Problem 6. Barry Simon. A comprehensive course in analysis.
up vote
0
down vote
favorite
Let $X$ compact Hausdorff space in wich every $left{xright}$ is a Baire set and let $mu$ be a Baire measure.
Define $mu_{pp}(A)=sum_{xin A}mu(left{xright})$
(a) Prove that $mu_{pp}$ is a pure point measure. I already proves!
(b) Prove that $mu_{c}:=mu-mu_{pp}$ is a nonnegative continuous measure.
I already prove!
(c)If $mu=nu_c+nu_{pp}$ with $nu_{pp}$ and $nu_{c}$ continuous, prove that $nu_{pp}=mu_{pp}$ and so $nu_{c}=mu_{c}$ (Correction. $nu_{pp}$ is a pure point)
How prove (c)?
functional-analysis measure-theory
asked Nov 21 at 5:32
eraldcoil
28119
add a comment |
up vote
0
down vote
favorite
Let $X$ compact Hausdorff space in wich every $left{xright}$ is a Baire set and let $mu$ be a Baire measure.
Define $mu_{pp}(A)=sum_{xin A}mu(left{xright})$
(a) Prove that $mu_{pp}$ is a pure point measure. I already proves!
(b) Prove that $mu_{c}:=mu-mu_{pp}$ is a nonnegative continuous measure.
I already prove!
(c)If $mu=nu_c+nu_{pp}$ with $nu_{pp}$ and $nu_{c}$ continuous, prove that $nu_{pp}=mu_{pp}$ and so $nu_{c}=mu_{c}$ (Correction. $nu_{pp}$ is a pure point)
How prove (c)?
functional-analysis measure-theory
asked Nov 21 at 5:32
eraldcoil
28119
You have not quoted c) correctly. If $nu_c$ and $nu_{pp}$ are both continuous than $mu$ itself is continuous.
– Kavi Rama Murthy
Nov 21 at 5:43
oh! yes. $nu_{pp}$ is pure point
– eraldcoil
Nov 21 at 5:51
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ compact Hausdorff space in wich every $left{xright}$ is a Baire set and let $mu$ be a Baire measure.
Define $mu_{pp}(A)=sum_{xin A}mu(left{xright})$
(a) Prove that $mu_{pp}$ is a pure point measure. I already proves!
(b) Prove that $mu_{c}:=mu-mu_{pp}$ is a nonnegative continuous measure.
I already prove!
(c)If $mu=nu_c+nu_{pp}$ with $nu_{pp}$ and $nu_{c}$ continuous, prove that $nu_{pp}=mu_{pp}$ and so $nu_{c}=mu_{c}$ (Correction. $nu_{pp}$ is a pure point)
How prove (c)?
functional-analysis measure-theory
asked Nov 21 at 5:32
eraldcoil
28119
Let $X$ compact Hausdorff space in wich every $left{xright}$ is a Baire set and let $mu$ be a Baire measure.
Define $mu_{pp}(A)=sum_{xin A}mu(left{xright})$
(a) Prove that $mu_{pp}$ is a pure point measure. I already proves!
(b) Prove that $mu_{c}:=mu-mu_{pp}$ is a nonnegative continuous measure.
I already prove!
(c)If $mu=nu_c+nu_{pp}$ with $nu_{pp}$ and $nu_{c}$ continuous, prove that $nu_{pp}=mu_{pp}$ and so $nu_{c}=mu_{c}$ (Correction. $nu_{pp}$ is a pure point)
How prove (c)?
functional-analysis measure-theory
functional-analysis measure-theory
asked Nov 21 at 5:32
eraldcoil
28119
asked Nov 21 at 5:32
eraldcoil
28119
edited Nov 21 at 5:53
asked Nov 21 at 5:32
eraldcoil
28119
asked Nov 21 at 5:32
eraldcoil
28119
asked Nov 21 at 5:32
eraldcoil
28119
28119
You have not quoted c) correctly. If $nu_c$ and $nu_{pp}$ are both continuous than $mu$ itself is continuous.
– Kavi Rama Murthy
Nov 21 at 5:43
oh! yes. $nu_{pp}$ is pure point
– eraldcoil
Nov 21 at 5:51
add a comment |
You have not quoted c) correctly. If $nu_c$ and $nu_{pp}$ are both continuous than $mu$ itself is continuous.
– Kavi Rama Murthy
Nov 21 at 5:43
oh! yes. $nu_{pp}$ is pure point
– eraldcoil
Nov 21 at 5:51
You have not quoted c) correctly. If $nu_c$ and $nu_{pp}$ are both continuous than $mu$ itself is continuous.
