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\centerline{\bf A counter--example in ergodic theory}
\medskip\centerline{\it P\'eter Major}
\centerline{\it Mathematical Institute of the Hungarian Academy of
Sciences}
\centerline{and}
\centerline{\it Bolyai College of the E\"otv\"os Lor\'and University,
Budapest}
\bigskip \noindent address: H--1364, Budapest, P.O.B. 127 \newline
(Mathematical Institute of the Hungarian Academy of Sciences) \newline
e-mail address: major\@math-inst.hu \newline
AMS Subject Classification (1991) 47A35
\bigskip\bigskip {\narrower \noindent {\bf Abstract:} We construct a
(non-integrable) function $f$ and two measure preserving, ergodic
transformations $\bold S$ and $\bold T$ on a measure space $(\Cal X,
\Cal A,\mu)$, $\mu(\Cal X)=1$, in such a way that the ergodic means
$\lim\limits_{n\to\infty}\dfrac1n\sum\limits_{k=1}^n f(\bold S^k x)$ and
$\lim\limits_{n\to\infty}\dfrac1n\sum\limits_{k=1}^n f(\bold T^k x)$
exist for almost all~$x$, they are finite constants not depending on $x$,
but these constants differ when we are averaging with respect to the
operators $\bold S$ and $\bold T$. This means that in the case of a
non-integrable function $f$ and an ergodic transformation $\bold T$ the
ergodic mean depends not only on the function $f$, but also on the
transformation~$\bold T$. The construction applies some probabilistic
arguments.\par}
 
\bigskip\bigskip \noindent
The aim of this paper is to construct a probability space $(\Cal
X, \Cal A, \mu)$, a function $f$ and two ergodic transformations
$\bold T$ and $\bold S$ on it such that
$$
\align
\lim_{n\to\infty}\frac 1n\sum_{k=1}^n \bold T^k f&=0\quad\text{a. s.}
\tag1\\
\lim_{n\to\infty}\frac 1n\sum_{k=1}^n \bold S^k f&=a\quad\text{a. s.}
\tag2
\endalign
$$
with some constant $0<a<\infty$. Here and in the sequel the same
symbol is used for a measure preserving transformation and the linear
operator it induces on measurable functions by composition. (Because of
the ergodic theorem such an example is possible only with a function
$f$ such that $\int f^+\,d\mu=\infty$ and $\int f^-\,d\mu= -\infty$.)
 
The question about the possibility of such an example was raised by
Zolt\'an Buczolich. He studied the problem that for how large class of
functions the classical notion of integral can be extended. He was
interested in the relation of this problem to the ergodic theorem
(see~[1]), and this was the reason for asking such a question. But we
hope that this question can also be interesting for its own sake.
 
We define the probability space $(\Cal X, \Cal A, \mu)$ as $\Cal
X=[0,1]^{\bold Z}$, where $\bold Z$ is the set of integers, $\Cal A$ is
the Borel $\sigma$-algebra on $\Cal X$ and $\mu=\lambda^{\bold Z}$,
the direct product of different copies
of the Lebesgue measure on $[0,1]$. Let $\bold T$ be the shift
operator to the left on $\Cal X$, i.e. for $x\in\Cal X$,
$x=(\dots,x_{-1},x_0,x_1,\dots)$
$$
\bold
T(\dots,x_{-1},x_0,x_1,\dots)=(\dots,x_{0},x_{1},x_2,\dots)\;.
$$
 
