# On the Lebesgue measure of the boundary of the evoluted set111LC, FR and DV
are partially supported by the Gruppo Nazionale per l’Analisi Matematica, la
Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta
Matematica (INdAM). FR is partially supported by the University of Padua under
the STARS Grants programme CONNECT: Control of Nonlocal Equations for Crowds
and Traffic models.

Francesco Boarotto<EMAIL_ADDRESS>Laura Caravenna
<EMAIL_ADDRESS>Francesco Rossi<EMAIL_ADDRESS>Davide
Vittone<EMAIL_ADDRESS>Dipartimento di Matematica “Tullio Levi-
Civita”, Università degli Studi di Padova, Via Trieste 63, 35121 Padova, Italy

###### Abstract

The evoluted set is the set of configurations reached from an initial set via
a fixed flow for all times in a fixed interval. We find conditions on the
initial set and on the flow ensuring that the evoluted set has negligible
boundary (i.e. its Lebesgue measure is zero). We also provide several
counterexample showing that the hypotheses of our theorem are close to sharp.

###### keywords:

evoluted set , attainable set , Lebesgue measure

###### MSC:

[2020]93B03, 93B27, 28A99

##  1 Introduction

The study of the attainable set from a point is a crucial problem in control
theory, starting from the classical orbit, Rashevsky-Chow and Krener theorems,
see [1, 12, 13]. If the initial state is not precisely identified, but lies in
a given set, the problem gets even more complicated. The goal of this article
is to study such problem in a first, simplified setting.

Here, we consider a fixed Lipschitz vector field $v(x)$ acting on sets via the
flow $\Phi^{v}_{t}$ it generates. Given an initial set $S$, we aim to describe
the evoluted set

$S^{t}:=\cup_{\tau\in[0,t]}\Phi^{v}_{\tau}(S),$ (1.1)

that is the set of points reached at times $\tau\in[0,t]$. It was studied e.g.
in [7, 14, 16], with the name “funnel” too. One can also observe that $S^{t}$
is the attainable set at time $t$ for the control problem

$\displaystyle\dot{x}(t)=u(t)v(x(t)),\qquad\qquad u(t)\in[0,1],\qquad\qquad
x(0)\in S.$ (1.2)

A first question needs to be answered:

If the initial set $S$ has a negligible boundary (i.e. its Lebesgue measure is
zero), does the evoluted set have a negligible boundary too?

The first, striking result of this article is to provide a negative answer to
this question. In Example 8 below, we exhibit a set $S\subset\mathbb{R}^{2}$
with negligible boundary and such that the boundary of the evoluted set
$S^{t}$ is not negligible. Even more surprisingly, the counterexample relies
on very low regularity of $S$, while the vector field $v$ is constant (so
extremely regular). Another counterexample was provided in [4].

We then turn our attention to find regularity properties of $S$ that ensure
that the evoluted set keeps having negligible boundary. Our main result is
based on the definition of Lebesgue point for the complement $S^{\complement}$
of a set $S$, that we recall here.

######  Definition 1 (Lebesgue point of a set).

Let $S\subset\mathbb{R}^{n}$ be a set. We say that $x\in\mathbb{R}^{n}$ is a
Lebesgue point of the complement $S^{\complement}$ if

$\lim_{r\to 0^{+}}\frac{\mathscr{L}^{n}(B(x,r)\setminus
S)}{\mathscr{L}^{n}(B(x,r))}=1.$ (1.3)

Our main result is given below. As customary, $\mathscr{H}^{k}$ denotes the
$k$-dimensional Hausdorff measure in $\mathbb{R}^{n}$.

######  Theorem 1.

If $v$ is a Lipschitz vector field in $\mathbb{R}^{n}$ and
$S\subset\mathbb{R}^{n}$ is any set such that $\mathscr{L}^{n}(\partial S)=0$
and $\mathscr{H}^{n-1}$-almost every point of its boundary $\partial S$ is not
a Lebesgue point of $S^{\complement}$, then
$\mathscr{L}^{n}(\partial(S^{t}))=0$ for every $t\in\mathbb{R}$.

Several corollaries and the complete proof are given in Section 4 below. In
particular, regularity hypotheses of Theorem 1 hold for several classes of
interest for $S$, such as Lipschitz or smooth domains. Moreover, the same
result holds for the evoluted set where initial and final times are excluded,
i.e. replacing $\tau\in(0,t)$ in (1.1).

