How to proof that $mathcal{P}(X)_{tau}=mathcal{P}(X)$ on the measure space $(X,mathcal{P}(X))$
$begingroup$
How to proof that $mathcal{P}(X)_{tau}=mathcal{P}(X)$ on the measure space $(X,mathcal{P}(X))$ when $tau$ is the counting measure.
My teacher has defined:
$$mathbb{N}_{tau}={Nsubseteq X; |; text{there exist }Pinmathcal{P}(X);text{so};Nsubseteq P;;text{and};tau(P)=0},$$
where I see that $mathbb{N}_{tau}={emptyset}$, but I don't know how to understand in terms of the Powerset.
I only seem to understand this intuitively, as if $mathcal{P}(X)$ contain all subsets of $X$, and $tau$ can be defined on all of the subsets, they must be equal to each other.
I have hard time understanding what my professor would like us to use in this scenario - I thought about defining $mathcal{P}(X)$ as point masses, however, I have to show that they contain the same sets.
Any help proofing this, would be greatly appreciated.
measure-theory
asked Dec 1 '18 at 6:57
FrederikFrederik
1009
$endgroup$
add a comment |
$begingroup$
How to proof that $mathcal{P}(X)_{tau}=mathcal{P}(X)$ on the measure space $(X,mathcal{P}(X))$ when $tau$ is the counting measure.
My teacher has defined:
$$mathbb{N}_{tau}={Nsubseteq X; |; text{there exist }Pinmathcal{P}(X);text{so};Nsubseteq P;;text{and};tau(P)=0},$$
where I see that $mathbb{N}_{tau}={emptyset}$, but I don't know how to understand in terms of the Powerset.
I only seem to understand this intuitively, as if $mathcal{P}(X)$ contain all subsets of $X$, and $tau$ can be defined on all of the subsets, they must be equal to each other.
I have hard time understanding what my professor would like us to use in this scenario - I thought about defining $mathcal{P}(X)$ as point masses, however, I have to show that they contain the same sets.
Any help proofing this, would be greatly appreciated.
measure-theory
asked Dec 1 '18 at 6:57
FrederikFrederik
1009
$endgroup$
$begingroup$
@saz Sorry, I forgot that - I have updated my question.
$endgroup$
– Frederik
Dec 1 '18 at 7:24
$begingroup$
the notation $mathcal P(X)_tau$ I guess it mean the $sigma$-algebra generated by $tau$, right? In this case it is enough to show that every subset of $X$ is measurable in the counting measure
$endgroup$
– Masacroso
Dec 1 '18 at 7:29
$begingroup$
Well, the definition of $mathcal{P}(X)_{tau}$ is still missing but probably you define $mathcal{P}(X)_{tau} := sigma(mathcal{P}(X),mathbb{N}_{tau})$. You write that you know that $mathbb{N}_{tau} = {emptyset}$... so what exactly do you not understand? If you know this, the assertion is immediate from the definition...
$endgroup$
– saz
Dec 1 '18 at 7:29
$begingroup$
@saz Sorry for the confusion in the question - I don't know how this can shown in any other way than intuitively.
$endgroup$
– Frederik
Dec 1 '18 at 7:39
$begingroup$
Thank you @Masacroso. You are right, it is the enlargement with respect to $tau$ (if I understand the idea of enlargement correctly).
$endgroup$
– Frederik
Dec 1 '18 at 7:39
add a comment |
$begingroup$
How to proof that $mathcal{P}(X)_{tau}=mathcal{P}(X)$ on the measure space $(X,mathcal{P}(X))$ when $tau$ is the counting measure.
My teacher has defined:
$$mathbb{N}_{tau}={Nsubseteq X; |; text{there exist }Pinmathcal{P}(X);text{so};Nsubseteq P;;text{and};tau(P)=0},$$
where I see that $mathbb{N}_{tau}={emptyset}$, but I don't know how to understand in terms of the Powerset.
