Measure Theory Problem: $lim_{ntoinfty} n cdot mu(E_n) = 0$

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$(mathbf X,M,mu)$ measure space, $f : mathbf X to [0;infty]$ a measurable function such that Integral $int_X f , dmu < infty$.



If $E_n = {x in mathbf X : n + 1 ge f(x) > n}$ then



$$lim_{ntoinfty} n cdot mu(E_n) = 0$$



I can't figure it out: I thought about the fact that $E_n$ are a partition of $X$, so I could use countable additivity there, but I don't know if $mu$ is a finite measure so I can't show any convergence. Thank you for your help!










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  • How can you rewrite $E_{n}$ using $f^{-1}$ ?
    – krirkrirk
    Sep 12 '15 at 18:49






  • 4




    Start with $infty>int_X f,dmu =sum_{n=0}^infty int_{E_{n}}f,dmuge sum_{n=0}^inftyint_{E_{n}}n,dmu=sum_{n=1}^infty ncdotmu(E_{n})$.
    – John Dawkins
    Sep 12 '15 at 18:53










  • Thank you guys! It always seems so trivial when you see the solution! Have a nice day :)
    – Tommaso Guerrini
    Sep 12 '15 at 18:54










  • To conceptually connect your comment (that you don't know if $mu$ is a finite measure) to the given solutions: We do know that $$nu(E):=int_E fdmu$$ is a finite measure on $M$. Then since $(E_n)_{n=1}^infty$ are pairwise disjoint, $nu(E_n)to 0$. But $nu(E_n)geqslant nmu(E_n)$.
    – bangs
    Nov 27 at 13:24
















1














$(mathbf X,M,mu)$ measure space, $f : mathbf X to [0;infty]$ a measurable function such that Integral $int_X f , dmu < infty$.



If $E_n = {x in mathbf X : n + 1 ge f(x) > n}$ then



$$lim_{ntoinfty} n cdot mu(E_n) = 0$$



I can't figure it out: I thought about the fact that $E_n$ are a partition of $X$, so I could use countable additivity there, but I don't know if $mu$ is a finite measure so I can't show any convergence. Thank you for your help!










share|cite|improve this question
























  • How can you rewrite $E_{n}$ using $f^{-1}$ ?
    – krirkrirk
    Sep 12 '15 at 18:49






  • 4




    Start with $infty>int_X f,dmu =sum_{n=0}^infty int_{E_{n}}f,dmuge sum_{n=0}^inftyint_{E_{n}}n,dmu=sum_{n=1}^infty ncdotmu(E_{n})$.
    – John Dawkins
    Sep 12 '15 at 18:53










  • Thank you guys! It always seems so trivial when you see the solution! Have a nice day :)
    – Tommaso Guerrini
    Sep 12 '15 at 18:54










  • To conceptually connect your comment (that you don't know if $mu$ is a finite measure) to the given solutions: We do know that $$nu(E):=int_E fdmu$$ is a finite measure on $M$. Then since $(E_n)_{n=1}^infty$ are pairwise disjoint, $nu(E_n)to 0$. But $nu(E_n)geqslant nmu(E_n)$.
    – bangs
    Nov 27 at 13:24














1












1








1







$(mathbf X,M,mu)$ measure space, $f : mathbf X to [0;infty]$ a measurable function such that Integral $int_X f , dmu < infty$.



If $E_n = {x in mathbf X : n + 1 ge f(x) > n}$ then



$$lim_{ntoinfty} n cdot mu(E_n) = 0$$



I can't figure it out: I thought about the fact that $E_n$ are a partition of $X$, so I could use countable additivity there, but I don't know if $mu$ is a finite measure so I can't show any convergence. Thank you for your help!










share|cite|improve this question















$(mathbf X,M,mu)$ measure space, $f : mathbf X to [0;infty]$ a measurable function such that Integral $int_X f , dmu < infty$.



If $E_n = {x in mathbf X : n + 1 ge f(x) > n}$ then



$$lim_{ntoinfty} n cdot mu(E_n) = 0$$



I can't figure it out: I thought about the fact that $E_n$ are a partition of $X$, so I could use countable additivity there, but I don't know if $mu$ is a finite measure so I can't show any convergence. Thank you for your help!







measure-theory






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share|cite|improve this question













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edited Nov 27 at 12:26









Martin Sleziak

44.7k7115270




44.7k7115270










asked Sep 12 '15 at 18:43









Tommaso Guerrini

1567




1567












  • How can you rewrite $E_{n}$ using $f^{-1}$ ?
    – krirkrirk
    Sep 12 '15 at 18:49






  • 4




    Start with $infty>int_X f,dmu =sum_{n=0}^infty int_{E_{n}}f,dmuge sum_{n=0}^inftyint_{E_{n}}n,dmu=sum_{n=1}^infty ncdotmu(E_{n})$.
    – John Dawkins
    Sep 12 '15 at 18:53










  • Thank you guys! It always seems so trivial when you see the solution! Have a nice day :)
    – Tommaso Guerrini
    Sep 12 '15 at 18:54










  • To conceptually connect your comment (that you don't know if $mu$ is a finite measure) to the given solutions: We do know that $$nu(E):=int_E fdmu$$ is a finite measure on $M$. Then since $(E_n)_{n=1}^infty$ are pairwise disjoint, $nu(E_n)to 0$. But $nu(E_n)geqslant nmu(E_n)$.
    – bangs
    Nov 27 at 13:24


















  • How can you rewrite $E_{n}$ using $f^{-1}$ ?
    – krirkrirk
    Sep 12 '15 at 18:49






  • 4




    Start with $infty>int_X f,dmu =sum_{n=0}^infty int_{E_{n}}f,dmuge sum_{n=0}^inftyint_{E_{n}}n,dmu=sum_{n=1}^infty ncdotmu(E_{n})$.
    – John Dawkins
    Sep 12 '15 at 18:53










  • Thank you guys! It always seems so trivial when you see the solution! Have a nice day :)
    – Tommaso Guerrini
    Sep 12 '15 at 18:54










  • To conceptually connect your comment (that you don't know if $mu$ is a finite measure) to the given solutions: We do know that $$nu(E):=int_E fdmu$$ is a finite measure on $M$. Then since $(E_n)_{n=1}^infty$ are pairwise disjoint, $nu(E_n)to 0$. But $nu(E_n)geqslant nmu(E_n)$.
    – bangs
    Nov 27 at 13:24
















How can you rewrite $E_{n}$ using $f^{-1}$ ?
– krirkrirk
Sep 12 '15 at 18:49




How can you rewrite $E_{n}$ using $f^{-1}$ ?
– krirkrirk
Sep 12 '15 at 18:49




4




4




Start with $infty>int_X f,dmu =sum_{n=0}^infty int_{E_{n}}f,dmuge sum_{n=0}^inftyint_{E_{n}}n,dmu=sum_{n=1}^infty ncdotmu(E_{n})$.
– John Dawkins
Sep 12 '15 at 18:53




Start with $infty>int_X f,dmu =sum_{n=0}^infty int_{E_{n}}f,dmuge sum_{n=0}^inftyint_{E_{n}}n,dmu=sum_{n=1}^infty ncdotmu(E_{n})$.
– John Dawkins
Sep 12 '15 at 18:53












Thank you guys! It always seems so trivial when you see the solution! Have a nice day :)
– Tommaso Guerrini
Sep 12 '15 at 18:54




Thank you guys! It always seems so trivial when you see the solution! Have a nice day :)
– Tommaso Guerrini
Sep 12 '15 at 18:54












To conceptually connect your comment (that you don't know if $mu$ is a finite measure) to the given solutions: We do know that $$nu(E):=int_E fdmu$$ is a finite measure on $M$. Then since $(E_n)_{n=1}^infty$ are pairwise disjoint, $nu(E_n)to 0$. But $nu(E_n)geqslant nmu(E_n)$.
– bangs
Nov 27 at 13:24




To conceptually connect your comment (that you don't know if $mu$ is a finite measure) to the given solutions: We do know that $$nu(E):=int_E fdmu$$ is a finite measure on $M$. Then since $(E_n)_{n=1}^infty$ are pairwise disjoint, $nu(E_n)to 0$. But $nu(E_n)geqslant nmu(E_n)$.
– bangs
Nov 27 at 13:24















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