Mixing
Showing a measure finitely additive but not countably additive
While reading Vector Measure from Diestel's book I find that Considering any Hahn-Banach extension $T$ to $L_{infty}[0,1]$ of point mass functional on $C[0,1]$ we can construct a measure $F$ defined on all Lebesgue measurable subsets of $[0,1]$ which satisfies $F(E)=T(chi_E)$ , where $Esubseteq [0,1]$ is Lebesgue measurable. Author said this measure is Finitely Additive but not Countably Additive.
I can not show that this measure fails to be Countably Additive, what I can guess is that if my point mass functional is $delta_x:C[0,1]rightarrow Bbb R,delta_x(f)=f(x)$ where $xin [0,1]$ then $||delta_x||=||T||=1$ and the measure $F$ has the property that $F(E)=0$ if and only if $xin E$. But I can not prove it. Am I right? Thanks in advance.
functional-analysis measure-theory operator-theory proof-explanation alternative-proof
asked Nov 20 '18 at 5:32
Mathlover
1326
add a comment |
While reading Vector Measure from Diestel's book I find that Considering any Hahn-Banach extension $T$ to $L_{infty}[0,1]$ of point mass functional on $C[0,1]$ we can construct a measure $F$ defined on all Lebesgue measurable subsets of $[0,1]$ which satisfies $F(E)=T(chi_E)$ , where $Esubseteq [0,1]$ is Lebesgue measurable. Author said this measure is Finitely Additive but not Countably Additive.
I can not show that this measure fails to be Countably Additive, what I can guess is that if my point mass functional is $delta_x:C[0,1]rightarrow Bbb R,delta_x(f)=f(x)$ where $xin [0,1]$ then $||delta_x||=||T||=1$ and the measure $F$ has the property that $F(E)=0$ if and only if $xin E$. But I can not prove it. Am I right? Thanks in advance.
functional-analysis measure-theory operator-theory proof-explanation alternative-proof
asked Nov 20 '18 at 5:32
Mathlover
1326
1
The basic example of the problem is $nu(A) = sum_{ n in A} frac{1}{n+1}$ which is a non-negative measure on $mathbb{N}$, $mu(A)= sum_{ n in A} frac{(-1)^n}{n+1}$ which is not a signed measure on $mathbb{N}$.
– reuns
Nov 20 '18 at 5:40
Can you explain in some details how do I prove my $F$ is not countably additive from your stated fact.
– Mathlover
Nov 20 '18 at 5:44
add a comment |
While reading Vector Measure from Diestel's book I find that Considering any Hahn-Banach extension $T$ to $L_{infty}[0,1]$ of point mass functional on $C[0,1]$ we can construct a measure $F$ defined on all Lebesgue measurable subsets of $[0,1]$ which satisfies $F(E)=T(chi_E)$ , where $Esubseteq [0,1]$ is Lebesgue measurable. Author said this measure is Finitely Additive but not Countably Additive.
I can not show that this measure fails to be Countably Additive, what I can guess is that if my point mass functional is $delta_x:C[0,1]rightarrow Bbb R,delta_x(f)=f(x)$ where $xin [0,1]$ then $||delta_x||=||T||=1$ and the measure $F$ has the property that $F(E)=0$ if and only if $xin E$. But I can not prove it. Am I right? Thanks in advance.
functional-analysis measure-theory operator-theory proof-explanation alternative-proof
asked Nov 20 '18 at 5:32
Mathlover
1326
While reading Vector Measure from Diestel's book I find that Considering any Hahn-Banach extension $T$ to $L_{infty}[0,1]$ of point mass functional on $C[0,1]$ we can construct a measure $F$ defined on all Lebesgue measurable subsets of $[0,1]$ which satisfies $F(E)=T(chi_E)$ , where $Esubseteq [0,1]$ is Lebesgue measurable. Author said this measure is Finitely Additive but not Countably Additive.
