$mu$ pure point measure if and only if $mu(A)=sum_{xin A} mu(left{xright})$
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$mu$ is a pure point measure if and only if for every $Ain mathcal{B}(X)$, $mathcal{B}(X)$ sigma-algebra Baire. $X$ compact hausdorff
$mu(A)=sum_{xin A} mu(left{xright})$
I have this
If $mu(A)=0$ is hold.
If $mu(A)>0$, then exists $xin A: mu(left{xright})>0$
Then $left{x:mu(A)>0right}=bigcup_{xin A} left{x:mu(left{xright})>0right}$ and I do not know how to continue ...
real-analysis functional-analysis measure-theory
asked Nov 17 at 3:10
eraldcoil
27619
add a comment |
up vote
0
down vote
favorite
$mu$ is a pure point measure if and only if for every $Ain mathcal{B}(X)$, $mathcal{B}(X)$ sigma-algebra Baire. $X$ compact hausdorff
$mu(A)=sum_{xin A} mu(left{xright})$
I have this
If $mu(A)=0$ is hold.
If $mu(A)>0$, then exists $xin A: mu(left{xright})>0$
Then $left{x:mu(A)>0right}=bigcup_{xin A} left{x:mu(left{xright})>0right}$ and I do not know how to continue ...
real-analysis functional-analysis measure-theory
asked Nov 17 at 3:10
eraldcoil
27619
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$mu$ is a pure point measure if and only if for every $Ain mathcal{B}(X)$, $mathcal{B}(X)$ sigma-algebra Baire. $X$ compact hausdorff
$mu(A)=sum_{xin A} mu(left{xright})$
I have this
If $mu(A)=0$ is hold.
If $mu(A)>0$, then exists $xin A: mu(left{xright})>0$
Then $left{x:mu(A)>0right}=bigcup_{xin A} left{x:mu(left{xright})>0right}$ and I do not know how to continue ...
real-analysis functional-analysis measure-theory
asked Nov 17 at 3:10
eraldcoil
27619
$mu$ is a pure point measure if and only if for every $Ain mathcal{B}(X)$, $mathcal{B}(X)$ sigma-algebra Baire. $X$ compact hausdorff
$mu(A)=sum_{xin A} mu(left{xright})$
I have this
If $mu(A)=0$ is hold.
If $mu(A)>0$, then exists $xin A: mu(left{xright})>0$
Then $left{x:mu(A)>0right}=bigcup_{xin A} left{x:mu(left{xright})>0right}$ and I do not know how to continue ...
real-analysis functional-analysis measure-theory
real-analysis functional-analysis measure-theory
asked Nov 17 at 3:10
eraldcoil
27619
asked Nov 17 at 3:10
eraldcoil
27619
asked Nov 17 at 3:10
eraldcoil
27619
asked Nov 17 at 3:10
eraldcoil
27619
asked Nov 17 at 3:10
eraldcoil
27619
27619
add a comment |
add a comment |
1 Answer
1
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2
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I think $mu$ is supposed to be a finite measure. ''If" part: ${x:mu{x}>frac 1 n}$ is finite for each $n$ and hence ${x:mu{x}>0}$ is an at most countable set , say ${x_1,x_2,...}$. Verify now that $mu =sum_n mu{x_n}delta_{x_n}$. (You need the fact that $mu{x}=0$ if $x notin {x_1,x_2,...}$). "Only if" part is obvious.
answered Nov 17 at 5:07
Kavi Rama Murthy
43.9k31852
1
Presumably it would be sufficient for $mu$ to be decomposable.
– copper.hat
Nov 19 at 0:27
thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
– eraldcoil
Nov 19 at 2:00
How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
– eraldcoil
Nov 19 at 2:22
1
@eraldcoil Yes, that is correct.
– Kavi Rama Murthy
Nov 19 at 5:09
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I think $mu$ is supposed to be a finite measure. ''If" part: ${x:mu{x}>frac 1 n}$ is finite for each $n$ and hence ${x:mu{x}>0}$ is an at most countable set , say ${x_1,x_2,...}$. Verify now that $mu =sum_n mu{x_n}delta_{x_n}$. (You need the fact that $mu{x}=0$ if $x notin {x_1,x_2,...}$). "Only if" part is obvious.
answered Nov 17 at 5:07
Kavi Rama Murthy
43.9k31852
1
Presumably it would be sufficient for $mu$ to be decomposable.
