A counterexample to the epsilon-delta criterion for Absolute Continuity of Measures












1












$begingroup$


Let $p>0$, and let $mu$ be a Borel measure on $[0,infty)$ defined by $mu(E)=int_Ex^pdlambda$ where $lambda$ denotes Lebesgue measure. Show that $mu$ is absolutely continuous with respect to $lambda$, but $mu$ does not meet the epsilon-delta criterion for absolute continuity, namely for every $epsilon>0$ there's a $delta>0$ such that $mu(E)<epsilon$ whenever $lambda(E)<delta$.



I've managed to prove that $mu$ is absolutely continuous with respect to $lambda$, but I'm not sure how to approach the second part. I basically need to find a sequence of sets $E_n$ (in $B([1,infty)$) such that $lambda(E_n)rightarrow 0$ but $int_{E_n}x^pdlambda$ does not go to zero. But I can't even think of such a sequence in the case where $p=1$, let alone the general case.






measure-theory lebesgue-integral absolute-continuity borel-measures radon-nikodym







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$endgroup$












  • $begingroup$
    The nastiness is occurring because the interval is unbounded.
    $endgroup$
    – ncmathsadist
    Dec 5 '18 at 15:34
















1












$begingroup$


Let $p>0$, and let $mu$ be a Borel measure on $[0,infty)$ defined by $mu(E)=int_Ex^pdlambda$ where $lambda$ denotes Lebesgue measure. Show that $mu$ is absolutely continuous with respect to $lambda$, but $mu$ does not meet the epsilon-delta criterion for absolute continuity, namely for every $epsilon>0$ there's a $delta>0$ such that $mu(E)<epsilon$ whenever $lambda(E)<delta$.



I've managed to prove that $mu$ is absolutely continuous with respect to $lambda$, but I'm not sure how to approach the second part. I basically need to find a sequence of sets $E_n$ (in $B([1,infty)$) such that $lambda(E_n)rightarrow 0$ but $int_{E_n}x^pdlambda$ does not go to zero. But I can't even think of such a sequence in the case where $p=1$, let alone the general case.






measure-theory lebesgue-integral absolute-continuity borel-measures radon-nikodym







share|cite|improve this question











$endgroup$












  • $begingroup$
    The nastiness is occurring because the interval is unbounded.
    $endgroup$
    – ncmathsadist
    Dec 5 '18 at 15:34














1












1








1


1



$begingroup$


Let $p>0$, and let $mu$ be a Borel measure on $[0,infty)$ defined by $mu(E)=int_Ex^pdlambda$ where $lambda$ denotes Lebesgue measure. Show that $mu$ is absolutely continuous with respect to $lambda$, but $mu$ does not meet the epsilon-delta criterion for absolute continuity, namely for every $epsilon>0$ there's a $delta>0$ such that $mu(E)<epsilon$ whenever $lambda(E)<delta$.



I've managed to prove that $mu$ is absolutely continuous with respect to $lambda$, but I'm not sure how to approach the second part. I basically need to find a sequence of sets $E_n$ (in $B([1,infty)$) such that $lambda(E_n)rightarrow 0$ but $int_{E_n}x^pdlambda$ does not go to zero. But I can't even think of such a sequence in the case where $p=1$, let alone the general case.






measure-theory lebesgue-integral absolute-continuity borel-measures radon-nikodym







share|cite|improve this question











$endgroup$




Let $p>0$, and let $mu$ be a Borel measure on $[0,infty)$ defined by $mu(E)=int_Ex^pdlambda$ where $lambda$ denotes Lebesgue measure. Show that $mu$ is absolutely continuous with respect to $lambda$, but $mu$ does not meet the epsilon-delta criterion for absolute continuity, namely for every $epsilon>0$ there's a $delta>0$ such that $mu(E)<epsilon$ whenever $lambda(E)<delta$.



I've managed to prove that $mu$ is absolutely continuous with respect to $lambda$, but I'm not sure how to approach the second part. I basically need to find a sequence of sets $E_n$ (in $B([1,infty)$) such that $lambda(E_n)rightarrow 0$ but $int_{E_n}x^pdlambda$ does not go to zero. But I can't even think of such a sequence in the case where $p=1$, let alone the general case.






measure-theory lebesgue-integral absolute-continuity borel-measures radon-nikodym




measure-theory lebesgue-integral absolute-continuity borel-measures radon-nikodym






share|cite|improve this question















share|cite|improve this question













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edited Dec 6 '18 at 14:27







Keshav Srinivasan

















asked Dec 5 '18 at 15:26









Keshav SrinivasanKeshav Srinivasan

2,08511443




2,08511443












  • $begingroup$
    The nastiness is occurring because the interval is unbounded.
    $endgroup$
    – ncmathsadist
    Dec 5 '18 at 15:34


















  • $begingroup$
    The nastiness is occurring because the interval is unbounded.
    $endgroup$
    – ncmathsadist
    Dec 5 '18 at 15:34
















$begingroup$
The nastiness is occurring because the interval is unbounded.
$endgroup$
– ncmathsadist
Dec 5 '18 at 15:34




