Show Regularity of two dimensional lebesgue measure












3












$begingroup$



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.










share|cite|improve this question











$endgroup$

















    3












    $begingroup$



    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.










    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$



      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.










      share|cite|improve this question











      $endgroup$





      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      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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