Show that $m(E)=1$.












2












$begingroup$


Let $E$ be measurable and $Esubset [0,1]$ .If for every $a,b$ with $0<a<b<1$ we have $m(Ecap (a,b))>frac{b-a}{4}$ then show that $m(E)=1$.



MY TRY::



Consider the sequence $a_n=frac{1}{n},b_n=1-frac{1}{n}$
Then $cup_{n=3}^infty (a_n,b_n)=(0,1)$



Now $E=Ecap (0,1)$



Then $m(E)=m(Ecap (cup_{n=3}^infty (a_n,b_n))=m(cup _{n=3}^infty (Ecap (a_n,b_n))=sum _{n=3}^infty m(Ecap (a_n,b_n))$



$gesum_{n=1}^infty dfrac{1-frac{2}{n}}{4}$



How to show that $sum_{n=1}^infty {1-frac{2}{n}}>4$



If I am able to show this I will be done.



Any help.






real-analysis measure-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's obvious that $sum(1-2/n)=+infty$, since the terms don't even tend to zero. (No, you're not done, there's a big error at the start.)
    $endgroup$
    – David C. Ullrich
    Dec 9 '18 at 14:52










  • $begingroup$
    Related math.stackexchange.com/q/2523417
    $endgroup$
    – Shashi
    Dec 9 '18 at 16:18
















2












$begingroup$


Let $E$ be measurable and $Esubset [0,1]$ .If for every $a,b$ with $0<a<b<1$ we have $m(Ecap (a,b))>frac{b-a}{4}$ then show that $m(E)=1$.



MY TRY::



Consider the sequence $a_n=frac{1}{n},b_n=1-frac{1}{n}$
Then $cup_{n=3}^infty (a_n,b_n)=(0,1)$



Now $E=Ecap (0,1)$



Then $m(E)=m(Ecap (cup_{n=3}^infty (a_n,b_n))=m(cup _{n=3}^infty (Ecap (a_n,b_n))=sum _{n=3}^infty m(Ecap (a_n,b_n))$



$gesum_{n=1}^infty dfrac{1-frac{2}{n}}{4}$



How to show that $sum_{n=1}^infty {1-frac{2}{n}}>4$



If I am able to show this I will be done.



Any help.






real-analysis measure-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's obvious that $sum(1-2/n)=+infty$, since the terms don't even tend to zero. (No, you're not done, there's a big error at the start.)
    $endgroup$
    – David C. Ullrich
    Dec 9 '18 at 14:52










  • $begingroup$
    Related math.stackexchange.com/q/2523417
    $endgroup$
    – Shashi
    Dec 9 '18 at 16:18














2












2








2


1



$begingroup$


Let $E$ be measurable and $Esubset [0,1]$ .If for every $a,b$ with $0<a<b<1$ we have $m(Ecap (a,b))>frac{b-a}{4}$ then show that $m(E)=1$.



MY TRY::



Consider the sequence $a_n=frac{1}{n},b_n=1-frac{1}{n}$
Then $cup_{n=3}^infty (a_n,b_n)=(0,1)$



Now $E=Ecap (0,1)$



Then $m(E)=m(Ecap (cup_{n=3}^infty (a_n,b_n))=m(cup _{n=3}^infty (Ecap (a_n,b_n))=sum _{n=3}^infty m(Ecap (a_n,b_n))$



$gesum_{n=1}^infty dfrac{1-frac{2}{n}}{4}$



How to show that $sum_{n=1}^infty {1-frac{2}{n}}>4$



If I am able to show this I will be done.



Any help.






real-analysis measure-theory







share|cite|improve this question











$endgroup$




Let $E$ be measurable and $Esubset [0,1]$ .If for every $a,b$ with $0<a<b<1$ we have $m(Ecap (a,b))>frac{b-a}{4}$ then show that $m(E)=1$.



MY TRY::



Consider the sequence $a_n=frac{1}{n},b_n=1-frac{1}{n}$
Then $cup_{n=3}^infty (a_n,b_n)=(0,1)$



Now $E=Ecap (0,1)$



Then $m(E)=m(Ecap (cup_{n=3}^infty (a_n,b_n))=m(cup _{n=3}^infty (Ecap (a_n,b_n))=sum _{n=3}^infty m(Ecap (a_n,b_n))$



$gesum_{n=1}^infty dfrac{1-frac{2}{n}}{4}$



How to show that $sum_{n=1}^infty {1-frac{2}{n}}>4$



If I am able to show this I will be done.



