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\begin{document}
\title{Invariant measures for\\
Anosov maps with small holes}
\author{N. Chernov$^{01}$,
R. Markarian$^{02}$ and S.~Troubetzkoy$^{01}$}
\date{\today}
\maketitle
\begin{abstract}
We study Anosov diffeomorphisms on surfaces
with small holes. The points that are mapped
into the holes disappear and never return.
In our previous paper \cite{CMT} we proved the
existence of a conditionally invariant measure
$\mu_+$. Here we show that the iterations of any
initially smooth measure, after renormalization,
converge to $\mu_+$. We construct the related
invariant measure on the repeller and prove that
it is ergodic and K-mixing. We prove the escape
rate formula, relating the escape rate to the
positive Lyapunov exponent and the entropy.
\end{abstract}
\footnotetext{$^1$ Department of Mathematics
University of Alabama in Birmingham, Birmingham, AL 35294, USA.}
\footnotetext{$^2$ Instituto de Matem\'atica y Estad\'{\i}stica
``Prof. Ing. Rafael Laguardia'' Facultad de Ingenier\'{\i}a,
Universidad de la Rep\'ublica, C.C. 30, Montevideo, Uruguay.}
\centerline{\em AMS classification numbers: 58F12, 58F15, 58F11}
\vspace*{1cm}\noindent {\em Keywords}:
Repellers, chaotic scattering theory, escape rates.
\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}
\section{Introduction}
\label{secI}
\setcounter{equation}{0}
A pictorial model of a chaotic dynamical system with holes (also
known as open dynamical systems) was proposed
by Pianigiani and Yorke \cite{PY}. Imagine a Sinai billiard table (with
dispersing boundary) in which the dynamics of the ball is strongly chaotic.
Let one or more holes be cut in the table, so that the ball can fall through.
One can also think of these holes as `pockets' at the corners of the
table. Let the initial position of the ball be chosen at random with some
probability distribution. Denote by $P(t)$ the probability that the ball
stays on the table for at least time $t,$ and if it does, by $\rho(t)$
its (normalized) distribution on the table at time $t.$ Some natural
questions are: at what rate does $P(t)$ converge to zero as $t \rightarrow
\infty$, what is the limit probability distribution $\lim_{t \rightarrow
\infty} \rho(t)$, and does it depend on the initial distribution $\rho(0)?$
These questions remain open.
Open billiards and other open Hamiltonian systems have become
very popular in physics under the name of chaotic scattering
theory in the past ten years. They have been
studied numerically and heuristically, see the survey \cite{GD} and
the references therein. This has prompted mathematicians to study
open systems as well. The first mathematical results
have dealt with the case when the underlying system is uniformly
hyperbolic and admits a finite Markov partition: expanding maps of
the interval \cite{PY,CMS},
horseshoes \cite{C86}, open billiard tables
with no eclipse \cite{LM}, and Anosov
diffeomorphisms with Markov holes \cite{CM1,CM2}.
In all these papers the holes are elements of
a Markov partition.
This is a continuation of the paper \cite{CMT} where we started the
study of Anosov diffeomorphisms with small open holes.
The relaxation of the Markov property of the holes is
our main objective of this article as well as \cite{CMT}.
The main result of \cite{CMT} was the existence and
uniqueness of the conditionally invariant measure $\mu_+$
with smooth distributions on unstable fibers. Here
we prove that the iterations of any initially smooth
measure, after renormalization, converge to $\mu_+$.
We also construct the related invariant measure,
$\bar{\mu}_+$, on the repeller, which turned out to be ergodic
and K-mixing. We then obtain an escape rate formula, relating
the escape rate to the Lyapunov exponent and the entropy.
Thus, the entire mathematical theory of open hyperbolic dynamical systems
is here extended from
examples with clear-cut Markov (`rectangular') repellers to Anosov
diffeomorphisms with quite general small open holes.
Our results have many promising physical applications to, e.g., open
Lorentz gases, billiard tables with holes and pockets,
and other models in the scattering theory.
This paper is closely connected to \cite{CMT}, even
though the main ideas here are quite different. We often
refer to \cite{CMT} for notations and technical results,
but we provide all necessary definitions here as well.
We make an additional technical assumption on the holes,
see Sect.~\ref{secPL},
but our principal theme is unchanged -- we work with
small open holes of quite general nature (not even
assuming the smoothness of their boundary).
\section{Statements of main results}
\label{secSMR}
\setcounter{equation}{0}
{\bf 2.1} Let $\hat{T}:\hat{M}\to\hat{M}$ be a topologically
transitive Anosov diffeomorphism of class $C^{1+\alpha}$ of
a compact Riemannian surface $\hat{M}$. Let $H\subset\hat{M}$
be an open set with a finite number of connected components.
We denote $M=\hat{M}\setminus H$. For any $n\geq 0$ we put
\be
M_n=\cap_{i=0}^n \hat{T}^iM\ \ \ {\rm and}\ \ \
M_{-n}=\cap_{i=0}^n \hat{T}^{-i}M,
\label{Mnn}
\ee
and also
\be
M_+=\cap_{n\geq 1} M_n,\ \ \
M_-=\cap_{n\geq 1} M_{-n},\ \ \
\Omega=M_+\cap M_-
\label{M+-}
\ee
Observe that all these sets are closed, $\hat{T}^{-1}M_+\subset M_+$,
$\hat{T}M_-\subset M_-$ and $\hat{T}\Omega=\hat{T}^{-1}\Omega=\Omega$.
The set $\Omega$ is called a repeller.
We refer to the connected components of $H$ as holes. We study
the dynamics outside the holes, i.e. the trajectories
that fall into $H$ disappear and never return.
We denote by $T$ the restriction of $\hat{T}$ on $M$, which
means that for any set $A\subset M$ and $n\geq 1$ we put
$T^nA=\hat{T}^n(A\cap M_{-n})$ and $T^{-n}A=\hat{T}^{-n}(A\cap M_n)$.
Let $W^u_x$ and $W^s_x$ be local unstable and stable
fibers through $x\in\hat{M}$. We denote by
$J^u_x$ and $J^s_x$ the Jacobians of the map $\hat{T}$
restricted to $W_x^u$ and $W_x^s$, respectively, at the
point $x$. We put
$$
\Lambda_{\min}=\min_{x\in\hat{M}} \{J^u_x,1/J^s_x\}\ \ \ \ {\rm and}\ \ \ \
\Lambda_{\max}=\max_{x\in\hat{M}} \{J^u_x,1/J^s_x\}
$$
%Let $\phi_0>0$ be the minimum angle between stable and
%unstable fibers in $\hat{M}$.
For any two points
$x,y\in W^s$ a holonomy map $h_{x,y}:\, W^u_x\to W^u_y$
is defined by sliding the points of $W^u_x$ along local stable fibers
(symmetrically, $h_{x,y}:\, W^s_x\to W^s_y$ is defined for $x,y\in W^u$).
A rectangle $R\subset\hat{M}$ is a
small subset such that for any $x,y\in R$ we
have $W^u_x\cap W^s_y\in R$. We consider only closed connected
rectangles. Those are bounded by two stable and two unstable
fibers (called stable and unstable sides of $R$). Segments of
local unstable and stable fibers inside $R$ that terminate, respectively,
on the stable and unstable sides of $R$ are called $R$-fibers.
Any subrectangle $R'\subset R$ whose stable (unstable) sides
are on the stable (unstable) sides of $R$ is called a u-subrectangle
(s-subrectangle).
Denote by $\mu_{\rm SBR}$ the unique Sinai-Bowen-Ruelle (SBR)
measure of the diffeomorphism $\hat{T}$, cf. \cite{Si72,Bo75,Ru}.
Its conditional distributions on local unstable fibers are
smooth (with H\"older continuous densities). Motivated by
this, we call the conditional distributions of $\mu_{\rm SBR}$
on unstable fibers u-SBR measures. Equivalently,
for any local unstable fiber $W^u$ its u-SBR measure
is a probability measure, $\nu_{W^u}$, on $W^u$ whose density
$\rho(x)$ with respect to the Riemannian length satisfies
the equation
\be
\frac{\rho(x)}{\rho(y)}=\lim_{n\to\infty}
\frac{J^u_{T^{-1}y}\cdots J^u_{T^{-n}y}}
{J^u_{T^{-1}x}\cdots J^u_{T^{-n}x}}
\label{rho}
\ee
The u-SBR measures are $\hat{T}$-invariant, i.e.
$\hat{T}_{\ast}\nu_{W^u}=\nu_{\hat{T}W^u}$.
For any $r>0$ we denote by $D_1(r)\geq 1$ the supremum of all
ratios $\rho(x)/\rho(y)$ in (\ref{rho}) for all $x,y\in W^u$ on all fibers
$W^u\subset\hat{M}$ of length $r$ (length always means the Riemannian length).
Next, $D_2(r)$ denotes the supremum of
all the Jacobians of holonomy maps $h_{x,y}$ for points $x,y
\in W^{u,s}$ at distance $\leq r$ (measured along $W^{u,s}$).
We put $D(r)=\max\{D_1(r),D_2(r)\}$. One can think of $D(r)$
as a general upper bound on distortions within the distance $r$
in $\hat{M}$. Obviously, $D(r)\to 1$ as $r\to 0$. \medskip
{\bf 2.2} We recall the assumptions on $H$ made in \cite{CMT}.
First, there is a constant $B_0>0$ such that
for any local unstable fiber $W^u$ and any local stable fiber
$W^s$ that intersect only one hole $H'$ (connected component of $H$)
the sets $W^u\setminus H'$ and $W^s\setminus H'$ consist of not
more than $B_0$ connected components.
Let $N_H$ be the number of holes. We denote
by $d_0(H)$ the minimum distance between the holes, if there
is more than one hole. We also assume that $d_0(H)$ is smaller
than a quarter of the length of the shortest closed geodesic
on $\hat{M}$. In the case $N_H=1$ this will be the definition of $d_0(H)$.
Let $d_0$ be any lower bound on $d_0(H)$, i.e. $d_0\leq d_0(H)$.
We fix $D=D(2d_0)$, which will be the only bound on
distortions that we use.
We assumed in \cite{CMT} that $\Lambda_{\min}>64D^2$,
which was not a restrictive assumption, because it can be always
fulfilled by taking higher iterates of $\hat{T}$.
We denote by $h$ the maximal size of holes defined as
follows. For any hole $H'\subset H$ its size is
$$
\sup_{x\in H'} \{{\rm diam}\, W^u_x\cap H',
\,{\rm diam}\, W^s_x\cap H'\}
$$
where the diameter is measured along the fibers $W^{u,s}_x$.
We will need $h$ to be small enough compared to $d_0$, i.e.
$h0$ such that $T_{\ast}\mu =\lambda\mu$.
The factor $\lambda$ is the {\it eigenvalue} of $\mu$.\medskip
Obviously, any conditionally
invariant measure $\mu$ is supported on $M_+$, and we have
$\lambda=||T_{\ast}\mu||=\mu(M_{-1})$.
We are interested in measures whose conditional distributions
on unstable fibers coincide with u-SBR measures. In addition,
we assume a certain natural balance between long and short
unstable fibers in terms of measures, see below.
First, for certain technical reasons it is convenient to limit
the length of unstable fibers in $M_m$ and $M_+$ by $d_0$.
This can be done as in \cite{CMT}, Sect.~2, by
subdividing longer unstable fibers into subfibers
of length between $d_0/2$ and $d_0$. This can be accomplished by
making a finite number of cuts in $M$ along some local stable
fibers, whose choice is not very important to us. Now, with
these additional cuts, any maximal unstable fiber $W^u\subset M_m$,
$m\geq 0$, and $W^u\subset M_+$, has length $\leq d_0$. We denote
by $|W^u|$ the length of $W^u$.
Now, for $m\geq 0$, we denote by ${\cal W}^u_m$ the set of
maximal unstable fibers $W^u\subset M_m$. For every
unstable fiber $W^u$ and $\varepsilon<|W^u|$ we denote by
$W^u(\varepsilon)\subset W^u$ the union of two subsegments
of $W^u$ of length $\varepsilon$ terminating at the endpoints
of $W^u$, and put $W^u(\varepsilon)=W^u$ if $\varepsilon\geq |W^u|$.
In other words, $W^u(\varepsilon)$ is the $\varepsilon$-neighborhood
of the endpoints of $W^u$ with respect to the length on that
curve. Let $U_{m,\varepsilon}=\cup_{W^u\in{\cal W}^u_m}
W^u(\varepsilon)$. Let ${\cal W}^u_{m,\varepsilon}=
\{W^u\in{\cal W}^u_m:\, |W^u|<\varepsilon\}$.
By replacing $m$ with $+$ in the above formulas, we
define ${\cal W}^u_+$, $U_{+,\varepsilon}$ and
${\cal W}^u_{+,\varepsilon}$.
Let $m\geq 0$. Denote by ${\cal M}_m$ the class of probability
measures supported on $M_m$, such that for any $\mu\in{\cal M}_m$\\
(M1) its conditional measures on unstable fibers $W^u\in{\cal W}^u_m$
coincide with u-SBR measures on those fibers;\\
(M2) for any $\varepsilon >0$ we have $\mu(U_{m,\varepsilon})
\leq C_1\varepsilon$.\\
Here $C_1=48D/d_0$ is the constant introduced in \cite{CMT}.
By replacing $m$ with $+$, we define the class of measures ${\cal M}_+$.
We call the above measures SBR-like measures.
\medskip
{\em Claim}.
For all $m\geq 0$ we have $T_+{\cal M}_m\subset {\cal M}_{m+1}$.
Also, $T_+{\cal M}_+\subset {\cal M}_+$.
\medskip
{\em Proof}. The preservation of the property (M1) is obvious.
That of (M2) follows from Theorem~2.1 in \cite{CMT},
along with the second remark after it
for measures supported on a finite union of unstable
fibers. Then taking weak limits of such measures automatically
extends this claim to all measures in ${\cal M}_m$
and ${\cal M}_+$. $\Box$ \medskip
The main result of \cite{CMT} is the following.
\begin{theorem}[\cite{CMT}]
There is a unique SBR-like conditionally invariant measure
$\mu_+\in {\cal M}_+$, i.e. the operator $T_+:{\cal M}_+\to{\cal M}_+$
has a unique fixed point, $\mu_+$.
\label{tmmain0}
\end{theorem}
The eigenvalue $\lambda_+$ of the measure $\mu_+$
satisfies the bound $\lambda_+\geq 1-C_2h$,
where $C_2=C_1D(\Lambda_{\max}/\Lambda_{\min}+1)$. Technically,
the uniqueness of $\mu_+$ was proved in \cite{CMT} under one
extra assumption, but that one was proved there (in the statements
5.10 and 5.11) based on the remaining assumptions, so we
drop it here. \medskip
{\em Convention}. We will use here the constants $C_1,\ldots,C_4$ and
$D_1,D_2$ introduced in \cite{CMT}. Generally, we will denote by
$C_i$ and $D_i$
constants determined by the given Anosov diffeomorphism
$\hat{T}$ and the parameters $d_0,D,B_0,N_H$, which we
call {\em global parameters} (as opposed to the size $h$
of the holes or any characteristics of
the shape of the holes). The same goes for
other constants: $a_i$, $b$, $\alpha_i$, $\beta_i$,
$\gamma_i$, $l_0$, $r_0$. All these constants will be
independent of the size $h$ of the holes or
the particular shape of the holes. We call such constants
{\em global constants}. (The only exception
is $g$ in Section~\ref{secMLUF}, which is proportional to $h$.)
\medskip
{\bf 2.4}
The SBR-like conditionally invariant measure $\mu_+$ plays
the same role in the theory of open systems (= systems with holes)
as SBR measures play in conservative systems: they are
the only physically observable measures. That means that
taking a large $N\gg 1$ and a point $x\in M_{-N}$ at
random (according to a smooth probability distribution),
the orbit $T^n x$, $1\leq n\leq N$, will be asymptotically
distributed according to $\mu_+$, as $N\to\infty$.
Equivalently, if $\mu_0$ is a smooth measure on $M$,
then the sequence $\{T^n_+\mu_0\}$ weakly converges,
as $n\to\infty$, to the conditionally invariant measure $\mu_+$.
First of all, if the conditional measures of $\mu_0$ on unstable
fibers $W^u\in{\cal W}_0$ have smooth densities, then the
densities of the measure
$T^n_+\mu_0$ on fibers $W^u\in{\cal W}_n$ approach those of
u-SBR measures exponentially fast in $n$, see \cite{CM1}. Therefore,
to find limit points of the sequence $\{T^n_+\mu_0\}$ we
can restrict ourselves to the measures $\mu_0\in{\cal M}_0$.
The weak convergence $T^n_+\mu_0\to\mu_+$
was previously proved for various open systems
with Markov `rectangular' holes, cf. \cite{CM1,CM2}. Here
we prove it in our context.
\begin{theorem}
For any measure $\mu_0\in{\cal M}_0$ the sequence of measures
$T_+^n\mu_0$ weakly converges, as $n\to\infty$, to the conditionally
invariant measure $\mu_+$. Moreover, the sequence of
measures $\lambda_+^{-n}\cdot T^n_{\ast}\mu_0$ weakly
converges to $\rho(\mu_0)\cdot\mu_+$, where the function
$\rho(\mu_0)$ is uniformly bounded on ${\cal M}_0$.
\label{tmmain1}
\end{theorem}
Observe that ${\cal M}_0$ contains any measure $\mu_0$ supported
on a single unstable fiber $W^u\subset M$ of length $\geq d_0/2$,
which coincides on it with the u-SBR measure $\nu_{W^u}$. The above
theorem also holds for measures supported on arbitrary short
single fibers, provided they are `eventually long', see
Corollary~\ref{crmain1}.
