text:
\section{Introduction}
\noindent

In this paper we present new methodological tools to investigate the
  well-posedness of the general transport equation
\begin{subequations}\label{1}
\begin{equation}\label{1a}
\partial_t f(\mathbf{x},t)+\mathscr{F}(\mathbf{x})\cdot \nabla_\mathbf{x}
f(\mathbf{x},t) =0 \qquad (\mathbf{x} \in\mathbf{\Omega}, \:t
> 0),\end{equation} supplemented by boundary condition
\begin{equation}\label{1b}
f_{|\Gamma_-}(\mathbf{y},t)=\psi_-(\mathbf{y},t), \qquad \qquad (\mathbf{y} \in \Gamma_-, t
>0),
\end{equation}
and the initial condition
\begin{equation}\label{1c}f(\mathbf{x},0)=f_0(\mathbf{x}), \qquad \qquad (\mathbf{x} \in \mathbf{\Omega}). \end{equation}\end{subequations}
Here $\mathbf{\Omega}$ is a sufficiently smooth open subset of $\mathbb{R}^N$,
$\Gamma_{\pm}$ are suitable boundaries of the phase space and $\psi_-$
is a given function of the trace space $L^1(\Gamma_-,\mathrm{d}\mu_-)$
corresponding to the boundary $\Gamma_-$ (see Section 2 for
details).

The present paper is part of a series of papers on transport
equations with general vector fields \cite{m2as,abl} and introduce
all the methodological tools that  allow us not only to solve the
initial-boundary problem \eqref{1} but also to treat in \cite{abl}
  the case of abstract boundary
conditions relying the incoming and outgoing fluxes, generalizing
the results of \cite{beals}.

The main novelty of our approach is that we assume $\mathbb{R}^N$ to
be endowed with a general positive Radon measure $\mu$. Here by a
Radon measure we understand a Borel measure (or its completions, see
\cite[p.~332]{Roy}) which is finite on compact sets. As we shall see
it further on, taking into account such general Radon measure $\mu$
leads to a large amount of technical difficulties, in particular in
the definition of trace spaces and in the derivation of Green's
formula. Moreover, for such a measure $\mu$, it is far from being
trivial to identify the vector field $\mathscr{F} \cdot \nabla_x$ to the
time derivative along the characteristic curves (as done in
\cite[Formulae (5.4) \& (5.5), p.392]{beals}): the main difficulty
stemming from the impossibility of applying classical convolution
arguments (and the so-called Friedrich's lemma). We overcome this
difficulty by introducing new mollification techniques  along the
characteristic curves. Let us explain in more details our general
assumptions:

\subsection{General assumption and motivations}
The transport coefficient $\mathscr{F}$ is a \textit{time independent}
vector field
$\mathscr{F} \::\:\mathbb{R}^N  \longrightarrow
\mathbb{R}^N$ which is (globally) \textit{Lipschitz-continuous} with Lipschitz
constant $\kappa > 0$, i.e.
\begin{equation}\label{lipsc}
|\mathscr{F}(\mathbf{x}_1)-\mathscr{F}(\mathbf{x}_2)|\leqslant \kappa |\mathbf{x}_1-\mathbf{x}_2| \qquad \text{ for any
} \:\mathbf{x}_1,\mathbf{x}_2 \in \mathbb{R}^N.\end{equation}
 Clearly, one can associate a flow $(T_t)_{t \in \mathbb{R}} $
to this field $\mathscr{F}$ (with the notations of Section \ref{sub:chara},
 $T_t=\mathbf{\Theta}(\cdot, t, 0)$)
and we make the following fundamental assumption (known as
\textit{\textbf{Liouville's Theorem}} whenever $\mu$ is the Lebesgue
measure) on $\mathscr{F}$:
\begin{hyp}\label{ass:h2} The measure $\mu$ is invariant under the
flow $(T_t)_{t \in \mathbb{R}}$, i.e. $\mu(T_t A) = \mu(A)$ for any
measurable subset  $A \subset \mathbb{R}^N$ and any $t \in
\mathbb{R}$.\end{hyp}

\begin{nb}  Notice that, whenever $\mu$ is the Lebesgue measure over
$\mathbb{R}^N$, it is well-known that Assumption \ref{ass:h2} is
equivalent to $ \mathrm{div}(\mathscr{F}(\mathbf{x}))=0$  for any $\mathbf{x} \in
\mathbb{R}^N.$ More generally, by virtue of \cite[Remark 3 \&
Proposition 4]{ambrosio}, Assumption \ref{ass:h2} holds for a
general
 Borel measure $\mu$ provided the  field $\mathscr{F}$ is locally
 integrable with respect to $\mu$ and
\textbf{\textit{divergence-free}} with respect to $\mu$ in the sense
that
$$\int_{\mathbb{R}^N}\mathscr{F}(T_t(\mathbf{x})) \cdot
\nabla_{\mathbf{x}}f(T_t(\mathbf{x}))\mathrm{d}\mu(\mathbf{x})=0, \qquad \qquad \forall t \in
\mathbb{R}$$ for any infinitely differentiable function $f$ with
compact support.
 \end{nb}

\noindent  A typical example of such a transport equation is the
so-called Vlasov equation for which:
\begin{enumerate}[i)\:] \item The phase space $\mathbf{\Omega}$ is given by the
cylindrical domain $\mathbf{\Omega}=\mathcal{D}\times \mathbb{R}^3 \subset
\mathbb{R}^6$ where $\mathcal{D }$ is a sufficiently
 smooth open subset of $\mathbb{R}^3$, referred to as the
\textit{position space}, while the so--called \textit{velocity
space} is here given by $\mathbb{R}^3.$ The measure $\mu$ is given
by $\mathrm{d}\mu(\mathbf{x})=\mathrm{d} x \mathrm{d}\beta(v)$ where $\beta $ is a suitable Radon
measure on $\mathbb{R}^3$, e.g. Lebesgue measure over $\mathbb{R}^3$
for continuous models or combination of Lebesgue measures over
suitable spheres for the multigroup model. \item For any $\mathbf{x}=(x,v)
\in \mathcal{D} \times \mathbb{R}^3$,
\begin{equation}\label{vlasov}\mathscr{F}(\mathbf{x})=(v,\mathbf{F}(x,v)) \in \mathbb{R}^6\end{equation} where
$\mathbf{F}=(\mathbf{F}_1,\mathbf{F}_2,\mathbf{F}_3)$ is a time independent force field over
$\mathcal{D} \times \mathbb{R}^3$ satisfying  Assumption
\ref{ass:h2} and (\ref{lipsc}). The free transport case,
investigated in \cite{voigt,labl}, corresponds to $\mathbf{F}=0$.
\end{enumerate}

The existence of solution to the transport equation \eqref{1a} is a
classical matter when considering the whole space $\mathbf{\Omega}=\mathbb{R}^N$.
In particular, the concept of renormalized solutions allows to
consider irregular transport coefficient $\mathscr{F}(\cdot)$ (see
\cite{diperna} and the recent contributions \cite{ambrosio,lions})
which is of particular relevance in fluid mechanics.

On the other hand, there are few results addressing the
initial-boundary value problem \eqref{1}, possibly due to
difficulties created by the boundary conditions \eqref{1b}. We
mention here the seminal works by C. Bardos \cite{bardos}, and by R.
Beals and V. Protopopescu \cite{beals} (see also \cite{proto,vdm}).
Let us however mention that the results of \cite{beals,proto}
introduce restrictive assumptions on the characteristics of the
equation. For instance, fields with 'too many' periodic trajectories
create serious difficulties. They are however covered in a natural
way by the theory presented here, see Examples \ref{exa1} \&
\ref{exa2}.

\subsection{Presentation of the results}

In this paper, we revisit and generalize the afore-mentioned results
to the general case $\mathbf{F}\neq 0$ and for a general Radon measure
$\mu$.  The latter, in particular, leads to numerous technical
problems such as e.g. determination of suitable measures $\mu_{\pm}$
over the `incoming' and `outgoing' parts $\Gamma_{\pm}$ of $\partial
\mathbf{\Omega}$. We provide here a general construction of these `trace
measures' generalizing, and making more precise, the results of
\cite{beals,proto}. This construction allows us to establish
Proposition \ref{prointegra} which allows to compute integrals over
$\mathbf{\Omega}$ via integration along the integral curves of
$\mathscr{F}(\cdot)$ coming from the boundary $\partial \mathbf{\Omega}$,
and which is free from some restrictive assumptions of \textit{op.
cit}. In particular, we present a new proof of the Green formula
clarifying and removing gaps of the proofs in \cite{beals,proto}. Of
course, the boundary condition \eqref{1b} we treat here is less
general than the abstract ones investigated in \cite{beals,proto}
but, as we already mentioned  it, the tools we introduce here will
allow us to generalize, in a subsequent paper \cite{abl}, the
results of the \textit{op. cited} by dealing with abstract boundary
conditions.

 Another major difficulty, when dealing with a
general Radon measure $\mu$, is to provide a precise definition of
the transport operator $\mathcal{T}_\mathrm{max}$ associated to \eqref{1}. It
appears quite natural to define the transport operator
$\mathcal{T}_\mathrm{max}$ (with its maximal domain on $L^1(\mathbf{\Omega},\mathrm{d}\mu)$) as a
\textit{{weak directional derivative along the characteristic
curves}} in the $L^1$-sense. However, it is not clear {\it a priori}
that any function $f$ for which  the weak directional derivative
exists in $L^1(\mathbf{\Omega},\mathrm{d}\mu)$ (with appropriate and minimal class of
test-functions)  admits a trace over $\Gamma_{\pm}$. With the aim of
proving  such a trace result, we provide here a new characterization
of the transport operator related to a \textit{{mild
representation}} of the solution to \eqref{1}. Namely, we prove
(Theorem \ref{representation}) that the domain $\Drond(\mathcal{T}_\mathrm{max})$
(as defined in Section 3), is precisely the set of functions $f \in
L^1(\mathbf{\Omega},\mathrm{d}\mu)$ that admits a representative which is
\textit{absolutely continuous along almost any characteristic
curve}.

Note that in the classical case when $\mu$ is the Lebesgue measure,
such a representation is known to be true \cite[Appendix]{diperna}.
Actually, in this case, one defines the domain $\Drond(\mathcal{T}_\mathrm{max})$
 as the set of all $f \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$ for which
the directional derivative $-\mathscr{F} \cdot \nabla f$ exists in the
distributional sense and belongs to $L^1(\mathbf{\Omega},\mathrm{d}\mu)$. Then, by
convolution arguments, it is well-known that the set $\ensuremath{\mathscr{C}}_0^1(\mathbf{\Omega})
\cap \Drond(\mathcal{T}_\mathrm{max})$ is dense in $\Drond(\mathcal{T}_\mathrm{max})$ for the
graph norm $\|f\|=\|f\|+\|\mathscr{F} \cdot \nabla f\|.$

The question is much more delicate for a   general Radon measure
$\mu$. Indeed, in such a case, the convolution argument used in the
case of the Lebesgue measure does not apply anymore. Our strategy to
prove the characterization of $\mathcal{T}_\mathrm{max}$ is also based on a
convolution argument but it uses  \textit{mollification technique
along the characteristic curves} as developed in Section 3. Such a
result shall allow us to obtain a rigorous derivation of Green's
formula, clarifying some results of \cite{beals}.
\subsection{Plan of the paper} The organization of the paper is as follows. In Section 2
we introduce main tools used throughout the paper and present the
aforementioned new results concerning \textit{integration over the
characteristic curves} of $\mathscr{F}$ as well as \textit{a new
construction of the boundary measures} over the `incoming' and
`outgoing' parts $\Gamma_{\pm}$ of $\partial \mathbf{\Omega}$ which generalizes and
clarifies that of \cite{beals,proto}. In Section 3 we provide a
construction of the maximal transport operator $\mathcal{T}_\mathrm{max}$. It
is defined in a weak sense, through its action on suitably defined
test functions. The fundamental result of this section shows that
any function in the domain $\Drond(\mathcal{T}_\mathrm{max})$ admits a
representation which is absolutely continuous along almost any
characteristic which, in turn, allows for existence of its traces on
the outgoing and incoming parts of the boundary. In Section 4 we
apply the results of Section 3 to prove well-posedness of the
time--dependent transport problem with no reentry boundary
conditions associated with $\mathcal{T}_\mathrm{max}$. Moreover, we consider
the corresponding stationary problem and, as a by-product, we
recover  a new proof of the \textit{Green formula}.

\section{Integration along the characteristics}
\subsection{\textbf{Characteristic curves}}\label{sub:chara}   A crucial role in our study
is played  by the characteristic curves associated to the
field $\mathscr{F}$.
Precisely, for any $\mathbf{x} \in \mathbb{R}^N$ and $t \in \mathbb{R}$,
consider the initial-value problem
\begin{equation}\label{chara}
\begin{cases}
\dfrac{\mathrm{d} }{\mathrm{d} s}\mathbf{X}(s)=\mathscr{F}(\mathbf{X}(s)), \qquad (s \in \mathbb{R});\\
\mathbf{X}(t)=\mathbf{x}.
\end{cases}\end{equation}
Since $\mathscr{F}$ is Lipschitz continuous on $\mathbb{R}^N$, Eq.
\eqref{chara} has a unique {\it global in time} solution and this
allows to define the flow--mapping $\mathbf{\Theta}$ : $\mathbb{R}^N \times
\mathbb{R}\times \mathbb{R} \to \mathbb{R}^N$, such that, for
$(\mathbf{x},t) \in \mathbb{R}^N \times \mathbb{R}$, the mapping:
$$\mathbf{X}(\cdot)\::\:s \in \mathbb
{R} \longmapsto \mathbf{\Theta}(\mathbf{x},t,s)$$ is the only solution of
Eq. \eqref{chara}.
Being concerned with solutions to the transport equation \eqref{1}
in the region $\mathbf{\Omega}$, we have to introduce the definition of stay
times of the characteristic curves in $\mathbf{\Omega}$:
\begin{defi} For any $\mathbf{x} \in \mathbf{\Omega}$, define
$\tau_{\pm}(\mathbf{x})=\inf \{s > 0\,;{\mathbf{\Theta}}(\mathbf{x},0,\pm s) \notin \mathbf{\Omega}\},$
with the convention that $\inf \varnothing=\infty,$ and set $\tau(\mathbf{x})
= \tau_+(\mathbf{x}) + \tau_-(\mathbf{x}). $\end{defi} In other words, given $\mathbf{x} \in
\mathbf{\Omega}$, $I_{\mathbf{x}}=(-\tau_-(\mathbf{x}),\tau_+(\mathbf{x}))$ is the maximal interval for which
${\mathbf{\Theta}}(\mathbf{x},0,s)$ lies in $\mathbf{\Omega}$ for any $s \in I_{\mathbf{x}}$ and
$\tau(\mathbf{x})$ is the length of the interval $I_{\mathbf{x}}$. Notice that $0\leqslant
\tau_{\pm}(\mathbf{x}) \leqslant \infty$. Thus, the function ${\mathbf{\Theta}}$
restricted to the set
$$\mathbf{\Lambda}:=\bigg\{(\mathbf{x},t,s)\,;\,\mathbf{x} \in \mathbf{\Omega},\,t \in
\mathbb{R}\,,\,s \in \left(t-\tau_-(\mathbf{x}),t+\tau_+(\mathbf{x})\right)\;\bigg\}$$
is such that ${\mathbf{\Theta}}(\mathbf{\Lambda})=\mathbf{\Omega}$.  Note that
here we \textit{\textbf{do not}} assume that the length of the
interval $I_{\mathbf{x}}=(-\tau_-(\mathbf{x}),\,\tau_+(\mathbf{x}))$ is
\textit{\textbf{finite}}. In particular, $I_{\mathbf{x}}=\mathbb{R}$ for any
stationary point $\mathbf{x}$ of $\mathscr{F}$, i.e. $\mathscr{F}(\mathbf{x})=0$. If $\tau(\mathbf{x})$ is
finite, then the function $\mathbf{X}\::\:s \in I_{\mathbf{x}} \longmapsto
{\mathbf{\Theta}}(\mathbf{x},0,s)$ is bounded since $\mathscr{F}$ is Lipschitz
continuous. Moreover, still by virtue of the Lipschitz continuity of
$\mathscr{F}$, the only case when $\tau_{\pm}(\mathbf{x})$ is finite is when
${\mathbf{\Theta}}(\mathbf{x},0,\pm s)$ reaches the boundary $\partial\mathbf{\Omega}$ so
that ${\mathbf{\Theta}}(\mathbf{x},0,\pm \tau_{\pm}(\mathbf{x})) \in
\partial\mathbf{\Omega}$. We note that, since $\mathscr{F}$ is Lipschitz
around each point of $\partial\mathbf{\Omega}$, the points of the set $\{\mathbf{y} \in
\partial\mathbf{\Omega}\,;\,\mathscr{F}(\mathbf{y})=0\}$ (introduced in
\cite{beals,proto}) are equilibrium points of the $\mathscr{F}$ and
cannot be reached in finite time.
 \begin{nb}
We emphasize that periodic trajectories which do not meet the
boundaries have $\tau_\pm =\infty$ and thus are treated as infinite
though geometrically they are bounded.
\end{nb}

Finally we mention that it is not difficult to prove that the
mappings $\tau_{\pm}$ : $\mathbf{\Omega} \to \mathbb{R}^+$ are lower
semicontinuous and therefore measurable, see e.g., \cite[p.\
301]{arloban}

The flow ${\mathbf{\Theta}}(\mathbf{x},t,s)$ defines, at each instant $t$, a
mapping of the phase space $\mathbf{\Omega}$ into $\mathbb{R}^N$. Through this
mapping, to each point $\mathbf{x}$ there corresponds the point
$\mathbf{x}_{s,t}={\mathbf{\Theta}}(\mathbf{x},t,s)$ reached at time $s$ by
the point which was at $\mathbf{x}$ at the `initial' time $t$.
The flow ${\mathbf{\Theta}}$, restricted to $\mathbf{\Lambda}$, has
the properties:
\begin{propo}\label{Phiprop} Let $\mathbf{x} \in \mathbf{\Omega}$ and $t \in \mathbb{R}$ be fixed. Then,
\begin{enumerate}[(i)\:]
\item ${\mathbf{\Theta}}(\mathbf{x},t,t)=\mathbf{x}.$ \item
${\mathbf{\Theta}}({\mathbf{\Theta}}(\mathbf{x},t,s_1),s_1,s_2)={\mathbf{\Theta}}(\mathbf{x},t,s_2),
\quad \forall  s_1,s_2 \in (t-\tau_-(\mathbf{x}),t+\tau_+(\mathbf{x})).$ \item
${\mathbf{\Theta}}(\mathbf{x},t,s)={\mathbf{\Theta}}(\mathbf{x},t-s,0)={\mathbf{\Theta}}(\mathbf{x},0,s-t),
\qquad \forall s \in (t-\tau_-(\mathbf{x}),t+\tau_+(\mathbf{x})).$ \item
$\left|{\mathbf{\Theta}}(\mathbf{x}_1,t,s)-{\mathbf{\Theta}}(\mathbf{x}_2,t,s)\right|
\leqslant \exp(\kappa |t-s|)|\mathbf{x}_1-\mathbf{x}_2|$ for any $\mathbf{x}_1,\mathbf{x}_2 \in \mathbf{\Omega}$, $s -
t \in I_{\mathbf{x}_1} \cap I_{\mathbf{x}_2}.$
\end{enumerate}\end{propo}

An important consequence of $(iii)$ above is that
${{\mathbf{\Theta}}}(\mathbf{x},0,s)={\mathbf{\Theta}}(\mathbf{x},-s,0)$ for any $\mathbf{x}
\in\mathbf{\Omega}$, $0 \leqslant s \leqslant \tau_+(\mathbf{x}).$ Therefore, from now on, to
shorten notations we shall denote
$${\mathbf{\Phi}}(\mathbf{x},t)={\mathbf{\Theta}}(\mathbf{x},0,t), \qquad \forall t \in \mathbb{R},$$
so that ${\mathbf{\Phi}}(\mathbf{x},-t)={\mathbf{\Theta}}(\mathbf{x},t,0)$, $t \in
\mathbb{R}.$ We define the incoming and outgoing part of the
boundary $\partial \mathbf{\Omega}$ through the flow ${\mathbf{\Phi}}$:
\begin{defi} The
\textit{incoming} $\Gamma_{-}$ and the \textit{outgoing} $\Gamma_+$
parts of the boundary $\partial \mathbf{\Omega}$ are defined by:
\begin{equation}\label{gammapm}
\Gamma_{\pm}:=\left\{\mathbf{y} \in \partial \mathbf{\Omega}\,;\exists \mathbf{x} \in \mathbf{\Omega},\, \tau_{\pm}(\mathbf{x})
< \infty \text{ and } \mathbf{y}={\mathbf{\Phi}}(\mathbf{x},\pm
\tau_{\pm}(\mathbf{x}))\,\right\}.\end{equation}
\end{defi}
Properties of ${\mathbf{\Phi}}$ and of $\tau_\pm$ imply that
$\Gamma_{\pm}$ are Borel sets. It is possible to extend the definition
of $\tau_{\pm}$ to $\Gamma_{\pm}$ as follows. If $\mathbf{x}\in\Gamma_-$ then we put
$\tau_-(\mathbf{x}) = 0$ and denote $\tau_+(\mathbf{x})$ the length of the integral
curve having $\mathbf{x}$ as its left end--point; similarly if
$\mathbf{x}\in\Gamma_+$ then we put $\tau_+(\mathbf{x}) = 0$ and denote $\tau_-(\mathbf{x})$ the
length of the integral curve having $\mathbf{x}$ as its right endpoint. Note
that this definition implies that $\tau_{\pm}$ are measurable over $\mathbf{\Omega}
\cup \Gamma_- \cup \Gamma_+$.