Any help.






real-analysis measure-theory




real-analysis measure-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 17:11









Dave L. Renfro

24.6k33981




24.6k33981










asked Dec 9 '18 at 9:06









Join_PhDJoin_PhD

3618




3618












  • $begingroup$
    It's obvious that $sum(1-2/n)=+infty$, since the terms don't even tend to zero. (No, you're not done, there's a big error at the start.)
    $endgroup$
    – David C. Ullrich
    Dec 9 '18 at 14:52










  • $begingroup$
    Related math.stackexchange.com/q/2523417
    $endgroup$
    – Shashi
    Dec 9 '18 at 16:18


















  • $begingroup$
    It's obvious that $sum(1-2/n)=+infty$, since the terms don't even tend to zero. (No, you're not done, there's a big error at the start.)
    $endgroup$
    – David C. Ullrich
    Dec 9 '18 at 14:52










  • $begingroup$
    Related math.stackexchange.com/q/2523417
    $endgroup$
    – Shashi
    Dec 9 '18 at 16:18
















$begingroup$
It's obvious that $sum(1-2/n)=+infty$, since the terms don't even tend to zero. (No, you're not done, there's a big error at the start.)
$endgroup$
– David C. Ullrich
Dec 9 '18 at 14:52




$begingroup$
It's obvious that $sum(1-2/n)=+infty$, since the terms don't even tend to zero. (No, you're not done, there's a big error at the start.)
$endgroup$
– David C. Ullrich
Dec 9 '18 at 14:52












$begingroup$
Related math.stackexchange.com/q/2523417
$endgroup$
– Shashi
Dec 9 '18 at 16:18




$begingroup$
Related math.stackexchange.com/q/2523417
$endgroup$
– Shashi
Dec 9 '18 at 16:18










2 Answers
2






active

oldest

votes


















3












$begingroup$

As already noted, this is immediate from the Lebesgue Differentiation Theorem. It's also easy to give a much more elementary proof.



If $I$ is an interval let $|I|$ denote the length of $I$. Let $F=[0,1]setminus E;$ we need to show $m(F)=0$.





Huge Hint: Suppose $FsubsetBbb R$ is measurable, $lambda>0$ and $m(Fcap I)le lambda |I|$ for every interval $I$. If $alpha>m(F)$ then $m(F)lelambdaalpha$.





Proof. Since Lebesgue measure is outer regular, or just "by definition of Lebesgue measure" depending on how it was defined, there exists a sequence $I_j$ of intervals such that $$Fsubsetbigcup I_j$$and $$sum|I_j|<alpha.$$
Now $$F=bigcup(Fcap I_j),$$hence $$m(F)lesum m(Fcap I_j)ledots..$$



It should be easy to finish the proof of the Huge Hint, and then to deduce that $m(F)=0$ if $lambda<1$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Unfortunately, your idea is not going to work, no matter if that sums is correct or not.



    Your crucial mistake is



    $$m(cup _{n=3}^infty (Ecap (a_n,b_n))=sum _{n=3}^infty m(Ecap (a_n,b_n))$$



    That formula for a measure of a union is valid only if the parts under the union are disjoint! This is absolutely not the case in your construction, as $(a_n,b_n) subset (a_{n+1},b_{n+1})$.



    That means you are measering the union of sets where each is a subset of the next, so for example, $$cup _{n=3}^4 (Ecap (a_n,b_n))=cup _{n=4}^4 (Ecap (a_n,b_n))= Ecap (a_4,b_4)$$



    and thus



    $$m(cup _{n=3}^4 (Ecap (a_n,b_n))= m(Ecap (a_4,b_4)) < m(Ecap (a_3,b_3)) + m(Ecap (a_4,b_4))$$



    I can't, at the moment, give a solution to the problem, but the way to do it must IMO go through increasinly smaller intervals, not, as you did, increasingly larger.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

      oldest

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      3












      $begingroup$

      As already noted, this is immediate from the Lebesgue Differentiation Theorem. It's also easy to give a much more elementary proof.