######  Remark 2.

The condition $\mathscr{L}^{n}(\partial S)=0$ in Theorem 1 seems natural,
observing that, when $v\equiv 0$, one has $S^{t}=S$ and
$\mathscr{L}^{n}(\partial(S^{t}))=\mathscr{L}^{n}(\partial S)$. Even with non-
vanishing vector fields, the thesis would not of course hold at $t=0$.

######  Remark 3.

The assumptions in Theorem 1 look satisfactory, as they are satisfied even by
open sets with wild (fractal) boundaries. In fact, the boundary of an open set
$S$ might have fractal Hausdorff dimension even if no point of such boundary
is a Lebesgue point of $S^{\complement}$, and thus necessarily
$\mathscr{L}^{n}(\partial S)=0$. A well-known example on the plane is the open
region $S\subset\mathbb{R}^{2}$ enclosed by the classical von Koch (closed
snowflake) curve (see e.g. [11]), whose Hausdorff dimension is $\log_{3}4$.

One can use similar fractal constructions to produce bounded open sets
$S\subset\mathbb{R}^{n}$ such that the Hausdorff dimension of $\partial S$ is
any desired number in $(n-1,n)$, but such that no point of $\partial S$ is a
Lebesgue point of $S^{\complement}$. Eventually, a countable union of such
sets produces open sets $S\subset\mathbb{R}^{n}$ such that $\partial S$ has
Hausdorff dimension $n$, but such that no point of $\partial S$ is a Lebesgue
point of $S^{\complement}$.

As for infinite-time horizon control problems, one might ask if Theorem 1 hold
for $t=+\infty$ too, i.e. by considering the union over $\tau\in[0,+\infty)$.
The answer is, surprisingly once again, negative. We provide counterexamples
in Examples 9–10.

The question about evoluted sets is even more interesting when it is
interpreted in terms of densities solving a Partial Differential Equation,
eventually with internal control. Assume to have an initial state that is a
probability measure on $S$, e.g. because the initial configuration is not
precisely identified, and consider $S$ to be some form of “safety region”. If
the measure is absolutely continuous with respect to the Lebesgue measure, is
it true that time modulations of the flow (1.2) do not concentrate mass along
the boundary, then eventually giving a non-zero probability of being close to
unsafe configurations? This problem has been addressed in [8, 9]. This is also
one of the main motivations and future applications of the result presented
here: to develop a theory describing the reachable set starting from a measure
under control action. See further results in this direction in [5, 6].

The results presented in this article can be seen as the final and most
general ones, among a list of contributions by the authors. Indeed:

  * 1.

in [9], we proved that $\mathscr{L}^{n}(\partial(S^{t}))=0$ under the stronger
condition of $S$ being open and satisfying the so-called uniform interior cone
condition. The proof relies on a Gronwall argument, adapted to evolution of
interior cones under flow action.

  * 2.

In [4], we prove the same property, but under the weaker hypothesis of $S$
being a $C^{1,1}$ domain. The proof is based on a careful study of the
evolution of the normal to the boundary under flow action.

  * 3.

In the present paper, we understand that the same result holds even just
assuming that $\mathscr{H}^{n-1}$-almost each point of the boundary is not a
Lebesgue point of the complementary set, see Theorem 1. The proof couples
measure theory with properties of evolution of non-Lebesgue points under flow
action, recalling that the flow is bi-Lipschitz.

The structure of the article is the following: in Section 2 we collect the
main definitions, as well as known results. Section 3 collects counterexamples
to questions introduced above, eventually showing sharpness of the hypotheses
in our main theorem. Section 4 presents the proof of our main result, that is
Theorem 1.

##  2 Definitions and preliminary results

In this article, we will use the following notation: $S$ is a general subset
of $\mathbb{R}^{n}$, with $S^{\complement}:=\mathbb{R}^{n}\setminus S$ being
its complement. We also define its topological closure $\bar{S}$, its interior
$\mathrm{int}(S)$ and its boundary $\partial
S:=\bar{S}\setminus\mathrm{int}(S)$.

We will deal with Lipschitz functions and vector fields. More precisely, we
will deal with _globally Lipschitz functions_ , i.e. functions for which there
exists $L>0$ such that the Lipschitz condition $|f(x)-f(y)|\leq L|x-y|$ holds
for all $x,y\in\mathbb{R}^{n}$. Most of the results can be translated to the
local setting, provided that more conditions ensuring some form of compactness
are added. See e.g. Corollary 14 below.

We will also use the following definition of Lipeomorphism, also known as bi-
Lipschitz function.

