Z_1(R)}\sum\Sb m'\in\Cal Z_1(R)\ ;\\
 0<\varphi(m')-\varphi(m)<\beta\endSb (\varphi(m')-\varphi(m))^{-\sigma}
=O(R^2(R^{\sigma-1}+\beta^{1-\sigma}))\ .$$
\endproclaim
\demo{Proof}
We may assume  $0\le\varphi(m)\le\pi/4$ , because  $\bold Z^2$  is invariant 
under the rotation by  $\pi j/2$ , $j=1,2,3$ , and because the lattice 
obtained by rotating  $\bold Z^2$  by  $\pi/4+\pi j/2$ , $j=1,2,3$  
is contained in the lattice  $(1/\sqrt{2})\bold Z^2$ . But in this case, 
we have  $\rho\equiv\tan\varphi(m)\in[0,1]$ .\par
Writing  $m'=(\mu_1,\mu_2)$ , we get
$$\align
&\ \sum\Sb m'\in\Cal Z_1(R)\ ;\\
 0<\varphi(m')-\varphi(m)<\beta\endSb (\varphi(m')-\varphi(m))^{-\sigma}
\tag 2.26\\
&\leq\sum_{1\le\mu_1\le2b_2R}\ \sum
_{\mu_2;\rho<\frac{\mu_2}{\mu_1}<\tan(\varphi(m)
+\beta)}\left\{\tan^{-1}\frac{\mu_2}{\mu_1}-\varphi(m)\right\}^{-\sigma}
\ .
\endalign$$
Since
$$\min\{\mu_2\ \vert\ \frac{\mu_2}{\mu_1}>\rho\}=[\rho\mu_1]+1\ ,\tag2.27$$
the inner summation on the right hand side of (2.26) is further estimated as
 below:
$$\align
&\ \sum_{\mu_2;\rho<\frac{\mu_2}{\mu_1}<\tan(\varphi(m)
+\beta)}\left\{\tan^{-1}\frac{\mu_2}{\mu_1}-\varphi(m)\right\}^{-\sigma}
\tag 2.28\\
&\leq\mu_1\int_{\rho}^{\tan(\varphi(m)+\beta)}(\tan^{-1}y-\varphi(m))
^{-\sigma}dy\\
&+\quad\frac{\{\rho\mu_1\}}{\mu_1}\left(\tan^{-1}
\frac{[\rho\mu_1]+1}{\mu_1}-\tan^{-1}\rho\right)^{-\sigma}\\
&\equiv F(\mu_1)+G(\mu_1)\ .
\endalign$$
Here  $[a]$  and $\{a\}$  are respectively the integer and the fractional 
parts of a real number  $a$ . Now we have
$$\align
\sum_{1\le\mu_1\le2b_2R}F(\mu_1)&=
\left(\sum_{1\le\mu_1\le2b_2R}\mu_1\right)
\int_{\varphi(m)}^{\varphi(m)+\beta}(\varphi-\varphi(m))^{-\sigma}
\frac{d\varphi}
{\cos^2\varphi} \tag2.29\\
&=O(R^2\beta^{1-\sigma})\ ,
\endalign$$
where we have used the assumption that  $0\le\varphi(m)\le\pi/4$  and that  
$\beta>0$  is small so that   $\cos^2\varphi$   is bounded away from  0  on 
the range of integration.\par
In order to estimate the sum of  $G(\mu_1)$  we set  $\rho=p/q$  with  
$p$  and  $q$  mutually prime integers. Since  $m\in\Cal Z_1(R)$ , and  
$0\le\rho\le1$ , we must have  $0\le p\le q\le2b_2R$ . Moreover for any 
integer  $\ell$ ,
$$\align
\left\{\{\rho n\}\ \vert\ n\in\bold N\right\}&=
\left\{\{\rho(q\ell+j)\}\ \vert\ j=1,\ldots,q\right\}\tag2.30\\
&=\{0,\frac1{q},\ldots,\frac{q-1}{q}\}\ .
\endalign$$
Then noting that
$$0\le\{\rho\mu_1\}\le1\ ;\ \frac{[\rho\mu_1]+1}{\mu_1}
=\rho+\frac{1-\{\rho\mu_1\}}{\mu_1} \tag2.31$$
hold, we can compute
$$\align
&\ \sum_{1\le\mu_1\le2b_2R}G(\mu_1) \tag2.32\\
&\leq \sum_{1\le j\le q-1}G(j)+\sum_{1\le\ell\le2b_2R/q}
\ \sum_{0\le j\le q-1}G(q\ell+j)\\
&\leq\sum_{1\le j\le q-1}\left\{\tan^{-1}(\rho+\frac{1-j/q}q)-\varphi(m)
\right\}^{-\sigma}\\
&+\quad\sum_{1\le\ell\le2b_2R/q}\ \sum_{0\le j\le q-1}
\frac1{q\ell}\left\{\tan^{-1}(\rho+\frac{1-j/q}{q(\ell+1)})-\varphi(m)
\right\}^{-\sigma}\\
&\leq q\int_0^1\left\{\tan^{-1}(\rho+\frac{1-x}q)-\varphi(m)\right\}
^{-\sigma}dx\\
&\quad+\sum_{1\le\ell\le2b_2R/q}\frac1{\ell}\int_0^1
\left\{\tan^{-1}(\rho+\frac{1-x}{q(\ell+1)})-\varphi(m)\right\}^{-\sigma}dx
\\
&=q^2\int_{\varphi(m)}^{\tan^{-1}(\rho+1/q)}(\varphi-\varphi(m))^{-\sigma}
\frac{d\varphi}{\cos^2\varphi}\\
&\quad+\sum_{1\le\ell\le2b_2R/q}\frac{q(\ell+1)}{\ell}\int_{\varphi(m)}
^{\tan^{-1}(\rho+1/q(\ell+1))}(\varphi-\varphi(m))^{-\sigma}
\frac{d\varphi}{\cos^2\varphi}\\
&=O(q^{\sigma+1}+R^{\sigma})\\
&=O(R^{\sigma+1})\ ,
\endalign$$
completing the proof of Lemma 1. 
\enddemo
\bigskip
Next we consider the case where  $\bold P$  
satisfies (II-2). By (2.11) and (2.12), 
there is a constant  $C>0$  such that the conditions
$$\frac1{2}b_1R\leq\vert m\vert\leq2b_2R$$
and
$$f(\varphi(m))\in I_m^R$$
imply the condition
$$\left\vert f(\varphi(m))-\frac{\vert m\vert}R\right\vert
\le\frac1{2}CR^{-2}\ .\tag2.33$$
By the Lipschitz continuity of  $f(\cdot)$ , we have the following 
implication for large  $R>0$ :
$$\align
&\ f(\varphi(m_j))\in I_{m_j}^R\ ,\ \frac1{2}b_1R\le\vert m_j\vert\le
2b_1R\ ,\ j=1,\ldots,k \tag2.34\\
&\Longrightarrow\left\vert\frac{\vert m_j\vert}R-
\frac{\vert m_{j-1}\vert}R\right\vert\le\frac{C}{R^2}+
b_3\vert\varphi(m_j)-\varphi(m_{j-1})\vert\ ,\ j=2,\ldots,k\ .
\endalign$$
On the other hand, it is easy to see that for some absolute constant  
$M>0$ ,
$$\min_{(m,m')\in\Cal Z_2(R)}(\varphi(m')-\varphi(m))\ge MR^{-2} 
\tag 2.35$$
for large  $R>0$ . Hence if we set  $K=b_3+C/M$ , the right hand side of 
(2.34) can be rewritten as
$$\left\vert \vert m_j\vert-\vert m_{j-1}\vert \right\vert
\leq KR \vert\varphi(m_j)-\varphi(m_{j-1})\vert\ ,\ j=2,\ldots,k\ .
\tag 2.36$$
>From this consideration and (1-8), we obtain
$$\align
&\ \sum_{\Cal Z'_k(\beta;R)}\bold P(f(\varphi(m_j))\in I_{m_j}^R\ ,\ 
j=1,\ldots,k) \tag 2.37\\
&\le \text{const.}\sum_{\Cal Z''_k(\beta;R)}R^{-2k}\prod_{j=2}^k
(\varphi(m_j)-\varphi(m_{j-1}))^{-\tau}\ ,
\endalign$$
where  $\Cal Z''_k(\beta;R)$  is the totality of  $(m_1,\ldots,m_k)\in
\Cal Z'_k(\beta;R)$  for which (2.36) holds. Then by Lemma 3 below, we 
will have
$$\varlimsup_{R\to\infty}
\sum_{\Cal Z''_k(\beta;R)}R^{-2k}\prod_{j=2}^k
(\varphi(m_j)-\varphi(m_{j-1}))^{-\tau}=O(\beta^{2-\tau})\ ,\ 
\beta\downarrow0\ , \tag 2.38$$
which completes the proof of Theorem 1.
\par
\bigskip
\proclaim{Lemma 2} 
Let $A>0$ , $S>0$ , $T>0$ and $R\geq1$ , where $T\geq\delta R$ and 
$S\geq\delta R$ for some $\delta>0$ . Let further $\beta>(AR)^{-1}$ . 
Then there is a constant $C>0$ such that 
$$\sum_{m\in\bold Z^2;
(AR)^{-1}<\varphi(m)-\varphi_0<\beta}
\ 1_{\{\vert\vert m\vert-S\vert\leq T(\varphi(m)-\varphi_0)\}}
(\varphi(m)-\varphi_0)^{-\tau}\leq CST\beta^{2-\tau}$$
for any $\varphi_0\in[0,\Theta)$ .
\endproclaim
\demo {Proof}
Divide the interval  $(\varphi_0+(AR)^{-1},\varphi_0+\beta]$  into 
subintervals  $\Delta_{\ell}$  of length  $(AR)^{-1}$ , 
where $\ell=1,\ldots,[AR\beta]-1$ , and an interval $\Delta_{[AR\beta]}$ 
of length $\leq(AR)^{-1}$ . 
Let  $N_{\ell}$  be the number of 
lattice points  $m\in\bold Z^2$  such that  $\varphi(m)\in\Delta_{\ell}$  
and that
$$\left\vert \vert m\vert-S \right\vert\leq 
T(\varphi(m)-\varphi_0)\ .$$
Then  $N_{\ell}\le \text{const.}R^{-2}ST\ell$  and we have
$$\align
&\ \sum_{m\in\bold Z^{2};(AR)^{-1}<\varphi(m)-\varphi_0<\beta}
1_{\{\vert\vert m\vert-S\vert\leq T(\varphi(m)-\varphi_0)\}}
(\varphi(m)-\varphi_0)^{-\tau}
\tag2.39\\
&=\sum_{\ell=1}^{[R\beta]}\sum_{m;\varphi(m)\in\Delta_{\ell}}
1_{\{\vert\vert m\vert-S\vert\leq T(\varphi(m)-\varphi_0)\}}
(\varphi(m)-\varphi_0)^{-\tau} \\
&\leq \sum_{\ell=1}^{[R\beta]}\sum_{m;\varphi(m)\in\Delta_i}
1_{\{\vert\vert m\vert-S\vert\leq T(\varphi(m)-\varphi_0)\}}
(iR^{-1})^{-\tau} \\
&\leq\text{const.}\sum_{\ell=1}^{[R\beta]}\frac{ST}{R^2}
\ell^{1-\tau}R^{\tau} \\
&\leq\text{const.}\frac{ST}{R^{2-\tau}}(R\beta)^{2-\tau} \\
&=CST\beta^{2-\tau}\ ,
\endalign$$
and this estimate is uniform in  $\varphi_0$ .
\enddemo
As a corollary of Lemma 2, we get the following
\proclaim{Lemma 3}  For large $R>0$ and small $\beta>0$ , we have
$$\max_{m'\in\Cal Z_1(R)}\sum_{m;(m',m)\in\Cal Z''_2(\beta;R)}
(\varphi(m)-\varphi(m'))^{-\tau}=O(KR^2\beta^{2-\tau})\ .$$
\endproclaim
\demo {Proof} 
By an elementary geometric consideration, we can show the existence of 
an absolute constant  $M'>0$  such that
$$\min_{m\in\Cal Z_1(R)}\min \Sb m';\varphi(m')>\varphi(m),\\
\vert\vert m'\vert-\vert m\vert\vert\le R
(\varphi(m')-\varphi(m)) \endSb (\varphi(m')-\varphi(m))
\geq M'R^{-1}\ .\tag 2.40$$
Hence if we take  $A^{-1}=M'$ , $S=\vert m'\vert$ , $T=KR$ 
and $\varphi_0=\varphi(m')$ in Lemma 2, then the condition 
$(AR)^{-1}<\varphi(m)-\varphi_0$ under the summation symbol can be dropped 
and we obtain 
the desired assertion.
\enddemo

\head {\bf \S3 Proof of Theorems 2 and 3} \endhead
In this section, we shall prove 
simultaneously Theorems 2' and 3' , which are equivalent variants 
of Theorems 2 and 3,  
admitting a technical lemma which will be proved in \S4. 
After that, we shall sketch how to transform the modified version 
of our theorems back to the original ones.
\par
In order to state the modified version of our theorems--which the author 
believs 
are in fact closer to the original problem raised by Berry and Tabor [BT]--
let us introduce the planar domain $\tilde{\Pi}_R(f)$ by
$$\align
\tilde{\Pi}_R(f)=\{x\in\bold R^2 &\vert\ 0\leq\varphi(x)\leq\Theta\ ,\tag3.1\\
& \sqrt{R+2c_1}f(\varphi(x))<\vert x\vert<\sqrt{R+2c_2}f(\varphi(x))\}\ .
\endalign$$
Then the area of $\tilde{\Pi}_R(f)$ is not only asymptotic to, but is 
equal to $\lambda(f)$ for all $R>0$ . 
Given positive numbers $0<a_1<a_2$ and $L>0$ , let $\mu_L^{a_1,a_2}(dR)$ 
be the uniform probability distribution on the interval $I_L=(a_1L,a_2L)$ . 
Finally we define 
$$\tilde{\xi}=\tilde{\xi}(R;f)=\sharp\{\tilde{\Pi}_R(f)\cap\bold Z^2\}\ ,
\tag 3.2$$
to be the number of lattice points in the domain $\tilde{\Pi}_R(f)$ . 
\par
Now we shall prove the following two theorems:
\bigskip
\proclaim{Theorem 2'} Suppose Conditions (I) and (II) hold. 
Then there is a sequence  $\bar{L}_n\to\infty$  
such that for  $\bold P$-almost all  $f\in\Cal X$ , the distribution of  
$\tilde{\xi}(\cdot;f)$  under  $\mu_{\bar{L}_n}^{a_1,a_2}$  converges to the 
Poisson 
distribution with parameter  $\lambda(f)$ .
\endproclaim
\bigskip
\proclaim{Theorem 3'} If we assume Condition (III) in addition to Conditions 
(I) and (II), then for  $\bold P$-almost all  $f\in\Cal X$ , the 
distribution of 
$\tilde{\xi}(\cdot;f)$  under  $\mu_L^{a_1,a_2}$  converges to the Poisson 
distribution 
with parameter  $\lambda(f)$  as  $L\to\infty$ .
\endproclaim
\bigskip
Let us turn to the proof of Theorem 3'.
\par
For  $k=1,2,\ldots$ , set
$$\bold E_L=\bold E_{L,k}(f)=\int_{I_L}\tilde{\xi}(R;f)\mu_L(dR)\ ,\tag3.3$$
where  $\mu_L=\mu_L^{a_1,a_2}$ , 
and
$$\tilde{\xi}_k=\tilde{\xi}_k(R;f)=\pmatrix \tilde{\xi} \\ k \endpmatrix 
1_{\{\tilde{\xi}\geq k\}}
=\sum_{n=k}^{\infty}\frac{n!}{k!(n-k)!}1_{\{\tilde{\xi}=n\}}\ .\tag 3.4$$
As was done in [M], Theorem 3' will be proved as soon as we have 
shown
$$\lim_{L\to\infty}\bold E_{L,k}(f)=\frac1{k!}\lambda(f)^k\ ,\ k=1,2,\ldots 
\tag3.5$$
for  $\bold P$ -almost all  $f\in\Cal X$ . Similarly, Theorem 2' will be 
proved 
as soon as we have shown the existence of a sequence  $\bar{L}_j\to\infty$
  such that (3.5) holds along this sequence for  $\bold P$ -almost all  
$f\in\Cal X$ .
\par
For a lattice point  $m\in\bold Z^2$ , let
$$D_m=D_m(f)=\{R>0\ \vert\ \tilde{\Pi}_R(f)\ni m\}\ .\tag3.6$$
By the definition (3.1) of $\tilde{\Pi}_R(f)$ , we have 
$$D_m=\left(\frac{\vert m\vert^2}{f(\varphi(m))^2}-2c_2,
 \frac{\vert m\vert^2}{f(\varphi(m))^2}-2c_1\right)\ .\tag 3.7$$
In particular, $\vert D_m\vert=2(c_2-c_1)$ . For convenience, we let 
$$\gamma_m=\gamma_m(f)=\frac{\vert m\vert^2}{f(\varphi(m))^2}\ .\tag 3.8$$
Now set  $L_n=2^n$ , $n=1,2,\ldots$ . Then if  $\tilde{\Pi}_R(f)\ni m$ , 
$R\in I_L$  
and  if $L_n\le L<L_{n+1}$ , then we will have
$$b_1\sqrt{a_1L_n}\leq\vert m\vert\leq2b_2\sqrt{a_2L_{n+1}}\ .\tag3.9$$
Under the same condition, it is also easy to see that if $m\ne m'$ but 
$\varphi(m)=\varphi(m')$ , then $D_m\cap D_{m'}=\phi$ . 
Hence if we define, as in \S2,
$$\align
\Cal Z_{k,n}=\{(m_1,\ldots,m_k)\in(\bold Z^2)^k\ \vert
& m_j\ \text{satisfies (3.9) for}\ j=1,\ldots,k\ \text{and} \tag3.10\\
&\quad 0\le\varphi(m_1)<\cdots<\varphi(m_k)\le\Theta\}\ ,
\endalign$$
then for large  $n$ and for $L_n\le L<L_{n+1}$ , we have 
$$\align
\bold E_{L,k} &=\sum_{(m_1,\ldots,m_k)\in\Cal Z_{k,n}}
\mu_L(m_1,\ldots,m_k\in\tilde{\Pi}_R(f)) \tag3.11\\
&=\frac{1}{\vert I_L\vert}\sum_{(m_1,\ldots,m_k)\in\Cal Z_{k,n}}
\vert I_L\cap D_{m_1}\cap\cdots\cap D_{m_k}\vert\ .
