Proof of a convergence of sets in the context of Finite Perimeter sets
$begingroup$
Let $E subset mathbb{R}^n$ be a set of finite perimeter that satisfies $ mathcal{L}^n (E) < infty$. Assume that $E$ is symmetric with respect to the hyperplane ${x_n = 0}$. We know that there exists a sequence $(E_h)_h$ of bounded open sets having polyhedral boundary and that are symmetric with respect to the hyperplane ${x_n = 0}$ such that
$$
E_h to E quad text{as } h to infty quad text{and} quad lim_{h to infty}P(E_h) =P(E).
$$
and
$$lim_{h to infty} P(E_h; F) = P(E; F) text{ for all } Fsubset mathbb{R}^n text{ that satisfies } P(E; partial F )=0.$$
For all $h in mathbb{N}$ and $z in mathbb{R}^{n-1}$ let us set
begin{gather*}
(E_h)_z := { t in mathbb{R} : (z,t ) in E_h }, quad (E)_z := { t in mathbb{R} : (z,t ) in E } \
m_h (z) := mathcal{L}^1 ((E_h)_z) text{ with }z in mathbb{R}^{n-1} , quad G_h := { z in mathbb{R}^{n-1} : m_h (z) > 0 } \
m (z) := mathcal{L}^1 (E_z) text{ with }z in mathbb{R}^{n-1} , quad G := { z in mathbb{R}^{n-1} : m(z) > 0 } .
end{gather*}
I have to prove that (up to a subsequence) $G_h to G $ as $ h to infty$ (which means $lim_{h to infty} mathcal{L}^{n-1} (G_h triangle G)= 0$).
My attempt: if we prove that
$$lim_{h to infty} mathcal{L}^n ( G setminus G_h) =0 text{ and } lim_{h to infty} mathcal{L}^n ( G_h setminus G) =0 $$
then I have finished.
By Fubini's Theorem (and using the fact that the vertical slices of these sets are intervals) it can be showed that
begin{equation*}
begin{split}
mathcal{L}^n( & E_h triangle E) = int_{mathbb{R}^{n-1}} mathcal{L}^1 bigl( (E_h triangle E)_z bigr) , dz = int_{mathbb{R}^{n-1}} mathcal{L}^1 bigl( {E_h}_z triangle {E}_z bigr) , dz = \
& = int_{mathbb{R}^{n-1}} left| mathcal{L}^1 ({E_h}_z) - mathcal{L}^1 ({E}_z) right| , dz = int_{mathbb{R}^{n-1}} left| m_h(z) - m(z) right| , dz.
end{split}
end{equation*}
Therefore passing to the limit as $h to infty$ we get
$$0 = lim_{h to infty} mathcal{L}^n ( {E_h} triangle E) = lim_{h to infty} int_{mathbb{R}^{n-1}} left| m_h(z) - m(z) right| , dz. $$
Up to a subsequence, for a.e $z in mathbb{R}^{n-1}$, $m_h(z) to m(z) $. This proves that
$$ lim_{h to infty} mathcal{L}^n ( G setminus G_h) =0,$$
Indeed for a.e. $z in G$, since $m_h(z) to m(z) $ and, by definition of $G$, $m(z) >0$ , we have that $m_h (z) > 0$ if $h$ is great enough, hence $z in G_h$ if $h$ is great enough.
But we I don't know how to prove that $lim_{h to infty} mathcal{L}^n ( G_h setminus G) =0$. The above argument can't be adapted to this case because I think there could exist $z notin G$ (ie $m(z)=0$) such that $m_h (z) to m(z)$ but $m_h(z) >0$ for all $h$ (thus $z in G_h $ for all $h$).
I've been trying for hours but I can't conclude!
