Equality of finite signed measures by showing that the integrals of every bounded continuous function...












1














In order to show the uniqueness of the Fourier coefficients of a signed measure, I need to show that :




For any two finite signed measures $nu_1, nu_2$ on $left([-pi, pi], mathcal{B}_{[-pi, pi]}right)$ such that : $int fmathrm dnu_1=int fmathrm dnu_2$ holds for every $f$ continuous (and bounded), we have: $nu_1=nu_2$.




This property holds for finite positive measures (Approximation of bounded measurable functions with continuous functions) because the equality of integrals for a continuous function implies that of integrals of a measurable bounded function.

The answers in this thread are based on the fact that for $mu, mu'$ probability measures, $mathcal{C}={Binmathcal{B}_{[-pi, pi]} : mu(B)=mu'(B)}$ is a $pi$-system.
Does this even hold here ?



EDIT: It does when $nu_1([-pi, pi])=nu_2([-pi, pi])$, (required to show that ${B∈mathcal{B}_{[−π,π]}:ν_1(B)=ν_2(B)}$ is a Dynkin system), which is true in my case, since $hat{ν_1}=hat{ν_2}$.






measure-theory signed-measures







share|cite|improve this question
























  • Do you know the Riesz–Markov–Kakutani representation theorem, stating that $mathcal M(X)simeq C(X)^*$?
    – Federico
    Dec 4 '18 at 21:45










  • Nope, that seems out of my reach for now. Plus the material I'm following didn't even feel the need to highlight the equality, so I thought it was going to be similar to the positive case.
    – deque
    Dec 4 '18 at 21:58


















1














In order to show the uniqueness of the Fourier coefficients of a signed measure, I need to show that :




For any two finite signed measures $nu_1, nu_2$ on $left([-pi, pi], mathcal{B}_{[-pi, pi]}right)$ such that : $int fmathrm dnu_1=int fmathrm dnu_2$ holds for every $f$ continuous (and bounded), we have: $nu_1=nu_2$.




This property holds for finite positive measures (Approximation of bounded measurable functions with continuous functions) because the equality of integrals for a continuous function implies that of integrals of a measurable bounded function.

The answers in this thread are based on the fact that for $mu, mu'$ probability measures, $mathcal{C}={Binmathcal{B}_{[-pi, pi]} : mu(B)=mu'(B)}$ is a $pi$-system.
Does this even hold here ?



EDIT: It does when $nu_1([-pi, pi])=nu_2([-pi, pi])$, (required to show that ${B∈mathcal{B}_{[−π,π]}:ν_1(B)=ν_2(B)}$ is a Dynkin system), which is true in my case, since $hat{ν_1}=hat{ν_2}$.






measure-theory signed-measures







share|cite|improve this question
























  • Do you know the Riesz–Markov–Kakutani representation theorem, stating that $mathcal M(X)simeq C(X)^*$?
    – Federico
    Dec 4 '18 at 21:45










  • Nope, that seems out of my reach for now. Plus the material I'm following didn't even feel the need to highlight the equality, so I thought it was going to be similar to the positive case.
    – deque
    Dec 4 '18 at 21:58
















1












1








1


1





In order to show the uniqueness of the Fourier coefficients of a signed measure, I need to show that :




For any two finite signed measures $nu_1, nu_2$ on $left([-pi, pi], mathcal{B}_{[-pi, pi]}right)$ such that : $int fmathrm dnu_1=int fmathrm dnu_2$ holds for every $f$ continuous (and bounded), we have: $nu_1=nu_2$.




This property holds for finite positive measures (Approximation of bounded measurable functions with continuous functions) because the equality of integrals for a continuous function implies that of integrals of a measurable bounded function.

The answers in this thread are based on the fact that for $mu, mu'$ probability measures, $mathcal{C}={Binmathcal{B}_{[-pi, pi]} : mu(B)=mu'(B)}$ is a $pi$-system.
Does this even hold here ?