\endalign$$
For $k=1$ , (3.5) can be proved without any reference to the randomness of 
$f(\cdot)$ . Indeed in this case,
$$\bold E_{L,1}(f)=\frac1{\vert I_L\vert}\sum_{m\in\Cal Z_{1,n}}
\vert I_L\cap D_m\vert\ .\tag 3.12$$
Since
$$D_m\cap I_L\ne\phi\Longleftrightarrow a_1L+2c_1<\gamma_m<a_2L+2c_2\ ,
\tag 3.13$$
and 
$$D_m\subset I_L\Longleftrightarrow a_1L+2c_2\le \gamma_m\le a_2L+2c_1\ ,
\tag 3.14$$
we see
$$\align
\bold E_{L,1}(f)&\le \frac1{\vert I_L\vert}\sum_{m\in\bold Z^2}
1_{(a_1L+2c_1,a_2L+2c_2)}(\gamma_m)\vert D_m\vert \tag 3.14\\
&= \frac{2(c_2-c_1)}{\vert I_L\vert}\sum_{m\in\bold Z^2}
1_{(a_1L+2c_1,a_2L+2c_2)}\left(\frac{\vert m\vert^2}{f(\varphi(m))^2}\right)
\ , 
\endalign$$
and similarly
$$\bold E_{L,1}(f)\geq\frac{2(c_2-c_1)}{\vert I_L\vert}\sum_{m\in\bold Z^2}
1_{[a_1L+2c_2,a_2L+2c_1]}\left(\frac{\vert m\vert^2}{f(\varphi(m))^2}\right)\ .
\tag 3.15$$
Define $\bar{\bold E}_{L,1}(f)$ and $\underline{\bold E}_{L,1}(f)$ to be the 
right hand sides of (3.14) and (3.15) respectively. 
\par
For each $\epsilon>0$ , consider the planar domain
$$\align
\Lambda_{\epsilon}&=\Lambda_{\epsilon}(f)\tag 3.17\\
&=\{x\in\bold R^2\ \vert \ 0\leq\varphi(x)\leq\Theta\ \text{and}\  a_1
f(\varphi(x))^2<\vert x\vert^2<(a_2+\epsilon)f(\varphi(x))^2\}\ .
\endalign$$
Since $f(\cdot)$ is a continuous function, the boundary of 
$\Lambda_{\epsilon}$ has 
zero Lesbesgue measure, and hence the indicator of $\Lambda_{\epsilon}$ is 
Riemann 
integrable. We have therefore
$$\align
\varlimsup_{L\to\infty}\bar{\bold E}_{L,1}(f)&\leq 
\varlimsup_{L\to\infty}\frac{2(c_2-c_1)}{a_2-a_1}
\left(\frac1{\sqrt{L}}\right)^2
\sum_{m\in\bold Z^2}1_{\Lambda_{\epsilon}}\left(\frac{m}{\sqrt{L}}\right) 
\tag 3.18\\
&=\frac{2(c_2-c_1)}{a_2-a_1}\times\text{the area of}\ \Lambda_{\epsilon} \\
&=\left(1+\frac{\epsilon}{a_2-a_1}\right)\lambda(f)\ .
\endalign$$
Letting $\epsilon\searrow0$ , we obtain
$$\varlimsup_{L\to\infty}\bar{\bold E}_{L,1}(f)\leq\lambda(f)\ .\tag 3.19$$
By a similar argument, we can also prove
$$\varliminf_{L\to\infty}\underline{\bold E}_{L,1}(f)\geq\lambda(f)\ .
\tag 3.20$$
So that
$$\lim_{L\to\infty}\bold E_{L,1}(f)=\lambda(f) \tag 3.21$$
for every $f\in\Cal X$ .
\par
Next we consider the case $k\geq2$ .
\par
If $L_n\leq L<L_{n+1}$ , then $I_L\subset[a_1L_n,a_2L_{n+1})$ . We divide 
the interval $[a_1L_n,a_2L_{n+1})$ into $q_n=n^{\eta}$ subintervals 
$[z_{i+1},z_i)$ 
having equal length, where $\eta>0$ is arbitrary. Obviously
$$z_i=a_1L_n+\frac{i}{q_n}(a_2L_{n+1}-a_1L_n)\ ,\ i=1,\ldots,q_n\ .\tag 3.22$$
If we let
$$\bar{A}_n(L)=\frac{a_1(L-L_n)+2c_1}{a_2L_{n+1}-a_1L_n}q_n\ ;\ 
\bar{B}_n(L)=1+\frac{a_2L-a_1L_n+2c_2}{a_2L_{n+1}-a_1L_n}q_n\ ,\tag 3.22$$
and
$$\underline{A}_n(L)=1+\frac{a_1(L-L_n)+2c_2}{a_2L_{n+1}-a_1L_n}q_n\ ;\ 
\underline{B}_n(L)=\frac{a_2L-a_1L_n+2c_1}{a_2L_{n+1}-a_1L_n}q_n\ ,\tag 3.23$$
then we have
$$\gamma_m\in[z_{i-1},z_i)\ ,\ D_m\cap I_L\ne\phi\Longrightarrow
 \bar{A}_n(L)<i<\bar{B}_n(L) \tag 3.24$$
and
$$\gamma_m\in[z_{i-1},z_i)\ ,\ \underline{A}_n(L)\leq i\leq\underline{B}_n(L)
\Longrightarrow D_m\subset I_L\ .\tag 3.25$$
Hence for $n$ large, we can write 
$$\align
\bold E_{k,L}(f)&=\frac1{\vert I_L\vert}\sum_{(m_1,\ldots,m_k)\in\Cal Z_{k,n}}
\vert I_L\cap D_{m_1}\cap\cdots\cap D_{m_k}\vert \tag 3.26 \\
&\leq \frac1{\vert I_L\vert}\sum_{\bar{A}_n(L)<i<\bar{B}_n(L)}
\sum_{m_1}1_{\{\gamma_{m_1}\in[z_{i-1},z_i)\}}\sum_{\Cal Z_{k,n}(m_1)}
\vert D_{m_1}\cap\cdots\cap D_{m_k}\vert\ ,
\endalign$$
and
$$\bold E_{k,L}(f)\geq\frac1{\vert I_L\vert}\sum_{\underline{A}_n(L)\leq i
\leq\underline{B}_n(L)}
\sum_{m_1}1_{\{\gamma_{m_1}\in[z_{i-1},z_i)\}}\sum_{\Cal Z_{k,n}(m_1)}
\vert D_{m_1}\cap\cdots\cap D_{m_k}\vert\ ,\tag 3.27$$
where
$$\Cal Z_{k,n}(m_1)\equiv\{(m_2,\ldots,m_k)\in(\bold Z^2)^{k-1}\vert
\ (m_1,m_2,\ldots,m_k)\in\Cal Z_{k,n}\}\ . \tag 3.28$$
As before, we define $\bar{E}_{k,L}^{(1)}(f)$ and 
$\underline{E}_{k,L}^{(1)}(f)$ 
to be the right hand sides of (3.26) and (3.27) respectively. We also define 
$\bar{E}_{k,L}^{(2)}(f)$ by
$$\align
\bar{E}_{k,L}^{(2)}(f)&= \frac1{\vert I_L\vert}\sum_{\bar{A}_n(L)<i<
\bar{B}_n(L)}
\sum_{m_1}1_{\{\gamma_{m_1}\in[z_{i-1},z_i)\}}\ \times \tag 3.29\\
& \times2(c_2-c_1)^k
\int_{\varphi(m_1)}^{\Theta}d\varphi_{2}\int_{\varphi_2}^{\Theta}d\varphi_{3}
\cdots
\int_{\varphi_{k-1}}^{\Theta}d\varphi_{k}\ f(\varphi_2)^2\cdots f(\varphi_k)^2
\ ,
\endalign$$
and $\underline{E}_{k,L}^{(2)}(f)$ by the same formula as (3.29) except that 
the condition 
under the summation symbol \lq\lq$\bar{A}_n(L)<i<\bar{B}_n(L)$\rq\rq is 
replaced by \lq\lq$\underline{A}_n(L)\leq i\leq\underline{B}_n(L)$\rq\rq .
\par
Now it is easy to prove 
$$\lim_{L\to\infty}\bar{E}_{k,L}^{(2)}(f)=\lim_{L\to\infty}
\underline{E}_{k,L}^{(2)}(f)
=\frac{\lambda(f)^k}{k!} \tag 3.30$$
for every $f\in\Cal X$ , so that we are left with estimating the differences
$$\bar{G}_{k,L}(f)\equiv\bar{E}_{k,L}^{(1)}(f)-\bar{E}_{k,L}^{(2)}(f)\ ;\ 
\underline{G}_{k,L}(f)\equiv\underline{E}_{k,L}^{(1)}(f)-
\underline{E}_{k,L}^{(2)}(f)\ . 
\tag 3.31$$
More precisely, writing
$$\align
\Gamma_n(f)\equiv
&\sup_{L_n\leq L<L_{n+1}}\max\{\vert\underline{G}_{k,L}(f)\vert,
\vert\bar{G}_{k,L}(f)\vert\} \tag 3.32 \\
&\leq \frac1{\vert I_{L_n}\vert}\sum_{i=1}^{q_n+1}
\left\vert\sum_{m_1}1_{\{\gamma_{m_1}\in[z_{i-1},z_i)\}}\right.\ \times \\
& \times\left\{\sum_{(m_2,\ldots,m_k)\in\Cal Z_{k,n}(m_1)}
\vert D_{m_1}D_{m_2}\cap\cdots\cap D_{m_k}\vert\right.\ - \\
& \ \left.-2(c_2-c_1)^k
\int_{\varphi(m_1)}^{\Theta}d\varphi_{2}\int_{\varphi_2}^{\Theta}d\varphi_{3}
\cdots
\int_{\varphi_{k-1}}^{\Theta}d\varphi_{k}\ f(\varphi_2)^2\cdots f(\varphi_k)^2
\right\}\left\vert\ \right. ,
\endalign$$
we shall prove
$$\sum_{n=1}^{\infty}\bold E_{\bold P}[\Gamma_n(\cdot)]=
\bold E_{\bold P}[\sum_{n=1}^{\infty}\Gamma_n(\cdot)]<\infty\ , \tag 3.33$$
which immediately gives
$$\lim_{L\to\infty}\max\{\vert\underline{G}_{k,L}(f)\vert,
\vert\bar{G}_{k,L}(f)\vert\}\leq\varlimsup_{n\to\infty}\Gamma_n(f)=0 
\tag 3.34$$
for $\bold P$-almost all $f\in\Cal X$ . 
\par
Let us finish the proof of (3.33), and hence of Theorem 3', admitting the 
following 
technical lemma:
\par
\proclaim{Lemma 4} For $n\geq1$ and $1\leq i\leq q_n+1$ , let
$$\align
U_{i,n}=\bold E_{\bold P}& \left[\left\vert
\sum_{m_1}1_{\{\gamma_{m_1}\in[z_{i-1},z_i)\}}\right.\right.
\left\{\sum_{(m_2,\ldots,m_k)\in\Cal Z_{k,n}(m_1)}
\vert D_{m_1}\cap D_{m_2}\cap\cdots\cap D_{m_k}\vert\ -\right. \\
& \left.\left.\left.-2(c_2-c_1)^k
\int_{\varphi(m_1)}^{\Theta}d\varphi_{2}\int_{\varphi_2}^{\Theta}d\varphi_{3}
\cdots
\int_{\varphi_{k-1}}^{\Theta}d\varphi_{k}\ f(\varphi_2)^2\cdots f(\varphi_k)^2
\right\}\right\vert^2\right]\ .\tag 3.35
\endalign$$
Let further $T^{-1}\leq a_1<a_2\leq T$ with $T\geq1$ . 
Then there is a contant $C_T>0$ 
which depends only on $T>0$ 
such that for each $n\geq1$ and $\beta>L_n^{-1/2}$ , 
$$\max_{1\leq i\leq q_n+1}U_{i,n}\leq C_T(\Psi_{2k}(\beta)+\sqrt{L_n}\beta^d)
L_n^{3/2}q_n^{-2}\ ,\tag 3.36$$
where $d=1-\sigma$ or $d=2-\tau$ according to which of the conditions (II-1) 
or 
(II-2) is satisfied by $\bold P$ .
\endproclaim
\bigskip
Take $\beta=\beta_n=An^{-1/{\nu_k}}$ in the above lemma. Then by (3.32) and 
(3.35), 
$$\align
\bold E[\Gamma_n(\cdot)]&\leq \frac1{\vert I_{L_n}\vert}
\sum_{i=1}^{q_n+1}\sqrt{U_{i,n}} \tag 3.37\\
&\leq \text{const.}L_n^{-1}q_n(\sqrt{\Psi_{2k}(\beta_n)}+L_n^{1/4}
\beta_n^{d/2})
L_n^{3/4}q_n^{-1} \\
&=\text{const.}(L_n^{-1/4}\sqrt{\Psi_{2k}(\beta_n)}+\beta_n^{d/2}) \\
&\leq \text{const.}(L_n^{-1/4}\exp(\frac12 A^{-\nu_k}n)+n^{-d/{2\nu_k}})\ .
\endalign$$
If we take $A>0$ sufficiently large, then the first term on the right hand side decays 
exponentially first, while $d/{2\nu_k}>1$ . 
Hence we have
$$\sum_{n=1}^{\infty}\bold E[\Gamma_n(\cdot)]<\infty\ ,\tag 3.38$$
which completes the proof of Theorem 3.
\par
When we do not assume Condition (III), $\Psi_{2k}(\beta)$ can be any positive 
function 
monotonically tending to $\infty$ as $\beta\searrow0$ . It is possible to 
choose a 
decreasing sequence $\{\beta_n\}$ tending to $0$ so slowly that 
$$L_n^{-1/2}\Psi_{2k}(\beta_n)\longrightarrow0 \tag 3.39$$
holds. Then choose a subsequence $\{n'\}$ such that 
$$\sum_{n'}(L_{n'}^{-1/4}\sqrt{\Psi_{2k}(\beta_{n'})}+\beta_{n'}^{d/2})<\infty
\ .\tag 3.40$$
Choose a number $\bar{L}_{n'}$ arbitrarily 
from the interval $[L_{n'},L_{n'+1})$ for each $n'$ . Then obvioiusly 
the sequence $\{\bar{L}_{n'}\}$ meets the requirement of Theorem 2' .
\par

\bigskip
Before closing this section, we sketch how to obtain Theorems 2 and 3 from 
what we have shown 
up to now.
\par
For this purpose, we first note that for any $f\in\Cal X$ and any $0<a_1<a_2$ ,
$$\lim_{L\to\infty}\mu_L^{a_1,a_2}(\{R>0\vert\tilde{\xi}(R;f)\ne\xi(\sqrt{R};f)
\})=0\ .
\tag 3.41$$
Indeed, $\tilde{\xi}(R;f)\ne\xi(\sqrt{R};f)$ implies
$$\{\tilde{\Pi}_R(f)\Delta\Pi_R(f)\}\cap\bold Z^2\ne\phi\ ,\tag 3.42$$
and since $a_1L\leq R\leq a_2L$ , $b_1\leq f(\cdot)\leq b_2$ , it holds that 
$$\align
& \mu_L^{a_1,a_2}(\{R>0\vert\tilde{\xi}(R;f)\ne\xi(\sqrt{R};f)\}) \tag 3.43\\
&\leq \mu_L^{a_1,a_2}(\{R>0\vert(\tilde{\Pi}_R(f)\Delta\Pi_R(f))\cap\bold Z^2
\ne\phi\}) \\
&\leq \sum_{m\in\Cal Z_{1,n}}\mu_L^{a_1,a_2}(\{R>0\vert\ 
m\in\tilde{\Pi}_R(f)\Delta\Pi_R(f)\})\ .
\endalign$$
If we set for each $m\in\bold Z^2$ with $0\leq\varphi(m)\leq\Theta$ ,
$$J_m=\{R>0\vert\ m\in\tilde{\Pi}_R(f)\Delta\Pi_R(f)\}\ ,\tag 3.44$$
and
$$J_m^{(i)}=\{R>0\vert\ \sqrt{R+2c_i}f(\varphi(m))\leq\vert m\vert\leq
(\sqrt{R}+\frac{c_i}{\sqrt{R}})f(\varphi(m))\}\ ,\ i=1,2\ ,\tag 3.45$$
then $J_m\subset J_m^{(1)}\cup J_m^{(2)}$ , and it suffices for our purpose to 
show
$$\lim_{L\to\infty}\sum_{m\in\Cal Z_{1,n}}\mu_L^{a_1,a_2}(J_m^{(i)})=0\ ,
\ i=1,2\ .\tag 3.46$$
Now it is easy to see that $R\in J_m^{(i)}$ implies
$$\frac{\vert m\vert^2}{f(\varphi(m))^2}-2c_i-\frac{c_1^2}{a_1L}\leq R\leq
\frac{\vert m\vert^2}{f(\varphi(m))^2}-2c_i\ ,\tag 3.47$$
and hence
$$\mu_L^{a_1,a_2}(J_m^{(i)})\leq\frac{c_i^2}{(a_2-a_1)a_1}\frac1{L^2}\ .
\tag 3.48$$
Since the number of lattice points in $\Cal Z_{1,n}$ is bounded by a constant 
times 
$L$ , we finally arrive at
$$\sum_{m\in\Cal Z_{1,n}}\mu_L^{a_1,a_2}(J_m^{(i)})=\Cal O(L^{-1}) \tag 3.49$$
as desired.