Thank you very much for any advice!!
real-analysis convergence lebesgue-measure geometric-measure-theory
asked Dec 7 '18 at 12:14
HermioneHermione
19619
$endgroup$
add a comment |
$begingroup$
Let $E subset mathbb{R}^n$ be a set of finite perimeter that satisfies $ mathcal{L}^n (E) < infty$. Assume that $E$ is symmetric with respect to the hyperplane ${x_n = 0}$. We know that there exists a sequence $(E_h)_h$ of bounded open sets having polyhedral boundary and that are symmetric with respect to the hyperplane ${x_n = 0}$ such that
$$
E_h to E quad text{as } h to infty quad text{and} quad lim_{h to infty}P(E_h) =P(E).
$$
and
$$lim_{h to infty} P(E_h; F) = P(E; F) text{ for all } Fsubset mathbb{R}^n text{ that satisfies } P(E; partial F )=0.$$
For all $h in mathbb{N}$ and $z in mathbb{R}^{n-1}$ let us set
begin{gather*}
(E_h)_z := { t in mathbb{R} : (z,t ) in E_h }, quad (E)_z := { t in mathbb{R} : (z,t ) in E } \
m_h (z) := mathcal{L}^1 ((E_h)_z) text{ with }z in mathbb{R}^{n-1} , quad G_h := { z in mathbb{R}^{n-1} : m_h (z) > 0 } \
m (z) := mathcal{L}^1 (E_z) text{ with }z in mathbb{R}^{n-1} , quad G := { z in mathbb{R}^{n-1} : m(z) > 0 } .
end{gather*}
I have to prove that (up to a subsequence) $G_h to G $ as $ h to infty$ (which means $lim_{h to infty} mathcal{L}^{n-1} (G_h triangle G)= 0$).
My attempt: if we prove that
$$lim_{h to infty} mathcal{L}^n ( G setminus G_h) =0 text{ and } lim_{h to infty} mathcal{L}^n ( G_h setminus G) =0 $$
then I have finished.
By Fubini's Theorem (and using the fact that the vertical slices of these sets are intervals) it can be showed that
begin{equation*}
begin{split}
mathcal{L}^n( & E_h triangle E) = int_{mathbb{R}^{n-1}} mathcal{L}^1 bigl( (E_h triangle E)_z bigr) , dz = int_{mathbb{R}^{n-1}} mathcal{L}^1 bigl( {E_h}_z triangle {E}_z bigr) , dz = \
& = int_{mathbb{R}^{n-1}} left| mathcal{L}^1 ({E_h}_z) - mathcal{L}^1 ({E}_z) right| , dz = int_{mathbb{R}^{n-1}} left| m_h(z) - m(z) right| , dz.
end{split}
end{equation*}
Therefore passing to the limit as $h to infty$ we get
$$0 = lim_{h to infty} mathcal{L}^n ( {E_h} triangle E) = lim_{h to infty} int_{mathbb{R}^{n-1}} left| m_h(z) - m(z) right| , dz. $$
Up to a subsequence, for a.e $z in mathbb{R}^{n-1}$, $m_h(z) to m(z) $. This proves that
$$ lim_{h to infty} mathcal{L}^n ( G setminus G_h) =0,$$
Indeed for a.e. $z in G$, since $m_h(z) to m(z) $ and, by definition of $G$, $m(z) >0$ , we have that $m_h (z) > 0$ if $h$ is great enough, hence $z in G_h$ if $h$ is great enough.
But we I don't know how to prove that $lim_{h to infty} mathcal{L}^n ( G_h setminus G) =0$. The above argument can't be adapted to this case because I think there could exist $z notin G$ (ie $m(z)=0$) such that $m_h (z) to m(z)$ but $m_h(z) >0$ for all $h$ (thus $z in G_h $ for all $h$).
I've been trying for hours but I can't conclude!
Thank you very much for any advice!!
real-analysis convergence lebesgue-measure geometric-measure-theory
asked Dec 7 '18 at 12:14
HermioneHermione
19619
$endgroup$
$begingroup$
Vertical slices of these sets aren't necessary intervals on $mathbb R$ (take a torus or a donut). You need $E_h$ convex in order to have $left(E_hright)_z$ an interval.
$endgroup$
– P De Donato
Jan 6 at 17:18
add a comment |
$begingroup$
Let $E subset mathbb{R}^n$ be a set of finite perimeter that satisfies $ mathcal{L}^n (E) < infty$. Assume that $E$ is symmetric with respect to the hyperplane ${x_n = 0}$. We know that there exists a sequence $(E_h)_h$ of bounded open sets having polyhedral boundary and that are symmetric with respect to the hyperplane ${x_n = 0}$ such that
$$
E_h to E quad text{as } h to infty quad text{and} quad lim_{h to infty}P(E_h) =P(E).