To define the function $f$ we introduce the sequences
$$
A_n=\[\frac{2^{n+10}}{n^2}\]\;,\quad n=1,2,\dots\;,
$$
$$
B_0=0,\quad B_n=2(A_1+\cdots+A_n),\quad n=1,2,\dots\;,
$$
where $[\cdot]$ denotes integer part. Put $f_k(u)=0$ for $k\le0$, \
$0\le u\le 1$, define the functions $f_k(u)$ for $B_{n-1}<k\le
B_n$, $n=1$, 2, \dots, on $[0,1]$ as
$$
\align
f_k(u)&=\left\{
\aligned
n^3&\quad\text{for }0\le u< 2^{-n}\\
0\hphantom{{}^3}&\quad\text{otherwise}
\endaligned        \right.
\qquad\text{if $k$ is odd}\\
f_k(u)&=\left\{
\aligned
-n^3&\quad\text{for }0\le u< 2^{-n}\\
0\hphantom{-^3}&\quad\text{otherwise}
\endaligned \right.
\qquad\text{if $k$ is even}\;,
\endalign
$$
and set
$$
f(x)=\sum_{k=-\infty}^\infty f_k(x_k)\;,\qquad
x=(\dots,x_{-1},x_0,x_1,\dots)\;.
$$
The relation
$$
\sum_{k=-\infty}^\infty \mu(f_k(x_k)\neq 0)=\sum_{n=1}^\infty 2A_n
2^{-n}\le2^{11}\sum_{n=1}^\infty \frac1{n^2}<\infty
$$
holds, hence the Borel--Cantelli lemma implies that for almost all $x$
with respect to the measure $\mu$ the sum defining $f(x)$ contains only
finitely many non-zero terms. Hence this sum is meaningful.
Moreover, since the sum defining $f(x)$ contains finitely many
non-zero terms with probability one, the rearrangements
of summation appearing in this article are legitimate.
 
First we show that the above defined $\bold T$ and $f$ satisfy (1), then
define an operator $\bold S$ which satisfies (2).
 
Write
$$
\align
\frac1n\sum_{k=1}^n \bold T^k f(x)
&=\frac1n \sum_{m=-\infty}^\infty H_{m,n}(x_m)=
\frac1n \sum_{m=-n+1}^{B_L}H_{m,n}(x_m) \\
& \qquad+
\frac1n \sum_{m=B_L+1}^{\infty}H_{m,n}(x_m)=Z_n^{(1)}(L)+Z_n^{(2)}(L)
\endalign
$$
with arbitrary $L>0$ and
$$
H_{m,n}(u)=\sum_{k=m+1}^{m+n} f_k(u)\;,\quad 0\le u\le 1\;.
$$
The relation
$$
\align
\mu\( Z^{(2)}_n(L)\ne 0\;\text{for some }n\ge 1\)&\le\sum_{k=L}
^\infty\sum_{m=B_k+1}^{B_{k+1}}\mu(x_m\in [0,2^{-k-1}]) \tag3\\
&=2\sum_{k=L}^\infty A_{k+1}2^{-k}<\frac{\text{const.}}L\;,\quad
\text{where }x=(x_m,\;m\in\bold Z)
\endalign
$$
holds.
 
Observe that the functions $H_{m,n}(x_m)$ are independent for fixed $n$,
and $H_{m,n}(u)=j(n+m)f_{n+m}(u)-j(m)f_{m+1}(u)$, where $j(m)=1$ if
$m$ is odd, and $j(m)=0$ if $m$ is even. Define the moments
$E_{n}^{(p)}(m)=\int H_{m,n}^p(x_m)\,d\mu(x)$ of the random variables
$H_{m,n}(x_m)$, $p=1$, 2 \dots. The identity $E_{n}^{(1)}(m)=
j(n+m)\ell^3(n+m)
2^{-\ell(n+m)}-j(m)\ell^3(m+1)2^{-\ell(m+1)}$ holds, where $\ell(m)$
denotes the number $l$ for which $B_{l-1}<m\le B_l$ if $m\ge1$, and
$\ell(m)=0$ if $m\le 0$. Also the inequalities $|E_{n}^{(p)}(m)|\le
C(p)$, $p=1$, 2, \dots, hold with some constant $C(p)$. These relations
together with the independence of the variables $H_{m,n}(x_m)$ imply
that
$$
\lim_{n\to\infty} E Z_n^{(1)}(L)=\lim_{n\to\infty}
\frac1n \sum_{m=-n+1}^{B_L} E^{(1)}_n(m)=0\;,\tag4
$$
and
$$
E\(Z^{(1)}_n(L)-EZ^{(1)}_n(L)\)^4\le\frac{\text{const.}}{n^2}\;.
$$
Hence
$$
\mu\(\left| Z^{(1)}_n(L)- EZ^{(1)}_n(L)\right|>\varepsilon\)=
\mu\(\left| Z^{(1)}_n(L)- EZ^{(1)}_n(L)\right|^4>\varepsilon^4\)
\le\frac{\text{const.}}{n^2\varepsilon^4}\;,
$$
and since the right-hand side of the last inequality is summable in $n$,
the Borel--Cantelli lemma implies that
$$
\lim_{n\to\infty}Z^{(1)}_n(L)-EZ^{(1)}_n(L)=0\quad\text{a. s.}\;.\tag5
$$
Since $L$ can be chosen arbitrary large, relations (3),~(4) and~(5) imply
relation~(1).
 