We also estimate the speed of convergence in this theorem.
Let $f$ be any continuous function on $M$.
Denote by $\delta_f(\varepsilon)=\sup\{|f(x)-f(y)|:\,
{\rm dist}(x,y)<\varepsilon\}$ its modulus of continuity.
Since $M$ is compact, $f(x)$ is uniformly continuous on $M$,
so $\delta_f(\varepsilon)\to 0$ as $\varepsilon\to 0$.
\begin{theorem}
Using the notation of the previous theorem, let $\mu_n=
T^n_+\mu_0$. For any continuous function $f\in C(M)$
\be
\left |\int_Mf(x)\, d\mu_n
-\int_Mf(x)\, d\mu_+\right |\leq {\rm const}\cdot
\left [\delta_f\left (e^{-a_1n^{1/2}}\right )
+||f||_{\infty}e^{-a_2n^{1/2}}\right ]
\label{intint2}
\ee
for some global constants $a_1,a_2>0$ and ${\rm const}>0$.
\label{tmmain2}
\end{theorem}
In particular, if $f(x)$ is a H\"older continuous function, then
$\delta_f(\varepsilon)=O(\varepsilon^a)$ with some $a>0$, and
we get the so-called stretched exponential convergence in the
last theorem.
\medskip
{\bf 2.5}
Next, since $M_+$ is invariant under $T^{-1}$, it makes sense
to look for an invariant measure for the transformation
$T:\Omega\to\Omega$ by taking a weak limit of $T^{-n}_{\ast}\mu_+$
as $n\to\infty$. Due to the conditional invariance of $\mu_+$,
$T^{-n}_{\ast}\mu_+$ simply coincides with the measure $\mu_+$
conditioned on $M_{-n}$ defined by
$$
\mu_+(A/M_{-n})=\mu_+(A\cap M_{-n})/\mu_+(M_{-n})
=\lambda_+^{-n}\cdot\mu_+(A\cap M_{-n})
$$
\begin{theorem}
The sequence $T_{\ast}^{-n}\mu_+$ weakly converges, as
$n\to\infty$, to a $T$-invariant probability measure,
called $\bar{\mu}_+$, supported on the repeller $\Omega$.
The measure $\bar{\mu}_+$ is ergodic and K-mixing.
\label{tmmain3}
\end{theorem}
We also estimate the speed of convergence in this theorem.
\begin{theorem}
In the notations of the previous theorem, let $\mu_{+,n}=
T^{-n}_{\ast}\mu_+$. For any continuous function $f\in C(M)$
\be
\left |\int_Mf(x)\, d\mu_{+,n}
-\int_Mf(x)\, d\bar{\mu}_+\right |\leq {\rm const}\cdot
\left [\delta_f\left (e^{-a_3n^{1/2}}\right )
+||f||_{\infty}e^{-a_4n^{1/2}}\right ]
\label{intint4}
\ee
for some global constants $a_3,a_4>0$ and ${\rm const}>0$.
\label{tmmain4}
\end{theorem}
We conjecture that our measure $\bar{\mu}_+$ is Bernoulli
and enjoys strong statistical properties (fast decay of
correlations and central limit theorem). It is quite clear,
though, that $\bar{\mu}_+$ is not (necessarily) a Gibbs measure,
because of the lack of a local product structure in its support.
The quantity $\gamma_+=-\ln\lambda_+$ is known as the {\em escape rate}.
It characterizes the rate of escape of the mass of any smooth
measure $\mu_0$ through holes under the iterations of $T$.
Denote by $\chi_+$ the positive Lyapunov exponent of the
ergodic measure $\bar{\mu}_+$ and by $h(\bar{\mu}_+)$ its
Kolmogorov-Sinai entropy. The following {\em escape rate formula}
relates these three quantities: $\gamma_+=\chi_+-h(\bar{\mu}_+)$.
It was previously proved for open systems
with Markov rectangular holes, cf. \cite{CM1,CM2}.
\begin{theorem}
$\gamma_+ = \chi_+ - h(\bar{\mu}_+)$.
\label{tmmain5}
\end{theorem}
Next we prove that the measures $\mu_+$, $\bar{\mu}_+$
and the values of $\gamma_+$, $h(\bar{\mu}_+)$ depend
continuously on the open hole $H$. For any open sets
$H',H''\subset\hat{M}$ consider the distance
\be
d(H',H''):=\min\{\varepsilon>0:\,
H'\setminus H''\subset H''_{\varepsilon}\ \ {\rm and}\ \
H''\setminus H'\subset H'_{\varepsilon}\}
\label{dHH}
\ee
where $H_{\varepsilon}$ is the $\varepsilon$-neighborhood
of $\partial H$ in $\hat{M}\setminus H$.
Let $H_n\subset\hat{M}$ be a sequence of open sets.
Assume that each $H_n$ satisfies our assumptions on $H$
with the same values of $d_0,B_0,N_H$, so for each $H_n$
the measures $\mu_+[H_n]$ and $\bar{\mu}_+[H_n]$ and
the quantities $\lambda_+[H_n]$ and $\gamma_+[H_n]$ are
well defined.
\begin{theorem}
If $d(H_n,H)\to 0$ as $n\to\infty$, then we have the weak
convergence $\mu_+[H_n]\to\mu_+$ and $\bar{\mu}_+[H_n]\to
\bar{\mu}_+$ and the convergence $\gamma_+[H_n]\to\gamma_+$
and $h(\bar{\mu}_+[H_n])\to h(\bar{\mu}_+)$.
\label{tmmain6}
\end{theorem}
\begin{corollary}
Let $H_1\supset H_2\supset\cdots$ be a decreasing sequence
of holes such that their intersection $\cap_n H_n$ consists
of isolated curves or isolated points, or is just empty. Then
$\gamma_+[H_n]\to 0$, as $n\to\infty$, and both measures
$\mu_+[H_n]$ and $\bar{\mu}_+[H_n]$ weakly converge to the
$\hat{T}$-invariant SBR measure $\mu_{\rm SBR}$.
\end{corollary}
Back to a single open set $H$.
By reversing the time, we can define the conditionally
invariant measure $\mu_-$ on $M_-$ for the map $T^{-1}$,
whose conditional distributions on stable fibers $W^s\subset M_-$
are smooth. It has an eigenvalue, $\lambda_-$. We can then define
the corresponding invariant measure $\bar{\mu}_-$ on the
repeller $\Omega$. Those also have all the properties
described in the above theorems. The measure $\bar{\mu}_-$
and the value of $\lambda_-$ are, generally, different from
$\bar{\mu}_+$ and $\lambda_+$, respectively. However, there
are important exceptions:
\begin{theorem}
If for every periodic point $x\in\Omega$, $T^kx=x$, we have
$|{\rm det}\, DT^k(x)|=1$, then $\bar{\mu}_+=\bar{\mu}_-$
and $\lambda_+=\lambda_-$. In particular, this happens
if the given Anosov diffeomorphism $\hat{T}$ preserves
an absolutely continuous measure.
\label{tmmain7}
\end{theorem}
The last theorem has potential applications to open Hamiltonian
systems, including billiards, which preserve smooth (Liouville) measures.
\section{Assumption on holes and preparatory lemmas}
\label{secPL}
\setcounter{equation}{0}
\noindent
{\bf Definition}. We say that an unstable fiber $W^u_1\subset\hat{M}$
is $d_0$-close to another fiber, $W^u_2\subset\hat{M}$ if for any
$x\in W^u_1$ there is a local stable fiber $W^s_x$ of length
$\leq d_0$ that meets $W^u_2$ in exactly one point.
In this case denote by $\tilde{h}: W^u_1\to W^u_2$ the holonomy map
defined by $\tilde{h}(x)= W^s_x\cap W^u_2$. \medskip
We will say that a local unstable fiber $W^u$ is $d_0$-close
to a hole $H'$ (a connected component of $H$) if for any
$x\in H'$ there is a local stable fiber $W^s_x$ of length
$\leq d_0$ that meets $W^u$ in exactly one point. Then we define
the holonomy projection of $H'$ onto $W^u$ by proj$_{W^u}(H')=
\{ W^s_x\cap W^u:\, x\in H'\}$. We put $|H'|_u=\sup_{W^u}
\left | {\rm proj}_{W^u}(H')\right |$, where
the supremum is taken over all local unstable
fibers $d_0$-close to $H'$. Now we put
\be
|H|_u=\sum_{H'}|H'|_u
\label{Hu}
\ee
Remember that our holes $H'\subset H$ are assumed to be
short (of size $h$) in stable and unstable directions.
The values $|H'|_u$ and $|H|_u$ are not necessarily small,
however. They can be large, for example, if the holes stretch
`diagonally', i.e., transversally to stable and unstable
directions on $\hat{M}$. We additionally assume here that\medskip
{\bf Assumption H0}. The diameter of every hole $H'\subset H$
in the Riemannian metric on $\hat{M}$ does not exceed $d_0/2$.
\medskip
Under this assumption, for any hole $H'$ there are local unstable
fibers $d_0$-close to it. It also follows that $|H'|_u\leq 3d_0$
and $|H|_u\leq 3N_Hd_0$.
Next, let $W^u_1,W^u_2$ be two local unstable fibers $d_0$-close
to a hole $H'$. We define
$$
d_{H'}(W^u_1,W^u_2)=\max\{
|(W^u_1\cap H')\setminus \tilde{h}^{-1}(W^u_2\cap H')|_1,
|(W^u_2\cap H')\setminus \tilde{h}(W^u_1\cap H')|_2\}
$$
where $|\cdot |_1$, $|\cdot |_2$ are the length measures
on $W^u_1$ and $W^u_2$, respectively. Clearly,
$d_{H'}(\cdot,\cdot )$ is a pseudo-metric on
the set of local unstable fibers $d_0$-close to $H'$.
Now let
$$
{\rm Var}_u(H')=\sup_{k,W^u_1,\cdots,W^u_k}
\left [ d_{H'}(W^u_1,W^u_2)+d_{H'}(W^u_2,W^u_3)+
\cdots +d_{H'}(W^u_{k-1},W^u_k)\right ]
$$
where the supremum is taken over all finite collections
of local unstable fibers $d_0$-close to $H'$ naturally
ordered in the stable direction.
Now, recall that any local stable and unstable fiber intersects
any hole $H'$ in at most $B_0-1$ open intervals on that fiber.
This implies the following.
\begin{lemma}
For any hole $H'$ we have Var$_u(H')\leq B_0D|H'|_u$,
so that $\sum_{H'}{\rm Var}_u(H')\leq B_0D|H|_u$.
\label{lmVarH}
\end{lemma}
Next, we prove two technical lemmas to be used later.
Let $W^u\subset\hat{M}$ be an unstable fiber, and
for any $n\geq 0$ let $W^u_{n,i}$ be
the connected components of $T^n(W^u\cap M_{-n})$.
Remember that the lengths of $W^u$ and all $W^u_{n,i}$
are bounded by $d_0$. Denote by $\nu$ the u-SBR measure on $W^u$
and $\nu_n=T^n_{\ast}\nu$ for $n\geq 0$.
Let $B\geq 1$ be an integer. For every $n\geq 0$ let
$\tilde{W}_{n,i,r}^u\subset W^u_{n,i}$ for $r=1,\ldots,B$
be some disjoint subsegments of the fiber $W^u_{n,i}$.
We assume that their total measure is less than 1/4:
\be
\tilde{s}_n:=\sum_i\sum_r\nu_n(\tilde{W}^u_{n,i,r})<1/4
\label{B13}
\ee
\begin{lemma}
Let $0\leq g<1/3$. Let $W^u$ be a fiber of length between $d_0/2$
and $d_0$. Then for every $n\geq 0$, every $i$, and any collection
of subfibers $\tilde{W}_{n,i,r}^u\subset W^u_{n,i}$ we have
\be
\sum_i\sum_r\left [\nu_n(\tilde{W}^u_{n,i,r})\right ]^{1-g}
\leq C_g \Lambda_{\max}^{ng}
\tilde{s}_n^{1-g}\left (\log_2\tilde{s}_n^{-1}\right )^g
\label{nug}
\ee
where $C_g=200D^{1+g}B^g$.
\label{lmnug}
\end{lemma}
%{\em Remark}. For $g=0$, we recover Theorem~2.1 in \cite{CMT}.
%\medskip
%{\em Remark}. We are interested only in small values of $g$
%($g\leq$const$\cdot h$), so that the factor $(2^{1-g}-1)^{-1}$
%will be uniformly bounded, say, by 2. \medskip
{\em Proof}. We start with the following obvious consequence
of the H\"older inequality:
\begin{sublemma}
Let $x_1,\ldots,x_m$ be positive real numbers.
Then $x_1^{1-g}+\cdots +x_m^{1-g}\leq (x_1+\cdots +x_m)^{1-g}m^g$.
\label{lmxxx}
\end{sublemma}
Let $\delta>0$. Observe that
\be
\#\{i:\, \delta < |W^u_{n,i}|\leq 2\delta\} \leq
C_1Dd_0\Lambda_{\max}^n\nu(W^u\cap M_{-n})
\leq 48D^2\Lambda_{\max}^n
\label{card}
\ee
Indeed, for any such $i$ we have $\left |T^{-n}W_{n,i}^u\right |
> \delta\Lambda_{\max}^{-n}$. On the other hand, $W^u_{n,i}$
coincides with the closure of $W^u_{n,i}(\delta)$
since $|W^u_{n,i}|\leq 2\delta$,
and so the total length of the
subfibers $T^{-n}W_{n,i}^u$, $i\geq 1$, of the fiber $W^u$
does not exceed $D|W^u|C_1\delta\nu(W^u\cap M_{-n})$ due
to Theorem~2.1 in \cite{CMT}. Finally, recall that $|W^u|
\leq d_0$, and the first bound in (\ref{card}) follows. In
the second bound we replaced $C_1$ by its value, $48D/d_0$,
and dropped $\nu(W^u\cap M_{-n})$, since that one was $<1$.
We now prove Lemma~\ref{lmnug}. First of all, we bound
the left hand side of (\ref{nug}) as follows:
\begin{eqnarray*}
\sum_{k=1}^\infty\sum_{i:\frac{d_0}{2^k}<
|W^u_{n,i}|\leq \frac{d_0}{2^{k-1}}}\sum_r
\left [\nu_n(\tilde{W}^u_{n,i,r})\right ]^{1-g}
&\leq&
\sum_{k=1}^\infty\left (\sum_{i:\frac{d_0}{2^k}<
|W^u_{n,i}|\leq\frac{d_0}{2^{k-1}}}\sum_r
\nu_n(\tilde{W}^u_{n,i,r})\right )^{1-g}\nonumber\\
&\times&
\left (B\times\#\left\{i:\, \frac{d_0}{2^k}< |W^u_{n,i}|
\leq\frac{d_0}{2^{k-1}}\right\}\right )^g \nonumber\\
&\leq&
\sum_{k=1}^\infty\left (\sum_{i:\frac{d_0}{2^k}<
|W^u_{n,i}|\leq\frac{d_0}{2^{k-1}}}\sum_r
\nu_n(\tilde{W}^u_{n,i,r})\right )^{1-g}\nonumber\\
&\times&
\left [ 48\, BD^2\Lambda_{\max}^n\right ]^g
\end{eqnarray*}
where we used Sublemma~\ref{lmxxx} and (\ref{card}).
Now, let $k_0\geq 1$. Then we have
\begin{eqnarray}
\sum_{k=1}^{k_0}\left (\sum_{i:\frac{d_0}{2^k}<
|W^u_{n,i}|\leq\frac{d_0}{2^{k-1}}}\sum_r
\nu_n(\tilde{W}^u_{n,i,r})\right )^{1-g}
&\leq&
\left (\sum_{i:\frac{d_0}{2^{k_0}} < |W^u_{n,i}|}\sum_r
\nu_n(\tilde{W}^u_{n,i,r})\right )^{1-g} \times k_0^g
\nonumber\\
&\leq&
\tilde{s}_n^{1-g}k_0^g
\label{big1}
\end{eqnarray}
where we again used Sublemma~\ref{lmxxx}.
Theorem~2.1 in \cite{CMT} implies that for any $k\geq k_0+1$
\begin{eqnarray*}
\sum_{i:\frac{d_0}{2^k}< |W^u_{n,i}|\leq\frac{d_0}{2^{k-1}}}\sum_r
\nu_n(\tilde{W}^u_{n,i,r}) &\leq &
\sum_{i:\frac{d_0}{2^k}< |W^u_{n,i}|\leq\frac{d_0}{2^{k-1}}}
\nu_n(W^u_{n,i}) \\
&\leq &
\sum_i \nu_n(W^u_{n,i}(d_0/2^k)) \\
&\leq &
48\, D/ 2^{k}
\end{eqnarray*}
Therefore,
\begin{eqnarray}
\sum_{k=k_0+1}^\infty\left (\sum_{i:\frac{d_0}{2^k}<
|W^u_{n,i}|\leq\frac{d_0}{2^{k-1}}}\sum_r
\nu_n(\tilde{W}^u_{n,i,r})\right )^{1-g} &\leq &
(48\, D)^{1-g}\cdot \left ( \sum_{k=k_0+1}^\infty 2^{-k(1-g)}\right )\nonumber\\
&\leq& (48\, D)^{1-g}\cdot 2^{-k_0(1-g)}\cdot (2^{1-g}-1)^{-1}\nonumber\\
&\leq& (48\, D)^{1-g}\cdot 2^{-k_0(1-g)}\cdot 2
\label{big2}
\end{eqnarray}
Here in the end we used the fact that $(2^{1-g}-1)^{-1}< 2$ since $g<1/3$.
We now fix $k_0$ so that $2^{-k_0-1}\leq \tilde{s}_n\leq 2^{-k_0}$.