Let us illustrate the above definition of $\Gamma_{\pm}$ by two
simple $2D$ examples:
\begin{exa}[\textit{\textbf{Harmonic oscillator in a rectangle}}]\label{exa1} Let $\mathbf{\Omega}=(-a,a) \times (-\xi,\xi)$ with $a,\xi >0$ and
let us consider the harmonic oscillator force field
\begin{equation}\label{harmonic}
\mathscr{F}(\mathbf{x})=(v,-\omega^2 x), \qquad \text{ for any } \mathbf{x}=(x,v) \in
\mathbf{\Omega}\end{equation} where $\omega
>0$. We take as $\mu$ the Lebesgue measure over $\mathbb{R}^2$
and, since $\mathscr{F}$ is divergence-free, Assumption
\ref{ass:h2} is fulfilled. In this case, for any $\mathbf{x}_0=(x_0,v_0)
\in \mathbf{\Omega}$, the solution $(x(t),v(t))={\mathbf{\Phi}}(\mathbf{x}_0,t)$ to the
characteristic equation $\frac{\mathrm{d}}{\mathrm{d}
t}\mathbf{X}(t)=\mathscr{F}\left(\mathbf{ X}(t)\right),$ $
\mathbf{X}(0)=\mathbf{x}_0$, given by
$${\mathbf{\Phi}}(\mathbf{x}_0,t)=\left(x_0\cos(\omega t)+\frac{v_0}{\omega}\sin(\omega
t)\,;\,-x_0\omega \sin(\omega t)+v_0 \cos(\omega t)\right),$$ is
such that
$$\omega^2 x^2(t)+v^2(t)=\omega^2 x_0^2+v_0^2, \qquad t \in (-\tau_-(\mathbf{x}_0),\tau_+(\mathbf{x}_0))$$ which means that the
integral curves associated to $\mathscr{F}$ are \textit{ellipses} centered
at $(0,0)$ and oriented in the counterclockwise direction. Now,
$$\partial \mathbf{\Omega}=\bigg(\{-a\} \times [-\xi,\xi]\bigg) \bigcup \bigg(\{a\} \times
[-\xi,\xi]\bigg) \bigcup \bigg([-a,a] \times \{-\xi\}\bigg) \bigcup
\bigg([-a,a] \times \{ \xi\}\bigg)$$ and it is easy to check that
$$\Gamma_{\pm}=\bigg(\{\pm a\} \times (-\xi,0]\bigg)
\bigcup \bigg(\{\mp a\} \times [0,\xi)\bigg) \bigcup \bigg([0,a)
\times \{\pm \xi\}\bigg) \bigcup \bigg((-a,0] \times \{\mp
\xi\}\bigg).$$ Notice that $\Gamma_{+} \cap
\Gamma_-=\{(a,0),\,(0,\xi),\,(-a,0),\,(0,-\xi)\}$ and $$\partial \mathbf{\Omega}
\setminus \left(\Gamma_+ \cup
\Gamma_-\right)=\{(a,\xi),(a,-\xi),(-a,\xi),(-a,\xi)\}$$ is a
discrete  set (of linear Lebesgue measure zero).
\end{exa}
\begin{exa}[\textit{\textbf{Hamonic oscillator in a
stadium}}]\label{exa2} Consider now the two-dimensional phase space
(where $\mathbb{R}^2$ is still endowed with the Lebesgue measure
$\mu$):
$${\mathbf{\Omega}}=\{\mathbf{x}=(x,v) \in \mathbb{R}^2\,;\,x^2+v^2 < 2 \:\text{
and } -1 < v < 1\}$$ and consider the harmonic oscillator force
field $\mathscr{F}$ given by \eqref{harmonic} with $\omega=1$ for
simplicity. Then, the integral curves associated to $\mathscr{F}$ are
 \textit{circles} centered at $(0,0)$ and oriented in the counterclockwise
direction. In this case, one can see that
$$\Gamma_{\pm}=\{(x,-1)\,;\,-1 < \pm\, x \leqslant 0\} \cup \{(x,1)\,;\,0\leqslant \pm
\,x<1\}.
$$
In particular, one sees that $\partial \mathbf{\Omega} \setminus \big(\Gamma_+
\cup \Gamma_-\big)=\left\{(x,v)\in \mathbb{R}^2\,;\,x^2+v^2=2\,;\,
-1 \leqslant v \leqslant 1\right\}$ is a 'big' part of the boundary $\partial
\mathbf{\Omega}$ (with positive linear Lebesgue measure). Notice also that
$\tau_+(\mathbf{x})=+\infty$ for any $\mathbf{x}=(x,v)$ with $x^2+v^2 <1$.\end{exa}
The main aim of the present discussion is to represent $\mathbf{\Omega}$ as a
collection of characteristics running between points of $\Gamma_-$
and $\Gamma_+$ so that the integral over $\mathbf{\Omega}$ can be split into
integrals over $\Gamma_-$ (or $\Gamma_+$) and along the
characteristics. However, at present we cannot do this in a precise
way since, in general, the sets $\Gamma_+$ and $\Gamma_- $ do not
provide a partition of $\partial \mathbf{\Omega}$ as there may be `too many'
characteristics which extend to infinity on either side. Since we
have not assumed $\mathbf{\Omega}$ to be bounded, $\Gamma_-$ or $\Gamma_+$ may be
empty and also we may have characteristics running from $-\infty$ to
$+\infty$ such as periodic ones. Thus, in general, characteristics
starting from $\Gamma_-$ or ending at $\Gamma_+$ would not fill the
whole $\mathbf{\Omega}$ and, to proceed, we have to construct an auxiliary set by
extending $\mathbf{\Omega}$ into the time domain and use the approach of
\cite{beals} which is explained below.

\subsection{\textbf{Integration along characteristics}} For any $0 < T < \infty$, we define the domain
$$\mathbf{\Omega}_T=\mathbf{\Omega} \times (0,T)$$
and the measure $\mathrm{d}\mu_T=\mathrm{d}\mu\otimes\mathrm{d} t$ on $\mathbf{\Omega}_T$.
Consider the vector field over $\mathbf{\Omega}_T$:
$$Y={\partial_t} + \mathscr{F}(\mathbf{x})\cdot \nabla_\mathbf{x}=\mathscr{A}(\xi)\cdot \nabla_{\xi}$$
where $\mathscr{A}(\xi)=(\mathscr{F}(\mathbf{x}),1)$ for any $\xi=(\mathbf{x},t)$. We can
define the characteristic curves of $\mathscr{A}$ as the solution
$\xi(s)=(\mathbf{X}(s),\theta(s))$ to the system $\dfrac{\mathrm{d} }{\mathrm{d}
s}\xi(s)=\mathscr{A}(\xi(s)),$ i.e.
\begin{equation*}\dfrac{\mathrm{d} }{\mathrm{d} s}\mathbf{X}(s)=\mathscr{F}(\mathbf{X}(s)), \:\:\:\dfrac{\mathrm{d} }{\mathrm{d} s}\theta(s)=1, \qquad (s \in \mathbb{R}),\end{equation*}
with $$\mathbf{X}(0)=\mathbf{x},\quad \theta(0)=t.$$ It is clear that the
solution $\xi(s)$ to the above system is given by
$$\mathbf{X}(s)={\mathbf{\Phi}}(\mathbf{x},s), \qquad \theta(s)=s+t,$$
and we  can define the flow of solution $\mathbf{\Psi}(\xi,s)=(
{\mathbf{\Phi}}(\mathbf{x},s),s+t)$ associated to $\mathscr{A}$ and the
existence times of the characteristic curves of $Y$ are defined, for
any $\xi=(\mathbf{x},t) \in \mathbf{\Omega}_T$, as
$$\ell_{\pm}(\xi)=\inf\{s >0,\left({\mathbf{\Phi}}(\mathbf{x},\pm s),\pm s+t\right) \notin
\mathbf{\Omega}_T\}.$$ The flow $\mathbf{\Psi}(\cdot,\cdot)$ enjoys,
\textit{mutatis mutandis}, the properties listed in Proposition
\ref{Phiprop} and $\mu_T$ is invariant under $\mathbf{\Psi}$.
Moreover, since $\mathscr{A}$ is clearly Lipschitz continuous on
$\overline{\mathbf{\Omega}_T}$, no characteristic of $Y$ can escape to infinity
in finite time. In other words, all characteristic curves of $Y$ now
have finite lengths. Indeed, if ${\mathbf{\Phi}}(\mathbf{x},\pm s)$ does not
reach $\partial \mathbf{\Omega}$, then the characteristic curve
$\mathbf{\Psi}(\xi,\pm s)$
 enters or leaves $\mathbf{\Omega}_T$ through the bottom
$\mathbf{\Omega} \times \{0\}$, or through the top $ \mathbf{\Omega} \times \{T\}$ of it.
Precisely, it is easy to verify that for $\xi=(\mathbf{x},t) \in\mathbf{\Omega}_T$ we
have
$$\ell_{+}(\xi)=\tau_+(\mathbf{x}) \wedge (T-t) \quad \text{ and } \quad
\ell_-(\xi)=\tau_-(\mathbf{x}) \wedge t,$$ where $\wedge$ denotes minimum.
This clearly implies $\sup\{\ell_{\pm}(\xi)\;;\,\xi \in \mathbf{\Omega}_T\,\}\leqslant T.$
Define now
$$\Sigma_{\pm,\,T}=\{\zeta \in \partial\mathbf{\Omega}_T\,;\,\exists \xi \in\mathbf{\Omega}_T\,\text{ such that }\:
\zeta=\mathbf{\Psi}(\xi,\pm \ell_{\pm}(\xi))\}.$$
 The definition of
$\Sigma_{\pm,\,T}$ is analogous to $\Gamma_{\pm}$ with the understanding
that now the charateristic curves correspond to the vector field
$\mathscr{A}$. In other words,
 $\Sigma_{-,\,T}$ (resp. $\Sigma_{+,\,T}$) is the subset of
$\partial\mathbf{\Omega}_T$ consisting of all left (resp. right) limits of
characteristic curves of $\mathscr{A}$ in $\mathbf{\Omega}_T$ whereas
$\Gamma_{-}$ (resp. $\Gamma_{+}$) is the subset of $\partial \mathbf{\Omega}$
consisting of all left (resp. right) limits of characteristic curves
of $\mathscr{F}$ in $\mathbf{\Omega}.$ The main difference (and the interest of such a
lifting to $\mathbf{\Omega}_T$) is the fact that \textit{each characteristic
curve of $\mathscr{A}$ does reach the boundaries $\Sigma_{\pm,\,T}$
in finite time}. The above formulae allow us to extend functions
$\ell_{\pm}$ to $\Sigma_{\pm,\,T}$ in the same way as we extended the
functions $\tau_\pm$ to $\Gamma_\pm$. With these considerations, we
can represent, up to a set of zero measure, the phase space $\mathbf{\Omega}_T$
as
\begin{equation}\begin{split}\label{parametre}
\mathbf{\Omega}_{T}&=\{\mathbf{\Psi}(\xi,s)\,;\,\xi \in \Sigma_{-,\,T}\,,\,0 < s
<\ell_+(\xi)\
\\
&=\{\mathbf{\Psi}(\xi,-s)\,;\,\xi \in \Sigma_{+,\,T}\,,\,0 < s
<\ell_-(\xi)\
.\end{split}\end{equation} With this realization we can prove the
following:
\begin{propo} Let $T > 0$ be fixed. There are unique positive Borel measures $\mathrm{d}\nu_{\pm}$
on $\Sigma_{\pm,T}$ such that $\mathrm{d}\mu_T=\mathrm{d}\nu_+\otimes\mathrm{d}
s=\mathrm{d}\nu_-\otimes\mathrm{d} s.$
\end{propo}
\begin{proof} For any $\delta>0$, define $\mathscr{E}_\delta$ as the set of all bounded Borel subsets
$E$ of $\Sigma_{-,T}$ such that $\ell_+(\xi) > \delta$ for any $\xi
\in E$. Let us now \textit{fix} $E \in \mathscr{E}_\delta$. For all
$0 < \sigma \leqslant \delta$ put
$$E_{\sigma} = \{\mathbf{\Psi}(\xi,s)\,;\,\xi \in E ,0 < s \leqslant \sigma \}.$$ Clearly
$E_{\sigma}$ is a measurable subset of $\mathbf{\Omega}_{T}$. Define the mapping
$h:\:\sigma \in (0,\delta] \mapsto h(\sigma)=\mu_T(E_{\sigma})$ with
$h(0) =0$. If $\sigma_1$ and $\sigma_2$ are two positive numbers
such that $ \sigma_1 + \sigma_2 \leqslant \delta$, then
$$E_{\sigma_1 + \sigma_2}\setminus E_{\sigma_1} = \{\mathbf{\Psi}(\xi,s)\,;\,\xi \in E ,\sigma_1 < s \leqslant \sigma_1 + \sigma_2
\}= \{\mathbf{\Psi}(\eta,\sigma_1)\,;\, \eta \in E_{\sigma_2}\}.$$
The properties of the flow $\mathbf{\Psi}$  (see Proposition
\ref{Phiprop}) ensure that the mapping $\eta \mapsto
\mathbf{\Psi}(\eta,\sigma_1)$ is one-to-one and measure preserving,
so that
$$\mu_T(E_{\sigma_1 + \sigma_2}\setminus E_{\sigma_1}) =
\mu_T(E_{\sigma_2})= h(\sigma_2).$$ Since $E_{\sigma_1 + \sigma_2} =
E_{\sigma_1} \cup (E_{\sigma_1 + \sigma_2}\setminus E_{\sigma_1}),$
we immediately obtain
\begin{equation}\label{cauchy}
h(\sigma_1 + \sigma_2) = h(\sigma_1) + h(\sigma_2) \qquad   \text{
for any } \qquad \sigma_1,\,\sigma_2 >0\,\text{ with }\sigma_1
+\sigma_2 \leqslant \delta .\end{equation} This is the well-known Cauchy
equation, though defined only on an interval of the real line. It
can be solved in a standard way using non-negativity instead of
continuity, yielding:
$$h(\sigma) = c_E \sigma \qquad \text{ for any } \quad 0 < \sigma
\leqslant \delta$$ where $c_E = h(\delta)/\delta$.
We define $\nu_-(E) = c_E$. It is not difficult to see that, with
the above procedure, the mapping $\nu_-(\cdot)$ defines a positive
measure on the ring $\mathscr{E}=\bigcup_{\delta
>0}\mathscr{E}_\delta$ of all the Borel subsets of $\Sigma_{-,T}$
on which the function $\ell_+(\xi)$ is bounded away from $0$. Such a
measure $\nu_-$ can be uniquely extended to the $\sigma$-algebra of
the Borel subsets of $\Sigma_{-,T}$ (see e.g. \cite[Theorem A, p.\
54]{Hal}). Consider now a Borel subset $E$ of $\Sigma_{-,T}$ and a
Borel subset $I$ of $\mathbb{R}^+$, such that for all $\xi \in E$
and $s \in I$ we have $0 < s < \ell_+(\xi)$. Then
$$E \times I = \{\mathbf{\Psi}(\xi,s)\,;\,\xi \in E , s \in I \} \subset
\mathbf{\Omega}_{T}.$$ Thanks to the definition of $\nu_-(\cdot)$,
 we can state that $\mu_T(E \times I)= \nu_-(E)\mathrm{meas}(I)$
where meas$(I)$ denotes the linear  Lebesgue measure of $I \subset
\mathbb{R}$. This shows that $\mathrm{d}\mu_T = \mathrm{d}\nu_-\otimes\mathrm{d} s.$
Similarly we can define a measure $\nu_+$ on $\Sigma_{+,T}$ and
prove that $\mathrm{d}\mu_T=\mathrm{d}\nu_+\otimes\mathrm{d} s.$ The uniqueness of the
measures $\mathrm{d}\nu_{\pm}$ is then obvious.\end{proof}
\begin{nb} Note that the above construction of the Borel measures $\mathrm{d}\nu_{\pm}$ differs from that of \cite[Lemmas
XI.3.1 \& 3.2]{proto}, \cite[Propositions 7 \& 8]{beals} which ,
moreover, only apply when $\mu$ is absolutely continuous with
respect to the Lebesgue measure. Our construction is much more
general and can also be generalized to the case of a non--divergence
force field $\mathscr{F}$, \cite{m2as}.\end{nb}
 Next, by  the cylindrical structure of $\mathbf{\Omega}_T$, and the representation of $\Sigma_{\pm,T}$ as
$$\Sigma_{-,\,T}=\left(\Gamma_- \times (0,T) \right) \cup\mathbf{\Omega} \times
\{0\}\quad \text{ and } \quad \Sigma_{+,\,T}=\left(\Gamma_+ \times
(0,T) \right) \cup\mathbf{\Omega} \times \{T\},$$ the measures $\mathrm{d}\nu_{\pm}$ over
$\Gamma_{\pm} \times (0,T)$ can be written as $\mathrm{d}\nu_{\pm}=\mathrm{d}\mu_{\pm} \otimes\mathrm{d}
t$, where $\mathrm{d}\mu_{\pm}$ are Borel measures on $\Gamma_{\pm}$.  This leads to
the following
\begin{lemme}\label{defmu+-} There are unique positive Borel measures $\mathrm{d}\mu_{\pm}$
on $\Gamma_{\pm}$ such that, for any $f \in L^1(\mathbf{\Omega}_T,\mathrm{d}\mu_T)$
\begin{equation}\label{beals1}\begin{split}
\int_{\mathbf{\Omega}_T}f(\mathbf{x},t)\mathrm{d}\mu_T(\mathbf{x},t)&=\int_0^T\mathrm{d}
t\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y}) \wedge t}
f({\mathbf{\Phi}}(\mathbf{y},-s),t-s)\mathrm{d} s\\
&\phantom{x}+\int_{\mathbf{\Omega}}\mathrm{d}\mu(\mathbf{x})\int_0^{\tau_-(\mathbf{x}) \wedge
T}f(\mathbf{\Phi}(\mathbf{x},-s),T-s)\mathrm{d} s,
\end{split}\end{equation} and
\begin{equation}\label{beals2}\begin{split}
\int_{\mathbf{\Omega}_T}f(\mathbf{x},t)\mathrm{d}\mu_T(\mathbf{x},t)&=\int_0^T\mathrm{d}
t\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y}) \wedge (T-t)}
f({\mathbf{\Phi}}(\mathbf{y},s),t+s)\mathrm{d} s\\
&\phantom{x}+\int_{\mathbf{\Omega}}\mathrm{d}\mu(\mathbf{x})\int_0^{\tau_+(\mathbf{x}) \wedge
T}f(\mathbf{\Phi}(\mathbf{x},s),s)\mathrm{d} s.
\end{split}\end{equation}\end{lemme}
The above fundamental result allows to compute integrals over the
cylindrical phase-space $\mathbf{\Omega}_T$ through integration along the
characteristic curves. Let us now generalize it to the phase space
$\mathbf{\Omega}$. Here the main difficulty stems from the fact that the
characteristic curves of the vector field $\mathscr{F}$ are no longer
assumed to be of finite length. In order to extend Lemma
\ref{defmu+-} to possibly infinite existence times, first we prove
the following:
\begin{lemme}
Let $T > 0$ be fixed. Then, $\tau_+(\mathbf{x}) < T$ for any $\mathbf{x} \in \mathbf{\Omega}$ if
and only if $\tau_-(\mathbf{x}) < T$ for any $\mathbf{x} \in \mathbf{\Omega}.$\end{lemme}
\begin{proof} It is easy to see that $\tau_+(\mathbf{x}) < T$ for any $\mathbf{x} \in \mathbf{\Omega}$ is equivalent to $\tau(\mathbf{x}) < T$ for any $\mathbf{x} \in \mathbf{\Omega}$ and this is also equivalent to
$\tau_-(\mathbf{x}) < T$ for  any $\mathbf{x} \in \mathbf{\Omega}$.\end{proof}

 Hereafter, the support of a measurable function $f$
defined on $\mathbf{\Omega}$ is defined as $\mathrm{Supp}f=\mathbf{\Omega} \setminus
\mathbf{\omega}$  where $\mathbf{\omega}$ is the maximal open subset
of $\mathbf{\Omega}$ on which $f$ vanishes $\mathrm{d} \mu$--almost everywhere.

