Are $sigma$-finite measures agreeing on a generating set equal?
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Suppose I have two measures $m,n$ and that they are both $sigma$-finite and agree on some semi-ring $R$ generating their $sigma$-algebras. Must these two measures then agree on the entire $sigma$-algebra?
Of course, the Caratheodory extension theorem states that if the two pre-measures given by the restriction of $m,n$ to $R$ are themselves $sigma$-finite, then their extension to the $sigma$-algebra is unique, but I suspect there might be a case where the restriction of a $sigma$-finite measure to the semi-ring might not remain $sigma$-finite.
real-analysis measure-theory
asked Dec 1 '18 at 13:53
Bar AlonBar Alon
484114
$endgroup$
add a comment |
$begingroup$
Suppose I have two measures $m,n$ and that they are both $sigma$-finite and agree on some semi-ring $R$ generating their $sigma$-algebras. Must these two measures then agree on the entire $sigma$-algebra?
Of course, the Caratheodory extension theorem states that if the two pre-measures given by the restriction of $m,n$ to $R$ are themselves $sigma$-finite, then their extension to the $sigma$-algebra is unique, but I suspect there might be a case where the restriction of a $sigma$-finite measure to the semi-ring might not remain $sigma$-finite.
real-analysis measure-theory
asked Dec 1 '18 at 13:53
Bar AlonBar Alon
484114
$endgroup$
$begingroup$
Let $mathcal{B}$ be the $sigma$-algebra. Then ${A in mathcal{B} : m(A) = n(A)}$ is a sigma-algebra containing $R$ and is thus $mathcal{B}$.
$endgroup$
– mathworker21
Dec 1 '18 at 13:58
$begingroup$
Don't we run into a problem because the measures may not agree on complements of sets of infinite measure?
$endgroup$
– Bar Alon
Dec 1 '18 at 14:23
add a comment |
$begingroup$
Suppose I have two measures $m,n$ and that they are both $sigma$-finite and agree on some semi-ring $R$ generating their $sigma$-algebras. Must these two measures then agree on the entire $sigma$-algebra?
Of course, the Caratheodory extension theorem states that if the two pre-measures given by the restriction of $m,n$ to $R$ are themselves $sigma$-finite, then their extension to the $sigma$-algebra is unique, but I suspect there might be a case where the restriction of a $sigma$-finite measure to the semi-ring might not remain $sigma$-finite.
real-analysis measure-theory
asked Dec 1 '18 at 13:53
Bar AlonBar Alon
484114
$endgroup$
Suppose I have two measures $m,n$ and that they are both $sigma$-finite and agree on some semi-ring $R$ generating their $sigma$-algebras. Must these two measures then agree on the entire $sigma$-algebra?
Of course, the Caratheodory extension theorem states that if the two pre-measures given by the restriction of $m,n$ to $R$ are themselves $sigma$-finite, then their extension to the $sigma$-algebra is unique, but I suspect there might be a case where the restriction of a $sigma$-finite measure to the semi-ring might not remain $sigma$-finite.
real-analysis measure-theory
real-analysis measure-theory
asked Dec 1 '18 at 13:53
Bar AlonBar Alon
484114
asked Dec 1 '18 at 13:53
Bar AlonBar Alon
484114
edited Dec 2 '18 at 15:29
Bar Alon
asked Dec 1 '18 at 13:53
Bar AlonBar Alon
484114
asked Dec 1 '18 at 13:53
Bar AlonBar Alon
484114
asked Dec 1 '18 at 13:53
Bar AlonBar Alon
484114
484114
$begingroup$
Let $mathcal{B}$ be the $sigma$-algebra. Then ${A in mathcal{B} : m(A) = n(A)}$ is a sigma-algebra containing $R$ and is thus $mathcal{B}$.
$endgroup$
– mathworker21
Dec 1 '18 at 13:58
$begingroup$
Don't we run into a problem because the measures may not agree on complements of sets of infinite measure?
$endgroup$
– Bar Alon
Dec 1 '18 at 14:23
add a comment |
$begingroup$
Let $mathcal{B}$ be the $sigma$-algebra. Then ${A in mathcal{B} : m(A) = n(A)}$ is a sigma-algebra containing $R$ and is thus $mathcal{B}$.
$endgroup$
– mathworker21
Dec 1 '18 at 13:58
$begingroup$
Don't we run into a problem because the measures may not agree on complements of sets of infinite measure?
$endgroup$
– Bar Alon
Dec 1 '18 at 14:23
$begingroup$
Let $mathcal{B}$ be the $sigma$-algebra. Then ${A in mathcal{B} : m(A) = n(A)}$ is a sigma-algebra containing $R$ and is thus $mathcal{B}$.
$endgroup$
– mathworker21
Dec 1 '18 at 13:58
$begingroup$
Let $mathcal{B}$ be the $sigma$-algebra. Then ${A in mathcal{B} : m(A) = n(A)}$ is a sigma-algebra containing $R$ and is thus $mathcal{B}$.
$endgroup$
– mathworker21
Dec 1 '18 at 13:58
$begingroup$
Don't we run into a problem because the measures may not agree on complements of sets of infinite measure?
$endgroup$
– Bar Alon
Dec 1 '18 at 14:23
$begingroup$
Don't we run into a problem because the measures may not agree on complements of sets of infinite measure?
