Given rectangle $L subseteq mathbb{R}^n$ st. $v(L) = r$, can we cover $L$ by countably many rectangles $Q_i$...












0












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Consider a rectangle $L = [a_1, b_1 ] times dots times [a_n, b_n]subseteq mathbb{R}^n$ such that $v(L) = r$, can we cover $L$ by countably many rectangles $Q_i$ such that $sum_{i=1}^{infty}v(Q_i) < r$?




Note that $v(L) = (b_1 - a_1)dots (b_n-a_n)$. Now intuitively I think the answer to this is no, however I'm not really sure how I could prove this, in the sense that if I suppose that I can find such a covering I don't see any way to reach a contradiction.






real-analysis measure-theory







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  • $begingroup$
    What do you mean by covering?
    $endgroup$
    – BigbearZzz
    Dec 13 '18 at 5:59










  • $begingroup$
    @BigbearZzz I mean that $L subseteq bigcup_{i=1}^{infty} Q_i$
    $endgroup$
    – Perturbative
    Dec 13 '18 at 6:00










  • $begingroup$
    I presume that by $v(A)$ you mean the Lebesgue measure of $A$?
    $endgroup$
    – BigbearZzz
    Dec 13 '18 at 6:03
















0












$begingroup$



Consider a rectangle $L = [a_1, b_1 ] times dots times [a_n, b_n]subseteq mathbb{R}^n$ such that $v(L) = r$, can we cover $L$ by countably many rectangles $Q_i$ such that $sum_{i=1}^{infty}v(Q_i) < r$?




Note that $v(L) = (b_1 - a_1)dots (b_n-a_n)$. Now intuitively I think the answer to this is no, however I'm not really sure how I could prove this, in the sense that if I suppose that I can find such a covering I don't see any way to reach a contradiction.






real-analysis measure-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you mean by covering?
    $endgroup$
    – BigbearZzz
    Dec 13 '18 at 5:59










  • $begingroup$
    @BigbearZzz I mean that $L subseteq bigcup_{i=1}^{infty} Q_i$
    $endgroup$
    – Perturbative
    Dec 13 '18 at 6:00










  • $begingroup$
    I presume that by $v(A)$ you mean the Lebesgue measure of $A$?
    $endgroup$
    – BigbearZzz
    Dec 13 '18 at 6:03














0












0








0





$begingroup$



Consider a rectangle $L = [a_1, b_1 ] times dots times [a_n, b_n]subseteq mathbb{R}^n$ such that $v(L) = r$, can we cover $L$ by countably many rectangles $Q_i$ such that $sum_{i=1}^{infty}v(Q_i) < r$?




Note that $v(L) = (b_1 - a_1)dots (b_n-a_n)$. Now intuitively I think the answer to this is no, however I'm not really sure how I could prove this, in the sense that if I suppose that I can find such a covering I don't see any way to reach a contradiction.






real-analysis measure-theory







share|cite|improve this question









$endgroup$





Consider a rectangle $L = [a_1, b_1 ] times dots times [a_n, b_n]subseteq mathbb{R}^n$ such that $v(L) = r$, can we cover $L$ by countably many rectangles $Q_i$ such that $sum_{i=1}^{infty}v(Q_i) < r$?




Note that $v(L) = (b_1 - a_1)dots (b_n-a_n)$. Now intuitively I think the answer to this is no, however I'm not really sure how I could prove this, in the sense that if I suppose that I can find such a covering I don't see any way to reach a contradiction.






real-analysis measure-theory




real-analysis measure-theory






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asked Dec 13 '18 at 5:55









PerturbativePerturbative

4,35811552




4,35811552












  • $begingroup$
    What do you mean by covering?
    $endgroup$
    – BigbearZzz
    Dec 13 '18 at 5:59










  • $begingroup$
    @BigbearZzz I mean that $L subseteq bigcup_{i=1}^{infty} Q_i$
    $endgroup$
    – Perturbative
    Dec 13 '18 at 6:00










  • $begingroup$
    I presume that by $v(A)$ you mean the Lebesgue measure of $A$?
    $endgroup$
    – BigbearZzz
    Dec 13 '18 at 6:03


















  • $begingroup$
    What do you mean by covering?
    $endgroup$
    – BigbearZzz
    Dec 13 '18 at 5:59