\begin{propo}\label{Prop10.10} Let $f \in L^1(\mathbf{\Omega},\mathrm{d} \mu)$. Assume that there exists $\tau_0>0$ such that
$\tau_{\pm}(\mathbf{x}) < \tau_0$ for any $\mathbf{x} \in \mathrm{Supp}(f).$ Then,
\begin{equation}\begin{split}
\int_{\mathbf{\Omega}}f(\mathbf{x})\mathrm{d}\mu(\mathbf{x})&=\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y})}f\left({\mathbf{\Phi}}(\mathbf{y},-s)\right)\mathrm{d}
s\\
&=\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}f({\mathbf{\Phi}}(\mathbf{y},s))\mathrm{d}
s.
\end{split}\end{equation}
\end{propo}
\begin{proof} For any $T > \tau_0$, define the  domain $\mathbf{\Omega}_{T}=\mathbf{\Omega} \times
(0,T)$. Since $T < \infty,$ it is clear that $f \in L^1(\mathbf{\Omega}_T,\mathrm{d}\mu
\mathrm{d} t)$ and, by \eqref{beals1}, we get
\begin{multline*}
T\int_{\mathbf{\Omega}}f(\mathbf{x})\mathrm{d}\mu(\mathbf{x})=\int_0^T\mathrm{d} t \int_{\Gamma_+}\mathrm{d}
\mu_+(\mathbf{y})\int_0^{t \wedge \tau_-(\mathbf{y})}f({\mathbf{\Phi}}(\mathbf{y},-s))\mathrm{d} s+\\
\int_{\mathbf{\Omega}}\mathrm{d}\mu(\mathbf{x})\int_0^{\tau_-(\mathbf{x})}f({\mathbf{\Phi}}(\mathbf{x},-s))\mathrm{d}
s.\end{multline*} Since the formula is valid for any $T > \tau_0$,
differentiating with respect to $T$ leads to the first assertion.
The second assertion is proved in the same way by using formula
\eqref{beals2}.
\end{proof}

To drop the finiteness assumption on $\tau_{\pm}(\mathbf{x})$, first we introduce
the  sets
$$\mathbf{\Omega}_{\pm}=\{\mathbf{x} \in \mathbf{\Omega}\,;\,\tau_{\pm}(\mathbf{x}) < \infty\}, \qquad
\mathbf{\Omega}_{\pm\infty}=\{\mathbf{x} \in \mathbf{\Omega}\,;\,\tau_{\pm}(\mathbf{x}) = \infty\},$$ and
$$\Gamma_{\pm\infty}=\{\mathbf{y} \in \Gamma_{\pm}\,;\, \tau_{\mp}(\mathbf{y}) = \infty\}.$$
Then
\begin{propo}\label{prointegra} Given $f \in L^1(\mathbf{\Omega},\mathrm{d} \mu)$, one has \begin{equation}\label{10.47}
\int_{\mathbf{\Omega}_{\pm}}f(\mathbf{x})\mathrm{d}\mu(\mathbf{x})
=\int_{\Gamma_\pm}\mathrm{d}\mu_\pm(\mathbf{y})\int_0^{\tau_\mp(\mathbf{y})}f\left({\mathbf{\Phi}}(\mathbf{y},\mp
s)\right)\mathrm{d} s,
\end{equation}
and
\begin{equation}\label{10.49}
\int_{\mathbf{\Omega}_{\pm} \cap
\mathbf{\Omega}_{\mp\infty}}f(\mathbf{x})\mathrm{d}\mu(\mathbf{x})=\int_{\Gamma_\pm\infty}\mathrm{d}\mu_\pm(\mathbf{y})\int_0^{\infty}f\left({\mathbf{\Phi}}(\mathbf{y},\mp
s)\right)\mathrm{d} s.
\end{equation}
\end{propo}
\begin{proof} Assume first $f \geqslant 0$. Let us fix $T >0$. It is clear that $\mathbf{x} \in \mathbf{\Omega}$ satisfies
$\tau_+(\mathbf{x}) < T$ if and only if $\mathbf{x}={\mathbf{\Phi}}(\mathbf{y},-s),$ with $\mathbf{y}
\in \Gamma_+$ and $0 < s < T \wedge \tau_-(\mathbf{y})$. Then, by Proposition
\ref{Prop10.10},
$$\int_{\{\tau_+(\mathbf{x}) <
T\}}f(\mathbf{x})\mathrm{d}\mu(\mathbf{x})=\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{T \wedge
\tau_-(\mathbf{y})} f({\mathbf{\Phi}}(\mathbf{y},-s))\mathrm{d} s.$$ Since $f \geqslant 0$, the
inner integral is increasing with $T$ and, using the monotone
convergence theorem, we let $T \to \infty$ to get
$$\int_{\mathbf{\Omega}_{+}}f(\mathbf{x})\mathrm{d}\mu(\mathbf{x})
=\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y})}f\left({\mathbf{\Phi}}(\mathbf{y},-s)\right)\mathrm{d}
s$$ which coincides with \eqref{10.47}.  We proceed in the same way
with integration on $\Gamma_-$ and get the second part of
\eqref{10.47}. Next we consider the set
$$\Delta=\{\mathbf{x} \in \mathbf{\Omega}\,;\,\mathbf{x}={\mathbf{\Phi}}(\mathbf{y},-s),\,\mathbf{y} \in
\mathbf{\Omega}_{+\infty},\,0<s<T\}.$$ Proposition \ref{Prop10.10} asserts that
$$\int_{\Delta}f(\mathbf{x})\mathrm{d}\mu(\mathbf{x})=\int_{\mathbf{\Omega}_{+\infty}}\mathrm{d}\mu_+(\mathbf{y})\int_0^Tf({\mathbf{\Phi}}(\mathbf{y},-s))\mathrm{d}
s.$$ Letting again $T \to \infty$, we get \eqref{10.49}. We extend
the results to arbitrary $f$ by linearity.\end{proof}

Finally, with the following, we show that it is possible to transfer
integrals over $\Gamma_-$ to $\Gamma_+$:
\begin{propo}
For any $\psi \in L^1(\Gamma_-,\mathrm{d}\mu_-)$,
\begin{equation}\label{10.51}
\int_{\Gamma_-\setminus
\Gamma_{-\infty}}\psi(\mathbf{y})\mathrm{d}\mu_-(\mathbf{y})=\int_{\Gamma_+\setminus
\Gamma_{+\infty}}\psi({\mathbf{\Phi}}(\mathbf{z},-\tau_-(\mathbf{z})))\mathrm{d}\mu_+(\mathbf{z}).\end{equation}
\end{propo}
\begin{proof} For any $\epsilon > 0$, let $f_{\epsilon}$ be the
function defined on $\mathbf{\Omega}_+ \cap \mathbf{\Omega}_-$ by
\begin{equation*}
\psi_{\epsilon}(\mathbf{x})=\begin{cases}
\dfrac{\psi({\mathbf{\Phi}}(\mathbf{x},-\tau_-(\mathbf{x})))}{\tau_+(\mathbf{x})+\tau_-(\mathbf{x})}
\quad &\text{ if } \quad \tau_-(\mathbf{x})+\tau_+(\mathbf{x}) > \epsilon,\\
0 \quad &\text{ else}.\end{cases}
\end{equation*}
Since $\psi_{\epsilon} \in L^1(\mathbf{\Omega}_+ \cap \mathbf{\Omega}_-,\mathrm{d} \mu)$, Eqs. \eqref{10.47}
and \eqref{10.49} give
\begin{equation*}\begin{split}
\int_{\mathbf{\Omega}_+ \cap \mathbf{\Omega}_-} \psi_{\epsilon}(\mathbf{x})\mathrm{d}\mu(\mathbf{x})&=\int_{\{\tau_+(\mathbf{y}) >
\epsilon\}\setminus
\Gamma_{-\infty}}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}\psi(\mathbf{y})\dfrac{\mathrm{d}
s}{\tau_+(\mathbf{y})}\\
&=\int_{\{\tau_+(\mathbf{y}) > \epsilon\}\setminus
\Gamma_{-\infty}}\psi(\mathbf{y})\mathrm{d}\mu_-(\mathbf{y}).\end{split}\end{equation*} In
the same way, \begin{equation*}\begin{split} \int_{\mathbf{\Omega}_+ \cap \mathbf{\Omega}_-}
\psi_{\epsilon}(\mathbf{x})\mathrm{d}\mu(\mathbf{x})&=\int_{\{\tau_-(\mathbf{y})
> \epsilon\}\setminus
\Gamma_{+\infty}}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y})}\psi({\mathbf{\Phi}}(\mathbf{y},-\tau_-(\mathbf{y})))\dfrac{\mathrm{d}
s}{\tau_-(\mathbf{y})}\\
&=\int_{\{\tau_-(\mathbf{y}) > \epsilon\}\setminus
\Gamma_{+\infty}}\psi({\mathbf{\Phi}}(\mathbf{y},-\tau_-(\mathbf{y})))\mathrm{d}\mu_-(\mathbf{y}),\end{split}\end{equation*}
which leads to
$$\int_{\{\tau_-(\mathbf{y}) > \epsilon\}\setminus
\Gamma_{+\infty}}\psi({\mathbf{\Phi}}(\mathbf{y},-\tau_-(\mathbf{y})))\mathrm{d}\mu_+(\mathbf{y})=\int_{\{\tau_+(\mathbf{y})
> \epsilon\}\setminus \Gamma_{-\infty}}\psi(\mathbf{y})\mathrm{d}\mu_-(\mathbf{y})$$
for any $\epsilon > 0$. Passing to the limit as $\epsilon \to 0$ we get the
conclusion.\end{proof}

We end this section with a technical result we shall need in the
sequel (see Lemma \ref{supp}):
\begin{propo}\label{support} Let $K$ be a compact subset of $\mathbf{\Omega}$.
Denote
$$K_\pm:=\left\{\mathbf{y} \in \Gamma_\pm\;;\;\exists  t_0
\in \mathbb{R}\;\text{ such that }  {\mathbf{\Phi}}(\mathbf{y},\pm t) \in
K\, \text{ for any } t \geqslant t_0\right\}.$$ Then $\mu_\pm(K_\pm)=0.$
\end{propo}
\begin{proof} Let $K$ be a fixed compact subset of $\mathbf{\Omega}$. Applying
Eq. \eqref{10.47} or \eqref{10.49} to the function
$f(\mathbf{x})=\chi_K(\mathbf{x})$, one has \begin{equation}\label{muK} \infty >
\mu(K) \geqslant \int_{K_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^\infty \chi_K(
{\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d} t.\end{equation} By definition, if $\mathbf{y} \in
K_-$, then for some $t_0 \in \mathbb{R}$,
$\chi_K({\mathbf{\Phi}}(\mathbf{y},t))=1$ for any $t \geqslant t_0$. Therefore,
$$\int_0^\infty \chi_K( {\mathbf{\Phi}}(\mathbf{y},t)) = \infty, \qquad \forall
\mathbf{y} \in K_-.$$ Inequality \eqref{muK} implies   that $\mu_-(K_-)=0$.
One proves the result for $K_+$ in the same way.
\end{proof}
\section{The maximal transport operator and trace
results}\label{sec:maxi}  The results of the previous section allow
us to define the (maximal) transport operator $\mathcal{T}_{\mathrm{max}}$ as
the weak derivative along the characteristic curves.   To be
precise, let us define the space of \textit{test functions}
$\mathfrak{Y}$ as follows:

\begin{defi}[\textit{\textbf{Test--functions}}] Let $\mathfrak{Y}$ be the set of all measurable and bounded
functions $\psi :\mathbf{\Omega} \to \mathbb{R}$ with compact support in $\mathbf{\Omega}$ and such
that, for any $\mathbf{x} \in \mathbf{\Omega}$, the mapping
$$s \in (-\tau_-(\mathbf{x}),\tau_+(\mathbf{x})) \longmapsto \psi({\mathbf{\Phi}}(\mathbf{x},s))$$
is continuously differentiable with
\begin{equation}\label{defi:deriva}\mathbf{x} \in \mathbf{\Omega} \longmapsto \dfrac{\mathrm{d}}{\mathrm{d}
s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0} \text{ measurable and
bounded}.\end{equation}
\end{defi}

\begin{nb}  Notice
that the class of test-functions $\mathfrak{Y}$ is not defined as a
subset of $L^\infty(\mathbf{\Omega},\mathrm{d}\mu)$; that is, we do not identify
functions equal $\mu$-almost everywhere. It is however a natural
question to know whether two test-functions coinciding $\mu$-almost
everywhere are such that there derivatives (defined by
\eqref{defi:deriva}) do coincide  $\mu$-almost everywhere. We
provide a positive answer to this question at the end of the paper
(see Appendix).
\end{nb}

An important property of test-functions is the following consequence of Proposition \ref{support}:
\begin{lemme}\label{supp} Let $\psi \in \mathfrak{Y}$ be given. For $\mu_\mp$-almost any $\mathbf{y} \in \Gamma_{\mp}$ there exists a
sequence $(t_n^\pm)_n$ (depending on $\mathbf{y}$) such that
$$\lim_{n \to \infty} t^\pm_n=\tau_\pm(\mathbf{y}) \qquad \text{ and } \quad
\psi(\mathbf{\Phi}(\mathbf{y},\pm t_n^\pm))=0 \quad \forall n \in \mathbb{N}.$$
\end{lemme}
\begin{proof} Let $\psi \in \mathfrak{Y}$ be given and let $K=\mathrm{Supp}(\psi)$. For any $\mathbf{y} \in \Gamma_-$
with $\tau_+(\mathbf{y}) < \infty$ one has $\mathbf{\Phi}(\mathbf{y},\tau_+(\mathbf{y})) \in
\Gamma_+$ and, since $K$ is compact in $\mathbf{\Omega}$,
$\psi(\mathbf{\Phi}(\mathbf{y},\tau_+(\mathbf{y}))=0$ and the existence of a sequence
$(t_n^+)_n$ converging to $\tau_+(\mathbf{y})$ with the above property is
clear. Now, Proposition \ref{support} applied to $K$ shows that
there exists a set $\Gamma_-' \subset \Gamma_-$ with $\mu_-(\Gamma
\setminus \Gamma_-')=0$  and such that, for any $\mathbf{y} \in \Gamma_-'$,
there is a sequence $(t_n^+)_n$ converging to $\infty$ such that
$\mathbf{\Phi}(\mathbf{y},t_n) \notin K$ for any $n \in \mathbb{N}$. This
proves the result. The statement for $\Gamma_+$ is proved in the
same way.\end{proof}

In the next step we define the transport operator
$(\mathcal{T}_\mathrm{max},\Drond(\mathcal{T}_\mathrm{max}))$.
\begin{defi}[\textit{\textbf{Transport operator $\mathcal{T}_\mathrm{max}$}}]
The domain of the maximal transport operator $\mathcal{T}_\mathrm{max}$ is
the set $\Drond(\mathcal{T}_\mathrm{max} )$ of all $f \in L^1(\mathbf{\Omega} ,\mathrm{d}\mu)$ for
which there exists $g \in L^1(\mathbf{\Omega} ,\mathrm{d}\mu)$ such that
$$\int_{\mathbf{\Omega} }g(\mathbf{x})\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x})=\int_{\mathbf{\Omega} }f(\mathbf{x})\dfrac{\mathrm{d}}{\mathrm{d}
  s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0} \mathrm{d}\mu(\mathbf{x}),\qquad \qquad \forall \psi \in
\mathfrak{Y}.$$ In this case, $g=:\mathcal{T}_\mathrm{max} f.$
\end{defi}
\begin{nb}\label{nb:weakderiv} Of course, in
some weak sense, $\mathcal{T}_{\mathrm{max}}f=-\mathscr{F} \cdot \nabla f$.
Precisely, for any $\varphi \in \ensuremath{\mathscr{C}}^1_0(\mathbf{\Omega})$, the following formula
holds:
$$\int_{\mathbf{\Omega}}\left(\mathscr{F}(\mathbf{x})\cdot \nabla
  \varphi(\mathbf{x})\right)f(\mathbf{x})\mathrm{d}\mu(\mathbf{x})=
  \int_{\mathbf{\Omega}}\mathcal{T}_{\mathrm{max}}f(\mathbf{x})\varphi(\mathbf{x})\mathrm{d}\mu(\mathbf{x}).$$
 \end{nb}

\subsection{\textbf{Fundamental representation formula: mild formulation}} Recall  that, if $f_1$ and $f_2$
are two functions defined over $\mathbf{\Omega}$, we say that $f_2$ is a {\it
representative of} $f_1$ if $\mu\{\mathbf{x} \in \mathbf{\Omega}\,;\,f_1(\mathbf{x}) \neq
f_2(\mathbf{x})\}=0$, i.e. when $f_1(\mathbf{x})=f_2(\mathbf{x})$ for $\mu$-almost every $\mathbf{x}
\in \mathbf{\Omega} $. The following fundamental result provides a characterization of  the
domain of $\Drond(\mathcal{T}_\mathrm{max})$:
\begin{theo}\label{representation}
Let $f \in L^1(\mathbf{\Omega},\mu)$. The following are equivalent:
\begin{enumerate}\item There exists $g \in L^1(\mathbf{\Omega},\mu)$ and
a representative $f^\sharp$ of $f$ such that, for $\mu$-almost every
$\mathbf{x} \in \mathbf{\Omega}$ and any $-\tau_-(\mathbf{x}) < t_1 \leqslant t_2 < \tau_+(\mathbf{x})$:
\begin{equation}\label{integralTM-}f^{\sharp}({\mathbf{\Phi}}(\mathbf{x},t_1))-f^{\sharp}({\mathbf{\Phi}}(\mathbf{x},t_2))
=\int_{t_1}^{t_2}
g({\mathbf{\Phi}}(\mathbf{x},s))\mathrm{d} s.
\end{equation}
\item $f \in \Drond(\mathcal{T}_\mathrm{max})$. In this case, $g=\mathcal{T}_\mathrm{max} f$.
\end{enumerate}
\end{theo}

 The proof of the theorem is made of several steps. The difficult
 part of the proof is the  implication $(2) \implies (1)$. It is carried out through
several technical lemmas based
 upon \textit{\textbf{mollification along the characteristic curves}}  (recall
 that, whenever $\mu$ is not absolutely continuous with respect to
 the Lebesgue measure, no global convolution argument is available). Let us make precise what this is all about. Consider
a sequence $(\varrho_n)_n$  of one dimensional mollifiers supported
in $[0,1]$, i.e. for any $n \in \mathbb{N}$, $\varrho_n \in
\ensuremath{\mathscr{C}}^{\infty}_0(\mathbb{R})$, $\varrho_n(s)=0$ if $s \notin
[0,1/n]$, $\varrho_n(s) \geqslant 0$  and $\int_0^{1/n}\varrho_n(s)\mathrm{d}
s=1.$ Then, for any $f \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$, define the (extended)  mollification:
$$\varrho_n \diamond f(\mathbf{x})=\int_0^{\tau_-(\mathbf{x})}
\varrho_n(s)f({\mathbf{\Phi}}(\mathbf{x},-s))\mathrm{d} s.$$
As we shall see later, such a definition corresponds precisely to a
time convolution  over any characteristic curves (see e.g. \eqref{equamoll}).
Note that, with such a
definition, it is not clear {\it a priori} that $\varrho_n \diamond
f$ defines a measurable function, finite almost everywhere. It is
proved in the following that actually such a function is integrable.
\begin{lemme} Given $f \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$, $\varrho_n \diamond f \in
  L^1(\mathbf{\Omega} ,\mathrm{d}\mu)$ for any $n \in \mathbb{N}$. Moreover,
\begin{equation}\label{contrac}
\|\varrho_n \diamond f\| \leqslant \|f\|,\qquad \forall f \in
L^1(\mathbf{\Omega},\mathrm{d}\mu), n \in \mathbb{N}.\end{equation}
\end{lemme}
\begin{proof} One considers, for a given $f \in
L^1(\mathbf{\Omega},\mathrm{d}\mu)$,  the extension of $f$ by zero outside $\mathbf{\Omega}$:
$$\overline{f}(\mathbf{x})=f(\mathbf{x}),\qquad \forall \mathbf{x} \in \mathbf{\Omega},\qquad
\overline{f}(\mathbf{x})=0 \quad \forall \mathbf{x} \in \mathbb{R}^N \setminus \mathbf{\Omega}.$$
Then $\overline{f} \in L^1(\mathbb{R}^N,\mathrm{d}\mu).$ Let us consider the
 transformation:
$$\Upsilon\::\:(\mathbf{x},s) \in \mathbb{R}^N \times \mathbb{R} \mapsto
\Upsilon(\mathbf{x},s)=({\mathbf{\Phi}}(\mathbf{x},-s),-s) \in \mathbb{R}^N \times
\mathbb{R}.$$ As a homeomorphism, $\Upsilon$ is measure preserving
for  pure Borel measures. It is also measure preserving for
completions of Borel measures (such as a Lebesgue measure) since
it is measure-preserving on Borel sets and the completion of a
measure is obtained by adding to the Borel $\sigma$-algebra all
sets contained in a measure-zero Borel sets, see \cite[Theorem
13.B, p.~55]{Hal}. Then, according to \cite[Theorem 39.B,
p.~162]{Hal}, the mapping
$$(\mathbf{x},s) \in \mathbb{R}^N \times \mathbb{R} \mapsto
\overline{f}({\mathbf{\Phi}}(\mathbf{x},-s))$$ is measurable as the
composition of $\Upsilon$ with the measurable function $(\mathbf{x},s)
\mapsto \overline{f}(\mathbf{x})$. Define now $\Lambda=\{(\mathbf{x},s)\,;\,\mathbf{x} \in
\mathbf{\Omega},\,0 < s < \tau_-(\mathbf{x})\},$ $\Lambda$ is a measurable subset of
$\mathbb{R}^N \times \mathbb{R}$. Therefore, the mapping
$$(\mathbf{x},s) \in \mathbb{R}^N \times \mathbb{R} \longmapsto
\overline{f}({\mathbf{\Phi}}(\mathbf{x},-s))\chi_{\Lambda}(\mathbf{x},s)
\varrho_n(s)$$ is measurable. Since $\varrho_n$ is compactly
supported, it is also integrable over $\mathbb{R}^N \times
\mathbb{R}$ and, according to Fubini's Theorem
\begin{equation*}
[\varrho_n \diamond f](\mathbf{x}):= \displaystyle
\int_{\mathbb{R}}\overline{f}({\mathbf{\Phi}}(\mathbf{x},-s))\chi_{\Lambda}(\mathbf{x},s)
\varrho_n(s)\mathrm{d} s = \displaystyle
\int_0^{\tau_-(\mathbf{x})}\varrho_n(s)f({\mathbf{\Phi}}(\mathbf{x},-s))\mathrm{d}
s\end{equation*} is finite for almost every $\mathbf{x} \in \mathbf{\Omega}$ the
 and the associated application $\varrho_n \diamond f$ is
 integrable.

