Lebesgue Stieltjes measure unique for invariance on $mathscr{B}_mathbb{R}$
$begingroup$
Exercsise: Let $mu$ be a Lebesgue-Stieltjes measure on $mathscr{B}_{mathbb{R}}$ invariant for the class of right half-closed intervals of $mathbb{R}$, so that, $mu(a+I)=mu(I)$, for all $ainmathbb{R}$ and $I=(x,y]$. Show that, in $mathscr{B}_mathbb{R}$, $mu=c.Leb$ where cin$mathbb{R}$ and Leb denotes the Lebesgue measure.
Attempted resolution:
Lets assume that $lambda$ is the Lebesgue measure and $nu$ is a measure defined on the same space such that $nu(a+I)=nu(I)$
If $mathscr{I}_n=(0,0+frac{1}{n}]$
Then $bigcap_{ngeqslant 1}mathscr{I}_n={0}$
As $nu$ is a measure then $nu(mathscr{I}_n+a)=nu(mathscr{I}_n)=lim_{ntoinfty}nu(mathscr{I}_n+a)=lim_{ntoinfty}nu(mathscr{I}_n)implies nu(a)=nu(0)$ and $ainmathbb{R}$ is an arbitrary point. In the Lebesgue measure it happens the same by definition of length $lambda(a)=lambda(0)=0$. But I do not think that this resemblance proves the measures to be equal.
Question:
What should I do to prove the statement?
Thanks in advance!
real-analysis measure-theory lebesgue-measure
asked Dec 22 '18 at 19:09
Pedro GomesPedro Gomes
1,8982721
$endgroup$
|
show 1 more comment
$begingroup$
Exercsise: Let $mu$ be a Lebesgue-Stieltjes measure on $mathscr{B}_{mathbb{R}}$ invariant for the class of right half-closed intervals of $mathbb{R}$, so that, $mu(a+I)=mu(I)$, for all $ainmathbb{R}$ and $I=(x,y]$. Show that, in $mathscr{B}_mathbb{R}$, $mu=c.Leb$ where cin$mathbb{R}$ and Leb denotes the Lebesgue measure.
Attempted resolution:
Lets assume that $lambda$ is the Lebesgue measure and $nu$ is a measure defined on the same space such that $nu(a+I)=nu(I)$
If $mathscr{I}_n=(0,0+frac{1}{n}]$
Then $bigcap_{ngeqslant 1}mathscr{I}_n={0}$
As $nu$ is a measure then $nu(mathscr{I}_n+a)=nu(mathscr{I}_n)=lim_{ntoinfty}nu(mathscr{I}_n+a)=lim_{ntoinfty}nu(mathscr{I}_n)implies nu(a)=nu(0)$ and $ainmathbb{R}$ is an arbitrary point. In the Lebesgue measure it happens the same by definition of length $lambda(a)=lambda(0)=0$. But I do not think that this resemblance proves the measures to be equal.
Question:
What should I do to prove the statement?
Thanks in advance!
real-analysis measure-theory lebesgue-measure
asked Dec 22 '18 at 19:09
Pedro GomesPedro Gomes
1,8982721
$endgroup$
$begingroup$
I am not certain, but I think Carathéodory's theorem implies that if you show $mu = c lambda$ for the class of right half-closed intervals, then the measures are equal on $mathscr{B}_{mathbb{R}}$.
$endgroup$
– angryavian
Dec 22 '18 at 19:46
$begingroup$
@angryavian Carathéodory's theorem assures the extension is unique, so I do not know at what extent linear transformations would be tolerated. But the measure $mu$ is Lebesgue-Stieltjes and not just Lebesgue. Extension is understood: when you have two algebraic structure $S$ you have an extension to $R$ if the measures in those different $mu(E)=mu'(E)forall E in S$.
$endgroup$
– Pedro Gomes
Dec 22 '18 at 20:00
$begingroup$
I deleted my answer since it was flawed a little bit.
$endgroup$
– Shashi
Dec 22 '18 at 22:20
$begingroup$
@Shashi Are you going to upload it again?
$endgroup$
– Pedro Gomes
Dec 22 '18 at 22:21
$begingroup$
@Shashi I need all the hints I can get.