$$
and
$$lim_{h to infty} P(E_h; F) = P(E; F) text{ for all } Fsubset mathbb{R}^n text{ that satisfies } P(E; partial F )=0.$$
For all $h in mathbb{N}$ and $z in mathbb{R}^{n-1}$ let us set
begin{gather*}
(E_h)_z := { t in mathbb{R} : (z,t ) in E_h }, quad (E)_z := { t in mathbb{R} : (z,t ) in E } \
m_h (z) := mathcal{L}^1 ((E_h)_z) text{ with }z in mathbb{R}^{n-1} , quad G_h := { z in mathbb{R}^{n-1} : m_h (z) > 0 } \
m (z) := mathcal{L}^1 (E_z) text{ with }z in mathbb{R}^{n-1} , quad G := { z in mathbb{R}^{n-1} : m(z) > 0 } .
end{gather*}
I have to prove that (up to a subsequence) $G_h to G $ as $ h to infty$ (which means $lim_{h to infty} mathcal{L}^{n-1} (G_h triangle G)= 0$).
My attempt: if we prove that
$$lim_{h to infty} mathcal{L}^n ( G setminus G_h) =0 text{ and } lim_{h to infty} mathcal{L}^n ( G_h setminus G) =0 $$
then I have finished.
By Fubini's Theorem (and using the fact that the vertical slices of these sets are intervals) it can be showed that
begin{equation*}
begin{split}
mathcal{L}^n( & E_h triangle E) = int_{mathbb{R}^{n-1}} mathcal{L}^1 bigl( (E_h triangle E)_z bigr) , dz = int_{mathbb{R}^{n-1}} mathcal{L}^1 bigl( {E_h}_z triangle {E}_z bigr) , dz = \
& = int_{mathbb{R}^{n-1}} left| mathcal{L}^1 ({E_h}_z) - mathcal{L}^1 ({E}_z) right| , dz = int_{mathbb{R}^{n-1}} left| m_h(z) - m(z) right| , dz.
end{split}
end{equation*}
Therefore passing to the limit as $h to infty$ we get
$$0 = lim_{h to infty} mathcal{L}^n ( {E_h} triangle E) = lim_{h to infty} int_{mathbb{R}^{n-1}} left| m_h(z) - m(z) right| , dz. $$
Up to a subsequence, for a.e $z in mathbb{R}^{n-1}$, $m_h(z) to m(z) $. This proves that
$$ lim_{h to infty} mathcal{L}^n ( G setminus G_h) =0,$$
Indeed for a.e. $z in G$, since $m_h(z) to m(z) $ and, by definition of $G$, $m(z) >0$ , we have that $m_h (z) > 0$ if $h$ is great enough, hence $z in G_h$ if $h$ is great enough.
But we I don't know how to prove that $lim_{h to infty} mathcal{L}^n ( G_h setminus G) =0$. The above argument can't be adapted to this case because I think there could exist $z notin G$ (ie $m(z)=0$) such that $m_h (z) to m(z)$ but $m_h(z) >0$ for all $h$ (thus $z in G_h $ for all $h$).
I've been trying for hours but I can't conclude!
Thank you very much for any advice!!
real-analysis convergence lebesgue-measure geometric-measure-theory
asked Dec 7 '18 at 12:14
HermioneHermione
19619
$endgroup$
Let $E subset mathbb{R}^n$ be a set of finite perimeter that satisfies $ mathcal{L}^n (E) < infty$. Assume that $E$ is symmetric with respect to the hyperplane ${x_n = 0}$. We know that there exists a sequence $(E_h)_h$ of bounded open sets having polyhedral boundary and that are symmetric with respect to the hyperplane ${x_n = 0}$ such that
$$
E_h to E quad text{as } h to infty quad text{and} quad lim_{h to infty}P(E_h) =P(E).
$$
and
$$lim_{h to infty} P(E_h; F) = P(E; F) text{ for all } Fsubset mathbb{R}^n text{ that satisfies } P(E; partial F )=0.$$
For all $h in mathbb{N}$ and $z in mathbb{R}^{n-1}$ let us set
begin{gather*}
(E_h)_z := { t in mathbb{R} : (z,t ) in E_h }, quad (E)_z := { t in mathbb{R} : (z,t ) in E } \
m_h (z) := mathcal{L}^1 ((E_h)_z) text{ with }z in mathbb{R}^{n-1} , quad G_h := { z in mathbb{R}^{n-1} : m_h (z) > 0 } \
m (z) := mathcal{L}^1 (E_z) text{ with }z in mathbb{R}^{n-1} , quad G := { z in mathbb{R}^{n-1} : m(z) > 0 } .
end{gather*}
I have to prove that (up to a subsequence) $G_h to G $ as $ h to infty$ (which means $lim_{h to infty} mathcal{L}^{n-1} (G_h triangle G)= 0$).