######  Definition 4.

Given $U,V\subset\mathbb{R}^{n}$, a map $F:U\to V$ is said to be a
Lipeomorphism if $F$ is a Lipschitz continuous bijective map from $U$ to $V$
and its inverse $F^{-1}$ is Lipschitz as well.

We now recall the definition of the flow $\Phi^{v}_{t}$ of a vector field $v$.

######  Definition 5 (Flow of a vector field).

Let $t\in\mathbb{R}$ and let $v$ be a Lipschitz vector field on
$\mathbb{R}^{n}$. The flow $\Phi_{t}^{v}:\mathbb{R}^{n}\to\mathbb{R}^{n}$
along $v$ at time $t$ is the map $\Phi_{t}^{v}(x_{0}):=x(t)$ that returns the
value at time $t$ of the (unique) solution to the Cauchy problem

$\left\\{\begin{aligned} \dot{x}(t)&=v(x(t)),\\\
x(0)&=x_{0}.\end{aligned}\right.$ (2.1)

######  Remark 6.

Consider a Lipschitz vector field $v$ and denote by $L$ its Lipschitz
constant. Then it is well-known that:

  1. 1.

The flow is invertible, and it holds $(\Phi^{v}_{t})^{-1}=\Phi^{v}_{-t}$;

  2. 2.

For all $T>0$ and $t\in[-T,T]$, the map
$\Phi^{v}_{t}:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is a Lipeomorphism with
Lipschitz constant $e^{LT}$. This is a consequence of Grönwall’s inequality.

  3. 3.

The map
$\Psi(x,t):=\Phi^{v}_{t}(x):\mathbb{R}^{n}\times\mathbb{R}\to\mathbb{R}^{n}$
is locally Lipschitz continuous. As a consequence, for
$N\subset\mathbb{R}^{n}\times\mathbb{R}$ the condition $\mathscr{H}^{n}(N)=0$
implies $\mathscr{H}^{n}(\Psi(N))=0$.

Our proof of Theorem 1 only relies on the above properties of the flow.

We are now ready to define the evoluted set $S^{t}$, that is the main object
that is studied in the article.

######  Definition 7 (Evoluted set).

Let $v$ be a Lipschitz vector field on $\mathbb{R}^{n}$ and $\Phi^{v}_{t}$ its
flow. Given a set $S\subset\mathbb{R}^{n}$ and $t\geq 0$, we define the
evoluted set

$S^{t}:=\bigcup_{\tau\in[0,t]}\Phi_{\tau}^{v}(S).$ (2.2)

##  3 Counterexamples

In this section, we provide counterexamples that show that the hypotheses of
the main Theorem 1 are close to sharpness. In Example 8, we will show that the
condition $\mathscr{L}^{n}(\partial S)=0$ does not ensure
$\mathscr{L}^{n}(\partial(S^{t}))=0$ even with $S$ open bounded and $v$ very
regular (indeed constant). In Example 9, we provide a variation of Example 8,
focusing on the infinite time horizon: we define an unbounded set $A_{u}$,
satisfying the hypothesis of Theorem 1, whose evoluted set $S^{+\infty}_{u}$
has non-neglibigle boundary.

In Example 10, very different from the previous ones, we prove again that the
evoluted set for $t=+\infty$ can have non-neglibigle boundary even when the
hypotheses of Theorem 1 are satisfied. We start with a bounded connected set
$S$, but with a vector field which is more complex and vanishes outside
$S^{\infty}$. An easy variation of this example would show that, for every
connected and simply connected bounded open set $\Omega\subset\mathbb{R}^{2}$,
there exist a Lipschitz vector field $v$ on $\mathbb{R}^{2}$ and a ball
$S=B(p,r)\subset\mathbb{R}^{2}$ such that $S^{\infty}=\Omega$.

######  Example 8 ($\mathscr{L}^{n}(\partial S)=0$ does not ensure
$\mathscr{L}^{n}(\partial(S^{t}))=0$).

Consider the sequence of positive natural numbers $i=1,2,\ldots$ and write
each of them as $i=2^{n}+k$ with $n$ the maximal natural exponent
$\lfloor\log_{2}(i)\rfloor$ and $k=0,\ldots,2^{n}-1$. Define the corresponding
sequence $(x_{i})_{i\geq 1}$ by $x_{i}=\frac{k}{2^{n}}$, where $n,k$ are given
above. It is easy to observe that $x_{i}$ is a dense sequence in $[0,1]$.