$endgroup$
– Pedro Gomes
Dec 22 '18 at 22:21
|
show 1 more comment
$begingroup$
Exercsise: Let $mu$ be a Lebesgue-Stieltjes measure on $mathscr{B}_{mathbb{R}}$ invariant for the class of right half-closed intervals of $mathbb{R}$, so that, $mu(a+I)=mu(I)$, for all $ainmathbb{R}$ and $I=(x,y]$. Show that, in $mathscr{B}_mathbb{R}$, $mu=c.Leb$ where cin$mathbb{R}$ and Leb denotes the Lebesgue measure.
Attempted resolution:
Lets assume that $lambda$ is the Lebesgue measure and $nu$ is a measure defined on the same space such that $nu(a+I)=nu(I)$
If $mathscr{I}_n=(0,0+frac{1}{n}]$
Then $bigcap_{ngeqslant 1}mathscr{I}_n={0}$
As $nu$ is a measure then $nu(mathscr{I}_n+a)=nu(mathscr{I}_n)=lim_{ntoinfty}nu(mathscr{I}_n+a)=lim_{ntoinfty}nu(mathscr{I}_n)implies nu(a)=nu(0)$ and $ainmathbb{R}$ is an arbitrary point. In the Lebesgue measure it happens the same by definition of length $lambda(a)=lambda(0)=0$. But I do not think that this resemblance proves the measures to be equal.
Question:
What should I do to prove the statement?
Thanks in advance!
real-analysis measure-theory lebesgue-measure
asked Dec 22 '18 at 19:09
Pedro GomesPedro Gomes
1,8982721
$endgroup$
Exercsise: Let $mu$ be a Lebesgue-Stieltjes measure on $mathscr{B}_{mathbb{R}}$ invariant for the class of right half-closed intervals of $mathbb{R}$, so that, $mu(a+I)=mu(I)$, for all $ainmathbb{R}$ and $I=(x,y]$. Show that, in $mathscr{B}_mathbb{R}$, $mu=c.Leb$ where cin$mathbb{R}$ and Leb denotes the Lebesgue measure.
Attempted resolution:
Lets assume that $lambda$ is the Lebesgue measure and $nu$ is a measure defined on the same space such that $nu(a+I)=nu(I)$
If $mathscr{I}_n=(0,0+frac{1}{n}]$
Then $bigcap_{ngeqslant 1}mathscr{I}_n={0}$
As $nu$ is a measure then $nu(mathscr{I}_n+a)=nu(mathscr{I}_n)=lim_{ntoinfty}nu(mathscr{I}_n+a)=lim_{ntoinfty}nu(mathscr{I}_n)implies nu(a)=nu(0)$ and $ainmathbb{R}$ is an arbitrary point. In the Lebesgue measure it happens the same by definition of length $lambda(a)=lambda(0)=0$. But I do not think that this resemblance proves the measures to be equal.
Question:
What should I do to prove the statement?
Thanks in advance!
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Dec 22 '18 at 19:09
Pedro GomesPedro Gomes
1,8982721
asked Dec 22 '18 at 19:09
Pedro GomesPedro Gomes
1,8982721
edited Dec 22 '18 at 19:43
Pedro Gomes
asked Dec 22 '18 at 19:09
Pedro GomesPedro Gomes
1,8982721
asked Dec 22 '18 at 19:09
Pedro GomesPedro Gomes
1,8982721
asked Dec 22 '18 at 19:09
Pedro GomesPedro Gomes
1,8982721
1,8982721
$begingroup$
I am not certain, but I think Carathéodory's theorem implies that if you show $mu = c lambda$ for the class of right half-closed intervals, then the measures are equal on $mathscr{B}_{mathbb{R}}$.
$endgroup$
– angryavian
Dec 22 '18 at 19:46
$begingroup$
@angryavian Carathéodory's theorem assures the extension is unique, so I do not know at what extent linear transformations would be tolerated. But the measure $mu$ is Lebesgue-Stieltjes and not just Lebesgue. Extension is understood: when you have two algebraic structure $S$ you have an extension to $R$ if the measures in those different $mu(E)=mu'(E)forall E in S$.
$endgroup$
– Pedro Gomes
Dec 22 '18 at 20:00
$begingroup$
I deleted my answer since it was flawed a little bit.
$endgroup$
– Shashi
Dec 22 '18 at 22:20
$begingroup$
@Shashi Are you going to upload it again?
