The crux of my question is that why $\mu(N^{(2)}) = 0$ when $N^{(2)}$ consists all points of a billiard that reflect infinitely often in finite time under a given flow (i.e. transformation) function. This is taken from the book Ergodic Theory by Cornfeld, Fomin and Sinai, Chapter 6 "Billiards", pages 138-142. I apologize in advance for the length of this question. You can always skip to the main question at the end.

Definitions: Suppose that $Q_0$ is a closed smooth Riemann manifold which is possibly noncompact. Suppose that $r$ smooth functions $f_i, i = 1,\dots,r$ are given on $Q_0$.

Suppose $\Gamma_q$ for $q \in f_i^{-1}(0)$ is the tangent space to $f_i^{-1}(0)$ at the point $q$. By $n(q)$ denote the unit normal vector to $\Gamma_q$ directed inside $Q$.

Denote by $M_0$ the unit tangent bundle over $Q_0$ so that the points of $M_0$ are of the form $x = (q, v), q \in Q_0, v \in S^{d - 1}, d = \mathrm{dim}(Q)$. Suppose that $\pi:M_0\to Q_0$ is the natural projection $\pi(q, v) = q, x = (q, v) \in M_0$. Define $M = \pi^{-1}(Q)$.

Note: While the authors do not say it explicitly, I am assuming that $\left<n(q), x\right>$ for $x \in M_0$ is defined implicitly as $\left<n(q), x\right> = \left<n(q), (q, v)\right> \equiv \left<n(q), \pi(q, v)\right> = \left<n(q), q\right>$.

Furthermore define the sets

$$\begin{cases}Q &= \{q \in Q_0\mid f_i(q) \geq 0, i = 1,\dots,r\}\\\ \partial Q_i &= f_i^{-1}(0)\cap Q, i = 1,\dots,r\\ \tilde{Q}_i &= \partial Q_i\setminus \cup_{k\neq i}\partial Q_k, i = 1,\dots,r\\ \partial \tilde{M}_i &= \pi^{-1}(\partial \tilde{Q}_i), i = 1,\dots,r\\ \partial \tilde{M} &= \bigcup_{i}\partial \tilde{M}_i\\M_1 &= \{x \in \partial \tilde{M}\mid \left<n(q), x\right> > 0, q = \pi(x)\}\\ \end{cases}$$

(Measures) Define the measure $\mu$ on $M_0$ by putting $d\mu = d\rho(q)d\omega_q$, where $d\rho(q)$ is the volume element in$Q_0$ generated by the Riemann metric, $\omega_q$ is the Lebesgue measure on the $(d - 1)$-dimensional sphere $S^{d-1}(q) = \pi^{-1}(x)$. With this formula we have that for any Borel set $A \subset M, \mu(A) = \int_{Q_0}d\rho(q)\int_{A\cap S^{d-1}(q)}d\omega_q(x)$. The same letter $\mu$ will denote the restriction of this measure to $M$. Furthermore it will be assumed that $\mu$ is normalized on $M$. Define also the measure $\mu_1$ on $\partial \tilde{M}$ by $d\mu_1(x) = d\rho_i(q)d\omega_q\left|\left<n(q), x\right>\right|, x \in \partial \tilde{M}_i$ and $d\rho(q)$ is the volume element induced by the Riemann metric on $\partial Q_i$

(Geodesics) Suppose a vector field $X$ generates a geodesic flow on $M_0$, where $X(x)$ is the tangent vector to $M_0$ at $x \in M_0$. By letting the same symbol $X$ denote the restriction of the vector field to $M$, $X$ determines the motion of our point ($x$) with unit velocity along geodesic lines in $M$. Take any $x \in \mathrm{int}(M)\setminus \cup_{i\neq j}N_{i, j}$ where $\mathrm{int}(M)$ denotes the set of interior points of $M$. According to the authors, there are two possible cases. 1.) The geodesic line constructed in the direction $x$ does not intersect the boundary $\partial Q$; 2.) The end point of a geodesic segment of some finite length $s$ is located at a point $q \in \bigcup_{i}\partial \tilde{Q}_i$.

(Movement) A point $x$ intersecting the boundary $\partial Q$ is reflected according the "incidence angle equals reflection angle" rule, i.e. $y' = y - 2\left<n(q), y\right>\cdot n(q)$ where $y$ is the tangent vector obtained from $x$ by parallel translation along the geodesic to the end point (i.e. the point of contant with $\partial Q$) of a segment.

Suppose that $N_{i, j}$ is the set of all interior points $x\in M$ such that the segment of the geodesic line constructed in the direction $x$ intersects $\partial Q$ on $\partial Q_i\cap \partial Q_j$.

(The set $N^{(2)}$) Denote by $N^{(2)}$ the set of points $x$ for which this construction leads to infinite number of reflections in finite time. Note: The authors do not really define what this means, so I am presuming that "infinite number of reflections in finite time" can mean e.g. that the number of reflections of a point $x$ is equal to $\frac{1}{C - t}$ at time $t < C < \infty$.

(The point $x^{-}$) Let $x \in \mathrm{int}(M)$. Denote by $x^{-}$ the nearest point in $\partial M$ to $x$ on the flow (i.e. billiard) trajectory so that $x = T^{\tau}x^{-}$ for some $\tau > 0$.

(The function $f(x)$) Before the lemma below, the first instance at which the authors refer to a function $f(x)$ (note no subscript) is when they define the transformation $T_1$ in the following way:

Define a one-parameter group of transformations $\{T^t\}$ on $M$ by setting $T^tx, x \in M, t \in \mathbb{R}$ equal to the tangent vector obtained by a translation of $x$ along the trajectory which it determines by a distance $t$. [...] The billiards $\{T^t\}$ are directly related to the transformation $T_1$ of the set $M_1$ defined in the following way: consider the geodesic segment in the direction of $x$ with origin $q = \pi(x)$ and end point at the first intersection with the boundary and reflect the tangent vector from the boundary at the end of the segment. The vector $y$ thus obtained will be put equal to $T_1x$. Clearly if $x$ is in the set where $\{T^t\}$ is defined, then $T_1x = T^{f(x)}x$, where $f(x)$ is the length of the geodesic segment.

Question: Consider the following lemma

Lemma $\mu(N^{(2)}) = 0$.

Proof: It is clear that if $x = T^{\tau}x^{-}$ and $x \in N^{(2)}$, then all the points $x' = T^tx^{-}$, $0 \leq t \leq f(x^{-})$, belong to the set $N^{(2)}$. Therefore $\mu(N^{(2)}) = \int_{N^{(2)}\cap M_1}f(x^{-})d\mu_1(x^{-})$. But the last integral vanishes since $N^{(2)}\cap M_1$ consists of all the points $x$ for which the sums $\sum_{k=0}^{n-1}f(T^k_1 x)$ remain bounded when $n\to \infty$. The lemma is proved. $\square$

What I don't understand is why "the sums $\sum_{k=0}^{n-1}f(T^k_1x)$ remain bounded" $\implies$ "the integral $\int_{N^{(2)}\cap M_1}f(x^{-})d\mu_1(x^{-})$ vanishes". But gut feeling for the implication is some combination of ergodicity theorems which force the integral to vanish or else some contradiction is met. But as the presented proof does not refer to any particular point in the book itself (or really to anything for that matter) I don't know how to begin to improve my intuition about the problem.



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