We are now ready to define the set $S\subset\mathbb{R}^{2}$ in Figure 1 as
follows:

$\displaystyle S:=\cup_{i=1,2,\ldots}B(c_{i},r_{i}),$ $\displaystyle
c_{i}=(x_{i},0),$ $\displaystyle r_{i}:=2^{-i-2}\ .$ (3.1)

Notice that the projection of $S$ on the $x$-axis is the relatively open set

$\displaystyle O=\cup_{i\geq 1}(x_{i}-r_{i},x_{i}+r_{i}),$
$\displaystyle\mathscr{L}^{1}(O)\leq\sum_{i=1}^{+\infty}2r_{i}=\frac{1}{2}\ .$
(3.2)

The boundary $\partial S$ is contained in the union of the circles $\partial
B(c_{i},r_{i})$ and the segment $[0,1]\times\\{0\\}$. In particular, it is
contained in a countable union of sets with zero $\mathscr{L}^{2}$-measure,
then it satisfies $\mathscr{L}^{2}(\partial S)=0$. Moreover, $\partial S$
contains the compact Cantor-like set $K\times\\{0\\}$, where

$\displaystyle K:=[0,1]\setminus O,$
$\displaystyle\mathscr{L}^{1}(K)=1-\mathscr{L}^{1}(O)\geq\frac{1}{2}\ .$ (3.3)

Indeed, each point in $[0,1]\times\\{0\\}$ is in the closure of $S$, since the
sequence $x_{i}$ is dense in $[0,1]$, while $S$ is open and thus coincides
with its interior. We now choose the constant vector field $v=(0,1)$. As the
projection of $S$ on the $x$-axis is $O$ then the boundary of the evoluted set
$\partial(S^{1})$ contains $K\times[0,1]$ and, in particular,
$\mathscr{L}^{2}(\partial(S^{1}))\geq\mathscr{L}^{1}(K)\cdot 1>0$.

Figure 1: The open set $S$ of Example 8, which evolves with the constant
vector field $v=(0,1)$.

######  Example 9 (Infinite time: initial unbounded set).

With notations of Example 8, define a new unbounded open set $A_{u}$ as
follows:

$\displaystyle A_{u}:=\cup_{i=1,2,\ldots}B(c_{i},r_{i})$ where now
$c_{i}=(x_{i},-i)$. (3.4)

The projection $A_{u}$ on the $x$ axis is again $O$: in particular
$\partial(A_{u}^{\infty})$ contains $K\times[0,1]$, thus
$\mathscr{L}^{2}(\partial(A_{u}^{\infty}))>0$. Remark that $A$ satisfies the
hypotheses of Theorem 1, thus $\mathscr{L}^{2}(\partial(A_{u}^{t}))=0$ for any
finite time $t\in\mathbb{R}$.

######  Example 10 (Infinite time: initial bounded set).

Consider an Osgood curve [15] in the plane, i.e. a Jordan curve
$\Gamma\subset\mathbb{R}^{2}\equiv\mathbb{C}$ such that
$\mathscr{L}^{2}(\Gamma)>0$, and let $\Omega$ be the bounded connected open
region enclosed by $\Gamma$. Let $D\subset\mathbb{R}^{2}\equiv\mathbb{C}$ be
the unit (open) disk centered at the origin. Since $\Omega$ is also simply
connected ([17, page 2]), by the Riemann Mapping Theorem [17, page 4] there
exists a biholomorphic map $G:D\to\Omega$. Moreover, by Carathéodory’s theorem
[17, Theorem 2.6] $G$ can be extended to a homeomorphism
$G:\overline{D}\to\overline{\Omega}$.

Let $g:[0,1)\to(0,+\infty)$ be a positive smooth function; $g$ will be chosen
later in such a way that, as $r\to 1^{-}$, both $g(r)$ and $g^{\prime}(r)$
decrease to 0 “rapidly enough”, in a way depending on $G$ only. Consider the
vector field $u(q):=g(|q|)q$ for $q\in D$; assuming also that $g$ is constant
on $[0,1/2]$, then $u\in C^{\infty}(D)$. Moreover, the flow of $u$ is radial
and it can be easily shown that, for every open neighborhood $U\subset D$ of
$0$, it holds $U^{\infty}=\cup_{t\geq 0}\Phi^{u}_{t}(U)=D$.