      If $I$ is an interval let $|I|$ denote the length of $I$. Let $F=[0,1]setminus E;$ we need to show $m(F)=0$.





      Huge Hint: Suppose $FsubsetBbb R$ is measurable, $lambda>0$ and $m(Fcap I)le lambda |I|$ for every interval $I$. If $alpha>m(F)$ then $m(F)lelambdaalpha$.





      Proof. Since Lebesgue measure is outer regular, or just "by definition of Lebesgue measure" depending on how it was defined, there exists a sequence $I_j$ of intervals such that $$Fsubsetbigcup I_j$$and $$sum|I_j|<alpha.$$
      Now $$F=bigcup(Fcap I_j),$$hence $$m(F)lesum m(Fcap I_j)ledots..$$



      It should be easy to finish the proof of the Huge Hint, and then to deduce that $m(F)=0$ if $lambda<1$.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        As already noted, this is immediate from the Lebesgue Differentiation Theorem. It's also easy to give a much more elementary proof.



        If $I$ is an interval let $|I|$ denote the length of $I$. Let $F=[0,1]setminus E;$ we need to show $m(F)=0$.





        Huge Hint: Suppose $FsubsetBbb R$ is measurable, $lambda>0$ and $m(Fcap I)le lambda |I|$ for every interval $I$. If $alpha>m(F)$ then $m(F)lelambdaalpha$.





        Proof. Since Lebesgue measure is outer regular, or just "by definition of Lebesgue measure" depending on how it was defined, there exists a sequence $I_j$ of intervals such that $$Fsubsetbigcup I_j$$and $$sum|I_j|<alpha.$$
        Now $$F=bigcup(Fcap I_j),$$hence $$m(F)lesum m(Fcap I_j)ledots..$$



        It should be easy to finish the proof of the Huge Hint, and then to deduce that $m(F)=0$ if $lambda<1$.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          As already noted, this is immediate from the Lebesgue Differentiation Theorem. It's also easy to give a much more elementary proof.



          If $I$ is an interval let $|I|$ denote the length of $I$. Let $F=[0,1]setminus E;$ we need to show $m(F)=0$.





          Huge Hint: Suppose $FsubsetBbb R$ is measurable, $lambda>0$ and $m(Fcap I)le lambda |I|$ for every interval $I$. If $alpha>m(F)$ then $m(F)lelambdaalpha$.





          Proof. Since Lebesgue measure is outer regular, or just "by definition of Lebesgue measure" depending on how it was defined, there exists a sequence $I_j$ of intervals such that $$Fsubsetbigcup I_j$$and $$sum|I_j|<alpha.$$
          Now $$F=bigcup(Fcap I_j),$$hence $$m(F)lesum m(Fcap I_j)ledots..$$



          It should be easy to finish the proof of the Huge Hint, and then to deduce that $m(F)=0$ if $lambda<1$.






          share|cite|improve this answer











          $endgroup$



          As already noted, this is immediate from the Lebesgue Differentiation Theorem. It's also easy to give a much more elementary proof.



          If $I$ is an interval let $|I|$ denote the length of $I$. Let $F=[0,1]setminus E;$ we need to show $m(F)=0$.





          Huge Hint: Suppose $FsubsetBbb R$ is measurable, $lambda>0$ and $m(Fcap I)le lambda |I|$ for every interval $I$. If $alpha>m(F)$ then $m(F)lelambdaalpha$.





          Proof. Since Lebesgue measure is outer regular, or just "by definition of Lebesgue measure" depending on how it was defined, there exists a sequence $I_j$ of intervals such that $$Fsubsetbigcup I_j$$and $$sum|I_j|<alpha.$$
          Now $$F=bigcup(Fcap I_j),$$hence $$m(F)lesum m(Fcap I_j)ledots..$$



          It should be easy to finish the proof of the Huge Hint, and then to deduce that $m(F)=0$ if $lambda<1$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 14:47

























          answered Dec 9 '18 at 14:41









          David C. UllrichDavid C. Ullrich

          59.9k43994




          59.9k43994























              2












              $begingroup$

              Unfortunately, your idea is not going to work, no matter if that sums is correct or not.