The operator $\bold S$ we shall construct is an appropriate conjugate
of $\bold T$. First we conjugate $\bold T$ with an operator $\bold
U_\pi$ which induces a permutation of the coordinates $x_k$. Then we
apply another conjugation with an operator $\bold G$ which places
the range of the functions $f_k$ to other intervals. We
describe this construction in more detail.
 
Let $\pi$ be a permutation of the positive integers $\bold
Z^+=\{1,2,\dots\}$. We denote its extension to $\bold Z$ defined by the
formula $\pi(k)=k$ for $k\le 0$ again by $\pi$ and define the
transformation $\bold U_\pi$ on $\Cal X$ as
$$
\bold U_\pi x=(x_{\pi(k)},\;k\in\bold Z)\,,\text{ for
}x=(x_k,\;k\in \bold Z)\,.
$$
For all $k\in\bold Z$ define a transformation $g_k\colon [0,1]\to [0,1]$
such that either $g_k(u)=u$ or it is an interval exchange transformation
defined in the following way: We consider two disjoint intervals
$I_k^{(1)}=[a(k),b(k))$ and $I_k^{(2)}=[c(k),d(k))$  such that
$0\le a(k)<b(k)\le1$, $0\le c(k)<d(k)\le1$, $b(k)-a(k)=d(k)-c(k)$, and
put
$$
g_k(u)=\cases
u&\text{if }u\in [0,1]\setminus(I_k^{(1)}\cup I_k^{(2)})\\
u+c(k)-a(k) &\text{if } u\in I_k^{(1)}\\
u+a(k)-c(k) &\text{if } u\in I_k^{(2)}
\endcases \;. \tag6
$$
Given the above defined set of functions $g_k(u)$, $k\in \bold Z$, we
introduce the following transformation $\bold G$ of $\Cal X$:
$$
\bold G(x)=\(g_k(x_k),\;k\in \bold Z\),\quad \text{for }x=(x_k,\;k\in
\bold Z)\;.
$$
 
We remark that $\bold G^{-1}=\bold G$ and define with the help of the
above considered permutation $\pi$ and $\bold G$ the transformation
$$
\bold S=\bold S(\pi,\, g_k,\,k\in \bold Z)=\bold G \bold U_\pi \bold T
\bold U_\pi^{-1}\bold G\;.
$$
Clearly,
$$
\bold S^k=\bold G \bold U_\pi
\bold T^k \bold U_\pi^{-1}\bold G\;.
$$
Observe that
$$
\bold S^kf(x)=\sum_{m=-\infty}^\infty
f_{\pi(k+\pi^{-1}(m))}
\bigl(g_{\pi(k+\pi^{-1}(m))}(g_m(x_m))\bigr)\quad\text{for }
x=(x_k,\;k\in \bold Z)\;,
$$
and this formula together with the relations $\pi(k)=k$ and
$f_k(u)\equiv 0$ for $k\le0$ imply that
$$
\frac1n\sum_{k=1}^n \bold S^kf(x)=\sum_{m=-n+1}^\infty
U_{m,n}(x_m)   \tag7
$$
with
$$
U_{m,n}(u)=\frac1n\sum_{k=1}^n f_{\pi(k+\pi^{-1}(m))}
\bigl(g_{\pi(k+\pi^{-1}(m))}(g_m(u))\bigr)\;.
$$
We want to construct the operator $\bold S$ in such a way that
$$
\lim_{n\to\infty}\sum_{m=1}^\infty
U_{m,n}(x_m)=0\quad\text{a. s.}\;,\tag8 $$
and
$$
\lim_{n\to\infty}\sum_{m=-n+1}^0
U_{m,n}(x_m)=a>0\quad\text{a. s.}\;.\tag9
$$
Relations (7), (8) and (9) imply (2). We want to guarantee (8) by
defining $\pi$ and $\bold G$ in such a way that for large $m$ the
probability of the set where $U_{n,m}(x_m)\neq0$ is negligibly small
and to deduce (9) from the law of large numbers. To accomplish this
goal we make such a construction where the cancellations between
different functions $f_k$ guarantee that $U_{m,n}(x_m)$ is strictly
positive and relatively small. Before this construction we show that
a transformation $\bold S$ defined in the above way is ergodic.
 