EDIT: It does when $nu_1([-pi, pi])=nu_2([-pi, pi])$, (required to show that ${B∈mathcal{B}_{[−π,π]}:ν_1(B)=ν_2(B)}$ is a Dynkin system), which is true in my case, since $hat{ν_1}=hat{ν_2}$.






measure-theory signed-measures







share|cite|improve this question















In order to show the uniqueness of the Fourier coefficients of a signed measure, I need to show that :




For any two finite signed measures $nu_1, nu_2$ on $left([-pi, pi], mathcal{B}_{[-pi, pi]}right)$ such that : $int fmathrm dnu_1=int fmathrm dnu_2$ holds for every $f$ continuous (and bounded), we have: $nu_1=nu_2$.




This property holds for finite positive measures (Approximation of bounded measurable functions with continuous functions) because the equality of integrals for a continuous function implies that of integrals of a measurable bounded function.

The answers in this thread are based on the fact that for $mu, mu'$ probability measures, $mathcal{C}={Binmathcal{B}_{[-pi, pi]} : mu(B)=mu'(B)}$ is a $pi$-system.
Does this even hold here ?



EDIT: It does when $nu_1([-pi, pi])=nu_2([-pi, pi])$, (required to show that ${B∈mathcal{B}_{[−π,π]}:ν_1(B)=ν_2(B)}$ is a Dynkin system), which is true in my case, since $hat{ν_1}=hat{ν_2}$.






measure-theory signed-measures




measure-theory signed-measures






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 22:42







deque

















asked Dec 4 '18 at 21:38









dequedeque

413111




413111












  • Do you know the Riesz–Markov–Kakutani representation theorem, stating that $mathcal M(X)simeq C(X)^*$?
    – Federico
    Dec 4 '18 at 21:45










  • Nope, that seems out of my reach for now. Plus the material I'm following didn't even feel the need to highlight the equality, so I thought it was going to be similar to the positive case.
    – deque
    Dec 4 '18 at 21:58




















  • Do you know the Riesz–Markov–Kakutani representation theorem, stating that $mathcal M(X)simeq C(X)^*$?
    – Federico
    Dec 4 '18 at 21:45










  • Nope, that seems out of my reach for now. Plus the material I'm following didn't even feel the need to highlight the equality, so I thought it was going to be similar to the positive case.
    – deque
    Dec 4 '18 at 21:58


















Do you know the Riesz–Markov–Kakutani representation theorem, stating that $mathcal M(X)simeq C(X)^*$?
– Federico
Dec 4 '18 at 21:45




Do you know the Riesz–Markov–Kakutani representation theorem, stating that $mathcal M(X)simeq C(X)^*$?
– Federico
Dec 4 '18 at 21:45












Nope, that seems out of my reach for now. Plus the material I'm following didn't even feel the need to highlight the equality, so I thought it was going to be similar to the positive case.
– deque
Dec 4 '18 at 21:58






Nope, that seems out of my reach for now. Plus the material I'm following didn't even feel the need to highlight the equality, so I thought it was going to be similar to the positive case.
– deque
Dec 4 '18 at 21:58












1 Answer
1






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oldest

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1














I suppose that both signed measures are finite, since otherwise we cannot expect that the integral over arbitrary continuous functions exists. You can approximate any interval $[a,b]$ by (for example) the sequence
$$g_n(x) := begin{cases} 1 & text{ if } x in [a,b] \ nx + (1-na) &text{ if } x in [a-1/n,a] \ -nx +(1+nb) & text{ if } x in [b,b+1/n] \ 0 & text{otherwise} end{cases}.$$
By the dominated convergence theorem (for signed measures) we get
$$nu_1([a,b]) =nu_2([a,b]).$$
Thus, both measures are equal on a $cap$-stable generator of the Borel-$sigma$-algebra. Now, you need to extend the uniqueness theorem for $sigma$-finite measures. In fact, the same proof applies. Note that $$mathcal{C}={Binmathcal{B}_{[-pi, pi]} : nu_1(B)=nu_2(B)}$$ is a $pi$-system containing a $cap$-stable generator with a sequence $E_n in mathcal{C}$ such that $E_n uparrow Omega$.