Then $k_0\leq \log_2 \tilde{s}_n^{-1}$, which we substitute in (\ref{big1}).
Also,the right hand side of (\ref{big2}) does not exceed
$$
(48\, D)^{1-g}\, 2^{1-g}\, \tilde{s}_n^{1-g}\cdot 2\leq
(192\, D)^{1-g}\, \tilde{s}_n^{1-g}\cdot
\left (\log_2\tilde{s}_n^{-1}\right )^g
$$
where we bounded the factor of $2^g$ by
$\left (\log_2 \tilde{s}_n^{-1}\right )^g$,
because $\tilde{s}_n<1/4$. Now,
combining all the above estimates gives Lemma~\ref{lmnug}. $\Box$ \medskip
\begin{lemma}
Let $R\subset\hat{M}$ be an arbitrary rectangle with at least
one stable $R$-fiber of length
between $d_0/2$ and $d_0$. Let $d^u_{\min}(R)$ and
$d^u_{\max}(R)$ be the minimum and maximum length
of unstable $R$-fibers, and suppose $d^u_{\max}(R)0$.
Here $k_1$ is the constant appearing in Theorem 3.6 of \cite{CMT}.
\label{lmTk1}
\end{lemma}
{\em Proof}. The upper bound follows from the requirement (M2) on
the class of measures ${\cal M}_0$, with any $C_5\geq C_1D^2$.
The lower bound follows from Lemma 5.9 in \cite{CMT}, see also
5.10 and 5.11 there. $\Box$\medskip
{\em Remark}. Examining the proof of Theorem 3.6 in \cite{CMT}
shows that the constant $k_1$ depends only on $\hat{T}$,
$d_0$, and $N_H$, so $k_1$ is a global constant. Also,
in this paper, for technical reasons, we restrict the lengths
of unstable fibers by $d_0$ rather than by $2d_0$, as we did
in \cite{CMT}. So, we need to replace $d_0$ in Theorem~3.6
of \cite{CMT} by $d_0/2$, which is clearly possible.
\section{Measures on long unstable fibers}
\label{secMLUF}
\setcounter{equation}{0}
Denote by ${\cal W}^u(d_0)$ the collection of all
unstable fibers in $\hat{M}$ of length between
$d_0\Lambda_{\max}^{-1}$ and $d_0$. This restriction is convenient
since for any unstable fiber $W^u\subset\hat{M}$ there
is an $n\in\ZZ$ such that $\hat{T}^nW^u\in{\cal W}^u(d_0)$.
\begin{theorem}
Let $W^u_1,W^u_2\in {\cal W}^u(d_0)$, and let $\nu_1,\nu_2$
be their u-SBR measures, respectively. Then for any $n\geq 0$
\be
01$ is a global constant.
\label{tmn12}
\end{theorem}
{\em Notations}. Put $\lambda_h=1-C_2h$, as in Theorem 2.2
of \cite{CMT}. Since $h$ is small, we can write
\be
\lambda_h = e^{-C_2'h}
\label{C2'}
\ee
with some $C_2'>C_2$ which is certainly close to $C_2$ for
small $h$. Put also
\be
g=C_2'h/\ln\Lambda_{\min}
\label{g}
\ee
Observe that $C_2'$ and $g$ are {\em not} global constants.
\medskip
{\em Remark}.
According to the above theorem, the sequence $\delta_n(W^u):=
\nu_{W^u}(W^u\cap M_{-n})$ has the same asymptotics for
all $W^u\in{\cal W}^u(d_0)$. Below we will show that
$\delta_n(W^u)\sim \lambda_+^n$, where $\lambda_+$ is
the eigenvalue of the conditionally invariant measure
$\mu_+$. For now, we fix a $W^u_0\in{\cal W}^u(d_0)$
and set $\delta_n=\nu_{W^u_0}(W^u_0\cap M_{-n})$.
Theorem 2.2 in \cite{CMT} implies that
for all $n\geq l\geq 0$
\be
\delta_n\geq\lambda_h^n\ \ \ \ \ {\rm and}\ \ \ \ \
\delta_{n-l}\leq \lambda_h^{-l}\delta_n
\label{nlnn}
\ee
The rest of this section is devoted to the proof of the above
theorem. Readers interested in getting quickly to the proofs
of the main results of this article can skip it. Only the
last remark in this section will be actually
used, just once, in the subsequent sections.
\medskip
{\em Proof}. Due to an obvious symmetry, it is enough to prove
the upper bound in (\ref{C6}).
There is an $m_0\geq 0$ and depending only on $d_0$
such that the fiber $\tilde{W}^u_2=\hat{T}^{m_0}W_2^u$
is long enough so that any fiber $W^u\in{\cal W}^u(d_0)$
is $d_0$-close to $\tilde{W}^u_2$, as defined in the
previous section.
Denote by $\tilde{h}_{W^u}:W^u\to \tilde{W}^u_2$
be the holonomy map, i.e. the sliding along local
stable fibers the distance $\leq d_0$.
Since $d_0$ is assumed to be less than
a quarter of the shortest closed geodesic in $\hat{M}$
\cite{CMT}, the rectangle with the unstable sides $W^u$ and
$\tilde{h}_{W^u}(W^u)\subset W^u_2$ is well defined (without overlaps).
Applying Theorem 2.2 in \cite{CMT} shows that it is enough
to prove the upper bound in (\ref{C6}) for $\tilde{W}^u_2$
instead of $W^u_2$, so we will simply assume that original fiber
$W^u_2$ is long enough, so that any fiber $W^u\subset{\cal W}^u(d_0)$
is $d_0$-close to $W^u_2$.
Note that dist$(x,\tilde{h}_{W^u}x)\leq d_0$
for any $x\in W^u$. Therefore, the jacobian of the map $\tilde{h}_{W^u}$
with respect to the length on $W^u$ and $W^u_2$ is bounded by $D$.
For $k\geq 1$ denote by $\tilde{h}_{W^u,k}
=\hat{T}^k\circ \tilde{h}\circ \hat{T}^{-k}$ the induced
holonomy map $\hat{T}^kW^u\to\hat{T}^kW^u_2$.
Since $T^k$ makes stable fibers shorter,
the jacobian of $\tilde{h}_{W^u,k}$ is also bounded by $D$.
Now, let $W^u_1\in{\cal W}^u(d_0)$ and $n\geq 0$.
We will define the generation gen$(x,n,W^u_1)$ for every point
$x\in W^u_1\cap M_{-n}$ as follows. If $\tilde{h}_{W^u_1}(x)\in
W^u_2\cap M_{-n}$, then gen$(x,n,W^u_1)=0$. If not, we set
$k=k(x)=\min\{ k'\geq 1:\, \hat{T}^{k'}(\tilde{h}_{W^u_1}(x))\in H\}$
(obviously, $k(x)\leq n$). Let $H'$ be the particular hole
(connected component of $H$) that contains the point
$\hat{T}^{k}(\tilde{h}_{W^u_1}(x))$, and let $W^u_{2,k,x}$ be
the component of $(\hat{T}^{k}W^u_2)\cap H'$ containing that point.
We denote the connected component of the set $\tilde{h}_{W^u_1,k}^{-1}
(W^u_{2,k,x})\setminus H'$ containing the point $T^kx$ by
$\tilde{W}^u_{1,k,x}$. This entire segment will be our
trouble, since its counterpart,
$\tilde{h}_{W^u_1,k}(\tilde{W}^u_{1,k,x})$ falls
through the hole $H'$.
Now, let $l=l(x)=\min\{l'>k=k(x):\, \hat{T}^{l'-k}
\tilde{W}^u_{1,k,x}\in {\cal W}^u(d_0)\}$, and let
$\hat{W}^u_{1,l,x}=\hat{T}^{l-k}\tilde{W}^u_{1,k,x}$.
Observe that
\be
l-k\leq \frac{\ln d_0-\ln|\tilde{W}^u_{1,k,x}|}{\ln\Lambda_{\min}}
\label{l-k}
\ee
(since $|\hat{W}^u_{1,l,x}|\leq d_0$) and
\be
\hat{T}^{l-k}(\tilde{W}^u_{1,k,x}\cap M_{-n+k})\subset
\hat{W}^u_{1,l,x}\cap M_{-n+l}
\ee
Now, if $l(x)\geq n$, we set gen$(x,n,W^u_1)=1$. Otherwise the equation
\be
{\rm gen}(x,n,W^u_1)=1+{\rm gen}(T^{l}x,n-l,\hat{W}^u_{1,l,x})
\label{genrec}
\ee
defines gen$(x,n,W^u_1)$ inductively on $n$ (for all fibers in
${\cal W}^u(d_0)$). In particular, gen$(x,n,W^u_1)=1$ if
$\tilde{h}_{\hat{W}^u_{1,l,x}}(\hat{T}^lx)\in W^u_2\cap M_{-n+l}$.
We now estimate $\nu_1(W^u_1\cap M_{-n})$.
Let $\nu_{2,W^u}$ be the measure on $W^u_2$
obtained by taking $\nu_{W^u}$ from $W^u$ under
the holonomy map $\tilde{h}_{W^u}$ to $W^u_2$. Clearly,
$d\nu_{2,W^u}/d\nu_2\leq D_3$ for some constant
$D_3=D_3(d_0)>0$. Therefore,
\be
\nu_1\{ x:\, {\rm gen}(x,n,W^1_u)=0\}
\leq \nu_{2,W^u_1}(W^u_2\cap M_{-n})
\leq D_3 \nu_2(W^u_2\cap M_{-n})
\label{gen0}
\ee
We now consider the points $x\in W^u_1\cap M_{-n}$ with
gen$(x,n,W^1_u)=1$. For every such point we have defined
the segment $\tilde{W}^u_{1,k,x}$ on the curve
$\hat{T}^kW^u_1$ with $k=k(x)\leq n$. Obviously,
for each $k=1,\ldots,n$ there is only a finite number of
distinct segments $\tilde{W}^u_{1,k,x}$ for
points $x$ with gen$(x,n,W^u_1)=1$.
We denote these segments by $\tilde{W}^u_{1,k,j}$,
$j\geq 1$. Denote by $\nu_{1,k}$ the u-SBR measure on
$\hat{T}^kW^u_1$ (the image of $\nu_1$ under $\hat{T}^k$).
Recall also that for every segment
$\tilde{W}^u_{1,k,j}$ we previously defined an
$l=l(k,j)$ by $l=\min\{l'>k:\, \hat{T}^{l'-k}
\tilde{W}^u_{1,k,j}\in {\cal W}^u(d_0)\}$. Put also
$\hat{W}^u_{1,l,j}=\hat{T}^{l-k}\tilde{W}^u_{1,k,j}$,
and we keep in mind that $l=l(k,j)$. \medskip
{\em Claim}. We have
\be
\nu_1\{x:\, {\rm gen}(x,n,W^u_1)=1\}
\leq \sum_{k=1}^n\sum_j\nu_{1,k}(\tilde{W}^u_{1,k,j})\cdot
D_3\,\lambda_h^{-l}\, \nu_2(W^u_2\cap M_{-n})
\label{gen1}
\ee
{\em Proof}.
First note that if $l=l(k,j)\geq n$, then
$\lambda_h^{-l} \nu_2(W^u_2\cap M_{-n})\geq 1$ in view of
Theorem 2.2 in \cite{CMT}. For those $k,j$ that $l< n$
we have
\begin{eqnarray*}
\nu_1\{x:\, {\rm gen}(x,n,W^u_1)=1\}
&\leq& \sum_{k=1}^n\sum_j\nu_{1,k}\left\{ y\in\tilde{W}^u_{1,k,j}:\,
\tilde{h}_{\hat{W}^u_{1,l,j}}(\hat{T}^{l-k}y)\in W^u_2\cap M_{-n+l}\right\}
\nonumber\\
&\leq& \sum_{k=1}^n\sum_j\nu_{1,k}(\tilde{W}^u_{1,k,j})\cdot
\nu_{\hat{W}^u_{1,l,j}}\left\{ z\in\hat{W}^u_{1,l,j}:\,
\tilde{h}_{\hat{W}^u_{1,l,j}}(z)\in W^u_2\cap M_{-n+l}\right\}
\nonumber\\
&\leq& \sum_{k=1}^n\sum_j\nu_{1,k}(\tilde{W}^u_{1,k,j})\cdot
D_3\,\,\nu_2(W^u_2\cap M_{-n+l})
\nonumber\\
&\leq& \sum_{k=1}^n\sum_j\nu_{1,k}(\tilde{W}^u_{1,k,j})\cdot
D_3\,\lambda_h^{-l}\, \nu_2(W^u_2\cap M_{-n})
\end{eqnarray*}
In the last step we used the inequality
\be
\nu_2(W^u_2\cap M_{-n+l}) \leq
\lambda_h^{-l}\, \nu_2(W^u_2\cap M_{-n})
\label{nln}
\ee
which follows from Theorem~2.2 in \cite{CMT}.
This proves the claim. \medskip
Due to (\ref{l-k}) we have
\be
\lambda_h^{-l}\leq\lambda_h^{-k}\cdot
\lambda_h^{\frac{\ln |\tilde{W}^u_{1,k,j}|-\ln d_0}{\ln\Lambda_{\min}}}
\label{lh1}
\ee
In order to estimate $|\tilde{W}^u_{1,k,j}|$ here, we observe that
\begin{eqnarray}
\nu_{1,k}(\tilde{W}^u_{1,k,j})=\nu_1(\hat{T}^{-k}\tilde{W}^u_{1,k,j})
&\leq& D\, |\hat{T}^{-k}\tilde{W}^u_{1,k,j}|/|W^u_1|\nonumber\\
&\leq& D\, |\tilde{W}^u_{1,k,j}|\Lambda_{\min}^{-k}/
(\Lambda_{\max}^{-1}d_0)
\label{115}
\end{eqnarray}
Therefore, $\ln |\tilde{W}^u_{1,k,j}|\geq\ln\nu_{1,k}(\tilde{W}^u_{1,k,j})
-\ln(Dd_0^{-1}\Lambda_{\max})+k\ln\Lambda_{\min}$, and so (\ref{lh1})
implies
\begin{eqnarray}
\lambda_h^{-l}
&\leq& \lambda_h^{-\frac{\ln(D\Lambda_{\max})}{\ln\Lambda_{\min}}}
\cdot \lambda_h^{\frac{\ln\left ( \nu_{1,k}(\tilde{W}^u_{1,k,j}\right )}
{\ln\Lambda_{\min}}}
\nonumber\\
&=& \exp\left (C_2'h\,\frac{\ln(D\Lambda_{\max})}{\ln\Lambda_{\min}}\right )
\cdot \exp\left (-C_2'h\,\frac{\ln(\nu_{1,k}(\tilde{W}^u_{1,k,j})}{\ln\Lambda_{\min}}\right )
\nonumber\\
&=& \left ( D\Lambda_{\max}\right )^g
\cdot \left [ \nu_{1,k}(\tilde{W}^u_{1,k,j})\right ]^{-g}
\end{eqnarray}
where we used (\ref{g}).
The main bound (\ref{gen1}) then yields
\be
\nu_1\{x:\, {\rm gen}(x)=1\}
\leq D_3\,(D\Lambda_{\max})^g \nu_2(W^u_2\cap M_{-n}) \cdot
\sum_{k=1}^n\sum_j\left [\nu_{1,k}(\tilde{W}^u_{1,k,j})\right ]^{1-g}
\label{gen1a}
\ee
In the spirit of notation in Lemma~\ref{lmnug}, we denote
by $W^u_{1,k,i}$ the components of $T^kW^u_1$. Each curve
$\tilde{W}^u_{1,k,j}$ lies on some component $W^u_{1,k,i}$,
and we denote this fact by a shorthand $j\in i$. Recall that
the components $W^u_{1,k,i}$ have length $\leq 2d_0$, so that every curve
$\tilde{h}_{W^u_1,k}(W^u_{1,k,i})\subset\hat{T}^kW^u_2$
has length $\leq 2Dd_0$ and thus can cross at most $2D+1$ holes.
The intersection of that curve with any hole consists of
$\leq B_0-1$ segments. Therefore, every curve $W^u_{1,k,i}$
contains no more than $(2D+1)(B_0-1)\leq 3DB_0$ subcurves
$\tilde{W}^u_{1,k,j}$, i.e. for any $i$ we have
\be
\#\{j:\, j\in i\}\leq 3DB_0
\label{numj}
\ee
In addition, the intersection of the curve
$\tilde{h}_{W^u_1,k}(W^u_{1,k,i})$ with any hole has total length
$\leq h$. Therefore, $|\tilde{h}_{W^u_1,k}(\tilde{W}^u_{1,k,i})
\cap H|\leq (2D+1)h\leq 3Dh$. Mapping this intersection back
on $W^u_{1,k,i}$ gives $\sum_{j\in i}|\tilde{W}^u_{1,k,j}|
\leq 3D^2h$, so that
$$
\sum_{j\in i}\nu_{1,k}(\tilde{W}^u_{1,k,j})
\leq D\nu_{1,k}(W_{1,k,i}(2D^2h))
$$
where $W_{1,k,i}(\varepsilon)$ stands for the $\varepsilon$-neighborhood
of the endpoints of $W_{1,k,i}$ within this curve.
Applying Theorem 2.1 in \cite{CMT} gives
\be
\sum_i\sum_{j\in i}\nu_{1,k}(\tilde{W}^u_{1,k,j})
\leq 2C_1D^3h
\leq \bar{D}h/d_0
\label{smallk}
\ee
where $\bar{D}=2C_1D^3d_0=96D^4$.