\par
Now fix $0<a_1<a_2$ and apply Theorem 3' with $a_1^2<a_2^2$ in use instead of 
$0<a_1<a_2$ . Then from the above argument, for $\bold P$-almost all 
$f\in\Cal X$ , 
one has
$$\lim_{L\to\infty}\frac1{L^2}\int_{a_1^2L^2}^{a_2^2L^2}
1_{\{\xi(\sqrt{R};f)=k\}}dR
=(a_2^2-a_1^2)p_k(f)\ ,
\tag 3.50$$
where we have set for brevity $p_k(f)=e^{-\lambda(f)}\lambda(f)/k!$ . On the 
other hand, the estimate in Lemma 4 is uniform in $a_1$ and $a_2$ such that 
$T^{-1}\leq a_1<a_2\leq T$ , for any $T\geq1$ . Hence 
with the same choice $\beta_n=An^{-1/{\nu_k}}$ as before, we see
$$\sum_{n=1}^{\infty}\int_{a_1<\alpha<T}d\alpha\ \bold E[\Gamma_n(\cdot)]<
\infty \tag 3.51$$
for any $T>a_1$ . This means that for $\bold P-$almost all $f\in\Cal X$ and 
for Lesbesgue 
almost all $\alpha>a_1$ , one has 
$$\lim_{L\to\infty}\frac1{L^2}\int_{a_1^2L^2}^{\alpha L^2}1_{\{\tilde{\xi}
(R;f)=k\}}dR=
\lim_{L\to\infty}\frac1{L^2}\int_{a_1^2L^2}^{\alpha L^2}
1_{\{\xi(\sqrt{R};f)=k\}}dR=
(\alpha-a_1^2)p_k(f)\ ,\ k\geq1\ .\tag 3.52$$ 
We now fix $f\in\Cal X$ for which (3.50) and (3.52) are valid, and letting 
$g_k(R)=1_{\{\xi(R;f)=k\}}$ for brevity, we 
further compute as follows:
$$\align
& \mu_L^{a_1,a_2}(\{R>0\vert\ \xi(R;f)=k\}) \tag 3.53\\
&=\frac1{(a_2-a_1)L}\int_{a_1L}^{a_2L}\ g_k(R)dR \\
&=\frac{1}{2(a_2-a_1)L}\int_{a_1^2L^2}^{a_2^2L^2}\ g_k(\sqrt{s})
\frac{ds}{\sqrt{s}} \\
&=\frac{1}{2(a_2-a_1)a_2}\frac1{L^2}
\int_{a_1^2L^2}^{a_2^2L^2}\ g_k(\sqrt{s})ds
+\frac{1}{2(a_2-a_1)}\int_{a_1}^{a_2}\frac{d\tau}{\tau^2}
\left\{\frac1{L^2}\int_{a_1^2L^2}^{\tau^2L^2}g(\sqrt{s})ds\right\}\ .
\endalign$$
The first term on the right hand side converges to 
$\frac{a_2^2-a_1^2}{2(a_2-a_1)a_2}p_k(f)$ as $L\to\infty$ because of 
(3.50). On the other hand, we have
$$\left\vert\frac{1}{L^2}\int_{a_1^2L^2}^{\tau^2L^2}g_k(\sqrt{s})ds\right\vert
\leq 
a_2^2-a_1^2 \tag 3.54$$
for all $\tau\in[a_1,a_2]$ and
$$\lim_{L\to\infty}\frac1{L^2}\int_{a_1^2L^2}^{\tau^2L^2}g(\sqrt{s})ds
=(\tau^2-a_1^2)p_k(f) \tag 3.55$$
for Lesbesgue almost all $\tau\in[a_1,a_2]$ because of (3.52). Now apply 
Lesbesgue's 
dominated convergence theorem to the second term on the right hand side, to 
obtain 
$$\align
& \lim_{L\to\infty}\mu_L^{a_1,a_2}(\{R>0\vert\ \xi(R;f)=k\}) \tag 3.56\\
&=\frac{a_2^2-a_1^2}{2(a_2-a_1)a_2}p_k(f)+
\frac1{2(a_2-a_1)}\int_{a_1}^{a_2}\frac{d\tau}{\tau^2}(\tau^2-a_1^2)p_k(f) \\
&=p_k(f)\ ,
\endalign$$
completing the proof of Theorem 3.
\par
Finally let $\{\bar{L}_n\}$ be the sequence taken according to Theorem 2'. 
We can repeat 
the above argument to see that the sequence $\{\sqrt{\bar{L}_n}\}$ meets the 
requirement of Theorem 2.

\head {\bf \S4 Proof of Lemma 4} \endhead
Let us fix an $i\in\{1,\ldots,1+q_n\}$ and a $T>1$ and assume 
$T^{-1}\leq a_1<a_2\leq T$ . We make the following definitions: 

$$\align
& Q^i(m_1,\ldots,m_k;m'_1,\ldots,m'_k)=Q^i(\bold m;\bold m') \tag 4.1\\
&=\bold E_{\bold P}[1_{\{\gamma_{m_1},\gamma'_{m_1}\in[z_{i-1},z_i)\}}
\vert D_{m_1}\cap\cdots\cap D_{m_k}\vert\ 
\vert D_{m'_1}\cap\cdots\cap D_{m'_k}\vert]\ ,
\endalign$$
where we write $(\bold m,\bold m')=(m_1,\ldots,m_k;m'_1,\ldots,m'_k)$ 
for brevity;
$$X^i_n=\sum_{(\bold m;\bold m')\in
\Cal Z_{k,n}\times\Cal Z_{k,n}}
Q^i(\bold m;\bold m')\ ;
\tag 4.2$$
$$\align
& R^i(m_1,\ldots,m_1;m'_1)=R^i(\bold m;m'_1) \tag 4.3\\
&=\bold E_{\bold P}[
1_{\{\gamma_{m_1},\gamma'_{m_1}\in[z_{i-1},z_i)\}}
\vert D_{m_1}\cap\cdots\cap D_{m_k}\vert\times\\
&\qquad\times\left.\left(\int_{\varphi(m'_1)}^{\Theta}d\varphi'_2
\int_{\varphi'_2}^{\Theta}d\varphi'_3\cdots
\int_{\varphi'_{k-1}}^{\Theta}d\varphi'_k\ 
f(\varphi'_2)^2\cdots f(\varphi'_k)^2\right)\right]\ ;
\endalign$$
$$Y^i_n=\sum_{(\bold m;m'_1)\in
\Cal Z_{k,n}\times\Cal Z_{1,n}}R^i(\bold m;m'_1)\ ;
\tag 4.4$$
$$\align
& S^i(m_1,m'_1)
=\bold E_{\bold P}[
1_{\{\gamma_{m_1},\gamma'_{m_1}\in[z_{i-1},z_i)\}}\times \tag 4.5\\
&\qquad\times\left(\int_{\varphi(m_1)}^{\Theta}d\varphi_2
\int_{\varphi_2}^{\Theta}d\varphi_3\cdots
\int_{\varphi_{k-1}}^{\Theta}d\varphi_k\ 
f(\varphi_2)^2\cdots f(\varphi_k)^2\right)\times\\
&\qquad\times\left.\left(\int_{\varphi(m'_1)}^{\Theta}d\varphi'_2
\int_{\varphi'_2}^{\Theta}d\varphi'_3\cdots
\int_{\varphi'_{k-1}}^{\Theta}d\varphi'_k\ 
f(\varphi'_2)^2\cdots f(\varphi'_k)^2\right)\right]\ ;
\endalign$$
$$Z^i_n=\sum_{(m_1,m'_1)\in
\Cal Z_{1,n}\times\Cal Z_{1,n}}S^i(m_1,m'_1)\ ;
\tag 4.6$$
Then
$$U_{i,n}=X_n^i-4(c_2-c_1)^kY_n^i+4(c_2-c_1)^{2k}Z_n^i\ . \tag 4.7$$
We shall prove that for some positive constant $C_T$  which depends only 
on $T$ , the following two 
inequlities hold as far as we have $T^{-1}\leq a_1<a_2\leq T$ :
$$\vert\ X_n^i-4(c_2-c_1)^{2k}Z_n^i\vert\leq 
C_T(L_n^{3/2}q_n^{-2}(\Psi_{2k}(\beta)+\sqrt{L_n}\beta^d))\ ;\tag4.8$$
and
$$\vert\ Y_n^i-2(c_2-c_1)^{k}Z_n^i\vert\leq 
C_T(L_n^{3/2}q_n^{-2}(\Psi_{2k}(\beta)+\sqrt{L_n}\beta^d))\ ,\tag4.9$$
from which the assertion of Lemma 4 follows immediately.
\par
At this point, we make the following convention. We denote by $C$ or 
$C_T$ positive constants, of which we are not interested in exact values, 
and which may differ from inequality to inequality. We distinguish $C_T$ 
from $C$ when it is depndent on $T$ , while $C$ does not depend on 
$a_1$ and $a_2$ . Similarly, when something is bounded by $C$ [resp. $C_T$] 
times something else, we shall say it is of $\Cal O(\cdot)$ 
[resp. $\Cal O_T(\cdot)$].
\par
In order to estimate $X^i_n$ , let us make the representation (4.1) of 
$Q^i(\bold m;\bold m')$ more explicit. To begin with, from the 
definition (3.8) of $\gamma_m$ , we see that
$$\gamma_{m_1}\in[z_{i-1},z_i)\ \Longleftrightarrow\ 
\frac{\vert m_1\vert}{\sqrt{z_i}}<f(\varphi(m_1))\leq
\frac{\vert m_1\vert}{\sqrt{z_{i-1}}}\ ,\tag 4.10$$
where
$$\frac{\vert m_1\vert}{\sqrt{z_{i-1}}}-\frac{\vert m_1\vert}{\sqrt{z_i}}
=\frac{\vert m_1\vert}{\sqrt{z_iz_{i-1}}}
\left(\frac{z_i-z_{i-1}}{\sqrt{z_i}+\sqrt{z_{i-1}}}\right)
=\Cal O((a_2/a_1)^{3/2}q_n^{-1})=\Cal O_T(q_n^{-1})\ ,\tag 4.11$$
if $m_1\in\Cal Z_{1,n}$ . 
Recall (3.9) and $a_1L_n\leq z_i\leq a_2L_{n+1}$ . 
Here we have simply written $\Cal O_T$ because 
the dependence of the estimate on $a_1$ and $a_2$ need not be made explicit. 
Next, writing $x_{+}=\max\{x,0\}=x\vee0$ for $x\in\bold R$ , we obtain 
from (3.7) and (3.8)
$$\vert D_{m_1}\cap\cdots\cap D_{m_k}\vert
=\left\{A-0\vee\max_{2\leq j\leq k}(\gamma_{m_j}-\gamma_{m_1})
+0\wedge\min_{2\leq j\leq k}(\gamma_{m_j}-\gamma_{m_1})\right\}_+\ ,
\tag 4.12$$
where we have let $A=2(c_2-c_1)$ for brevity. 
Letting $y_s=f(\varphi(m_s))$ and $y'_t=f(\varphi(m'_t))$ , 
we can write, when $\varphi(m_s)$ , $\varphi(m'_t)$ , $s,t=1,\ldots,k$ 
are all different,
$$\align
& Q^i(\bold m;\bold m')\tag 4.13\\
&=\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
\int_0^{\infty}dy_2\cdots\int_0^{\infty}dy_k\int_0^{\infty}dy'_2\cdots
\int_0^{\infty}dy'_k
\times\\
&\times\left\{A-0\vee\max_{2\leq s\leq k}
\left(\frac{\vert m_s\vert^2}{y_s^2}-\frac{\vert m_1\vert^2}{y_1^2}\right)
+0\wedge\min_{2\leq s\leq k}
\left(\frac{\vert m_s\vert^2}{y_s^2}-\frac{\vert m_1\vert^2}{y_1^2}\right)
\right\}_+\times\\
&\times\left\{A-0\vee\max_{2\leq t\leq}
\left(\frac{\vert m'_t\vert^2}{y_t^{\prime2}}
-\frac{\vert m'_1\vert^2}{y_1^{\prime2}}\right)
+0\wedge\min_{2\leq t\leq}
\left(\frac{\vert m'_t\vert^2}{y_t^{\prime2}}
-\frac{\vert m'_1\vert^2}{y_1^{\prime2}}\right)
\right\}_+\times\\
&\times p_{2k}(y_1,\ldots,y_k,y'_1,\ldots,y'_k\ \vert\ 
\varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_1),\ldots,\varphi(m'_k))
\endalign$$
Note that $p_{2k}$ , as a function of $y's$ , is 
supported on the set of $(y_1,\ldots,y_k,y'_1,\ldots,y'_k)$ such that 
$b_1\leq y_s,y_t\leq b_2$ , $s,t=1,\ldots,k$ .
\par
Let us make the following change of variables:
$$\zeta_s=\frac{\vert m_s\vert^2}{y_s^2}-\frac{\vert m_1\vert^2}{y_1^2}\ ;\ 
\zeta'_s=
\frac{\vert m'_s\vert^2}{y_s^{\prime2}}-\frac{\vert m'_1\vert^2}{y_1^{\prime2}}
\ ,\ s=2,\ldots,k\ .\tag 4.14$$
Then we have
$$y_s=\frac{\vert m_s\vert}{\ell_s}\ ,\ y'_s=\frac{\vert m'_s\vert}{\ell'_s}\ ,
\quad s=2,\ldots,k\tag 4.15$$
and
$$dy_s=-\frac12\frac{\vert m_s\vert}{\ell^3_s}d\zeta_s\ ,\ 
dy'_s=-\frac12\frac{\vert m_s\vert}{\ell^{\prime 3}_s}d\zeta'_s\ ,
\quad s=2,\ldots,k\tag 4.16$$
where we have defined
$$\ell_s=\ell_s(y_1,m_1,\zeta_s)=
\frac{\vert m_1\vert}{y_1}\sqrt{1+\frac{y^2_1}{\vert m_1\vert^2}\zeta_s}\ ,
\tag 4.17$$
and similarly for $\ell'_s=\ell'_s(y'_1,m'_1,\zeta'_s)$ . (4.13) can now be 
written as follows:
$$\align
& Q^i(\bold m;\bold m')\tag 4.18\\
&=\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
\int_{-A}^{A}d\zeta_2
\cdots\int_{-A}^{A}d\zeta_k
\int_{-A}^{A}d\zeta'_2\cdots
\int_{-A}^{A}d\zeta'_k\times\\
&\times\left\{A-(0\vee\max_{2\leq s\leq k}\zeta_s)
+(0\wedge\min_{2\leq s\leq k}\zeta_s)
\right\}_+
\left\{A-(0\vee\max_{2\leq t\leq k}\zeta_s)
+(0\wedge\min_{2\leq t\leq k}\zeta_s)
\right\}_+\times\\
&\times\prod_{s=2}^k\left\{\left(
\frac12\left\vert\frac{m_s}{\ell_s}\right\vert\right)
\left(
\frac12\left\vert\frac{m'_s}{\ell'_s}\right\vert\right)
\frac1{\ell^2_s}\frac1{\ell^{\prime 2}_s}\right\}\times\\
&\times p_{2k}(y_1,\left\vert\frac{m_2}{\ell_2}\right\vert,
\ldots,\left\vert\frac{m_k}{\ell_k}\right\vert,y'_1,
\left\vert\frac{m'_2}{\ell'_2}\right\vert,\ldots,
\left\vert\frac{m'_k}{\ell'_k}\right\vert\ \vert\ 
\varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_1),\ldots,\varphi(m'_k))
\endalign$$
Here the variables $\zeta$'s range in  $(-A,A)$ 
because the integrand vanishes if at least one of them is outside of 
this interval.
\par
Given $\beta>L_n^{-1/2}$ , define a subset $\Cal W_n(\beta)$ of 
$\Cal Z_{k,n}\times\Cal Z_{k,n}$ to be the totality of those 
$(\bold m;\bold m')\mathbreak=(m_1,\ldots,m_k;m'_1,\ldots,m'_k)$ 
such that
$$\varphi(m_s)-\varphi(m_{s-1})\geq\beta\ ,\ 
\varphi(m'_s)-\varphi(m'_{s-1})\geq\beta\ ,\ s=2,\ldots,k \tag 4.19$$
and that
$$\vert\varphi(m_s)-\varphi(m'_t)\vert\geq\beta\ ,\ s,t=1,\ldots,k\ .
\tag 4.20$$
Then
$$X_n^i=\left(\sum_{\Cal W_n(\beta)}\ +
\ \sum_{(\Cal Z_{k,n}\times\Cal Z_{k,n})\setminus\Cal W_n(\beta)}\right)
Q^i(\bold m;\bold m')\ . \tag 4.21$$
Let $X_n^i(\beta)$ be the first sum on the right hand side of (4.21). 
If for given $m_1,m'_1\in\Cal Z_{1,n}$ with 
$\vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta$ one defines
$$\align
&\ \Cal W_n(\beta;m_1,m'_1) \tag 4.22 \\
&=\{(m_2,\ldots,m_k;m'_2,\ldots,m'_k)\ ;\ 
(m_1,m_2,\ldots,m_k;m'_1,m'_2,\ldots,m'_k)\in\Cal W_n(\beta)\}\ ,
\endalign$$
then
$$X_n^i(\beta)=\sum\Sb m_1,m'_1\in\Cal Z_{1,n}
\\ \vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta\endSb
\sum_{\Cal W_n(\beta;m_1,m'_1)}Q^i(m_1,\ldots,m_k;m'_1,\ldots,m'_k)\ .
\tag 4.23$$
Fix $m_1,m'_1\in\Cal Z_{1,n}$ such that
$$\vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta \tag 4.24$$
and $y_1$ , $y'_1$ such that
$$\frac{\vert m_1\vert}{\sqrt{z_i}}\leq y_1\leq
\frac{\vert m_1\vert}{\sqrt{z_{i-1}}}\ ;\ 
\frac{\vert m'_1\vert}{\sqrt{z_i}}\leq y'_1\leq
\frac{\vert m'_1\vert}{\sqrt{z_{i-1}}}\ . \tag 4.25$$
Take also $\zeta_s$ , $\zeta'_s$ , $s=2,\ldots,k$ from the interval 
$(-A,A)$ .
Then for some positive constant $C_T>0$ , it holds that
$$C_T^{-1}L_n^{1/2}\leq\ell_s\ ,\ \ell'_s\leq C_T L_n^{1/2}\ ,\ s=2,\ldots,k
\ . \tag 4.26$$
Now we shall evaluate approximately the sum
$$\align
& \sum\Sb (m_2,\ldots,m_k;m'_2,\ldots,m'_k)
 \\ \in\Cal W_n(\beta;m_1,m'_1) \endSb
\ \prod_{s=2}^k\left\{\left(
\frac12\left\vert\frac{m_s}{\ell_s}\right\vert\right)
\left(
\frac12\left\vert\frac{m'_s}{\ell'_s}\right\vert\right)
\frac1{\ell^2_s}\frac1{\ell^{\prime 2}_s}\right\}\times \tag 4.27\\
&\times p_{2k}(y_1,\left\vert\frac{m_2}{\ell_2}\right\vert,
\ldots,\left\vert\frac{m_k}{\ell_k}\right\vert,y'_1,
\left\vert\frac{m'_2}{\ell'_2}\right\vert,\ldots,
\left\vert\frac{m'_k}{\ell'_k}\right\vert\ \vert\ 
\varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_1),\ldots,\varphi(m'_k))\ ,
\endalign$$
the approximation being uniform in $\zeta_2,\ldots,\zeta_k$ , 
$\zeta'_2,\ldots,\zeta'_k$ . Note that the summand appears as a part of 
the integrand in (4.18).
\par
For this purpose, we introduce a function
$$\align
F&=F_{m_1,m'_1,y_1,y'_1}(x_2,\ldots,x_k;x'_2,\ldots,x'_k) \tag 4.28\\
&\equiv\prod_{s=2}^k(\frac{\vert x_s\vert}{2}\frac{\vert x'_s\vert}{2})
p_{2k}(y_1,\vert x_2\vert,\ldots,\vert x_k\vert,y'_1,\vert x'_2\vert,
\ldots,\vert x'_k\vert\ \left\vert\ \right. \\
&\quad \varphi(m_1),\varphi(x_2),
\ldots,\varphi(x_k),\varphi(m'_1),\varphi(x'_2),
\ldots,\varphi(x'_k))
\endalign$$
which is defined on the set
$$\align
& \Cal D_{m_1,m'_1}\equiv\{(x_2,\ldots,x_k;x'_2,\ldots,x'_k)
\in(\bold R^2)^{2(k-1)}\ \left\vert\ b_1\leq\vert x_s\vert,\vert x'_s\vert
\leq b_2\ ,\right. \tag 4.29 \\
&\qquad \varphi(m_1)<\varphi(x_2)<\cdots<
\varphi(x_k)\ ;\ \varphi(m'_1)<\varphi(x'_2)<
\cdots<\varphi(x'_k))\}\ .
\endalign$$
It is easy to see that for some $C>0$ ,
$$\left\Vert\frac{\partial F}{\partial x_j}\right\Vert\ ,\ 
\left\Vert\frac{\partial F}{\partial x'_j}\right\Vert\ \leq\ 
C(\psi_{2k}(\theta)+\Psi_{2k}(\theta)) \tag 4.30$$
holds on $\Cal D_{m_1,m'_1}$ , 
where $\theta$ is the minimum of $\varphi(x_2)-\varphi(m_1)$ , 
$\varphi(x_3)-\varphi(x_2)\ldots$ etc..
\par
In view of Conditions (II) and (III) , we may suppose 
$\psi_{2k}(\theta)<\Psi_{2k}(\theta)$ , so that (4.30) is actually 
$$\left\Vert\frac{\partial F}{\partial x_j}\right\Vert\ ,\ 
\left\Vert\frac{\partial F}{\partial x'_j}\right\Vert\ \leq\ 
C\Psi_{2k}(\theta)\ . \tag 4.31$$
The sum (4.27) can then be viewed as a Riemann sum approximation for the 
integral of the function $F$ , and (4.26), (4.31) can be used to estimate its 
accuracy. Namely (4.27) is equal to
$$\align
& \sum\Sb (m_2,\ldots,m_k;m'_2,\ldots,m'_k)
 \\ \in\Cal W_n(\beta;m_1,m'_1) \endSb
\left(\prod_{s=2}^k \frac1{\ell^2_s}\frac1{\ell^{\prime 2}_s}\right)
F_{m_1,m'_1,y_1,y'_1}(\frac{m_2}{\ell_2},\ldots,\frac{m_k}{\ell_k},
\frac{m'_2}{\ell'_2},\ldots,\frac{m'_k}{\ell'_k}) \tag 4.32 \\
&=\int dx_2\cdots\int dx_k\int dx'_2\cdots\int dx'_k 
F_{m_1,m'_1,y_1,y'_1}(x_2,\ldots,x_k,x'_2,\ldots,x'_k)\\
&\quad+\Cal O_T(L_n^{-1/2}\Psi_{2k}(\beta))\ ,
\endalign$$
where the integration ranges over those $(x_2,\ldots,x_k,x'_2,\ldots,x'_k)
\in\Cal D_{m_1,m'_1}$ for which
$$\align
& \varphi(x_2)-\varphi(m_1)\geq\beta\ ,
\ \varphi(x_s)-\varphi(x_{s-1})\geq\beta\ ,\ s=3,\ldots,k\ ; \tag 4.33 \\
& \varphi(x'_2)-\varphi(m'_1)\geq\beta\ ,
\ \varphi(x'_t)-\varphi(x'_{t-1})\geq\beta\ ,\ t=3,\ldots,k\ ;\\
& \vert\varphi(x_s)-\varphi(x'_t)\vert\geq\beta\ ,\ s,t=2,\ldots,k\ .
\endalign$$
Transforming each of $x_s$ and $x'_s$ into the polar coordinates, the above 
integral becomes
$$\align
& \int d\varphi_2\cdots\int d\varphi_k\int d\varphi'_2
\cdots\int d\varphi'_k
\int_0^{\infty} dr_2\cdots\int_0^{\infty} dr_k
\int_0^{\infty} dr'_2\cdots\int_0^{\infty} dr'_k
\times \tag 4.34\\
&\times \prod_{s=2}^k\left(\frac{r_s^2}2\frac{r_s^{\prime 2}}2\right)\times \\
&\times p_{2k}(y_1,r_2,\ldots,r_k,y'_1,r'_2,\ldots,r'_k\ \vert\ 
\varphi(m_1),\varphi_2,\ldots,\varphi_k,\varphi(m'_1),
\varphi'_2,\ldots,\varphi'_k)\ ,
\endalign$$
where the integration ranges over those $\varphi_s$ , 
$\varphi'_s$ such that
$$\align
& \varphi(x_2)-\varphi(m_1)\geq\beta\ ,
\ \varphi(x_s)-\varphi(x_{s-1})\geq\beta\ ,\ s=3,\ldots,k\ ; \tag 4.35 \\
& \varphi(x'_2)-\varphi(m'_1)\geq\beta\ ,
\ \varphi(x'_t)-\varphi(x'_{t-1})\geq\beta\ ,\ t=3,\ldots,k\ ;
\endalign$$
and that
$$\vert\varphi_s-\varphi'_t\vert\geq\beta\ ,\ s,t=1,\ldots,k\ .\tag4.36$$
\par
If we integrate out with respect to the variables $r_s$ , $r'_s$ , 
$s=2,\ldots,k$ , we see without difficulty that the difference made by 
letting $\beta=0$ in the range of integration of (4.34) is less than
$$C\beta p_2(y_1,y'_1\ \vert\ \varphi(m_1),\varphi(m'_1)\ .\tag 4.37$$
We have thus approximated the sum (4.27) uniformly in the parameters 
$\zeta_2,\ldots,\zeta_k$ , $\zeta'_2,\ldots,\zeta'_k$ . In order to 
finish the estimation of $X^i_n(\beta)$ (recall (4.23) for the definition), 
we need to get an approximation of the sum of 
$Q^i(m_1,\ldots,m_k;m'_1,\ldots,m'_k)$ over $\Cal W(\beta;m_1,m'_1)$ . 
This is done by integrating the sum (4.27) over parameters $y_1$ , 
$y'_1$ , $\zeta_2,\ldots,\zeta_k$ , $\zeta'_2,\ldots,\zeta'_k$ . By noting 
the above mentioned uniformity of the integral-approximation in 
$\zeta_s$ and $\zeta'_s$ , and by using the formula
$$\int_{-A}^{A}d\zeta_2\cdots
\int_{-A}^{A}d\zeta_k
\{A-(0\vee\zeta_2\vee\cdots\vee\zeta_k)+
(0\wedge\zeta_2\wedge\cdots\wedge\zeta_k)\}_+
=A^k\ ,\ k\geq2\ ,\tag 4.38$$
which we shall prove as Lemma 5 at the end of this section, we obtain
$$\align
& X_n^i(\beta)=\sum\Sb m_1,m'_1\in\Cal Z_{1,n}
\\ \vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta\endSb
A^{2k}
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
\times \tag 4.39\\
&\times\left\{\idotsint_{\varphi(m_1)<\varphi_2<\cdots<\varphi_k}
d\varphi_2\cdots d\varphi_k
\idotsint_{\varphi(m'_1)<\varphi'_2<\cdots<\varphi'_k}
d\varphi'_2\cdots d\varphi'_k\times \right.\\
&\times\int_0^{\infty}dr_2\cdots\int_0^{\infty}dr_k
\int_0^{\infty}dr'_2\cdots\int_0^{\infty}dr'_k 
\prod_{s=2}^k\left(\frac{r_s^2}2\frac{r^{\prime 2}_s}2\right)\times\\
&\times p_{2k}(y_1,r_2,\ldots,r_k,y'_1,r'_2,\ldots,r'_k\ \vert\ 
\varphi(m_1),\varphi_2,\ldots,\varphi_k,\varphi(m'_1),
\varphi'_2,\ldots,\varphi'_k) +\\
&\left.\quad+\Cal O_T(L_n^{-1/2}\Psi_{2k}(\beta)+\beta p_2(y_1,y'_1\vert
\varphi(m_1),\varphi(m'_1))\right\} \\
&=4(c_2-c_1)^{2k}\sum\Sb m_1,m'_1\in\Cal Z_{1,n}
\\ \vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta\endSb S^i(m_1,m'_1)
+\Cal O_T(L_n^{3/2}q_n^{-2}\Psi_{2k}(\beta)) \\
&+\Cal O\left(\sum\Sb m_1,m'_1\in\Cal Z_{1,n}
\\ \vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta\endSb \beta
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
p_2(y_1,y'_1\ \vert\ \varphi(m_1),\varphi(m'_1))\right)\ ,
\endalign$$
where we have used (4.11) and $\sharp(\Cal Z_{1,n}\times\Cal Z_{1,n})=
\Cal O_T(L_n)$ 
to get the bound of the first error term. See also (4.5) for the definition 
of $S^i(m_1,m'_1)$ . In order to 
estimate the second error term, let us first consider the case where our 
probability measure $\bold P$ satisfies Condition (II-1). Then the sum 
$$\sum\Sb m_1,m'_1\in\Cal Z_{1,n}
\\ \vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta\endSb \beta
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
p_2(y_1,y'_1\ \vert\ \varphi(m_1),\varphi(m'_1)) \tag 4.40$$
is bounded by
$$\align 
& C\beta q_n^{-2}\sum\Sb m_1,m'_1\in\Cal Z_{1,n}
\\ \vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta\endSb
\vert\varphi(m_1)-\varphi(m'_1)\vert^{-\sigma} \tag 4.41 \\
&\leq C\beta\beta^{-\sigma}q_n^{-2}\sum_{m_1,m'_1\in\Cal Z_{1,n}}\ 1 \\
&\leq C_T L_n^2q_n^{-2}\beta^{1-\sigma}\ .
\endalign$$
\par
Next suppose that $\bold P$ satisfies Condition (II-2). Then we have 
$$p_2(y_1,y'_1\ \vert\ \varphi(m_1),\varphi(m'_1))\leq 
C1_{\{\vert y_1-y'_1\vert\leq b_3\vert\varphi(m_1)-\varphi(m'_1)\vert\}}
\vert\varphi(m_1)-\varphi(m'_1)\vert^{-\tau}\ . \tag 4.42$$
Then the sum (4.40) is bounded by 
$$\align
& C\beta\int_{b_1}^{b_2}dy_1\int_{b_1}^{b_2}dy'_1
\sum_{m_1,m'_1\in\Cal Z_{1,n}}
1_{\{y_1\sqrt{z_{i-1}}\leq\vert m_1\vert y_1\leq\sqrt{z_i}\}}\times \tag 4.43\\
&\quad\times1_{\{y'_1\sqrt{z_{i-1}}\leq\vert m'_1\vert\leq y'_1\sqrt{z_i}\}}
1_{\{\vert y_1-y'_1\vert\vee(b_3\beta)\leq
\vert\varphi(m_1)-\varphi(m'_1)\vert \}}\times \\
&\quad\times\vert\varphi(m_1)-\varphi(m'_1)\vert^{-\tau}\ .
\endalign$$
For fixed $y_1,y'_1$ and $m_1$ , we shall estimate 
$$\sum_{m'_1\in\Cal Z_{1,n}}
1_{\{y'_1\sqrt{z_{i-1}}\leq\vert m'_1\vert\leq y'_1\sqrt{z_i}\}}
1_{\{\vert y_1-y'_1\vert\vee(b_3\beta)\leq
\vert\varphi(m_1)-\varphi(m'_1)\vert \}}
\vert\varphi(m_1)-\varphi(m'_1)\vert^{-\tau}\ .\tag 4.44$$
It is sufficient to do this assuming $\varphi(m_1)>\varphi(m'_1)$ . 
To this end, let us divide the interval $[\beta\vee(\vert y_1-y'_1\vert/b_3),
\Theta]$ into subintervals $\Delta_j$ of length $L_n^{-1/2}$ . Since the 
number of those $m'_1\in\Cal Z_{1,n}$ for which 
$y'_1\sqrt{z_{i-1}}\leq\vert m'_1\vert\leq y'_1\sqrt{z_i}$ and 
$\varphi(m'_1)\in\Delta_j$ hold is bounded by $C_T L_n^{1/2}q_n^{-1}$ , 
the sum (4.44) is less than 
$$\align
& C_T L_n^{1/2}q_n^{-1}\sum_{j=1}^{\infty}
\{jL_n^{-1/2}+(\beta\vee\frac{\vert y_1-y'_1\vert}{b_3})\}^{-\tau} \tag 4.45\\
&\leq C_T L_nq_n^{-1}\int_{\beta\vee(\frac{\vert y_1-y'_1\vert}{b_3})}^{\infty}
t^{-\tau}dt \\
&=C_T L_nq_n^{-1}(\beta^{1-\tau})\wedge(\vert y_1-y'_1\vert^{1-\tau})\ .
\endalign$$
Inserting this estimate into (4.43), 
we obtain the following new bound of (4.40):
$$\align
& C_T\beta\int_{b_1}^{b_2}dy_1\int_{b_1}^{b_2}dy'_1
\left(\sum_{m_1\in\Cal Z_{1,n}}
1_{\{y_1\sqrt{z_{i-1}}\leq\vert m_1\vert\leq y_1\sqrt{z_i}\}}
\right) L_nq_n^{-1}\{\beta^{1-\tau}\wedge\vert y_1-y'_1\vert^{1-\tau}\} 
\tag4.46\\
&\leq C_T\beta(L_nq_n^{-1})^2\int_{b_1}^{b_2}dy_1\int_{b_1}^{b_2}dy'_1
\{\beta^{1-\tau}\wedge\vert y_1-y'_1\vert^{1-\tau}\} \\
&=C_T L_n^2q_n^{-2}\beta^{2-\tau}\ .
\endalign$$
Returning to (4.39), we have thus proved
$$X_n^i(\beta)=4(c_2-c_1)^{2k}\sum\Sb m_1,m'_1\in\Cal Z_{1,n}
\\ \vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta\endSb S^i(m_1,m'_1)+
\Cal O_T(L_n^{3/2}q_n^{-2}\Psi_{2k}(\beta)+L_n^2q_n^{-2}\beta^d)\ ,
\tag 4.47$$
where $d=1-\sigma$ or $d=2-\tau$ . (Recall $A=2(c_2-c_1)$ .)
\par
To finish the discussion on $X_n^i(\beta)$ , we estimate the error
$$\sum\Sb m_1,m'_1\in\Cal Z_{1,n}
\\ \vert\varphi(m_1)-\varphi(m'_1)\vert<\beta\endSb S^i(m_1,m'_1) \tag 4.48$$
which is made by dropping the condition 
$\vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta$ under the summation symbol 
in (4.47). For this purpose, we divide the sum (4.48) into two parts, namely 
we rewrite (4.48) as 
$$\sum_{0<\vert\varphi(m_1)-\varphi(m'_1)\vert<\beta}S^i(m_1,m'_1)+
\sum_{\varphi(m_1)=\varphi(m'_1)}S^i(m_1,m'_1)\equiv J_1+J_2 \tag 4.49$$
when Condition (II-1) holds, or
$$\sum_{L_n^{-1/2}\leq\vert\varphi(m_1)-\varphi(m'_1)\vert<\beta}S^i(m_1,m'_1)+
\sum_{\vert\varphi(m_1)-\varphi(m'_1)\vert<L_n^{-1/2}}S^i(m_1,m'_1)
\equiv K_1+K_2 \tag 4.50$$
when Condition (II-2) holds.
\par
Note that if $\varphi(m_1)\ne\varphi(m'_1)$
$$\align
S^i(m_1,m'_1)&\leq C\bold P(\gamma_{m_1},\gamma_{m'_1}\in[z_{i-1},z_i)) 
\tag 4.51 \\
&=C \int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
p_2(y_1,y'_1\ \vert\ \varphi(m_1),\varphi(m'_1))\ .
\endalign$$
Suppose that Condition (II-1) is satisfied. Then by (4-11), 
$$S(m_1,m'_1)\leq C_T q_n^{-2}\vert\varphi(m_1)-\varphi(m'_1)\vert^{-\sigma}
\tag 4.51$$
with $\sigma\in(0,1)$ . Hence applying Lemma 1 with $R=L_n^{1/2}$ and 
$\beta>L_n^{-1/2}$ , we see
$$J_1\leq C_T L_n^2q_n^{-2}\beta^{1-\sigma}\ . \tag 4.52$$
On the other hand, if $\varphi(m_1)=\varphi(m'_1)$ , one will have
$$S(m_1,m'_1)\leq C\int_{b_1}^{b_2}dy_1
1_{\{y_1\sqrt{z_{i-1}}\leq\vert m_1\vert,\vert m'_1\vert \leq y_1\sqrt{z_i}\}}
\ . \tag 4.53$$
Since $y_1(\sqrt{z_{i}}-\sqrt{z_{i-1}})=\Cal O_T(L_n^{1/2}q_n^{-1})$ , we have
$$J_2\leq C_T L_n^{3/2}q_n^{-2}\ . \tag 4.54$$
Next suppose that Condition (II-2) holds. Then if 
$\varphi(m_1)\ne\varphi(m'_1)$ ,
$$\align
S(m_1,m'_1)&\leq C\int_{b_1}^{b_2}dy_1\int_{b_1}^{b_2}dy'_1
1_{\{y_1\sqrt{z_{i-1}}\leq\vert m_1\vert \leq y_1\sqrt{z_i}\}}
1_{\{y'_1\sqrt{z_{i-1}}\leq\vert m'_1\vert \leq y'_1\sqrt{z_i}\}}\times 
\tag4.55\\
&\quad\times 
1_{\{\vert y_1-y'_1\vert\leq b_3\vert\varphi(m_1)-\varphi(m'_1)\vert\}}
\vert\varphi(m_1)-\varphi(m'_1)\vert^{-\tau}\ .