For completeness, we should give reasons for the application of the dominated convergence theorem: By Jordan's decomposition theorem, you can write for a signed measure $mu = mu_1 - mu_2$ with measures $mu_1$ and $mu_2$. (Moreover, there exists a 'minimal decomposition' - called the Jordan decompisition.)






share|cite|improve this answer























  • Of course the measures involved here are finite, I edited that, sorry. The only problem remaining is that I can't prove that $mathcal{C}$ is stable under intersection. I managed to do it when the measures are of probability, but since they don't have the same total mass, I couldn't do it
    – deque
    Dec 5 '18 at 8:43










  • In the uniqueness theorem for $sigma$-finite measures you need that there exists a sequence $(E_n)$ in $mathcal{C}$ with $E_n uparrow Omega$. Of course, we need this here, too! And it is satisfied as we can take the constant function $f=1$ to get $nu_1(Omega) = nu_2(Omega)$.
    – p4sch
    Dec 5 '18 at 8:50










  • By "uniqueness theorem for σ-finite measures" you mean the monotone class theorem right (although it doesn't require the existence of such a sequence)? Because that was the first approach I tried, but as I said, I couldn't prove that $mathcal{C}$ is closed under intersections.
    – deque
    Dec 5 '18 at 8:58






  • 1




    I mean this theorem. I called it 'uniqueness theorem for $sigma$-finite measures'.
    – p4sch
    Dec 5 '18 at 9:12












  • This theorem still requires to show that $mathcal{C}$ is closed under intersections though :/ Can you help me with that please ?
    – deque
    Dec 5 '18 at 11:47













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I suppose that both signed measures are finite, since otherwise we cannot expect that the integral over arbitrary continuous functions exists. You can approximate any interval $[a,b]$ by (for example) the sequence
$$g_n(x) := begin{cases} 1 & text{ if } x in [a,b] \ nx + (1-na) &text{ if } x in [a-1/n,a] \ -nx +(1+nb) & text{ if } x in [b,b+1/n] \ 0 & text{otherwise} end{cases}.$$
By the dominated convergence theorem (for signed measures) we get
$$nu_1([a,b]) =nu_2([a,b]).$$
Thus, both measures are equal on a $cap$-stable generator of the Borel-$sigma$-algebra. Now, you need to extend the uniqueness theorem for $sigma$-finite measures. In fact, the same proof applies. Note that $$mathcal{C}={Binmathcal{B}_{[-pi, pi]} : nu_1(B)=nu_2(B)}$$ is a $pi$-system containing a $cap$-stable generator with a sequence $E_n in mathcal{C}$ such that $E_n uparrow Omega$.



For completeness, we should give reasons for the application of the dominated convergence theorem: By Jordan's decomposition theorem, you can write for a signed measure $mu = mu_1 - mu_2$ with measures $mu_1$ and $mu_2$. (Moreover, there exists a 'minimal decomposition' - called the Jordan decompisition.)






share|cite|improve this answer























  • Of course the measures involved here are finite, I edited that, sorry. The only problem remaining is that I can't prove that $mathcal{C}$ is stable under intersection. I managed to do it when the measures are of probability, but since they don't have the same total mass, I couldn't do it
    – deque
    Dec 5 '18 at 8:43










  • In the uniqueness theorem for $sigma$-finite measures you need that there exists a sequence $(E_n)$ in $mathcal{C}$ with $E_n uparrow Omega$. Of course, we need this here, too! And it is satisfied as we can take the constant function $f=1$ to get $nu_1(Omega) = nu_2(Omega)$.
    – p4sch
    Dec 5 '18 at 8:50










  • By "uniqueness theorem for σ-finite measures" you mean the monotone class theorem right (although it doesn't require the existence of such a sequence)? Because that was the first approach I tried, but as I said, I couldn't prove that $mathcal{C}$ is closed under intersections.
    – deque
    Dec 5 '18 at 8:58






  • 1




    I mean this theorem. I called it 'uniqueness theorem for $sigma$-finite measures'.
    – p4sch
    Dec 5 '18 at 9:12












  • This theorem still requires to show that $mathcal{C}$ is closed under intersections though :/ Can you help me with that please ?
    – deque
    Dec 5 '18 at 11:47


