On the other hand, between any component $W^u_{1,k,i}\subset
T^kW^u_1$ and its counterpart $\tilde{h}_{W^u_1,k}(W^u_{1,k,i})
\subset \hat{T}^kW^u_2$ there is no other parts of $\hat{T}^kW_1^u$
or $\hat{T}^kW^u_2$. Therefore, Lemma~\ref{lmVarH} applies
and gives
$$
\sum_j|\tilde{W}^u_{1,k,j}|
\leq B_0D|H|_u
$$
so that using (\ref{115})
\be
\sum_j\nu_{1,k}(\tilde{W}^u_{1,k,j})
\leq B_0D^2|H|_ud_0^{-1}\Lambda_{\max}\Lambda_{\min}^{-k}
\label{largek}
\ee
Now we fix $k_{\ast}$ so that $\Lambda_{\min}^{-k_{\ast}-1}\leq h/|H|_u\leq
\Lambda_{\min}^{-k_{\ast}}$, i.e.
$$
k_{\ast}\approx \ln (|H|_u/h)/\ln\Lambda_{\min}
$$
We will use (\ref{smallk}) for $k< k_{\ast}$ and (\ref{largek})
for $k\geq k_{\ast}$.
We can now apply Lemma~\ref{lmnug} with $B=3DB_0$,
cf. (\ref{numj}). But first, just to simplify our
calculations, we will assume that $h$ (and hence, $g$)
is small enough, so that, e.g., $(\bar{D}h/d_0)^{-g}=1+o(1)
\leq 2$ and $\left [\log_2(\bar{D}h/d_0)^{-1}\right ]^g
=1+o(1)\leq 2$, $(BD)^g=1+o(1)\leq 2$, etc.
Then Lemma~\ref{lmnug} combined with (\ref{numj}) and
(\ref{smallk}) yields
\begin{eqnarray*}
\sum_{k=1}^{k_{\ast}-1}\sum_j
\left [\nu_{1,k}(\tilde{W}^u_{1,k,j})\right ]^{1-g}
&\leq& 800\, D\sum_{k=1}^{k_{\ast}-1}\Lambda_{\max}^{kg}\,
(\bar{D}h/d_0)^{1-g}\, \left [\log_2(\bar{D}h/d_0)^{-1}\right ]^g
\nonumber\\
&\leq& 3200\, D\bar{D}\cdot\frac{h}{d_0}\,
\frac{\Lambda_{\max}^{k_{\ast}g}-1}{\Lambda_{\max}^g-1}
\end{eqnarray*}
Note that both $g$ and $k_{\ast}g$ approach zero as $h\to 0$.
Thus, if $h$ is sufficiently small, we have
$$
\frac{\Lambda_{\max}^{k_{\ast}g}-1}{\Lambda_{\max}^g-1}
= k_{\ast} + o(k_{\ast})\leq 2k_{\ast}
$$
and hence
\be
\sum_{k=1}^{k_{\ast}-1}\sum_j
\left [\nu_{1,k}(\tilde{W}^u_{1,k,j})\right ]^{1-g}
\leq D' (h/d_0)\, \ln (|H|_u/h)
\label{kast1}
\ee
where $D'=10^6D^5/\ln\Lambda_{\min}$.
Next, using Lemma~\ref{lmnug} and (\ref{largek}) gives
$$
\sum_{k=k_{\ast}}^{\infty}\sum_j
\left [\nu_{1,k}(\tilde{W}^u_{1,k,j})\right ]^{1-g}
\leq
C_g\sum_{k=k_{\ast}}^{\infty}
\Lambda_{\max}^{kg}\,
\left (\frac{B_0D^2\Lambda_{\max}|H|_u}{\Lambda_{\min}^{k}d_0}\right)^{1-g}
\, \left [\log_2
\left (\frac{\Lambda_{\min}^{k}d_0}{B_0D^2\Lambda_{\max}|H|_u}\right)\right ]^g
$$
We again assume that $h$ is small enough, then it is
simple estimation that
\be
\sum_{k=k_{\ast}}^{\infty}\sum_j
\left [\nu_{1,k}(\tilde{W}^u_{1,k,j})\right ]^{1-g}
\leq D'' (h/d_0)\, \ln (|H|_u/h)
\label{kast2}
\ee
where $D''=2000\,D^3B_0\Lambda_{\max}$. (A crucial
point here is to observe that $t=\Lambda_{\max}^g/\Lambda_{\min}^{1-g}
<1$, and $\sum_{k_{\ast}}^\infty k^gt^k < \int_{k_{\ast}}^\infty
xt^x\, dx<2k_{\ast}t^{k_{\ast}}\ln t^{-1}$ with $t^{k_{\ast}}
\leq 2h/|H|_u$.)
Combining (\ref{g}), (\ref{gen1a}), (\ref{kast1}) and (\ref{kast2})
(and assuming $h$ is small enough) gives
\be
\nu_1\{x:\, {\rm gen}(x,n,W^u_1)=1\}
\leq D_3D_4\, \nu_2(W^u_2\cap M_{-n}) \cdot (h/d_0)\, \ln (|H|_u/h)
\label{gen1d}
\ee
with $D_4=2(D'+D'')$. We assume $h$ is so small that
\be
h_1\stackrel{\rm def}{=}(h/d_0)\, \ln (|H|_u/h)0$, and $G\subset M$ be
an arbitrary subset such that for every maximal unstable fiber
$W^u\in{\cal W}^u_m$ the intersection $W^u\cap G$ is a union of
no more than $2B$ subfibers of $W^u$, and each of
those subfibers has length $\leq\varepsilon$.
One can think of $G=U_{m,\varepsilon}$, as an example, in which $B=1$.
Loosely speaking, the set $G$ is $(2B\varepsilon)$-thick in
the unstable direction.
\begin{theorem}
For any $\mu\in{\cal M}_m$,
$B\geq 1$, $\varepsilon >0$, and any set $G$ described above,
we have for all $n\geq 0$
\be
\mu(G\cap M_{-n})\leq C_8 B\delta_n\varepsilon^{1-g}
\label{muC8g}
\ee
where $C_8=2C_1C_7D$.
\label{tmGMn}
\end{theorem}
{\em Proof}. Consider the factor measure $\mu^f$ induced by
$\mu$ on the set of maximal unstable fibers ${\cal W}^u_m$. Then
\begin{eqnarray}
\mu(U_{m,B\varepsilon})
&=&\int_{{\cal W}^u_m}\nu_{W^u}(W^u\cap U_{m,B\varepsilon})\, d\mu^f(W^u)\nonumber\\
&\geq&\mu\left (\cup W^u:\, W^u\in {\cal W}^u_{m,2B\varepsilon}\right )\nonumber\\
&+& \int_{{\cal W}^u\setminus {\cal W}^u_{m,2B\varepsilon}}
D^{-1}(2B\varepsilon/|W^u|) d\mu^f(W^u)
\label{muf}
\end{eqnarray}
Now, let
\be
F_{\mu}(y)=\mu\left (\cup W^u:\, W^u\in {\cal W}^u_{m,y}\right )
\label{Fmu0}
\ee
The function $F_{\mu}(y)$ is a kind of distribution function
of the length of maximal fibers $W^u\in {\cal W}^u_m$ with respect
to the given measure $\mu\in{\cal M}_m$.
Then (\ref{muf}) and the properties (M1)-(M2) of
$\mu$ imply that for any $\varepsilon>0$ we have
\be
F_{\mu}(2B\varepsilon)+
\int_{2B\varepsilon}^{d_0}D^{-1}(2B\varepsilon/y)\,dF_{\mu}(y)
\leq C_1B\varepsilon
\label{Fmu}
\ee
In particular,
\be
F_{\mu}(y)\leq \frac 12 C_1y
\label{Fmuy}
\ee
In the following estimate, we apply Theorem~\ref{tmsmall}
to fibers of length $>2B\varepsilon$ and a subfiber of
length $2B\varepsilon$, and Corollary~\ref{crsmall}
to fibers smaller than $2B\varepsilon$:
\begin{eqnarray}
\mu(G\cap M_{-n})
&\leq&\int_{2B\varepsilon}^{d_0}2BC_7\delta_n
\varepsilon^{1-g}/y\, dF_{\mu}(y)
+\int_0^{2B\varepsilon}C_7\delta_ny^{-g}\, dF_{\mu}(y)
\nonumber\\
&=&
C_7D\delta_n\varepsilon^{-g}\int_{2B\varepsilon}^{d_0}
D^{-1}(2B\varepsilon/y)\, dF_{\mu}(y)
+C_7\delta_n(2B\varepsilon)^{-g}F_{\mu}(2B\varepsilon)
\nonumber\\
&+& C_7g\delta_n
\int_0^{2B\varepsilon}y^{-1-g}F_{\mu}(y)\, dy
\label{GMn}
\end{eqnarray}
Here we applied the integration by parts to the integral from 0
to $2B\varepsilon$.
Due to (\ref{Fmuy}),
the last integral in (\ref{GMn}) does not exceed
$C_1B\varepsilon^{1-g}/(1-g)$. To the first two terms on the
right hand side of (\ref{GMn}) we apply (\ref{Fmu}) and get
$$
\mu(G\cap M_{-n})\leq
C_7D\delta_n\varepsilon^{-g}C_1B\varepsilon
+C_1C_7Bg\delta_n\varepsilon^{1-g}/(1-g)
$$
For $h$ small enough, we have $g/(1-g)d_0\Lambda_{\max}^{-1}\right )
\nonumber\\
&\geq& C_6^{-1}\delta_n\,\mu\left (\cup W^u:\,
|W^u|>d_0\Lambda_{\max}^{-1}\right )
\label{muMn}
\end{eqnarray}
Here we used Theorem~\ref{tmn12} (recall that $|W^u|\leq d_0$,
$\forall W^u\in{\cal W}^u$).
Using the property (M2) with $\varepsilon=(2C_1)^{-1}$ gives
$$
\mu\left (\cup W^u:\, |W^u|>C_1^{-1}\right )\geq 1/2
$$
Since $C_1=48D/d_0$ and $\Lambda_{\max}\geq
\Lambda_{\min}\geq 64D^2$, cf. Section 2.2, then
$$
\mu\left (\cup W^u:\, |W^u|>d_0\Lambda_{\max}^{-1}\right )
\geq 1/2
$$
so that (\ref{muMn}) implies
\be
\mu(M_{-n})\geq \delta_n/(2C_6)
\label{mudelta2}
\ee
Lastly, recall that
$\mu_+(M_{-n})=||T^n_{\ast}\mu_+||=\lambda_+^n$.
Combining (\ref{mudelta1}) and (\ref{mudelta2}) gives the following:
\begin{corollary}
For any $\mu\in{\cal M}_m$ and $n\geq 0$ we have
\be
C_9^{-2}\lambda_+^n\leq\mu(M_{-n})\leq C_9^2\lambda_+^n
\label{muC9}
\ee
and
\be
C_9^{-1}\lambda_+^n\leq\delta_n\leq C_9\lambda_+^n
\label{delC9}
\ee
with $C_9=\max\{2C_6,d_0C_8\}$.
The last bound holds true for any $\delta_n(W^u)$,
$W^u\in{\cal W}^u(d_0)$.
\label{crC9}
\end{corollary}
This also gives a corollary to Lemma~\ref{lmTk1}:
\begin{corollary}
Using notation of Lemma~\ref{lmTk1}, we have for any $n\geq k_1$
$$
C_{10}^{-1}\lambda_+^{n-k_1}d^u_{\max}(R)\leq
\left ( T_{\ast}^{n}\mu\right )(R)
\leq C_{10}\lambda_+^{n-k_1}d^u_{\min}(R)
$$
with $C_{10}=C_5C_9^2$.
\label{crC10}
\end{corollary}
Recall that for any $\varepsilon>0$ we denote by
$H_{\varepsilon}\subset M$ the
$\varepsilon$-neighborhood of $\partial H$ in $M$.
\begin{theorem}
Let $\varepsilon>0$ and put
$$
q_{\varepsilon}=\ln(D^{-1}d_0/\varepsilon)/\ln\Lambda_{\max}
$$
Then for any $\mu\in{\cal M}_m$ and $q\geq
q_{\varepsilon}$, $n\geq 0$, we have
\be
\left (T^q_+\mu\right )(H_{\varepsilon}\cap M_{-n})\leq
C_{11}\varepsilon^{b(1-4g)}\left (T^q_+\mu\right )(M_{-n})
\label{Heps}
\ee
where $b=\ln\Lambda_{\min}/(2\ln\Lambda_{\max})$ and
$C_{11}>0$ is a global constant.
\label{tmHeps}
\end{theorem}
{\em Proof}. Assume for a moment that $\partial H$ consists
of a finite number of smooth curves transversal to the unstable
direction. Then the set $H_{\varepsilon}$ is at most
$(B_0\varepsilon)$-thick in the unstable direction, and
a combination of Theorem~\ref{tmGMn} and (\ref{mudelta2})
applied to the measure $T^q_+\mu$
imply (\ref{Heps}) with $b=1/2$ and all $q\geq 0$.
Similarly, if $\partial H$ consists of curves which
either are transversal to the unstable direction or
have curvature larger than that of unstable fibers,
the set $H_{\varepsilon}$ is
(const$\cdot\varepsilon^{1/2}$)-thick in the unstable
direction and we again get (\ref{Heps}) with $b=1/2$
and all $q\geq 0$.
There are, however, reasons why the theorem cannot always
hold for all $q\geq 0$. For example, let a hole $H'\subset H$
be a rectangle of size $h$ in the unstable direction,
and $\mu$ be supported on a single unstable fiber
of length $d_0$ that is $(\varepsilon/2)$-close
to an unstable side of $H'$, then $\mu(H_{\varepsilon})\geq h/d_0$
independently of $\varepsilon$.
Now, in the general case, we consider maximal connected
unstable fibers $W^u\subset H_{\varepsilon}$ and call
them long if $|W^u|\geq\varepsilon^b$.
To any long fiber $W^u\subset H_{\varepsilon}$
we associate the rectangle $R=R(W^u)$ such that (i)
the fiber $W^u$ is an $R$-fiber \cite{CMT}
(i.e., it terminates on the stable sides of $R$)
and (ii) the minimal distance from
$W^u$ to each unstable side of $R(W^u)$, measured
along stable fibers, is $2\varepsilon$ (the maximal
distance is then $<2D\varepsilon$). We now
pick any long fiber $W^u_1\subset H_{\varepsilon}$,
then any long fiber $W^u_2\subset H_{\varepsilon}
\setminus R(W^u_1)$, etc. In that way we will
find a finite collection of rectangle $R(W^u_r)$,
$r=1,\ldots,\bar{r}$ such that the residual set $H_{\varepsilon}
\setminus\left (\cup_r R(W^u_r)\right )$ consists
of short fibers. That residual set is then at most
$(2B_0\varepsilon^b)$-thick in the unstable direction,
and the above argument applies giving the right estimate
for it. It remains to consider the set $\cup_r R(W^u_r)$.
The regularity of our holes $H'\subset H$ (the fact
that any
local stable/unstable fiber intersects any hole
in no more than $B_0$ intervals) implies that
$$
\sum_{r=1}^{\bar{r}} |W^u_r|\leq B_1|H|_u
$$
with some $B_1={\rm const}(B_0)>0$, cf. (\ref{Hu}). Therefore,
\be
\bar{r}\leq B_1|H|_u\varepsilon^{-b}
\label{barr}
\ee
For any $r=1,\ldots,\bar{r}$ the rectangle
$R_r'=\hat{T}^{-q_{\varepsilon}}R(W^u_r)$ has
length $\leq 4d_0$ in the stable direction
and $\leq |W^u_r|\cdot\Lambda_{\min}^{-q_{\varepsilon}}\leq
|H|_u(D\varepsilon/d_0)^{2b}$ in the unstable
direction. Let $\mu'=T^{q-q_{\varepsilon}}_+\mu$
(note that $\mu'\in{\cal M}_{m+q-q_{\varepsilon}}$).
Observe that
$$
\left (T^q_+\mu\right )(R(W^u_r)\cap M_{-n})=
\left (T^{q_\varepsilon}_+\mu'\right )(R(W^u_r)\cap M_{-n})=
\frac{\left (T^{q_\varepsilon}_\ast\mu'\right )(R(W^u_r)\cap M_{-n})}
{||T^{q_{\varepsilon}}_\ast\mu'||}\leq
\frac{\mu'(R_r'\cap M_{-n-q_{\varepsilon}})}
{||T^{q_{\varepsilon}}_\ast\mu'||}
$$
Applying Theorem~\ref{tmGMn} with $B=1$ to $R_r'$ gives
$$
\mu'(R_r'\cap M_{-n-q_{\varepsilon}})\leq
C_8\delta_{n+q_{\varepsilon}}|H|_u^{1-g}(D\varepsilon/d_0)^{2b(1-g)}\leq
2C_8\delta_{n}|H|_u(D\varepsilon/d_0)^{2b(1-g)}
$$
According to (\ref{mudelta2}), (\ref{nlnn}), (\ref{g}), and (\ref{C2'}),
we have
$$
||T^{q_{\varepsilon}}_\ast\mu'||=\mu'(M_{-q_{\varepsilon}})
\geq\delta_{q_{\varepsilon}}/(2C_6)\geq (2C_6)^{-1}e^{-C_2'hq_{\varepsilon}}
\geq (2C_6)^{-1}(D\varepsilon/d_0)^{2bg}
$$
Combining the above estimates gives
$$
\left (T^q_+\mu\right )(R(W^u_r)\cap M_{-n})\leq
4C_6C_8\delta_n|H|_u(D\varepsilon/d_0)^{2b(1-2g)}
$$
Together with (\ref{barr}) this yields
$$
\left (T^q_+\mu\right )\left(\left [\cup_r R(W^u_r)\right ]\cap M_{-n}\right )\leq
4C_6C_8DB_1\delta_n|H|_u^2\varepsilon^{b(1-4g)}/d_0^{2b(1-2g)}
$$
Theorem~\ref{tmHeps} is now proved. $\Box$\medskip
Let $W^u\subset\hat{M}$ be an unstable fiber.
