Fourier uniqueness theorem
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I am reading Theorem 26.2 in the book "Probability and Measure" (2nd ed.) by Patrick Billingsley. In the proof of that theorem, the author claims that:
Fact Let $mu$ and $nu$ be probability measures on the Borel subsets of $mathbb{R}$, and set
$$A=left{;left(a,bright];middle|;mu({a})=mu({b})=nu({a})=nu({b})=0right}.$$
Then $sigma(A)$ coincides with the Borel sigma-algebra on $mathbb{R}$.
I have been thinking about this result, but have absolutely no clues on this. It does not seem straightforward. So my question is:
Question What is a clean way to prove this?
Any hint/help is highly appreciated.
probability-theory probability-distributions fourier-transform characteristic-functions
asked Nov 26 at 6:19
weirdo
413210
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up vote
1
down vote
favorite
I am reading Theorem 26.2 in the book "Probability and Measure" (2nd ed.) by Patrick Billingsley. In the proof of that theorem, the author claims that:
Fact Let $mu$ and $nu$ be probability measures on the Borel subsets of $mathbb{R}$, and set
$$A=left{;left(a,bright];middle|;mu({a})=mu({b})=nu({a})=nu({b})=0right}.$$
Then $sigma(A)$ coincides with the Borel sigma-algebra on $mathbb{R}$.
I have been thinking about this result, but have absolutely no clues on this. It does not seem straightforward. So my question is:
Question What is a clean way to prove this?
Any hint/help is highly appreciated.
probability-theory probability-distributions fourier-transform characteristic-functions
asked Nov 26 at 6:19
weirdo
413210
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am reading Theorem 26.2 in the book "Probability and Measure" (2nd ed.) by Patrick Billingsley. In the proof of that theorem, the author claims that:
Fact Let $mu$ and $nu$ be probability measures on the Borel subsets of $mathbb{R}$, and set
$$A=left{;left(a,bright];middle|;mu({a})=mu({b})=nu({a})=nu({b})=0right}.$$
Then $sigma(A)$ coincides with the Borel sigma-algebra on $mathbb{R}$.
I have been thinking about this result, but have absolutely no clues on this. It does not seem straightforward. So my question is:
Question What is a clean way to prove this?
Any hint/help is highly appreciated.
probability-theory probability-distributions fourier-transform characteristic-functions
asked Nov 26 at 6:19
weirdo
413210
I am reading Theorem 26.2 in the book "Probability and Measure" (2nd ed.) by Patrick Billingsley. In the proof of that theorem, the author claims that:
Fact Let $mu$ and $nu$ be probability measures on the Borel subsets of $mathbb{R}$, and set
$$A=left{;left(a,bright];middle|;mu({a})=mu({b})=nu({a})=nu({b})=0right}.$$
Then $sigma(A)$ coincides with the Borel sigma-algebra on $mathbb{R}$.
I have been thinking about this result, but have absolutely no clues on this. It does not seem straightforward. So my question is:
Question What is a clean way to prove this?
Any hint/help is highly appreciated.
probability-theory probability-distributions fourier-transform characteristic-functions
probability-theory probability-distributions fourier-transform characteristic-functions
asked Nov 26 at 6:19
weirdo
413210
asked Nov 26 at 6:19
weirdo
413210
asked Nov 26 at 6:19
weirdo
413210
asked Nov 26 at 6:19
weirdo
413210
asked Nov 26 at 6:19
weirdo
413210
413210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
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accepted
Hint: $mu {x}=nu {x}=0$ for all but countable number of $x$. So there is a countable set ${t_1,t_2,...}$ such that the given class contains all intervals $(a,b]$ whose end points are not in this countable set. Can you express any interval in terms of these using countable unions /intersections? (The complement of any countable set is dense). Will be glad to provide more details if needed.
answered Nov 26 at 6:28
Kavi Rama Murthy
43.6k31751
Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
– weirdo
Nov 26 at 7:30
1
@weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
– Kavi Rama Murthy
Nov 26 at 7:32
I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
– weirdo
Nov 26 at 7:59
1
@weirdo You are right.
