Lebesgue outer measure of a union of 2 sets












0












$begingroup$


Let $Esubsetmathbb{R}$.



Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$



Prove that $m^*(Ecup F)le m^*(E)+m^*(F)$ where $E$ and $F$ are any subsets of $mathbb{R}$.



My attempt:



My approach is to find a sequence of intervals ${K_n}$ of length $le m^*(E)+m^*(F)$ that covers $Ecup F$. Then we could say $m^*(Ecup F)lesum_{i=1}^inftyell(K_n)le m^*(E)+m^*(F)$.



Let ${I_n}$ be a sequence of intervals that covers $E$ such that $m^*(E)=sum_{n=1}^inftyell(I_n)$.



$Esubsetbigcup_{n=1}^infty I_n$



Let ${J_n}$ be a sequence of intervals that covers $F$ such that $m^*(F)=sum_{n=1}^inftyell(J_n)$.



$Fsubsetbigcup_{n=1}^infty J_n$



Let $K_n$ be the smallest interval that contains both $I_n$ and $J_n$. We claim that ${K_n}$ covers $Ecup F$. The trouble is that $sum_{n=1}^inftyell(K_n)notle m^*(E)+m^*(F)$. Is it possible to construct a suitable sequence of intervals from ${I_n}$ and ${J_n}$ that covers $Ecup F$ and has the appropriate measure?






real-analysis







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$endgroup$












  • $begingroup$
    I don't see why such ${I_n}$ should exist. The inf might not be achieved. This is true, for example, when $E = {x}$ is a single point.
    $endgroup$
    – mathworker21
    Dec 2 '18 at 21:02










  • $begingroup$
    @mathworker21 In that case, wouldn't the inf just be 0? Sorry, I don't know what you mean?
    $endgroup$
    – Thomas
    Dec 2 '18 at 21:39






  • 1




    $begingroup$
    @mathworker21 Oh, you mean there's no sequence of intervals with length 0 that covers $E={x}$!
    $endgroup$
    – Thomas
    Dec 2 '18 at 22:48
















0












$begingroup$


Let $Esubsetmathbb{R}$.



Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$



Prove that $m^*(Ecup F)le m^*(E)+m^*(F)$ where $E$ and $F$ are any subsets of $mathbb{R}$.



My attempt:



My approach is to find a sequence of intervals ${K_n}$ of length $le m^*(E)+m^*(F)$ that covers $Ecup F$. Then we could say $m^*(Ecup F)lesum_{i=1}^inftyell(K_n)le m^*(E)+m^*(F)$.



Let ${I_n}$ be a sequence of intervals that covers $E$ such that $m^*(E)=sum_{n=1}^inftyell(I_n)$.



$Esubsetbigcup_{n=1}^infty I_n$



Let ${J_n}$ be a sequence of intervals that covers $F$ such that $m^*(F)=sum_{n=1}^inftyell(J_n)$.



$Fsubsetbigcup_{n=1}^infty J_n$



Let $K_n$ be the smallest interval that contains both $I_n$ and $J_n$. We claim that ${K_n}$ covers $Ecup F$. The trouble is that $sum_{n=1}^inftyell(K_n)notle m^*(E)+m^*(F)$. Is it possible to construct a suitable sequence of intervals from ${I_n}$ and ${J_n}$ that covers $Ecup F$ and has the appropriate measure?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't see why such ${I_n}$ should exist. The inf might not be achieved. This is true, for example, when $E = {x}$ is a single point.
    $endgroup$
    – mathworker21
    Dec 2 '18 at 21:02










  • $begingroup$
    @mathworker21 In that case, wouldn't the inf just be 0? Sorry, I don't know what you mean?
    $endgroup$
    – Thomas
    Dec 2 '18 at 21:39






  • 1




    $begingroup$
    @mathworker21 Oh, you mean there's no sequence of intervals with length 0 that covers $E={x}$!
    $endgroup$
    – Thomas
    Dec 2 '18 at 22:48














0












0








0





$begingroup$


Let $Esubsetmathbb{R}$.



Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$



Prove that $m^*(Ecup F)le m^*(E)+m^*(F)$ where $E$ and $F$ are any subsets of $mathbb{R}$.



My attempt:



My approach is to find a sequence of intervals ${K_n}$ of length $le m^*(E)+m^*(F)$ that covers $Ecup F$. Then we could say $m^*(Ecup F)lesum_{i=1}^inftyell(K_n)le m^*(E)+m^*(F)$.



Let ${I_n}$ be a sequence of intervals that covers $E$ such that $m^*(E)=sum_{n=1}^inftyell(I_n)$.



$Esubsetbigcup_{n=1}^infty I_n$



Let ${J_n}$ be a sequence of intervals that covers $F$ such that $m^*(F)=sum_{n=1}^inftyell(J_n)$.



$Fsubsetbigcup_{n=1}^infty J_n$



Let $K_n$ be the smallest interval that contains both $I_n$ and $J_n$. We claim that ${K_n}$ covers $Ecup F$. The trouble is that $sum_{n=1}^inftyell(K_n)notle m^*(E)+m^*(F)$. Is it possible to construct a suitable sequence of intervals from ${I_n}$ and ${J_n}$ that covers $Ecup F$ and has the appropriate measure?






real-analysis







share|cite|improve this question











$endgroup$




Let $Esubsetmathbb{R}$.



Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$



Prove that $m^*(Ecup F)le m^*(E)+m^*(F)$ where $E$ and $F$ are any subsets of $mathbb{R}$.