I only seem to understand this intuitively, as if $mathcal{P}(X)$ contain all subsets of $X$, and $tau$ can be defined on all of the subsets, they must be equal to each other.
I have hard time understanding what my professor would like us to use in this scenario - I thought about defining $mathcal{P}(X)$ as point masses, however, I have to show that they contain the same sets.
Any help proofing this, would be greatly appreciated.
measure-theory
asked Dec 1 '18 at 6:57
FrederikFrederik
1009
$endgroup$
How to proof that $mathcal{P}(X)_{tau}=mathcal{P}(X)$ on the measure space $(X,mathcal{P}(X))$ when $tau$ is the counting measure.
My teacher has defined:
$$mathbb{N}_{tau}={Nsubseteq X; |; text{there exist }Pinmathcal{P}(X);text{so};Nsubseteq P;;text{and};tau(P)=0},$$
where I see that $mathbb{N}_{tau}={emptyset}$, but I don't know how to understand in terms of the Powerset.
I only seem to understand this intuitively, as if $mathcal{P}(X)$ contain all subsets of $X$, and $tau$ can be defined on all of the subsets, they must be equal to each other.
I have hard time understanding what my professor would like us to use in this scenario - I thought about defining $mathcal{P}(X)$ as point masses, however, I have to show that they contain the same sets.
Any help proofing this, would be greatly appreciated.
measure-theory
measure-theory
asked Dec 1 '18 at 6:57
FrederikFrederik
1009
asked Dec 1 '18 at 6:57
FrederikFrederik
1009
edited Dec 1 '18 at 7:21
Frederik
asked Dec 1 '18 at 6:57
FrederikFrederik
1009
asked Dec 1 '18 at 6:57
FrederikFrederik
1009
asked Dec 1 '18 at 6:57
FrederikFrederik
1009
1009
$begingroup$
@saz Sorry, I forgot that - I have updated my question.
$endgroup$
– Frederik
Dec 1 '18 at 7:24
$begingroup$
the notation $mathcal P(X)_tau$ I guess it mean the $sigma$-algebra generated by $tau$, right? In this case it is enough to show that every subset of $X$ is measurable in the counting measure
$endgroup$
– Masacroso
Dec 1 '18 at 7:29
$begingroup$
Well, the definition of $mathcal{P}(X)_{tau}$ is still missing but probably you define $mathcal{P}(X)_{tau} := sigma(mathcal{P}(X),mathbb{N}_{tau})$. You write that you know that $mathbb{N}_{tau} = {emptyset}$... so what exactly do you not understand? If you know this, the assertion is immediate from the definition...
$endgroup$
– saz
Dec 1 '18 at 7:29
$begingroup$
@saz Sorry for the confusion in the question - I don't know how this can shown in any other way than intuitively.
$endgroup$
– Frederik
Dec 1 '18 at 7:39
$begingroup$
Thank you @Masacroso. You are right, it is the enlargement with respect to $tau$ (if I understand the idea of enlargement correctly).
$endgroup$
– Frederik
Dec 1 '18 at 7:39
add a comment |
$begingroup$
@saz Sorry, I forgot that - I have updated my question.
$endgroup$
– Frederik
Dec 1 '18 at 7:24
$begingroup$
the notation $mathcal P(X)_tau$ I guess it mean the $sigma$-algebra generated by $tau$, right? In this case it is enough to show that every subset of $X$ is measurable in the counting measure
$endgroup$
– Masacroso
Dec 1 '18 at 7:29
$begingroup$
Well, the definition of $mathcal{P}(X)_{tau}$ is still missing but probably you define $mathcal{P}(X)_{tau} := sigma(mathcal{P}(X),mathbb{N}_{tau})$. You write that you know that $mathbb{N}_{tau} = {emptyset}$... so what exactly do you not understand? If you know this, the assertion is immediate from the definition...
$endgroup$
– saz
Dec 1 '18 at 7:29
$begingroup$
@saz Sorry for the confusion in the question - I don't know how this can shown in any other way than intuitively.