\endalign$$
Hence we have
$$\align
K_1&\leq C\int_{b_1}^{b_2}dy_1\int_{b_1}^{b_2}dy'_1
1_{\{\vert y_1-y'_1\vert\leq\beta\}}
\sum_{m_1\in\Cal Z_{1,n}}
1_{\{y_1\sqrt{z_{i-1}}\leq\vert m_1\vert\leq y_1\sqrt{z_i}\}}\times \tag 4.56\\
&\quad\times\sum_{m'_1\in\Cal Z_{1,n}}
1_{\{y'_1\sqrt{z_{i-1}}\leq\vert m'_1\vert\leq y'_1\sqrt{z_i}\}}
1_{\{\vert y_1-y'_1\vert\vee(b_3 L_n^{-1/2})\leq
\vert\varphi(m_1)-\varphi(m'_1)\vert \}} \\
&\qquad\times\vert\varphi(m_1)-\varphi(m'_1)\vert^{-\tau}\ .
\endalign$$
Let us estimate the summation over $m'_1$ . We can assume $\varphi(m'_1)
>\varphi(m_1)$ . As we did in the estimation of the sum (4.44) , we divide 
the interval $[L_n^{-1/2}\vee(\frac1{b_3}\vert y_1-y'_1\vert),\Theta]$ into 
subintervals of length $L_n^{-1/2}$ , and obtain
$$\align
& \sum_{m'_1\in\Cal Z_{1,n}}
1_{\{y'_1\sqrt{z_{i-1}}\leq\vert m'_1\vert\leq y'_1\sqrt{z_i}\}}
1_{\{\vert y_1-y'_1\vert\vee(b_3 L_n^{-1/2})\leq
\vert\varphi(m_1)-\varphi(m'_1)\vert \}}
\vert\varphi(m_1)-\varphi(m'_1)\vert^{-\tau} \tag 4.57 \\
&\leq C_T L_n^{1/2}q_n^{-1}\sum_{j=0}^{\infty}
\{(L_n^{-1/2}\vee\frac1{b_3}\vert y_1-y'_1\vert)+jL_n^{-1/2}\}^{-\tau} \\
&\leq C_T L_n q_n^{-1}(L_n^{-1/2}\vee\frac1{b_3}\vert y_1-y'_1\vert)^{1-\tau}
\ .\endalign$$
Inserting this inequality into (4.56), one easily get the bound
$$K_1\leq C_T L_n^2q_n^{-2}\beta^{2-\tau}\ . \tag 4.58$$
We now turn to estimating $K_2$ . By $\sqrt{z_i}\leq C_T L_n^{1/2}$ and 
$\vert\varphi(m_1)-\varphi(m'_1)\vert\leq L_n^{-1/2}$ , one easily sees 
that there is a constant $C_T>0$ such that the condition,
$$f(\varphi(m'_1))\sqrt{z_{i-1}}\leq\vert m'_1\vert \leq 
f(\varphi(m'_1))\sqrt{z_i} \tag 4.59$$
implies
$$f(\varphi(m_1))\sqrt{z_{i-1}}-C_T\leq\vert m'_1\vert \leq 
f(\varphi(m_1))\sqrt{z_i}+C_T\ . \tag 4.60$$
For each fixed $m_1\in\Cal Z_{1,n}$ , the number of $m'_1$ which satisfies 
the condition (4.60) and $\vert\varphi(m_1)-\varphi(m'_1)\vert<L_n^{-1/2}$ 
is again bounded by $C_T L_n^{1/2}q_n^{-1}$ . Hence
$$\align
\sum \Sb m_1,m'_1\in\Cal Z_{1,n} \\
\vert\varphi(m_1)-\varphi(m'_1)\vert<L_n^{-1/2} \endSb S(m_1,m'_1)
&\leq C_TL_n^{1/2}q_n^{-1}\sum_{m_1\in\Cal Z_{1,n}}
\bold P(\gamma_{m_1}\in[z_{i-1},z_i)) \tag 4.61 \\
&\leq C_T L_n^{3/2}q_n^{-2}\ .
\endalign$$
We have thus shown that the sum (4.48) is bounded by 
$$C_T(L_n^{3/2}q_n^{-2}+L_n^2q_n^{-2}\beta^d) \tag 4.62$$
with $d=1-\sigma$ or $d=2-\tau$ . Since $1=\Cal O(\Psi_{2k}(\beta))$ , we 
finally arrive at 
$$X_n^i(\beta)=4(c_2-c_1)^{2k}
\sum_{m_1,m'_1\in\Cal Z_{1,n}}S(m_1,m'_1)
+\Cal O_T(L_n^{3/2}q_n^{-2}(\Psi_{2k}(\beta)+L_n^{1/2}\beta^d))\ .
\tag 4.63$$
We are still left with the task of estimating
$$X_n^i-X_n^i(\beta)=\sum_{(\Cal Z_{1,n}\times\Cal Z_{1,n})\setminus
\Cal W_n(\beta)}Q^i(\bold m;\bold m')\ . \tag 4.64$$
Again we divide this sum into two parts, in a slightly different ways 
according to which of the conditions (II-1) or (II-2) is satisfied. Namely 
we let
$$X_n^i-X_n^i(\beta)\equiv X_n^{\prime}(\beta)+X_n^{\prime\prime}
=\left(\sum_{\Cal W'_n(\beta)}+\sum_{\Cal W_n^{\prime\prime}}
\right)Q^i(\bold m;\bold m')\ ,\tag 4.65$$
where we define, in case Condition (II-1) is satisfied,
$$\Cal W_n^{\prime\prime}\equiv\{(\bold m;\bold m')\in
\Cal Z_{k,n}\times\Cal Z_{k,n}\ \vert\ 
\varphi(m_s)=\varphi(m'_t)\ 
\text{for some}\ s,t=1,\ldots,k\}\ , \tag 4.66$$
and define, in case Condition (II-2) is satisfied,
$$\Cal W_n^{\prime\prime}\equiv\{(\bold m;\bold m)\in
\Cal Z_{k,n}\times\Cal Z_{k,n}\ \vert\ 
\vert\varphi(m_s)-\varphi(m'_t)\vert<M'L_n^{-1/2}\ 
\text{for some}\ s,t=1,\ldots,k\}\ , \tag 4.67$$
the constant $M'>0$ being chosen later.
\par
In either case, we let
$$\Cal W'_n(\beta)=(\Cal Z_{1,n}\times\Cal Z_{1,n})\setminus
(\Cal W_n(\beta)\cup \Cal W_n^{\prime\prime})\ . \tag 4.68$$
Recall $\Cal W_n(\beta)$ was defined just before (4.21).
\par
Suppose (II-1) is satisfied. We shall estimate $X'_n(\beta)$ . By 
(4.66) and (4.68), $X'_n(\beta)$ is the sum of 
$Q^i(\bold m;\bold m')$ over those $(\bold m;\bold m')
\in\Cal Z_{k,n}\times\Cal Z_{k,n}$ 
such that $\varphi(m_s)$ and $\varphi(m'_t)$ are all different but that 
at least one of the following is less than $\beta$ :
$$\align
& \varphi(m_s)-\varphi(m_{s-1})\ ,\ \varphi(m'_s)-\varphi(m'_{s-1})\ , 
s=2,\ldots,k\ ; \tag 4.69\\ 
& \vert\varphi(m_s)-\varphi(m'_t)\vert\ ,\ s,t=1,2,\ldots,k\ .
\endalign$$
Now let $\bar{m}_1\ldots,\bar{m}_{2k}$ be the rearrangement of 
$m_1,\ldots,m_k$ , $m'_1,\ldots,m'_k$ in the increasing order of 
$\varphi(\cdot)$ , namely
$$\varphi(\bar{m}_1)<\cdots<\varphi(\bar{m}_{2k})\ ,\ 
\{\bar{m}_1\ldots,\bar{m}_{2k}\}=\{m_1,\ldots,m_k,m'_1,\ldots,m'_k\}\ . 
\tag 4.70$$
There are $(2k)!/(k!)^2$ ways of dividing a given 
$\{\bar{m}_1\ldots,\bar{m}_{2k}\}$ into two classes $\{m_1,\ldots,m_k\}$ 
and $\{m'_1,\ldots,m'_k\}$ , a typical one, let us call it $\sigma$,
 being as follows:
$$\align
& \varphi(m_1)<\cdots<\varphi(m_r)<\varphi(m'_1)<
\varphi(\bar{m}_{r+2})<\cdots<\varphi(\bar{m}_{2k})\ ; \tag 4.71 \\ 
& \{\bar{m}_{r+2},\ldots,\bar{m}_{2k}\}=\{m_{r+1},\ldots,m_k,m'_2,\ldots,
m'_k\}\ . 
\endalign$$
Accordingly $\Cal W'_n(\beta)$ is divided into $(2k)!/(k!)^2$ disjoint 
classes:
$$\Cal W'_n(\beta)=\bigcup_{\sigma}\Cal W'_{n,\sigma}(\beta)\ ,\tag 4.72$$
and obviously it is sufficient to estimate the sum $X'_{n,\sigma}(\beta)$ 
of $Q^i(m_1,\ldots,m_k;m'_1,\ldots,m'_k)$ over one of these classes. Recalling 
the definition (3.10) of $\Cal Z_{k,n}$ , of which $\Cal W'_n(\beta)$ is a 
subset, and the definition (4.17) of $\ell_s$ , we see that 
$\vert m_s\vert/\ell_s$ and $\vert m'_s\vert/\ell'_s$ are of $\Cal O_T(1)$ , 
and hence 
$$\align
& X'_n(\beta) \tag 4.73\\
&\leq C_T L_n^{-2(k-1)}\sum_{\Cal W'_n(\beta)}
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
\int_{-A}^{A}d\zeta_2\cdots\int_{-A}^{A}d\zeta_k
\int_{-A}^{A}d\zeta'_2\cdots\int_{-A}^{A}d\zeta'_k\times \\
&\times p_{2k}(y_1,\left\vert\frac{m_2}{\ell_2}\right\vert,
\ldots,\left\vert\frac{m_k}{\ell_k}\right\vert,y'_1,
\left\vert\frac{m'_2}{\ell'_2}\right\vert,\ldots,
\left\vert\frac{m'_k}{\ell'_k}\right\vert\ \vert
\varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_1),\ldots,\varphi(m'_k))\ .
\endalign$$
At this point, we assume that $(m_1,\ldots,m_k;m'_1,\ldots,m'_k)$ are arranged 
like (4.71) and set
$$\bar{y}_1=y_1\ ,\ \bar{y}_j=\frac{\vert m_s\vert}{\ell_s}\ ;\ 
\bar{\zeta}_j=\zeta_s\quad \text{if}\ \bar{m}_j=m_s\ \text{for some}\ 
s=2,\ldots,k \tag 4.74$$
and
$$\bar{y}_{r+1}=y'_1\ ,\ \bar{y}'_j=\frac{\vert m'_t\vert}{\ell'_t}\ ;\ 
\bar{\zeta}_j=\zeta'_t\quad \text{if}\ \bar{m}_j=m'_t\ \text{for some}\ 
t=2,\ldots,k \tag 4.75$$
Then by Condition (II-1), we get
$$\align
& X'_n(\beta) \tag 4.76\\
&\leq C_T L_n^{-2(k-1)}\sum_{\Cal W'_{n,\sigma}(\beta)}
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{-A}^{A}d\zeta_2\cdots
\int_{-A}^{A}d\zeta_r\times\\
&\times\int_{\vert m'_1\vert/\sqrt{z_i}}^
{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
\int_{-A}^{A}d\bar{\zeta}_{r+2}\cdots
\int_{-A}^{A}d\bar{\zeta}_{2k}\times \\
&\times p_{2k}(y_1,\bar{y}_2,\ldots,\bar{y}_r,y'_1,
\bar{y}_{r+2},\ldots,\bar{y}_{2k}\ \vert
\varphi(m_1),\ldots,\varphi(m_r),
\varphi(m'_1),\ldots,\varphi(\bar{m}_{r+2}),
\ldots,\varphi(\bar{m}_{2k})) \\
&\leq C_T L_n^{-2(k-1)}\sum \Sb \bar{m}_1\ldots,\bar{m}_{2k},\\
0<\bar{m}_j-\bar{m}_{j-1}<\beta_j \endSb 
q_n^{-2}\prod_{j=2}^{2k}(\varphi(\bar{m}_j)-\varphi(\bar{m}_{j-1}))
^{-\sigma}\ ,
\endalign$$
where $\beta_j=\beta$ or $\beta_j=\Theta$ and at least one of $\beta_j$ is 
equal to $\beta$ .
We now apply Lemma 1 with $R=L_n^{1/2}$ , successively to the summation with 
respect to $\bar{m}_j$ , $j=2k,2k-1,\ldots,2$ , to obtain
$$X'_n(\beta)\leq C_T q_n^{-2}L_n^2\prod_{j=2}^{2k}\bar{\beta}_j^{1-\sigma}
\leq C_T L_n^2q_n^{-2}\beta^{1-\sigma}\ . \tag 4.77$$
Next let us estimate $X_n^{\prime\prime}$ , still assuming Condition (II-1). 
By definition, we can write
$$X_n^{\prime\prime}=\sum_{\ell=1}^k\ \sum_{1\leq s_1<\cdots<s_{\ell}\leq k}\ 
\sum_{1\leq t_1<\cdots<t_{\ell}\leq k} 
\sum \Sb (\bold m;\bold m')\in
\Cal Z_{k,n}\times\Cal Z_{k,n};\\ \varphi(m_{s_j})=\varphi(m_{t_j}),
j=1,\ldots,\ell \endSb Q^i(\bold m;\bold m')\ . \tag 4.78$$
Let us denote the last sum by $X_n^{\prime\prime}(s_1,\ldots,s_{\ell};
t_1,\ldots,t_{\ell})$ . We shall prove
$$X_n^{\prime\prime}(s_1,\ldots,s_{\ell};t_1,\ldots,t_{\ell})
\leq C_T L_n^{3/2}q_n^{-2}\ , \tag 4.79$$
which is sufficient for our purpose. We first consider the special case of 
$s_j=t_j=j$ , $j=1,\ldots,\ell$ . Letting as before $y_s=f(\varphi(m_s))$ , 
$s=1,\ldots,k$ and $y'_t=f(\varphi(m'_t))$ , $t=\ell+1,\ldots,k$ , we can 
write, noting $f(\varphi(m_j))=f(\varphi(m'_j))$ , $j=1,\ldots,\ell$ ,
$$\align
& Q^i(\bold m;\bold m') \tag 4.80 \\
&=\int_{(\frac{\vert m_1\vert}{\sqrt{z_i}},
\frac{\vert m_1\vert}{\sqrt{z_{i-1}}})\cap
(\frac{\vert m'_1\vert}{\sqrt{z_i}},
\frac{\vert m'_1\vert}{\sqrt{z_{i-1}}})}dy_1
\int dy_2\cdots \int dy_k\int dy'_{\ell+1}\cdots\int dy'_k\times \\
&\quad\times\left[A-\{0\vee\max_{2\leq s\leq k}
(\frac{\vert m_s\vert^2}{y_s^2}-\frac{\vert m_1\vert^2}{y_1^2})\}
+\{0\wedge\min_{2\leq s\leq k}
(\frac{\vert m_s\vert^2}{y_s^2}-\frac{\vert m_1\vert^2}{y_1^2})\}\right]_
+\times \\
&\quad\times\left[A-\{0\vee\max_{2\leq t\leq \ell}
(\frac{\vert m'_t\vert^2}{y_t^2}-\frac{\vert m'_1\vert^2}{y_1^2})
\vee\max_{\ell<t\leq k}
(\frac{\vert m'_t\vert^2}{y_t^{\prime 2}}
-\frac{\vert m'_1\vert^2}{y_1^{\prime 2}})\}\right. \\
&+\left.\{0\wedge\min_{2\leq t\leq \ell}
(\frac{\vert m'_t\vert^2}{y_t^2}-\frac{\vert m'_1\vert^2}{y_1^2})
\wedge\min_{\ell<t\leq k}
(\frac{\vert m'_t\vert^2}{y_t^{\prime 2}}
-\frac{\vert m'_1\vert^2}{y_1^{\prime 2}})
\}\right]_+\times \\
&\quad\times p_{2k-\ell}(y_1,\ldots,y_k,y'_{\ell+1},\ldots,y'_k\ \vert\ 
\varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_{\ell+1}),\ldots,
\varphi(m'_k))\ .
\endalign$$
If we make the change of variables
$$\zeta_s=\frac{\vert m_s\vert^2}{y_s^2}-\frac{\vert m_1\vert^2}{y_1^2}\ ,
\ s=2,\ldots,k\ ;\ 
\zeta'_t=\frac{\vert m'_t\vert^2}{y_t^{\prime 2}}
-\frac{\vert m'_1\vert^2}{y_1^{\prime 2}}\ ,
t=\ell+1,\ldots,k\ ,\tag 4.81$$
and if we note that for $t=2,\ldots,\ell$ ,
$$\frac{\vert m'_t\vert^2}{y_t^{2}}
-\frac{\vert m'_1\vert^2}{y_1^{2}}=
\left(\frac{\zeta_t}{\vert m_t\vert^2}+
\frac{\vert m_1\vert^2}{y_1^2\vert m_t\vert^2}\right)
\vert m'_t\vert^2-
\frac{\vert m'_1\vert^2}{y_1^2}\ , \tag 4.82$$
we can further rewrite (4.80) as
$$\align
& Q^i(\bold m;\bold m') \tag 4.83 \\
&=\int_{(\frac{\vert m_1\vert}{\sqrt{z_i}},
\frac{\vert m_1\vert}{\sqrt{z_{i-1}}})\cap
(\frac{\vert m'_1\vert}{\sqrt{z_i}},
\frac{\vert m'_1\vert}{\sqrt{z_{i-1}}})}dy_1
\int_{-A}^A d\zeta_2\ldots\int_{-A}^A d\zeta_k\int_{-A}^A d\zeta'_{\ell+1}
\ldots\int_{-A}^A d\zeta'_k\times\\
&\quad\times\prod_{s=2}^k\left(
\frac12\left\vert\frac{m_s}{\ell_s}\right\vert\right)\frac1{\ell^2_s}
\prod_{t=\ell+1}^k
\left(\frac12\left\vert\frac{m'_t}{\ell'_t}\right\vert\right)
\frac1{\ell^{\prime 2}_t}\times \\
&\times[A-(0\vee\max_{2\leq s\leq k}\zeta_s)
+(0\wedge\min_{2\leq s\leq k}\zeta_s)]_+\times \\
&\times\left[A-
\left\{0\vee\max_{2\leq t\leq\ell}\left(
\left(\frac{\zeta_t}{\vert m_t\vert^2}+
\frac{\vert m_1\vert^2}{y_1^2\vert m_t\vert^2}\right)
\vert m'_t\vert^2-
\frac{\vert m'_1\vert^2}{y_1^2}\right)\vee\max_{\ell<t\leq k}\zeta'_t\right\}
\right. \\
&\left.+\left\{0\wedge\min_{2\leq t\leq\ell}\left(
\left(\frac{\zeta_t}{\vert m_t\vert^2}+
\frac{\vert m_1\vert^2}{y_1^2\vert m_t\vert^2}\right)
\vert m'_t\vert^2-
\frac{\vert m'_1\vert^2}{y_1^2}\right)\wedge\min_{\ell<t\leq k}\zeta'_t\right\}
\right]_+\times \\
&\times p_{2k-\ell}(y_1,\left\vert\frac{m_2}{\ell_2}\right\vert,\ldots,
 \left\vert\frac{m_k}{\ell_k}\right\vert,
 \left\vert\frac{m'_{\ell+1}}{\ell'_{\ell+1}}\right\vert,\ldots,
\left\vert\frac{m'_k}{\ell'_k}\right\vert
\vert
\varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_{\ell+1}),\ldots,
\varphi(m'_k))\ .