For every $n\geq 0$ and $x\in W^u\cap M_{-n}$ denote by
$W^u_n(x)$ the smooth component of $T^n(W^u\cap M_{-n})$
containing $T^nx$ and by $|W^u_n(x)|$ its length.
For every $x\in W^u\cap M_{-n}$ let $l_n(x)=\min\{
l\in [0,n]:\, |W^u_l(x)|\geq d_0/2\}$, and if $|W^u_l(x)|0$
such that for every $n\geq m\geq 0$
$$
\nu_{W^u}\{x\in W^u\cap M_{-n}:\, l_n(x)\geq m\}
\leq C_{13}\,\beta_1^m\lambda_+^n/|W^u|
$$
\label{lmWuln}
\end{lemma}
{\em Proof}. For each $i=0,\ldots,m$ let $W^u_i=\{x\in W^u\cap M_{-i}:\,
l_i(x)=i\}$. The points $x\in W^u_i$ have their first $i-1$ images under
$T$ in short components (of length $q$. Since
$$
M_{-n+q}\setminus M^{(k)}_{-n+q} =
\cup_{i=0}^{n-q} \hat{T}^{-i}(H^{(k)}\setminus H) \subset
\cup_{i=0}^{n-q} \hat{T}^{-i}H_{\varepsilon_k}
$$
we have the following estimate:
\begin{eqnarray}
\mu_q\left (M_{-n+q}\setminus M^{(k)}_{-n+q}\right )
&\leq& \sum_{i=0}^{n-q}\mu_q(M_{-n+q}\cap\hat{T}^{-i}
H_{\varepsilon_k})
\nonumber\\
&\leq& \sum_{i=0}^{n-q} C_{11}\varepsilon_k^{b(1-4g)}
\mu_q(M_{-n+q})
\nonumber\\
&=& C_{11}(n-q+1)\varepsilon_k^{b(1-4g)}\mu_q(M_{-n+q})
\label{mu0k}
\end{eqnarray}
Here we used the following estimate, based on (\ref{qk}) and
Theorem~\ref{tmHeps}:
\begin{eqnarray}
\mu_q\left (M_{-n+q}\cap\hat{T}^{-i}H_{\varepsilon_k}\right )
&=&
\mu_q\left [\hat{T}^{-i}\left (M_{-n+q+i}\cap H_{\varepsilon_k}\right )\cap M_{-i}\right ]
\nonumber\\
&=&
\left ( T_{\ast}^i\mu_q\right )\left ( M_{-n+q+i}\cap H_{\varepsilon_k}\right )
\nonumber\\
&=&
\left (T_+^i\mu_q\right )
\left (M_{-n+q+i}\cap H_{\varepsilon_k}\right )\cdot ||T_{\ast}^i\mu_q||
\nonumber\\
&\leq&
C_{11}\varepsilon_k^{b(1-4g)}\left (T_+^i\mu_q\right )(M_{-n+q+i})\cdot ||T_{\ast}^i\mu_q||
\nonumber\\
&=&
C_{11}\varepsilon_k^{b(1-4g)}\cdot\left (T_{\ast}^i\mu_q\right )(M_{-n+q+i})
\nonumber\\
&=&
C_{11}\varepsilon_k^{b(1-g)}\mu_q(M_{-n+q})
\nonumber
\end{eqnarray}
for any $i=0,1,\ldots,n-q$. The following bound is symmetric
to (\ref{mu0k}), its proof is similar:
\be
\mu_q^{(k)}\left (M_{-n+q}^{(k)}\setminus M_{-n+q}\right )
\leq C_{11}(n-q+1)\varepsilon_k^{b(1-4g)}\mu_q^{(k)}(M_{-n+q}^{(k)})
\label{mu0kk}
\ee
Observe that the measure $\mu_n$ is supported on $M_{n-q}$ and
the measure $\mu_n^{(k)}$ is supported on $M_{n-q}^{(k)}$.
On the common part of their supports,
$M_{n-q}\cap M_{n-q}^{(k)}$, these two measures are
proportional, so their conditional measures on that
common part coincide, we call that conditional measure
$\hat{\mu}_+^{(k)}$. Therefore,
\be
\mu_n=(1-\sigma_n)\hat{\mu}_n^{(k)} + \sigma_n\mu'
\label{sigma1}
\ee
and
\be
\mu_n^{(k)}=(1-\sigma_n^{(k)})\hat{\mu}_n^{(k)} + \sigma_n^{(k)}\mu''
\label{sigma2}
\ee
for some $\sigma_n,\sigma_n^{(k)}>0$ and some
probability measures $\mu',\mu''$ supported on
$M_{n-q}\setminus M_{n-q}^{(k)}$ and
$M_{n-q}^{(k)}\setminus M_{n-q}$, respectively.
Moreover, it follows from
(\ref{mu0k}) and (\ref{mu0kk}) that
\be
\sigma_n+\sigma_n^{(k)}\leq
2C_{11}n\varepsilon_k^{b(1-4g)}\leq C_{12}n\alpha_1^k
\label{sigma3}
\ee
with $C_{12}=2C_{11}D_5^{b/2}$ and $\alpha_1=\Lambda_{\min}^{-b/2}<1$
(recall that $g$, as well as $h$, are very small, so we can
assume that $g<1/8$). Therefore,
for any bounded function $f(x)$ on $M$ we have
\be
\left |\int_Mf(x)\, d\mu_n
-\int_Mf(x)\, d\mu_n^{(k)}\right |
\leq 2||f||_{\infty}(\sigma_n+\sigma_n^{(k)})
\leq 2||f||_{\infty}C_{12}n\alpha_1^k
\label{sigma4}
\ee
\noindent
{\bf Definition}.
We say that two sequences of probability measures, $\{\mu_n'\}$
and $\{\mu_n''\}$ on $M$ are (asymptotically) equivalent
(denoted by $\mu_n'\sim\mu_n''$) if for any continuous function
$f(x)$ on $M$
$$
\lim_{n\to\infty}\left |\int_Mf(x)\, d\mu_n'
-\int_Mf(x)\, d\mu_n''\right |=0
$$
Note that if $\mu_n'\sim\mu_n''$ and $\mu_n'$ converges
weakly to a limit measure, $\mu_{\infty}$, then $\mu_n''$
also converges weakly to $\mu_{\infty}$.
\begin{proposition}
Let $k=k_n$ be a sequence that grows faster than any
logarithmic function but slower than any linear function:
$k_n=o(n)$ and $1/k_n=o(1/\ln n)$ as $n\to\infty$. Then
$\mu_n\sim\mu_n^{(k_n)}$.
\label{prmunmunk}
\end{proposition}
{\em Proof}.
For large $n$ we have $n>q=k_n+l_0$, so that
(\ref{sigma1})-(\ref{sigma4}) hold, and $n\alpha_1^{k_n}
\to 0$ as $n\to\infty$. $\Box$
\medskip
Our next goal is to show that
there is a sequence $k_n\to\infty$ such that $k_n=o(n)$ and
$1/k_n=o(1/\ln n)$, for which $\mu^{(k_n)}_n\sim\mu_+^{(k_n)}$.
We say that a sequence of numbers $\{L_k\}$ majorizes
another sequence $\{M_k\}$ if $L_k\geq M_k$, $\forall k\geq 1$.
The above goal will be achieved if we show
the following:\medskip
{\em There is a sequence $\{M_k\}$
such that for any majorizing sequence $\{L_k\}$ the sequence
of measures $\mu_{L_k}^{(k)}=[T_+^{(k)}]^{L_k-k-l_0}\mu_{k+l_0}$
is equivalent to $\mu_+^{(k)}$, i.e.
\be
\lim_{k\to\infty}\left |\int_Mf(x)\, d\mu_{L_k}^{(k)}
-\int_Mf(x)\, d\mu_+^{(k)}\right |=0
\label{intint}
\ee
for every $f\in C(M)$. The sequence
$\{M_k\}$ has to grow faster than any linear function $Ak$,
$A>0$, and slower than any exponential function $e^{ak}$, $a>0$. }\medskip
The weak convergence of $\mu_L^{(k)}$ to $\mu_+^{(k)}$, as
$L\to\infty$, for all $k\geq k_0$ was proved in \cite{CMT}.
It implies that for any continuous function $f\in C(M)$ there is
a sequence $M_k=M_k(f)$ such that (\ref{intint}) holds for any
majorizing sequence $\{L_k\}$. All
we need to prove is that one can choose $M_k$ uniformly
in $f$ and so that $M_k=o(e^{ak})$, $\forall a>0$.
Recall that the measures $\mu_L^{(k)}$ and $\mu_+^{(k)}$ are
supported on $M_{L-q}^{(k)}\subset M^{(k)}$ and $M_+^{(k)}\subset M^{(k)}$,
respectively. Recall that diam$R\leq \varepsilon_k=D_5\Lambda_{\min}^{-k}$,
$\forall R\in{\cal R}^{(k)}$. The oscillation of
$f(x)$ on any rectangle $R\in {\cal R}^{(k)}$
does not exceed $\delta_f(\varepsilon_k)$, and
$\delta_f(\varepsilon_k)\to 0$ as $k\to\infty$.
For any two probability measures, $\mu'$ and $\mu''$, on $M^{(k)}$ let
$$
|\mu'-\mu''|_k:=\sum_{R\in{\cal R}^{(k)}}
\left |\mu'(R)-\mu''(R)\right |
$$
%(this is the distance in variation between two probability distributions
%on the atoms of ${\cal R}^{(k)}$ induced by these two measures).
It is then a simple calculation that
\be
\left |\int_Mf(x)\, d\mu_{L}^{(k)}-\int_Mf(x)\, d\mu_+^{(k)}\right |
\leq 2\delta_f(\varepsilon_k)+||f||_{\infty}\cdot
\left | \mu_L^{(k)}-\mu_+^{(k)}\right |_k
\label{intL+}
\ee
Therefore, to establish (\ref{intint})
it is enough to prove the following:
\begin{proposition}
There is a sequence $M_k\to\infty$ such that for any majorizing
sequence $L_k$
$$
\lim_{k\to\infty} \left |\mu_{L_k}^{(k)}-\mu_+^{(k)}\right |_k = 0
$$
The sequence $\{M_k\}$ has to grow faster than any linear function $Ak$,
$A>0$, and slower than any exponential function $e^{ak}$, $a>0$.
\label{prmu+L}
\end{proposition}
We will also sharpen this proposition as follows:
\begin{proposition}
There are global constants $r\geq 1$, $C_{14}>0$ and
$\alpha_2\in (0,1)$ such that for all $k\geq k_0$ and
$L\geq rk^2+k+l_0$
$$
|\mu_L^{(k)}-\mu_+^{(k)}|\leq C_{14}\alpha_2^k
$$
\label{prmukL}
\end{proposition}
These two propositions are only concerned with properties of
conditionally invariant measures $\mu_+^{(k)}$
for the Anosov maps $T^{(k)}$ with rectangular holes, $H^{(k)}$.
So, we can apply the finite-dimensional approximations
and matrix techniques developed in \cite{C86,CM1,CM2}.
A complete proof of Proposition~\ref{prmukL} is provided
in Section~\ref{secPPP}, while \ref{prmu+L} immediately
follows from \ref{prmukL}.
\section{Proofs of the main theorems}
\label{secPMT}
\setcounter{equation}{0}
{\em Proof of Theorem~\ref{tmmain1}}. To prove the weak convergence
of $T^n_+\mu_0$ to $\mu_+$, it is enough to
combine Proposition~\ref{prmunmunk}, (\ref{intint}) and the weak
convergence of $\mu_+^{(k)}$ to $\mu_+$ proved in \cite{CMT} for
the specific rectangular holes $H^{(k)}$ described in the first paragraph
of the previous section.
The second statement of the theorem will follow from Corollary~\ref{crmulam}
below. $\Box$\medskip
{\em Proof of Theorem~\ref{tmmain2}}. For any large $n$, i.e.
$n\geq n_0:=rk_0^2+k_0+l_0$, we take
$k=k_n=\max\{k:\, n\geq rk^2+k+l_0\}$. Then we combine (\ref{sigma4})
and (\ref{intL+}) with Proposition~\ref{prmukL},
in which we set $L=n$. As a result,
\be
\left |\int_Mf(x)\, d\mu_n
-\int_Mf(x)\, d\mu_+^{(k_n)}\right |\leq {\rm const}\cdot
\left [\delta_f\left (e^{-a_1n^{1/2}}\right )
+||f||_{\infty}e^{-a_2n^{1/2}}\right ]
\label{intint5}
\ee
with some global constants $a_1,a_2>0$ and const$>0$.
Since this bound holds for any $\mu_0\in{\cal M}_0$,
it holds, in particular, for the conditionally invariant measure
$\mu_+$, for which we have $(\mu_+)_n=T_+^n\mu_+=\mu_+$. Now, we
combine the above bound for $\mu_n=T_+^n\mu_0$ with that same
bound for $\mu_+=T^n_+\mu_+$ and use the triangle inequality.
That concludes the proof of Theorem~\ref{tmmain2} for
$n\geq n_0$. To enforce that theorem for all $n\geq 0$,
it is enough to adjust the constant coefficient in
(\ref{intint2}). $\Box$\medskip
{\em Remark}. For each $k\geq k_0$ put $n=rk^2+k+l_0$
in (\ref{intint5}), and then combining it with
the just proved Theorem~\ref{tmmain2} yields
\be
\left |\int_Mf(x)\, d\mu_+^{(k)}
-\int_Mf(x)\, d\mu_+\right |\leq {\rm const}\cdot
\left [\delta_f\left (e^{-a_1k}\right )
+||f||_{\infty}e^{-a_2k}\right ]
\label{intint6}
\ee
Note that the measures $\mu_+^{(k)}$ here correspond to any
rectangular holes $H^{(k)}$ that properly approximate $H$.
\medskip
Let $\mu_0\in{\cal M}_0$ and $\mu_n=T^n_+\mu_0$. Recall that
$||T_{\ast}\mu_n||=\mu_n(M_{-1})$.
\begin{lemma}
There are global constants $C_{15},a_5>0$ such that $\forall n\geq 0$
$$
\exp(-C_{15}e^{-a_5n^{1/2}})\leq
\mu_n(M_{-1})/\lambda_+
\leq \exp (C_{15}e^{-a_5n^{1/2}})
$$
\label{lmmulam}
\end{lemma}
{\em Proof}. Let again $n\geq n_0$ and
$k=k_n=\max\{k:\, n\geq rk^2+k+l_0\}$.
Then (\ref{sigma1}-\ref{sigma3}), and Proposition~\ref{prmukL}
allow us to compare the measures $\mu_n$ and $\mu_+=(\mu_+)_n$ with
the measure $\mu_+^{(k)}$ and imply that
$|\mu_n-\mu_+|_k\leq Ce^{-ak}$ with some global constants
$C,a>0$. Therefore,
$$
\left |\sum_{R\subset M_{-1}}\mu_n(R)-
\sum_{R\subset M_{-1}}\mu_+(R)\right |
\leq Ce^{-ak}
$$
where the sums are taken over $R\in{\cal R}^{(k)}$.
Furthermore,
$$
\sum_{R\cap\partial M_{-1}\neq\emptyset}\mu_n(R)
\leq [T_{\ast}\mu_n](H_{\varepsilon_k'})\leq C_{11}
\left (\varepsilon_k'\right )^{b(1-4g)}
$$
with $\varepsilon_k'=\varepsilon_k\Lambda_{\max}$
(observe that if $R\cap\partial M_{-1}\neq\emptyset$,
then $R$ lies in an $\varepsilon_k$-neighborhood of
$\hat{T}^{-1}(\partial H)$, and so $\hat{T}R\subset
H_{\varepsilon_k'}$).
The second bound follows from (\ref{Heps}), by setting $n=0$ there.
The above estimate also holds for $\mu_+$. Hence,
$$
\left | \mu_n(M_{-1})-\mu_+(M_{-1})\right |
\leq Ce^{-ak}+2C_{11}\left (\varepsilon_k'\right )^{b(1-4g)}
$$
for all $n\geq n_0$.
Lemma~\ref{lmmulam} is then proven for $n\geq n_0$.
To enforce it for all $n\geq 0$, it is enough to adjust
the value of $C_{15}$. $\Box$\medskip
Now recall that $||T^n_{\ast}\mu_0||=\mu_0(M_{-n})=\prod_{i=0}^{n-1}
\mu_i(M_{-1})$.
\begin{corollary}
For any measure $\mu_0\in{\cal M}_0$ there is a finite positive limit
\be
\lim_{n\to\infty}\mu_0(M_{-n})/\lambda_+^n:= \rho(\mu_0)
\label{rhomu}
\ee
Furthermore,
\be
00$.
In addition, $\forall n\geq 0$
\be
|\mu_0(M_{-n})/\lambda_+^n-\rho(\mu_0)|\leq
\gamma_n':=C_{15}C_{16}\sum_{j=n}^{\infty}e^{-a_5j^{1/2}}
\label{gamma'}
\ee
\label{crmulam}
\end{corollary}
Now, let $W^u\subset\hat{M}$ be an unstable fiber and $\nu$
its u-SBR measure. We will use the functions $l_n(x)$, $n\geq 0$, on
$W^u$ introduced before Lemma~\ref{lmWuln}. In terms of these
functions, the fiber $W^u$ is said to be eventually long, cf.