My attempt:



My approach is to find a sequence of intervals ${K_n}$ of length $le m^*(E)+m^*(F)$ that covers $Ecup F$. Then we could say $m^*(Ecup F)lesum_{i=1}^inftyell(K_n)le m^*(E)+m^*(F)$.



Let ${I_n}$ be a sequence of intervals that covers $E$ such that $m^*(E)=sum_{n=1}^inftyell(I_n)$.



$Esubsetbigcup_{n=1}^infty I_n$



Let ${J_n}$ be a sequence of intervals that covers $F$ such that $m^*(F)=sum_{n=1}^inftyell(J_n)$.



$Fsubsetbigcup_{n=1}^infty J_n$



Let $K_n$ be the smallest interval that contains both $I_n$ and $J_n$. We claim that ${K_n}$ covers $Ecup F$. The trouble is that $sum_{n=1}^inftyell(K_n)notle m^*(E)+m^*(F)$. Is it possible to construct a suitable sequence of intervals from ${I_n}$ and ${J_n}$ that covers $Ecup F$ and has the appropriate measure?






real-analysis




real-analysis






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share|cite|improve this question













share|cite|improve this question




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edited Dec 2 '18 at 21:04







Thomas

















asked Dec 2 '18 at 20:59









ThomasThomas

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  • $begingroup$
    I don't see why such ${I_n}$ should exist. The inf might not be achieved. This is true, for example, when $E = {x}$ is a single point.
    $endgroup$
    – mathworker21
    Dec 2 '18 at 21:02










  • $begingroup$
    @mathworker21 In that case, wouldn't the inf just be 0? Sorry, I don't know what you mean?
    $endgroup$
    – Thomas
    Dec 2 '18 at 21:39






  • 1




    $begingroup$
    @mathworker21 Oh, you mean there's no sequence of intervals with length 0 that covers $E={x}$!
    $endgroup$
    – Thomas
    Dec 2 '18 at 22:48


















  • $begingroup$
    I don't see why such ${I_n}$ should exist. The inf might not be achieved. This is true, for example, when $E = {x}$ is a single point.
    $endgroup$
    – mathworker21
    Dec 2 '18 at 21:02










  • $begingroup$
    @mathworker21 In that case, wouldn't the inf just be 0? Sorry, I don't know what you mean?
    $endgroup$
    – Thomas
    Dec 2 '18 at 21:39






  • 1




    $begingroup$
    @mathworker21 Oh, you mean there's no sequence of intervals with length 0 that covers $E={x}$!
    $endgroup$
    – Thomas
    Dec 2 '18 at 22:48
















$begingroup$
I don't see why such ${I_n}$ should exist. The inf might not be achieved. This is true, for example, when $E = {x}$ is a single point.
$endgroup$
– mathworker21
Dec 2 '18 at 21:02




$begingroup$
I don't see why such ${I_n}$ should exist. The inf might not be achieved. This is true, for example, when $E = {x}$ is a single point.
$endgroup$
– mathworker21
Dec 2 '18 at 21:02












$begingroup$
@mathworker21 In that case, wouldn't the inf just be 0? Sorry, I don't know what you mean?
$endgroup$
– Thomas
Dec 2 '18 at 21:39




$begingroup$
@mathworker21 In that case, wouldn't the inf just be 0? Sorry, I don't know what you mean?
$endgroup$
– Thomas
Dec 2 '18 at 21:39




1




1




$begingroup$
@mathworker21 Oh, you mean there's no sequence of intervals with length 0 that covers $E={x}$!
$endgroup$
– Thomas
Dec 2 '18 at 22:48




$begingroup$
@mathworker21 Oh, you mean there's no sequence of intervals with length 0 that covers $E={x}$!
$endgroup$
– Thomas
Dec 2 '18 at 22:48










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$begingroup$

Suppose $I^E_k$ covers $E$ and $I^F_l$ covers $F$. Then
$I^E_k,I^F_l,...$ are a cover for $E cup F$ and
$m^* (E cup F) le sum_k l(I^E_k) + sum_l l(I^F_k)$.



Now take the $inf$ over the
covers $I^E_k$ and $I^F_l$.






share|cite|improve this answer









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    $begingroup$

    Suppose $I^E_k$ covers $E$ and $I^F_l$ covers $F$. Then
    $I^E_k,I^F_l,...$ are a cover for $E cup F$ and
    $m^* (E cup F) le sum_k l(I^E_k) + sum_l l(I^F_k)$.



    Now take the $inf$ over the
    covers $I^E_k$ and $I^F_l$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Suppose $I^E_k$ covers $E$ and $I^F_l$ covers $F$. Then
      $I^E_k,I^F_l,...$ are a cover for $E cup F$ and
      $m^* (E cup F) le sum_k l(I^E_k) + sum_l l(I^F_k)$.



      Now take the $inf$ over the
      covers $I^E_k$ and $I^F_l$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Suppose $I^E_k$ covers $E$ and $I^F_l$ covers $F$. Then
        $I^E_k,I^F_l,...$ are a cover for $E cup F$ and
        $m^* (E cup F) le sum_k l(I^E_k) + sum_l l(I^F_k)$.



        Now take the $inf$ over the
        covers $I^E_k$ and $I^F_l$.






        share|cite|improve this answer









        $endgroup$



        Suppose $I^E_k$ covers $E$ and $I^F_l$ covers $F$. Then
        $I^E_k,I^F_l,...$ are a cover for $E cup F$ and
        $m^* (E cup F) le sum_k l(I^E_k) + sum_l l(I^F_k)$.



        Now take the $inf$ over the
        covers $I^E_k$ and $I^F_l$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 21:12









        copper.hatcopper.hat

        127k559160




        127k559160






























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