The difference in the two arguments regarding completion of dirac-measure
$begingroup$
Given the space $(mathbb{R},mathbb{B},mu)$ where $mathbb{B}$ is the $sigma$-algebra on $X$ and $delta$ is the dirac-measure (1 if $ain Binmathbb{B}$ and zero otherwise), I cannot understand the difference in the following two arguments.
The completion would be:
$$mathbb{B}_{delta_{a}}={Bcup N;|;Binmathbb{B},Ninmathbb{N}_{delta_{a}}}subseteq{Bcup {a};|;Binmathbb{B},anotin B}=mathcal{P}(X)$$
however, why is it wrong to say:
$mathbb{B}(mathbb{R})cup mathcal{P}(mathbb{R}$ ${a}}=mathcal{P}(X),$
as $mathbb{N}_{delta_{a}}={Nsubseteq mathbb{R}|Nsubseteq mathbb{R}$ ${a}}?$
measure-theory
asked Dec 3 '18 at 7:37
FrederikFrederik
1009
$endgroup$
add a comment |
$begingroup$
Given the space $(mathbb{R},mathbb{B},mu)$ where $mathbb{B}$ is the $sigma$-algebra on $X$ and $delta$ is the dirac-measure (1 if $ain Binmathbb{B}$ and zero otherwise), I cannot understand the difference in the following two arguments.
The completion would be:
$$mathbb{B}_{delta_{a}}={Bcup N;|;Binmathbb{B},Ninmathbb{N}_{delta_{a}}}subseteq{Bcup {a};|;Binmathbb{B},anotin B}=mathcal{P}(X)$$
however, why is it wrong to say:
$mathbb{B}(mathbb{R})cup mathcal{P}(mathbb{R}$ ${a}}=mathcal{P}(X),$
as $mathbb{N}_{delta_{a}}={Nsubseteq mathbb{R}|Nsubseteq mathbb{R}$ ${a}}?$
measure-theory
asked Dec 3 '18 at 7:37
FrederikFrederik
1009
$endgroup$
add a comment |
$begingroup$
Given the space $(mathbb{R},mathbb{B},mu)$ where $mathbb{B}$ is the $sigma$-algebra on $X$ and $delta$ is the dirac-measure (1 if $ain Binmathbb{B}$ and zero otherwise), I cannot understand the difference in the following two arguments.
The completion would be:
$$mathbb{B}_{delta_{a}}={Bcup N;|;Binmathbb{B},Ninmathbb{N}_{delta_{a}}}subseteq{Bcup {a};|;Binmathbb{B},anotin B}=mathcal{P}(X)$$
however, why is it wrong to say:
$mathbb{B}(mathbb{R})cup mathcal{P}(mathbb{R}$ ${a}}=mathcal{P}(X),$
as $mathbb{N}_{delta_{a}}={Nsubseteq mathbb{R}|Nsubseteq mathbb{R}$ ${a}}?$
measure-theory
asked Dec 3 '18 at 7:37
FrederikFrederik
1009
$endgroup$
Given the space $(mathbb{R},mathbb{B},mu)$ where $mathbb{B}$ is the $sigma$-algebra on $X$ and $delta$ is the dirac-measure (1 if $ain Binmathbb{B}$ and zero otherwise), I cannot understand the difference in the following two arguments.
The completion would be:
$$mathbb{B}_{delta_{a}}={Bcup N;|;Binmathbb{B},Ninmathbb{N}_{delta_{a}}}subseteq{Bcup {a};|;Binmathbb{B},anotin B}=mathcal{P}(X)$$
however, why is it wrong to say:
$mathbb{B}(mathbb{R})cup mathcal{P}(mathbb{R}$ ${a}}=mathcal{P}(X),$
as $mathbb{N}_{delta_{a}}={Nsubseteq mathbb{R}|Nsubseteq mathbb{R}$ ${a}}?$
measure-theory
measure-theory
asked Dec 3 '18 at 7:37
FrederikFrederik
1009
asked Dec 3 '18 at 7:37
FrederikFrederik
1009
asked Dec 3 '18 at 7:37
FrederikFrederik
1009
asked Dec 3 '18 at 7:37
FrederikFrederik
1009
asked Dec 3 '18 at 7:37
FrederikFrederik
1009
1009
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $mathcal A $ and $mathcal B $ are two families of sets there is a difference between $mathcal A cup mathcal B $ and ${Acup B: Ain mathcal A, B in mathcal B }$. So you cannot write the completion as $mathcal B (mathbb R) cup mathcal P (mathbb Rsetminus {a})$.
answered Dec 3 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
56.2k42158
$endgroup$
$begingroup$
If understand this correctly, this ${Acup B: Ain mathcal A, B in mathcal B }$ will generate a set of subsets, where each subset in the set contains the union between $A$ and $B$, where $Acup B$ will generate a set containing the union of the sets in $A$ and $B$ as subsets. So the last-mentioned will not have the union of $A$ and $B$ in each subset?