$endgroup$
– Pedro Gomes
Dec 22 '18 at 22:21
$begingroup$
@Shashi I need all the hints I can get.
$endgroup$
– Pedro Gomes
Dec 22 '18 at 22:21
|
show 1 more comment
$begingroup$
I am not certain, but I think Carathéodory's theorem implies that if you show $mu = c lambda$ for the class of right half-closed intervals, then the measures are equal on $mathscr{B}_{mathbb{R}}$.
$endgroup$
– angryavian
Dec 22 '18 at 19:46
$begingroup$
@angryavian Carathéodory's theorem assures the extension is unique, so I do not know at what extent linear transformations would be tolerated. But the measure $mu$ is Lebesgue-Stieltjes and not just Lebesgue. Extension is understood: when you have two algebraic structure $S$ you have an extension to $R$ if the measures in those different $mu(E)=mu'(E)forall E in S$.
$endgroup$
– Pedro Gomes
Dec 22 '18 at 20:00
$begingroup$
I deleted my answer since it was flawed a little bit.
$endgroup$
– Shashi
Dec 22 '18 at 22:20
$begingroup$
@Shashi Are you going to upload it again?
$endgroup$
– Pedro Gomes
Dec 22 '18 at 22:21
$begingroup$
@Shashi I need all the hints I can get.
$endgroup$
– Pedro Gomes
Dec 22 '18 at 22:21
$begingroup$
I am not certain, but I think Carathéodory's theorem implies that if you show $mu = c lambda$ for the class of right half-closed intervals, then the measures are equal on $mathscr{B}_{mathbb{R}}$.
$endgroup$
– angryavian
Dec 22 '18 at 19:46
$begingroup$
I am not certain, but I think Carathéodory's theorem implies that if you show $mu = c lambda$ for the class of right half-closed intervals, then the measures are equal on $mathscr{B}_{mathbb{R}}$.
$endgroup$
– angryavian
Dec 22 '18 at 19:46
$begingroup$
@angryavian Carathéodory's theorem assures the extension is unique, so I do not know at what extent linear transformations would be tolerated. But the measure $mu$ is Lebesgue-Stieltjes and not just Lebesgue. Extension is understood: when you have two algebraic structure $S$ you have an extension to $R$ if the measures in those different $mu(E)=mu'(E)forall E in S$.
$endgroup$
– Pedro Gomes
Dec 22 '18 at 20:00
$begingroup$
@angryavian Carathéodory's theorem assures the extension is unique, so I do not know at what extent linear transformations would be tolerated. But the measure $mu$ is Lebesgue-Stieltjes and not just Lebesgue. Extension is understood: when you have two algebraic structure $S$ you have an extension to $R$ if the measures in those different $mu(E)=mu'(E)forall E in S$.
$endgroup$
– Pedro Gomes
Dec 22 '18 at 20:00
$begingroup$
I deleted my answer since it was flawed a little bit.
$endgroup$
– Shashi
Dec 22 '18 at 22:20
$begingroup$
I deleted my answer since it was flawed a little bit.
$endgroup$
– Shashi
Dec 22 '18 at 22:20
$begingroup$
@Shashi Are you going to upload it again?
$endgroup$
– Pedro Gomes
Dec 22 '18 at 22:21
$begingroup$
@Shashi Are you going to upload it again?
$endgroup$
– Pedro Gomes
Dec 22 '18 at 22:21
$begingroup$
@Shashi I need all the hints I can get.
$endgroup$
– Pedro Gomes
Dec 22 '18 at 22:21
$begingroup$
@Shashi I need all the hints I can get.
$endgroup$
– Pedro Gomes
Dec 22 '18 at 22:21
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If $nu (A)=infty$ for every non-empty Borel set then $nu$ has the invariance property but it is not constant times Lebesgue measure. If you assume that $nu (E) <infty$ for bounded Borel sets then the fact that $nu (a)=nu (0)$ for all $a$ implies that $nu (a)=0$ for all $a$, so $nu$ is actually translation invariant on the class of all intervals from which we can deduce that it is translation invariant.
