show that $sum_{n geq 0}n mu(|f| > n) < +infty implies f in L^1$












0












$begingroup$


the series being convergent we have $lim_{n to infty }n mu(|f| > n) = 0$



because $ n to infty $ then $mu(|f| > n)$ must tend to $0$ with an atleast $O(n^{1+epsilon})$ speed , right ?



so $mu(|f| > +infty) = 0$ and therefore $f in L^1$



is my work correct ?






integration measure-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    there are measure spaces with measure $mu$ and functions $f$ such that $mu(|f| = infty) = 0$ but $f not in L^1$ (e.g. $frac1x$ on $(0,infty)$ with the lebesgue measure). You need to do more than show that $mu(|f| = infty) = 0$.
    $endgroup$
    – Rhys Steele
    Jan 9 at 19:01












  • $begingroup$
    @RhysSteele what if i write $X = cup_{n geq 0} A_n = cup_{n geq 0} {|f| > n} $ then $int|f|dmu leq sum_{n geq 0} int_{A_n}|f|dmu $ can this lead me somewhere ?
    $endgroup$
    – rapidracim
    Jan 9 at 19:11






  • 1




    $begingroup$
    nearly all of your work is wrong. in addition to what Rhys said, $mu(|f| > n)$ can be like $frac{1}{nlog^2 n}$
    $endgroup$
    – mathworker21
    Jan 9 at 19:17








  • 1




    $begingroup$
    Is $mu$ finite? Because one can have $mu(X)=infty$ and $fequiv frac{1}{2}$, and this will not be true.
    $endgroup$
    – Keen-ameteur
    Jan 9 at 20:19
















0












$begingroup$


the series being convergent we have $lim_{n to infty }n mu(|f| > n) = 0$



because $ n to infty $ then $mu(|f| > n)$ must tend to $0$ with an atleast $O(n^{1+epsilon})$ speed , right ?



so $mu(|f| > +infty) = 0$ and therefore $f in L^1$



is my work correct ?






integration measure-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    there are measure spaces with measure $mu$ and functions $f$ such that $mu(|f| = infty) = 0$ but $f not in L^1$ (e.g. $frac1x$ on $(0,infty)$ with the lebesgue measure). You need to do more than show that $mu(|f| = infty) = 0$.
    $endgroup$
    – Rhys Steele
    Jan 9 at 19:01












  • $begingroup$
    @RhysSteele what if i write $X = cup_{n geq 0} A_n = cup_{n geq 0} {|f| > n} $ then $int|f|dmu leq sum_{n geq 0} int_{A_n}|f|dmu $ can this lead me somewhere ?
    $endgroup$
    – rapidracim
    Jan 9 at 19:11






  • 1




    $begingroup$
    nearly all of your work is wrong. in addition to what Rhys said, $mu(|f| > n)$ can be like $frac{1}{nlog^2 n}$
    $endgroup$
    – mathworker21
    Jan 9 at 19:17








  • 1




    $begingroup$
    Is $mu$ finite? Because one can have $mu(X)=infty$ and $fequiv frac{1}{2}$, and this will not be true.
    $endgroup$
    – Keen-ameteur
    Jan 9 at 20:19














0












0








0





$begingroup$


the series being convergent we have $lim_{n to infty }n mu(|f| > n) = 0$



because $ n to infty $ then $mu(|f| > n)$ must tend to $0$ with an atleast $O(n^{1+epsilon})$ speed , right ?



so $mu(|f| > +infty) = 0$ and therefore $f in L^1$



is my work correct ?






integration measure-theory







share|cite|improve this question









$endgroup$




the series being convergent we have $lim_{n to infty }n mu(|f| > n) = 0$



because $ n to infty $ then $mu(|f| > n)$ must tend to $0$ with an atleast $O(n^{1+epsilon})$ speed , right ?



so $mu(|f| > +infty) = 0$ and therefore $f in L^1$



is my work correct ?






integration measure-theory




integration measure-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 18:50









rapidracimrapidracim

1,6881319




1,6881319








  • 1




    $begingroup$
    there are measure spaces with measure $mu$ and functions $f$ such that $mu(|f| = infty) = 0$ but $f not in L^1$ (e.g. $frac1x$ on $(0,infty)$ with the lebesgue measure). You need to do more than show that $mu(|f| = infty) = 0$.
    $endgroup$
    – Rhys Steele
    Jan 9 at 19:01