My attempt: if we prove that
$$lim_{h to infty} mathcal{L}^n ( G setminus G_h) =0 text{ and } lim_{h to infty} mathcal{L}^n ( G_h setminus G) =0 $$
then I have finished.
By Fubini's Theorem (and using the fact that the vertical slices of these sets are intervals) it can be showed that
begin{equation*}
begin{split}
mathcal{L}^n( & E_h triangle E) = int_{mathbb{R}^{n-1}} mathcal{L}^1 bigl( (E_h triangle E)_z bigr) , dz = int_{mathbb{R}^{n-1}} mathcal{L}^1 bigl( {E_h}_z triangle {E}_z bigr) , dz = \
& = int_{mathbb{R}^{n-1}} left| mathcal{L}^1 ({E_h}_z) - mathcal{L}^1 ({E}_z) right| , dz = int_{mathbb{R}^{n-1}} left| m_h(z) - m(z) right| , dz.
end{split}
end{equation*}
Therefore passing to the limit as $h to infty$ we get
$$0 = lim_{h to infty} mathcal{L}^n ( {E_h} triangle E) = lim_{h to infty} int_{mathbb{R}^{n-1}} left| m_h(z) - m(z) right| , dz. $$
Up to a subsequence, for a.e $z in mathbb{R}^{n-1}$, $m_h(z) to m(z) $. This proves that
$$ lim_{h to infty} mathcal{L}^n ( G setminus G_h) =0,$$
Indeed for a.e. $z in G$, since $m_h(z) to m(z) $ and, by definition of $G$, $m(z) >0$ , we have that $m_h (z) > 0$ if $h$ is great enough, hence $z in G_h$ if $h$ is great enough.
But we I don't know how to prove that $lim_{h to infty} mathcal{L}^n ( G_h setminus G) =0$. The above argument can't be adapted to this case because I think there could exist $z notin G$ (ie $m(z)=0$) such that $m_h (z) to m(z)$ but $m_h(z) >0$ for all $h$ (thus $z in G_h $ for all $h$).
I've been trying for hours but I can't conclude!
Thank you very much for any advice!!
real-analysis convergence lebesgue-measure geometric-measure-theory
real-analysis convergence lebesgue-measure geometric-measure-theory
asked Dec 7 '18 at 12:14
HermioneHermione
19619
asked Dec 7 '18 at 12:14
HermioneHermione
19619
edited Dec 8 '18 at 14:11
Hermione
asked Dec 7 '18 at 12:14
HermioneHermione
19619
asked Dec 7 '18 at 12:14
HermioneHermione
19619
asked Dec 7 '18 at 12:14
HermioneHermione
19619
19619
$begingroup$
Vertical slices of these sets aren't necessary intervals on $mathbb R$ (take a torus or a donut). You need $E_h$ convex in order to have $left(E_hright)_z$ an interval.
$endgroup$
– P De Donato
Jan 6 at 17:18
add a comment |
$begingroup$
Vertical slices of these sets aren't necessary intervals on $mathbb R$ (take a torus or a donut). You need $E_h$ convex in order to have $left(E_hright)_z$ an interval.
$endgroup$
– P De Donato
Jan 6 at 17:18
$begingroup$
Vertical slices of these sets aren't necessary intervals on $mathbb R$ (take a torus or a donut). You need $E_h$ convex in order to have $left(E_hright)_z$ an interval.
$endgroup$
– P De Donato
Jan 6 at 17:18
$begingroup$
Vertical slices of these sets aren't necessary intervals on $mathbb R$ (take a torus or a donut). You need $E_h$ convex in order to have $left(E_hright)_z$ an interval.
$endgroup$
– P De Donato
Jan 6 at 17:18
add a comment |
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$begingroup$
Vertical slices of these sets aren't necessary intervals on $mathbb R$ (take a torus or a donut). You need $E_h$ convex in order to have $left(E_hright)_z$ an interval.
$endgroup$
– P De Donato
Jan 6 at 17:18