$mu$ pure point measure if and only if $mu(A)=sum_{xin A} mu(left{xright})$
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
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
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
$mu$ 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
26119
26119
add a comment |
add a comment |
1 Answer
1
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.
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.
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.
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.
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
43k31751
43k31751
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001926%2fmu-pure-point-measure-if-and-only-if-mua-sum-x-in-a-mu-left-x-right%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown