Measure Theory Problem: $lim_{ntoinfty} n cdot mu(E_n) = 0$
$(mathbf X,M,mu)$ 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
edited Nov 27 at 12:26
Martin Sleziak
44.7k7115270
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
add a comment |
$(mathbf X,M,mu)$ 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
edited Nov 27 at 12:26
Martin Sleziak
44.7k7115270
asked Sep 12 '15 at 18:43
Tommaso Guerrini
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
add a comment |
$(mathbf X,M,mu)$ 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
edited Nov 27 at 12:26
Martin Sleziak
44.7k7115270
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
$(mathbf X,M,mu)$ 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
measure-theory
edited Nov 27 at 12:26
Martin Sleziak
44.7k7115270
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
edited Nov 27 at 12:26
Martin Sleziak
44.7k7115270
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
edited Nov 27 at 12:26
Martin Sleziak
44.7k7115270
edited Nov 27 at 12:26
Martin Sleziak
44.7k7115270
edited Nov 27 at 12:26
Martin Sleziak
44.7k7115270
44.7k7115270
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
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
add a comment |
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
add a comment |
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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