  • $begingroup$
    @BigbearZzz I mean that $L subseteq bigcup_{i=1}^{infty} Q_i$
    $endgroup$
    – Perturbative
    Dec 13 '18 at 6:00










  • $begingroup$
    I presume that by $v(A)$ you mean the Lebesgue measure of $A$?
    $endgroup$
    – BigbearZzz
    Dec 13 '18 at 6:03
















$begingroup$
What do you mean by covering?
$endgroup$
– BigbearZzz
Dec 13 '18 at 5:59




$begingroup$
What do you mean by covering?
$endgroup$
– BigbearZzz
Dec 13 '18 at 5:59












$begingroup$
@BigbearZzz I mean that $L subseteq bigcup_{i=1}^{infty} Q_i$
$endgroup$
– Perturbative
Dec 13 '18 at 6:00




$begingroup$
@BigbearZzz I mean that $L subseteq bigcup_{i=1}^{infty} Q_i$
$endgroup$
– Perturbative
Dec 13 '18 at 6:00












$begingroup$
I presume that by $v(A)$ you mean the Lebesgue measure of $A$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 6:03




$begingroup$
I presume that by $v(A)$ you mean the Lebesgue measure of $A$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 6:03










2 Answers
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$begingroup$

No, of course this follows from the theory of Lebesgue measure.



If you don't want to appeal to that, then consider this. One can expand
the $Q_i$ slightly so they become open but still the sum of their measures
is less than $v(L)$. As $L$ is compact, Heine-Borel shows that a finite
number of the $Q_i$ cover $L$. Now it is quite elementary that
$sum_{i=1}^n v(Q_i)ge v(L)$ for a finite covering of rectangles.






share|cite|improve this answer









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    0












    $begingroup$

    I'll suppose that $v(A)$ coincides with the $n$-dimensional Lebesgue measure of $A$. Assuming we can find such a cover, then
    $$
    v(L) le vleft(bigcup_{i=1}^infty Q_iright)
    $$
    since $v$ is monotonic and $Lsubset cup_{i=1}^infty Q_i$. However, $v$ is also subadditive, i.e. $vleft(bigcup_{i=1}^infty A_iright) le sum_{i=1}^infty v(A_i)$ (and equality holds when $A_i$ are disjoint), hence
    $$
    r=v(L) le vleft(bigcup_{i=1}^infty Q_iright) le sum_{i=1}^infty v(Q_i) <r.
    $$
    This is a contradiction.



    Edit: If your definition of the function $v$ only involves rectangular set, then it is possible to extend $v$ so that its domain is a pretty large class of subsets of $Bbb R^n$ called the measurable sets. You can look up Caratheodory extension theorem if you want to know precisely what I meant.






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      1












      $begingroup$

      No, of course this follows from the theory of Lebesgue measure.



      If you don't want to appeal to that, then consider this. One can expand
      the $Q_i$ slightly so they become open but still the sum of their measures
      is less than $v(L)$. As $L$ is compact, Heine-Borel shows that a finite
      number of the $Q_i$ cover $L$. Now it is quite elementary that
      $sum_{i=1}^n v(Q_i)ge v(L)$ for a finite covering of rectangles.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        No, of course this follows from the theory of Lebesgue measure.



        If you don't want to appeal to that, then consider this. One can expand
        the $Q_i$ slightly so they become open but still the sum of their measures
        is less than $v(L)$. As $L$ is compact, Heine-Borel shows that a finite
        number of the $Q_i$ cover $L$. Now it is quite elementary that
        $sum_{i=1}^n v(Q_i)ge v(L)$ for a finite covering of rectangles.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          No, of course this follows from the theory of Lebesgue measure.



          If you don't want to appeal to that, then consider this. One can expand
          the $Q_i$ slightly so they become open but still the sum of their measures
          is less than $v(L)$. As $L$ is compact, Heine-Borel shows that a finite
          number of the $Q_i$ cover $L$. Now it is quite elementary that
          $sum_{i=1}^n v(Q_i)ge v(L)$ for a finite covering of rectangles.






          share|cite|improve this answer









          $endgroup$



          No, of course this follows from the theory of Lebesgue measure.