Section~3 in \cite{CMT},
if $l_n(x)\leq n$ for some $n\geq 0$ and some $x\in W^u\cap M_{-n}$,
otherwise the fiber $W^u$ is said to be forever short.
\begin{theorem}
For every local unstable fiber $W^u\subset\hat{M}$ there is a finite limit
\be
\lim_{n\to\infty}\nu(W^u\cap M_{-n})/\lambda_+^n:= \rho(W^u)
\label{rhoW}
\ee
which is positive if and only if the fiber $W^u$ is eventually long.
In addition,
\be
|\nu(W^u\cap M_{-n})/\lambda_+^n-\rho(W^u)|\leq
\gamma_n''/|W^u|
\label{rhoWa}
\ee
where
\be
\gamma_n'':=C_{17}\sum_{j=0}^n\gamma_{n-j}'\beta_1^j\, +\,
C_{18}\sum_{j=n}^{\infty}\beta_1^j\leq C_{19}e^{-a_6n^{1/2}}
\label{rhoWb}
\ee
with some global constants $a_6,C_{17},C_{18},C_{19}>0$ and
$\gamma_n'$ defined in (\ref{gamma'}).
\label{tmrhoW}
\end{theorem}
{\em Remark}. Due to Corollaries~\ref{crsmall} and
\ref{crC9}, we have $\nu(W^u\cap M_{-n})/\lambda_+^n
\leq C_{20}|W^u|^{-g}$ for every $n\geq 0$,
and so $\rho(W^u)\leq C_{20}|W^u|^{-g}$. Here $C_{20}=C_7C_9$\medskip
{\em Remark}. If $W^u\subset M_+$, then $\rho(T^{-n}W^u)
=\lambda_+^{-n}\rho(W^u)$ for all $n\geq 0$. \medskip
{\em Proof}. For fibers $W^u$ of length between $d_0/2$ and $d_0$,
we have $\nu\in{\cal M}_0$, thus the theorem follows from
Corollary~\ref{crmulam}, and we actually get
a slightly better estimate: $\gamma_n''=d_0\gamma_n'$.
Assume now that $|W^u|0$
\be
\int_{{\cal W}_{+,\varepsilon}^u}\rho(W^u)\, d\mu_+^f
\leq C_{20}\int_0^{\varepsilon}y^{-g}\, dF_{\mu_+}(y)
\leq C_{20}C_1\varepsilon^{1-g}
\label{inteps}
\ee
Observe that the above bound also holds if we substitute
$\rho_n(W^u)$ for $\rho(W^u)$.
We now prove (\ref{intint4}), which will imply the weak
convergence of $\mu_{+,n}$ to $\bar{\mu}_+$.
For every $n\geq 1$ we take $\varepsilon=\varepsilon_n=
e^{-a_7n^{1/2}}$ with $a_7=a_6/2$. Then according to
(\ref{rhoWb}), $\gamma_n''/\varepsilon
\leq C_{19}e^{-a_7n^{1/2}}$. We put, for brevity,
${\cal W}^{u,c}_{+,\varepsilon}={\cal W}^u_+
\setminus{\cal W}^u_{+,\varepsilon}$.
We have
\begin{eqnarray}
\left |\int_Mf\, d\mu_{+,n}
-\int_Mf\, d\bar{\mu}_+\right | &=&
\left |\int_{{\cal W}^u_+}d\mu_{+,n}^f\int_{W^u}f\, d\nu_{W^u,n}
-\int_{{\cal W}^u_+}d\bar{\mu}_+^f\int_{W^u}f\, d\bar{\nu}_{W^u}\right |\nonumber\\
&\leq &
\left |\int_{{\cal W}^u_{+,\varepsilon}}d\mu_{+,n}^f\int_{W^u}f\, d\nu_{W^u,n}\right |
+\left |\int_{{\cal W}^u_{+,\varepsilon}}d\bar{\mu}_+^f\int_{W^u}f\, d\bar{\nu}_{W^u}\right |\nonumber\\
& + &
\left |\int_{{\cal W}^{u,c}_{+,\varepsilon}}d\mu_{+,n}^f\int_{W^u}f\, d\nu_{W^u,n}
-\int_{{\cal W}^{u,c}_{+,\varepsilon}}d\bar{\mu}_+^f\int_{W^u}f\, d\bar{\nu}_{W^u}\right |
\label{gross1}
\end{eqnarray}
The first two integrals on the right hand side of (\ref{gross1})
do not exceed
$$
2||f||_{\infty}C_{20}\int_0^{\varepsilon}y^{-g}\, dF_{\mu_+}(y)
\leq 2||f||_{\infty}C_{20}C_1\varepsilon^{1/2}
$$
according to (\ref{inteps}).
Using (\ref{rhoW2}) and (\ref{rhoWu}) we see that
the last term on the right hand side of (\ref{gross1})
does not exceed $I_1+I_2$, where
$$
I_1 = \int_{{\cal W}^{u,c}_{+,\varepsilon}}
\left |\int_{W^u}f(x)\, d\nu_{W^u,n}
-\int_{W^u}f(x)\, d\bar{\nu}_{W^u}\right |\,\rho(W^u)\, d\mu_+^f
$$
and
$$
I_2 = \int_{{\cal W}^{u,c}_{+,\varepsilon}}\left |\int_{W^u}f(x)\, d\nu_{W^u,n} \right |\cdot
\left |\rho_n(W^u)-\rho(W^u)\right |\, d\mu_+^f
$$
The bound (\ref{rhoWa}) immediately implies that
$I_2\leq ||f||_{\infty}\gamma_n''/\varepsilon$.
We now bound $I_1$. Observe that $\rho(W^u)=0$ for forever short
fibers, so they can be ignored. Let $W^u\in{\cal W}^{u,c}_{+,\varepsilon}$
be an eventually long fiber, $|W^u|>\varepsilon$.
Partition it into some subfibers $W^u_i$, $i\geq 1$,
of length between $\varepsilon/2$ and $\varepsilon$. On each
$W^u_i$, pick a point $x_i\in W^u_i$. Then for any probability
measure $\nu$ on $W^u$ we have
$$
\left |\int_{W^u}f(x)\, d\nu\, -\, \sum_if(x_i)\nu(W_i)\right |
\leq \delta_f(\varepsilon)
$$
Therefore,
\begin{eqnarray}
\left |\int_{W^u}f\, d\nu_{W^u,n}
-\int_{W^u}f\, d\bar{\nu}_{W^u}\right |
&\leq&
\left |\sum_if(x_i)\left (\nu_{W^u,n}(W^u_i)-\bar{\nu}_{W^u}(W^u_i)\right )\right |
+2\delta_f(\varepsilon)\nonumber\\
&\leq&
||f||_{\infty}\sum_i\left |\nu_{W^u,n}(W^u_i)-\bar{\nu}_{W^u}(W^u_i)\right |
+2\delta_f(\varepsilon)
\label{FF}
\end{eqnarray}
For each subfiber $W^u_i$, we have
$$
\nu_{W^u,n}(W^u_i)=
\frac{\nu_{W^u}(W^u_i)\, \nu_{W^u_i}(W^u_i\cap M_{-n})}
{\nu_{W^u}(W^u\cap M_{-n})}
$$
Using (\ref{rhoW2}) and (\ref{rhoW1}) yield
\begin{eqnarray*}
\left |\nu_{W^u,n}(W^u_i)-\bar{\nu}_{W^u}(W^u_i)\right | &=&
\left |\nu_{W^u}(W^u_i)\frac{\rho_n(W^u_i)}{\rho_n(W^u)}\, - \,
\nu_{W^u}(W^u_i)\frac{\rho(W^u_i)}{\rho(W^u)}\right |\\
&=& \nu_{W^u}(W^u_i)
\left |\frac{\rho_n(W^u_i)\rho(W^u)-\rho_n(W^u)\rho(W^u_i)}
{\rho_n(W^u)\rho(W^u)}\right |
\end{eqnarray*}
Now, using (\ref{rhoWa})
\begin{eqnarray*}
\left |\rho_n(W^u_i)\rho(W^u)-\rho_n(W^u)\rho(W^u_i)\right |&=&
\left |\rho_n(W^u_i)(\rho(W^u)-\rho_n(W^u))
+\rho_n(W^u)(\rho_n(W^u_i)-\rho(W^u_i))\right |\\
&\leq&\rho_n(W^u_i)\gamma_n''/|W^u|
+2\rho_n(W^u)\gamma_n''/\varepsilon
\end{eqnarray*}
Hence,
$$
\left |\nu_{W^u,n}(W^u_i)-\bar{\nu}_{W^u}(W^u_i)\right | \leq
\nu_{W^u}(W^u_i)\cdot\frac{\rho_n(W^u_i)\gamma_n''}{\rho_n(W^u)\rho(W^u)\varepsilon}
\, + \,
\nu_{W^u}(W^u_i)\cdot\frac{2\gamma_n''}{\varepsilon\rho(W^u)}
$$
Note that
$$
\sum_i\nu_{W^u}(W^u_i)=\nu_{W^u}(W^u)=1
$$
and
$$
\sum_i\nu_{W^u}(W^u_i)\rho_n(W^u_i)/\rho_n(W^u)=\nu_{W^u,n}(W^u)=1
$$
Therefore,
$$
\sum_i\left |\nu_{W^u,n}(W^u_i)-\bar{\nu}_{W^u}(W^u_i)\right |
\leq \frac{3\gamma_n''}{\varepsilon\rho(W^u)}
$$
Combining this with (\ref{FF}) and substituting in the
integral formula for $I_1$ give the bound
\begin{eqnarray*}
I_1&\leq& 3||f||_{\infty}\gamma_n''/\varepsilon
+2\delta_f(\varepsilon)\int_{{\cal W}^{u,c}_{+,\varepsilon}}
\rho(W^u)\,d\mu_+^f\\
&\leq& 3||f||_{\infty}\gamma_n''/\varepsilon
+2\delta_f(\varepsilon)\,\Big ( d_0^{-g}+gC_1d_0^{1-g}\Big )
\end{eqnarray*}
where we used (\ref{C22}).
The proof of Theorem~\ref{tmmain4} is completed. This implies
the weak convergence of the measures $\mu_{+,n}$ to $\bar{\mu}_+$.
In particular, $\bar{\mu}_+$, and hence, $\bar{\mu}_+^f$, are
probability measures.
We now continue the proof of Theorem~\ref{tmmain3}.
The invariance of the measure $\bar{\mu}_+$ follows from the
above weak convergence.
To prove the ergodicity of $\bar{\mu}_+$ we employ the classical
Hopf technique of connecting any two generic points of $\Omega$
by a finite chain of stable and unstable fibers, see, e.g.,
\cite{Brin} for a self contained introduction to the
Hopf technique in Hopf's original setting. The crucial
step in this technique is to ensure some analogue of
the absolute continuity of the foliations of $\Omega$
by stable and unstable fibers. Precisely, we need to show
that for any two nearby parallel unstable fibers
$W^u_1,W^u_2\subset M_+$
the holonomy map (sliding along stable fibers) is not completely
singular with respect to the conditional measures
$\bar{\nu}_{W^u_1},\bar{\nu}_{W^u_2}$.
Let $W^u_1,W^u_2\subset M_+$ be two unstable fibers of
lengths $\geq d_0/4$, and let $\nu_1,\nu_2$ be their u-SBR
measures, respectively. Assume that $W^u_1$ is $d_0$-close
to $W^u_2$, see Sect.~\ref{secPL}. By shortening $W^u_2$, if
necessary, we can make it $d_0$-close to $W^u_1$, too.
In particular, the holonomy
map $\tilde{h}$ (sliding along stable fibers of length $\leq d_0$)
will be then a bijection of $W^u_1$ to $W^u_2$. For any $n\geq 0$
let $\tilde{W}^u_{1,n}=\{x\in W^u_1\cap M_{-n}:\, \tilde{h}(x)\in
W^u_2\cap M_{-n}\}$, and $\tilde{W}^u_{2,n}:=\tilde{h}(\tilde{W}^u_{1,n})
\subset W^u_2\cap M_{-n}$. The sets $\tilde{W}^u_{1,n}$ and $\tilde{W}^u_{2,n}$
are finite unions of closed subfibers in $W^u_1,W^u_2$, respectively,
and the jacobian of $\tilde{h}:\tilde{W}^u_{1,n}\to \tilde{W}^u_{2,n}$
is uniformly bounded away from zero and infinity (by $D^{-1}$ and $D$).
The remark in the end of Section~\ref{secMLUF} shows that
$\nu_i(\tilde{W}^u_{i,n})\geq \nu_i(W^u_i\cap M_{-n})/2$, i.e.
\be
\nu_{i,n}(\tilde{W}^u_{i,n})=\nu_i(\tilde{W}^u_{i,n}/M_{-n})\geq 1/2
\label{12}
\ee
for $i=1,2$ and all $n\geq 0$. \medskip
{\em Claim}. The holonomy map $\tilde{h}:W^u_1\to W^u_2$
is not singular with respect to the measures $\bar{\nu}_{W^u_i}$,
$i=1,2$.\medskip
{\em Proof}. By way of contradiction, assume that $\exists
A\subset W^u_1$ such that $\bar{\nu}_{W^u_1}(A)=0$
and $\bar{\nu}_{W^u_2}(\tilde{h}(A))=1$. Then $\forall\delta>0$
there is a countable union of disjoint open subintervals $I_i\subset
W^u_1$ such that $A\subset\cup I_i$ and $\bar{\nu}_{W^u_1}(\cup I_i)
<\delta$. Since $\bar{\nu}_{W^u_2}(\tilde{h}(\cup I_i))=1$,
we can find a finite subunion $G_j=\cup_{i=1}^jI_i$ for some
$j<\infty$ such that $\bar{\nu}_{W^u_2}(\tilde{h}(G_j))\geq 0.99$.
Recall that $\nu_{i,n}$ weakly converge, as $n\to\infty$,
to $\bar{\nu}_{W^u_i}$ for $i=1,2$.
Since $\tilde{h}(G_j)$ is an open subset of $W^u_2$,
there is an $n_0\geq 1$ such that $\nu_{2,n}(\tilde{h}(G_j))
\geq 0.98$ for all $n\geq n_0$. Due to (\ref{12}),
we have
$$
\nu_{2,n}(\tilde{h}(G_j\cap \tilde{W}^u_{1,n})) =
\nu_{2,n}(\tilde{h}(G_j)\cap \tilde{W}^u_{2,n}) \geq 0.48
$$
for all $n\geq n_0$. Hence,
$\nu_{1,n}(G_j\cap\tilde{W}^u_{1,n})\geq 0.48\, (C_6D)^{-1}>0$,
where we used Theorem~\ref{tmn12}. Hence,
$\nu_{1,n}(G_j)\geq 0.48\, (C_6D)^{-1}$ for all $n\geq n_0$.
Consider the closure $\bar{G}_j$ of $G_j$. The weak convergence
$\nu_{1,n}\to\bar{\nu}_{W^u_1}$ implies that
$\bar{\nu}_{W^u_1}(\bar{G}_j)\geq 0.48\, (C_6D)^{-1}$.
Note that $\bar{\nu}_{W^u_1}$ is a nonatomic measure,
cf. (\ref{rhoW1}) and the first remark after Theorem~\ref{tmrhoW}.
Hence, $\bar{\nu}_{W^u_1}(G_j)\geq 0.48\, (C_6D)^{-1}$,
a contradiction that proves the claim. \medskip
This allows us to link $W^u_1$ and $W^u_2$ by stable
fibers and make one Hopf chain that contains both $W^u_1$ and $W^u_2$.
Next, the unstable fibers $W^u\subset M_+$ of length
$\geq d_0/4$ are rather dense in $\hat{M}$. As it was
shown in \cite{CMT} (Lemma 3.4), any stable fiber $W^s
\subset\hat{M}$ of length $d_2$ (with $d_2\sim h$) crosses
at least one unstable fiber $W^u\subset M_+$ whose endpoints
are the distance $\geq d_0/3$ away from the point of
intersection, $W^s\cap W^u$. Therefore, any two
unstable fibers $W^u_1,W^u_2\subset M_+$ of length $\geq d_0/4$
can be connected by one finite Hopf chain of stable and
unstable fibers, so they all belong (mod 0) in one ergodic
component of the measure $\bar{\mu}_+$.
Consider now short unstable fibers $W^u\subset M_+$.
If $W^u$ is forever short then $\rho(W^u)=0$, and
hence the union of such fibers has zero $\bar{\mu}_+$ measure.
If $W^u$ is eventually long, then, in the same way,
only its parts that become
long under the iterations of $T$ can carry
positive mass of $\bar{\mu}_+$.
Those parts, however, belong (mod 0) to the same ergodic
component as above, since that component is $T$-invariant.
The ergodicity of $\bar{\mu}_+$ is proved.
It is standard that the K-property on each ergodic
component follows from the existence of stable and unstable
fibers at a.e. point, see, e.g., \cite{KS} and \cite{LY} (Theorem~B).
Since our measure $\bar{\mu}_+$ is ergodic,
it is K-mixing. $\Box$\medskip
{\em Remark}. Due to (\ref{Heps}),
\be
\mu_{+,n}(H_{\varepsilon})
\leq C_{11}\varepsilon^{b/2}\ \ \ \
\forall \varepsilon>0\ {\rm and}\ n\geq 0
\label{muHeps}
\ee
Hence, $\bar{\mu}_+(H_{\varepsilon})\leq C_{11}\varepsilon^{b/2}$.
It is then a standard application of Borel-Cantelli lemma
to show that at $\bar{\mu}_+$-a.e. point $x\in\Omega$
there are fibers $W^u_x\subset M_+$ and $W^s_x\subset M_-$
of nonzero length. Furthermore, for any $\varepsilon>0$ the
set of points whose fibers $W^u_x\subset M_+$ and $W^s_x
\subset M_-$ are shorter than $\varepsilon$ has
$\bar{\mu}_+$-measure $\leq{\rm const}\cdot\varepsilon^{b/2}$.