Consider now the push-forward of $u$, i.e., the smooth vector field $v$ on
$\Omega$ defined by $v(G(q)):=dG_{q}(u(q))$. We will later prove that $v$ is
Lipschitz continuous on $\Omega$, hence it can be extended to a Lipschitz
vector field in $\mathbb{R}^{2}$. The flow of $v$ on $\Omega$ is conjugated to
the flow of $u$; in particular, choosing $S:=B(G(0),\delta)\subset\Omega$ and
setting $U:=G^{-1}(S)$, we have that

$S^{\infty}=\bigcup_{t\geq 0}\Phi^{v}_{t}(S)=G\left(\bigcup_{t\geq
0}\Phi^{u}_{t}(U)\right)=G(D)=\Omega.$

Therefore
$\mathscr{L}^{2}(\partial(S^{\infty}))=\mathscr{L}^{2}(\partial\Omega)>0$ and
we have only to prove that $v$ is Lipschitz continuous on $\Omega$. To this
end, on differentiating the equality

$v(p)=g(|G^{-1}(p)|)\;\nabla G(G^{-1}(p))\>G^{-1}(p)$

(where we use matrix notation) we obtain the estimate

$\begin{split}|\nabla v(p)|\leq&\;|g^{\prime}(|q|)|\,\,|\nabla
G^{-1}(p)|\,\,|\nabla G(q)|\,\,|q|\\\
&\;+g(|q|)\,\Big{[}\;|\nabla^{2}G(q)|\,\,|\nabla G^{-1}(p)|\,\,|q|+|\nabla
G(q)|\,\,|\nabla G^{-1}(p)|\;\Big{]},\end{split}$ (3.5)

where $q:=G^{-1}(p)$. Let us show that the quantity on the right hand side of
(3.5) can be bounded uniformly in $p\in\Omega$ with a proper choice of the
function $g$; this will prove the Lipschitz continuity of $v$.

For $r\in(0,1)$ define

$\displaystyle m_{0}(r):=1+\max_{q\in\overline{B(0,r)}}\Big{\\{}$
$\displaystyle|\nabla G^{-1}(G(q))|\,\,|\nabla
G(q)|\,\,|q|+|\nabla^{2}G(q)|\,\,|\nabla G^{-1}(G(q))|\,\,|q|$
$\displaystyle+|\nabla G(q)|\,\,|\nabla G^{-1}(G(q))|\Big{\\}}$

and observe that (3.5) gives

$|\nabla
v(p)|\leq\big{(}|g^{\prime}(|q_{p}|)|+g(|q_{p}|)\big{)}\>m_{0}(|q_{p}|)\qquad\forall\>p\in\Omega.$
(3.6)

The function $m_{0}$ is continuous and non-decreasing on $(0,1)$; fix a
smooth, non-decreasing function $m_{1}:(0,1)\to\mathbb{R}$ such that
$m_{1}\geq m_{0}\geq 1$ and consider the smooth and decreasing function

$g_{1}(r):=\int_{r}^{1}\frac{1}{m_{1}(\rho)}\>d\rho,\qquad r\in(0,1).$

Eventually, let $g:[0,1)\to\mathbb{R}$ be a positive, non-increasing smooth
function such that $g$ is constant on $[0,1/2]$ and $g=g_{1}$ on $[2/3,1)$; in
particular, for every $r\in[2/3,1)$ we have

$\begin{split}&|g^{\prime}(r)|=\frac{1}{m_{1}(r)}\leq\frac{1}{m_{0}(r)}\\\
&|g(r)|=\int_{r}^{1}\frac{1}{m_{1}(\rho)}d\rho\leq\frac{1-r}{m_{1}(r)}\leq\frac{1}{m_{0}(r)}\end{split}$
(3.7)

due to the monotonicity of $m_{1}$. From (3.6) and (3.7) we deduce

$\displaystyle|\nabla
v(p)|\leq\left(\max_{[0,2/3]}\\{|g^{\prime}|+|g|\\}\right)m_{0}(2/3)\quad$
$\displaystyle\text{if }p\in G(B(0,2/3)),$ $\displaystyle|\nabla v(p)|\leq 2$
$\displaystyle\text{if }p\in\Omega\setminus G(B(0,2/3))$

and the Lipschitz continuity of $v$ follows.

##  4 Proof of the main result

In this section, we prove our main theorem. We first prove two auxiliary
results.

######  Proposition 11.

If $\Phi:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is a Lipeomorphism and
$S\subset\mathbb{R}^{n}$, then $\Phi(x)$ is a Lebesgue point of
$(\Phi(S))^{\complement}$ if and only if $x$ is a Lebesgue point of
$S^{\complement}$.