– copper.hat
Nov 19 at 0:27
thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
– eraldcoil
Nov 19 at 2:00
How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
– eraldcoil
Nov 19 at 2:22
1
@eraldcoil Yes, that is correct.
– Kavi Rama Murthy
Nov 19 at 5:09
add a comment |
up vote
2
down vote
accepted
I think $mu$ is supposed to be a finite measure. ''If" part: ${x:mu{x}>frac 1 n}$ is finite for each $n$ and hence ${x:mu{x}>0}$ is an at most countable set , say ${x_1,x_2,...}$. Verify now that $mu =sum_n mu{x_n}delta_{x_n}$. (You need the fact that $mu{x}=0$ if $x notin {x_1,x_2,...}$). "Only if" part is obvious.
answered Nov 17 at 5:07
Kavi Rama Murthy
43.9k31852
1
Presumably it would be sufficient for $mu$ to be decomposable.
– copper.hat
Nov 19 at 0:27
thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
– eraldcoil
Nov 19 at 2:00
How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
– eraldcoil
Nov 19 at 2:22
1
@eraldcoil Yes, that is correct.
– Kavi Rama Murthy
Nov 19 at 5:09
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I think $mu$ is supposed to be a finite measure. ''If" part: ${x:mu{x}>frac 1 n}$ is finite for each $n$ and hence ${x:mu{x}>0}$ is an at most countable set , say ${x_1,x_2,...}$. Verify now that $mu =sum_n mu{x_n}delta_{x_n}$. (You need the fact that $mu{x}=0$ if $x notin {x_1,x_2,...}$). "Only if" part is obvious.
answered Nov 17 at 5:07
Kavi Rama Murthy
43.9k31852
I think $mu$ is supposed to be a finite measure. ''If" part: ${x:mu{x}>frac 1 n}$ is finite for each $n$ and hence ${x:mu{x}>0}$ is an at most countable set , say ${x_1,x_2,...}$. Verify now that $mu =sum_n mu{x_n}delta_{x_n}$. (You need the fact that $mu{x}=0$ if $x notin {x_1,x_2,...}$). "Only if" part is obvious.
answered Nov 17 at 5:07
Kavi Rama Murthy
43.9k31852
answered Nov 17 at 5:07
Kavi Rama Murthy
43.9k31852
answered Nov 17 at 5:07
Kavi Rama Murthy
43.9k31852
answered Nov 17 at 5:07
Kavi Rama Murthy
43.9k31852
43.9k31852
1
Presumably it would be sufficient for $mu$ to be decomposable.
– copper.hat
Nov 19 at 0:27
thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
– eraldcoil
Nov 19 at 2:00
How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
– eraldcoil
Nov 19 at 2:22
1
@eraldcoil Yes, that is correct.
– Kavi Rama Murthy
Nov 19 at 5:09
add a comment |
1
Presumably it would be sufficient for $mu$ to be decomposable.
– copper.hat
Nov 19 at 0:27
thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
– eraldcoil
Nov 19 at 2:00
How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
– eraldcoil
Nov 19 at 2:22
1
@eraldcoil Yes, that is correct.
– Kavi Rama Murthy
Nov 19 at 5:09
1
1
Presumably it would be sufficient for $mu$ to be decomposable.
– copper.hat
Nov 19 at 0:27
Presumably it would be sufficient for $mu$ to be decomposable.
– copper.hat
Nov 19 at 0:27
thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
– eraldcoil
Nov 19 at 2:00
thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
– eraldcoil
Nov 19 at 2:00
How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
– eraldcoil
Nov 19 at 2:22
How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
– eraldcoil
Nov 19 at 2:22
1
1
@eraldcoil Yes, that is correct.
– Kavi Rama Murthy
Nov 19 at 5:09
@eraldcoil Yes, that is correct.
– Kavi Rama Murthy
Nov 19 at 5:09
add a comment |
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