\endalign$$
In order to estimate the summation over those 
$(\bold m;\bold m')
\in\Cal Z_{k,n}\times\Cal Z_{k,n}$ such that $\varphi(m_j)=\varphi(m'_j)$ , 
$j=1,\ldots,\ell$ , we make the following observation.
\par
First, we have
$$\ell_s\geq C_T\sqrt{L_n}\quad ,\quad 
\left\vert\frac{m_s}{\ell_s}\right\vert\leq 
C_T\ ,\ s=2,\ldots,k \tag 4.84$$
and
$$\ell'_t\geq C_T\sqrt{L_n}\quad ,
\quad \left\vert\frac{m'_t}{\ell'_t}\right\vert\leq 
C_T\ ,\ t=\ell+1,\ldots,k \tag 4.85$$
Moreover $Q^i(\bold m;\bold m')$ must vanish unless for 
$t=2,\ldots,\ell$ and for some value of $y_1$ and $\zeta_t$ , one has
$$\left\vert\left(\frac{\zeta_t}{\vert m_t\vert^2}+
\frac{\vert m_1\vert^2}{y_1^2\vert m_t\vert^2}\right)
\vert m'_t\vert^2-
\frac{\vert m'_1\vert^2}{y_1^2}\right\vert<A \ . \tag 4.86$$
It is easy to see that for given values of $m_1$ , $m_t$ , $m'_1$ 
and $y_1$ , $\zeta_t$ , the 
number of $m'_t$ which satisfies this condition and $\varphi(m'_t)=
\varphi(m_t)$ is at most one. Hence if we make summation over 
$m'_2,\ldots,m'_{\ell}$ first, then 
$$\align
& X_n^{\prime\prime}(1,\ldots,\ell;1,\ldots,\ell) \tag 4.87 \\
&\leq C_T(L_n^{-1})^{2k-\ell-1}
\sum \Sb m_1,\ldots,m_k;m'_1,\\ m'_{\ell+1},\ldots,m'_k \endSb
\int_{(\frac{\vert m_1\vert}{\sqrt{z_i}},
\frac{\vert m_1\vert}{\sqrt{z_{i-1}}})\cap
(\frac{\vert m'_1\vert}{\sqrt{z_i}},
\frac{\vert m'_1\vert}{\sqrt{z_{i-1}}})}dy_1\times \\
&\quad \times\int_{-A}^{A} 
d\zeta_2\cdots\int_{-A}^{A} d\zeta_k\int_{-A}^{A} d\zeta'_{\ell+1}
\cdots\int_{-A}^{A} d\zeta'_k\times\\
&\quad\times p_{2k-\ell}(y_1,\ldots,y_k,y'_{\ell+1},\ldots,y'_k\ \vert\ 
\varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_{\ell+1}),\ldots,
\varphi(m'_k))\ .
\endalign$$
Now we arrange $m_2,\ldots,m_k$ and $m'_{\ell+1},\ldots,m'_k$ in the 
increasing order of $\varphi$ , and call them $\bar{m}_2,\ldots,
\bar{m}_{2k-\ell}$ , namely
$$\varphi(m_1)=\varphi(m'_1)<\varphi(\bar{m}_2)<\cdots<
\varphi(\bar{m}_{2k-\ell})\ . \tag 4.88$$
By Condition (II-1), we then have
$$\align
& X_n^{\prime\prime}(1,\ldots,\ell;1,\ldots,\ell) \tag 4.89 \\
&\leq C_T(L_n^{-1})^{2k-\ell-1}
\sum \Sb (m_1,m'_1,\bar{m}_2,\ldots,\bar{m}_{2k-\ell})
\in\Cal Z_{2k-\ell,n} \\ \varphi(m'_1)=\varphi(m_1) \endSb 
\ \int_{(\frac{\vert m_1\vert}{\sqrt{z_i}},
\frac{\vert m_1\vert}{\sqrt{z_{i-1}}})\cap
(\frac{\vert m'_1\vert}{\sqrt{z_i}},
\frac{\vert m'_1\vert}{\sqrt{z_{i-1}}})}dy_1 \times\\
&\quad\times \prod_{s=3}^{2k-\ell}(\varphi(\bar{m}_s)-\varphi(\bar{m}_{s-1}))
^{-\sigma}(\varphi(\bar{m}_2)-\varphi(m_1))^{-\sigma}\ ,
\endalign$$
Applying Lemma 1 successively to the summation over 
$\bar{m}_{2k-\ell},\bar{m}_{2k-\ell-1},\ldots,\bar{m}_2$ , we arrive at
$$X_n^{\prime\prime}(1,\ldots,\ell;1,\ldots,\ell)
\leq C_T\sum\Sb m_1,m'_1\in\Cal Z_{1,n}\\ \varphi(m'_1)=\varphi(m_1) \endSb
\left[\frac{\vert m_1\vert\wedge\vert m'_1\vert}{\sqrt{z_{i-1}}}-
\frac{\vert m_1\vert\vee\vert m'_1\vert}{\sqrt{z_i}}\right]_+\ . 
\tag 4.90$$
The summand vanishes unless
$$\frac{\sqrt{z_{i-1}}}{\sqrt{z_i}}\vert m_1\vert<
\vert m'_1\vert<\frac{\sqrt{z_i}}{\sqrt{z_{i-1}}}\vert m_1\vert\ .
\tag 4.91$$
But for each $m_1\in\Cal Z_{1,n}$ , the number of $m'_1$ which satisfies 
this condition and $\varphi(m'_1)=\varphi(m_1)$ is bounded by 
$C_T L_n^{1/2}q_n^{-1}$ . On the other hand, each summand is bounded by 
$$\frac{\vert m_1\vert}{\sqrt{z_{i-1}}}-\frac{\vert m_1\vert}{\sqrt{z_i}}
=\left(\frac{\sqrt{z_i}}{\sqrt{z_{i-1}}}-1\right)
\frac{\vert m_1\vert}{\sqrt{z_i}}
\leq C_T q_n^{-1}\ , \tag 4.92$$
hence we finally obtain
$$X_n^{\prime\prime}(1,\ldots,\ell;1,\ldots,\ell)
\leq C_T(L_n^{1/2}q_n^{-1})q_n^{-1}\sum_{m_1\in\Cal Z_{1,n}}1
\leq C_T L_n^{3/2}q_n^{-2}\ , \tag 4.93$$
as desired.
The general $X_n^{\prime\prime}(s_1,\ldots,s_{\ell};t_1,\ldots,t_{\ell})$ 
can be estimated in essentially the same way. Indeed, if 
$\gamma_{m_1},\gamma_{m'_1}\in[z_{i-1},z_i)$ and if 
$\vert D_{m_1}\cap\cdots\cap D_{m_k}\vert>0$ , 
$\vert D_{m'_1}\cap\cdots\cap D_{m'_k}\vert>0$ , then we must have
$$\vert\gamma_{m_{s_1}}-\gamma_{m_1}\vert<A\ ,\ 
\vert\gamma_{m'_{t_1}}-\gamma_{m'_1}\vert<A\ , \tag 4.94$$
which means
$$\gamma_{m_1},\gamma_{m'_1}\in\left(z_{i-1}-A,z_i+A\right)\ . 
\tag 4.95$$
Hence
$$Q^i(\bold m;\bold m')
\leq \bold E_{\bold P}[1_{\gamma_{m_1},\gamma_{m'_1}
\in(z_{i-1}-A,z_i+A)}
\vert D_{m_1}\cap\cdots\cap D_{m_k}\vert
\vert D_{m'_1}\cap\cdots\cap D_{m'_k}\vert]\ .\tag 4.96$$
The right hand side can be treated similarly as above, except that the role 
of 
$$(m_1,\ldots,m_{\ell}\ ;\ m'_1,\ldots,m'_{\ell})$$ 
is now played by 
$$(m_{s_1},\ldots,m_{s_{\ell}}\ ;\ m'_{t_1},\ldots,m'_{t_{\ell}})\ ,$$
 and 
$z_{i-1}$ , $z_i$ are replaced by $z_{i-1}-A$ 
and $z_i+A$ respectively. 
Thus we obtain (4.79) again. 
\par
Combining (4.63), (4.77) and (4.79), and noting $1=\Cal O(\Psi_{2k}(\beta))$ 
as $\beta\searrow0$ , we now conclude
$$X_n^i=4(c_2-c_1)^{2k}Z_n^i+\Cal O(L_n^{3/2}q_n^{-2}(\Psi_{2k}(\beta)+
L_n^{1/2}\beta^d)) \tag 4.97$$
under Condition (II-1).
\par
Next we assume Condition (II-2) and estimate $X'_n(\beta)$ and 
$X_n^{\prime\prime}$ again. This time, $X'_n(\beta)$ is the sum of 
$Q^i(\bold m;\bold m')$ over those 
$(\bold m;\bold m')\in\Cal Z_{k,n}\times\Cal Z_{k,n}$ 
such that
$$\vert\varphi(m_s)-\varphi(m'_t)\vert\geq ML_n^{-1/2}\ ,\ s,t=1,\ldots,k
 \tag 4.98$$
hold and that at least one of $\varphi(m_s)-\varphi(m_{s-1})$ , 
$\varphi(m'_s)-\varphi(m'_{s-1})$ or $\vert\varphi(m_s)-\varphi(m'_t)\vert$ 
is less than $\beta$ .
\par
Let us take $M'>0$ sufficiently small. Then as can be seen from the proof of 
Lemma 3, as far as we have 
$(\bold m;\bold m')\in\Cal Z_{k,n}\times\Cal Z_{k,n}$ , 
the condition
$$\varphi(m_s)-\varphi(m_{s-1})\geq M'L_n^{-1/2}\ ;\ 
\varphi(m'_s)-\varphi(m'_{s-1})\geq M'L_n^{-1/2}\ ,\ s=2,\ldots,k \tag 4.99$$
is automatically satisfied.
\par
As before, let $\Cal W'_{n,\sigma}(\beta)$ be the subset of 
$\Cal W'_n(\beta)$ consisting of those $(\bold m;\bold m')$ 
which can be arranged like (4.71) .
\par
Let us define, for $j=1,\ldots,2k$ ,
$$\bar{y}_j=\cases
y_1 &\quad \bar{m}_j=m_1 \\
y'_1 &\quad \bar{m}_j=m'_1 \\
\vert m_s\vert/\ell_s &\quad \bar{m}_j=m_s\  \text{for some}\ s=2,\ldots,k\\
\vert m'_t\vert/\ell'_t&\quad \bar{m}_j=m'_t\  \text{for some}\ t=2,\ldots,k
\endcases\tag 4.100$$
Then if we note that
$$C_T^{-1}L_n^{1/2}\leq \ell_s,\ell'_s\leq C_T L_n^{1/2} \tag 4.101$$
for some constant $C_T\geq1$ , we see from (4.18) and Condition (II-2), that
$$\align
& Q^i(\bold m;\bold m') \tag 4.102 \\
&\leq C_T L_n^{-2(k-1)}
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{-A}^{A}d\zeta_2\cdots 
\int_{-A}^{A}d\zeta_r
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
\times \\
&\quad\times \int_{-A}^{A}d\bar{\zeta}_{r+1}
\cdots\int_{-A}^{A}d\bar{\zeta}_{2k}
\prod_{j=2}^{2k}
1_{\{\vert \bar{y}_j-\bar{y}_{j-1}\vert\leq b_3(\varphi(\bar{m}_j)-
\varphi(\bar{m}_{j-1})\}}(\varphi(\bar{m}_j)-\varphi(\bar{m}_{j-1}))^{-\tau}\ ,
\endalign$$
where $\bar{\zeta}_j=\zeta_s$ or $\bar{\zeta}_j=\zeta'_t$ according to 
$\bar{m}_j=m_s$ or $\bar{m}_j=m'_t$ .
\par
Let $X'_{n,\sigma}(\beta)$ be the sum of $Q^i(\bold m;\bold m')$
 over $\Cal W'_{n,\sigma}(\beta)$ . Noting (4.99), we apply Lemma 2 
 successively to the summation over $\bar{m}_{2k},\ldots,\bar{m}_{r+2}$ , 
 and integrate out over $d\bar{\zeta}_j$ , to obtain 
$$\align
& X'_{n,\sigma}(\beta)
\leq C_T L_n^{-r+1}\left(\prod_{j=r+2}^{2k}\bar{\beta}_j^{2-\tau}\right)
\times \tag 4.103\\
&\times \sum_{m_1\in\Cal Z_{1,n}}
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\prod_{s=2}^r \int_{-A}^{A}d\zeta_s
\sum_{m_s}1_{\{\vert \bar{y}_s-\bar{y}_{s-1}\vert\leq b_3(\varphi(\bar{m}_s)-
\varphi(\bar{m}_{s-1})\}}
(\varphi(\bar{m}_s)-\varphi(\bar{m}_{s-1})^{-\tau}\times \\
&\times\int_{b_1}^{b_2}dy'_1\sum_{m'_1}
1_{\{y'_1\sqrt{z_{i-1}}\leq\vert m'_1\vert\leq y'_1\sqrt{z_i}\}}
1_{\{\vert \bar{y}_s-\bar{y}_{s-1}\vert\leq b_3(\varphi(\bar{m}_s)-
\varphi(\bar{m}_{s-1})\}}
(\varphi(\bar{m}_s)-\varphi(\bar{m}_{s-1})^{-\tau}\ ,
\endalign$$
where as before $\bar{\beta}_j=\beta$ or $\bar{\beta}_j=\Theta$ .
\par
We can apply the same method which we used in obtaining (4.57), namely 
we see without difficulty that
$$\align
& \sum_{m'_1}1_{\{y'_1\sqrt{z_{i-1}}\leq\vert m'_1\vert\leq y'_1\sqrt{z_i}\}}
1_{\{\vert \bar{y}_s-\bar{y}_{s-1}\vert\leq b_3(\varphi(\bar{m}_s)-
\varphi(\bar{m}_{s-1}))\}}\times 
(\varphi(\bar{m}_s)-\varphi(\bar{m}_{s-1}))^{-\tau} \\
&\leq C_T q_n^{-1}L_n^{1/2}(L_n^{\tau/2}\wedge b_3^{-1}
\vert \bar{y}_r-y'_1\vert)^{1-\tau}\ .\tag 4.104
\endalign$$
Integrating with respect to $y'_1$ on the region $\{\vert y'_1-y_r\vert<
\bar{\beta}_{r+1}\}$ , we get
$$\align
& \int dy'_1
\sum_{m'_1}1_{\{y'_1\sqrt{z_{i-1}}\leq\vert m'_1\vert\leq y'_1\sqrt{z_i}\}}
1_{\{\vert \bar{y}_s-\bar{y}_{s-1}\leq b_3(\varphi(\bar{m}_s)-
\varphi(\bar{m}_{s-1}))\}}
(\varphi(\bar{m}_s)-\varphi(\bar{m}_{s-1}))^{-\tau} \tag 4.105\\
&\leq C_T q_n^{-1}L_n(L_n^{\tau/2-1}+\bar{\beta}_{r+1}^{2-\tau}) \\
&\leq C_T q_n^{-1}L_n\bar{\beta}_{r+1}^{2-\tau}\ ,
\endalign$$
because of $\beta\geq L_n^{-1/2}$ .
\par
We insert the estimate (4.102), which is uniform in $m_1,\ldots,m_r$ , 
$y_1$ and in $\zeta_2,\ldots,\zeta_r$ , 
apply Lemma 2 successively to the summation 
over $m_r,\ldots,m_2$ , and then integrate out over $\zeta_2,\ldots,\zeta_r$ , 
to obtain
$$\align
X'_{n,\sigma}(\beta) 
&\leq C_T L_n^{-r+1}L^{r-1}
\left(\prod_{j=r+2}^{2k}\bar{\beta}_j^{2-\tau}\right)
q_n^{-1}L_n
\sum_{m_1\in\Cal Z_{1,n}}
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}\ 1\ dy_1 
\tag 4.106\\
&\leq C_T L_n^2q_n^{-2}\left(\prod_{j=r+2}^{2k}\bar{\beta}_j^{2-\tau}\right)\\
&\leq C_T L_n^2q_n^{-2}\beta^{2-\tau}\ ,
\endalign$$
because at least one of $\bar{\beta}_j$ 's is $\beta$ .
\par
This estimate being similar for all arrangement (4.71), we can now conclude
$$X'_n(\beta)\leq C_T L_n^2q_n^{-2}\beta^{2-\tau}\ . \tag 4.107$$
Finally, we turn to the estimate of $X_n^{\prime\prime}$ under Condition 
(II-2), to finish the estimate of $X_n$ .