  • $begingroup$
    @RhysSteele what if i write $X = cup_{n geq 0} A_n = cup_{n geq 0} {|f| > n} $ then $int|f|dmu leq sum_{n geq 0} int_{A_n}|f|dmu $ can this lead me somewhere ?
    $endgroup$
    – rapidracim
    Jan 9 at 19:11






  • 1




    $begingroup$
    nearly all of your work is wrong. in addition to what Rhys said, $mu(|f| > n)$ can be like $frac{1}{nlog^2 n}$
    $endgroup$
    – mathworker21
    Jan 9 at 19:17








  • 1




    $begingroup$
    Is $mu$ finite? Because one can have $mu(X)=infty$ and $fequiv frac{1}{2}$, and this will not be true.
    $endgroup$
    – Keen-ameteur
    Jan 9 at 20:19














  • 1




    $begingroup$
    there are measure spaces with measure $mu$ and functions $f$ such that $mu(|f| = infty) = 0$ but $f not in L^1$ (e.g. $frac1x$ on $(0,infty)$ with the lebesgue measure). You need to do more than show that $mu(|f| = infty) = 0$.
    $endgroup$
    – Rhys Steele
    Jan 9 at 19:01












  • $begingroup$
    @RhysSteele what if i write $X = cup_{n geq 0} A_n = cup_{n geq 0} {|f| > n} $ then $int|f|dmu leq sum_{n geq 0} int_{A_n}|f|dmu $ can this lead me somewhere ?
    $endgroup$
    – rapidracim
    Jan 9 at 19:11






  • 1




    $begingroup$
    nearly all of your work is wrong. in addition to what Rhys said, $mu(|f| > n)$ can be like $frac{1}{nlog^2 n}$
    $endgroup$
    – mathworker21
    Jan 9 at 19:17








  • 1




    $begingroup$
    Is $mu$ finite? Because one can have $mu(X)=infty$ and $fequiv frac{1}{2}$, and this will not be true.
    $endgroup$
    – Keen-ameteur
    Jan 9 at 20:19








1




1




$begingroup$
there are measure spaces with measure $mu$ and functions $f$ such that $mu(|f| = infty) = 0$ but $f not in L^1$ (e.g. $frac1x$ on $(0,infty)$ with the lebesgue measure). You need to do more than show that $mu(|f| = infty) = 0$.
$endgroup$
– Rhys Steele
Jan 9 at 19:01






$begingroup$
there are measure spaces with measure $mu$ and functions $f$ such that $mu(|f| = infty) = 0$ but $f not in L^1$ (e.g. $frac1x$ on $(0,infty)$ with the lebesgue measure). You need to do more than show that $mu(|f| = infty) = 0$.
$endgroup$
– Rhys Steele
Jan 9 at 19:01














$begingroup$
@RhysSteele what if i write $X = cup_{n geq 0} A_n = cup_{n geq 0} {|f| > n} $ then $int|f|dmu leq sum_{n geq 0} int_{A_n}|f|dmu $ can this lead me somewhere ?
$endgroup$
– rapidracim
Jan 9 at 19:11




$begingroup$
@RhysSteele what if i write $X = cup_{n geq 0} A_n = cup_{n geq 0} {|f| > n} $ then $int|f|dmu leq sum_{n geq 0} int_{A_n}|f|dmu $ can this lead me somewhere ?
$endgroup$
– rapidracim
Jan 9 at 19:11




1




1




$begingroup$
nearly all of your work is wrong. in addition to what Rhys said, $mu(|f| > n)$ can be like $frac{1}{nlog^2 n}$
$endgroup$
– mathworker21
Jan 9 at 19:17






$begingroup$
nearly all of your work is wrong. in addition to what Rhys said, $mu(|f| > n)$ can be like $frac{1}{nlog^2 n}$
$endgroup$
– mathworker21
Jan 9 at 19:17






1




1




$begingroup$
Is $mu$ finite? Because one can have $mu(X)=infty$ and $fequiv frac{1}{2}$, and this will not be true.
$endgroup$
– Keen-ameteur
Jan 9 at 20:19




$begingroup$
Is $mu$ finite? Because one can have $mu(X)=infty$ and $fequiv frac{1}{2}$, and this will not be true.
$endgroup$
– Keen-ameteur
Jan 9 at 20:19










1 Answer
1






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oldest

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1












$begingroup$

You need the hypothesis that $mu({|f| > 0}) < +infty$.