answered Dec 23 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
$endgroup$
$begingroup$
How would I conclude the argument? I need to prove the constant times Lebesgue is equal to $mu$. Should I use Radon-Nikodym? How did you deduce $nu(a)=0$? Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 23 '18 at 11:28
$begingroup$
@PedroGomes Suppose $nu (a) >0$ .If $nu (A)<infty$ then you can have atmost $[frac {nu (A)} {nu(0)}]+1$ points in $A$ because if $a_1,a_2,..,a_k$ are distinct points of $A$ then $mu (A) geq nu(a_1)+nu(a_2)+...+nu(a_k) =knu(0)$. In particular any open interval has a finite number of points. This contradcition shows that $nu(a)nu (0)=0$ for all $a$. The last part is a well known theorem Lebesgue measure is, up to a constant factor, the only transaltion invariant Borel measure finite on compact sets.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 11:41
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $nu (A)=infty$ for every non-empty Borel set then $nu$ has the invariance property but it is not constant times Lebesgue measure. If you assume that $nu (E) <infty$ for bounded Borel sets then the fact that $nu (a)=nu (0)$ for all $a$ implies that $nu (a)=0$ for all $a$, so $nu$ is actually translation invariant on the class of all intervals from which we can deduce that it is translation invariant.
answered Dec 23 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
$endgroup$
$begingroup$
How would I conclude the argument? I need to prove the constant times Lebesgue is equal to $mu$. Should I use Radon-Nikodym? How did you deduce $nu(a)=0$? Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 23 '18 at 11:28
$begingroup$
@PedroGomes Suppose $nu (a) >0$ .If $nu (A)<infty$ then you can have atmost $[frac {nu (A)} {nu(0)}]+1$ points in $A$ because if $a_1,a_2,..,a_k$ are distinct points of $A$ then $mu (A) geq nu(a_1)+nu(a_2)+...+nu(a_k) =knu(0)$. In particular any open interval has a finite number of points. This contradcition shows that $nu(a)nu (0)=0$ for all $a$. The last part is a well known theorem Lebesgue measure is, up to a constant factor, the only transaltion invariant Borel measure finite on compact sets.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 11:41
add a comment |
$begingroup$
If $nu (A)=infty$ for every non-empty Borel set then $nu$ has the invariance property but it is not constant times Lebesgue measure. If you assume that $nu (E) <infty$ for bounded Borel sets then the fact that $nu (a)=nu (0)$ for all $a$ implies that $nu (a)=0$ for all $a$, so $nu$ is actually translation invariant on the class of all intervals from which we can deduce that it is translation invariant.
answered Dec 23 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
$endgroup$
$begingroup$
How would I conclude the argument? I need to prove the constant times Lebesgue is equal to $mu$. Should I use Radon-Nikodym? How did you deduce $nu(a)=0$? Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 23 '18 at 11:28
$begingroup$
@PedroGomes Suppose $nu (a) >0$ .If $nu (A)<infty$ then you can have atmost $[frac {nu (A)} {nu(0)}]+1$ points in $A$ because if $a_1,a_2,..,a_k$ are distinct points of $A$ then $mu (A) geq nu(a_1)+nu(a_2)+...+nu(a_k) =knu(0)$. In particular any open interval has a finite number of points. This contradcition shows that $nu(a)nu (0)=0$ for all $a$. The last part is a well known theorem Lebesgue measure is, up to a constant factor, the only transaltion invariant Borel measure finite on compact sets.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 11:41
add a comment |
$begingroup$
If $nu (A)=infty$ for every non-empty Borel set then $nu$ has the invariance property but it is not constant times Lebesgue measure. If you assume that $nu (E) <infty$ for bounded Borel sets then the fact that $nu (a)=nu (0)$ for all $a$ implies that $nu (a)=0$ for all $a$, so $nu$ is actually translation invariant on the class of all intervals from which we can deduce that it is translation invariant.
answered Dec 23 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
$endgroup$
If $nu (A)=infty$ for every non-empty Borel set then $nu$ has the invariance property but it is not constant times Lebesgue measure. If you assume that $nu (E) <infty$ for bounded Borel sets then the fact that $nu (a)=nu (0)$ for all $a$ implies that $nu (a)=0$ for all $a$, so $nu$ is actually translation invariant on the class of all intervals from which we can deduce that it is translation invariant.