\par
Recall that this time, $X_n^{\prime\prime}$ is the sum of 
$Q^i(\bold m;\bold m')$ over those 
$(\bold m;\bold m')\in\Cal Z_{k,n}\times\Cal Z_{k,n}$ 
for which (4.99) holds but $\vert\varphi(m_s)-\varphi(m'_t)\vert<M'L_n^{-1/2}$ 
for some $s,t=1,\ldots,k$ . Analogously to (4.79), we let 
$$X_n^{\prime\prime}(s_1,\ldots,s_{\ell},s_{\ell+1},\ldots,s_r
;t_1,\ldots,t_{\ell},t_{\ell+1},\ldots,t_r)$$
 to be the sum of 
$Q^i(\bold m;\bold m')$ over those 
$(\bold m;\bold m')\in\Cal Z_{k,n}\times\Cal Z_{k,n}$ 
such that for some $1\leq s_1<\cdots<s_{\ell}\leq k$ , 
$1\leq s_{\ell+1}<\cdots<s_r\leq k$ , $1\leq t_1<\cdots<t_{\ell}\leq k$ and 
$1\leq t_{\ell+1}<\cdots<t_r\leq k$ , it holds that
$$\varphi(m_{s_j})=\varphi(m'_{t_j})\ ,\ j=1,\ldots,\ell \tag 4.108$$
and that
$$0<\vert\varphi(m_{s_j})-\varphi(m'_{t_j})\vert<M'L_n^{-1/2}\ ,
\ j=\ell+1,\ldots,r\ . \tag 4.109$$
(Possibly one of the equalities $\ell=0$ or $r=\ell$ may holds, but not 
both.) We shall treat in detail the case of $\ell\geq1$ , $r\geq\ell$ 
and $s_j=t_j=j$ , $j=1,\ldots,r$ , and give a brief sketch of the other 
cases.
\par
Since $\varphi(m_s)=\varphi(m'_s)$ for $s=1,\ldots,\ell$ , 
$Q^i(\bold m;\bold m')$ has the representation (4.83) and 
summing over $m'_2,\ldots,m'_{\ell}$ , we obtain
$$\align
& X_n^{\prime\prime}(1,\ldots,r;1,\ldots,r) \tag 4.110 \\
&\leq C_T
\sum \Sb m_1,\ldots,m_k;m'_1,\\ m'_{\ell},\ldots,m'_k \endSb
\int_{(\frac{\vert m_1\vert}{\sqrt{z_i}},
\frac{\vert m_1\vert}{\sqrt{z_{i-1}}})\cap
(\frac{\vert m'_1\vert}{\sqrt{z_i}},
\frac{\vert m'_1\vert}{\sqrt{z_{i-1}}})}dy_1\times\\
&\quad\times\int_{-A}^A d\zeta_2\cdots\int_{-A}^A d\zeta_k
\int_{-A}^A d\zeta'_{\ell+1}
\cdots\int_{-A}^A d\zeta'_k
\prod_{s=2}^k\left(
\frac12\left\vert\frac{m_s}{\ell_s}\right\vert\right)\frac1{\ell^2_s}
\prod_{t=\ell+1}^k
\left(\frac12\left\vert\frac{m'_t}{\ell'_t}\right\vert\right)
\frac1{\ell^{\prime 2}_t}\times \\
&\quad \times p_{2k-\ell}(y_1,
\left\vert\frac{m_2}{\ell_2}\right\vert,\ldots,
\left\vert\frac{m_k}{\ell_k}\right\vert
,\left\vert\frac{m'_{\ell+1}}{\ell'_{\ell+1}}\right\vert
,\ldots,\left\vert\frac{m'_k}{\ell'_k}\right\vert\vert
\varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_{\ell+1}),\ldots,
\varphi(m'_k))\ .
\endalign$$
By Condition (II-2) and by 
$\vert\varphi(m_j)-\varphi(m'_j)\vert<M'L_n^{-1/2}$ which holds for 
$j=\ell+1,\ldots,r$ , the summand vanishes if the condition
$$\left\vert\frac{\vert m'_j\vert}{\ell'_j}-\frac{\vert m_j\vert}{\ell_j}
\right\vert<b_3M'L_n^{-1/2}\ ,\ j=\ell+1,\ldots,r \tag 4.111$$
fails to hold for all $y_1$ , $\zeta_j$ and $\zeta'_j$ .
\par
Since
$$\ell_j
=\frac{\vert m_1\vert}{y_1}\sqrt{1+\frac{y^2_1}{\vert m_1\vert^2}\zeta_j}
\quad ;\quad
\ell'_j
=\frac{\vert m'_1\vert}{y_1}\sqrt{1+\frac{y^2_1}{\vert m_1\vert^2}\zeta'_j}\ ,
\tag 4.112$$
we see
$$\frac{\ell'_j}{\ell_j}=\frac{\vert m'_1\vert}{\vert m_1\vert}
+\Cal O_T(L_n^{-1}) \tag 4.113$$
uniformly in $y_1\in(b_1,b_2)$ and in $\zeta_j,\zeta'_j\in(-A,A)$ . 
Hence if (4.111) ever holds for some $y_1$ , $\zeta_j$ 
and $\zeta'_j$ , then there is a constant $C$ such that
$$\left\vert\vert m'_j\vert-\frac{\vert m'_1\vert}{\vert m_1\vert}
\vert m_j\vert\right\vert<C\ .\tag 4.114$$
Now if we perform the integration with respect to $\zeta_j$ , 
$j=\ell+1,\ldots,r$ , then by (4.16), the variables of integration transform
 back into $y'_{\ell+1},\ldots,y'_r$ , and since $p_{2k-\ell}(\cdot)$ 
is a probability density, the summand of (4.110) is bounded by
$$\align
& \int_{(\frac{\vert m_1\vert}{\sqrt{z_i}},
\frac{\vert m_1\vert}{\sqrt{z_{i-1}}})\cap
(\frac{\vert m'_1\vert}{\sqrt{z_i}},
\frac{\vert m'_1\vert}{\sqrt{z_{i-1}}})}dy_1
\int_{-A}^{A} d\zeta_2\cdots
\int_{-A}^{A} d\zeta_k
\int_{-A}^{A} d\zeta'_{r+1}
\cdots\int_{-A}^{A} d\zeta'_k\times \tag 4.115\\
&\quad\times \prod_{s=2}^k\left(
\frac12\left\vert\frac{m_s}{\ell_s}\right\vert\right)\frac1{\ell^2_s}
\prod_{t=r+1}^k
\left(\frac12\left\vert\frac{m'_t}{\ell'_t}\right\vert\right)
\frac1{\ell^{\prime 2}_t}\prod_{j=\ell+1}^r
1_{\{\vert\vert m'_j\vert-\frac{\vert m'_1\vert}{\vert m_1\vert}
\vert m_j\vert\vert<C\}}\times\\
&\times p_{2k-r}(y_1,
\left\vert\frac{m_2}{\ell_2}\right\vert,\ldots,
\left\vert\frac{m_k}{\ell_k}\right\vert
,\left\vert\frac{m'_{r+1}}{\ell'_{r+1}}\right\vert
,\ldots,\left\vert\frac{m'_k}{\ell'_k}\right\vert\vert
\varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_{r+1}),\ldots,
\varphi(m'_k))\ .
\endalign$$
For given values of $m_1,m'_1,m'_j\in\Cal Z_{1,n}$ , the number of $m'_j$ 
which satisfy (4.114) and $\varphi(m'_j)=\varphi(m_j)$ is bounded by 
a constant which is independent of $m_1,m'_1,m'_j$ and also of $L_n$ . 
Hence if we make summation over $m'_{\ell+1},\ldots,m'_r$ first, then by 
(4.110) and (4.115),
$$\align
& X_n^{\prime\prime}(1,\ldots,r;1,\ldots,r) \tag 4.116 \\
&\leq C_T
\sum \Sb m_1,\ldots,m_k;m'_1,\\ m'_{r+1},\ldots,m'_k \endSb L_n^{-(2k-r-1)}
\int_{(\frac{\vert m_1\vert}{\sqrt{z_i}},
\frac{\vert m_1\vert}{\sqrt{z_{i-1}}})\cap
(\frac{\vert m'_1\vert}{\sqrt{z_i}},
\frac{\vert m'_1\vert}{\sqrt{z_{i-1}}})}dy_1\times \\
&\quad \times\int_{-A}^{A} d\zeta_2\cdots
\int_{-A}^{A} d\zeta_k
\int_{-A}^{A} d\zeta'_{r+1}
\ldots\int_{-A}^A d\zeta'_k\times\\
&\quad \times p_{2k-r}(y_1,
\left\vert\frac{m_2}{\ell_2}\right\vert,\ldots,
\left\vert\frac{m_k}{\ell_k}\right\vert
,\left\vert\frac{m'_{r+1}}{\ell'_{r+1}}\right\vert
,\ldots,\left\vert\frac{m'_k}{\ell'_k}\right\vert\vert
\varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_{r+1}),\ldots,
\varphi(m'_k))\ .
\endalign$$
Again we arrange $m_2,\ldots,m_k;m'_{r+1},\ldots,m'_k$ in the increasing 
order of $\varphi$ , and relabel them as $\bar{m}_2,\ldots,\bar{m}_{2k-r}$ .
\par
Introducing $\bar{y}_j$ by (4.100), we then have a bound of the summand of 
(4.116) similar to (4.102). Applying Lemma 2 successively to the summation 
over $\bar{m}_{2k-r},\ldots,\bar{m}_2$ , we arrive at
$$\align
X_n^{\prime\prime}(1,\ldots,r;1,\ldots,r)
&\leq C_T 
\sum \Sb m_1,m'_1\in\Cal Z_{1,n},\\ \varphi(m_1)=\varphi(m'_1) \endSb
\left[\frac{\vert m_1\vert\wedge\vert m'_1\vert}{\sqrt{z_{i-1}}}-
\frac{\vert m_1\vert\vee\vert m'_1\vert}{\sqrt{z_i}}\right]_+  \tag 4.117\\
&\leq C_T L_n^{3/2}q_n^{-2}\ ,
\endalign$$
as before.
\par
Next suppose $\ell\geq1$ and that for some $1\leq s_1<\cdots<s_{\ell}\leq k$ 
and $1\leq t_1<\cdots<t_{\ell}\leq k$ one has $\varphi(m_{s_j})=
\varphi(m'_{t_j})$ , while for some other $1\leq s_{\ell+1}
<\cdots<s_r\leq k$ and $1\leq t_{\ell+1}<\cdots<t_r\leq k$ , one has 
$0<\vert\varphi(m_{s_j})-\varphi(m'_{t_j})\vert<ML_n^{-1/2}$ . We can use 
(4.96) to reduce the analysis of this case to the case of 
$s_j=t_j=j$ , $j=1,\ldots,r$ . 
\par
In case of $\ell=0$ , $r>0$ , we assume, without loss of generality, that 
$s_j=t_j=j$ , $j=1,\ldots,r$ holds.
\par
This time, $\varphi(m_s)$ , $\varphi(m'_t)$ , $s,t=1,\ldots,k$ are all 
different, and we have the representation (4.18) for 
$Q^i(\bold m;\bold m')$ . Now in (4.18), $y'_1$ can vary 
only within the interval $\{\vert y'_1-y_1\vert<b_3ML_n^{-1/2}\}$ , 
from which we deduce, instead of (4.113),
$$\frac{\ell'_j}{\ell_j}=\frac{\vert m'_1\vert}{\vert m_1\vert}
+\Cal O_T(L_n^{-1/2})\ .\tag 4.118$$
But this still gives the following condition on $m_j$ :
$$\left\vert\vert m'_j\vert-\frac{\vert m'_1\vert}{\vert m_1\vert}
\vert m_j\vert\right\vert<C\ .\tag 4.119$$
for the non-vanishing of $Q^i(m_1,\ldots,m_k;m'_1,\ldots,m'_k)$ .
\par
For given values of $m_1$ , $m'_1$ and $m_j$ , the number of $m'_j$ satisfying 
(4.119) and $\vert\varphi(m'_j)-\varphi(m'_j)\vert<M'L_n^{-1/2}$ is 
bounded by some constant which is independent of $m_1$ , $m'_1$ , $m_j$ and 
$L_n$ . Hence the same argument which lead to (4.116) yields
$$\align
& X_n^{\prime\prime}(1,\ldots,r;1,\ldots,r) \tag 4.120 \\
&\leq C_T 
\sum \Sb m_1,m'_1\in\Cal Z_{1,n},\\ \vert\varphi(m_1)-\varphi(m'_1)
\vert<M'L_n^{-1/2} \endSb 
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
p_2(y_1,y'_1\ \vert\ \varphi(m_1),\varphi(m'_1)) \\
&\leq C_T \sum \Sb m_1,m'_1\in\Cal Z_{1,n},\\ \vert\varphi(m_1)-\varphi(m'_1)
\vert<M'L_n^{-1/2} \endSb
S(m_1,m'_1)\ .
\endalign$$
But this sum was already estimated in (4.61). Namely we have
$$X_n^{\prime\prime}(1,\ldots,r;1,\ldots,r)\leq C_T L_n^{3/2}q_n^{-2} 
\tag 4.121$$
once again. thus we have proved (4.97) under Condition (II-2).
\par
Let us finish the proof of Lemma 4 by estimating $Y_n^i$ . Since no new 
technical difficulty arises, we shall only sketch the outline, omitting the 
repetition of detailed argument. 
\par
To begin with, we divide the summation (4.4) into two parts: $Y_n^i=Y_n^{
\prime\prime}+(Y_n^i-Y_n^{\prime\prime})$ , where we define
$$Y_n^{\prime\prime}=\sum \Sb \varphi(m'_1)=\varphi(m_s) \\
\text{for some}\ s=1,\ldots,k \endSb R^i(\bold m;m'_1) 
\tag 4.122$$
in case Condition (II-1) holds, and
$$Y_n^{\prime\prime}=\sum \Sb \vert\varphi(m'_1)-\varphi(m_1)\vert
<ML_n^{-1/2} \\
\text{for some}\ s=1,\ldots,k \endSb R^i(\bold m;m'_1) 
\tag 4.123$$
in case Condition (II-2) holds.
\par
Suppose Condition (II-1) holds, and let $\varphi(m'_1)=\varphi(m_s)$ . 
By the same consideration which lead us to (4.96), we see
$$R^i(\bold m;m'_1)\leq C\bold E_{\bold P}[
1_{\{\gamma_{m_s},\gamma_{m'_1}\in(z_{i-1}-A,z_i+A)\}}
\vert D_{m_1}\cap\cdots\cap D_{m_k}\vert]\ , \tag 4.124$$
and it is easy to obtain
$$\align
Y_n^{\prime\prime}&=\sum_{s=1}^k
\sum_{\varphi(m'_1)=\varphi(m_s)}R^i(\bold m;m'_1) \tag 4.125 \\
&\leq C \sum_{s=1}^k\sum_{\varphi(m'_1)=\varphi(m_s)}
\bold E_{\bold P}[
1_{\{\gamma_{m_s},\gamma_{m'_1}\in(z_{i-1}-A,z_i+A)\}}
\vert D_{m_1}\cap\cdots\cap D_{m_k}\vert] \\
&\leq C L_n^{3/2}q_n^{-2}\ ,
\endalign$$
as before.
\par
In case Condition (II-2) holds, we further divide 
$Y_n^{\prime\prime}$ into summtion over those 
$(\bold m;m'_1)$ for which $\varphi(m'_1)=\varphi(m_s)$ for some 
$s=1,\ldots,k$ and over those for which $0<\vert\varphi(m'_1)-\varphi(m_1)\vert
<ML_n^{-1/2}$ for some $s$ . Recalling what we did in estimating 
$X_n^{\prime\prime}$ under Condition (II-1), there is no difficulty in 
obtaining (4.125) again. 
\par
Now define a subset $\Cal V_n(\beta)$ of $\Cal Z_{k,n}\times\Cal Z_{1,n}$ 
to be the totality of those $(m_1,\ldots,m_k,m'_1)$ such that 
$$\varphi(m_s)-\varphi(m_{s-1})\geq\beta\ ,\ s=2,\ldots,k \tag 4.126$$
and that
$$\vert\varphi(m'_1)-\varphi(m_s)\vert\geq\beta\ ,\ s=2,\ldots,k\ . 
\tag 4.127$$
Corresponding to this, we let
$$Y_n(\beta)\equiv\sum_{\Cal V_n(\beta)}R^i(\bold m;m'_1) \tag 4.128$$
and
$$Y'_n(\beta)\equiv Y_n-Y_n^{\prime\prime}-Y_n(\beta)\ . \tag 4.129$$
To estimate $Y'_n(\beta)$ , we use the bound
$$\align
R^i(\bold m;m'_1)
&\leq C\bold E_{\bold P}[
1_{\{\gamma_{m_s},\gamma_{m'_1}\in[z_{i-1},z_i))\}}
\vert D_{m_1}\cap\cdots\cap D_{m_k}\vert] \tag 4.130\\
%&=C \int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
%\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
%\int_{-A}^{A} d\zeta_2
%\ldots\int_{-A}^{A} d\zeta_k\times \\
%&\quad\times[A-(0\vee\max_{2\leq s\leq k}\zeta_s)
%+(0\wedge\min_{2\leq s\leq k}\zeta_s)]_+\times
%\prod_{s=2}^k\left(
%\frac12\left\vert\frac{m_s}{\ell_s}\right\vert\right)\frac1{\ell^2_s}\times \\
%&\quad\times p_{k+1}(y_1,
%\left\vert\frac{m_2}{\ell_2}\right\vert,\ldots,
%\left\vert\frac{m_k}{\ell_k}\right\vert,
%y'_1\ \vert\ \varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_1))\\
&\leq CL_n^{-2(k-1)}\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
\int_{-A}^A d\zeta_2\ldots\int_{-A}^A d\zeta_k\times \\
&\quad\times p_{k+1}(y_1,
\left\vert\frac{m_2}{\ell_2}\right\vert,\ldots,
\left\vert\frac{m_k}{\ell_k}\right\vert,
y'_1\ \vert\ \varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_1))\ .