Let $E_n = { n < |f| leqslant n+1} $ and $F_n = { |f| > n}$. Since $E_n subset F_n$ we have $mu(E_n) leqslant mu(F_n)$ and since $sum_{n geqslant 0}nmu(F_n) < +infty$, by the comparison test it follows that
$$sum_{n geqslant 0}mu(E_n),,, sum_{n geqslant 0}nmu(E_n) < +infty$$



This also implies that



$$sum_{n geqslant 0}(n+1)mu(E_n) < +infty$$



Thus,



$$int|f|,dmu = sum_{n geqslant 0} int_{E_n}|f|, dmu leqslant sum_{n geqslant 0} int_{E_n}(n+1) , dmu = sum_{n geqslant 0} (n+1)mu(E_n) < +infty, $$



and $f in L^1$.






share|cite|improve this answer











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    $begingroup$

    You need the hypothesis that $mu({|f| > 0}) < +infty$.



    Let $E_n = { n < |f| leqslant n+1} $ and $F_n = { |f| > n}$. Since $E_n subset F_n$ we have $mu(E_n) leqslant mu(F_n)$ and since $sum_{n geqslant 0}nmu(F_n) < +infty$, by the comparison test it follows that
    $$sum_{n geqslant 0}mu(E_n),,, sum_{n geqslant 0}nmu(E_n) < +infty$$



    This also implies that



    $$sum_{n geqslant 0}(n+1)mu(E_n) < +infty$$



    Thus,



    $$int|f|,dmu = sum_{n geqslant 0} int_{E_n}|f|, dmu leqslant sum_{n geqslant 0} int_{E_n}(n+1) , dmu = sum_{n geqslant 0} (n+1)mu(E_n) < +infty, $$



    and $f in L^1$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      You need the hypothesis that $mu({|f| > 0}) < +infty$.



      Let $E_n = { n < |f| leqslant n+1} $ and $F_n = { |f| > n}$. Since $E_n subset F_n$ we have $mu(E_n) leqslant mu(F_n)$ and since $sum_{n geqslant 0}nmu(F_n) < +infty$, by the comparison test it follows that
      $$sum_{n geqslant 0}mu(E_n),,, sum_{n geqslant 0}nmu(E_n) < +infty$$



      This also implies that



      $$sum_{n geqslant 0}(n+1)mu(E_n) < +infty$$



      Thus,



      $$int|f|,dmu = sum_{n geqslant 0} int_{E_n}|f|, dmu leqslant sum_{n geqslant 0} int_{E_n}(n+1) , dmu = sum_{n geqslant 0} (n+1)mu(E_n) < +infty, $$



      and $f in L^1$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        You need the hypothesis that $mu({|f| > 0}) < +infty$.



        Let $E_n = { n < |f| leqslant n+1} $ and $F_n = { |f| > n}$. Since $E_n subset F_n$ we have $mu(E_n) leqslant mu(F_n)$ and since $sum_{n geqslant 0}nmu(F_n) < +infty$, by the comparison test it follows that
        $$sum_{n geqslant 0}mu(E_n),,, sum_{n geqslant 0}nmu(E_n) < +infty$$



        This also implies that



        $$sum_{n geqslant 0}(n+1)mu(E_n) < +infty$$



        Thus,



        $$int|f|,dmu = sum_{n geqslant 0} int_{E_n}|f|, dmu leqslant sum_{n geqslant 0} int_{E_n}(n+1) , dmu = sum_{n geqslant 0} (n+1)mu(E_n) < +infty, $$



        and $f in L^1$.






        share|cite|improve this answer











        $endgroup$



        You need the hypothesis that $mu({|f| > 0}) < +infty$.



        Let $E_n = { n < |f| leqslant n+1} $ and $F_n = { |f| > n}$. Since $E_n subset F_n$ we have $mu(E_n) leqslant mu(F_n)$ and since $sum_{n geqslant 0}nmu(F_n) < +infty$, by the comparison test it follows that
        $$sum_{n geqslant 0}mu(E_n),,, sum_{n geqslant 0}nmu(E_n) < +infty$$



        This also implies that



        $$sum_{n geqslant 0}(n+1)mu(E_n) < +infty$$



        Thus,



        $$int|f|,dmu = sum_{n geqslant 0} int_{E_n}|f|, dmu leqslant sum_{n geqslant 0} int_{E_n}(n+1) , dmu = sum_{n geqslant 0} (n+1)mu(E_n) < +infty, $$



        and $f in L^1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 21 at 0:41

























        answered Jan 21 at 0:26









        RRLRRL

        50.7k42573




        50.7k42573






























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