I can not show that this measure fails to be Countably Additive, what I can guess is that if my point mass functional is $delta_x:C[0,1]rightarrow Bbb R,delta_x(f)=f(x)$ where $xin [0,1]$ then $||delta_x||=||T||=1$ and the measure $F$ has the property that $F(E)=0$ if and only if $xin E$. But I can not prove it. Am I right? Thanks in advance.
functional-analysis measure-theory operator-theory proof-explanation alternative-proof
functional-analysis measure-theory operator-theory proof-explanation alternative-proof
asked Nov 20 '18 at 5:32
Mathlover
1326
asked Nov 20 '18 at 5:32
Mathlover
1326
asked Nov 20 '18 at 5:32
Mathlover
1326
asked Nov 20 '18 at 5:32
Mathlover
1326
asked Nov 20 '18 at 5:32
Mathlover
1326
1326
1
The basic example of the problem is $nu(A) = sum_{ n in A} frac{1}{n+1}$ which is a non-negative measure on $mathbb{N}$, $mu(A)= sum_{ n in A} frac{(-1)^n}{n+1}$ which is not a signed measure on $mathbb{N}$.
– reuns
Nov 20 '18 at 5:40
Can you explain in some details how do I prove my $F$ is not countably additive from your stated fact.
– Mathlover
Nov 20 '18 at 5:44
add a comment |
1
The basic example of the problem is $nu(A) = sum_{ n in A} frac{1}{n+1}$ which is a non-negative measure on $mathbb{N}$, $mu(A)= sum_{ n in A} frac{(-1)^n}{n+1}$ which is not a signed measure on $mathbb{N}$.
– reuns
Nov 20 '18 at 5:40
Can you explain in some details how do I prove my $F$ is not countably additive from your stated fact.
– Mathlover
Nov 20 '18 at 5:44
1
1
The basic example of the problem is $nu(A) = sum_{ n in A} frac{1}{n+1}$ which is a non-negative measure on $mathbb{N}$, $mu(A)= sum_{ n in A} frac{(-1)^n}{n+1}$ which is not a signed measure on $mathbb{N}$.
– reuns
Nov 20 '18 at 5:40
The basic example of the problem is $nu(A) = sum_{ n in A} frac{1}{n+1}$ which is a non-negative measure on $mathbb{N}$, $mu(A)= sum_{ n in A} frac{(-1)^n}{n+1}$ which is not a signed measure on $mathbb{N}$.
– reuns
Nov 20 '18 at 5:40
Can you explain in some details how do I prove my $F$ is not countably additive from your stated fact.
– Mathlover
Nov 20 '18 at 5:44
Can you explain in some details how do I prove my $F$ is not countably additive from your stated fact.
– Mathlover
Nov 20 '18 at 5:44
add a comment |
1 Answer
1
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Suppose $F$ is countably additive. Then it is absolutely continuous w.r.t. Lebesgue measure $m$. Let $g=frac {dF} {dm}$. Then $f(x)=Tf=int_0^{1} f(y)g(y), dy$ for all $f in C[0,1]$ which means $int_0^{1} f(y)g(y), dy=int_0^{1} f(y), ddelta_x (y)$ and this implies $g(y), dy =delta_x(dy)$ which is a contradiction.
answered Nov 20 '18 at 5:46
Kavi Rama Murthy
50.4k31854
1
How can I prove $F<<m$.
– Mathlover
Nov 20 '18 at 6:04
1
If $E$ has Lebesgue measure $0$ then $I_E$ is the zero element of $L^{infty}$, so $T(I_E)=0$.
– Kavi Rama Murthy
Nov 20 '18 at 6:08
1
I think you contradict by showing that the continuous linear functionals $frightarrow int_0^1fgdm$ and $frightarrow int_0^1fddelta_x$ are same on the dense set $C[0,1]$ of $L^1[0,1]$ , hence these functionals are equal on $L^1[0,1]$ , hence $int _Egdm=int_Eddelta_x=delta_x(E)implies delta_x<<m$ and which is impossible - am I right?
– Mathlover
Nov 20 '18 at 7:00
1
@UserD You are right, but I just used (without proof ) the well known result that if $int f, dmu=int f, dnu$ for all continuous $f$ then $mu =nu$ on the Borel sigma algebra. What you are doing is to provide a proof of this fact.
– Kavi Rama Murthy
Nov 20 '18 at 7:17
add a comment |
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Suppose $F$ is countably additive. Then it is absolutely continuous w.r.t. Lebesgue measure $m$. Let $g=frac {dF} {dm}$. Then $f(x)=Tf=int_0^{1} f(y)g(y), dy$ for all $f in C[0,1]$ which means $int_0^{1} f(y)g(y), dy=int_0^{1} f(y), ddelta_x (y)$ and this implies $g(y), dy =delta_x(dy)$ which is a contradiction.
answered Nov 20 '18 at 5:46
Kavi Rama Murthy
50.4k31854
1
How can I prove $F<<m$.