answered Dec 23 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
answered Dec 23 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
answered Dec 23 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
answered Dec 23 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
64.5k42665
$begingroup$
How would I conclude the argument? I need to prove the constant times Lebesgue is equal to $mu$. Should I use Radon-Nikodym? How did you deduce $nu(a)=0$? Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 23 '18 at 11:28
$begingroup$
@PedroGomes Suppose $nu (a) >0$ .If $nu (A)<infty$ then you can have atmost $[frac {nu (A)} {nu(0)}]+1$ points in $A$ because if $a_1,a_2,..,a_k$ are distinct points of $A$ then $mu (A) geq nu(a_1)+nu(a_2)+...+nu(a_k) =knu(0)$. In particular any open interval has a finite number of points. This contradcition shows that $nu(a)nu (0)=0$ for all $a$. The last part is a well known theorem Lebesgue measure is, up to a constant factor, the only transaltion invariant Borel measure finite on compact sets.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 11:41
add a comment |
$begingroup$
How would I conclude the argument? I need to prove the constant times Lebesgue is equal to $mu$. Should I use Radon-Nikodym? How did you deduce $nu(a)=0$? Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 23 '18 at 11:28
$begingroup$
@PedroGomes Suppose $nu (a) >0$ .If $nu (A)<infty$ then you can have atmost $[frac {nu (A)} {nu(0)}]+1$ points in $A$ because if $a_1,a_2,..,a_k$ are distinct points of $A$ then $mu (A) geq nu(a_1)+nu(a_2)+...+nu(a_k) =knu(0)$. In particular any open interval has a finite number of points. This contradcition shows that $nu(a)nu (0)=0$ for all $a$. The last part is a well known theorem Lebesgue measure is, up to a constant factor, the only transaltion invariant Borel measure finite on compact sets.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 11:41
$begingroup$
How would I conclude the argument? I need to prove the constant times Lebesgue is equal to $mu$. Should I use Radon-Nikodym? How did you deduce $nu(a)=0$? Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 23 '18 at 11:28
$begingroup$
How would I conclude the argument? I need to prove the constant times Lebesgue is equal to $mu$. Should I use Radon-Nikodym? How did you deduce $nu(a)=0$? Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 23 '18 at 11:28
$begingroup$
@PedroGomes Suppose $nu (a) >0$ .If $nu (A)<infty$ then you can have atmost $[frac {nu (A)} {nu(0)}]+1$ points in $A$ because if $a_1,a_2,..,a_k$ are distinct points of $A$ then $mu (A) geq nu(a_1)+nu(a_2)+...+nu(a_k) =knu(0)$. In particular any open interval has a finite number of points. This contradcition shows that $nu(a)nu (0)=0$ for all $a$. The last part is a well known theorem Lebesgue measure is, up to a constant factor, the only transaltion invariant Borel measure finite on compact sets.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 11:41
$begingroup$
@PedroGomes Suppose $nu (a) >0$ .If $nu (A)<infty$ then you can have atmost $[frac {nu (A)} {nu(0)}]+1$ points in $A$ because if $a_1,a_2,..,a_k$ are distinct points of $A$ then $mu (A) geq nu(a_1)+nu(a_2)+...+nu(a_k) =knu(0)$. In particular any open interval has a finite number of points. This contradcition shows that $nu(a)nu (0)=0$ for all $a$. The last part is a well known theorem Lebesgue measure is, up to a constant factor, the only transaltion invariant Borel measure finite on compact sets.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 11:41
add a comment |
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$begingroup$
I am not certain, but I think Carathéodory's theorem implies that if you show $mu = c lambda$ for the class of right half-closed intervals, then the measures are equal on $mathscr{B}_{mathbb{R}}$.
$endgroup$
– angryavian
Dec 22 '18 at 19:46
$begingroup$
@angryavian Carathéodory's theorem assures the extension is unique, so I do not know at what extent linear transformations would be tolerated. But the measure $mu$ is Lebesgue-Stieltjes and not just Lebesgue. Extension is understood: when you have two algebraic structure $S$ you have an extension to $R$ if the measures in those different $mu(E)=mu'(E)forall E in S$.
$endgroup$
– Pedro Gomes
Dec 22 '18 at 20:00
$begingroup$
I deleted my answer since it was flawed a little bit.
$endgroup$
– Shashi
Dec 22 '18 at 22:20
$begingroup$
@Shashi Are you going to upload it again?
$endgroup$
– Pedro Gomes
Dec 22 '18 at 22:21
$begingroup$
@Shashi I need all the hints I can get.
$endgroup$
– Pedro Gomes
Dec 22 '18 at 22:21