$endgroup$
– Frederik
Dec 3 '18 at 8:14
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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$begingroup$
If $mathcal A $ and $mathcal B $ are two families of sets there is a difference between $mathcal A cup mathcal B $ and ${Acup B: Ain mathcal A, B in mathcal B }$. So you cannot write the completion as $mathcal B (mathbb R) cup mathcal P (mathbb Rsetminus {a})$.
answered Dec 3 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
56.2k42158
$endgroup$
$begingroup$
If understand this correctly, this ${Acup B: Ain mathcal A, B in mathcal B }$ will generate a set of subsets, where each subset in the set contains the union between $A$ and $B$, where $Acup B$ will generate a set containing the union of the sets in $A$ and $B$ as subsets. So the last-mentioned will not have the union of $A$ and $B$ in each subset?
$endgroup$
– Frederik
Dec 3 '18 at 8:14
add a comment |
$begingroup$
If $mathcal A $ and $mathcal B $ are two families of sets there is a difference between $mathcal A cup mathcal B $ and ${Acup B: Ain mathcal A, B in mathcal B }$. So you cannot write the completion as $mathcal B (mathbb R) cup mathcal P (mathbb Rsetminus {a})$.
answered Dec 3 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
56.2k42158
$endgroup$
$begingroup$
If understand this correctly, this ${Acup B: Ain mathcal A, B in mathcal B }$ will generate a set of subsets, where each subset in the set contains the union between $A$ and $B$, where $Acup B$ will generate a set containing the union of the sets in $A$ and $B$ as subsets. So the last-mentioned will not have the union of $A$ and $B$ in each subset?
$endgroup$
– Frederik
Dec 3 '18 at 8:14
add a comment |
$begingroup$
If $mathcal A $ and $mathcal B $ are two families of sets there is a difference between $mathcal A cup mathcal B $ and ${Acup B: Ain mathcal A, B in mathcal B }$. So you cannot write the completion as $mathcal B (mathbb R) cup mathcal P (mathbb Rsetminus {a})$.
answered Dec 3 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
56.2k42158
$endgroup$
If $mathcal A $ and $mathcal B $ are two families of sets there is a difference between $mathcal A cup mathcal B $ and ${Acup B: Ain mathcal A, B in mathcal B }$. So you cannot write the completion as $mathcal B (mathbb R) cup mathcal P (mathbb Rsetminus {a})$.
answered Dec 3 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
56.2k42158
answered Dec 3 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
56.2k42158
answered Dec 3 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
56.2k42158
answered Dec 3 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
56.2k42158
56.2k42158
$begingroup$
If understand this correctly, this ${Acup B: Ain mathcal A, B in mathcal B }$ will generate a set of subsets, where each subset in the set contains the union between $A$ and $B$, where $Acup B$ will generate a set containing the union of the sets in $A$ and $B$ as subsets. So the last-mentioned will not have the union of $A$ and $B$ in each subset?
$endgroup$
– Frederik
Dec 3 '18 at 8:14
add a comment |
$begingroup$
If understand this correctly, this ${Acup B: Ain mathcal A, B in mathcal B }$ will generate a set of subsets, where each subset in the set contains the union between $A$ and $B$, where $Acup B$ will generate a set containing the union of the sets in $A$ and $B$ as subsets. So the last-mentioned will not have the union of $A$ and $B$ in each subset?
$endgroup$
– Frederik
Dec 3 '18 at 8:14
$begingroup$
If understand this correctly, this ${Acup B: Ain mathcal A, B in mathcal B }$ will generate a set of subsets, where each subset in the set contains the union between $A$ and $B$, where $Acup B$ will generate a set containing the union of the sets in $A$ and $B$ as subsets. So the last-mentioned will not have the union of $A$ and $B$ in each subset?
$endgroup$
– Frederik
Dec 3 '18 at 8:14
$begingroup$
If understand this correctly, this ${Acup B: Ain mathcal A, B in mathcal B }$ will generate a set of subsets, where each subset in the set contains the union between $A$ and $B$, where $Acup B$ will generate a set containing the union of the sets in $A$ and $B$ as subsets. So the last-mentioned will not have the union of $A$ and $B$ in each subset?
$endgroup$
– Frederik
Dec 3 '18 at 8:14
add a comment |
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