$begingroup$
The nastiness is occurring because the interval is unbounded.
$endgroup$
– ncmathsadist
Dec 5 '18 at 15:34










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hopefully the case $p=1$ will help:



$$lambda([a,b])= b-a$$
and
$$
mu([a,b]) = frac{1}{2}(b^2-a^2) = frac{1}{2}(b-a)(b+a)
$$



Taking, for example, $a=3^n$ and $b=3^n+frac{1}{2^n}$ will give you sets $E_n$ which you can show the desired properties of.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, the $p=1$ is clear. But how do you generalize it, given that you can't factor in the general case?
    $endgroup$
    – Keshav Srinivasan
    Dec 6 '18 at 5:59










  • $begingroup$
    Are you sure? E.g. $5^3 -4^3 = y^2-x^2$ for certain $y,x$. I haven't fully worked it out but it might work just as well, although you'd have to think of a different inequality.
    $endgroup$
    – user25959
    Dec 6 '18 at 12:12










  • $begingroup$
    I'm thinking of the case when $p$ is not an integer.
    $endgroup$
    – Keshav Srinivasan
    Dec 6 '18 at 14:26












  • $begingroup$
    It doesn't have to be. Think about how to write $x^p$ as a square.no integer assumption needed.
    $endgroup$
    – user25959
    Dec 6 '18 at 15:44



















1












$begingroup$

The issue is more or less that $x^{p+1}$ is not a uniformly continuous function: it grows faster and faster as $x to infty$, which means that $mu(I)$ can be large even when $lambda(I)$ is small if $I$ is located far to the right.



Being more precise, consider $delta>0$ arbitrary and then try computing $mu([delta^{-1/p},delta^{-1/p}+delta])$.






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









    1












    $begingroup$

    Hopefully the case $p=1$ will help:



    $$lambda([a,b])= b-a$$
    and
    $$
    mu([a,b]) = frac{1}{2}(b^2-a^2) = frac{1}{2}(b-a)(b+a)
    $$



    Taking, for example, $a=3^n$ and $b=3^n+frac{1}{2^n}$ will give you sets $E_n$ which you can show the desired properties of.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks, the $p=1$ is clear. But how do you generalize it, given that you can't factor in the general case?
      $endgroup$
      – Keshav Srinivasan
      Dec 6 '18 at 5:59










    • $begingroup$
      Are you sure? E.g. $5^3 -4^3 = y^2-x^2$ for certain $y,x$. I haven't fully worked it out but it might work just as well, although you'd have to think of a different inequality.
      $endgroup$
      – user25959
      Dec 6 '18 at 12:12










    • $begingroup$
      I'm thinking of the case when $p$ is not an integer.
      $endgroup$
      – Keshav Srinivasan
      Dec 6 '18 at 14:26












    • $begingroup$
      It doesn't have to be. Think about how to write $x^p$ as a square.no integer assumption needed.
      $endgroup$
      – user25959
      Dec 6 '18 at 15:44
















    1












    $begingroup$

    Hopefully the case $p=1$ will help:



    $$lambda([a,b])= b-a$$
    and
    $$
    mu([a,b]) = frac{1}{2}(b^2-a^2) = frac{1}{2}(b-a)(b+a)
    $$



    Taking, for example, $a=3^n$ and $b=3^n+frac{1}{2^n}$ will give you sets $E_n$ which you can show the desired properties of.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks, the $p=1$ is clear. But how do you generalize it, given that you can't factor in the general case?
      $endgroup$
      – Keshav Srinivasan
      Dec 6 '18 at 5:59










    • $begingroup$
      Are you sure? E.g. $5^3 -4^3 = y^2-x^2$ for certain $y,x$. I haven't fully worked it out but it might work just as well, although you'd have to think of a different inequality.
      $endgroup$
      – user25959
      Dec 6 '18 at 12:12










    • $begingroup$
      I'm thinking of the case when $p$ is not an integer.
      $endgroup$
      – Keshav Srinivasan
      Dec 6 '18 at 14:26












    • $begingroup$
      It doesn't have to be. Think about how to write $x^p$ as a square.no integer assumption needed.
      $endgroup$
      – user25959
      Dec 6 '18 at 15:44














    1












    1








    1





    $begingroup$

    Hopefully the case $p=1$ will help:



    $$lambda([a,b])= b-a$$
    and
    $$
    mu([a,b]) = frac{1}{2}(b^2-a^2) = frac{1}{2}(b-a)(b+a)
    $$



    Taking, for example, $a=3^n$ and $b=3^n+frac{1}{2^n}$ will give you sets $E_n$ which you can show the desired properties of.






    share|cite|improve this answer









    $endgroup$



    Hopefully the case $p=1$ will help:



    $$lambda([a,b])= b-a$$
    and
    $$
    mu([a,b]) = frac{1}{2}(b^2-a^2) = frac{1}{2}(b-a)(b+a)
    $$



