Show Regularity of two dimensional lebesgue measure
$begingroup$
I want to show the regularity of the $n$-dimensional lebesgue measure.
I figured, that if I can show it for the two-dimensional case, I should be able to generalize this result to the $n$-dimensional case.
More precisely I want to show that
begin{gather*}
m^n(A)=inf{m^n(V)mid Asubset V, V text{ open}} \
m^n(V)=sup{m^n(K)mid Ksubset V, K text{ compact}}.
end{gather*}
Prelims
It should be noted that
I know that the one dimensional lebesgue measure is regular. So I am hoping to be able to utilize this fact.
The way the $n$-dimensional lebesgue measure was formulated as the product measure of $n$ one-dimensional lebesgue measures. In particular, for the two dimensional case the formulation for $m^2$ I know is the following
Consider all measurable rectangles $Atimes Bin mathcal{B}timesmathcal{B}$, where $mathcal{B}$ is the Borel $sigma$-algebra on $mathbb{R}$. Consider the algebra of finite unions of disjoint measurable rectangles and define a premeasure $mu_{0}$ over this algebra as follows.
$$
mu_{0}left(bigcup_{i=1}^{k}(A_{i}times B_{I})right)
=sum_{i=1}^{n}m(A_{i})m(B_{i})$$
Then, using the caratheodory extension theorem, we get a measure on $mathcal{B}otimes mathcal{B}=sigma(mathcal{B}times mathcal{B})$.
The reason why I am saying all this is because I am aware that there are many formulations of the $n$-dimensional lebesgue measure, however I would like answers that appeal to this construction of the lebesgue measure (even though I am aware that all the formulations give the same thing, note I only have awareness, the only formulation I know is the one described above).
My attempt
My attempt for outer regularity:
I have tried first showing outer regularity on measurable rectangles.
Consider a measurable rectangles $Atimes Bin mathcal{B}times mathcal{B}$.
Then, since the lebesgue measure is outer regular, for $epsilon>0$ there exist open sets in mathbb{R} $V_{A}$ and $V_{B}$ such that $m(V_{A}-A)<epsilon$ and $m(V_{B}-B)<epsilon$.
Then, we have that
$$
m^2(V_{A}times V_{B}-Atimes B
)<epsilon^2 + epsilon m(A) + epsilon m(B).
$$
Provided $m(A)$ and $m(B)$ are finite, we can choose epsilon to be arbitrarily small and hence make $m^2(V_{A}times V_{B}-Atimes B)$ arbitrarily small.
I am concerned if one of $m(A)$ or $m(B)$ is infinity and the other is zero. In such a case $m^2(Atimes B)$ is defined to be zero and we cannot use the above argument to show outer regularity, what should I do?
Supposing that all worked though, I can see how this would work for general set in $mathcal{B}otimes mathcal{B}$. Since every element of $mathcal{B}otimes mathcal{B}$ is the countable union of measurable rectangles we can find open rectangles of the form $V_{i}times W_{i}in mathcal{B}times mathcal{B}$ such that
$$
m^2(V_{i}times W_{i}-A_{i}times B_{i})<frac{epsilon}{2^{i}}.
$$
My attempt for inner regularity:
Since there is a countable basis for the topology of $mathbb{R}times mathbb{R}$ we know that any open set in $mathbb{R}times mathbb{R}$ can be expressed as the countable union of elements of the form $V_{i}times W_{i}$ where $V_{i}$ and $W_{i}$ are in the countable basis for the topology of $mathbb{R}$.
Let $Uin mathbb{R}times mathbb{R}$ be an open set and let $U=bigcup_{i=1}^{infty}V_{i}times W_{i}$.
For each $V_{i}times W_{i}$ we can use a similar trick we used in the proof of outer regularity to find $K_{i}^1$ and $K_{i}^2$, which are compact sets, such that
$$
m^2(V_{i}times W_{i}-K_{i}^1times K_{i}^2)
<frac{epsilon}{2^{i}},
$$
which is all good, but the set $bigcup_{i=1}^{infty}K_{i}^1times K_{i}^2$ is not necessarily compact anymore.
Any help on how to solve this question is greatly appreciated. Sorry my attempt and explanation are a bit long.
measure-theory lebesgue-measure
$endgroup$
add a comment |
$begingroup$
I want to show the regularity of the $n$-dimensional lebesgue measure.