– Mathlover
Nov 20 '18 at 6:04
1
If $E$ has Lebesgue measure $0$ then $I_E$ is the zero element of $L^{infty}$, so $T(I_E)=0$.
– Kavi Rama Murthy
Nov 20 '18 at 6:08
1
I think you contradict by showing that the continuous linear functionals $frightarrow int_0^1fgdm$ and $frightarrow int_0^1fddelta_x$ are same on the dense set $C[0,1]$ of $L^1[0,1]$ , hence these functionals are equal on $L^1[0,1]$ , hence $int _Egdm=int_Eddelta_x=delta_x(E)implies delta_x<<m$ and which is impossible - am I right?
– Mathlover
Nov 20 '18 at 7:00
1
@UserD You are right, but I just used (without proof ) the well known result that if $int f, dmu=int f, dnu$ for all continuous $f$ then $mu =nu$ on the Borel sigma algebra. What you are doing is to provide a proof of this fact.
– Kavi Rama Murthy
Nov 20 '18 at 7:17
add a comment |
Suppose $F$ is countably additive. Then it is absolutely continuous w.r.t. Lebesgue measure $m$. Let $g=frac {dF} {dm}$. Then $f(x)=Tf=int_0^{1} f(y)g(y), dy$ for all $f in C[0,1]$ which means $int_0^{1} f(y)g(y), dy=int_0^{1} f(y), ddelta_x (y)$ and this implies $g(y), dy =delta_x(dy)$ which is a contradiction.
answered Nov 20 '18 at 5:46
Kavi Rama Murthy
50.4k31854
1
How can I prove $F<<m$.
– Mathlover
Nov 20 '18 at 6:04
1
If $E$ has Lebesgue measure $0$ then $I_E$ is the zero element of $L^{infty}$, so $T(I_E)=0$.
– Kavi Rama Murthy
Nov 20 '18 at 6:08
1
I think you contradict by showing that the continuous linear functionals $frightarrow int_0^1fgdm$ and $frightarrow int_0^1fddelta_x$ are same on the dense set $C[0,1]$ of $L^1[0,1]$ , hence these functionals are equal on $L^1[0,1]$ , hence $int _Egdm=int_Eddelta_x=delta_x(E)implies delta_x<<m$ and which is impossible - am I right?
– Mathlover
Nov 20 '18 at 7:00
1
@UserD You are right, but I just used (without proof ) the well known result that if $int f, dmu=int f, dnu$ for all continuous $f$ then $mu =nu$ on the Borel sigma algebra. What you are doing is to provide a proof of this fact.
– Kavi Rama Murthy
Nov 20 '18 at 7:17
add a comment |
Suppose $F$ is countably additive. Then it is absolutely continuous w.r.t. Lebesgue measure $m$. Let $g=frac {dF} {dm}$. Then $f(x)=Tf=int_0^{1} f(y)g(y), dy$ for all $f in C[0,1]$ which means $int_0^{1} f(y)g(y), dy=int_0^{1} f(y), ddelta_x (y)$ and this implies $g(y), dy =delta_x(dy)$ which is a contradiction.
answered Nov 20 '18 at 5:46
Kavi Rama Murthy
50.4k31854
Suppose $F$ is countably additive. Then it is absolutely continuous w.r.t. Lebesgue measure $m$. Let $g=frac {dF} {dm}$. Then $f(x)=Tf=int_0^{1} f(y)g(y), dy$ for all $f in C[0,1]$ which means $int_0^{1} f(y)g(y), dy=int_0^{1} f(y), ddelta_x (y)$ and this implies $g(y), dy =delta_x(dy)$ which is a contradiction.
answered Nov 20 '18 at 5:46
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 5:46
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 5:46
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 5:46
Kavi Rama Murthy
50.4k31854
50.4k31854
1
How can I prove $F<<m$.
– Mathlover
Nov 20 '18 at 6:04
1
If $E$ has Lebesgue measure $0$ then $I_E$ is the zero element of $L^{infty}$, so $T(I_E)=0$.
– Kavi Rama Murthy
Nov 20 '18 at 6:08
1
I think you contradict by showing that the continuous linear functionals $frightarrow int_0^1fgdm$ and $frightarrow int_0^1fddelta_x$ are same on the dense set $C[0,1]$ of $L^1[0,1]$ , hence these functionals are equal on $L^1[0,1]$ , hence $int _Egdm=int_Eddelta_x=delta_x(E)implies delta_x<<m$ and which is impossible - am I right?
– Mathlover
Nov 20 '18 at 7:00
1
@UserD You are right, but I just used (without proof ) the well known result that if $int f, dmu=int f, dnu$ for all continuous $f$ then $mu =nu$ on the Borel sigma algebra. What you are doing is to provide a proof of this fact.
– Kavi Rama Murthy
Nov 20 '18 at 7:17
add a comment |
1
How can I prove $F<<m$.
– Mathlover
Nov 20 '18 at 6:04
1
If $E$ has Lebesgue measure $0$ then $I_E$ is the zero element of $L^{infty}$, so $T(I_E)=0$.
– Kavi Rama Murthy
Nov 20 '18 at 6:08
1
I think you contradict by showing that the continuous linear functionals $frightarrow int_0^1fgdm$ and $frightarrow int_0^1fddelta_x$ are same on the dense set $C[0,1]$ of $L^1[0,1]$ , hence these functionals are equal on $L^1[0,1]$ , hence $int _Egdm=int_Eddelta_x=delta_x(E)implies delta_x<<m$ and which is impossible - am I right?
– Mathlover
Nov 20 '18 at 7:00
1
@UserD You are right, but I just used (without proof ) the well known result that if $int f, dmu=int f, dnu$ for all continuous $f$ then $mu =nu$ on the Borel sigma algebra. What you are doing is to provide a proof of this fact.
– Kavi Rama Murthy
Nov 20 '18 at 7:17
1
1
How can I prove $F<<m$.
– Mathlover
Nov 20 '18 at 6:04
How can I prove $F<<m$.
– Mathlover
Nov 20 '18 at 6:04
1
1
If $E$ has Lebesgue measure $0$ then $I_E$ is the zero element of $L^{infty}$, so $T(I_E)=0$.
– Kavi Rama Murthy
Nov 20 '18 at 6:08
If $E$ has Lebesgue measure $0$ then $I_E$ is the zero element of $L^{infty}$, so $T(I_E)=0$.
– Kavi Rama Murthy
Nov 20 '18 at 6:08
1
1
I think you contradict by showing that the continuous linear functionals $frightarrow int_0^1fgdm$ and $frightarrow int_0^1fddelta_x$ are same on the dense set $C[0,1]$ of $L^1[0,1]$ , hence these functionals are equal on $L^1[0,1]$ , hence $int _Egdm=int_Eddelta_x=delta_x(E)implies delta_x<<m$ and which is impossible - am I right?
– Mathlover
Nov 20 '18 at 7:00
I think you contradict by showing that the continuous linear functionals $frightarrow int_0^1fgdm$ and $frightarrow int_0^1fddelta_x$ are same on the dense set $C[0,1]$ of $L^1[0,1]$ , hence these functionals are equal on $L^1[0,1]$ , hence $int _Egdm=int_Eddelta_x=delta_x(E)implies delta_x<<m$ and which is impossible - am I right?
– Mathlover
Nov 20 '18 at 7:00
1
1
@UserD You are right, but I just used (without proof ) the well known result that if $int f, dmu=int f, dnu$ for all continuous $f$ then $mu =nu$ on the Borel sigma algebra. What you are doing is to provide a proof of this fact.
– Kavi Rama Murthy
Nov 20 '18 at 7:17
@UserD You are right, but I just used (without proof ) the well known result that if $int f, dmu=int f, dnu$ for all continuous $f$ then $mu =nu$ on the Borel sigma algebra. What you are doing is to provide a proof of this fact.
– Kavi Rama Murthy
Nov 20 '18 at 7:17
add a comment |
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The basic example of the problem is $nu(A) = sum_{ n in A} frac{1}{n+1}$ which is a non-negative measure on $mathbb{N}$, $mu(A)= sum_{ n in A} frac{(-1)^n}{n+1}$ which is not a signed measure on $mathbb{N}$.
– reuns
Nov 20 '18 at 5:40
Can you explain in some details how do I prove my $F$ is not countably additive from your stated fact.
– Mathlover
Nov 20 '18 at 5:44