\medskip
Next, the estimates on convergence in Theorem~\ref{tmmain4}
are independent of $h$ and the shape of the holes.
Therefore, the weak convergence of the measures $\mu^{(k)}_{+,n}:=
[T^{(k)}]^{-n}_{\ast}\mu_+^{(k)}$ to $\bar{\mu}_+^{(k)}$,
proved in \cite{CM1},\cite{CM2} is actually
{\em uniform in $k$} for all $k\geq k_0$. Hence, for any sequence
$n_k\to\infty$ as $k\to\infty$ the sequences of measures
$\{\mu^{(k)}_{+,n_k}\}$ is equivalent to $\bar{\mu}_+^{(k)}$,
i.e. $\mu^{(k)}_{+,n_k}\sim\bar{\mu}_+^{(k)}$ as defined
in the previous section.
Recall that the sequence $\mu_+^{(k)}$ converges, as $k\to\infty$,
to $\mu_+$, cf. \cite{CMT}. Since $T^{-1}$ acts smoothly on $M_1$,
the measures $\mu_{+,n}^{(k)}=T^{-n}_{\ast}\mu_+^{(k)}$ remain close
to $\mu_{+,n}=T^{-n}_{\ast}\mu_+$ in the weak topology for
sufficiently small $n$ (compared to $k$). It is, however, clear, that
$n$ may be taken larger as $k$ grows, i.e. there is a sequence
$n_k\to\infty$ such that $\mu_{+,n_k}^{(k)}\sim\mu_{+,n_k}$
as $k\to\infty$.
Combining the above observations with Theorem~\ref{tmmain3} gives
\begin{corollary}
The sequence of measures $\bar{\mu}_+^{(k)}$ supported on the repellers
$\Omega^{(k)}\subset\Omega$ weakly converges, as $k\to\infty$,
to the measure $\bar{\mu}_+$ supported on $\Omega$. Note that all these
measures are $\hat{T}$-invariant, ergodic and K-mixing.
\label{crbarmu}
\end{corollary}
{\em Proof of Theorem~\ref{tmmain5}}. We start with a formula
for the Kolmogorov-Sinai entropy of smooth hyperbolic maps
on surfaces:
\be
h(\mu)=\chi_+(\mu)\cdot\lim_{\varepsilon\to 0}
\frac{\log \mu^u(B^u(x,\varepsilon))}{\log\varepsilon}
\ee
where $\mu$ is an ergodic invariant measure, $\chi_+(\mu)$ is
its positive Lyapunov exponent, $x$ is a $\mu$-generic
point, $\mu^u$ is the conditional measure induced by $\mu$ on the
local unstable fiber $W^u_x$, and $B^u(x,\varepsilon)$ is
the $\varepsilon$-ball in $W^u_x$ centered at $x$ (i.e.,
the interval on $W^u_x$ of length $2\varepsilon$ centered at $x$).
This formula was proved by Ledrappier and Young, even in
the nonuniformly hyperbolic case, see pp. 545, 559 in
Part~II of \cite{LY}.
For any $\bar{\mu}_+$-generic point $x\in\Omega$ there is
a sequence $n_i\to\infty$ such that the points $x_i:=
T^{n_i}x$ have long unstable fibers in $M_+$, i.e.
the distance from $x_i$ to the endpoints of
$W^u_{x_i}\subset M_+$ is at least $d_0/8$.
For every sufficiently large $i$ we find an $\varepsilon_i>0$
such that $W^u_i:=T^{n_i}B^u(x,\varepsilon_i)$ is
a subsegment of $W^u_{x_i}$ of length $\geq d/8$.
Due to the standard distortion estimates,
$$
(8D)^{-1}\leq
\frac{d_0/\varepsilon_i}{J^u_xJ^u_{Tx}\cdots J^u_{T^{n_i-1}x}}
\leq 8D
$$
Next, let $W^u\subset M_+$ be the maximal unstable fiber containing $x$.
Then
\begin{eqnarray*}
\bar{\nu}_{W^u}(B^u(x,\varepsilon_i)) &=&
\nu_{W^u}(B^u(x,\varepsilon_i))\,\rho(B^u(x,\varepsilon_i))/\rho(W^u) \\
&=&
\nu_{W^u}(B^u(x,\varepsilon_i))\,\lambda_+^{-n_i}
\rho(T^{n_i}B^u(x,\varepsilon_i))/\rho(W^u)
\end{eqnarray*}
where we used (\ref{rhoW1}) and the second remark after Theorem~\ref{tmrhoW}.
Observe that $\nu_{W^u}(B^u(x,\varepsilon_i))\sim \varepsilon_i/|W^u|$,
and $\rho(T^{n_i}B^u(x,\varepsilon_i))\sim 1$. Therefore,
\begin{eqnarray*}
h(\bar{\mu}_+)&=&\chi_+\cdot\lim_{i\to \infty}
\frac{\log\varepsilon_i-\log\lambda_+^{n_i}}{\log\varepsilon_i}\\
&=&\chi_+\cdot\left (1-\lim_{i\to \infty}
\frac{\log\lambda_+}{n_i^{-1}\log\varepsilon_i}\right )\\
&=&\chi_+\cdot\left (1-\lim_{i\to \infty}
\frac{\gamma_+}{n_i^{-1}\log J_x^u\cdots J^u_{T^{n_i-1}x}}\right )\\
&=&\chi_+\cdot\left (1-
\frac{\gamma_+}{\chi_+}\right )
\end{eqnarray*}
Theorem~\ref{tmmain5} is proved. $\Box$\medskip
\begin{corollary}[see \cite{Ma}]
Let $G_{\bar{\mu}_+}\subset\Omega$ be the set of $\bar{\mu}_+$-generic
points $x\in\Omega$. Then the Hausdorff dimension, denoted by
$\delta^u$, of $G_{\bar{\mu}_+}\cap W^u(x)$ is independent
of $x\in\Omega$, and
$$
\delta^u=h(\bar{\mu}_+)/\chi_+=
\lim_{\varepsilon\to 0}
\frac{\log \mu^u(B^u(x,\varepsilon))}{\log\varepsilon}
$$
\end{corollary}
{\em Proof of Theorem~\ref{tmmain6}}. Let now, for a moment,
$H^{(k)}$ be rectangular holes approximating $H$ `from outside'
as described in the first paragraph of Section~\ref{secIM}.
They satisfy $d(H^{(k)},H)\leq\varepsilon_k/2$,
see (\ref{HHeps}).
Hence, $d(H^{(k)},H_n)\leq d(H_n,H)+\varepsilon_k/2$.
Let $k=k_n=\max\{k\geq k_0:\, \varepsilon_k\geq 2d(H_n,H)\}$.
Obviously, $k_n\to\infty$ as $n\to\infty$.
Then $d(H^{(k_n)},H_n)\leq \varepsilon_{k_n}$, so the
rectangular holes $H^{(k_n)}$ properly approximate $H_n$.
Hence, we can apply (\ref{intint6}) with
$\mu_+$ replaced by $\mu_+[H_n]$ and $k=k_n$.
This proves the weak convergence $\mu_+[H_n]\to\mu_+$.
The convergence $\bar{\mu}_+[H_n]\to\bar{\mu}_+$ is
proved exactly as Corollary~\ref{crbarmu}.
Now, $\lambda_+[H_n]=1-\mu_+[H_n](\hat{T}^{-1}H_n)$.
The weak convergence $\mu_+[H_n]\to\mu_+$, the assumption
$d(H_n,H)\to 0$, and the bound (\ref{muHeps}), in which
one sets $n=0$, imply
that $\lambda_+[H_n]\to\lambda_+$, and hence
$\gamma_+[H_n]\to\gamma_+$.
The escape rate formula holds for all measures
$\bar{\mu}_+[H_n]$ and implies the convergence
of the entropies $h(\bar{\mu}_+[H_n])\to h(\bar{\mu}_+)$.
Observe that, generally, the entropy
$h(\mu)$ in not (!) a continuous function of the invariant
measure $\mu$ for $T$. $\Box$\medskip
{\em Proof of Theorem~\ref{tmmain7}}. That theorem was
proved in \cite{CM1} for Anosov diffeomorphisms with
rectangular holes. We can apply it to $T^{(k)}$, because
$\Omega^{(k)}\subset\Omega$, so the necessary assumptions
are fulfilled on $\Omega^{(k)}$. Thus, we get
$\bar{\mu}^{(k)}_+=\bar{\mu}^{(k)}_-$ and $\lambda^{(k)}_+
=\lambda^{(k)}_-$ for all $k\geq k_0$. Taking the limit as
$k\to\infty$ proves Theorem~\ref{tmmain7}. $\Box$
\section{Proof of Proposition 6.3}
\label{secPPP}
\setcounter{equation}{0}
The matrix techniques for proving the weak convergence of
measures is based on the following.
Let $\xi^{(1)}<\xi^{(2)}<\cdots$ be an increasing sequence of finite
partitions of the underlying space, $\xi^{(k)}=\{A_1^{(k)},
\ldots,A_{m_k}^{(k)}\}$, that converges to a partition
into single points. Then one can represent any measure
$\mu$ by a sequence of (row) vectors $p^{(k)}$ with
components $p^{(k)}_i=\mu(A_i^{(k)})$, $1\leq i\leq m_k$.
The norm of the measure is given by $|\mu|=|p^{(k)}|
=\sum_ip^{(k)}_i$. Then, under certain regularity conditions,
the weak convergence of a sequence of measures, $\mu_n\to
\mu_{\infty}$, is equivalent to the componentwise convergence
of the sequence of vectors $p^{(k)}(\mu_n)\to p^{(k)}(\mu_{\infty})$,
as $n\to\infty$, for all $k\geq 1$. Similarly, the transformation
of measures $\mu'=T_{\ast}\mu$ is equivalent to the right
multiplication by matrices, $p^{(k)}(\mu')=p^{(k)}(\mu)\Pi^{(k)}(\mu)$,
with components
$$
\Pi^{(k)}_{ij}(\mu)=\mu\left ( A_i^{(k)}\cap T^{-1}A_j^{(k)}\right )
/\mu(A^{(k)}_i)
$$
which is a nonnegative substochastic matrix for every $k$.
More details of our techniques can be found in \cite{CM1,CM2,Sen}.
Fix a $k\geq k_0$.
We represent any measure $\mu$ on $M^{(k)}$ by the row vector
$p(\mu)$ with components $\{\mu(R):\, R\in{\cal R}^{(k)}\}$.
We always assume, for simplicity, that $\mu(\cup_R\partial R)=0$,
so that $|\mu|=|p(\mu)|$.
For $n\geq q$ denote by $p_n=\{p_n(R)\}$
the vector representing the measure $[T^{(k)}_{\ast}]^{n-q}\mu_q$,
$q=k+l_0$ (this is not a probability measure for $n>q$, of course).
The normalized vector, $p_n/|p_n|$, will then represent the
measure $\mu^{(k)}_n$.
Recall that the conditional distributions of $\mu_n$ on
unstable $R$-fibers, $R\in{\cal R}^{(k)}$, coincide
with u-SRB measures on those fibers. Then
the sequence of vectors $p_n$ can be well approximated by
the product of the vector $p_{n_0}$ (representing $\mu_{n_0}$)
for some $n_0\geq q$ and
certain substochastic matrices defined as follows. In every
rectangle $R\in {\cal R}^{(k)}$ pick an arbitrary unstable
$R$-fiber $U(R)$. For any $m\geq 1$ define the substochastic
matrix $\Pi_m$ with components
\be
\Pi_m(R',R'')=\nu_{U(R')}
\left ( U(R')\cap [T^{(k)}]^{-m}(R''\cap M_m^{(k)})\right )
\label{Pim}
\ee
where $\nu_{U(R')}$ is the u-SBR measure on $U(R')$.
Since the rectangles in ${\cal R}^{(k)}$ are exponentially
small (in $k$), then for any $n\geq m+q$ and $R''$
$$
e^{-C\alpha^k}\leq
\frac{\sum_{R'}p_{n-m}(R')\, \Pi_m(R',R'')}{p_n(R'')}\leq e^{C\alpha^k}
$$
and for any $R',R''$ and $m_1,m_2\geq 1$
$$
e^{-C\alpha^k}\leq
\frac{\sum_{R'''}\Pi_{m_1}(R',R''')\, \Pi_{m_2}(R''',R'')}
{\Pi_{m_1+m_2}(R',R'')}\leq e^{C\alpha^k}
$$
for some global constants $C>0$ and $\alpha\in (0,1)$.
Let $\bar{m}=m_1+\cdots+m_t\leq n-q$ and put
$$
\hat{p}_n=p_{n-\bar{m}}\Pi_{m_1}\cdots\Pi_{m_t}
$$
Then, for any $R\in{\cal R}^{(k)}$
$$
e^{-Ct\alpha^k}\leq
\frac{\hat{p}_n(R)}{p_n(R)}\leq e^{Ct\alpha^k}
$$
As a result, $e^{-Ct\alpha^k}\leq |\hat{p}_n|/|p_n|\leq e^{Ct\alpha^k}$,
so that the normalized vectors $\hat{p}_n/|\hat{p}_n|$ and
$p_n/|p_n|$ are close as well:
\be
e^{-2Ct\alpha^k}\leq
\frac{\hat{p}_n(R)/|\hat{p}_n|}{p_n(R)/|p_n|}\leq e^{2Ct\alpha^k}
\label{ee}
\ee
According to (\ref{ee}), the vectors $\hat{p}_n/|\hat{p}_n|$
approximately represent the measure $\mu_n^{(k)}$
as long as $t\ll\alpha^{-k}$. In our further considerations, $t$
will be actually equal to $k$, so $\hat{p}_n$ will always
approximate $p_n$ well enough. In fact, we will find an integer
$r\geq 1$ such that such that the matrix $\Pi_{rk}$ has good
mixing properties uniformly in $k$, see below.
Then for any $n\geq rk^2+q$ we approximate $\mu_n^{(k)}$ by
\be
\hat{p}_n:=p_{n-rk^2}\Pi_{rk}^k
\label{hatp}
\ee
Due to (\ref{ee}), for any $n\geq rk^2+q$
\be
\sum_{R\in{\cal R}^{(k)}}
\left | \frac{p_n(R)}{|p_n|}-
\frac{\hat{p}_n(R)}{|\hat{p}_n|}\right |
\leq e^{2Ck\alpha^k}-1=O(k\alpha^k)
\label{eee}
\ee
{\em Remark}. The vector $p_n/|p_n|$ represents the measure
$\mu_n^{(k)}$. According to (\ref{eee}), the vector
$\hat{p}_n/|\hat{p}_n|$ approximately represents the same
measure. The vector $\hat{p}_n/|\hat{p}_n|$ will not
change if we normalize the vector $p_{n-rk^2}$ in (\ref{hatp}),
i.e. if we assume that $p_{n-rk^2}$ represents the measure
$\mu_{n-rk^2}^{(k)}$ rather than
$[T^{(k)}_{\ast}]^{n-rk^2-q}\mu_q$.
\medskip
We will need certain matrix estimates similar to those in
\cite{Sen}. Denote by ${\cal A},{\cal B},{\cal C}$
nonnegative $N\times N$ matrices
and $X,Y$ nonnegative row vectors of length $N$ with the
norm $|X|=\sum x_i$, the distance $|X-Y|=\sum |x_i-y_i|$,
and the normed distance
$$
||X-Y||=\sum_{i=1}^N\left | \frac{x_i}{|X|}-\frac{y_i}{|Y|}\right |
$$
Let $\tilde{X}=X+X'$ and $\tilde{Y}=Y+Y'$, where
$\tilde{X},X',\tilde{Y},Y'$ are some nonnegative vectors.
Denote $\varepsilon_X=|X'|/|\tilde{X}|$ and
$\varepsilon_Y=|Y'|/|\tilde{Y}|$.
The following estimate is a result of direct calculations:
\be
||\tilde{X}-\tilde{Y}||\leq ||X-Y||+2(\varepsilon_{X}+\varepsilon_Y)
\label{mat1}
\ee
Let $1\leq J\leq N$. Assume that
$x_j>0$ iff $j\leq J$ and $y_j>0$ iff $j\leq J$. Let
$$
d(X,Y)=\ln\left (\frac{\max_{1\leq i\leq J}(x_i/y_i)}
{\min_{1\leq i\leq J}(x_i/y_i)}\right )=
\max_{1\leq i,j\leq J}\ln\left (\frac{x_iy_j}{x_jy_i}\right )
$$
be the projective distance between vectors, based on their
first $J$ components.
Obviously, $x_i/x_j\leq e^{d(X,Y)}y_i/y_j$ for all $1\leq i,j\leq J$.
Summing over $i=1,\ldots, J$ gives $y_j/|Y|\leq e^{d(X,Y)}x_j/|X|$,
hence
\be
||X-Y||\leq e^{d(X,Y)}-1
\label{mat2}
\ee
Note that
$d(X,Y)=0$ (hence, $||X-Y||=0$) iff $X=\lambda Y$ for some $\lambda>0$.
Now, let $1\leq I\leq J$. Let ${\cal A}={\cal B}+{\cal C}$,
where ${\cal B}$ and ${\cal C}$ are nonnegative matrices.
For the matrix ${\cal B}$, assume that\\
(B) $b_{ij}>0$ iff $i\in\{1,\ldots,I\}\cup I'$ and $j\in\{1,\ldots,J\}$,\\
where $I'\subset\{J+1,\ldots, N\}$ is an arbitrary subset of indices.
Denote by ${\cal B}_n^{\ast}$ the $J\times J$ principal minor
of the matrix ${\cal B}^n$, $n\geq 1$. Observe that ${\cal B}_n^{\ast}
=({\cal B}_1^{\ast})^n$.