          If you don't want to appeal to that, then consider this. One can expand
          the $Q_i$ slightly so they become open but still the sum of their measures
          is less than $v(L)$. As $L$ is compact, Heine-Borel shows that a finite
          number of the $Q_i$ cover $L$. Now it is quite elementary that
          $sum_{i=1}^n v(Q_i)ge v(L)$ for a finite covering of rectangles.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 6:00









          Lord Shark the UnknownLord Shark the Unknown

          104k1160132




          104k1160132























              0












              $begingroup$

              I'll suppose that $v(A)$ coincides with the $n$-dimensional Lebesgue measure of $A$. Assuming we can find such a cover, then
              $$
              v(L) le vleft(bigcup_{i=1}^infty Q_iright)
              $$
              since $v$ is monotonic and $Lsubset cup_{i=1}^infty Q_i$. However, $v$ is also subadditive, i.e. $vleft(bigcup_{i=1}^infty A_iright) le sum_{i=1}^infty v(A_i)$ (and equality holds when $A_i$ are disjoint), hence
              $$
              r=v(L) le vleft(bigcup_{i=1}^infty Q_iright) le sum_{i=1}^infty v(Q_i) <r.
              $$
              This is a contradiction.



              Edit: If your definition of the function $v$ only involves rectangular set, then it is possible to extend $v$ so that its domain is a pretty large class of subsets of $Bbb R^n$ called the measurable sets. You can look up Caratheodory extension theorem if you want to know precisely what I meant.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                I'll suppose that $v(A)$ coincides with the $n$-dimensional Lebesgue measure of $A$. Assuming we can find such a cover, then
                $$
                v(L) le vleft(bigcup_{i=1}^infty Q_iright)
                $$
                since $v$ is monotonic and $Lsubset cup_{i=1}^infty Q_i$. However, $v$ is also subadditive, i.e. $vleft(bigcup_{i=1}^infty A_iright) le sum_{i=1}^infty v(A_i)$ (and equality holds when $A_i$ are disjoint), hence
                $$
                r=v(L) le vleft(bigcup_{i=1}^infty Q_iright) le sum_{i=1}^infty v(Q_i) <r.
                $$
                This is a contradiction.



                Edit: If your definition of the function $v$ only involves rectangular set, then it is possible to extend $v$ so that its domain is a pretty large class of subsets of $Bbb R^n$ called the measurable sets. You can look up Caratheodory extension theorem if you want to know precisely what I meant.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I'll suppose that $v(A)$ coincides with the $n$-dimensional Lebesgue measure of $A$. Assuming we can find such a cover, then
                  $$
                  v(L) le vleft(bigcup_{i=1}^infty Q_iright)
                  $$
                  since $v$ is monotonic and $Lsubset cup_{i=1}^infty Q_i$. However, $v$ is also subadditive, i.e. $vleft(bigcup_{i=1}^infty A_iright) le sum_{i=1}^infty v(A_i)$ (and equality holds when $A_i$ are disjoint), hence
                  $$
                  r=v(L) le vleft(bigcup_{i=1}^infty Q_iright) le sum_{i=1}^infty v(Q_i) <r.
                  $$
                  This is a contradiction.



                  Edit: If your definition of the function $v$ only involves rectangular set, then it is possible to extend $v$ so that its domain is a pretty large class of subsets of $Bbb R^n$ called the measurable sets. You can look up Caratheodory extension theorem if you want to know precisely what I meant.






                  share|cite|improve this answer











                  $endgroup$



                  I'll suppose that $v(A)$ coincides with the $n$-dimensional Lebesgue measure of $A$. Assuming we can find such a cover, then
                  $$
                  v(L) le vleft(bigcup_{i=1}^infty Q_iright)
                  $$
                  since $v$ is monotonic and $Lsubset cup_{i=1}^infty Q_i$. However, $v$ is also subadditive, i.e. $vleft(bigcup_{i=1}^infty A_iright) le sum_{i=1}^infty v(A_i)$ (and equality holds when $A_i$ are disjoint), hence
                  $$
                  r=v(L) le vleft(bigcup_{i=1}^infty Q_iright) le sum_{i=1}^infty v(Q_i) <r.
                  $$
                  This is a contradiction.



                  Edit: If your definition of the function $v$ only involves rectangular set, then it is possible to extend $v$ so that its domain is a pretty large class of subsets of $Bbb R^n$ called the measurable sets. You can look up Caratheodory extension theorem if you want to know precisely what I meant.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 13 '18 at 17:09

























                  answered Dec 13 '18 at 6:10









                  BigbearZzzBigbearZzz

                  8,72621652




                  8,72621652






























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