$lim_{k to infty}v(E_k) = v(E)$












0












$begingroup$


$E_1,E_2, dots$ are measurable sets, $E subset E_i$ for all $i$ is also a measurable set and bounded.



In addition we are given that for all $k in mathbb N$ and for all $x in E_k$ there is a $y in E$ such that $|x-y| < frac{1}{k}$



We are asked to prove that $lim_{k to infty}v(E_k) = v(E)$, where $v$ stands for volume.



I understand the idea, $E_k$ gets closer and closer to $E$, until the maximum distance between $E_k$ and $E$ becomes infinitesimally small, but I don't know how to write a formal proof for this idea.



We are working with jordan measure here (as in, boundary is negligible)






calculus measure-theory volume







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I guess you are missing some condition. Take for example $$ E_k=(-1-1/k; 1+1/k)$$ then both $[-1;1]$ and $ [-1;1]cap mathbb{Q}$ are valide candidates for $E$, but they do not have the same measure.
    $endgroup$
    – Severin Schraven
    Jan 18 at 22:11












  • $begingroup$
    Maybe you want $E$ to be open? Then you could squeeze $$nu(E)leq nu(E_k) leq nu(E^{(1/k)})$$ where $$E^{(1/k)}={ x : d(x, E)leq 1/k}$$
    $endgroup$
    – Severin Schraven
    Jan 18 at 22:20












  • $begingroup$
    Nope, it's exactly as I wrote it. $E$ is not known to be anything but bounded and jordan measurable.
    $endgroup$
    – Rick Joker
    Jan 18 at 22:21










  • $begingroup$
    Why must $E$ be open to use squeeze? and why is the volume of $E^{(1/k)}$ approach zero?
    $endgroup$
    – Rick Joker
    Jan 18 at 22:24








  • 1




    $begingroup$
    @RickJoker You should be aware that when people say "measurable" they tend to mean "Lebesgue measurable" or measurable in an abstract measure space. Jordan measure isn't a measure in the modern sense of the word, and so if you ask a question about Jordan measurable sets, you need to specify that (and not just say measurable).
    $endgroup$
    – Aaron
    Jan 18 at 22:30
















0












$begingroup$


$E_1,E_2, dots$ are measurable sets, $E subset E_i$ for all $i$ is also a measurable set and bounded.



In addition we are given that for all $k in mathbb N$ and for all $x in E_k$ there is a $y in E$ such that $|x-y| < frac{1}{k}$



We are asked to prove that $lim_{k to infty}v(E_k) = v(E)$, where $v$ stands for volume.



I understand the idea, $E_k$ gets closer and closer to $E$, until the maximum distance between $E_k$ and $E$ becomes infinitesimally small, but I don't know how to write a formal proof for this idea.



We are working with jordan measure here (as in, boundary is negligible)






calculus measure-theory volume







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I guess you are missing some condition. Take for example $$ E_k=(-1-1/k; 1+1/k)$$ then both $[-1;1]$ and $ [-1;1]cap mathbb{Q}$ are valide candidates for $E$, but they do not have the same measure.
    $endgroup$
    – Severin Schraven
    Jan 18 at 22:11












  • $begingroup$
    Maybe you want $E$ to be open? Then you could squeeze $$nu(E)leq nu(E_k) leq nu(E^{(1/k)})$$ where $$E^{(1/k)}={ x : d(x, E)leq 1/k}$$
    $endgroup$
    – Severin Schraven
    Jan 18 at 22:20












  • $begingroup$
    Nope, it's exactly as I wrote it. $E$ is not known to be anything but bounded and jordan measurable.
    $endgroup$
    – Rick Joker
    Jan 18 at 22:21










  • $begingroup$
    Why must $E$ be open to use squeeze? and why is the volume of $E^{(1/k)}$ approach zero?
    $endgroup$
    – Rick Joker
    Jan 18 at 22:24








  • 1




    $begingroup$
    @RickJoker You should be aware that when people say "measurable" they tend to mean "Lebesgue measurable" or measurable in an abstract measure space. Jordan measure isn't a measure in the modern sense of the word, and so if you ask a question about Jordan measurable sets, you need to specify that (and not just say measurable).
    $endgroup$
    – Aaron
    Jan 18 at 22:30














0












0








0





$begingroup$


$E_1,E_2, dots$ are measurable sets, $E subset E_i$ for all $i$ is also a measurable set and bounded.



In addition we are given that for all $k in mathbb N$ and for all $x in E_k$ there is a $y in E$ such that $|x-y| < frac{1}{k}$



We are asked to prove that $lim_{k to infty}v(E_k) = v(E)$, where $v$ stands for volume.