– Kavi Rama Murthy
Nov 26 at 8:00
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Hint: $mu {x}=nu {x}=0$ for all but countable number of $x$. So there is a countable set ${t_1,t_2,...}$ such that the given class contains all intervals $(a,b]$ whose end points are not in this countable set. Can you express any interval in terms of these using countable unions /intersections? (The complement of any countable set is dense). Will be glad to provide more details if needed.
answered Nov 26 at 6:28
Kavi Rama Murthy
43.6k31751
Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
– weirdo
Nov 26 at 7:30
1
@weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
– Kavi Rama Murthy
Nov 26 at 7:32
I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
– weirdo
Nov 26 at 7:59
1
@weirdo You are right.
– Kavi Rama Murthy
Nov 26 at 8:00
add a comment |
up vote
4
down vote
accepted
Hint: $mu {x}=nu {x}=0$ for all but countable number of $x$. So there is a countable set ${t_1,t_2,...}$ such that the given class contains all intervals $(a,b]$ whose end points are not in this countable set. Can you express any interval in terms of these using countable unions /intersections? (The complement of any countable set is dense). Will be glad to provide more details if needed.
answered Nov 26 at 6:28
Kavi Rama Murthy
43.6k31751
Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
– weirdo
Nov 26 at 7:30
1
@weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
– Kavi Rama Murthy
Nov 26 at 7:32
I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
– weirdo
Nov 26 at 7:59
1
@weirdo You are right.
– Kavi Rama Murthy
Nov 26 at 8:00
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Hint: $mu {x}=nu {x}=0$ for all but countable number of $x$. So there is a countable set ${t_1,t_2,...}$ such that the given class contains all intervals $(a,b]$ whose end points are not in this countable set. Can you express any interval in terms of these using countable unions /intersections? (The complement of any countable set is dense). Will be glad to provide more details if needed.
answered Nov 26 at 6:28
Kavi Rama Murthy
43.6k31751
Hint: $mu {x}=nu {x}=0$ for all but countable number of $x$. So there is a countable set ${t_1,t_2,...}$ such that the given class contains all intervals $(a,b]$ whose end points are not in this countable set. Can you express any interval in terms of these using countable unions /intersections? (The complement of any countable set is dense). Will be glad to provide more details if needed.
answered Nov 26 at 6:28
Kavi Rama Murthy
43.6k31751
answered Nov 26 at 6:28
Kavi Rama Murthy
43.6k31751
answered Nov 26 at 6:28
Kavi Rama Murthy
43.6k31751
answered Nov 26 at 6:28
Kavi Rama Murthy
43.6k31751
43.6k31751
Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
– weirdo
Nov 26 at 7:30
1
@weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
– Kavi Rama Murthy
Nov 26 at 7:32
I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
– weirdo
Nov 26 at 7:59
1
@weirdo You are right.
– Kavi Rama Murthy
Nov 26 at 8:00
add a comment |
Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
– weirdo
Nov 26 at 7:30
1
@weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
– Kavi Rama Murthy
Nov 26 at 7:32
I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
– weirdo
Nov 26 at 7:59
1
@weirdo You are right.
– Kavi Rama Murthy
Nov 26 at 8:00
Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
– weirdo
Nov 26 at 7:30
Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
– weirdo
Nov 26 at 7:30
1
1
@weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
– Kavi Rama Murthy
Nov 26 at 7:32
@weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
– Kavi Rama Murthy
Nov 26 at 7:32
I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
– weirdo
Nov 26 at 7:59
I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
– weirdo
Nov 26 at 7:59
1
1
@weirdo You are right.
– Kavi Rama Murthy
Nov 26 at 8:00
@weirdo You are right.
– Kavi Rama Murthy
Nov 26 at 8:00
add a comment |
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