 Let us prove now \eqref{contrac}. Since $|\varrho_n \diamond f| \leqslant \varrho_n \diamond |f|$, to
  show that $\varrho_n \diamond f \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$, it suffices to deal
  with a \textit{nonnegative function} $f \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$. One sees easily that,
  for any $\mathbf{y} \in \Gamma_-$ and any $0 < t < \tau_+(\mathbf{y})$,
\begin{equation}\label{equamoll}(\varrho_n \diamond f)({\mathbf{\Phi}}(\mathbf{y},t))=\int_0^t
\varrho_n(s)f({\mathbf{\Phi}}(\mathbf{y},t-s))\mathrm{d} s=\int_0^t \varrho_n(t-s)
f({\mathbf{\Phi}}(\mathbf{y},s))\mathrm{d} s.\end{equation} Thus,
\begin{equation*}\begin{split}
\int_0^{\tau_+(\mathbf{y})}[\varrho_n \diamond f]({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d} t
&=\int_0^{\tau_+(\mathbf{y})}\mathrm{d} s
\int_s^{\tau_+(\mathbf{y})}\varrho_n(s)f({\mathbf{\Phi}}(\mathbf{y},t-s))\mathrm{d} t\\
&=\int_0^{\tau_+(\mathbf{y}) \wedge 1/n}\varrho_n(s)\mathrm{d}
s\int_0^{\tau_+(\mathbf{y})-s}f({\mathbf{\Phi}}(\mathbf{y},r))\mathrm{d} r.
\end{split}\end{equation*}
Therefore, \begin{equation*}\begin{split} 0 \leqslant
\int_0^{\tau_+(\mathbf{y})}[\varrho_n \diamond f]({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d} t
&\leqslant \int_0^{1/n}\varrho_n(s)\mathrm{d}
s\int_0^{\tau_+(\mathbf{y})}f({\mathbf{\Phi}}(\mathbf{y},r))\mathrm{d}
r\\
&=\int_0^{\tau_+(\mathbf{y})}f({\mathbf{\Phi}}(\mathbf{y},r))\mathrm{d} r,\qquad \forall \mathbf{y}
\in \Gamma_-,\:n \in \mathbb{N}\end{split}\end{equation*} so that
$$\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}[\varrho_n \diamond
f]({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d} t \leqslant
\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}
f({\mathbf{\Phi}}(\mathbf{y},r))\mathrm{d} r.$$ This proves, thanks to Proposition
\ref{prointegra}, that
\begin{equation}\label{intO_-}\int_{\mathbf{\Omega}_-}[\varrho_n \diamond f]\mathrm{d}\mu \leqslant
\int_{\mathbf{\Omega}_-}f\mathrm{d}\mu.\end{equation} Now, in the same way:
\begin{equation*}\begin{split}
\int_{\mathbf{\Omega}_+ \cap \mathbf{\Omega}_{-\infty}}&[\varrho_n \diamond f](\mathbf{x})\mathrm{d}\mu(\mathbf{x})=
\int_{\Gamma_{+\infty}}\mathrm{d}\mu_+(\mathbf{y})\int_0^\infty [\varrho_n \diamond
f]({\mathbf{\Phi}}(\mathbf{y},-t))\mathrm{d} t\\
&=\int_{\Gamma_{+\infty}}\mathrm{d}\mu_+(\mathbf{y})\int_0^\infty\mathrm{d}
t\int_0^\infty\varrho_n(s)f({\mathbf{\Phi}}(\mathbf{y},-s-t))\mathrm{d} s\\
&=\int_{\Gamma_{+\infty}}\mathrm{d}\mu_+(\mathbf{y})\int_0^\infty\mathrm{d}
t\int_t^\infty\varrho_n(r-t)f({\mathbf{\Phi}}(\mathbf{y},-r))\mathrm{d}
r.\end{split}\end{equation*} so that
\begin{equation*}\begin{split}
\int_{\mathbf{\Omega}_+ \cap \mathbf{\Omega}_{-\infty}}[\varrho_n \diamond f](\mathbf{x})\mathrm{d}\mu(\mathbf{x})
&=\int_{\Gamma_{+\infty}}\mathrm{d}\mu_+(\mathbf{y})\int_0^\infty
f({\mathbf{\Phi}}(\mathbf{y},-r))\mathrm{d}
r\int_0^r\varrho_n(r-t)\mathrm{d} t\\
&\leqslant \int_{\Gamma_{+\infty}}\mathrm{d}\mu_+(\mathbf{y})\int_0^\infty
f({\mathbf{\Phi}}(\mathbf{y},-r))\mathrm{d} r
\end{split}\end{equation*} i.e.
\begin{equation}\label{intO_+}
\int_{\mathbf{\Omega}_+ \cap \mathbf{\Omega}_{-\infty}}\varrho_n \diamond f(\mathbf{x})\mathrm{d}\mu(\mathbf{x})\leqslant
\int_{\mathbf{\Omega}_+ \cap \mathbf{\Omega}_{-\infty}} f(\mathbf{x})\mathrm{d}\mu(\mathbf{x}).\end{equation} Finally
\begin{equation*}\begin{split}
\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}}[\varrho_n \diamond
f](\mathbf{x})\mathrm{d}\mu(\mathbf{x})&= \int_{\mathbf{\Omega}_{+\infty} \cap
\mathbf{\Omega}_{-\infty}}\mathrm{d}\mu(\mathbf{x})\int_0^\infty\varrho_n(s)f({\mathbf{\Phi}}(\mathbf{x},-s))\mathrm{d} s\\
&=\int_0^\infty\varrho_n(s)\mathrm{d} s\int_{\mathbf{\Omega}_{+\infty}\cap
\mathbf{\Omega}_{-\infty}}f({\mathbf{\Phi}}(\mathbf{x},-s))\mathrm{d} \mu(\mathbf{x}).
\end{split}\end{equation*} Now, from Assumption \ref{ass:h2}, for any $s
\geqslant 0,$
$$\int_{\mathbf{\Omega}_{+\infty}\cap
\mathbf{\Omega}_{-\infty}}f({\mathbf{\Phi}}(\mathbf{x},-s))\mathrm{d}
\mu(\mathbf{x})=\int_{\mathbf{\Omega}_{+\infty}\cap \mathbf{\Omega}_{-\infty}}f(\mathbf{x})\mathrm{d} \mu(\mathbf{x}),$$ so
that
\begin{equation}\label{intOinfty} \int_{\mathbf{\Omega}_{+\infty} \cap
\mathbf{\Omega}_{-\infty}}[\varrho_n \diamond
f](\mathbf{x})\mathrm{d}\mu(\mathbf{x})=\int_{\mathbf{\Omega}_{+\infty}\cap \mathbf{\Omega}_{-\infty}}f(\mathbf{x})\mathrm{d}
\mu(\mathbf{x}).\end{equation} Combining \eqref{intO_-}, \eqref{intO_+} and
\eqref{intOinfty}, one finally gets $\|\varrho_n \diamond f\|  \leqslant
\|f\| .$\end{proof}

As it is the case for classical convolution, the family $(\varrho_n
\diamond f)_n$ approximates $f$ in $L^1$-norm:
\begin{propo}\label{approx} Given $f \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$,
\begin{equation}\label{convergconv}
\lim_{n \to \infty}\int_{\mathbf{\Omega} }\bigg|(\varrho_n \diamond f)(\mathbf{x})
-f(\mathbf{x})\bigg|\mathrm{d}\mu(\mathbf{x})=0.\end{equation}
\end{propo}
\begin{proof}
 According to \eqref{contrac} and from the density of $\mathscr{C}_0(\mathbf{\Omega})$ in $L^1(\mathbf{\Omega},\mathrm{d}\mu)$, it
suffices to prove the result for any $f$ continuous over $\mathbf{\Omega}$ and
compactly supported. Splitting $f$ into positive  and negative
parts, $f=f^+-f^-$, one can also assume $f$ to be nonnegative.  From
the continuity of both $f$ and $\mathbf{\Phi}(\cdot,\cdot)$, one has
$$\mathscr{K}_n:=\mathrm{Supp}(\varrho_n \diamond f)=\overline{\bigg\{\mathbf{x} \in \mathbf{\Omega}\,,\,\exists s_0
\in \mathrm{Supp}(\varrho_n) \text{ such that } \mathbf{\Phi}(\mathbf{x},-s_0) \in
\mathrm{Supp}(f)\bigg\}}.$$  Moreover, it is easily seen that
$\mathscr{K}_{n+1} \subset \mathscr{K}_n$ for any $n \geqslant 1$.
Finally, it is clear that $$\mathscr{K}_1 \subset \{\mathbf{x} \in
\overline{\mathbf{\Omega}}\,;\exists \mathbf{y} \in \mathrm{Supp}(f)\,\text{ with } |\mathbf{x}-\mathbf{y}|
\leqslant d\}$$ where $d=\sup\{|\mathbf{\Phi}(\mathbf{x}, s)-\mathbf{x}|\,;\,0 \leqslant s
\leqslant 1\,; \mathbf{x} \in \mathrm{Supp}(f)\} < \infty.$ Therefore,
$\mathscr{K}_1$ is compact. Set now
$$\mathcal{O}_n:=\mathscr{K}_n \cup \mathrm{Supp}(f) \qquad \text{
and } \qquad
 \mathcal{O}_n^-=\{\mathbf{x} \in \mathcal{O}_n\,;\,\tau_-(\mathbf{x}) <1/n\}.$$
Noticing that $\mu(\mathcal{O}_1)$ is finite, one can see easily
that $\lim_n\mu(\mathcal{O}_n^-)=0$. Since  $\sup_{\mathbf{x} \in \mathbf{\Omega}}
|\varrho_n \diamond f(\mathbf{x})| \leqslant \sup_{\mathbf{x} \in \mathbf{\Omega}} |f(\mathbf{x})|,$ for any
$\varepsilon > 0$, there exists $n_0 \geqslant 1$ such that
$$\int_{\mathcal{O}_n^-}|f(\mathbf{x})|\mathrm{d}\mu(\mathbf{x}) \leqslant \varepsilon, \quad \text{ and }
\quad \int_{\mathcal{O}_n^-}|\varrho_n \diamond f(\mathbf{x})|\mathrm{d}\mu(\mathbf{x}) \leqslant
\varepsilon \qquad \forall n \geqslant n_0.$$ Now, noticing that $
\mathrm{Supp}(\varrho_n \diamond f -f) \subset \mathcal{O}_n $, one
has for any $n \geqslant n_0$,
$$\int_{\mathbf{\Omega}}|\varrho_n \diamond f-f|\mathrm{d}\mu=\int_{\mathcal{O}_n}|\varrho_n
\diamond f-f| \leqslant 2\varepsilon + \int_{\mathcal{O}_n \setminus
  \mathcal{O}_n^-}|\varrho_n \diamond f-f|\mathrm{d}\mu.$$
For any $\mathbf{x} \in \mathcal{O}_n \setminus \mathcal{O}_n^-$, since
$\varrho$ is supported in $[0,1/n]$, one has
\begin{equation*}\begin{split}
[\varrho_n \diamond f](\mathbf{x})-f(\mathbf{x}) &=\int_0^{1/n}\varrho_n(s)
 f({\mathbf{\Phi}}(\mathbf{x}, -s))\mathrm{d}
s-f(\mathbf{x})\\
&=\int_0^{1/n}\varrho_n(s)\left(f({\mathbf{\Phi}}(\mathbf{x},
-s))-f(\mathbf{x})\right)\mathrm{d} s.\end{split}\end{equation*} Note that, thanks
to Gronwall's lemma,
$$|\mathbf{\Phi}(\mathbf{x},-s)-\mathbf{x}| \leqslant \frac{L}{\kappa}(\exp(k s)-1) \leqslant
\frac{L}{\kappa}(\exp(\kappa/n)-1),\qquad \forall \mathbf{x} \in
\mathcal{O}_1, s \in (0,1/n)$$ where $L=\sup\{|\mathscr{F}(\mathbf{x})|,\mathbf{x} \in
\mathcal{O}_1\}.$ Since $f$ is   uniformly continuous on
$\mathcal{O}_1$, it follows that
$$\lim_{n \to \infty} \sup \bigg\{|f(\mathbf{\Phi}(\mathbf{x},-s)-f(\mathbf{x})|\,;\,\mathbf{x} \in
\mathcal{O}_1,\;s \in (0,1/n)\bigg\}=0$$ from which we deduce  that
there exists some $n_1 \geqslant 0$, such that $|\varrho_n \diamond
f(\mathbf{x})-f(\mathbf{x})| \leqslant \varepsilon$ for any $\mathbf{x} \in \mathcal{O}_n
\setminus \mathcal{O}_n^-$ and any $n \geqslant n_1$. One obtains then,
for any $n \geqslant n_1$,
$$\int_{\mathbf{\Omega}}|\varrho_n \diamond f-f|\mathrm{d}\mu \leqslant 2\varepsilon +\varepsilon
\mu(\mathcal{O}_n \setminus \mathcal{O}_n^-) \leqslant 2\varepsilon
+\varepsilon \mu(\mathcal{O}_1)$$ which proves the
result.\end{proof}

We saw that, for a given $f \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$,
$\varrho_n \diamond f$ is also integrable ($n \in \mathbb{N}$).
Actually, we shall see that $\varrho_n \diamond f$ is even more
regular than $f$:
\begin{lemme}\label{moll} Given $f \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$, set $f_n=\varrho_n \diamond f$, $n \in \mathbb{N}$.
Then, $f_n \in \Drond(\mathcal{T}_\mathrm{max})$ with
\begin{equation*}
[\mathcal{T}_\mathrm{max}
f_n](\mathbf{x})=-\int_0^{\tau_-(\mathbf{x})}\varrho_n'(s)f({\mathbf{\Phi}}(\mathbf{x},-s))\mathrm{d}
s, \qquad \mathbf{x} \in \mathbf{\Omega}.
\end{equation*}
\end{lemme}
\begin{proof} Set $g_n(\mathbf{x})=-\int_0^{\tau_-(\mathbf{x})}\varrho_n'(s)f({\mathbf{\Phi}}(\mathbf{x},-s))\mathrm{d}
s,$ $\mathbf{x} \in \mathbf{\Omega}.$ It is easy to see that $g_n \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$.
Now, given $\psi \in \mathfrak{Y}$, let us consider the quantity
$$I=\int_{\mathbf{\Omega} }f_n(\mathbf{x})\dfrac{\mathrm{d}}{\mathrm{d}
  s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0} \mathrm{d}\mu(\mathbf{x}).$$
One has to prove that $I=\int_{\mathbf{\Omega}}g_n(\mathbf{x})\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x})$. We split
the above integral over $\mathbf{\Omega}$ into three integrals $I_-$, $I_+$ and
$I_\infty$ over $\mathbf{\Omega}_-$, $\mathbf{\Omega}_{-\infty}\cap \mathbf{\Omega}_+$ and
$\mathbf{\Omega}_{+\infty}\cap \mathbf{\Omega}_{-\infty}$ respectively.
 Recall that, for any $\mathbf{x}
\in \mathbf{\Omega}_-$, there is some $\mathbf{y} \in \Gamma_-$ and some $t \in
(0,\tau_+(\mathbf{y}))$ such that $\mathbf{x}={\mathbf{\Phi}}(\mathbf{y},t)$. In such a case
\begin{equation}\label{lip}
\dfrac{\mathrm{d}}{\mathrm{d}
s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0}=\dfrac{\mathrm{d}}{\mathrm{d}
t}\psi({\mathbf{\Phi}}(\mathbf{y},t)).\end{equation}  Then, according to
Prop. \ref{prointegra} and Eq. \eqref{equamoll}:
\begin{equation}\label{i-}\begin{split}
I_-&=\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}f_n({\mathbf{\Phi}}(\mathbf{y},t))\frac{\mathrm{d}}{\mathrm{d}
  t}\psi({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d} t\\
  &=\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})} \frac{\mathrm{d}}{\mathrm{d}
  t}\psi({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d}
  t\int_0^t\varrho_n(t-s)f({\mathbf{\Phi}}(\mathbf{y},s))\mathrm{d} s\\
  &=\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})} f({\mathbf{\Phi}}(\mathbf{y},s))\mathrm{d} s \int_s^{\tau_+(\mathbf{y})}\frac{\mathrm{d}}{\mathrm{d}
  t}\psi({\mathbf{\Phi}}(\mathbf{y},t))
   \varrho_n(t-s) \mathrm{d} t.
  \end{split}\end{equation}
 Let us now investigate more carefully this last integral. Let $\mathbf{y} \in \Gamma_-$ be fixed.
 If $\tau_+(\mathbf{y}) < \infty$ then, since $\psi$ is compactly supported, we have $\psi(\mathbf{\Phi}(\mathbf{y},\tau_+(\mathbf{y})))=0$ and integration by part (together with $\varrho_n(0)=0$) leads to
 $$\int_s^{\tau_+(\mathbf{y})}\frac{\mathrm{d}}{\mathrm{d}
  t}\psi({\mathbf{\Phi}}(\mathbf{y},t))
   \varrho_n(t-s) \mathrm{d} t=-\int_s^{\tau_+(\mathbf{y})}\varrho_n'(t-s)\psi(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d} t.$$
If now $\tau_+(\mathbf{y}) > \infty$, then, since $\varrho_n$ is supported in $[0,1/n]$, one has
\begin{equation*}\begin{split}
      \int_s^{\tau_+(\mathbf{y})}\frac{\mathrm{d}}{\mathrm{d}
  t}\psi({\mathbf{\Phi}}(\mathbf{y},t))
   \varrho_n(t-s) \mathrm{d} t&=\int_s^{s+1/n}\frac{\mathrm{d}}{\mathrm{d}
  t}\psi({\mathbf{\Phi}}(\mathbf{y},t))
   \varrho_n(t-s) \mathrm{d} t\\
   &=-\int_s^{\tau_+(y)}\varrho_n'(t-s)\psi(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d} t
                 \end{split}\end{equation*}
Finally, we obtain,
\begin{equation*}\begin{split}I_-&=-\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})} f({\mathbf{\Phi}}(\mathbf{y},s))\mathrm{d} s  \int_s^{\tau_+(\mathbf{y})} \psi({\mathbf{\Phi}}(\mathbf{y},t))
   \varrho_n'(t-s) \mathrm{d} t \\
&=-\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}
\psi({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d}
  t\int_0^t\varrho_n'(s)f({\mathbf{\Phi}}(\mathbf{y},t- s))\mathrm{d} s.
   \end{split}\end{equation*}
Using again Prop. \ref{prointegra}, we finally get
$$I_-=\int_{\mathbf{\Omega}_-}g_n(\mathbf{x})\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x}).$$
One proves in the same way that
$$I_+=\int_{\mathbf{\Omega}_+ \cap \mathbf{\Omega}_{-\infty} }f_n(\mathbf{x})\dfrac{\mathrm{d}}{\mathrm{d}
  s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0} \mathrm{d}\mu(\mathbf{x})=\int_{\mathbf{\Omega}_+\cap
  \mathbf{\Omega}_{-\infty}}g_n(\mathbf{x})\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x}).$$
It remains to consider $I_\infty=\int_{\mathbf{\Omega}_{+\infty} \cap
\mathbf{\Omega}_{-\infty} }f_n(\mathbf{x})\frac{\mathrm{d}}{\mathrm{d}
  s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\big|_{s=0} \mathrm{d}\mu(\mathbf{x}).$ One has
\begin{equation*}\begin{split}
I_\infty&=\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}}\frac{\mathrm{d}}{\mathrm{d}
  s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0}\mathrm{d}\mu(\mathbf{x})\int_0^\infty
  \varrho_n(t)f({\mathbf{\Phi}}(\mathbf{x},-t))\mathrm{d} t\\
  &=\int_0^\infty \varrho_n(t)\mathrm{d} t\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}}\frac{\mathrm{d}}{\mathrm{d}
  s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0}f({\mathbf{\Phi}}(\mathbf{x},-t))\mathrm{d}
  \mu(\mathbf{x}).
\end{split}\end{equation*}
For any $\mathbf{x} \in \mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}$ and any $t \geqslant 0$,
setting $\mathbf{y}={\mathbf{\Phi}}(\mathbf{x},-t)$, one has $\mathbf{y} \in \mathbf{\Omega}_{-\infty}
\cap \mathbf{\Omega}_{+\infty}$ and $\frac{\mathrm{d}}{\mathrm{d}
  s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\big |_{s=0}=\frac{\mathrm{d}}{\mathrm{d}
  t}\psi({\mathbf{\Phi}}(\mathbf{y},t))$ from which Liouville's Theorem (Assumption \ref{ass:h2}) yields
$$\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}}\frac{\mathrm{d}}{\mathrm{d}
  s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0}f({\mathbf{\Phi}}(\mathbf{x},-t))\mathrm{d}
  \mu(\mathbf{x})=\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}}\frac{\mathrm{d}}{\mathrm{d}
  t}\psi({\mathbf{\Phi}}(\mathbf{y},t)) f(\mathbf{y})\mathrm{d}
  \mu(\mathbf{y}).$$
Therefore,
\begin{eqnarray*}I_\infty&=&\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}} f(\mathbf{y})\mathrm{d}
  \mu(\mathbf{y})\int_0^\infty \varrho_n(t)\frac{\mathrm{d}}{\mathrm{d}
  t}\psi({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d} t\\&=&-\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}} f(\mathbf{y})\mathrm{d}
  \mu(\mathbf{y})\int_0^\infty \varrho_n'(t) \psi({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d} t\\
  &=&-\int_0^\infty \varrho_n'(t)\mathrm{d} t\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}} f(\mathbf{y})\psi({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d}
  \mu(\mathbf{y}).
\end{eqnarray*} Arguing as above, one can "turn back" to the $\mathbf{x}$
variable  to get
$$\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}} f(\mathbf{y})
\psi({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d}\mu(\mathbf{y})=\int_{\mathbf{\Omega}_{+\infty} \cap
\mathbf{\Omega}_{-\infty}} f({\mathbf{\Phi}}(\mathbf{x},-t))\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x}),$$ i.e.
$$I_\infty=-\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}}\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x})\int_0^\infty \varrho_n'(t)f({\mathbf{\Phi}}(\mathbf{x},-t))\mathrm{d}
t=\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}}\psi(\mathbf{x})g_n(\mathbf{x})\mathrm{d}\mu(\mathbf{x})$$
and the Lemma is proven.\end{proof}
\begin{nb}
Notice that Proposition \ref{approx} together with Lemma \ref{moll}
prove that $\Drond(\mathcal{T}_\mathrm{max})$ is a dense subset of
$L^1(\mathbf{\Omega},\mathrm{d}\mu)$.
\end{nb} Now, whenever $f \in \Drond(\mathcal{T}_\mathrm{max} )$, one has the following
more precise result:
\begin{propo}\label{moll2}
If $f \in \Drond(\mathcal{T}_\mathrm{max}  )$, then
\begin{equation}\label{claimdiamond}
[\mathcal{T}_\mathrm{max}  (\varrho_n \diamond f)](\mathbf{x})=[\varrho_n \diamond
\mathcal{T}_\mathrm{max} f](\mathbf{x}),\qquad \qquad (\mathbf{x} \in \mathbf{\Omega}\,,\,n \in
\mathbb{N}).\end{equation}\end{propo}