I figured, that if I can show it for the two-dimensional case, I should be able to generalize this result to the $n$-dimensional case.
More precisely I want to show that
begin{gather*}
m^n(A)=inf{m^n(V)mid Asubset V, V text{ open}} \
m^n(V)=sup{m^n(K)mid Ksubset V, K text{ compact}}.
end{gather*}
Prelims
It should be noted that
I know that the one dimensional lebesgue measure is regular. So I am hoping to be able to utilize this fact.
The way the $n$-dimensional lebesgue measure was formulated as the product measure of $n$ one-dimensional lebesgue measures. In particular, for the two dimensional case the formulation for $m^2$ I know is the following
Consider all measurable rectangles $Atimes Bin mathcal{B}timesmathcal{B}$, where $mathcal{B}$ is the Borel $sigma$-algebra on $mathbb{R}$. Consider the algebra of finite unions of disjoint measurable rectangles and define a premeasure $mu_{0}$ over this algebra as follows.
$$
mu_{0}left(bigcup_{i=1}^{k}(A_{i}times B_{I})right)
=sum_{i=1}^{n}m(A_{i})m(B_{i})$$
Then, using the caratheodory extension theorem, we get a measure on $mathcal{B}otimes mathcal{B}=sigma(mathcal{B}times mathcal{B})$.
The reason why I am saying all this is because I am aware that there are many formulations of the $n$-dimensional lebesgue measure, however I would like answers that appeal to this construction of the lebesgue measure (even though I am aware that all the formulations give the same thing, note I only have awareness, the only formulation I know is the one described above).
My attempt
My attempt for outer regularity:
I have tried first showing outer regularity on measurable rectangles.
Consider a measurable rectangles $Atimes Bin mathcal{B}times mathcal{B}$.
Then, since the lebesgue measure is outer regular, for $epsilon>0$ there exist open sets in mathbb{R} $V_{A}$ and $V_{B}$ such that $m(V_{A}-A)<epsilon$ and $m(V_{B}-B)<epsilon$.
Then, we have that
$$
m^2(V_{A}times V_{B}-Atimes B
)<epsilon^2 + epsilon m(A) + epsilon m(B).
$$
Provided $m(A)$ and $m(B)$ are finite, we can choose epsilon to be arbitrarily small and hence make $m^2(V_{A}times V_{B}-Atimes B)$ arbitrarily small.
I am concerned if one of $m(A)$ or $m(B)$ is infinity and the other is zero. In such a case $m^2(Atimes B)$ is defined to be zero and we cannot use the above argument to show outer regularity, what should I do?
Supposing that all worked though, I can see how this would work for general set in $mathcal{B}otimes mathcal{B}$. Since every element of $mathcal{B}otimes mathcal{B}$ is the countable union of measurable rectangles we can find open rectangles of the form $V_{i}times W_{i}in mathcal{B}times mathcal{B}$ such that
$$
m^2(V_{i}times W_{i}-A_{i}times B_{i})<frac{epsilon}{2^{i}}.
$$
My attempt for inner regularity:
Since there is a countable basis for the topology of $mathbb{R}times mathbb{R}$ we know that any open set in $mathbb{R}times mathbb{R}$ can be expressed as the countable union of elements of the form $V_{i}times W_{i}$ where $V_{i}$ and $W_{i}$ are in the countable basis for the topology of $mathbb{R}$.
Let $Uin mathbb{R}times mathbb{R}$ be an open set and let $U=bigcup_{i=1}^{infty}V_{i}times W_{i}$.
For each $V_{i}times W_{i}$ we can use a similar trick we used in the proof of outer regularity to find $K_{i}^1$ and $K_{i}^2$, which are compact sets, such that
$$
m^2(V_{i}times W_{i}-K_{i}^1times K_{i}^2)
<frac{epsilon}{2^{i}},
$$
which is all good, but the set $bigcup_{i=1}^{infty}K_{i}^1times K_{i}^2$ is not necessarily compact anymore.
Any help on how to solve this question is greatly appreciated. Sorry my attempt and explanation are a bit long.
measure-theory lebesgue-measure
$endgroup$
add a comment |
$begingroup$
I want to show the regularity of the $n$-dimensional lebesgue measure.