###### Proof.

Fix $L>0$ and $M>0$ such that for every $y,z\in\mathbb{R}^{n}$

$\|\Phi(y)-\Phi(z)\|\leq
L\|y-z\|\quad\text{and}\quad\|\Phi^{-1}(y)-\Phi^{-1}(z)\|\leq M\|y-z\|.$

Let us show that if $\Phi(x)$ is a Lebesgue point of $(\Phi(S))^{\complement}$
then $x$ is a Lebesgue point of $S^{\complement}$. The other implication
follows exchanging $\Phi$, $\Phi^{-1}$, $\Phi(S)$, $S$. For $r>0$, it holds

$0\leq{\mathscr{L}^{n}(S\cap B(x,\tfrac{r}{L}))}\leq
M^{n}\mathscr{L}^{n}(\Phi(S\cap B(x,\tfrac{r}{L})))\leq
M^{n}\mathscr{L}^{n}(\Phi(S)\cap B(\Phi(x),r)).$ (4.1)

As a consequence, if $\limsup_{r\to 0^{+}}\mathscr{L}^{n}(\Phi(S)\cap
B(\Phi(x),r))/\mathscr{L}^{n}(B(\Phi(x),r))=0$, then $\limsup_{r\to
0^{+}}\mathscr{L}^{n}(S\cap B(x,r))/\mathscr{L}^{n}(B(x,r))=0$. This proves
the statement. ∎

######  Lemma 12.

If $v$ is a Lipschitz vector field in $\mathbb{R}^{n}$ and
$S\subset\mathbb{R}^{n}$ is any set, then

$\text{for every $t>0$}\qquad\partial(S^{t})\subset(\partial
S)^{t}\qquad\text{and }\qquad\overline{(\partial S)^{t}}=(\partial S)^{t}.$
(4.2)

###### Proof.

We first observe that if $C\subseteq\mathbb{R}^{n}$ is closed, then $C^{t}$ is
closed. Consider any sequence $p_{n}$ in $C^{t}$ converging to some
$\overline{p}$. By definition $p_{n}=\Phi^{v}_{\tau_{n}}(x_{n})$ with
$|\tau_{n}|\leq|t|$ and $x_{n}\in C$. Recall that the inverse map
$\Psi(p,-\tau)=(\Phi^{v}_{\tau})^{-1}(p)$ is continuous in both argument.
Thus, eventually passing to a subsequence, let $(\tau_{n})_{n}$ converge to
some $\overline{\tau}$, then $x_{n}=\Psi(p_{n},-\tau_{n})$ converges to
$\overline{x}:=\Psi(\overline{p},-\overline{\tau})$. As $x_{n}\in C$ and $C$
is closed, we conclude that
$\overline{p}=\Phi^{v}_{\overline{\tau}}(\overline{x})\in C^{t}$.

In particular, $(\partial S)^{t}$ and $(\overline{S})^{t}$ are closed since
$\partial S$ and $\overline{S}$ are closed. This already proves the second
statement.

We now prove that $\partial(S^{t})\subset(\partial S)^{t}$. Since
$\Phi^{v}_{\tau}$ is an homeomorphism, then
$(\mathrm{int}(S))^{t}=\cup_{\tau\in[0,t]}\Phi^{v}_{\tau}(\mathrm{int}(S))$ is
open, being the union of open sets, so that
$(\mathrm{int}(S))^{t}\subset\mathrm{int}(S^{t})$. Since
$S^{t}\subset(\overline{S})^{t}$ by definition of evoluted set and
$(\overline{S})^{t}$ is closed, then
$\partial(S^{t})\subset(\overline{S})^{t}$. From
$(\mathrm{int}(S))^{t}\subset\mathrm{int}(S^{t})$ and
$\partial(S^{t})\subset(\overline{S})^{t}$ we get

$\partial(S^{t})=\partial(S^{t})\setminus\mathrm{int}(S^{t})\subset\partial(S^{t})\setminus(\mathrm{int}(S))^{t}\subset(\overline{S})^{t}\setminus(\mathrm{int}(S))^{t}\subset(\partial
S)^{t}\ .$

The last inclusion is just a consequence of the definition of evoluted set. ∎

######  Remark 13.

When we consider the case $t=+\infty$, we lack precisely the property of Lemma
12, see Example 10. Indeed, $(\partial S)^{\infty}$ is not anymore closed and
$\partial(S^{\infty})$, even though it is still subset of the closure of
$(\partial S)^{\infty}$, might contain parts of the closure of $(\partial
S)^{\infty}$ of positive measure.