– Kavi Rama Murthy
Nov 21 at 5:43
You have not quoted c) correctly. If $nu_c$ and $nu_{pp}$ are both continuous than $mu$ itself is continuous.
– Kavi Rama Murthy
Nov 21 at 5:43
oh! yes. $nu_{pp}$ is pure point
– eraldcoil
Nov 21 at 5:51
oh! yes. $nu_{pp}$ is pure point
– eraldcoil
Nov 21 at 5:51
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let $C={x:mu {x} neq 0}$ and $D={x:nu_{pp} {x} neq 0}$, Verify the following for any measurable set $E$: $$nu_c(E)=nu_c(Esetminus Ccup D)=mu (Esetminus Ccup D)-nu_{pp}(Esetminus Ccup D)$$ $$=mu_c (Esetminus Ccup D)+mu_{pp} (Esetminus Ccup D)-nu_{pp}(Esetminus Ccup D)$$ $$=mu_c (Esetminus Ccup D)+0+0=mu_c(E).$$ This proves that $nu_c=mu_c$ and it follows automatically that $mu_{pp}=nu_{pp}$.
answered Nov 21 at 6:03
Kavi Rama Murthy
44.7k31852
Why $mu_{pp}( (Esetminus C)cup D)=0$? I have this. Sup. $mu_{pp}( (Esetminus C)cup D)>0$ then exists $xin (Esetminus C)cup D$ such that $mu(x)>0$. If $xin Esetminus C$ then $xnotin C$, implies $mu(x)=0$ a contradiction. If $xin D.$ What is the contradiction?
– eraldcoil
Nov 25 at 7:33
@eraldcoil By $Esetminus Ccup D$ I meant $Esetminus (Ccup D)$, not $(Esetminus C)cup D$
– Kavi Rama Murthy
Nov 25 at 11:38
ah ok! thanks. I see $nu{pp}(Esetminus (Ccup D))=0$ but if $nu{pp}(Esetminus (Ccup D))>0$ then exits $xin Esetminus (Ccup D): nu_{pp}(x)>0$ a contradiction because $xin D^c$. Similary, $mu_{pp}(Esetminus (Ccup D))=0$. But I still do not see $nu_c(E)=nu_c(Esetminus (Ccup D))$
– eraldcoil
Nov 25 at 21:27
@eraldcoil $C$ and $D$ are both countable sets and $nu_c$ is a continuous measure. Hence $nu_c(Ccup D)=0$.
– Kavi Rama Murthy
Nov 25 at 23:10
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
Let $C={x:mu {x} neq 0}$ and $D={x:nu_{pp} {x} neq 0}$, Verify the following for any measurable set $E$: $$nu_c(E)=nu_c(Esetminus Ccup D)=mu (Esetminus Ccup D)-nu_{pp}(Esetminus Ccup D)$$ $$=mu_c (Esetminus Ccup D)+mu_{pp} (Esetminus Ccup D)-nu_{pp}(Esetminus Ccup D)$$ $$=mu_c (Esetminus Ccup D)+0+0=mu_c(E).$$ This proves that $nu_c=mu_c$ and it follows automatically that $mu_{pp}=nu_{pp}$.
answered Nov 21 at 6:03
Kavi Rama Murthy
44.7k31852
Why $mu_{pp}( (Esetminus C)cup D)=0$? I have this. Sup. $mu_{pp}( (Esetminus C)cup D)>0$ then exists $xin (Esetminus C)cup D$ such that $mu(x)>0$. If $xin Esetminus C$ then $xnotin C$, implies $mu(x)=0$ a contradiction. If $xin D.$ What is the contradiction?
– eraldcoil
Nov 25 at 7:33
@eraldcoil By $Esetminus Ccup D$ I meant $Esetminus (Ccup D)$, not $(Esetminus C)cup D$
– Kavi Rama Murthy
Nov 25 at 11:38
ah ok! thanks. I see $nu{pp}(Esetminus (Ccup D))=0$ but if $nu{pp}(Esetminus (Ccup D))>0$ then exits $xin Esetminus (Ccup D): nu_{pp}(x)>0$ a contradiction because $xin D^c$. Similary, $mu_{pp}(Esetminus (Ccup D))=0$. But I still do not see $nu_c(E)=nu_c(Esetminus (Ccup D))$
– eraldcoil
Nov 25 at 21:27
@eraldcoil $C$ and $D$ are both countable sets and $nu_c$ is a continuous measure. Hence $nu_c(Ccup D)=0$.