The transformation $\bold S$ is measure preserving. Given a
measurable set $A\subset \Cal X$ define the set $B$ such that
$A=\bold G \bold U_\pi B$. Then $\bold S^{-1} A=\bold G \bold U_\pi
\bold T^{-1} B$. Hence
$A=\bold S^{-1} A$ if and only if $B=\bold T^{-1} B$. The latter
relation can hold only if $\mu(B)$, hence $\mu(A)$ equals zero or
one because of the ergodicity of the operator $\bold T$.
 
To define the permutation $\pi$ and functions $g_k$ such that the
operator $\bold S$ given with their help satisfies relations (8) and~(9)
we introduce some sequences.
 
Put $M_n=n^32^{-n}$ and
$$
\align
P_1&=A_1,\quad P_n=2\left\lceil\frac
{P_{n-1}}2\frac{M_{n-1}}{M_n}\right\rceil\;,\quad n=2,3,\dots\;,\\
R_0&=0,\quad R_n=B_{n-1}+P_n\;,\quad
S_n=R_{n}+2(A_{n}-P_{n})=B_n-P_n\;,\quad n=1,2,\dots\;,
\endalign
$$
where $\lceil x\rceil$ denotes the smallest integer larger than or equal
to $x$. Simple
induction shows that $P_n\le A_n$ for all positive integers $n$, and
$P_n< A_n$ for $n\ge2$. Really,
this relation holds for $n=1$, and by induction
$$
P_n\le2\(\frac{A_{n-1}}2\frac{M_{n-1}}{M_n}+1\)
=\frac{2(n-1)^3}{n^3}A_{n-1}+2\le2^{n+10}\frac{(n-1)}{n^3}+2 <A_n
$$
for $n>1$. This relation implies that $R_n\le S_n\le B_n\le R_{n+1}$ for
all $n=1$, 2, \dots. Now we define the value of the permutation $\pi$
for $1\le k\le P_1=S_1$, \ $R_n<k\le S_n$, $n\ge2$, and $S_n< k\le
R_{n+1}$, $n\ge1$, together with some functions $b(k)$ and $j(k)$ that
we need for the definition of the functions $g_k(u)$. Since $R_1=S_1$,
these sets cover $\bold Z^+$.
 
For $1\le k\le P_1$ let $\pi(k)=2k-1$, \ $j(k)=0$ and
$b(k)=0$. For $R_{n}<k\le S_n$, $n\ge2$, put $j(k)=0$, \ $b(k)=n$
and define
$$
\align
\pi(k)&={k+P_n}\quad\text{if $k-R_n$ is odd}\\
\pi(k)&={k-P_n}\quad\text{if $k-R_n$ is even.}
\endalign
$$
(The numbers $\pi(k)$, $k$ and $k-R_n$ have the same parity.) We have
$$
\{\pi(k),\;1\le k\le P_1\}=\{k,\; 1\le k<B_1\}\cap\{\text{odd
numbers}\}
$$
and
$$
\align
\{\pi(k),\;R_n<k\le S_n\}&=[\{k,\;{B_{n-1}}+2P_n+1\le k\le
{B_{n}}-1\}\cap\{\text{odd numbers}\}] \\
&\qquad\bigcup[\{k,\;{B_{n-1}}+2\le k\le {B_{n}}-2P_n\}
\cap\{\text{even numbers}\}]\;.
\endalign
$$
For $S_n< k\le R_{n+1}$, $n\ge1$, we define $\pi(k)$ as a map from
$K^{(n)}=[S_n+1,R_{n+1}]$ to $J_1^{(n)}\cup J_2^{(n)}$ with
$$
\align
J_1^{(n)}&=[B_n+1,\,B_n+2P_{n+1}-1]\cap\{\text{odd numbers}\}\\
J_2^{(n)}&=[B_n-2P_n+2,\,B_n]\cap\{\text{even numbers}\}\;.
\endalign
$$
The set $J_1^{(n)}$ has cardinality $P_{n+1}$ and $J_2^{(n)}$ cardinality
$P_n$. The cardinality of $K^{(n)}$ equals $R_{n+1}-S_n=P_n+P_{n+1}$.
We define the functions $\pi(k)$, \ $b(k)$ and $j(k)$, \ $S_n<k\le
R_{n+1}$, by induction. The function $j(k)$ equals either  0 or 1 and
$b(k)$ takes the values $n$ or $n+1$. Let us assume that we have already
defined these functions for all $S_n<k\le L$ with some $L\ge S_n+1$, but
not for $L<k\le R_{n+1}$. Put
$$
K_1^{(n)}(L)=\{k\:\;S_n<k\le L,\,
b(k)=n+1\}
$$
and
$$
K_2^{(n)}(L)=\{k\:\;S_n<k\le L,\, b(k)=n\}\;.
$$
Set
$$
\Pi_1^{(n)}(L)=\max\limits_{k\in K_1^{(n)}(L)}\pi(k)\quad\text{and}
\quad \Pi_2^{(n)}(L)=\max\limits_{k\in K_2^{(n)}(L)}\pi(k)\;.
$$
If $M_{n+1}|K_1^{(n)}(L)|\ge M_{n}(|K_2^{(n)}(L)|+1)$, where
$|K_i^{(n)}(L)|$ denotes the cardinality of the set $K_i^{(n)}(L)$, then
define
$\pi(L+1)=\Pi_2^{(n)}(L)+2$, $j(L+1)=0$ and $b(L+1)=n$. In the other
case define $\pi(L+1)=\Pi_1^{(n)}(L)+2$, \ $\pi(L+2)=\Pi_1^{(n)}(L)+4$,
$j(L+1)=0$, $j(L+2)=1$ and $b(L+1)=b(L+2)=n+1$. If the set
$K_1^{(n)}(L)$ or $K_2^{(n)}(L)$ is empty, then we define
$\Pi_1^{(n)}(L)=B_n-1$ and $\Pi_2^{(n)}(L)=B_n-2P_n$ respectively.
 