Before proving this result, we need the following very simple lemma:
\begin{lemme}\label{lemme:prop4.5} For any $\psi \in \mathfrak{Y}$ and any $n \in
\mathbb{N}$, define
$$\chi_n(\mathbf{x})=
\int_0^{\tau_+(\mathbf{x})}\varrho_n(s)\psi({\mathbf{\Phi}}(\mathbf{x},s))\mathrm{d} s, \qquad
\mathbf{x} \in \mathbf{\Omega}.$$ Then, $\chi_n$ belongs to $\mathfrak{Y}$.
\end{lemme}
\begin{proof} Since $\tau_+$ is measurable and $\varrho_n$ is
compactly supported, it is easy to see that $\chi_n$ is measurable
and bounded over $\mathbf{\Omega}$. Now, for any $\mathbf{x} \in \mathbf{\Omega}$, and any $t \in
(\tau_-(\mathbf{x}),\tau_+(\mathbf{x}))$, one has
$$\chi_n({\mathbf{\Phi}}(\mathbf{x},t))=\int_t^{\tau_+(\mathbf{x})}
\varrho_n(s-t)\psi({\mathbf{\Phi}}(\mathbf{x},s))\mathrm{d} s.$$
It is clear then from the properties of $\varrho_n$ that the mapping
$t \in (-\tau_-(\mathbf{x}),\tau_+(\mathbf{x})) \mapsto \chi_n({\mathbf{\Phi}}(\mathbf{x},t))$
is continuously differentiable  with
\begin{equation}\label{derivativechin}
\dfrac{\mathrm{d} }{\mathrm{d} t}
\chi_n({\mathbf{\Phi}}(\mathbf{x},t))=-\int_t^{\tau_+(\mathbf{x})}\varrho_n'(s-t)\psi({\mathbf{\Phi}}(\mathbf{x},s))\mathrm{d}
s =\int_t^{\tau_+(\mathbf{x})} \varrho_n(s-t) \dfrac{\mathrm{d} }{\mathrm{d}
s}\left[\psi({\mathbf{\Phi}}(\mathbf{x},s))\right]\mathrm{d} s.
\end{equation}
In particular, for $t=0$, one gets
$$\dfrac{\mathrm{d} }{\mathrm{d} t}
\chi_n({\mathbf{\Phi}}(\mathbf{x},t))\bigg|_{t=0}=-\int_0^{\tau_+(\mathbf{x})}\varrho_n'(s)\psi({\mathbf{\Phi}}(\mathbf{x},s))\mathrm{d}
s.$$ Since $\varrho_n'$ is compactly supported and $\psi \in
\mathfrak{Y}$, the application $\mathbf{x}\in \mathbf{\Omega} \longmapsto \dfrac{\mathrm{d} }{\mathrm{d}
t} \chi_n({\mathbf{\Phi}}(\mathbf{x},t))\big|_{t=0}$ is measurable and
bounded.\end{proof}

\begin{proof1} We use the notations of Lemma \ref{moll}. Since $\varrho_n
\diamond \mathcal{T}_\mathrm{max}  f \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$, it suffices to show
that
$$\int_{\mathbf{\Omega}  }f_n(\mathbf{x})\dfrac{\mathrm{d}}{\mathrm{d}
s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0}\mathrm{d}\mu(\mathbf{x})=\int_{\mathbf{\Omega}}\psi(\mathbf{x})
[\varrho_n \diamond \mathcal{T}_\mathrm{max}  f](\mathbf{x})\mathrm{d}\mu(\mathbf{x}), \qquad \forall
\psi \in \mathfrak{Y}.$$ Here again, we shall deal separately with
the integrals over $\mathbf{\Omega}_-$, $\mathbf{\Omega}_{+}\cap \mathbf{\Omega}_{-\infty}$ and
$\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}.$
 Let $\chi_n$ be defined as in Lemma \ref{lemme:prop4.5}, as we
 already saw it (see \eqref{derivativechin}), for any $\mathbf{y} \in \Gamma_-,$ and any $0<
s<\tau_+(\mathbf{y})$,
$\frac{\mathrm{d}}{\mathrm{d}
s}\chi_n({\mathbf{\Phi}}(\mathbf{y},s))=\int_s^{\tau_+(\mathbf{y})}\varrho_n(t-s)\frac{\mathrm{d}}{\mathrm{d}
t}[\psi({\mathbf{\Phi}}(\mathbf{y},t))]\mathrm{d} t.$ Consequently, according to
\eqref{i-},
\begin{equation*}\begin{split} \int_{\mathbf{\Omega}_-
} f_n(\mathbf{x})&\dfrac{\mathrm{d}}{\mathrm{d}
s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0}\mathrm{d}\mu(\mathbf{x})=\int_{\Gamma_-}\mathrm{d}\mu(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}f({\mathbf{\Phi}}(\mathbf{y},r))
\dfrac {\mathrm{d}}{\mathrm{d} r}\chi_n({\mathbf{\Phi}}(\mathbf{y},r))\mathrm{d} r\\
&=\int_{\mathbf{\Omega}_-}f(\mathbf{x})\dfrac{\mathrm{d}}{\mathrm{d}
s}\chi_n({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0}\mathrm{d}\mu(\mathbf{x})
=\int_{{\mathbf{\Omega}}_-}\chi_n(\mathbf{x})[\mathcal{T}_\mathrm{max} f](\mathbf{x})\mathrm{d}\mu(\mathbf{x})
\end{split}\end{equation*} where, for the two last identities, we
used \eqref{lip} and the fact that $\chi_n \in \mathfrak{Y}.$ Now,
using Prop. \ref{prointegra}
\begin{equation*} \begin{split}\int_{\mathbf{\Omega}_-}&\chi_n(\mathbf{x})[\mathcal{T}_\mathrm{max}  f](\mathbf{x})\mathrm{d}\mu(\mathbf{x})=\int_{\mathbf{\Omega}_-}[\mathcal{T}_\mathrm{max}  f](\mathbf{x})\mathrm{d}\mu(\mathbf{x})
\int_0^{\tau_+(\mathbf{x})}\varrho_n(r)\psi({\mathbf{\Phi}}(\mathbf{x},r))\mathrm{d} r\\
&=\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}\psi({\mathbf{\Phi}}(\mathbf{y},s))\mathrm{d}
s\int_0^s\varrho_n(s-t)[\mathcal{T}_\mathrm{max} f]({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d}
t.
\end{split}\end{equation*}
Therefore, Eq. \eqref{equamoll} leads to
\begin{equation*}\label{TMO-}\begin{split}
\int_{\mathbf{\Omega}_-} \chi_n(\mathbf{x})[\mathcal{T}_\mathrm{max}
f](\mathbf{x})\mathrm{d}\mu(\mathbf{x})&=\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}\psi({\mathbf{\Phi}}(\mathbf{y},s))[\varrho_n
\diamond \mathcal{T}_\mathrm{max} f]({\mathbf{\Phi}}(\mathbf{y},s))\mathrm{d} s\\
&=\int_{\mathbf{\Omega}_-}\psi(\mathbf{x})\left[\varrho_n \diamond
\mathcal{T}_{\mathrm{max}}f\right](\mathbf{x})\mathrm{d}\mu(\mathbf{x}).
\end{split}\end{equation*} The integrals over $\mathbf{\Omega}_+
\cap \mathbf{\Omega}_{-\infty}$ and $\mathbf{\Omega}_{-\infty}\cap \mathbf{\Omega}_{+\infty}$ are evaluated
in the same way.
\end{proof1}

We are in position to prove the following
\begin{propo} Let $f \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$ and $f_n=\varrho_n \diamond f$, $n \in \mathbb{N}$. Then, for $\mu_-$-- a. e.
$\mathbf{y} \in \Gamma_-$,
\begin{equation}\label{fny-}
f_n({\mathbf{\Phi}}(\mathbf{y},s))-f_n({\mathbf{\Phi}}(\mathbf{y},t))=\int_s^t
[\mathcal{T}_\mathrm{max} f_n]({\mathbf{\Phi}}(\mathbf{y},r))\mathrm{d} r \qquad \forall 0 <
s < t < \tau_+(\mathbf{y}).\end{equation} In the same way, for almost every
$\mathbf{z} \in \Gamma_+$,
$$f_n({\mathbf{\Phi}}(\mathbf{z},-s))-f_n({\mathbf{\Phi}}(\mathbf{z},-t))=\int_s^t \mathcal{T}_\mathrm{max}
f_n({\mathbf{\Phi}}(\mathbf{z},-r))\mathrm{d} r, \qquad \forall 0 < s < t <
\tau_-(\mathbf{z}).$$ \label{prop312}
\end{propo}
\begin{proof} We focus only on \eqref{fny-}, the second assertion following the same lines. Since $f \in L^1(\mathbf{\Omega}_-,\mathrm{d}\mu)$,
  Proposition  \ref{prointegra} implies that the integral
$\int_0^{\tau_+(\mathbf{y})} |f({\mathbf{\Phi}}(\mathbf{y},r))|\mathrm{d} r$ exists and is
finite for $\mu_-$-almost every $\mathbf{y} \in \Gamma_-$. Therefore, for
$\mu_-$-almost every $\mathbf{y} \in \Gamma_-$ and  any $0 < t < \tau_+(\mathbf{y})$,
the quantities $\int_0^t \varrho_n(t-s) f({\mathbf{\Phi}}(\mathbf{y},s))\mathrm{d}
s$ and $\int_0^t \varrho_n'(t-s) f({\mathbf{\Phi}}(\mathbf{y},s))\mathrm{d} s$ are
well-defined and finite. Moreover, thanks to Eq. \eqref{equamoll}
Lemma \ref{moll}, they are respectively equal to
$f_n({\mathbf{\Phi}}(\mathbf{y},t))$ and $[\mathcal{T}_\mathrm{max}
f_n]({\mathbf{\Phi}}(\mathbf{y},t)).$ In particular, the mapping $t \in
(0,\tau_+(\mathbf{y})) \mapsto [\mathcal{T}_\mathrm{max} f_n]({\mathbf{\Phi}}(\mathbf{y},t))
\in \mathbb{R}$ is continuous. It is then easy to see that, for any
$0 < s < t < \tau_+(\mathbf{y})$
\begin{equation*}\begin{split}
\int_s^t &[\mathcal{T}_\mathrm{max} f_n]( {\mathbf{\Phi}}(\mathbf{y},r))\mathrm{d}
r=-\int_s^t \mathrm{d} r \int_0^r
\varrho_n'(r-u) f({\mathbf{\Phi}}(\mathbf{y},u))\mathrm{d}  u\\
&=-\int_0^s f({\mathbf{\Phi}}(\mathbf{y},u))\mathrm{d} u \int_s^t \varrho_n'(r-u) \mathrm{d}
r - \int_s^t f({\mathbf{\Phi}}(\mathbf{y},u))\mathrm{d} u \int_u^t \varrho_n'(r-u)
\mathrm{d}
r\\
&=-\int_0^t f({\mathbf{\Phi}}(\mathbf{y},u))\varrho_n(t-u)\mathrm{d} u+\int_0^s
f({\mathbf{\Phi}}(\mathbf{y},u))\varrho_n(s-u)\mathrm{d} u,
\end{split}
\end{equation*}
which is nothing but \eqref{fny-}.
\end{proof}
As a consequence, one gets the following result :
\begin{propo}\label{f+-}
For any $f \in \Drond(\mathcal{T}_\mathrm{max})$, there exists some functions
$\widetilde{f}_{\pm} \in L^1(\mathbf{\Omega}_\pm,\mathrm{d}\mu)$ such that
$\widetilde{f}_\pm(\mathbf{x})=f(\mathbf{x})$ for $\mu$- almost every $\mathbf{x} \in
\mathbf{\Omega}_\pm$ and, for $\mu_-$--almost every $\mathbf{y} \in \Gamma_-$:
\begin{equation}\label{fy-}
\widetilde{f }_-({\mathbf{\Phi}}(\mathbf{y},s))-\widetilde{f
}_-({\mathbf{\Phi}}(\mathbf{y},t))=\int_s^t [\mathcal{T}_\mathrm{max} f]
({\mathbf{\Phi}}(\mathbf{y},r))\mathrm{d} r \qquad \forall 0 < s < t <
\tau_+(\mathbf{y}),\end{equation} while, for $\mu_+$--almost every $\mathbf{z} \in
\Gamma_+$:
\begin{equation*}
\widetilde{f }_+({\mathbf{\Phi}}(\mathbf{z},-s))-\widetilde{f
}_+({\mathbf{\Phi}}(\mathbf{z},-t))=\int_s^t [\mathcal{T}_\mathrm{max} f ]
({\mathbf{\Phi}}(\mathbf{z},-r))\mathrm{d} r \qquad \forall 0 < s < t < \tau_-(\mathbf{z}).
\end{equation*}
\end{propo}
\begin{proof} Define, for any $n \geqslant 1$, $f_n=\varrho_n \diamond
f$, so that, from Propositions \ref{moll2} and \ref{approx},
$\lim_{n \to \infty}\|f_n -f\|  + \|\mathcal{T}_\mathrm{max} f_n
-\mathcal{T}_\mathrm{max} f\| =0.$ In particular, $$\lim_{n \to
\infty}\int_{\mathbf{\Omega}_-}\left|f_n(\mathbf{x})-f(\mathbf{x})\right|+\left|\,[\mathcal{T}_\mathrm{max}
f_n](\mathbf{x})-[\mathcal{T}_\mathrm{max} f ](\mathbf{x})\right|\mathrm{d}\mu(\mathbf{x})=0.$$
 Then
Eq. \eqref{10.47} yields
\begin{multline*}
\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}\left|f_n({\mathbf{\Phi}}(\mathbf{y},s))-f({\mathbf{\Phi}}(\mathbf{y},s))\right|\mathrm{d}
s \\
+ \int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}\left|[\mathcal{T}_\mathrm{max}
f_n](\mathbf{\Phi}(\mathbf{y},s))- [\mathcal{T}_\mathrm{max}
f]({\mathbf{\Phi}}(\mathbf{y},s))\right|\mathrm{d} s \underset{n\to
\infty}{\longrightarrow} 0\end{multline*} since $\mathcal{T}_\mathrm{max} f$
and $\mathcal{T}_\mathrm{max} f_n$ both belong to $L^1(\mathbf{\Omega},\mathrm{d}\mu)$.
Consequently, for almost every $\mathbf{y} \in \Gamma_-$ (up to a
subsequence, still denoted by $f_n$) we get
\begin{equation*}
\begin{cases}
f_n({\mathbf{\Phi}}(\mathbf{y},\cdot)) \longrightarrow f({\mathbf{\Phi}}(\mathbf{y},\cdot))\\
\mathcal{T}_\mathrm{max} f_n ({\mathbf{\Phi}}(\mathbf{y},\cdot )) \longrightarrow
[\mathcal{T}_\mathrm{max} f ]({\mathbf{\Phi}}(\mathbf{y},\cdot )) \quad \text{ in }
\quad L^1((0,\tau_+(\mathbf{y}))\,,\mathrm{d} s)\end{cases}\end{equation*} as $n \to
\infty$. Let us fix $\mathbf{y} \in \Gamma_-$ for which this holds. Passing
again to a subsequence, we may assume that $f_n({\mathbf{\Phi}}(\mathbf{y},s
))$ converges (pointwise) to $f({\mathbf{\Phi}}(\mathbf{y},s ))$ for almost
every $s \in (0,\tau_+(\mathbf{y}))$. Let us fix such a $s_0$. Then,
$$f_n({\mathbf{\Phi}}(\mathbf{y},s_0))-f_n({\mathbf{\Phi}}(\mathbf{y},s ))=\int_{s_0}^s[\mathcal{T}_\mathrm{max} f_n]({\mathbf{\Phi}}(\mathbf{y},r ))\mathrm{d}
r \qquad \forall s \in (0,\tau_+(\mathbf{y})).$$ Now, the right-hand-side has
a limit as $n \to \infty$ so that the first term on the left-hand
side also must converge as $n \to \infty.$ Thus, for any $s \in
(0,\tau_+(\mathbf{y}))$, the limit
$$\lim_{n \to \infty}f_n({\mathbf{\Phi}}(\mathbf{y},s))=\widetilde{f}_-({\mathbf{\Phi}}(\mathbf{y},s ))$$
exists and, for any $0 < s < \tau_+(\mathbf{y})$
\begin{equation*}\label{d/ds}
\widetilde{f}_-({\mathbf{\Phi}}(\mathbf{y},s
))=\widetilde{f}_-({\mathbf{\Phi}}(\mathbf{y},s_0 ))-\int_{s_0}^s
[\mathcal{T}_\mathrm{max} f]({\mathbf{\Phi}}(\mathbf{y},r ))\mathrm{d} r.\end{equation*}  It
is easy to check then that $\widetilde{f}_-(\mathbf{x})=f(\mathbf{x})$ for almost
every $\mathbf{x} \in \mathbf{\Omega}_-$. The same arguments lead to the existence of
$\widetilde{f}_+$.
\end{proof}
The above result shows that the mild formulation of Theorem
\ref{representation} is fulfilled for any $\mathbf{x} \in \mathbf{\Omega}_- \cup \mathbf{\Omega}_+$.
It remains to deal with $\mathbf{\Omega}_\infty:=\mathbf{\Omega}_{-\infty}\cap \mathbf{\Omega}_{+\infty}.$
\begin{propo}\label{f+-infinity} Let $f \in \Drond(\mathcal{T}_\mathrm{max})$. Then, there exists a set $\mathcal{O}
\subset \mathbf{\Omega}_\infty$ with $\mu(\mathcal{O})=0$ and a function
$\widetilde{f}$ defined on $\{\mathbf{z}={\mathbf{\Phi}}(\mathbf{x},t),\,\mathbf{x} \in
\mathbf{\Omega}_\infty \setminus \mathcal{O},\,t \in \mathbb{R}\}$ such that
$f(\mathbf{x})=\widetilde{f}(\mathbf{x})$ $\mu$-almost every $\mathbf{x} \in \mathbf{\Omega}_\infty$ and
$$\widetilde{f}({\mathbf{\Phi}}(\mathbf{x},s))-\widetilde{f}({\mathbf{\Phi}}(\mathbf{x},t))=\int_s^t
[\mathcal{T}_\mathrm{max} f ]({\mathbf{\Phi}}(\mathbf{x}, r))\mathrm{d} r, \qquad \forall\,
\mathbf{x} \in \mathbf{\Omega}_\infty \setminus \mathcal{O},\: s < t.$$
\end{propo}
\begin{proof} Since $(\mathbf{x},t) \mapsto
(\mathbf{z},t)=(\mathbf{\Phi}(\mathbf{x},t),t)$ is a measurable and measure
preserving mapping from $\mathbf{\Omega}_\infty \times \mathbb{R}$ onto itself,
Propositions \ref{approx} and \ref{moll2} give
\begin{eqnarray}&&\lim_{n \to
\infty}\int_{\mathbf{\Omega}_\infty}\mathrm{d}\mu(\mathbf{x})\int_{I_k}\left|f_n(\mathbf{\Phi}(\mathbf{x},t))-f(\mathbf{\Phi}(\mathbf{x},t))\right|dt=0\label{jeden}\\
&&\lim_{n \to \infty}
\int_{\mathbf{\Omega}_\infty}\mathrm{d}\mu(\mathbf{x})\int_{I_k}\left|\mathcal{T}_\mathrm{max}
f_n(\mathbf{\Phi}(\mathbf{x},t))-\mathcal{T}_\mathrm{max}
f(\mathbf{\Phi}(\mathbf{x},t))\right|\mathrm{d} t =0,\label{dwa} \end{eqnarray} for
any $I_k=[-k,k]$, $k \in \mathbb{N}$. This shows, in particular,
that there is (a maximal) $\mathcal{E} \subset \mathbf{\Omega}_\infty$ with
$\mu(\mathcal{E})=0$ such that, for almost every $\mathbf{x} \in \mathbf{\Omega}_\infty
\setminus \mathcal{E}$ and any bounded interval $I \subset
\mathbb{R}$:
$$\int_{I} |f(\mathbf{\Phi}(\mathbf{x},t))|\mathrm{d} t + \int_{I}|[\mathcal{T}_\mathrm{max} f ]
(\mathbf{\Phi}(\mathbf{x},t))|\mathrm{d} t < \infty$$ and we can argue as in
Proposition \ref{prop312} that
$$f_n(\mathbf{\Phi}(\mathbf{x},s))-f_n(\mathbf{x})=-\int_0^s \mathcal{T}_\mathrm{max}
f_n(\mathbf{\Phi}(\mathbf{x},r))\mathrm{d} r, \qquad \forall s \in \mathbb{R}.$$
Proposition \ref{approx} yields the existence of a subsequence
$(f_{n_p})_p$ and a $\mu$-null set $A_0$ with $\mathcal{E} \subset
A_0 \subset \mathbf{\Omega}_\infty$ such that
$$\lim_{p \to \infty}f_{n_p}(\mathbf{x})=f(\mathbf{x}), \qquad \forall \mathbf{x} \in
\mathcal{\mathbf{\Omega}_\infty} \setminus \mathcal{A}_0.$$ Now, for any $k \in
\mathbb{N}$,
$$\lim_{p \to
\infty} \int_{\mathbf{\Omega}_\infty}\mathrm{d}\mu(\mathbf{x})\int_{I_k}\left|\mathcal{T}_\mathrm{max}
f_{n_p}(\mathbf{\Phi}(\mathbf{x},t))-\mathcal{T}_\mathrm{max}
f(\mathbf{\Phi}(\mathbf{x},t))\right|\mathrm{d} t =0$$ so that, there is a
subsequence (depending on $k$) and a $\mu$-null set $A_k$ with $A_0
\subset A_k \subset \mathbf{\Omega}_\infty$ such that
$$\lim_{p_{(k)} \to\infty}\int_{I_k}\left|\mathcal{T}_\mathrm{max} f_{n_{p_{(k)}}}(\mathbf{\Phi}(\mathbf{x},t))-\mathcal{T}_\mathrm{max}
f(\mathbf{\Phi}(\mathbf{x},t))\right|\mathrm{d} t=0, \qquad \forall \mathbf{x} \in
\mathbf{\Omega}_\infty\setminus A_k.$$ Let $\mathbf{x} \in \mathbf{\Omega}_\infty \setminus A_k$ and
$|s| < k$ be fixed. From
$$f_{n_{p_{(k)}}}(\mathbf{\Phi}(\mathbf{x},s))-f_{n_{p_{(k)}}}(\mathbf{x})=-\int_0^s \mathcal{T}_\mathrm{max}
f_{n_{p_{(k)}}}(\mathbf{\Phi}(\mathbf{x},r))\mathrm{d} r$$ we deduce that the limit
$\lim_{p_{(k)} \to \infty}f_{n_{p_{(k)}}}(\mathbf{\Phi}(\mathbf{x},s))$
exists and is equal to  $$\lim_{p_{(k)} \to
\infty}f_{n_{p_{(k)}}}(\mathbf{\Phi}(\mathbf{x},s))=f(\mathbf{x})-\int_0^s
\mathcal{T}_\mathrm{max}
 f(\mathbf{\Phi}(\mathbf{x},r))\mathrm{d} r.$$
 We define then $\widetilde{f}$ by
 $$\widetilde{f}(\mathbf{\Phi}(\mathbf{x},s))=\lim_{p_{(k)} \to
\infty}f_{n_{p_{(k)}}}(\mathbf{\Phi}(\mathbf{x},s)), \qquad \mathbf{x} \in \mathbf{\Omega}_\infty
\setminus A_k,\:|s| < k$$ and defining $\mathcal{O}=\bigcup_{k \geqslant
1} A_k$, we get the result.
\end{proof}

