How to proof that $mathcal{P}(X)_{tau}=mathcal{P}(X)$ on the measure space $(X,mathcal{P}(X))$












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    @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


















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    @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
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 7:21







Frederik

















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




















  • $begingroup$
    @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












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