For any nonnegative vector $\tilde{X}=\{\tilde{x}_i\}$ we put
$\tilde{X}_n=\tilde{X}{\cal A}^n$, $X_n=\tilde{X}{\cal B}^n$ and
$X_n'=\tilde{X}_n-X_n$. We say that $\tilde{X}$ is
admissible if $\tilde{x}_i>0$ for at least one
$i\in\{1,\ldots,I\}\cup I'$.
In that case $x^{(1)}_j>0$, $\forall j\leq J$, where
$x^{(1)}_j$ are the components of $X_1=\tilde{X}{\cal B}$.
Therefore, for any two admissible vectors $\tilde{X}$,
$\tilde{Y}$ we have $d(X_n,Y_n)<\infty$, $\forall n\geq 1$.
It is known that $d(X_n,Y_n)\leq
d(X_{n-1},Y_{n-1})$, cf. \cite{Sen}. Let
$$
\tau ({\cal B})=\sup_{X\neq \lambda Y}\frac{d(X{\cal B},Y{\cal B})}{d(X,Y)}
$$
where the supremum is taken over
vectors $X,Y$ whose components $x_j,y_j$ are positive
if and only if $j\leq J$. $\tau({\cal B})$ is called Birkhoff
contraction coefficient. It is known that $\tau({\cal B}^n)\leq [\tau({\cal B})]^n$,
cf. \cite{Sen}. There is an explicit formula for $\tau({\cal B})$,
see \cite{Sen}, pp. 101--106:
$$
\tau ({\cal B})=\frac{1-\varphi^{1/2}}{1+\varphi^{1/2}},
\ \ \ \ \
\varphi=\varphi({\cal B})=\min_{i,j\leq I,\ k,l\leq J}\frac{b_{ik}b_{jl}}{b_{jk}b_{il}}
$$
In particular, $\tau({\cal B})=0$ iff the $J\times J$ principal minor
of ${\cal B}$ coincides with that of the matrix $U^TV$, where $U$ and $V$ are
some nonnegative row vectors such that $u_i>0$ iff $i\leq I$
and $v_j>0$ iff $j\leq J$, and $U^T$ is the transpose of $U$.
Assume that $\exists\beta_0>0$ such that
\be
0<\beta_0\leq\frac{b_{ik}b_{jl}}{b_{jk}b_{il}}\leq\beta_0^{-1}
\label{beta0}
\ee
for all $i,j\leq I$ and $k,l\leq J$. Then $\tau({\cal B})\leq
\beta:=(1-\sqrt{\beta_0})/(1+\sqrt{\beta_0})<1$, and hence
$$
d(X_n,Y_n)\leq d(X_2,Y_2)\cdot \beta^{n-2}
$$
It is also a direct calculation that
$$
d(X_2,Y_2)\leq d(X_1{\cal B},Y_1{\cal B})\leq 2\ln\beta_0^{-1}
$$
for any two admissible vectors $\tilde{X}$, $\tilde{Y}$.
Summarizing (\ref{mat1}), (\ref{mat2}) and (\ref{beta0}) gives
\begin{eqnarray}
||\tilde{X}_n-\tilde{Y}_n||
&\leq & \left (\beta_0^{-2\beta^{n-2}}-1\right )
+ 2(\varepsilon_{X_n}+\varepsilon_{Y_n})\nonumber\\
&\leq & {\rm const}(\beta_0)\cdot\beta^{n}
+ 2(\varepsilon_{X_n}+\varepsilon_{Y_n})
\label{XnYn}
\end{eqnarray}
To prove (\ref{beta0}), it is enough to show that there exist two nonnegative
vectors $U$ and $V$ such that
$u_i>0$ iff $i\leq I$ and $v_j>0$ iff $j\leq J$
and the components $b_{ij}$
of the matrix ${\cal B}$ admit the decoupling $b_{ij}\sim u_iv_j$,
i.e. for some $\gamma_0>0$
\be
0<\gamma_0<\frac{b_{ij}}{{u_i}{v_j}}\leq \gamma_0^{-1}
\label{gamma0}
\ee
for all $i\leq I$, $j\leq J$. Then (\ref{beta0})
follows with $\beta_0=\gamma_0^4$.
Next, we will define the matrices
${\cal A}$ and ${\cal B}$ related to the transition
matrix $\Pi_m$ in (\ref{Pim}) and show (\ref{gamma0}).
For brevity, we will suppress the superscript $(k)$
in ${\cal R}^{(k)}$, $T^{(k)}$, and $M^{(k)}$.
Let $R\in{\cal R}$ and $U$ be an unstable $R$-fiber.
Consider the functions $l_n(x)$, $n\geq 0$, on the fiber $U$
as defined before Lemma~\ref{lmWuln}.
Let $S$ be a stable $R$-fiber. In a similar way,
define functions $l_n^-(x)$, $n\geq 0$, on $S$.
For every $n\geq 0$ and $x\in S\cap M_{n}$ denote by
$S_n(x)$ the smooth component of $T^{-n}(S\cap M_{n})$
containing $T^{-n}x$.
For every $x\in S\cap M_{-n}$ let $l^-_n(x)=\min\{
l\in [0,n]:\, |S_l(x)|\geq d_0/2\}$ (if $|S_l(x)| r_0 > 1$ to be specified below. We set
${\cal A}=\Pi_{rk}$, see (\ref{Pim}),
and will now define ${\cal B}$ so that ${\cal C}={\cal A}-{\cal B}$
will be a nonnegative matrix. For any two rectangles
$R',R''\in {\cal R}$ we define ${\cal B}(R',R'')$ as follows.
Let $U(R')$ be the unstable $R'$-fiber fixed in (\ref{Pim}).
Let $\tilde{U}(R')=\{x\in U(R')\cap M_{-rk}:\, l_{rk}(x)\leq r_0k\}$.
Now, let $S(R'')$ be an arbitrary stable $R''$-fiber.
Let $\tilde{S}(R'')=\{x\in S(R'')\cap M_{rk}:\, l^-_{rk}(x)\leq r_0k\}$.
Now, let $\tilde{R}''\subset R''$ be
the union of unstable $R''$-fibers that cross $\tilde{S}(R'')$.
Obviously, $\tilde{R}''$ is a finite union of u-subrectangles
in $R''$. Observe that $T^{-rk}\tilde{R}''\cap R'$ consists
of a finite number of s-subrectangles in $R'$. We now define
\be
{\cal B}(R',R'')=\nu_{U(R')}[\tilde{U}(R')\cap
T^{-rk}(\tilde{R}'')]
\label{BRR}
\ee
Comparing this to (\ref{Pim}) shows that ${\cal B}(R',R'')\leq
\Pi_{rk}(R',R'')={\cal A}(R',R'')$, so that the matrix ${\cal C}={\cal A}-{\cal B}$ is, indeed,
nonnegative.
We now prove (\ref{gamma0}). Let again $R',R''\in{\cal R}$.
Denote by $\tilde{U}_i$, $i\geq 1$, all distinct curves
$U_{l_{rk}(x)}(x)$ for points $x\in\tilde{U}(R')$. If
$\tilde{U}_i=U_{l_{rk}(x)}(x)$ for some $x$, we put $l_i=l_{rk}(x)$,
the iteration associated with $\tilde{U}_i$.
There are finitely many of those curves, and their lengths are
$\geq d_0/2$ and $\leq d_0$.
Observe that $T^{-l_i}\tilde{U}_i$, $i\geq 1$, are
disjoint subsegments of $U(R')$, and their union covers
$\tilde{U}(R')$.
Similarly, we define stable fibers $\tilde{S}_j$, $j\geq 1$,
and $l^-_j\leq r_0k$ such that $T^{l^-_j}\tilde{S}_j$
are disjoint subsegments of $S(R'')$, their union covers
$\tilde{S}(R'')$, and the length of every $\tilde{S}_j$
is between $d_0/2$ and $d_0$. Denote by $\tilde{R}_j''$
the u-subrectangle in $T^{-l^-_j}R''$ that consists
of unstable fibers crossing the curve $\tilde{S}_j$.
Let $d^u_j$ be the maximum length of unstable $\tilde{R}_j''$-fibers.
Assume that $r>2r_0$, so that $rk>2r_0k+k_1$ for all
sufficiently large $k$.
Denote by $\nu_i$ the u-SBR measure
on the fiber $\tilde{U}_i$ and put
$w_i=\nu_{U(R')}(T^{-l_i}\tilde{U}_i)$.
In the following, we use Corollary~\ref{crC10}:
\begin{eqnarray}
{\cal B}(R',R'')
&=&\sum_{i,j}w_i\cdot
\left [ T_{\ast}^{rk-l_i-l^-_j}\nu_i\right ] (\tilde{R}_j'')\nonumber\\
&\sim &\sum_{i,j} w_i\cdot
[\lambda_+^{(k)}]^{rk-l_i-l^-_j-k_1}\cdot d^u_j \nonumber\\
&= &\left (\sum_i w_i\cdot
[\lambda_+^{(k)}]^{rk-l_i}\right )\left (\sum_j
[\lambda_+^{(k)}]^{-l^-_j-k_1}\cdot d^u_j\right ) \nonumber\\
\end{eqnarray}
where $a\sim b$ means that the ratio $a/b$ is bounded
above and below by two positive global constants,
in this case the constants are $C_{10}$ and $C_{10}^{-1}$.
This proves (\ref{gamma0}) with $\gamma_0=C_{10}^{-1}$. \medskip
{\em Remark}. Observe that if $\tilde{U}(R')=\emptyset$ or $\tilde{S}(R'')
=\emptyset$, then ${\cal B}(R',R'')=0$. Otherwise ${\cal B}(R',R'')>0$,
as the above calculation shows. We can number the rectangles
$R_i\in{\cal R}$, $1\leq i\leq N$, so that $\tilde{S}(R_i)
\neq\emptyset$ iff $i\leq J$ for some $J\leq N$, and furthermore,
for $i\leq J$ we require $\tilde{U}(R_i)\neq\emptyset$ iff
$i\leq I$ for some $I\leq J$. Then the matrix ${\cal B}$ will
satisfy our early assumption (B), if only
$I\geq 1$, i.e. if there is at least one rectangle $R$
with $\tilde{U}(R)\neq\emptyset$ and $\tilde{S}(R)\neq\emptyset$
simultaneously.
This will be ensured by our choice of $r_0$ and $r$ below.
Here we just observe that if $I=0$, nonetheless,
then ${\cal B}^2=0$, so $X_n=\tilde{X}{\cal B}^n$
will be zero vectors for all $n\geq 2$.
\medskip
We now turn to the proof og Proposition~\ref{prmukL}.
Let $L\geq rk^2+q$.
Let $\tilde{X}$ represent the measure $\mu_{L-rk^2}^{(k)}$
(cf. the remark after (\ref{eee})),
and let $\tilde{X}_k=\tilde{X}\Pi_{rk}^k$. Then the vector
$\tilde{X}_k/|\tilde{X}_k|$ approximately represents
the measure $\mu_{L}^{(k)}$, according to (\ref{eee}).
Also, let $\tilde{Y}$ represent the measure $\mu_+^{(k)}$.
Then $\tilde{Y}_k/|\tilde{Y}_k|$ approximates the same
measure $\mu_+^{(k)}$. Therefore, due to (\ref{eee})
$$
\left |\mu_{L}^{(k)}-\mu_+^{(k)}\right |_k =
||\tilde{X}_k-\tilde{Y}_k|| + O(k\alpha^k)
$$
Here and further on, $a_k=O(b_k)$ means that $a_k\leq Cb_k$,
with some global constant $C>0$.
Next, we use (\ref{XnYn}) with $n=k$. If $I\geq 1$, then
the just proven bound (\ref{gamma0}) ensures that the first
term on the right hand side of (\ref{XnYn}) is exponentially
small in $k$, i.e.
$$
\left |\mu_{L}^{(k)}-\mu_+^{(k)}\right |_k
\leq 2(\varepsilon_{X_k}+\varepsilon_{Y_k})+
O(\beta^k) + O(k\alpha^k)
$$
with $\beta=(1-\gamma_0^2)/(1+\gamma_0^2)<1$.
If $I=0$, however, then $X_k=Y_k=0$, and the
first term in (\ref{XnYn}) can be simply
omitted, but then $\varepsilon_{X_k}=\varepsilon_{Y_k}=1$.
Therefore, it remains to estimate the quantities $\varepsilon_{X_k}$,
$\varepsilon_{Y_k}$. We will show that they are exponentially
small in $k$ by choosing $r_0$ and $r$ properly. In particular,
that will imply that $I\geq 1$, i.e. the matrix ${\cal B}$ will
be nontrivial (${\cal B}^2\neq 0$).
Again, for brevity we will suppress the superscript $(k)$
in ${\cal R}^{(k)}$, $T^{(k)}$ and $M^{(k)}$.
Let $R\in{\cal R}$ and $U$ be an unstable $R$-fiber.
Denote by $\nu_U$ the u-SBR measure on $U$.
\begin{lemma}
There is a global constant $C_{21}>0$ such that
for every $n\geq m\geq 0$
$$
\nu_U\{x\in U\cap M_{-n}:\, l_n(x)\geq m\}
\leq C_{13}\,\beta_1^m[\lambda_+^{(k)}]^n/|U|
\leq C_{21}\,\beta_1^m[\lambda_+^{(k)}]^n\Lambda_{\max}^k
$$
\label{lmloss1}
\end{lemma}
{\em Proof}. The first bound is claimed in Lemma~\ref{lmWuln}
for the original holes $H$, so, it applies here provided
$$
\beta_1 > B_0\Lambda_{\min}^{-1}/\lambda^{(k)}_+
$$
for all sufficiently large $k$, which is true, see (\ref{beta1}).
The second inequality in the lemma follows from the obvious bound
$|U|\geq {\rm const}\cdot\Lambda_{\max}^{-k}$. $\Box$
\begin{corollary}
By setting $m=0$ we get
$\nu_U (U\cap M_{-n})\leq C_{13}\,[\lambda_+^{(k)}]^n/|U|
\leq C_{21}\,[\lambda_+^{(k)}]^n\Lambda_{\max}^k$.
\label{crloss1}
\end{corollary}
Now, let $R\in{\cal R}$ and $S$ be a stable $R$-fiber. Let
$m\geq 1$ and $S_m=\{x\in S\cap M_m:\, l_m^-(x)=m+1\}$.
These are points in $S$ whose first $m$ backward images
(under $T^{-1}$) are in short components (of length $0$
$$
\left (T_{\ast}^m\mu\right )(G_m\cap M_{-n})\leq
C_{22}\,\beta_1^m[\lambda_+^{(k)}]^{n+m}\Lambda_{\max}^{2k}
$$
\label{lmloss2}
\end{lemma}
{\em Proof}. Let $R\in{\cal R}$ and $d^u(R)$, $d^s(R)$ be the maximum
lengths of unstable and stable $R$-fibers, respectively.
The set $T^{-m}R_m$ consists of at most $B_0^m$
connected rectangles,
denote them by $R_{m,i}$, $1\leq i\leq B_0^m$.
Each $R_{m,i}$ has length in the stable direction $2r_0$,
so that $rk>2r_0k+k_1$ for all sufficiently large $k$.
We now estimate the value $\varepsilon_{X_k}=|\tilde{X}_k
-X_k|/|\tilde{X}_k|$, i.e. the relative difference between
the vectors $\tilde{X}_k$ and $X_k$.
Recall that $\tilde{X}$ represents the probability
measure $\mu^{(k)}_{L-rk^2}$ and that $\tilde{X}_k=
\tilde{X}\Pi_{rk}^k$, cf. Remark after (\ref{eee}).
Observe that
$$
|\tilde{X}_k|
=||T_{\ast}^{rk^2}\mu^{(k)}_{L-rk^2}||\cdot \left (1+O(e^{Ck\alpha^k})\right )
\sim [\lambda_+^{(k)}]^{rk^2}
$$
We now make the last crucial observation.
The difference between $\tilde{X}_k$ and $X_k$ results from
the differences between the matrices ${\cal A}=\Pi_{rk}$ and ${\cal B}$,
compare (\ref{Pim}) setting $m=rk$ there and (\ref{BRR}).
We have defined the components ${\cal B}(R',R'')$ by removing
certain parts from $R'$ and $R''$ that entered the
definition of $\Pi_{rk}={\cal A}$. Those parts are precisely
described by Lemmas~\ref{lmloss1} and \ref{lmloss2},
where one sets $m=r_0k$ and $n=rk,2rk,\ldots,rk^2$.
Our choice of $r_0$ ensures that the
relative losses incurred by the removal of those parts
from all $R',R''\in{\cal R}$
are exponentially small in $k$ (the losses are
bounded by const$\cdot\beta_2^k$).
The vector $X_k=\tilde{X}{\cal B}^k$ suffers these losses
every time we replace ${\cal A}$ by ${\cal B}$ in the product
$\tilde{X}_k=\tilde{X}{\cal A}^k$. This happens
$k$ times during $rk^2$ iterations of $T$, so
the total relative difference between $\tilde{X}_k$ and $X_k$
will be bounded by const$\cdot k\beta_2^k$. The argument
for the vector $\tilde{Y}_k$ is the same. Therefore,
$$
\varepsilon_{X_k}+\varepsilon_{Y_k}\leq{\rm const}\cdot k\beta_2^k
$$
which is exponentially small in $k$.
This completes the proof of Proposition~\ref{prmukL}. $\Box$
\medskip
{\bf Acknowledgements}. We are grateful to H. Hu, who suggested
to us Manning's entropy formula for the proof of the escape
rate formula. N.~Chernov was partially supported by NSF grant
DMS-9654622. R.~Markarian acknowledges the support of
CONICYT and CSIC, Universidad de la Rep\'ublica, Uruguay.
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