Before the proof of Theorem \ref{representation}, we have to
establish existence of the trace on $\Gamma_-$.
\begin{propo}\label{trace} Let $f$ satisfies condition (1) of Theorem \ref{representation}.
Then  $$\lim_{t \to 0+}f^\sharp({\mathbf{\Phi}}(\mathbf{y},t))$$ exists  for
almost every $\mathbf{y} \in \Gamma_-$. Similarly, $\lim_{t \to
0+}f^\sharp({\mathbf{\Phi}}(\mathbf{y},-t))$ exists for almost every $\mathbf{y} \in
\Gamma_+$.
\end{propo}
\begin{proof} First we note that there is $\widetilde{\Omega}_-\subset \Omega_-$ with
 $\mu(\Omega_-\setminus\widetilde{\Omega}_-)=0$ such that (\ref{integralTM-}) is valid any $\mathbf{x} \in \widetilde{\Omega}_-$. Let
$\widetilde{\Gamma}_-=\{\mathbf{y}\in \Gamma_-;\; \mathbf{y} =
\mathbf{\Phi}(\mathbf{x},\tau_-(\mathbf{x})), \mathbf{x} \in \widetilde{\Omega}_-\}$. It is
easy to see that $\mu_-(\Gamma_-\setminus \widetilde{\Gamma}_-) =0$.
Indeed, otherwise, by (\ref{10.47}), there would be a subset of
$\Omega_-$ of positive $\mu$-measure, not intersecting
$\widetilde{\Omega}_-$, which would contradict (\ref{integralTM-}).
Consequently,  any $\mathbf{x}\in \widetilde{\Omega}_-$ can be written as
$\mathbf{x} = \Phi(\mathbf{y}, \tau_-(\mathbf{y}))$, $\mathbf{y} \in \widetilde{\Gamma}_-$ and
(\ref{integralTM-}) can be recast as
\begin{equation}
f^\sharp({\mathbf{\Phi}}(\mathbf{y},t))-f^{\sharp}({\mathbf{\Phi}}(\mathbf{y},t_0))=\int_{t}^{t_0}g(\mathbf{\Phi}(\mathbf{y},s)\mathrm{d}
s. \label{nowe}
\end{equation}
 for almost any $\mathbf{y} \in \Gamma_-$, where $0<t\leqslant t_0<
\tau_+(\mathbf{y})$. Using again (\ref{10.47}), $s\mapsto
g({\mathbf{\Phi}}(\mathbf{y},s)$ is integrable on $(0,\tau_+(\mathbf{y})$ for almost
any $\mathbf{y}\in \Gamma_-$. Consequently, for almost every $\mathbf{y}\in
\Gamma_-$ we can pass to the limit in (\ref{nowe}) with $t\to 0$; it
is easy to check that this limit does not depend on $t_0$. The
existence of $\lim_{t\to 0+}f^\sharp({\mathbf{\Phi}}(\mathbf{y},-t))$ for a.
e. $\mathbf{y} \in \Gamma_+$ follows by the same argument.
\end{proof}

The above proposition allows to define the trace operators.
\begin{defi}
For any $f \in \Drond(\mathcal{T}_{\mathrm{max}})$, define the \textit{traces}
$\mathsf{B}^{\pm}f$ by
\begin{equation*}
\mathsf{B}^+f(\mathbf{y}):=\lim_{t \to 0+}f^\sharp({\mathbf{\Phi}}(\mathbf{y},-t)) \qquad
\text{ and } \qquad \mathsf{B}^-f(\mathbf{y}):=\lim_{t \to
0+}f^\sharp({\mathbf{\Phi}}(\mathbf{y},t))
\end{equation*} for any $\mathbf{y} \in \Gamma_\pm$ for which the limits
exist, where   $f^\sharp$ is a suitable representative of $f$.
\end{defi}

\begin{preuve2}
To prove that $(2) \implies (1)$,  given $f \in
\Drond(\mathcal{T}_\mathrm{max})$, set
\begin{equation*}
f^\natural(\mathbf{x})=\begin{cases} \widetilde{f}_-(\mathbf{x})  \quad &\text{ if }
\mathbf{x}
\in \mathbf{\Omega}_-,\\
 \widetilde{f}_+(\mathbf{x})  \quad &\text{ if } \mathbf{x}
\in \mathbf{\Omega}_+ \cap \mathbf{\Omega}_{-\infty},\\
 \widetilde{f}(\mathbf{x}) \quad &\text{ if } \mathbf{x}
\in \mathbf{\Omega}_{-\infty} \cap \mathbf{\Omega}_{+\infty},\\
\end{cases}
\end{equation*}
where $\widetilde{f}_{\pm}$ are given by Proposition \ref{f+-} while
$\widetilde{f}$ is provided by Prop. \ref{f+-infinity}. Then, it is
clear that for any $\mathbf{x} \in
  \mathbf{\Omega}$ and any $-\tau_-(\mathbf{x}) < t_1 \leqslant t_2 <
  \tau_+(\mathbf{x})$
  $$f^\sharp({\mathbf{\Phi}}(\mathbf{x},t_1))-f^\sharp({\mathbf{\Phi}}(\mathbf{x},t_2))=\int_{t_1}^{t_2} [\mathcal{T}_\mathrm{max} f ]({\mathbf{\Phi}}(\mathbf{x},s))\mathrm{d}
  s$$
and \eqref{integralTM-} is proven.

Let us now prove that $(1) \implies (2)$. Let us fix $\psi \in
\mathfrak{Y}$, one has
\begin{equation*}\begin{split}
\int_{\mathbf{\Omega}_-} f(\mathbf{x}) \dfrac{\mathrm{d}}{\mathrm{d} s}
\psi(\mathbf{\Phi}(\mathbf{x},s))\big|_{s=0}\mathrm{d}\mu(\mathbf{x})&=\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}
 f(\mathbf{\Phi}(\mathbf{y},t))  \dfrac{\mathrm{d}}{\mathrm{d} t}\psi(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d} t\\
&=\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}
f^\sharp(\mathbf{\Phi}(\mathbf{y},t)) \dfrac{\mathrm{d}}{\mathrm{d}
t}\psi(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d} t.
\end{split}
\end{equation*}
Notice that since both $\int_{\mathbf{\Omega}_-}f(\mathbf{x}) \dfrac{\mathrm{d}}{\mathrm{d} s}
\psi(\mathbf{\Phi}(\mathbf{x},s))\big|_{s=0}\mathrm{d}\mu(\mathbf{x})$ and $\int_{\mathbf{\Omega}_-}
\psi(\mathbf{x})g(\mathbf{x})\mathrm{d}\mu(\mathbf{x})$ exist, Proposition \ref{prointegra} and
Fubini's Theorem, the integrals $$\int_0^{\tau_+(\mathbf{y})}
f^\sharp(\mathbf{\Phi}(\mathbf{y},t)) \dfrac{\mathrm{d}}{\mathrm{d}
t}\psi(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d} t \quad \text{ and } \quad
\int_0^{\tau_+(\mathbf{y})} g(\mathbf{\Phi}(\mathbf{y},t)) \psi(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d}
t$$ are well-defined for $\mu_-$-almost every $\mathbf{y} \in \Gamma_-$. Let
us prove that these two integrals coincide for almost-every $\mathbf{y} \in
\Gamma_-$. According to Lemma \ref{supp}, for almost every $\mathbf{y} \in
\Gamma_-$, there is a sequence $(t_n)_n$ (depending on $\mathbf{y}$) such
that $\psi(\mathbf{\Phi}(\mathbf{y},t_n))=0$ and $t_n \to \tau_+(\mathbf{y})$. Thus,
$$\int_0^{\tau_+(\mathbf{y})}
f^\sharp(\mathbf{\Phi}(\mathbf{y},t)) \dfrac{\mathrm{d}}{\mathrm{d}
t}\psi(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d} t=\lim_{n \to \infty}\int_0^{t_n}
f^\sharp(\mathbf{\Phi}(\mathbf{y},t)) \dfrac{\mathrm{d}}{\mathrm{d}
t}\psi(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d} t$$ and
$$\int_0^{\tau_+(\mathbf{y})} g(\mathbf{\Phi}(\mathbf{y},t)) \psi(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d}
t=\lim_{n \to
\infty}\int_0^{t_n}\psi(\mathbf{\Phi}(\mathbf{y},t))g(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d}
t.$$ Further, for almost every $\mathbf{y} \in \Gamma_-$, according to
\eqref{integralTM-},
$$f^\sharp(\mathbf{\Phi}(\mathbf{y},t))=\mathsf{B}^-f(\mathbf{y})-\int_{0}^t
g(\mathbf{\Phi}(\mathbf{y},r))\mathrm{d} r, \qquad \forall t \in (0,\tau_+(\mathbf{y})).$$
Integration by parts, using the fact that
$\psi(\mathbf{\Phi}(\mathbf{y},0))=\psi(\mathbf{\Phi}(\mathbf{y},t_n))=0$ for any
$n$,  leads to
$$\int_0^{t_n}
f^\sharp(\mathbf{\Phi}(\mathbf{y},t)) \dfrac{\mathrm{d}}{\mathrm{d}
t}\psi(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d}
t=\int_0^{t_n}g(\mathbf{\Phi}(\mathbf{y},t))\psi(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d} t.$$
Consequently, for $\mu_-$ almost every $\mathbf{y} \in \Gamma_-$:
\begin{equation}\label{tfini}\int_{0}^{\tau_+(\mathbf{y})}
f^\sharp(\mathbf{\Phi}(\mathbf{y},t)) \dfrac{\mathrm{d}}{\mathrm{d}
t}\psi(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d}
t=\int_{0}^{\tau_+(\mathbf{y})}\psi(\mathbf{\Phi}(\mathbf{y},s))g(\mathbf{\Phi}(\mathbf{y},t))
\mathrm{d} t.\end{equation} Finally, we get
\begin{equation}\label{og-}\begin{split}
\int_{\mathbf{\Omega}_-} f(\mathbf{x}) \dfrac{\mathrm{d}}{\mathrm{d} s}
\psi(\mathbf{\Phi}(\mathbf{x},s))\big|_{s=0}\mathrm{d}\mu(\mathbf{x})&=\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}\psi(\mathbf{\Phi}(\mathbf{y},t))g(\mathbf{\Phi}(\mathbf{y},t))\mathrm{d}
t \\
&=\int_{\mathbf{\Omega}_-} g( \mathbf{x}) \psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x}).
\end{split}
\end{equation}  Using
now parametrization over $\Gamma_+$, we prove in the same way  that
\begin{equation}\label{og+}\int_{\mathbf{\Omega}_+ \cap \mathbf{\Omega}_{-\infty}} f(\mathbf{x}) \dfrac{\mathrm{d}}{\mathrm{d} s}
\psi(\mathbf{\Phi}(\mathbf{x},s))\big|_{s=0}\mathrm{d}\mu(\mathbf{x})=\int_{\mathbf{\Omega}_+ \cap
\mathbf{\Omega}_{-\infty}}g(\mathbf{x}) \psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x}).\end{equation} It remains now
to evaluate $A:=\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}} f(\mathbf{x})
\frac{\mathrm{d}}{\mathrm{d} s} \psi(\mathbf{\Phi}(\mathbf{x},s))\big|_{s=0}\mathrm{d}\mu(\mathbf{x}).$
According to Assumption \ref{ass:h2}  $$A =\int_{\mathbf{\Omega}_{+\infty} \cap
\mathbf{\Omega}_{-\infty}}
 f^\sharp(\mathbf{\Phi}(\mathbf{x},t))  \dfrac{\mathrm{d}}{\mathrm{d} t}
\psi(\mathbf{\Phi}(\mathbf{x},t))\mathrm{d}\mu(\mathbf{x}), \qquad \forall t \in
\mathbb{R}.$$ Let us integrate the above identity over $(0,1)$, so
that
$$  A =\int_{\mathbf{\Omega}_{-\infty} \cap \mathbf{\Omega}_{+\infty}}\mathrm{d}\mu(\mathbf{x})\int_0^1 f^\sharp(\mathbf{\Phi}(\mathbf{x},t))  \dfrac{\mathrm{d}}{\mathrm{d} t}
\psi(\mathbf{\Phi}(\mathbf{x},t)) \mathrm{d} t.$$ Let us fix $\mathbf{x} \in \mathbf{\Omega}_{-\infty}
\cap \mathbf{\Omega}_{+\infty}$. For any $t \in (0,1)$, one  has
 $f^\sharp(\mathbf{\Phi}(\mathbf{x},t))=f^\sharp(\mathbf{x})-\int_0^t g(\mathbf{\Phi}(\mathbf{x},s))\mathrm{d} s$
and integration by parts yields
\begin{multline*}
\int_0^1 f^\sharp(\mathbf{\Phi}(\mathbf{x},t))\dfrac{\mathrm{d} }{\mathrm{d}
t}\psi(\mathbf{\Phi}(\mathbf{x},t))\mathrm{d}
t=\int_0^1 \psi(\mathbf{\Phi}(\mathbf{x},t))g(\mathbf{\Phi}(\mathbf{x},t))\mathrm{d} t -\psi(\mathbf{x})f^\sharp(\mathbf{x})\\
+\psi(\mathbf{\Phi}(\mathbf{x},1))\bigg(f^\sharp(\mathbf{x})
 -\int_0^1
g(\mathbf{\Phi}(\mathbf{x},s))\mathrm{d} s\bigg)\\
=\int_0^1 \psi(\mathbf{\Phi}(\mathbf{x},t))g(\mathbf{\Phi}(\mathbf{x},t))\mathrm{d} t
+\psi(\mathbf{\Phi}(\mathbf{x},1))f^\sharp(\mathbf{\Phi}(\mathbf{x},1))-\psi(\mathbf{x})f^\sharp(\mathbf{x})
\end{multline*}
where we used again \eqref{integralTM-}. Integrating over
$\mathbf{\Omega}_{-\infty} \cap \mathbf{\Omega}_{+\infty}$ we see from Liouville's Theorem
(Assumption \ref{ass:h2}) that
$$\int_{\mathbf{\Omega}_{-\infty} \cap \mathbf{\Omega}_{+\infty}}
 \psi(\mathbf{\Phi}(\mathbf{x},1))f^\sharp(\mathbf{\Phi}(\mathbf{x},1))\mathrm{d}\mu(\mathbf{x})=\int_{\mathbf{\Omega}_{-\infty} \cap
 \mathbf{\Omega}_{+\infty}}\psi(\mathbf{x})f^\sharp(\mathbf{x})\mathrm{d}\mu(\mathbf{x}),$$ i.e.
$$A=\int_{\mathbf{\Omega}_{-\infty} \cap \mathbf{\Omega}_{+\infty}}\mathrm{d}\mu(\mathbf{x})\int_0^1\psi(\mathbf{\Phi}(\mathbf{x},t))g(\mathbf{\Phi}(\mathbf{x},t))\mathrm{d}
t$$ which, thanks to Liouville's Theorem, is nothing but
\begin{equation}\label{og+infty}
\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty}} f(\mathbf{x}) \frac{\mathrm{d}}{\mathrm{d} s}
\psi(\mathbf{\Phi}(\mathbf{x},s))\big|_{s=0}\mathrm{d}\mu(\mathbf{x})=\int_{\mathbf{\Omega}_{-\infty}
\cap \mathbf{\Omega}_{+\infty}} g(\mathbf{x})\,\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x}).\end{equation} Combining
\eqref{og-}, \eqref{og+} and \eqref{og+infty}, we obtain
$$\int_\mathbf{\Omega} f(\mathbf{x})\frac{\mathrm{d}}{\mathrm{d} s}
\psi(\mathbf{\Phi}(\mathbf{x},s))\big|_{s=0}\mathrm{d}\mu(\mathbf{x})=\int_\mathbf{\Omega} g(\mathbf{x})
\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x}), \qquad \forall \psi \in \mathfrak{Y}$$ which
exactly means that $f \in \Drond(\mathcal{T}_\mathrm{max})$ with
$g=\mathcal{T}_\mathrm{max}$ and the proof is complete.\end{preuve2}

\begin{cor}\label{cortrace} Traces $B^{\pm}f$ on $\Gamma_\pm$ can be defined for any
$f \in \Drond(\mathcal{T}_{\mathrm{max}})$. For $\mu_-$- almost any $\mathbf{y} \in
\Gamma_-$ we have
$$\mathsf{B}^-f(\mathbf{y})=f^\sharp({\mathbf{\Phi}}(\mathbf{y},t))+\int_0^t[\mathcal{T}_{\mathrm{max}}f]({\mathbf{\Phi}}(y, s))\mathrm{d}
s, \qquad \forall t \in (0,\tau_+(y)),$$ where $f^\sharp$ is a
suitable representative of $f$. An analogous formula holds for
$\mathsf{B}^+f$.\end{cor}