– Kavi Rama Murthy
Nov 25 at 23:10
add a comment |
up vote
2
down vote
accepted
Let $C={x:mu {x} neq 0}$ and $D={x:nu_{pp} {x} neq 0}$, Verify the following for any measurable set $E$: $$nu_c(E)=nu_c(Esetminus Ccup D)=mu (Esetminus Ccup D)-nu_{pp}(Esetminus Ccup D)$$ $$=mu_c (Esetminus Ccup D)+mu_{pp} (Esetminus Ccup D)-nu_{pp}(Esetminus Ccup D)$$ $$=mu_c (Esetminus Ccup D)+0+0=mu_c(E).$$ This proves that $nu_c=mu_c$ and it follows automatically that $mu_{pp}=nu_{pp}$.
answered Nov 21 at 6:03
Kavi Rama Murthy
44.7k31852
Why $mu_{pp}( (Esetminus C)cup D)=0$? I have this. Sup. $mu_{pp}( (Esetminus C)cup D)>0$ then exists $xin (Esetminus C)cup D$ such that $mu(x)>0$. If $xin Esetminus C$ then $xnotin C$, implies $mu(x)=0$ a contradiction. If $xin D.$ What is the contradiction?
– eraldcoil
Nov 25 at 7:33
@eraldcoil By $Esetminus Ccup D$ I meant $Esetminus (Ccup D)$, not $(Esetminus C)cup D$
– Kavi Rama Murthy
Nov 25 at 11:38
ah ok! thanks. I see $nu{pp}(Esetminus (Ccup D))=0$ but if $nu{pp}(Esetminus (Ccup D))>0$ then exits $xin Esetminus (Ccup D): nu_{pp}(x)>0$ a contradiction because $xin D^c$. Similary, $mu_{pp}(Esetminus (Ccup D))=0$. But I still do not see $nu_c(E)=nu_c(Esetminus (Ccup D))$
– eraldcoil
Nov 25 at 21:27
@eraldcoil $C$ and $D$ are both countable sets and $nu_c$ is a continuous measure. Hence $nu_c(Ccup D)=0$.
– Kavi Rama Murthy
Nov 25 at 23:10
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $C={x:mu {x} neq 0}$ and $D={x:nu_{pp} {x} neq 0}$, Verify the following for any measurable set $E$: $$nu_c(E)=nu_c(Esetminus Ccup D)=mu (Esetminus Ccup D)-nu_{pp}(Esetminus Ccup D)$$ $$=mu_c (Esetminus Ccup D)+mu_{pp} (Esetminus Ccup D)-nu_{pp}(Esetminus Ccup D)$$ $$=mu_c (Esetminus Ccup D)+0+0=mu_c(E).$$ This proves that $nu_c=mu_c$ and it follows automatically that $mu_{pp}=nu_{pp}$.
answered Nov 21 at 6:03
Kavi Rama Murthy
44.7k31852
Let $C={x:mu {x} neq 0}$ and $D={x:nu_{pp} {x} neq 0}$, Verify the following for any measurable set $E$: $$nu_c(E)=nu_c(Esetminus Ccup D)=mu (Esetminus Ccup D)-nu_{pp}(Esetminus Ccup D)$$ $$=mu_c (Esetminus Ccup D)+mu_{pp} (Esetminus Ccup D)-nu_{pp}(Esetminus Ccup D)$$ $$=mu_c (Esetminus Ccup D)+0+0=mu_c(E).$$ This proves that $nu_c=mu_c$ and it follows automatically that $mu_{pp}=nu_{pp}$.
answered Nov 21 at 6:03
Kavi Rama Murthy
44.7k31852
answered Nov 21 at 6:03
Kavi Rama Murthy
44.7k31852
answered Nov 21 at 6:03
Kavi Rama Murthy
44.7k31852
answered Nov 21 at 6:03
Kavi Rama Murthy
44.7k31852
44.7k31852
Why $mu_{pp}( (Esetminus C)cup D)=0$? I have this. Sup. $mu_{pp}( (Esetminus C)cup D)>0$ then exists $xin (Esetminus C)cup D$ such that $mu(x)>0$. If $xin Esetminus C$ then $xnotin C$, implies $mu(x)=0$ a contradiction. If $xin D.$ What is the contradiction?
– eraldcoil
Nov 25 at 7:33
@eraldcoil By $Esetminus Ccup D$ I meant $Esetminus (Ccup D)$, not $(Esetminus C)cup D$
– Kavi Rama Murthy
Nov 25 at 11:38
ah ok! thanks. I see $nu{pp}(Esetminus (Ccup D))=0$ but if $nu{pp}(Esetminus (Ccup D))>0$ then exits $xin Esetminus (Ccup D): nu_{pp}(x)>0$ a contradiction because $xin D^c$. Similary, $mu_{pp}(Esetminus (Ccup D))=0$. But I still do not see $nu_c(E)=nu_c(Esetminus (Ccup D))$
– eraldcoil
Nov 25 at 21:27
@eraldcoil $C$ and $D$ are both countable sets and $nu_c$ is a continuous measure. Hence $nu_c(Ccup D)=0$.