$endgroup$
– Frederik
Dec 1 '18 at 7:39
$begingroup$
Thank you @Masacroso. You are right, it is the enlargement with respect to $tau$ (if I understand the idea of enlargement correctly).
$endgroup$
– Frederik
Dec 1 '18 at 7:39
$begingroup$
@saz Sorry, I forgot that - I have updated my question.
$endgroup$
– Frederik
Dec 1 '18 at 7:24
$begingroup$
@saz Sorry, I forgot that - I have updated my question.
$endgroup$
– Frederik
Dec 1 '18 at 7:24
$begingroup$
the notation $mathcal P(X)_tau$ I guess it mean the $sigma$-algebra generated by $tau$, right? In this case it is enough to show that every subset of $X$ is measurable in the counting measure
$endgroup$
– Masacroso
Dec 1 '18 at 7:29
$begingroup$
the notation $mathcal P(X)_tau$ I guess it mean the $sigma$-algebra generated by $tau$, right? In this case it is enough to show that every subset of $X$ is measurable in the counting measure
$endgroup$
– Masacroso
Dec 1 '18 at 7:29
$begingroup$
Well, the definition of $mathcal{P}(X)_{tau}$ is still missing but probably you define $mathcal{P}(X)_{tau} := sigma(mathcal{P}(X),mathbb{N}_{tau})$. You write that you know that $mathbb{N}_{tau} = {emptyset}$... so what exactly do you not understand? If you know this, the assertion is immediate from the definition...
$endgroup$
– saz
Dec 1 '18 at 7:29
$begingroup$
Well, the definition of $mathcal{P}(X)_{tau}$ is still missing but probably you define $mathcal{P}(X)_{tau} := sigma(mathcal{P}(X),mathbb{N}_{tau})$. You write that you know that $mathbb{N}_{tau} = {emptyset}$... so what exactly do you not understand? If you know this, the assertion is immediate from the definition...
$endgroup$
– saz
Dec 1 '18 at 7:29
$begingroup$
@saz Sorry for the confusion in the question - I don't know how this can shown in any other way than intuitively.
$endgroup$
– Frederik
Dec 1 '18 at 7:39
$begingroup$
@saz Sorry for the confusion in the question - I don't know how this can shown in any other way than intuitively.
$endgroup$
– Frederik
Dec 1 '18 at 7:39
$begingroup$
Thank you @Masacroso. You are right, it is the enlargement with respect to $tau$ (if I understand the idea of enlargement correctly).
$endgroup$
– Frederik
Dec 1 '18 at 7:39
$begingroup$
Thank you @Masacroso. You are right, it is the enlargement with respect to $tau$ (if I understand the idea of enlargement correctly).
$endgroup$
– Frederik
Dec 1 '18 at 7:39
add a comment |
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$begingroup$
@saz Sorry, I forgot that - I have updated my question.
$endgroup$
– Frederik
Dec 1 '18 at 7:24
$begingroup$
the notation $mathcal P(X)_tau$ I guess it mean the $sigma$-algebra generated by $tau$, right? In this case it is enough to show that every subset of $X$ is measurable in the counting measure
$endgroup$
– Masacroso
Dec 1 '18 at 7:29
$begingroup$
Well, the definition of $mathcal{P}(X)_{tau}$ is still missing but probably you define $mathcal{P}(X)_{tau} := sigma(mathcal{P}(X),mathbb{N}_{tau})$. You write that you know that $mathbb{N}_{tau} = {emptyset}$... so what exactly do you not understand? If you know this, the assertion is immediate from the definition...
$endgroup$
– saz
Dec 1 '18 at 7:29
$begingroup$
@saz Sorry for the confusion in the question - I don't know how this can shown in any other way than intuitively.
$endgroup$
– Frederik
Dec 1 '18 at 7:39
$begingroup$
Thank you @Masacroso. You are right, it is the enlargement with respect to $tau$ (if I understand the idea of enlargement correctly).
$endgroup$
– Frederik
Dec 1 '18 at 7:39