I understand the idea, $E_k$ gets closer and closer to $E$, until the maximum distance between $E_k$ and $E$ becomes infinitesimally small, but I don't know how to write a formal proof for this idea.



We are working with jordan measure here (as in, boundary is negligible)






calculus measure-theory volume







share|cite|improve this question











$endgroup$




$E_1,E_2, dots$ are measurable sets, $E subset E_i$ for all $i$ is also a measurable set and bounded.



In addition we are given that for all $k in mathbb N$ and for all $x in E_k$ there is a $y in E$ such that $|x-y| < frac{1}{k}$



We are asked to prove that $lim_{k to infty}v(E_k) = v(E)$, where $v$ stands for volume.



I understand the idea, $E_k$ gets closer and closer to $E$, until the maximum distance between $E_k$ and $E$ becomes infinitesimally small, but I don't know how to write a formal proof for this idea.



We are working with jordan measure here (as in, boundary is negligible)






calculus measure-theory volume




calculus measure-theory volume






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 22:31







Rick Joker

















asked Jan 18 at 21:56









Rick JokerRick Joker

285214




285214








  • 1




    $begingroup$
    I guess you are missing some condition. Take for example $$ E_k=(-1-1/k; 1+1/k)$$ then both $[-1;1]$ and $ [-1;1]cap mathbb{Q}$ are valide candidates for $E$, but they do not have the same measure.
    $endgroup$
    – Severin Schraven
    Jan 18 at 22:11












  • $begingroup$
    Maybe you want $E$ to be open? Then you could squeeze $$nu(E)leq nu(E_k) leq nu(E^{(1/k)})$$ where $$E^{(1/k)}={ x : d(x, E)leq 1/k}$$
    $endgroup$
    – Severin Schraven
    Jan 18 at 22:20












  • $begingroup$
    Nope, it's exactly as I wrote it. $E$ is not known to be anything but bounded and jordan measurable.
    $endgroup$
    – Rick Joker
    Jan 18 at 22:21










  • $begingroup$
    Why must $E$ be open to use squeeze? and why is the volume of $E^{(1/k)}$ approach zero?
    $endgroup$
    – Rick Joker
    Jan 18 at 22:24








  • 1




    $begingroup$
    @RickJoker You should be aware that when people say "measurable" they tend to mean "Lebesgue measurable" or measurable in an abstract measure space. Jordan measure isn't a measure in the modern sense of the word, and so if you ask a question about Jordan measurable sets, you need to specify that (and not just say measurable).
    $endgroup$
    – Aaron
    Jan 18 at 22:30














  • 1




    $begingroup$
    I guess you are missing some condition. Take for example $$ E_k=(-1-1/k; 1+1/k)$$ then both $[-1;1]$ and $ [-1;1]cap mathbb{Q}$ are valide candidates for $E$, but they do not have the same measure.
    $endgroup$
    – Severin Schraven
    Jan 18 at 22:11












  • $begingroup$
    Maybe you want $E$ to be open? Then you could squeeze $$nu(E)leq nu(E_k) leq nu(E^{(1/k)})$$ where $$E^{(1/k)}={ x : d(x, E)leq 1/k}$$
    $endgroup$
    – Severin Schraven
    Jan 18 at 22:20












  • $begingroup$
    Nope, it's exactly as I wrote it. $E$ is not known to be anything but bounded and jordan measurable.
    $endgroup$
    – Rick Joker
    Jan 18 at 22:21










  • $begingroup$
    Why must $E$ be open to use squeeze? and why is the volume of $E^{(1/k)}$ approach zero?
    $endgroup$
    – Rick Joker
    Jan 18 at 22:24








  • 1




    $begingroup$
    @RickJoker You should be aware that when people say "measurable" they tend to mean "Lebesgue measurable" or measurable in an abstract measure space. Jordan measure isn't a measure in the modern sense of the word, and so if you ask a question about Jordan measurable sets, you need to specify that (and not just say measurable).
    $endgroup$
    – Aaron
    Jan 18 at 22:30








1




1




$begingroup$
I guess you are missing some condition. Take for example $$ E_k=(-1-1/k; 1+1/k)$$ then both $[-1;1]$ and $ [-1;1]cap mathbb{Q}$ are valide candidates for $E$, but they do not have the same measure.
$endgroup$
– Severin Schraven
Jan 18 at 22:11






$begingroup$
I guess you are missing some condition. Take for example $$ E_k=(-1-1/k; 1+1/k)$$ then both $[-1;1]$ and $ [-1;1]cap mathbb{Q}$ are valide candidates for $E$, but they do not have the same measure.
$endgroup$
– Severin Schraven
Jan 18 at 22:11