– Kavi Rama Murthy
Nov 25 at 23:10
add a comment |
Why $mu_{pp}( (Esetminus C)cup D)=0$? I have this. Sup. $mu_{pp}( (Esetminus C)cup D)>0$ then exists $xin (Esetminus C)cup D$ such that $mu(x)>0$. If $xin Esetminus C$ then $xnotin C$, implies $mu(x)=0$ a contradiction. If $xin D.$ What is the contradiction?
– eraldcoil
Nov 25 at 7:33
@eraldcoil By $Esetminus Ccup D$ I meant $Esetminus (Ccup D)$, not $(Esetminus C)cup D$
– Kavi Rama Murthy
Nov 25 at 11:38
ah ok! thanks. I see $nu{pp}(Esetminus (Ccup D))=0$ but if $nu{pp}(Esetminus (Ccup D))>0$ then exits $xin Esetminus (Ccup D): nu_{pp}(x)>0$ a contradiction because $xin D^c$. Similary, $mu_{pp}(Esetminus (Ccup D))=0$. But I still do not see $nu_c(E)=nu_c(Esetminus (Ccup D))$
– eraldcoil
Nov 25 at 21:27
@eraldcoil $C$ and $D$ are both countable sets and $nu_c$ is a continuous measure. Hence $nu_c(Ccup D)=0$.
– Kavi Rama Murthy
Nov 25 at 23:10
Why $mu_{pp}( (Esetminus C)cup D)=0$? I have this. Sup. $mu_{pp}( (Esetminus C)cup D)>0$ then exists $xin (Esetminus C)cup D$ such that $mu(x)>0$. If $xin Esetminus C$ then $xnotin C$, implies $mu(x)=0$ a contradiction. If $xin D.$ What is the contradiction?
– eraldcoil
Nov 25 at 7:33
Why $mu_{pp}( (Esetminus C)cup D)=0$? I have this. Sup. $mu_{pp}( (Esetminus C)cup D)>0$ then exists $xin (Esetminus C)cup D$ such that $mu(x)>0$. If $xin Esetminus C$ then $xnotin C$, implies $mu(x)=0$ a contradiction. If $xin D.$ What is the contradiction?
– eraldcoil
Nov 25 at 7:33
@eraldcoil By $Esetminus Ccup D$ I meant $Esetminus (Ccup D)$, not $(Esetminus C)cup D$
– Kavi Rama Murthy
Nov 25 at 11:38
@eraldcoil By $Esetminus Ccup D$ I meant $Esetminus (Ccup D)$, not $(Esetminus C)cup D$
– Kavi Rama Murthy
Nov 25 at 11:38
ah ok! thanks. I see $nu{pp}(Esetminus (Ccup D))=0$ but if $nu{pp}(Esetminus (Ccup D))>0$ then exits $xin Esetminus (Ccup D): nu_{pp}(x)>0$ a contradiction because $xin D^c$. Similary, $mu_{pp}(Esetminus (Ccup D))=0$. But I still do not see $nu_c(E)=nu_c(Esetminus (Ccup D))$
– eraldcoil
Nov 25 at 21:27
ah ok! thanks. I see $nu{pp}(Esetminus (Ccup D))=0$ but if $nu{pp}(Esetminus (Ccup D))>0$ then exits $xin Esetminus (Ccup D): nu_{pp}(x)>0$ a contradiction because $xin D^c$. Similary, $mu_{pp}(Esetminus (Ccup D))=0$. But I still do not see $nu_c(E)=nu_c(Esetminus (Ccup D))$
– eraldcoil
Nov 25 at 21:27
@eraldcoil $C$ and $D$ are both countable sets and $nu_c$ is a continuous measure. Hence $nu_c(Ccup D)=0$.
– Kavi Rama Murthy
Nov 25 at 23:10
@eraldcoil $C$ and $D$ are both countable sets and $nu_c$ is a continuous measure. Hence $nu_c(Ccup D)=0$.
– Kavi Rama Murthy
Nov 25 at 23:10
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%2f3007304%2fproblem-6-barry-simon-a-comprehensive-course-in-analysis%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
You have not quoted c) correctly. If $nu_c$ and $nu_{pp}$ are both continuous than $mu$ itself is continuous.
– Kavi Rama Murthy
Nov 21 at 5:43
oh! yes. $nu_{pp}$ is pure point
– eraldcoil
Nov 21 at 5:51