I figured, that if I can show it for the two-dimensional case, I should be able to generalize this result to the $n$-dimensional case.
More precisely I want to show that
begin{gather*}
m^n(A)=inf{m^n(V)mid Asubset V, V text{ open}} \
m^n(V)=sup{m^n(K)mid Ksubset V, K text{ compact}}.
end{gather*}
Prelims
It should be noted that
I know that the one dimensional lebesgue measure is regular. So I am hoping to be able to utilize this fact.
The way the $n$-dimensional lebesgue measure was formulated as the product measure of $n$ one-dimensional lebesgue measures. In particular, for the two dimensional case the formulation for $m^2$ I know is the following
Consider all measurable rectangles $Atimes Bin mathcal{B}timesmathcal{B}$, where $mathcal{B}$ is the Borel $sigma$-algebra on $mathbb{R}$. Consider the algebra of finite unions of disjoint measurable rectangles and define a premeasure $mu_{0}$ over this algebra as follows.
$$
mu_{0}left(bigcup_{i=1}^{k}(A_{i}times B_{I})right)
=sum_{i=1}^{n}m(A_{i})m(B_{i})$$
Then, using the caratheodory extension theorem, we get a measure on $mathcal{B}otimes mathcal{B}=sigma(mathcal{B}times mathcal{B})$.
The reason why I am saying all this is because I am aware that there are many formulations of the $n$-dimensional lebesgue measure, however I would like answers that appeal to this construction of the lebesgue measure (even though I am aware that all the formulations give the same thing, note I only have awareness, the only formulation I know is the one described above).
My attempt
My attempt for outer regularity:
I have tried first showing outer regularity on measurable rectangles.
Consider a measurable rectangles $Atimes Bin mathcal{B}times mathcal{B}$.
Then, since the lebesgue measure is outer regular, for $epsilon>0$ there exist open sets in mathbb{R} $V_{A}$ and $V_{B}$ such that $m(V_{A}-A)<epsilon$ and $m(V_{B}-B)<epsilon$.
Then, we have that
$$
m^2(V_{A}times V_{B}-Atimes B
)<epsilon^2 + epsilon m(A) + epsilon m(B).
$$
Provided $m(A)$ and $m(B)$ are finite, we can choose epsilon to be arbitrarily small and hence make $m^2(V_{A}times V_{B}-Atimes B)$ arbitrarily small.
I am concerned if one of $m(A)$ or $m(B)$ is infinity and the other is zero. In such a case $m^2(Atimes B)$ is defined to be zero and we cannot use the above argument to show outer regularity, what should I do?
Supposing that all worked though, I can see how this would work for general set in $mathcal{B}otimes mathcal{B}$. Since every element of $mathcal{B}otimes mathcal{B}$ is the countable union of measurable rectangles we can find open rectangles of the form $V_{i}times W_{i}in mathcal{B}times mathcal{B}$ such that
$$
m^2(V_{i}times W_{i}-A_{i}times B_{i})<frac{epsilon}{2^{i}}.
$$
My attempt for inner regularity:
Since there is a countable basis for the topology of $mathbb{R}times mathbb{R}$ we know that any open set in $mathbb{R}times mathbb{R}$ can be expressed as the countable union of elements of the form $V_{i}times W_{i}$ where $V_{i}$ and $W_{i}$ are in the countable basis for the topology of $mathbb{R}$.
Let $Uin mathbb{R}times mathbb{R}$ be an open set and let $U=bigcup_{i=1}^{infty}V_{i}times W_{i}$.
For each $V_{i}times W_{i}$ we can use a similar trick we used in the proof of outer regularity to find $K_{i}^1$ and $K_{i}^2$, which are compact sets, such that
$$
m^2(V_{i}times W_{i}-K_{i}^1times K_{i}^2)
<frac{epsilon}{2^{i}},
$$
which is all good, but the set $bigcup_{i=1}^{infty}K_{i}^1times K_{i}^2$ is not necessarily compact anymore.
Any help on how to solve this question is greatly appreciated. Sorry my attempt and explanation are a bit long.
measure-theory lebesgue-measure
$endgroup$
I want to show the regularity of the $n$-dimensional lebesgue measure.
I figured, that if I can show it for the two-dimensional case, I should be able to generalize this result to the $n$-dimensional case.
More precisely I want to show that
begin{gather*}
m^n(A)=inf{m^n(V)mid Asubset V, V text{ open}} \
m^n(V)=sup{m^n(K)mid Ksubset V, K text{ compact}}.
end{gather*}
Prelims
It should be noted that
I know that the one dimensional lebesgue measure is regular. So I am hoping to be able to utilize this fact.
The way the $n$-dimensional lebesgue measure was formulated as the product measure of $n$ one-dimensional lebesgue measures. In particular, for the two dimensional case the formulation for $m^2$ I know is the following
Consider all measurable rectangles $Atimes Bin mathcal{B}timesmathcal{B}$, where $mathcal{B}$ is the Borel $sigma$-algebra on $mathbb{R}$. Consider the algebra of finite unions of disjoint measurable rectangles and define a premeasure $mu_{0}$ over this algebra as follows.
$$
mu_{0}left(bigcup_{i=1}^{k}(A_{i}times B_{I})right)
=sum_{i=1}^{n}m(A_{i})m(B_{i})$$
Then, using the caratheodory extension theorem, we get a measure on $mathcal{B}otimes mathcal{B}=sigma(mathcal{B}times mathcal{B})$.
The reason why I am saying all this is because I am aware that there are many formulations of the $n$-dimensional lebesgue measure, however I would like answers that appeal to this construction of the lebesgue measure (even though I am aware that all the formulations give the same thing, note I only have awareness, the only formulation I know is the one described above).
My attempt
My attempt for outer regularity:
I have tried first showing outer regularity on measurable rectangles.
Consider a measurable rectangles $Atimes Bin mathcal{B}times mathcal{B}$.
Then, since the lebesgue measure is outer regular, for $epsilon>0$ there exist open sets in mathbb{R} $V_{A}$ and $V_{B}$ such that $m(V_{A}-A)<epsilon$ and $m(V_{B}-B)<epsilon$.
Then, we have that
$$
m^2(V_{A}times V_{B}-Atimes B
)<epsilon^2 + epsilon m(A) + epsilon m(B).
$$
Provided $m(A)$ and $m(B)$ are finite, we can choose epsilon to be arbitrarily small and hence make $m^2(V_{A}times V_{B}-Atimes B)$ arbitrarily small.
I am concerned if one of $m(A)$ or $m(B)$ is infinity and the other is zero. In such a case $m^2(Atimes B)$ is defined to be zero and we cannot use the above argument to show outer regularity, what should I do?
Supposing that all worked though, I can see how this would work for general set in $mathcal{B}otimes mathcal{B}$. Since every element of $mathcal{B}otimes mathcal{B}$ is the countable union of measurable rectangles we can find open rectangles of the form $V_{i}times W_{i}in mathcal{B}times mathcal{B}$ such that
$$
m^2(V_{i}times W_{i}-A_{i}times B_{i})<frac{epsilon}{2^{i}}.
$$
My attempt for inner regularity:
Since there is a countable basis for the topology of $mathbb{R}times mathbb{R}$ we know that any open set in $mathbb{R}times mathbb{R}$ can be expressed as the countable union of elements of the form $V_{i}times W_{i}$ where $V_{i}$ and $W_{i}$ are in the countable basis for the topology of $mathbb{R}$.
Let $Uin mathbb{R}times mathbb{R}$ be an open set and let $U=bigcup_{i=1}^{infty}V_{i}times W_{i}$.
For each $V_{i}times W_{i}$ we can use a similar trick we used in the proof of outer regularity to find $K_{i}^1$ and $K_{i}^2$, which are compact sets, such that
$$
m^2(V_{i}times W_{i}-K_{i}^1times K_{i}^2)
<frac{epsilon}{2^{i}},
$$
which is all good, but the set $bigcup_{i=1}^{infty}K_{i}^1times K_{i}^2$ is not necessarily compact anymore.
Any help on how to solve this question is greatly appreciated. Sorry my attempt and explanation are a bit long.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
edited Jan 10 at 16:13
Viktor Glombik
1,0931527
1,0931527
asked Jun 18 '17 at 9:13
DamoDamo
486210
486210
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