Lemma \ref{defmu+-} provides the existence of Borel measures
$\mathrm{d}\mu_{\pm}$ on $\Gamma_{\pm}$, which allow us to define the natural trace
spaces associated to Problem \eqref{1}, namely,
$$L^1_{\pm}:=L^1(\Gamma_\pm,\mathrm{d}\mu_\pm).$$
However, the traces $B^\pm f$, $f \in \Drond(\mathcal{T}_{\mathrm{max}})$, not
necessarily belong to $L^1_{\pm}.$

\section{Well-posedness for initial and boundary- value problems}

\subsection{Absorption semigroup}\label{sec:exis}
From now on, we will denote $X=L^1(\mathbf{\Omega},\mathrm{d}\mu)$ endowed with its
natural norm $\|\cdot\|_X$. Let $\mathcal{T}_0$ be the free streaming
operator with \textit{no re--entry boundary conditions}:
$$\mathcal{T}_0\psi=\mathcal{T}_{\mathrm{max}}\psi, \qquad \text{ for any }
\psi \in \Drond(\mathcal{T}_0),$$ where the domain $\Drond(\mathcal{T}_0)$ is defined by
$$\Drond(\mathcal{T}_0)=\{\psi \in \Drond(\mathcal{T}_{\mathrm{max}})\,;\,\mathsf{B}^-\psi=0\}.$$
We state the following generation result:
\begin{theo}\label{uot} The operator $(\mathcal{T}_0,\Drond(\mathcal{T}_0))$ is the generator of a nonnegative $C_0$-semigroup
of contractions $(U_0(t))_{t \geq 0}$ in $L^1(\mathbf{\Omega},\mathrm{d}\mu)$ given by
$$U_0(t)f(\mathbf{x})=f({\mathbf{\Phi}}(\mathbf{x},-t))\chi_{\{t <
\tau_-(\mathbf{x})\}}(\mathbf{x}), \qquad (\mathbf{x} \in \mathbf{\Omega},\:f \in X),$$ where $\chi_A$
denotes the characteristic function of a set $A$.
\end{theo}
\begin{proof} The proof is divided into three steps:

$\bullet$ {\it Step 1.} Let us first check that the family of
operators $(U_0(t))_{t \geq 0}$ is a nonnegative contractive $C_0$-semigroup in
$X$. Thanks to Proposition \ref{Phiprop}, we can prove that, for any
$f \in X$ and any $t \geqslant 0$, the mapping $U_0(t)f\::\mathbf{\Omega} \to
\mathbb{R}$ is measurable and the semigroup properties $U_0(0)f=f$
and $U_0(t)U_0(s)f=U_0(t+s)f$ $(t,s \geqslant 0)$ hold. Let us now show
that $\|U_0(t)f\|_X \leqslant \|f\|_X$. We have
$$\|U_0(t)f\|_X=\int_{\mathbf{\Omega}_+}|U_0(t)f|\mathrm{d}\mu+\int_{\mathbf{\Omega}_-\cap
\mathbf{\Omega}_{+\infty}}|U_0(t)f|\mathrm{d}\mu+\int_{\mathbf{\Omega}_{-\infty}\cap\mathbf{\Omega}_{+\infty}}|U_0(t)f|\mathrm{d}\mu.$$
Propositions \ref{prointegra} and \ref{Phiprop} yield
\begin{equation*}\begin{split}
\int_{\mathbf{\Omega}_+}|U_0(t)f|\mathrm{d}\mu
&=\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y})}
|U_0(t)f({\mathbf{\Phi}}(\mathbf{y},-s))|\mathrm{d} s\\
&= \int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\max(0,\tau_-(\mathbf{y})-t)}
|f({\mathbf{\Phi}}(\mathbf{y},-s-t))|\mathrm{d} s\\
&\leqslant \int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_t^{\max(t,\tau_-(\mathbf{y}))}
|f({\mathbf{\Phi}}(\mathbf{y},-r))|\mathrm{d} r \leqslant \int_{\mathbf{\Omega}_+}|f|\mathrm{d}\mu.
\end{split}\end{equation*}
In the same way we obtain
\begin{equation*}
\int_{\mathbf{\Omega}_- \cap \mathbf{\Omega}_{+\infty}}|U_0(t)f| \mathrm{d}\mu
=\int_{\Gamma_{-\infty}}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\infty}
|U_0(t)f({\mathbf{\Phi}}(\mathbf{y}, s))|\mathrm{d} s= \int_{\mathbf{\Omega}_- \cap
\mathbf{\Omega}_{+\infty}}|f|\mathrm{d}\mu,
\end{equation*}
and
\begin{equation*}
\int_{\mathbf{\Omega}_{-\infty}\cap \mathbf{\Omega}_{+\infty}}|U_0(t)f|\mathrm{d}\mu =
\int_{\mathbf{\Omega}_{-\infty}\cap \mathbf{\Omega}_{+\infty}}|f|\mathrm{d}\mu.\end{equation*} This
proves contractivity of $U_0(t)$. Let us now show that $U_0(t)f$ is
continuous, i.e.
$$\lim_{t \to 0}\|U_0(t)f-f\|_X=0.$$
It is enough to show that this property holds for any $f \in
\ensuremath{\mathscr{C}}_0(\mathbf{\Omega})$. In this case, $\lim_{t\to 0}U_0(t)f(\mathbf{x})=f(\mathbf{x})$ for any
$\mathbf{x} \in \mathbf{\Omega}.$ Moreover, $\sup_{\mathbf{x} \in \mathbf{\Omega}}|U_0(t)f(\mathbf{x})| \leqslant \sup_{\mathbf{x}
\in \mathbf{\Omega}}|f(\mathbf{x})|$ and the support of $U_0(t)f$ is bounded, so that the
Lebesgue dominated convergence theorem leads to the result. This
proves that $(U_0(t))_{t \geq 0}$ is a $C_0$-semigroup of contractions in $X$. Let
$\mathcal{A}_0$ denote its generator.

$\bullet$  \textit{Step 2.} To show that $\Drond(\mathcal{A}_0) \subset
\Drond(\mathcal{T}_0)$,   fix $f \in \Drond(\mathcal{A}_0)$, $\lambda > 0$ and $g=(\lambda-
\mathcal{A}_0)f.$ Then,
\begin{equation}f(\mathbf{x})=\int_0^{\tau_-(\mathbf{x})}\exp(-\lambda
t)\,g({\mathbf{\Phi}}(\mathbf{x},-t))\mathrm{d} t, \qquad (\mathbf{x} \in \mathbf{\Omega}). \label{ho}
\end{equation}
  To prove
that $f \in \Drond(\mathcal{T}_\mathrm{max})$ with $\mathcal{T}_\mathrm{max}f=\mathcal{A}_0f$, it
suffices to prove that
$$\int_{\mathbf{\Omega}
}(\lambda f(\mathbf{x})-g(\mathbf{x}))\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x})=\int_{\mathbf{\Omega} }f(\mathbf{x})\dfrac{\mathrm{d}}{\mathrm{d}
  s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0} \mathrm{d}\mu(\mathbf{x}),\qquad \qquad \forall \psi \in
\mathfrak{Y}.$$ Let us fix $\psi \in \mathfrak{Y}$, set
$\varphi(\mathbf{x}):=\frac{\mathrm{d}}{\mathrm{d}
  s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\big |_{s=0}$ and write
\begin{multline*}
\int_{\mathbf{\Omega}}  f(\mathbf{x})\varphi(\mathbf{x})\mathrm{d}\mu(\mathbf{x})=\int_{\mathbf{\Omega}_+}
f(\mathbf{x})\varphi(\mathbf{x})\mathrm{d}\mu(\mathbf{x})+\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_-}
f(\mathbf{x})\varphi(\mathbf{x})\mathrm{d}\mu(\mathbf{x})\\+\int_{\mathbf{\Omega}_{+\infty} \cap \mathbf{\Omega}_{-\infty} }
f(\mathbf{x})\varphi(\mathbf{x})\mathrm{d}\mu(\mathbf{x})=I_1+I_2+I_3.
\end{multline*}
We first deal with $I_1$. For any $\mathbf{y} \in \Gamma_+$ and $t \in
(0,\tau_-(\mathbf{y}))$ we have $\varphi({\mathbf{\Phi}}(\mathbf{y},-t))=-\frac{\mathrm{d}}{\mathrm{d}
t}\psi({\mathbf{\Phi}}(\mathbf{y},-t))$ and
$f({\mathbf{\Phi}}(\mathbf{y},-t))=\int_t^{\tau_-(\mathbf{y})}\exp(-\lambda
(s-t))g({\mathbf{\Phi}}(\mathbf{y},-s))\mathrm{d} s.$  Then, by Proposition
\ref{prointegra},
\begin{equation*}\begin{split}
I_1&=-\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y})}\frac{\mathrm{d}}{\mathrm{d}
t}\psi({\mathbf{\Phi}}(\mathbf{y},-t))\mathrm{d} t \int_t^{\tau_-(\mathbf{y})}\exp(-\lambda
(s-t))g({\mathbf{\Phi}}(\mathbf{y},-s))\mathrm{d} s\\
&=-\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y})}g({\mathbf{\Phi}}(\mathbf{y},-s))\mathrm{d}
s\int_0^s \exp(-\lambda (s-t))\frac{\mathrm{d}}{\mathrm{d}
t}\left(\psi({\mathbf{\Phi}}(\mathbf{y},-t ))\right)\mathrm{d} t\\
&=\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y})}g({\mathbf{\Phi}}(\mathbf{y},-s )) \times \\
& \phantom{\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})}\times\left\{ \lambda \int_0^s
\exp(-\lambda (s-t))\psi({\mathbf{\Phi}}(\mathbf{y},-t))\mathrm{d} t -
\psi({\mathbf{\Phi}}(\mathbf{y},-s))\right\}\mathrm{d} s\end{split}
\end{equation*}
where we used the fact that $\psi(\mathbf{\Phi}(\mathbf{y},0))=0$ for
 any $\mathbf{y} \in \Gamma_+$ since $\psi$ is compactly supported.
Thus
\begin{equation*}\begin{split}
I_1&=\lambda\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y})}\psi(\mathbf{\Phi}(\mathbf{y},-t))\mathrm{d}
t \int_t^{\tau_-(\mathbf{y})}\exp(-\lambda(s-t))g(\mathbf{\Phi}(\mathbf{y},-s))\mathrm{d} s
\\
&\phantom{++++++++++++}
-\int_{\Gamma_-}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y})}g(\mathbf{\Phi}(\mathbf{y},-s))\psi(\mathbf{\Phi}(\mathbf{y},-s))\mathrm{d}
s\\
&=
\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y})}\psi(\mathbf{\Phi}(\mathbf{y},-t))\big(\lambda
f(\mathbf{\Phi}(\mathbf{y},-t))-g(\mathbf{\Phi}(\mathbf{y},-t))\big)\mathrm{d} t.
\end{split}
\end{equation*}
 Using again
Proposition \ref{prointegra}, we obtain
\begin{equation}\label{I1}I_1=\int_{\mathbf{\Omega}_+}\left(\lambda f(\mathbf{x})-g(\mathbf{x})\right)\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x}).\end{equation}
Arguing in a similar way, we prove that
\begin{equation}\label{I2}
I_2=-\int_{\mathbf{\Omega}_- \cap \mathbf{\Omega}_{+\infty}}\left(\lambda
f(\mathbf{x})-g(\mathbf{x})\right)\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x}).\end{equation} Finally, since
$$f(\mathbf{x})=\int_0^{\infty}\exp(-\lambda t)g\left({\mathbf{\Phi}}(\mathbf{x},-t)\right)\mathrm{d} t
\qquad \text{ for any } \quad\mathbf{x} \in \mathbf{\Omega}_{-\infty} \cap
\mathbf{\Omega}_{+\infty},$$ one has
\begin{equation*}\begin{split}
I_3&=\int_{\mathbf{\Omega}_{-\infty} \cap
\mathbf{\Omega}_{+\infty}}\varphi(\mathbf{x})\mathrm{d}\mu(\mathbf{x})\int_0^\infty
\exp(-\lambda t) g({\mathbf{\Phi}}(\mathbf{x},-t))\mathrm{d} t\\
&=\int_0^\infty \exp(-\lambda t)\mathrm{d} t \int_{\mathbf{\Omega}_{-\infty} \cap
\mathbf{\Omega}_{+\infty}}\varphi(\mathbf{x})g({\mathbf{\Phi}}(\mathbf{x},-t))\mathrm{d}\mu(\mathbf{x}).\end{split}\end{equation*}
Now, Assumption \ref{ass:h2}  asserts that
$$\int_{\mathbf{\Omega}_{-\infty} \cap
\mathbf{\Omega}_{+\infty}}\varphi(\mathbf{x})g({\mathbf{\Phi}}(\mathbf{x},-t))\mathrm{d}\mu(\mathbf{x})=\int_{\mathbf{\Omega}_{-\infty}
\cap \mathbf{\Omega}_{+\infty}}g(\mathbf{x})\varphi({\mathbf{\Phi}}(\mathbf{x}, t))\mathrm{d}\mu(\mathbf{x}),
\qquad \forall t \geqslant 0,$$ and, since
$\varphi({\mathbf{\Phi}}(\mathbf{x},t))=\frac{\mathrm{d} }{\mathrm{d} t}
\psi({\mathbf{\Phi}}(\mathbf{x},t))$,  finally
\begin{equation*}\begin{split}
I_3&=\int_{\mathbf{\Omega}_{-\infty} \cap
\mathbf{\Omega}_{+\infty}}g(\mathbf{x})\mathrm{d}\mu(\mathbf{x})\int_0^{\infty}\exp(-\lambda
t)\dfrac{\mathrm{d}}{\mathrm{d}
t}\left(\psi({\mathbf{\Phi}}(\mathbf{x} ,t))\right)\mathrm{d} t\\
&=-\int_{\mathbf{\Omega}_{-\infty} \cap \mathbf{\Omega}_{+\infty}}g(\mathbf{x})\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x}) +
\lambda \int_{\mathbf{\Omega}_{-\infty} \cap
\mathbf{\Omega}_{+\infty}}g(\mathbf{x})\mathrm{d}\mu(\mathbf{x})\int_0^\infty \exp(-\lambda
t)\psi({\mathbf{\Phi}}(\mathbf{x}, t))\mathrm{d} t.\end{split}\end{equation*} Using
again Assumption \ref{ass:h2}, this finally gives
\begin{equation}\label{I3}
I_3=-\int_{\mathbf{\Omega}_{-\infty} \cap \mathbf{\Omega}_{+\infty}}\left(g(\mathbf{x})-\lambda
f(\mathbf{x})\right)\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x}).\end{equation} Combining
\eqref{I1}--\eqref{I3} leads to
$$\int_{\mathbf{\Omega} }f(\mathbf{x})\dfrac{\mathrm{d}}{\mathrm{d}
  s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\bigg|_{s=0} \mathrm{d}\mu(\mathbf{x})=-\int_{\mathbf{\Omega}}\left(g(\mathbf{x})-\lambda
f(\mathbf{x})\right)\psi(\mathbf{x})\mathrm{d}\mu(\mathbf{x})$$ which proves that $f \in
\Drond(\mathcal{T}_{\mathrm{max}})$ and $(\lambda- \mathcal{T}_{\mathrm{max}})f=g.$ Next,
for $\mathbf{y} \in \Gamma_-$ and $0 < t < \tau_+(\mathbf{y})$ we write
$t=\tau_-\left({\mathbf{\Phi}}(\mathbf{y},t)\right)$ and, by Proposition
\ref{Phiprop} and (\ref{ho}), we obtain
\begin{equation}\label{fy+}
f({\mathbf{\Phi}}(\mathbf{y},t)) = \int_0^t\exp(-\lambda
(t-s))\,g({\mathbf{\Phi}}(\mathbf{y},s))\mathrm{d} s.
\end{equation} Consequently, $\lim_{t \to
0^+}f({\mathbf{\Phi}}(\mathbf{y},t))=0$  a.e. $\mathbf{y} \in \Gamma_-,$ i.e.
$\mathsf{B}^-f=0$ so that $f \in \Drond(\mathcal{T}_0)$ and $\mathcal{A}_0f=\mathcal{T}_0f=\lambda f-g$.