$endgroup$
– Bar Alon
Dec 1 '18 at 14:23
add a comment |
1 Answer
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It turns out that the answer is no.
Consider the two measures $m,n$ defined on the borel $sigma$-algebra $mathcal{B}(mathbb{R})$ as follows:
$$m(A)=|Acapmathbb{Q}|,quad n(A)=|Acap(mathbb{Q}cup{sqrt{2}})|$$
Since $mathbb{Q}$ is countable, they are both $sigma$-finite. Additionally, they agree on the semi-ring $R$ of all half-open intervals (for any $a,bin mathbb{R}$: $m([a,b))=infty=n([a,b))$), and $R$ indeed generates $mathcal{B}(mathbb{R})$.
On the other hand $m({sqrt{2}})not =n({sqrt{2}})$ which is clearly a borel set.
Note that in the above example $m, n$ are indeed no longer $sigma$-finite when restricted to $R$.
answered Dec 2 '18 at 15:27
Bar AlonBar Alon
484114
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add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It turns out that the answer is no.
Consider the two measures $m,n$ defined on the borel $sigma$-algebra $mathcal{B}(mathbb{R})$ as follows:
$$m(A)=|Acapmathbb{Q}|,quad n(A)=|Acap(mathbb{Q}cup{sqrt{2}})|$$
Since $mathbb{Q}$ is countable, they are both $sigma$-finite. Additionally, they agree on the semi-ring $R$ of all half-open intervals (for any $a,bin mathbb{R}$: $m([a,b))=infty=n([a,b))$), and $R$ indeed generates $mathcal{B}(mathbb{R})$.
On the other hand $m({sqrt{2}})not =n({sqrt{2}})$ which is clearly a borel set.
Note that in the above example $m, n$ are indeed no longer $sigma$-finite when restricted to $R$.
answered Dec 2 '18 at 15:27
Bar AlonBar Alon
484114
$endgroup$
add a comment |
$begingroup$
It turns out that the answer is no.
Consider the two measures $m,n$ defined on the borel $sigma$-algebra $mathcal{B}(mathbb{R})$ as follows:
$$m(A)=|Acapmathbb{Q}|,quad n(A)=|Acap(mathbb{Q}cup{sqrt{2}})|$$
Since $mathbb{Q}$ is countable, they are both $sigma$-finite. Additionally, they agree on the semi-ring $R$ of all half-open intervals (for any $a,bin mathbb{R}$: $m([a,b))=infty=n([a,b))$), and $R$ indeed generates $mathcal{B}(mathbb{R})$.
On the other hand $m({sqrt{2}})not =n({sqrt{2}})$ which is clearly a borel set.
Note that in the above example $m, n$ are indeed no longer $sigma$-finite when restricted to $R$.
answered Dec 2 '18 at 15:27
Bar AlonBar Alon
484114
$endgroup$
add a comment |
$begingroup$
It turns out that the answer is no.
Consider the two measures $m,n$ defined on the borel $sigma$-algebra $mathcal{B}(mathbb{R})$ as follows:
$$m(A)=|Acapmathbb{Q}|,quad n(A)=|Acap(mathbb{Q}cup{sqrt{2}})|$$
Since $mathbb{Q}$ is countable, they are both $sigma$-finite. Additionally, they agree on the semi-ring $R$ of all half-open intervals (for any $a,bin mathbb{R}$: $m([a,b))=infty=n([a,b))$), and $R$ indeed generates $mathcal{B}(mathbb{R})$.
On the other hand $m({sqrt{2}})not =n({sqrt{2}})$ which is clearly a borel set.
Note that in the above example $m, n$ are indeed no longer $sigma$-finite when restricted to $R$.
answered Dec 2 '18 at 15:27
Bar AlonBar Alon
484114
$endgroup$
It turns out that the answer is no.
Consider the two measures $m,n$ defined on the borel $sigma$-algebra $mathcal{B}(mathbb{R})$ as follows:
$$m(A)=|Acapmathbb{Q}|,quad n(A)=|Acap(mathbb{Q}cup{sqrt{2}})|$$
Since $mathbb{Q}$ is countable, they are both $sigma$-finite. Additionally, they agree on the semi-ring $R$ of all half-open intervals (for any $a,bin mathbb{R}$: $m([a,b))=infty=n([a,b))$), and $R$ indeed generates $mathcal{B}(mathbb{R})$.
On the other hand $m({sqrt{2}})not =n({sqrt{2}})$ which is clearly a borel set.
Note that in the above example $m, n$ are indeed no longer $sigma$-finite when restricted to $R$.
answered Dec 2 '18 at 15:27
Bar AlonBar Alon
484114
answered Dec 2 '18 at 15:27
Bar AlonBar Alon
484114
answered Dec 2 '18 at 15:27
Bar AlonBar Alon
484114
answered Dec 2 '18 at 15:27
Bar AlonBar Alon
484114
484114
add a comment |
add a comment |
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$begingroup$
Let $mathcal{B}$ be the $sigma$-algebra. Then ${A in mathcal{B} : m(A) = n(A)}$ is a sigma-algebra containing $R$ and is thus $mathcal{B}$.
$endgroup$
– mathworker21
Dec 1 '18 at 13:58
$begingroup$
Don't we run into a problem because the measures may not agree on complements of sets of infinite measure?
$endgroup$
– Bar Alon
Dec 1 '18 at 14:23