              Your crucial mistake is



              $$m(cup _{n=3}^infty (Ecap (a_n,b_n))=sum _{n=3}^infty m(Ecap (a_n,b_n))$$



              That formula for a measure of a union is valid only if the parts under the union are disjoint! This is absolutely not the case in your construction, as $(a_n,b_n) subset (a_{n+1},b_{n+1})$.



              That means you are measering the union of sets where each is a subset of the next, so for example, $$cup _{n=3}^4 (Ecap (a_n,b_n))=cup _{n=4}^4 (Ecap (a_n,b_n))= Ecap (a_4,b_4)$$



              and thus



              $$m(cup _{n=3}^4 (Ecap (a_n,b_n))= m(Ecap (a_4,b_4)) < m(Ecap (a_3,b_3)) + m(Ecap (a_4,b_4))$$



              I can't, at the moment, give a solution to the problem, but the way to do it must IMO go through increasinly smaller intervals, not, as you did, increasingly larger.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Unfortunately, your idea is not going to work, no matter if that sums is correct or not.



                Your crucial mistake is



                $$m(cup _{n=3}^infty (Ecap (a_n,b_n))=sum _{n=3}^infty m(Ecap (a_n,b_n))$$



                That formula for a measure of a union is valid only if the parts under the union are disjoint! This is absolutely not the case in your construction, as $(a_n,b_n) subset (a_{n+1},b_{n+1})$.



                That means you are measering the union of sets where each is a subset of the next, so for example, $$cup _{n=3}^4 (Ecap (a_n,b_n))=cup _{n=4}^4 (Ecap (a_n,b_n))= Ecap (a_4,b_4)$$



                and thus



                $$m(cup _{n=3}^4 (Ecap (a_n,b_n))= m(Ecap (a_4,b_4)) < m(Ecap (a_3,b_3)) + m(Ecap (a_4,b_4))$$



                I can't, at the moment, give a solution to the problem, but the way to do it must IMO go through increasinly smaller intervals, not, as you did, increasingly larger.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Unfortunately, your idea is not going to work, no matter if that sums is correct or not.



                  Your crucial mistake is



                  $$m(cup _{n=3}^infty (Ecap (a_n,b_n))=sum _{n=3}^infty m(Ecap (a_n,b_n))$$



                  That formula for a measure of a union is valid only if the parts under the union are disjoint! This is absolutely not the case in your construction, as $(a_n,b_n) subset (a_{n+1},b_{n+1})$.



                  That means you are measering the union of sets where each is a subset of the next, so for example, $$cup _{n=3}^4 (Ecap (a_n,b_n))=cup _{n=4}^4 (Ecap (a_n,b_n))= Ecap (a_4,b_4)$$



                  and thus



                  $$m(cup _{n=3}^4 (Ecap (a_n,b_n))= m(Ecap (a_4,b_4)) < m(Ecap (a_3,b_3)) + m(Ecap (a_4,b_4))$$



                  I can't, at the moment, give a solution to the problem, but the way to do it must IMO go through increasinly smaller intervals, not, as you did, increasingly larger.






                  share|cite|improve this answer









                  $endgroup$



                  Unfortunately, your idea is not going to work, no matter if that sums is correct or not.



                  Your crucial mistake is



                  $$m(cup _{n=3}^infty (Ecap (a_n,b_n))=sum _{n=3}^infty m(Ecap (a_n,b_n))$$



                  That formula for a measure of a union is valid only if the parts under the union are disjoint! This is absolutely not the case in your construction, as $(a_n,b_n) subset (a_{n+1},b_{n+1})$.



                  That means you are measering the union of sets where each is a subset of the next, so for example, $$cup _{n=3}^4 (Ecap (a_n,b_n))=cup _{n=4}^4 (Ecap (a_n,b_n))= Ecap (a_4,b_4)$$



                  and thus



                  $$m(cup _{n=3}^4 (Ecap (a_n,b_n))= m(Ecap (a_4,b_4)) < m(Ecap (a_3,b_3)) + m(Ecap (a_4,b_4))$$



                  I can't, at the moment, give a solution to the problem, but the way to do it must IMO go through increasinly smaller intervals, not, as you did, increasingly larger.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 9:25









                  IngixIngix

                  3,464146




                  3,464146






























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