We are now ready to prove our main result.

###### Proof of Theorem 1.

The case $t=0$ is trivial. Moreover, since
$\Phi^{v}_{\tau}=\Phi^{-v}_{-\tau}$, one can always assume $t>0$. Let us
define $I:=\mathrm{int}(S)$. By assumption we can write

$\partial S=N\cup B,$ (4.3)

where $\mathscr{H}^{n-1}(N)=0$ and

$\displaystyle B:=$ $\displaystyle\\{p\in\partial S:p\text{ is not a Lebesgue
point of }S^{\complement}\\}$ $\displaystyle=$ $\displaystyle\\{p\in\partial
I:p\text{ is not a Lebesgue point of }I^{\complement}\\}.$

The last equality follows from the fact that, since $\mathscr{L}^{n}(\partial
S)=0$, the Lebesgue points of $S^{\complement}$ are exactly those of
$I^{\complement}$. Since $\mathscr{H}^{n}(N\times\mathbb{R})=0$, Remark 6
implies that

$\mathscr{L}^{n}(N^{t})=\mathscr{L}^{n}(\Psi(N\times[0,t]))=0.$ (4.4)

We now prove that

$\mathscr{L}^{n}(B^{t}\setminus I^{t})=0.$ (4.5)

Fix $p\in B^{t}\setminus I^{t}$; by definition, there exist $x\in
B\subset\partial I$ and $\tau\in[0,t]$ such that $p=\Phi^{v}_{\tau}(x)$. Since
$x$ is not a Lebesgue point of $I^{\complement}$, then $p$ is not a Lebesgue
point of the open set $(\Phi^{v}_{\tau}(I))^{\complement}$. In particular

$\displaystyle\liminf_{r\to 0^{+}}\frac{\mathscr{L}^{n}((B^{t}\setminus
I^{t})\cap B(p,r))}{\mathscr{L}^{n}(B(p,r))}$ $\displaystyle\leq\liminf_{r\to
0^{+}}\frac{\mathscr{L}^{n}(B(p,r)\setminus I^{t})}{\mathscr{L}^{n}(B(p,r))}$
$\displaystyle\leq\liminf_{r\to
0^{+}}\frac{\mathscr{L}^{n}(B(p,r)\setminus\Phi^{v}_{\tau}(I))}{\mathscr{L}^{n}(B(p,r))}\
<\ 1.$

This proves that $B^{t}\setminus I^{t}$ has no Lebesgue points, thus (4.5)
follows.

It is now easy to conclude the proof: since $I^{t}$ is the union of the open
sets $\Phi^{v}_{\tau}(I)$, then $I^{t}$ is open and
$I^{t}\subset\mathrm{int}(S^{t})$. Using Lemma 12 and the decomposition (4.3),
the inclusions

$\partial(S^{t})=\partial(S^{t})\setminus\mathrm{int}(S^{t})\subset\partial(S^{t})\setminus
I^{t}\subset(\partial S)^{t}\setminus I^{t}\subset N^{t}\cup(B^{t}\setminus
I^{t}),$

together with (4.4) and (4.5), imply that
$\mathscr{L}^{n}(\partial(S^{t}))=0$. ∎

Let us state some consequences of our main result.

######  Corollary 14.

Let $S$ be any set that satisfies one of the following conditions:

  1. 1.

$\mathscr{H}^{n-1}$-almost every point of $\partial S$ is not a Lebesgue point
of $S^{\complement}$, as in Theorem 1;

  2. 2.

$S$ is a Lipschitz domain (see e.g. [10, Definition 4.4]);

  3. 3.

$S$ is a weakly Lipschitz domain (see e.g. [3, Definition 2.1]);

  4. 4.

$S$ is a set of finite perimeter for which $\mathscr{H}^{n-1}(\partial
S\setminus\partial_{*}S)=0$, where $\partial_{*}S$ denotes the reduced
boundary (see e.g. [2, Definition 3.54]).

Let $v$ satisfy one of the following conditions:

  1. (i)

$v$ is autonomous and globally Lipschitz, as in Theorem 1;

  2. (ii)

$v$ is autonomous locally Lipschitz continuous and with sub-linear growth
(i.e., $\|v(x)\|\leq M(1+\|x\|)$ for every $x\in\mathbb{R}^{n}$), with $S$
moreover bounded;

  3. (iii)

$v$ is a uniformly Lipschitz continuous and bounded time-dependent vector
field $v=v(x,t):\mathbb{R}^{n}\times\mathbb{R}\to\mathbb{R}^{n}$.

Then, $\mathscr{L}^{n}(\partial(S^{t}))=0$ for every $t\in\mathbb{R}$.

###### Proof.

It is easy to prove that each of the hypotheses on $S$ implies that
$\mathscr{H}^{n-1}$-almost every point of its boundary $\partial S$ is not a
Lebesgue point of $S^{\complement}$. For the conditions 2–3 this is a direct
consequence of the definition, by recalling that a Lipeomorphism maps an open
set $S$ onto an open set $\Phi(S)$, and the boundary $\partial S$ into the
boundary $\partial\Phi(S)$. For condition 4, it is enough to recall that
$\lim_{r\to 0^{+}}\mathscr{L}^{n}(S\cap B(x,r))/\mathscr{L}^{n}(B(x,r))=1/2$
for every $x\in\partial_{*}S$. See e.g. [2, Theorem 3.61].

As for the hypotheses on $v$, the non-autonomous case is similar to the
autonomous one, since the flow satisfies the same regularity hypotheses as
well as Grönwall’s inequality. For case (ii), it is sufficient to observe that
boundedness of $S$ and sublinearity of $v$ imply boundedness of $S^{t}$. Thus,
one can change $v$ outside $\overline{S^{t}}$ and recover the same result. ∎

######  Remark 15.

It is worth noticing that in the counterexamples to the negligibility of
$\partial S^{t}$ provided in Examples 8-9 and in [4], the open set $S$ has
finite perimeter but $\mathscr{H}^{n-1}(\partial S\setminus\partial_{*}S)>0$.
In fact, in these examples (many) points of $\partial S\setminus\partial_{*}S$
are Lebesgue points of $S^{\complement}$.

######  Remark 16.

The corollary above also includes our previous results related to this
problem. We proved in [4] that the statement holds when $S$ is bounded
$C^{1,1}$, while we proved in [9, Lemma 1.5] that it holds when $S$ is open
bounded and satisfying the uniform interior cone condition with $v$ of class
$C^{1}$.

######  Corollary 17.

Consider the evoluted set on the open time interval
$B^{t}:=\cup_{\tau\in(0,t)}\Phi_{\tau}^{v}(S)$. The statements of Theorem 1
and Corollary 14 hold in this case too.

###### Proof.

It is sufficient to observe that

$\partial(B^{t})\subset\partial(S^{t})\cup\partial
S\cup\partial(\Phi^{v}_{t}(S))$ (4.6)

implies

$\mathscr{L}^{n}(\partial(B^{t}))\leq\mathscr{L}^{n}(\partial(S^{t}))+\mathscr{L}^{n}(\partial
S)+\mathscr{L}^{n}(\partial(\Phi^{v}_{t}(S))=0,$

where we used our main result, as well as Proposition 11 and the fact that
$\mathscr{L}^{n}(\partial S)=0$. We also need to prove (4.6). First observe
that a point $b\in\partial(B^{t})$ is both the limit of a sequence
$b_{n}=\Phi_{\tau_{n}}^{v}(x_{n})$ with $(x_{n},\tau_{n})\in S\times(0,t)$ and
of a sequence $y_{n}\not\in S^{t}$. Since $\tau_{n}$ is bounded, we can
consider a subsequence (that we do not relabel) so that
$\tau_{n}\to\tau\in[0,t]$. We have three possibilities:

  * 1.

If $\tau\in(0,\tau)$ we already have the thesis, since $b$ is the limit of
both $y_{n}\not\in S^{t}$ and $b_{n}=\Phi_{\tau_{n}}^{v}(x_{n})$ which
definitively belongs to $S^{t}$.

  * 2.

If $\tau=0$, then $x_{n}=\Psi(b_{n},-\tau_{n})$ and $b_{n}=\Psi(b_{n},0)$ have
the same limit $b=\Psi(b,0)$ by continuity of $\Psi$ in $B(b,1)\times[-1,1]$.
Then $b$ is the limit of $x_{n}\in S$ and $y_{n}\not\in S^{t}\supset S$, then
$b\in\partial S$.

  * 3.

The case $\tau=t$ is similar to the previous case $\tau=0$: one can prove that
$b\in\partial\Phi^{v}_{t}(S)$.

∎

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