    Taking, for example, $a=3^n$ and $b=3^n+frac{1}{2^n}$ will give you sets $E_n$ which you can show the desired properties of.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 5 '18 at 15:34









    user25959user25959

    1,573816




    1,573816












    • $begingroup$
      Thanks, the $p=1$ is clear. But how do you generalize it, given that you can't factor in the general case?
      $endgroup$
      – Keshav Srinivasan
      Dec 6 '18 at 5:59










    • $begingroup$
      Are you sure? E.g. $5^3 -4^3 = y^2-x^2$ for certain $y,x$. I haven't fully worked it out but it might work just as well, although you'd have to think of a different inequality.
      $endgroup$
      – user25959
      Dec 6 '18 at 12:12










    • $begingroup$
      I'm thinking of the case when $p$ is not an integer.
      $endgroup$
      – Keshav Srinivasan
      Dec 6 '18 at 14:26












    • $begingroup$
      It doesn't have to be. Think about how to write $x^p$ as a square.no integer assumption needed.
      $endgroup$
      – user25959
      Dec 6 '18 at 15:44


















    • $begingroup$
      Thanks, the $p=1$ is clear. But how do you generalize it, given that you can't factor in the general case?
      $endgroup$
      – Keshav Srinivasan
      Dec 6 '18 at 5:59










    • $begingroup$
      Are you sure? E.g. $5^3 -4^3 = y^2-x^2$ for certain $y,x$. I haven't fully worked it out but it might work just as well, although you'd have to think of a different inequality.
      $endgroup$
      – user25959
      Dec 6 '18 at 12:12










    • $begingroup$
      I'm thinking of the case when $p$ is not an integer.
      $endgroup$
      – Keshav Srinivasan
      Dec 6 '18 at 14:26












    • $begingroup$
      It doesn't have to be. Think about how to write $x^p$ as a square.no integer assumption needed.
      $endgroup$
      – user25959
      Dec 6 '18 at 15:44
















    $begingroup$
    Thanks, the $p=1$ is clear. But how do you generalize it, given that you can't factor in the general case?
    $endgroup$
    – Keshav Srinivasan
    Dec 6 '18 at 5:59




    $begingroup$
    Thanks, the $p=1$ is clear. But how do you generalize it, given that you can't factor in the general case?
    $endgroup$
    – Keshav Srinivasan
    Dec 6 '18 at 5:59












    $begingroup$
    Are you sure? E.g. $5^3 -4^3 = y^2-x^2$ for certain $y,x$. I haven't fully worked it out but it might work just as well, although you'd have to think of a different inequality.
    $endgroup$
    – user25959
    Dec 6 '18 at 12:12




    $begingroup$
    Are you sure? E.g. $5^3 -4^3 = y^2-x^2$ for certain $y,x$. I haven't fully worked it out but it might work just as well, although you'd have to think of a different inequality.
    $endgroup$
    – user25959
    Dec 6 '18 at 12:12












    $begingroup$
    I'm thinking of the case when $p$ is not an integer.
    $endgroup$
    – Keshav Srinivasan
    Dec 6 '18 at 14:26






    $begingroup$
    I'm thinking of the case when $p$ is not an integer.
    $endgroup$
    – Keshav Srinivasan
    Dec 6 '18 at 14:26














    $begingroup$
    It doesn't have to be. Think about how to write $x^p$ as a square.no integer assumption needed.
    $endgroup$
    – user25959
    Dec 6 '18 at 15:44




    $begingroup$
    It doesn't have to be. Think about how to write $x^p$ as a square.no integer assumption needed.
    $endgroup$
    – user25959
    Dec 6 '18 at 15:44











    1












    $begingroup$

    The issue is more or less that $x^{p+1}$ is not a uniformly continuous function: it grows faster and faster as $x to infty$, which means that $mu(I)$ can be large even when $lambda(I)$ is small if $I$ is located far to the right.



    Being more precise, consider $delta>0$ arbitrary and then try computing $mu([delta^{-1/p},delta^{-1/p}+delta])$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The issue is more or less that $x^{p+1}$ is not a uniformly continuous function: it grows faster and faster as $x to infty$, which means that $mu(I)$ can be large even when $lambda(I)$ is small if $I$ is located far to the right.



      Being more precise, consider $delta>0$ arbitrary and then try computing $mu([delta^{-1/p},delta^{-1/p}+delta])$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The issue is more or less that $x^{p+1}$ is not a uniformly continuous function: it grows faster and faster as $x to infty$, which means that $mu(I)$ can be large even when $lambda(I)$ is small if $I$ is located far to the right.



        Being more precise, consider $delta>0$ arbitrary and then try computing $mu([delta^{-1/p},delta^{-1/p}+delta])$.






        share|cite|improve this answer









        $endgroup$



        The issue is more or less that $x^{p+1}$ is not a uniformly continuous function: it grows faster and faster as $x to infty$, which means that $mu(I)$ can be large even when $lambda(I)$ is small if $I$ is located far to the right.



        Being more precise, consider $delta>0$ arbitrary and then try computing $mu([delta^{-1/p},delta^{-1/p}+delta])$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 15:33









        IanIan

        67.7k25388




        67.7k25388






























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