$begingroup$
Maybe you want $E$ to be open? Then you could squeeze $$nu(E)leq nu(E_k) leq nu(E^{(1/k)})$$ where $$E^{(1/k)}={ x : d(x, E)leq 1/k}$$
$endgroup$
– Severin Schraven
Jan 18 at 22:20






$begingroup$
Maybe you want $E$ to be open? Then you could squeeze $$nu(E)leq nu(E_k) leq nu(E^{(1/k)})$$ where $$E^{(1/k)}={ x : d(x, E)leq 1/k}$$
$endgroup$
– Severin Schraven
Jan 18 at 22:20














$begingroup$
Nope, it's exactly as I wrote it. $E$ is not known to be anything but bounded and jordan measurable.
$endgroup$
– Rick Joker
Jan 18 at 22:21




$begingroup$
Nope, it's exactly as I wrote it. $E$ is not known to be anything but bounded and jordan measurable.
$endgroup$
– Rick Joker
Jan 18 at 22:21












$begingroup$
Why must $E$ be open to use squeeze? and why is the volume of $E^{(1/k)}$ approach zero?
$endgroup$
– Rick Joker
Jan 18 at 22:24






$begingroup$
Why must $E$ be open to use squeeze? and why is the volume of $E^{(1/k)}$ approach zero?
$endgroup$
– Rick Joker
Jan 18 at 22:24






1




1




$begingroup$
@RickJoker You should be aware that when people say "measurable" they tend to mean "Lebesgue measurable" or measurable in an abstract measure space. Jordan measure isn't a measure in the modern sense of the word, and so if you ask a question about Jordan measurable sets, you need to specify that (and not just say measurable).
$endgroup$
– Aaron
Jan 18 at 22:30




$begingroup$
@RickJoker You should be aware that when people say "measurable" they tend to mean "Lebesgue measurable" or measurable in an abstract measure space. Jordan measure isn't a measure in the modern sense of the word, and so if you ask a question about Jordan measurable sets, you need to specify that (and not just say measurable).
$endgroup$
– Aaron
Jan 18 at 22:30










2 Answers
2






active

oldest

votes


















0












$begingroup$

I believe I may have found a solution.



I will attempt to prove that $lim_{k to infty}v(E_ksetminus E) = 0$ which is identical.



(*) Consider $x in E_k setminus E$. The closest point $y in E$ to it can't be an interior point, so infact $y in partial E$ and $|x-y| < frac{1}{k}$.



Let $epsilon > 0$, and $k$ such that $(frac{2}{k})^n < epsilon$.



Since $partial E$ is negligible, we can cover it by countably many boxes with sum of volumes less than epsilon. Cover $partial E$ by boxes with side length $frac{2}{k}$.



This is also a cover of $E_ksetminus E$ because of (*), and each box has volume $(frac{2}{k})^n < epsilon$, so each box is negligible, and so we actually found a countable union of negligible boxes that cover $E_k setminus E$, so it's negligible, which means it has zero volume.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Let $E^{(k)}$ denote ${x:|x-y| <frac 1 k text {for some}, y in E}$. Since the boundary of $E$ has measure $0$ it follows that $nu (E^{(k)}) to nu (E)$. [Indeed, $E^{(k)}$'s are decreasing and $cap_k E^{(k)}$ lies between $E$ and $overline {E}$]. Since $E_k subset E^{(k)}$ we get $nu (E_k) leq nu( E^{(k)})$ so $lim sup nu (E_k) leq nu(E)$. Since $E subset E^{(k)}$ for all $k$ we are done.






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      I believe I may have found a solution.



      I will attempt to prove that $lim_{k to infty}v(E_ksetminus E) = 0$ which is identical.



      (*) Consider $x in E_k setminus E$. The closest point $y in E$ to it can't be an interior point, so infact $y in partial E$ and $|x-y| < frac{1}{k}$.



      Let $epsilon > 0$, and $k$ such that $(frac{2}{k})^n < epsilon$.



      Since $partial E$ is negligible, we can cover it by countably many boxes with sum of volumes less than epsilon. Cover $partial E$ by boxes with side length $frac{2}{k}$.



      This is also a cover of $E_ksetminus E$ because of (*), and each box has volume $(frac{2}{k})^n < epsilon$, so each box is negligible, and so we actually found a countable union of negligible boxes that cover $E_k setminus E$, so it's negligible, which means it has zero volume.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        I believe I may have found a solution.



        I will attempt to prove that $lim_{k to infty}v(E_ksetminus E) = 0$ which is identical.



        (*) Consider $x in E_k setminus E$. The closest point $y in E$ to it can't be an interior point, so infact $y in partial E$ and $|x-y| < frac{1}{k}$.



        Let $epsilon > 0$, and $k$ such that $(frac{2}{k})^n < epsilon$.



        Since $partial E$ is negligible, we can cover it by countably many boxes with sum of volumes less than epsilon. Cover $partial E$ by boxes with side length $frac{2}{k}$.



        This is also a cover of $E_ksetminus E$ because of (*), and each box has volume $(frac{2}{k})^n < epsilon$, so each box is negligible, and so we actually found a countable union of negligible boxes that cover $E_k setminus E$, so it's negligible, which means it has zero volume.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          I believe I may have found a solution.



          I will attempt to prove that $lim_{k to infty}v(E_ksetminus E) = 0$ which is identical.



          (*) Consider $x in E_k setminus E$. The closest point $y in E$ to it can't be an interior point, so infact $y in partial E$ and $|x-y| < frac{1}{k}$.



          Let $epsilon > 0$, and $k$ such that $(frac{2}{k})^n < epsilon$.



          Since $partial E$ is negligible, we can cover it by countably many boxes with sum of volumes less than epsilon. Cover $partial E$ by boxes with side length $frac{2}{k}$.



          This is also a cover of $E_ksetminus E$ because of (*), and each box has volume $(frac{2}{k})^n < epsilon$, so each box is negligible, and so we actually found a countable union of negligible boxes that cover $E_k setminus E$, so it's negligible, which means it has zero volume.






          share|cite|improve this answer









          $endgroup$



          I believe I may have found a solution.



          I will attempt to prove that $lim_{k to infty}v(E_ksetminus E) = 0$ which is identical.



          (*) Consider $x in E_k setminus E$. The closest point $y in E$ to it can't be an interior point, so infact $y in partial E$ and $|x-y| < frac{1}{k}$.



          Let $epsilon > 0$, and $k$ such that $(frac{2}{k})^n < epsilon$.



          Since $partial E$ is negligible, we can cover it by countably many boxes with sum of volumes less than epsilon. Cover $partial E$ by boxes with side length $frac{2}{k}$.



          This is also a cover of $E_ksetminus E$ because of (*), and each box has volume $(frac{2}{k})^n < epsilon$, so each box is negligible, and so we actually found a countable union of negligible boxes that cover $E_k setminus E$, so it's negligible, which means it has zero volume.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 18 at 23:24









          Rick JokerRick Joker

          285214




          285214























              0












              $begingroup$

              Let $E^{(k)}$ denote ${x:|x-y| <frac 1 k text {for some}, y in E}$. Since the boundary of $E$ has measure $0$ it follows that $nu (E^{(k)}) to nu (E)$. [Indeed, $E^{(k)}$'s are decreasing and $cap_k E^{(k)}$ lies between $E$ and $overline {E}$]. Since $E_k subset E^{(k)}$ we get $nu (E_k) leq nu( E^{(k)})$ so $lim sup nu (E_k) leq nu(E)$. Since $E subset E^{(k)}$ for all $k$ we are done.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Let $E^{(k)}$ denote ${x:|x-y| <frac 1 k text {for some}, y in E}$. Since the boundary of $E$ has measure $0$ it follows that $nu (E^{(k)}) to nu (E)$. [Indeed, $E^{(k)}$'s are decreasing and $cap_k E^{(k)}$ lies between $E$ and $overline {E}$]. Since $E_k subset E^{(k)}$ we get $nu (E_k) leq nu( E^{(k)})$ so $lim sup nu (E_k) leq nu(E)$. Since $E subset E^{(k)}$ for all $k$ we are done.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let $E^{(k)}$ denote ${x:|x-y| <frac 1 k text {for some}, y in E}$. Since the boundary of $E$ has measure $0$ it follows that $nu (E^{(k)}) to nu (E)$. [Indeed, $E^{(k)}$'s are decreasing and $cap_k E^{(k)}$ lies between $E$ and $overline {E}$]. Since $E_k subset E^{(k)}$ we get $nu (E_k) leq nu( E^{(k)})$ so $lim sup nu (E_k) leq nu(E)$. Since $E subset E^{(k)}$ for all $k$ we are done.






                  share|cite|improve this answer









                  $endgroup$



                  Let $E^{(k)}$ denote ${x:|x-y| <frac 1 k text {for some}, y in E}$. Since the boundary of $E$ has measure $0$ it follows that $nu (E^{(k)}) to nu (E)$. [Indeed, $E^{(k)}$'s are decreasing and $cap_k E^{(k)}$ lies between $E$ and $overline {E}$]. Since $E_k subset E^{(k)}$ we get $nu (E_k) leq nu( E^{(k)})$ so $lim sup nu (E_k) leq nu(E)$. Since $E subset E^{(k)}$ for all $k$ we are done.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 18 at 23:28









                  Kavi Rama MurthyKavi Rama Murthy

                  76.4k53370




                  76.4k53370






























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