\endalign$$
Now we make summation over those $(m_1,\ldots,m_k;m'_1)\in
\Cal Z_{k,n}\times\Cal Z_{1,n}$ satisfying
$$\varphi(m_s)-\varphi(m_{s-1})<\beta\ ,\ \text{for some}\ s=2,\ldots,k 
\tag 4.131$$
or
$$0<\vert\varphi(m'_1)-\varphi(m_s)\vert<\beta\ ,\ \text{for some}
\ s=1,\ldots,k \tag 4.132$$
in case Condition (II-1) holds, or over those 
$(m_1,\ldots,m_k;m'_1)\in\Cal Z_{k,n}\times\Cal Z_{1,n}$ satisfying 
$$ML_n^{-1/2}\leq\vert\varphi(m'_1)-\varphi(m_s)\vert<\beta\ \text{for some}
\ s=1,\ldots,k \tag 4.133$$
in case Condition (II-2) holds. By the same argument as in the estiamtion 
of $X'_n(\beta)$ , we then obtain
$$Y'_n(\beta)\leq CL_n^2q_n^{-2}\beta^d\ .\tag 4.134$$
Let us proceed to the estimation of $Y_n(\beta)$ . We use the following 
representation of $R^i(\bold m;m'_1)$ :
$$\align
& R^i(\bold m;m'_1) \tag 4.135 \\
%&=\int_{\varphi(m'_1)}^{\Theta}d\varphi'_2 
%\int_{\varphi'_2}^{\Theta}d\varphi'_3
%\cdots\int_{\varphi'_{k-1}}^{\Theta}d\varphi'_k\\
%&\quad\times\bold E_{\bold P}
%[1_{\{\gamma_{m_s},\gamma_{m'_1}\in[z_{i-1},z_i))\}}
%\cap D_{m_1}\cap\cdots\vert D_{m_k}\vert
%f(\varphi'_2)^2\cdots f(\varphi'_k)^2]  \\
&=\int_{\varphi(m'_1)}^{\Theta}d\varphi'_2 
\int_{\varphi'_2}^{\Theta}d\varphi'_3
\cdots\int_{\varphi'_{k-1}}^{\Theta}d\varphi'_k
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
\times \\
&\quad\times \int_{-2(c_2-c_1)}^{2(c_2-c_1)} d\zeta_2\cdots
\int_{-2(c_2-c_1)}^{2(c_2-c_1)} d\zeta_k
\int dy'_2\cdots\int dy'_k(y_2^{\prime 2}\cdots y_k^{\prime 2})\times \\
&\quad\times[2(c_2-c_1)-(0\vee\max_{2\leq s\leq k}\zeta_s)
+(0\wedge\min_{2\leq s\leq k}\zeta_s)]_+
\prod_{s=2}^k\left(
\frac12\left\vert\frac{m_s}{\ell_s}\right\vert\right)\frac1{\ell^2_s}\times \\
&\quad\times p_{2k}(y_1,
\left\vert\frac{m_2}{\ell_2}\right\vert,\ldots,
\left\vert\frac{m_k}{\ell_k}\right\vert,
y'_1,y'_2,\ldots,y'_k\ \vert\ \varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_1),
\varphi'_2,\ldots,\varphi'_k)\ .
\endalign$$
Now we decompose the integral with respect to $\varphi'_2,\ldots,\varphi'_k$ 
into to parts. Namely we set
$$D(m'_1)=\{(\varphi'_2,\ldots,\varphi'_k)\ \vert\ 
\varphi(m'_1)<\varphi'_2<\cdots<\varphi'_k<\Theta\} \tag 4.136$$
and
$$\align
D(\bold m;m'_1;\beta)
=\{&(\varphi'_2,\ldots,\varphi'_k)\in D(m_1)\ \vert\ 
\varphi'_2-\varphi(m'_1)\geq\beta\ ,\tag 4.137\\
&\qquad\varphi'_s-\varphi'_{s-1}\geq\beta\ ,\ s=3,\ldots,\ 
\vert\varphi'_t-\varphi(m_s)\vert\geq\beta\ ,\forall s,\forall t\}\ ;
\endalign$$
$$D'(\bold m;m'_1;\beta)=D(m'_1)\setminus D(m_1,\ldots,m_k;m'_1;\beta)\ .\tag 4.138$$
Let $R_{\beta}^i(\bold m;m'_1)$ and $R_{\beta}^{i\prime}
(\bold m;m'_1)$ be defined by the right hand side of (4.135) with 
the integration over $\varphi'_2,\ldots,\varphi'_k$ ranging in 
$D(\bold m;m'_1;\beta)$ and $D'(\bold m;m'_1;\beta)$ 
respectively. If we integrate with respect to $y'_2,\ldots,y'_k$ first, 
then the resulting integral is bounded with respect to 
$\varphi'_2,\ldots,\varphi'_k$ . Hence we obviously have
$$\align
R_{\beta}^{i\prime}(\bold m;m'_1)
&\leq C_T\beta L_n^{-(k-1)}
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
\times \tag 4.139\\
&\qquad\times 
p_{k+1}(y_1,\left\vert\frac{m_2}{\ell_2}\right\vert,
\ldots,\left\vert\frac{m_k}{\ell_k}\right\vert,
y'_1\ \vert\ \varphi(m_1),\ldots,\varphi(m_k),\varphi(m'_1))\ ,
\endalign$$
and hence it is easy to see that
$$\sum_{\Cal V_n(\beta)}R_{\beta}^{i\prime}(\bold m;m'_1)
\leq C\beta L_n^2q_n^{-2}\ , \tag 4.140$$
namely
$$Y_n(\beta)=\sum_{\Cal V_n(\beta)}R_{\beta}^i(\bold m;m'_1)
+\Cal O(\beta L_nq_n^{-2})\ .\tag 4.141$$
At this point, we introduce a function
$$\align
& V(x_2,\ldots,x_k) \tag 4.142 \\
&=V(x_2,\ldots,x_k;m_1,m'_1;y_1,y'_1,\ldots,y'_k;\varphi'_2,\ldots,\varphi'_k)
 \\
&\equiv p_{2k}(y_1,\vert x_2\vert,\ldots,\vert x_2\vert,y'_1,y'_2,\ldots,y'_k
\ \vert\ \varphi(m_1),\varphi(x_2),\ldots,\varphi(x_k),\varphi(m'_1),
\varphi'_2,\ldots,\varphi'_k)\times\\
&\quad\times\prod_{s=2}^k(\frac12\vert x_s\vert) 
\endalign$$
which is defined on
$$\align
\Cal D_{m_1}\equiv\{(x_2,\ldots,x_k)\in(\bold R^2)^{k-1}\ &\vert\ 
b_1\leq\vert x_s\vert\leq b_2,s=2,\ldots,k\ ;\\
&\varphi(m_1)<\varphi(x_2)<\cdots<\varphi(x_k)\}\ .\tag 4.143
\endalign$$
Then we can write the first term on the right hand side of (4.141) as follows:
$$\align
& \sum_{\Cal V_n(\beta)}R_{\beta}^i(\bold m;m'_1) \tag 4.144 \\
&=\sum_{m_1,m'_1}\idotsint_{D(m'_1)}d\varphi'_2\cdots\varphi'_k
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
\int_{-A}^A d\zeta_2\ldots\int_{-A}^A d\zeta_k \times \\
&\quad\times 
\int dy'_2\cdots\int dy'_k(y_2^{\prime 2}\cdots y_k^{\prime 2})\times \\
&\quad\times[A-(0\vee\max_{2\leq s\leq k}\zeta_s)
+(0\wedge\min_{2\leq s\leq k}\zeta_s)]_+\times \\
&\quad\times\sum_{m_2,\ldots,m_k}1_{D(m_1,\ldots,m_k;m'_1;\beta)}
V\left(\left\vert\frac{m_2}{\ell_2}\right\vert,\ldots,
\left\vert\frac{m_k}{\ell_k}\right\vert\right)\prod_{s=2}^k\frac1{\ell_s^2}\ .
\endalign$$
As before, the summation over $(m_2,\ldots,m_k)$ is nothing but the Riemann 
sum approximation of the integral of $V$ , namely we have
$$\align
& \sum_{m_2,\ldots,m_k}1_{D(m_1,\ldots,m_k;m'_1;\beta)}
V\left(\left\vert\frac{m_2}{\ell_2}\right\vert,\ldots,
\left\vert\frac{m_k}{\ell_k}\right\vert\right)
\prod_{s=2}^k\frac1{\ell_s^2} \tag 4.145\\
&=\idotsint_{\Cal D_{\beta}(\varphi'_2,\ldots,\varphi'_k)}
dx_2\cdots dx_k V(x_2,\ldots,x_k)+\Cal O(L_n^{-1/2}\Psi_{2k}(\beta))\ ,
\endalign$$
where we have defined the range of integration by
$$\Cal D_{\beta}(\varphi'_2,\ldots,\varphi'_k)\equiv
\{(x_2,\ldots,x_k)\in\Cal D_{m_1}\ \vert\ 
\vert\varphi(x_s)-\varphi'_t\vert\geq\beta\ ,\forall s,\forall t\}\ .
\tag 4.146$$
Inserting this approximation into (4.144) and using (4.38)
, we obtain 
$$\align
& \sum_{\Cal V_n(\beta)}R_{\beta}^i(\bold m;m'_1) \tag 4.147 \\
%&=\sum_{m_1,m'_1}\idotsint_{D(m'_1)}d\varphi'_2\cdots\varphi'_k
%\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
%\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
%\int_{-A}^A d\zeta_2\ldots\int_{-A}^A d\zeta_k \times \\
%&\quad\times 
%\int dy'_2\cdots\int dy'_k(y_2^{\prime 2}\cdots y_k^{\prime 2})\times \\
%&\quad\times[A-(0\vee\max_{2\leq s\leq k}\zeta_s)
%+(0\wedge\min_{2\leq s\leq k}\zeta_s)]_+\times \\
%&\quad\times\left\{\idotsint_{\Cal D_{\beta}(\varphi'_2,\ldots,\varphi'_k)}
%dx_2\cdots dx_k
%\prod_{s=2}^k(\frac12\vert x_s\vert) 
%p_{2k}(y_1,\vert x_2\vert,\ldots,\vert x_2\vert,y'_1,y'_2,\ldots,y'_k
%\ \vert\right.\\
%&\left.\qquad\vert\ \varphi(m_1),
%\varphi(x_2),\ldots,\varphi(x_k),\varphi(m'_1),
%\varphi'_2,\ldots,\varphi'_k)
%+\Cal O(L_n^{-1/2}\Psi_{2k}(\beta)\right\} \\
&=\sum_{m_1,m'_1}\idotsint_{D(m'_1)}d\varphi'_2\cdots\varphi'_K
\int_{\vert m_1\vert/\sqrt{z_i}}^{\vert m_1\vert/\sqrt{z_{i-1}}}dy_1
\int_{\vert m'_1\vert/\sqrt{z_i}}^{\vert m'_1\vert/\sqrt{z_{i-1}}}dy'_1
\times \\
&\quad\times A^k
\int dy'_2\cdots\int dy'_k(y_2^{\prime 2}\cdots y_k^{\prime 2})\times \\
&\quad\times 2^{-(k-1)}
\int dy_2\cdots\int dy_k(y_2^2\cdots y_n^2)
\idotsint_{\tilde{\Cal D}_{\beta}(\varphi'_2,\ldots,\varphi'_k)}
d\varphi_2\cdots\varphi_k\times \\
&\quad\times p_{2k}(y_1,\ldots,y_k,y'_1\ldots,y'_k\vert
\varphi(m_1),\varphi_2,\ldots,\varphi_k,\varphi(m'_1),
\varphi'_2,\ldots,\varphi'_k) \\
&\quad+\Cal O(L_n^{3/2}q_n^{-2}\Psi_{2k}(\beta))\ ,
\endalign$$
where we have set
$$\align
& \tilde{\Cal D}_{\beta}(\varphi'_2,\ldots,\varphi'_k) \tag 4.148 \\
&=\{(\varphi_2,\ldots,\varphi_k)\ \vert\ 
\varphi(m_1)<\varphi(x_2)<\cdots<\varphi(x_k)\ ;\ 
\vert\varphi_s-\varphi'_t\vert\geq\beta\}\ .
\endalign$$
Now (4.147) is further rewritten as
$$\align
& \sum_{\Cal V_n(\beta)}R_{\beta}^i(\bold m;m'_1) \tag 4.149 \\
&=\sum \Sb m_1,m'_1; \\ \vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta \endSb
2(c_2-c_1)^k\bold E_{\bold P}\left[
1_{\{\gamma_{m_s},\gamma_{m'_1}\in[z_{i-1},z_i))\}}\times\right. \\
&\quad\times \idotsint d\varphi'_2\cdots\varphi'_k
f(\varphi'_2)^2\cdots f(\varphi'_k)^2\times \\
&\quad\left.\times \idotsint_{\tilde{\Cal D}_{\beta}
(\varphi'_2,\ldots,\varphi'_k)}
d\varphi_2\cdots\varphi_k
f(\varphi_2)^2\cdots f(\varphi_k)^2\right]
+\Cal O(L_n^{3/2}q_n^{-2}\Psi_{2k}(\beta))
\\
&=2(c_2-c_1)^k
\sum \Sb m_1,m'_1; \\ \vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta \endSb
S(m_1,m'_1) \\
&\quad+\Cal O\left(L_n^{3/2}q_n^{-2}\Psi_{2k}(\beta)+
\sum \Sb m_1,m'_1; \\ \vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta \endSb
\beta\bold P(\gamma_{m_1},\gamma_{m'_1}\in[z_{i-1},z_i))\right)\ .
\endalign$$
Since we already have the estimate
$$\sum \Sb m_1,m'_1; \\ \vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta \endSb
\beta\bold P(\gamma_{m_1},\gamma_{m'_1}\in[z_{i-1},z_i))
=\Cal O(L_n^2q_n^{-2}\beta^d) $$
(see (4.41) and (4.46)), we finally arrive at
$$\align
Y_n&=Y_n^{\prime\prime}+(Y_n-Y_n^{\prime\prime}) \tag 4.150 \\
&=Y_n^{\prime\prime}+Y_n(\beta)+Y'_n(\beta) \\
&=\sum_{\Cal V_n(\beta)}R_{\beta}^i(m_1,\ldots,m_k;m'_1)
+\Cal O(\beta L_n^2q_n^{-2})+Y'_n(\beta)+Y_n^{\prime\prime} \\
&=2(c_2-c_1)^k
\sum \Sb m_1,m'_1; \\ \vert\varphi(m_1)-\varphi(m'_1)\vert\geq\beta \endSb
S(m_1,m'_1)+
\Cal O(L_n^{3/2}q_n^{-2}\Psi_{2k}(\beta)+L_n^2q_n^{-2}\beta^d) \\
&=2(c_2-c_1)^k\sum_{m_1,m'_1}S(m_1,m'_1)+
\Cal O(L_n^{3/2}q_n^{-2}(\Psi_{2k}(\beta)+L_n^{1/2}\beta^d))\ .
\endalign$$
We have thus proved 
$$\align
U_{i,n}&=X_n^i-4(c_2-c_1)^kY_n^i+4(c_2-c_1)^{2k}Z_n^i \tag 4.151 \\
&=4(c_2-c_1)^{2k}Z_n^i-8(c_2-c_1)^{2k}Z_n^i4(c_2-c_1)^{2k}Z_n^i
+\Cal O(L_n^{3/2}q_n^{-2}(\Psi_{2k}(\beta)+L_n^{1/2}\beta^d)) \\
&=\Cal O(L_n^{3/2}q_n^{-2}(\Psi_{2k}(\beta)+L_n^{1/2}\beta^d))\ ,
\endalign$$
completing the proof of Lemma 4.
\par
Before closing this section, we give a proof of formula (4.38) as a lemma.
\bigskip
\proclaim{Lemma 5}
For $A>0$ and $k\geq2$ ,
$$\int_{-A}^Ad\zeta_2\cdots \int_{-A}^Ad\zeta_k
[A-\{0\vee\max_{2\leq s\leq k}\zeta_s\}+
\{0\wedge\min_{2\leq s\leq k}\zeta_s\}]_+=A^k\ .$$
\endproclaim
\demo {Proof} 
We denote the left hand side of the above formula by $I_k$ . Moreover, if we 
set $\bar{\tau}_k=0\vee\max_{2\leq s\leq k}\zeta_s$ and 
$\underline{\tau}_k=0\wedge\min_{2\leq s\leq k}\zeta_s$ for brevity, 
then we can compute
$$\align
I_k&=\idotsint_{\{\bar{\tau}_{k-1}-\underline{\tau}_{k-1}<A\}}
d\zeta_2\cdots d\zeta_{k-1}
\left[\int_{\bar{\tau}_{k-1}-A}^{\underline{\tau}_{k-1}}
\{A-(\bar{\tau}_{k-1}-\zeta_k)\}d\zeta_k\right. \tag 4.152 \\
&\ \left.+\{A-(\bar{\tau}_{k-1}-\underline{\tau}_{k-1})\}(\bar{\tau}_{k-1}
-\underline{\tau}_{k-1})
+\int_{\bar{\tau}_{k-1}}^{\underline{\tau}_{k-1}+A}
\{A-(\zeta_k-\underline{\tau}_{k-1})\}d\zeta_k\right] \\
&=\idotsint_{\{\bar{\tau}_{k-1}-\underline{\tau}_{k-1}<A\}}
d\zeta_2\cdots d\zeta_{k-1} \\
&\ \left[\int_0^{A-\bar{\tau}_{k-1}+\underline{\tau}_{k-1}}sds+
\int_0^{A-\bar{\tau}_{k-1}+\underline{\tau}_{k-1}}sds+
\{A-(\bar{\tau}_{k-1}-\underline{\tau}_{k-1}\}(\bar{\tau}_{k-1}-
\underline{\tau}_{k-1})\right]\\
&=A\idotsint d\zeta_2\cdots d\zeta_{k-1}\{A-(\bar{\tau}_{k-1}-\underline{\tau}
_{k-1})\}_+ \\
&=AI_{k-1}\ .
\endalign$$
Thus we obtain
$$I_k=A^{k-2}I_2=A^{k-2}\int d\zeta_2(A-\vert\zeta_2\vert)_+=A^k\ .
\tag 4.153$$
%\qed
\enddemo

\remark{Acknowledgement} 
The author warmly thanks Professors Ya. Sinai and I. Kubo for their 
interest in this work and for continually encouraging him to publish it. 
He also thanks Professor T. Funaki, who made him available a copy of 
Professor Sinai's note prior to publication and Professor P. Major, who 
sent him a reprint of his paper, with both of which he started this 
work. Finally, the author is very grateful to the referee for reviewing 
a technical paper like this, and for giving him valuable suggestions.
\endremark
\bigskip
\head References \endhead
\roster
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(1992)
\item"[Mi]" Minami, N.: Level statistics for quantum Hamiltonians--Some 
preliminary ideas toward mathematical justification of the theory of 
Berry and Tabor. to appear in the {\it Proceedings of the Second Congress 
ISAAC}
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\endroster

\enddocument