 $\bullet$ \textit{Step 3.} Now let us show the converse
inclusion $\Drond(\mathcal{T}_0) \subset \Drond(\mathcal{A}_0)$. Let $f \in \Drond(\mathcal{T}_0)$.
Changing possibly $f$ on a set of zero measure, we may write
$f=f^\sharp$, where $f^\sharp$ is the representative of $f$ given by
Theorem \ref{representation}. Then, for any $\mathbf{x} \in \mathbf{\Omega}$ and any $0
\leqslant t<\tau_-(\mathbf{x})$
$$f({\mathbf{\Phi}}(\mathbf{x},-t))-f(\mathbf{x})=\int_0^t
[\mathcal{T}_\mathrm{max}f]({\mathbf{\Phi}}(\mathbf{x},-r))\mathrm{d} r$$ which, according to
the explicit expression of $U_0(t)$,  means that
\begin{equation}\label{uot0}
U_0(t)f(\mathbf{x})-f(\mathbf{x})=\int_0^t\,U_0(r)\mathcal{T}_{\mathrm{max}}f(\mathbf{x})\mathrm{d}
r\end{equation}  for any  $\mathbf{x} \in \mathbf{\Omega}$ and $t < \tau_-(\mathbf{x}).$  Letting
$t$ converge towards $\tau_-(\mathbf{x})$ we obtain
$$f(\mathbf{x})=-\int_0^{\tau_-(\mathbf{x})}[\mathcal{T}_\mathrm{max}f]({\mathbf{\Phi}}(\mathbf{x},-r))\mathrm{d} r.$$
In particular,  Eq. \eqref{uot0} holds true for any $\mathbf{x} \in \mathbf{\Omega}$ and
any $t \geqslant \tau_-(\mathbf{x}).$ Arguing exactly as in \cite[p. 38]{voigt},
the pointwise identity \eqref{uot0} represents the $X$--integral,
i.e, $U_0(t)f-f=\int_0^t \,U_0(r)\mathcal{T}_{\mathrm{max}}f \mathrm{d} r$ in
$L^1(\mathbf{\Omega},\mathrm{d}\mu)$. Consequently, $f \in \Drond(\mathcal{A}_0)$ with
$\mathcal{A}_0f=\mathcal{T}_{\mathrm{max}}f.$\end{proof}

\subsection{\textbf{Green's formula}}The above result allows us to treat more general boundary-value
problem:
\begin{theo}\label{Theo10.43} Let $u \in L^1_-$ and $g \in X$ be
given. Then the function
\begin{equation*}
f(\mathbf{x})=\int_0^{\tau_-(\mathbf{x})}\exp(-\lambda t)\,g({\mathbf{\Phi}}(\mathbf{x},-t))\mathrm{d}
t +\chi_{\{\tau_-(\mathbf{x}) < \infty\}} \exp(-\lambda \tau_-(\mathbf{x}))
u({\mathbf{\Phi}}(\mathbf{x},-\tau_-(\mathbf{x})))\end{equation*} is a \textbf{
unique} solution $f \in \Drond(\mathcal{T}_{\mathrm{max}})$ of the boundary value
problem:
\begin{equation}\label{BVP1}
\begin{cases}
(\lambda- \mathcal{T}_{\mathrm{max}})f=g,\\
\mathsf{B}^-f=u,
\end{cases}
\end{equation}
where $\lambda > 0.$ Moreover, $\mathsf{B}^+f \in L^1_+$ and
\begin{equation}\label{10.1156} \|\mathsf{B}^+f\|_{L^1_+} + \lambda \|f\|_X \leqslant
\|u\|_{L^1_-} + \|g\|_X,\end{equation} with equality sign if $g \geqslant
0$ and $u \geqslant 0$.
\end{theo}
\begin{proof} Let us write $f=f_1+f_2$ with $f_1(\mathbf{x})=\int_0^{\tau_-(\mathbf{x})}\exp(-\lambda t)\,g({\mathbf{\Phi}}(\mathbf{x},-t))\mathrm{d}
t,$ and
$$f_2(\mathbf{x})=\chi_{\{\tau_-(\mathbf{x}) < \infty\}} \exp(-\lambda \tau_-(\mathbf{x})) u\big({\mathbf{\Phi}}\left(\mathbf{x},-\tau_-(\mathbf{x})\right)\big), \qquad \mathbf{x} \in
\mathbf{\Omega}.$$ According to Theorem \ref{uot}, $f_1=(\lambda- \mathcal{T}_0)^{-1}g,$
i.e. $f_1 \in \Drond(\mathcal{T}_{\mathrm{max}})$ with $(\lambda-
\mathcal{T}_{\mathrm{max}})f_1=g$ and $\mathsf{B}^-f_1=0$. Therefore, to prove that
$f$ is a solution of \eqref{BVP1} it suffices to check that $f_2 \in
\Drond(\mathcal{T}_{\mathrm{max}})$, $(\lambda- \mathcal{T}_{\mathrm{max}})f_2=0$ and
$\mathsf{B}^-f_2=u.$ It is easy to see that $f_2 \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$ (see
also \eqref{f-2uu}). To prove that $f_2 \in \Drond(\mathcal{T}_\mathrm{max})$ one
argues as in the proof of Theorem \ref{uot}. Precisely, let $\psi
\in \mathfrak{Y}$,
 noticing that $f_2$ vanishes outside $\mathbf{\Omega}_-$, one has thanks to
 \eqref{f2-u}
\begin{equation*}\begin{split}
\int_{\mathbf{\Omega}}f_2(\mathbf{x})\dfrac{\mathrm{d} }{\mathrm{d}
s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\big|_{s=0}\mathrm{d}\mu(\mathbf{x})
&=\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}f_2({\mathbf{\Phi}}(\mathbf{y},t))\frac{\mathrm{d}
}{\mathrm{d} t} \psi({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d} t\\
&=\int_{\Gamma_-} u(\mathbf{y})\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})} \exp(-\lambda t)
\frac{\mathrm{d} }{\mathrm{d} t} \psi({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d} t.
\end{split}
\end{equation*}
For almost every $\mathbf{y} \in \Gamma_-$, we compute the integral over
$(0,\tau_+(\mathbf{y}))$ by parts, which yields $f_2 \in \Drond(\mathcal{T}_\mathrm{max} )$
with $\mathcal{T}_\mathrm{max}f_2=\lambda f_2.$ Also,
\begin{equation}\label{f2-u}f_2({\mathbf{\Phi}}(\mathbf{y},t))=\exp(-\lambda t)u(\mathbf{y}), \qquad  \mathbf{y} \in
\Gamma_-,\;\;0<t<\tau_+(\mathbf{y})\end{equation} from which we see that
$\mathsf{B}^- f_2=u.$

Consequently, $f$ is a solution to \eqref{BVP1}. To prove that the
solution is unique, it is sufficient to prove that the only solution
$h \in \Drond(\mathcal{T}_{\mathrm{max}})$ to
 $(\lambda- \mathcal{T}_{\mathrm{max}})h=0, \mathsf{B}^-h=0, $
is $h=0$. This follows from the fact that such a solution $h$
actually belongs to $\Drond(\mathcal{T}_0)$ if $\lambda \in \varrho(\mathcal{T}_0)$.
Finally, it remains to prove \eqref{10.1156}. For simplicity, we
denote the representative of $f_i$, $i=1,2$, defined in Proposition
\ref{trace}, with the same letter. Using \eqref{f2-u} and the fact
that $f_2$ vanishes on $\mathbf{\Omega}_{-\infty}$,  from \eqref{10.47} we get
\begin{equation}\label{f-2uu}\begin{split}
\lambda \int_{\mathbf{\Omega}}|f_2|\mathrm{d}\mu&=\lambda
\int_{\mathbf{\Omega}_-}|f_2|\mathrm{d}\mu=\lambda \int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}e^{-\lambda t}|u(\mathbf{y})|\mathrm{d} t\\
&=\int_{\Gamma_-}|u(\mathbf{y})|\left(1-e^{-\lambda
\tau_+(\mathbf{y})}\right)\mathrm{d}\mu_-(\mathbf{y}).\end{split}\end{equation} Define $h$ :
$\mathbf{y} \in \Gamma_- \longmapsto h(\mathbf{y})=|u(\mathbf{y})|e^{-\lambda \tau_+(\mathbf{y})}.$
It is clear that $h$ vanishes on $\Gamma_{-\infty}$ and $h(\mathbf{y}) \leqslant
|u(\mathbf{y})|$ for a.e. $\mathbf{y} \in \Gamma_-$. In particular, $h \in L^1_-$ and,
according to \eqref{10.51},
\begin{equation*}\begin{split}
\int_{\Gamma_-}&h(\mathbf{y})\mathrm{d}\mu_-(\mathbf{y}) =\int_{\Gamma_- \setminus
\Gamma_{-\infty}}h(\mathbf{y})\mathrm{d}\mu_-(\mathbf{y})=\int_{\Gamma_+ \setminus
\Gamma_{+\infty}}h({\mathbf{\Phi}}(z,-\tau_-(z)))\mathrm{d}\mu_+(z)\\
&=\int_{\Gamma_+\setminus \Gamma_{+\infty}}e^{-\lambda
\tau_-(z)}|u({\mathbf{\Phi}}(z,-\tau_-(z)))|\mathrm{d}\mu_+(z)=\int_{\Gamma_+}|\mathsf{B}^+f_2(z)|\mathrm{d}\mu_+(z)=\|\mathsf{B}^+f_2\|_{L^1_+}.
\end{split}\end{equation*}
Combining this with \eqref{f-2uu} leads to
\begin{equation}\label{B+f-2}\lambda \|f_2\|_X+\|\mathsf{B}^+f_2\|_{L^1_+}=\|u\|_{L^1_-}.\end{equation} Now, let us show that $\mathsf{B}^+f_1 \in
L^1_+$ and $\|\mathsf{B}^+f_1\|_{L^1_+} + \lambda \|f_1\|_X \leqslant \|g\|_X.$ For
any $\mathbf{y} \in \Gamma_+$ and $0 < t < \tau_-(\mathbf{y}),$ we see, as above,
that $f_1({\mathbf{\Phi}}(\mathbf{y},-t))=\int_t^{\tau_-(\mathbf{y})}\exp(-\lambda
(s-t))g({\mathbf{\Phi}}(\mathbf{y},-s))\mathrm{d} s.$ This shows that
$$\mathsf{B}^+f_1(\mathbf{y})=\lim_{t \to
0^+}f_1({\mathbf{\Phi}}(\mathbf{y},-t))=\int_0^{\tau_-(\mathbf{y})}\exp(-\lambda
s))g({\mathbf{\Phi}}(\mathbf{y},-s))\mathrm{d} s.$$ According to Proposition
\ref{prointegra},
$$\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y})}\left|g({\mathbf{\Phi}}(\mathbf{y},-s))\right|\mathrm{d}
s=\int_{\mathbf{\Omega}_+}|g|\,\mathrm{d}\mu$$ which, since $\exp(-\lambda
(s-t))|g({\mathbf{\Phi}}(\mathbf{y},-s))| \leqslant |g({\mathbf{\Phi}}(\mathbf{y},-s))|,$
implies $\mathsf{B}^+f_1 \in L^1_+.$ Let us now assume $g \geqslant 0$. Then $f_1
\geqslant 0$ and hence
$$\lambda \|f_1\|=\lambda \int_{\mathbf{\Omega}}f_1\,\mathrm{d}\mu=\lambda \int_{\mathbf{\Omega}_+}f_1\,\mathrm{d}\mu+\lambda \int_{\mathbf{\Omega}_-\cap \mathbf{\Omega}_{+\infty}}f_1\,\mathrm{d}\mu
+\lambda \int_{\mathbf{\Omega}_{-\infty} \cap \mathbf{\Omega}_{+\infty}}f_1\,\mathrm{d}\mu.$$ Using
similar arguments to those used in the study of $f_2$, we have
$$\lambda \int_{\mathbf{\Omega}_+}f_1\,\mathrm{d}\mu=\int_{\Gamma_+}\mathrm{d}\mu_+(\mathbf{y})\int_0^{\tau_-(\mathbf{y})}g({\mathbf{\Phi}}(\mathbf{y},-t))\left(1-\exp(-\lambda t)\right)\mathrm{d} t,$$
 which,  by Proposition \ref{prointegra}, implies
 $\lambda
 \int_{\mathbf{\Omega}_+}f_1\,\mathrm{d}\mu=\int_{\mathbf{\Omega}_+}g\,\mathrm{d}\mu-\int_{\Gamma_+}\mathsf{B}^+f_1\,\mathrm{d}\mu_+.$
Similar argument shows that $\lambda \int_{\mathbf{\Omega}_- \cap
\mathbf{\Omega}_{+\infty}}f_1\,\mathrm{d}\mu= \int_{\mathbf{\Omega}_- \cap \mathbf{\Omega}_{+\infty}} g\,\mathrm{d}\mu,$
while the equality $$ \lambda \int_{\mathbf{\Omega}_{-\infty} \cap
\mathbf{\Omega}_{+\infty}}f_1\,\mathrm{d}\mu=\int_{\mathbf{\Omega}_{-\infty} \cap
\mathbf{\Omega}_{+\infty}}g\,\mathrm{d}\mu,$$ is a direct consequence of the invariance
of $\mu$ with respect to ${\mathbf{\Phi}}(\cdot,t)$. This shows that
$\lambda \|f\|_X=\|g\|_X-\|\mathsf{B}^+f\|_{L^1_+}$ for $g \geqslant 0$. In
general, defining
$$F_1(\mathbf{x})=\int_0^{\tau_-(\mathbf{x})}\exp(-\lambda s)\,\left|g({\mathbf{\Phi}}(\mathbf{x},-s)\right|\mathrm{d} s,\qquad \mathbf{x} \in
\mathbf{\Omega},$$ we obtain $\|\mathsf{B}^+f_1\|_{L^1_+}+\lambda \|f_1\|_X \leqslant
\|B^+F_1\|_{L^1_+}+\lambda \|F_1\|_X=\|g\|_X$ which, combined with
\eqref{B+f-2}, gives \eqref{10.1156}.
\end{proof}
\begin{nb} Notice that, in order to get the existence and uniqueness
of the solution $f$ to \eqref{BVP1}, it is not necessary for $u$ to
belong to $L^1(\Gamma_-,\mathrm{d}\mu_-)$. Indeed, we only have to make sure
that  $f_2 \in L^1(\mathbf{\Omega},\mathrm{d}\mu)$, i.e., from \eqref{f-2uu},
$\int_{\Gamma_-}|u(\mathbf{y})|\left(1-e^{-\lambda
\tau_+(\mathbf{y})}\right)\mathrm{d}\mu_-(\mathbf{y}) < \infty.$ Of course, to get
\eqref{10.1156}, the assumption $u \in L^1(\Gamma_-,\mathrm{d}\mu_-)$ is
necessary.\end{nb} Let us note that, with the notation of Theorem
\ref{Theo10.43}, we have
\begin{equation}\label{green}
\int_{\Gamma_+}\mathsf{B}^+f\mathrm{d}\mu_++\lambda
\int_{\mathbf{\Omega}}f\,\mathrm{d}\mu=\int_{\Gamma_-}u\,\mathrm{d}\mu_-+\int_{\mathbf{\Omega}}g\,\mathrm{d}\mu.
\end{equation}
Indeed, for nonnegative $u$ and $g$, \eqref{10.1156} turns out to be
precisely \eqref{green}. Then, for arbitrary $u \in L^1_-$ and $g \in
X$, we get \eqref{green} by splitting functions into positive and
negative parts. This leads to the following generalization of
Green's formula:
\begin{propo}[\textit{\textbf{Green's formula}}]\label{propgreen}
Let $f \in \Drond(\mathcal{T}_{\mathrm{max}})$ satisfies $\mathsf{B}^-f \in L^1_-.$ Then
$\mathsf{B}^+f \in L^1_+$ and
$$\int_{\mathbf{\Omega}}\mathcal{T}_{\mathrm{max}}f\mathrm{d}\mu=\int_{\Gamma_-}\mathsf{B}^-f\mathrm{d}\mu_-
-\int_{\Gamma_+}\mathsf{B}^+f\,\mathrm{d}\mu_+$$
\end{propo}
\begin{proof} For given $f \in \Drond(\mathcal{T}_{\mathrm{max}})$, we obtain the result by setting
$u=\mathsf{B}^-f \in L^1_-$ and $g=(\lambda- \mathcal{T}_{\mathrm{max}})f \in X$ in Eq.
\eqref{green}.\end{proof}

\begin{nb} If $\mathrm{d}\mu$ is the Lebesgue measure on $\mathbb{R}^N$, the above formula leads to a better understanding of the measures $\mathrm{d}\mu_{\pm}$.
Indeed, comparing it to the classical Green's formula (see e.g.
\cite{bardos}), we see that the restriction of $\mathrm{d}\mu_{\pm}$ on the set
$\Sigma_{\pm}=\{\mathbf{y} \in
\partial\mathbf{\Omega}\,;\,\pm \mathscr{F}(\mathbf{y}) \cdot n(\mathbf{y}) > 0\}$  equals
$$\left|\mathscr{F}(\mathbf{y}) \cdot n(\mathbf{y})\right|\mathrm{d}\gamma(\mathbf{y}),$$
where $\mathrm{d}\gamma(\cdot)$ is the surface Lebesgue measure on
$\partial\mathbf{\Omega}$.
\end{nb}

\section*{Appendix: About the class of test-functions}\setcounter{equation}{0}
\renewcommand{\theequation}{A.\arabic{equation}}

We answer in this Appendix a natural question concerning the
definition of the class of test-functions $\mathfrak{Y}$. Precisely,
we prove that two test-functions equal $\mu$--almost everywhere are
such that their derivatives (in the sense of \eqref{defi:deriva})
also coincide $\mu$-almost everywhere. To prove our claim, it
clearly suffices to prove that, given $\psi \in \mathfrak{Y}$ such
that $\psi(\mathbf{x})=0$ for $\mu$-a. e. $\mathbf{x} \in \mathbf{\Omega}$, one has
$\varphi(\mathbf{x})=0$ for $\mu$-a. e. $\mathbf{x} \in \mathbf{\Omega}$ where
$\varphi(\mathbf{x})=\frac{\mathrm{d}}{\mathrm{d} s}\psi({\mathbf{\Phi}}(\mathbf{x},s))\big|_{s=0}.$
Let
$$E:=\big\{\mathbf{x} \in \mathbf{\Omega}\,;\,\psi(\mathbf{x})=0 \text{ and } \varphi(\mathbf{x}) \neq 0\big\}.$$
It is clear that $E$ is measurable and that one has to prove that
$\mu(E)=0$. It is no loss of generality to assume that $E$
\textit{is bounded}. We observe that for any $\mathbf{x} \in E$, there
exists $\delta_\mathbf{x}
>0$ such that
\begin{equation}\label{01}\psi({\mathbf{\Phi}}(\mathbf{x},t)) \neq 0, \qquad \forall\, 0 < |t| < \delta_\mathbf{x}.\end{equation}
Let us split $E$ as follows
$$E=\big(E \cap \mathbf{\Omega}_-\big) \cup \big(E \cap \mathbf{\Omega}_+ \cap \mathbf{\Omega}_{-\infty}\big) \cup \big(E \cap \mathbf{\Omega}_{+\infty}
\cap \mathbf{\Omega}_{-\infty}\big):=E_- \cup E_+ \cup E_\infty$$ and prove that
$\mu(E_-)=\mu(E_+)= \mu(E_\infty)=0$.

\begin{enumerate}
\item First consider $E_-$. Since $\psi(\mathbf{x})=0$ for $\mu$-a. e. $\mathbf{x} \in \mathbf{\Omega}_-$ and using the fact that
any $\mathbf{x} \in \mathbf{\Omega}_-$ can be written as  $\mathbf{x}={\mathbf{\Phi}}(\mathbf{y},t)$ for
some $\mathbf{y} \in \Gamma_-$ and $0 < t < \tau_+(\mathbf{y})$, we observe that, for
$\mu_-$ a. e. $\mathbf{y} \in \Gamma_-$, $\psi({\mathbf{\Phi}}(\mathbf{y},t))=0$ for
almost every (in the sense of the Lebesgue measure in $\mathbb{R}$)
$0 < t < \tau_+(\mathbf{y})$. For such a $\mathbf{y} \in \Gamma_-$, continuous
differentiability of $t \mapsto \psi({\mathbf{\Phi}}(\mathbf{y},t))$ implies
$\psi({\mathbf{\Phi}}(\mathbf{y},t))=0$ for {\it any } $0 < t < \tau_+(\mathbf{y})$.
This means, according to \eqref{01} that, for $\mu_-$-a. e. $\mathbf{y} \in
\Gamma_-$, ${\mathbf{\Phi}}(\mathbf{y},t) \notin E$ for any $0 < t <
\tau_+(\mathbf{y})$. Since
$$\mu(E \cap \mathbf{\Omega}_-)=\int_{\Gamma_-}\mathrm{d}\mu_-(\mathbf{y})\int_0^{\tau_+(\mathbf{y})}\chi_{E}({\mathbf{\Phi}}(\mathbf{y},t))\mathrm{d} t$$
we see that $\mu(E_-)=0.$

\item In the same way, using $\Gamma_+$ instead of
$\Gamma_-$, we show that $\mu(E \cap \mathbf{\Omega}_+ \cap \mathbf{\Omega}_{-\infty})=0$.

\item It remains to prove that $\mu(E_\infty)=0$. In accordance with \eqref{01},
we define for, any $n \in \mathbb{N}$,
    $$E_n:=\bigg\{\mathbf{x} \in E_\infty\,;\,\delta_\mathbf{x} \geqslant 1/n\bigg\}=\bigg\{\mathbf{x} \in E_\infty\,;\,\psi({\mathbf{\Phi}}(\mathbf{x},t)) \neq 0, \:\: \forall\, 0 < |t| < 1/n\bigg\}.$$
 According to Assumption \ref{ass:h2}, it is easy to see that $\mu(E_n)=0$ for any $n \in \mathbb{N}$ since $\psi(\mathbf{x})=0$ for $\mu$-a.e. $\mathbf{x} \in \mathbf{\Omega}.$ Moreover, $E_1 \subset E_2 \subset \ldots \subset E_n \subset E_{n+1} \subset \ldots $, and
 $$\bigcap_{n=1}^\infty \bigg(E_\infty \setminus E_n\bigg)= \varnothing.$$
 Since we assumed $\mu(E) < \infty$,  we have $\mu(E_\infty \setminus E_1) < \infty$ and
 $\lim_{n \to \infty} \mu\big(E_\infty \setminus E_n\big)=0$. Writing
 $E_\infty=E_n \cup \big(E_\infty \setminus E_n\big)$, we see that $\mu(E_\infty)=0.$
\end{enumerate}

redpajama_set_name: RedPajamaArXiv