We have to show that the above definition is correct, i.e.\ the
iteration can be stopped in such a way that
$\Pi_1^{(n)}(L)=B_n+2P_{n+1}-1$ and $\Pi_2^{(n)}(L)=B_n$, what
is equivalent to saying that $|K_1^{(n)}(L)|=P_{n+1}$ and
$|K_2^{(n)}(L)|=P_n$.
To prove the correctness of this definition first we make the following
observation:
$$
\aligned
P_n M_n=2\frac{P_n}2\frac{M_n}{M_{n+1}} M_{n+1}&\le
P_{n+1}M_{n+1}\\ &<2\(\frac{P_n}2\frac{M_n}{M_{n+1}}+1\)M_{n+1}
=P_nM_n+2M_{n+1}\;.
\endaligned \tag10
$$
Let $\bar L$ be a value for which one of the relations
$|K_1^{(n)}(\bar L)|=P_{n+1}$  or $|K_2^{(n)}(\bar L)|=P_n$ holds.
It is enough to show that if the cardinality of  the other set is less
than the corresponding number, then in the next step of the
iteration this set increases.
 
If $|K_1^{(n)}(\bar L)|=P_{n+1}$ and $|K_2^{(n)}(\bar L)|<P_n$, then
$|K_2^{(n)}(\bar L)|\le P_n-1$ and $|K_1^{(n)}(\bar L)|M_{n+1}
\allowmathbreak=M_{n+1}P_{n+1}$. Hence by the left side of relation
(10),
$$
M_n(|K_2^{(n)}(\bar L)|+1)\le P_n M_n\le P_{n+1}
M_{n+1}=M_{n+1}|K_1^{(n)}(\bar L)|\;,
$$
and the value of $|K_2^{(n)}(\cdot)|$ is increasing in the next step of
the iteration.
 
If $|K_2^{(n)}(\bar L)|=P_n$ and $|K_1^{(n)}(\bar L)|<P_{n+1}$, then
$|K_1^{(n)}(\bar
L)|\le P_{n+1}-2$. Hence we get, using the right-hand side of (10), that
$$
M_{n+1}|K_1^{(n)}(\bar L)|\le M_{n+1}(P_{n+1}-2)<P_n
M_n<M_n(|K_2^{(n)}(\bar L)|+1)\;.
$$
This implies that the value of $|K_1^{(n)}(\cdot)|$ is increasing in the
next step of the iteration. Hence the above definition of the functions
$\pi(\cdot)$, $b(k)$ and $j(k)$ is meaningful.
 
It is not difficult to check that the above defined  function $\pi$ is
an isomorphism from $\bold Z^+$ to $\bold Z^+$. Hence we can use it as
the permutation $\pi$ in the definition of $\bold S$. We define the
functions $g_k$ needed in this definition in the following way: Let
$g_k(u)=u$ if $k\le 0$ or $j(k)=0$. If $j(k)=1$ and $b(k)=n$, then let
$g_k(u)$ be the interval exchange transformation described in
formula (6) with the intervals $I^{(1)}_k=[0,2^{-n})$ and
$I^{(2)}_k=[2^{-n},2^{-n+1})$. (We defined the function $j(k)$ in order
to decide whether we want to make an interval exchange transformation
$g_k(u)$ in the $k$-th coordinate. The function $b(k)$ identifies in
which interval $(B_{n-1}, B_{n}]$ the number $\pi(k)$ lies.)
 
Now we turn to the proof of relation (8). First we show that if
$k\in (B_{n-1}, B_n]$, $n\ge2$, then $\pi(k)\in
(B_{n-2},B_{n+1}]$. This follows from the following observations:
If $k\in (R_n,S_{n}]$ then $k\in (B_{n-1},B_n]$ and $\pi(k)\in
(B_{n-1}, B_n]$, and if $k\in (S_n,R_{n+1}]$ then $k\in
(B_{n-1},B_{n+1}]$ and $\pi(k)\in J_1^{(n)}\cup J_2^{(n)}\subset
(B_{n-1},B_{n+1}]$. This relation implies that
$\pi(k)> B_{n-1}$ and $\pi^{-1}(k)> B_{n-1}$ for $k> B_n$. Hence we
get that if $m> B_L$ with some $L>1$ then $\pi(k+\pi^{-1}(m))> B_{L-2}$
for all $k\ge 0$, and
$$
U_{m,n}(u)=0\quad\text{if } m> B_L\text{ and }2^{-L+3}\le u\le
1\;.
$$
For $1\le m\le B_L$ write
$$
\align
U_{m,n}(u)&=\frac1n\sum_{k=1}^{B_L} f_{\pi(k+\pi^{-1}(m))}
\bigl(g_{\pi(k+\pi^{-1}(m))}(g_m(u))\bigr)\\
&\qquad+\frac1n\sum_{k=B_L+1}^n f_{\pi(k+\pi^{-1}(m))}
\bigl(g_{\pi(k+\pi^{-1}(m))}(g_m(u))\bigr) \\
&=\Sigma_L^1(m,n,u)+\Sigma_L^2(m,n,u)\;.
\endalign
$$
Then
$$
\Sigma_L^1(m,n,u)<\frac{C(L)}n\;,
$$
and
$$
\Sigma_L^2(m,n,u)=0\quad\text{if }2^{-L+3}\le g_m(u)\le 1\;.
$$
Hence
$$
\align
&\sum_{m=B_L+1}^\infty U_{m,n}(x_m)=0\quad\text{on the set}\\
&\qquad D_L=\{x=(x_j,\;j\in\bold Z),\; x_j\ge 2^{-(p+L)+3}\text{ for
}B_{L+p}< j\le B_{L+p+1},\;\;p=0,1,\dots\}\;,
\endalign
$$
$$
\align
&\sum_{m=1}^{B_L} \Sigma_L^2(m,n,x_m)=0\quad\text{on the set}\\
&\qquad\qquad E_L=\{x=(x_j,\;j\in\bold Z),\;g_j(x_j)\ge 2^{-L+3}
\text{ for }\;1\le j\le B_L\}
\endalign
$$
and
$$
\lim_{n\to\infty}\sum_{m=1}^{B_L}\Sigma_L^1(m,n,u)=0\;.
$$
Since
$$
\align
\mu(\Cal X\setminus D_L)&\le 2\sum_{p=0}^\infty A_{p+L}
2^{-(p+L)+3}<\text{const.\,}\frac1L\;,\\
\mu(\Cal X\setminus E_L)&\le
2^{-L+3}B_L<\text{const.\,}\frac1{L^2}\;,
\endalign
$$
and $L$ can be chosen arbitrary large, the above relations imply (8).
 
Since $\pi(m)=m$ and $g_m(u)=u$ for $m\le0$, the expression
$U_{m,n}(u)$ is simpler in this case. We can write
$$
 U_{m,n}(u)=\frac1n\sum_{k=1}^n f_{\pi(k+m)}
\bigl(g_{\pi(k+m)}(u)\bigr)=\frac1n\sum_{l=1}^{ n+m}
h_l(u)=\frac{H_{n+m}(u)}n\quad\text{for } 1-n\le m\le0
$$
with
$$
h_l(u)=f_{\pi(l)}(g_{\pi(l)}(u))\quad\text{and}
\quad H_L(u)=\sum_{l=1}^L h_l(u)\;, \quad (H_L(u)=0\text { for }L\le
0)\;.
$$
Let us study the expression $H_L(u)$. It follows from the definition of
the functions $\pi(k)$ and $g_k(u)$ that
$$
H_L(u)=\cases
L&\text{for }0\le u< \frac12\\
0&\text{for } \frac12\le u\le1
\endcases
\qquad\text {if }1\le L\le P_1\;,
$$
$$
H_L(u)-H_{R_n}(u)=\cases
n^3&\text{if }0\le u< 2^{-n}\text{ and  $L-R_n$ is odd}\\
0&\text{if } 2^{-n} \le u\le1 \text{ or $L-R_n$ is even}
\endcases
\quad\text{if }R_n\le L\le S_{n},\;\;n\ge2\;.
$$
In particular,
$$
H_{S_{n}}(u)-H_{R_n}(u)=0,\quad \text{for all } 0\le u
\le1,\;n\ge2 \;.
$$
We have
$$
H_{R_{n+1}}(u)-H_{S_n}(u)=\cases
\frac12 P_{n+1}(n+1)^3-P_n n^3&\quad\text{if }0\le u< 2^{-n}\\
0&\quad\text{if } 2^{-n} \le u\le1
\endcases\;,%\qquad n\ge1  \;.
$$
and the functions
$$
P_{n,L}(u)=H_{L}(u)-H_{S_n}(u)\;,\quad S_n<L\le R_{n+1}\;,\qquad
n\ge1\;,
$$
satisfy the identity
$$
P_{n,L}(u)=\cases
0&\text{if }2^{-n}\le u\le1\\
\frac12|K_1^{(n)}(L)|(n+1)^3-|K_2^{(n)}(L)| n^3&\text{if }0\le u<2^{-n}\\
&\quad\text{ and $|K_1^{(n)}({L)}|$ is even}\\
\frac12\(|K_1^{(n)}(L)|+1\)(n+1)^3-|K_2^{(n)}(L)| n^3&\text{if }0\le
u< 2^{-(n+1)}\\
&\quad\text{ and $|K_1^{(n)}(L)|$ is odd}\\
\frac12\(|K_1^{(n)}(L)|-1\)(n+1)^3-|K_2^{(n)}(L)| n^3&\text{if
}2^{-(n+1)}\le u< 2^{-n}\\
&\quad\text{ and $|K_1^{(n)}(L)|$ is odd}
\endcases\;,
$$
where the functions $K_i^{(n)}(L)$, $i=1$, 2,
appeared in the definition of the function $\pi(k)$. To bound the
function $P_{N,L}(u)$, first we prove the following relation for
$S_n<L\le R_{n+1}$ by induction:
$$
0\le M_{n+1}|K_1^{(n)}(L)|-M_n |K_1^{(n)}(L)|<
\left\{ \aligned
&M_n+M_{n+1}
\quad \text{if $|K_1^{(n)}(L)|$ is odd}\\
&M_n+2M_{n+1}
\quad \text{if $|K_1^{(n)}(L)|$ is even}
\endaligned \right.         \;.       \tag11
$$
Indeed, relation (11) holds for $L=S_{n}+1$. Taking into consideration
the inequality which decides which one of the sets $K_i^{(n)}(L)$,
$i=1$,~2, increases in the next step of the iteration we get the proof
of relation (11) by separating the two cases by induction in the
following way. If
$$
M_n\le M_{n+1}|K_1^{(n)}(L)|-M_n|K_2^{(n)}(L)|<M_n+2M_{n+1}\;,
$$
then $|K_1^{(n)}(L+1)|=|K_1^{(n)}(L)|$ and
$|K_2^{(n)}(L+1)|=|K_2^{(n)}(L)|+1$, hence relation (11) holds for
$L+1$. If
$$
0\le M_{n+1}|K_1^{(n)}(L)|-M_n|K_2^{(n)}(L)|<M_n\quad \text {and
$|K_1^{(n)}(L)|$ is even}\;,
$$
then $|K_1^{(n)}(L+i)|=|K_1^{(n)}(L)|+i$ and $|K_2^{(n)}(L+i)|
=|K_2^{(n)}(L)|$, $i=1$,~2. Hence, in this case relation~(11) holds
for $L+1$ and $L+2$.
 
Since $n^3=2^nM_n$, relation (11) implies that
$$
0\le P_{n,L}(u)\le 3(n+1)^3\;.
$$
The above inequalities imply that for $B_p<L\le B_{p+1}$
$$
0\le H_L(u)\le \cases
0&\text{if }\frac12\le u\le 1\\
\text{const.\,}s^4&\text{for }2^{-s-1}\le u<2^{-s}\quad\text {for }1\le
s\le p \\
\text{const.\,}p^4&\text{for } 0\le u< 2^{-p-1}
\endcases \;.
$$
 
Define the moments of the random variables $H_L(x_m)$
$$
E^{(p)}_L(m)=E^{(p)}_L=\int H_L^p(x_m)\,d\mu(x)=\int_0^1
H_L^p(u)\,du\;,\quad\text{ for }L\ge1 \text { and }m\le 0\;.
$$
Because of the bound given for the functions $H_L(u)$
$$
0\le E_L^{(p)}<C(p)<\infty\quad\text{for all $p=1$, 2, \dots}
$$
with some appropriate constant $C(p)$. In particular, the first moment
$E^{(1)}_L$ satisfies the relation
$$
\lim_{L\to\infty}E^{(1)}_L=E=A_1+\sum_{k=1}^\infty
(P_{k+1}M_{k+1}-P_kM_k)>0\;. \tag12
$$
Also the relation $E=\lim\limits_{L\to\infty}E^{(1)}_L<\infty$ holds,
since $E=A_1+\lim\limits_{k\to\infty}P_kM_k$, and
$$
P_{k+1}M_{k+1}=\left\lceil \frac {P_k}2+\frac
{M_k}{M_{k+1}}\right\rceil M_{k+1}\le P_k M_k+M_k\;,
$$
hence $\sup\limits_k P_kM_k\le P_0M_0+\sum\limits_{k=1}^\infty
M_k<\infty$.
 
Introduce the random variables $\xi_L(m)=H_L(x_{-m})-E_L^{(1)}$, $m=0$,
1, 2, \dots, on the probability space $(\Cal X,\Cal A, \mu)$. The
random variables $\xi_{L(m)}(m)$, \ $m=1$, 2, \dots, are independent for
an arbitrary sequence $L(m)$, \ $E\xi_L(m)=0$ and $E\xi^4_L(m)\le
C<\infty$ for all $L\ge 0$ and $m$. Hence
$$
E\(\frac {\xi_n(0)+\cdots +\xi_1(n-1)}n\)^4<\frac{\text{const.}}{n^2}\;,
$$
and
$$
\mu\(\left|\frac {\xi_n(0)+\cdots
+\xi_1(n-1)}n\right|>\varepsilon\)<\frac{\text{const.}}
{n^2\varepsilon^4} \quad\text{for all }\varepsilon>0\;.
$$
Since the right-hand side of the last expression is summable in $n$ for
all $\varepsilon>0$,
$$
\lim_{n\to\infty}\frac {\xi_n(0)+\cdots+\xi_1(n-1)}n=0\quad\text{a. s.}
\;.\tag13
$$
 
Relations (12) and (13) imply that
$$
\lim_{n\to\infty}\sum_{m=-n+1}^0 U_{m,n}(x_m)=\lim_{n\to\infty}
\frac1n\sum_{m=0}^{n-1}
\xi_{n-m}(m)+\lim_{n\to\infty}\frac1n\sum_{m=o}^{n-1}
E_{n-m}^{(1)}=E>0\quad \text{a. s.}\;.
$$
Relation (9), hence relation (2) is proved.
\bigskip\noindent{\bf Reference:}\smallskip
\item{[1]} Buczolich, Z: Arithmetic averages of rotations of measurable
functions. To appear in {\it Journal of Ergodic Theory and Dynamical
Systems}
 
 
\bye
 
 
 
 
 
 