1














I suppose that both signed measures are finite, since otherwise we cannot expect that the integral over arbitrary continuous functions exists. You can approximate any interval $[a,b]$ by (for example) the sequence
$$g_n(x) := begin{cases} 1 & text{ if } x in [a,b] \ nx + (1-na) &text{ if } x in [a-1/n,a] \ -nx +(1+nb) & text{ if } x in [b,b+1/n] \ 0 & text{otherwise} end{cases}.$$
By the dominated convergence theorem (for signed measures) we get
$$nu_1([a,b]) =nu_2([a,b]).$$
Thus, both measures are equal on a $cap$-stable generator of the Borel-$sigma$-algebra. Now, you need to extend the uniqueness theorem for $sigma$-finite measures. In fact, the same proof applies. Note that $$mathcal{C}={Binmathcal{B}_{[-pi, pi]} : nu_1(B)=nu_2(B)}$$ is a $pi$-system containing a $cap$-stable generator with a sequence $E_n in mathcal{C}$ such that $E_n uparrow Omega$.



For completeness, we should give reasons for the application of the dominated convergence theorem: By Jordan's decomposition theorem, you can write for a signed measure $mu = mu_1 - mu_2$ with measures $mu_1$ and $mu_2$. (Moreover, there exists a 'minimal decomposition' - called the Jordan decompisition.)






share|cite|improve this answer























  • Of course the measures involved here are finite, I edited that, sorry. The only problem remaining is that I can't prove that $mathcal{C}$ is stable under intersection. I managed to do it when the measures are of probability, but since they don't have the same total mass, I couldn't do it
    – deque
    Dec 5 '18 at 8:43










  • In the uniqueness theorem for $sigma$-finite measures you need that there exists a sequence $(E_n)$ in $mathcal{C}$ with $E_n uparrow Omega$. Of course, we need this here, too! And it is satisfied as we can take the constant function $f=1$ to get $nu_1(Omega) = nu_2(Omega)$.
    – p4sch
    Dec 5 '18 at 8:50










  • By "uniqueness theorem for σ-finite measures" you mean the monotone class theorem right (although it doesn't require the existence of such a sequence)? Because that was the first approach I tried, but as I said, I couldn't prove that $mathcal{C}$ is closed under intersections.
    – deque
    Dec 5 '18 at 8:58






  • 1




    I mean this theorem. I called it 'uniqueness theorem for $sigma$-finite measures'.
    – p4sch
    Dec 5 '18 at 9:12












  • This theorem still requires to show that $mathcal{C}$ is closed under intersections though :/ Can you help me with that please ?
    – deque
    Dec 5 '18 at 11:47
















1












1








1






I suppose that both signed measures are finite, since otherwise we cannot expect that the integral over arbitrary continuous functions exists. You can approximate any interval $[a,b]$ by (for example) the sequence
$$g_n(x) := begin{cases} 1 & text{ if } x in [a,b] \ nx + (1-na) &text{ if } x in [a-1/n,a] \ -nx +(1+nb) & text{ if } x in [b,b+1/n] \ 0 & text{otherwise} end{cases}.$$
By the dominated convergence theorem (for signed measures) we get
$$nu_1([a,b]) =nu_2([a,b]).$$
Thus, both measures are equal on a $cap$-stable generator of the Borel-$sigma$-algebra. Now, you need to extend the uniqueness theorem for $sigma$-finite measures. In fact, the same proof applies. Note that $$mathcal{C}={Binmathcal{B}_{[-pi, pi]} : nu_1(B)=nu_2(B)}$$ is a $pi$-system containing a $cap$-stable generator with a sequence $E_n in mathcal{C}$ such that $E_n uparrow Omega$.



For completeness, we should give reasons for the application of the dominated convergence theorem: By Jordan's decomposition theorem, you can write for a signed measure $mu = mu_1 - mu_2$ with measures $mu_1$ and $mu_2$. (Moreover, there exists a 'minimal decomposition' - called the Jordan decompisition.)






share|cite|improve this answer














I suppose that both signed measures are finite, since otherwise we cannot expect that the integral over arbitrary continuous functions exists. You can approximate any interval $[a,b]$ by (for example) the sequence
$$g_n(x) := begin{cases} 1 & text{ if } x in [a,b] \ nx + (1-na) &text{ if } x in [a-1/n,a] \ -nx +(1+nb) & text{ if } x in [b,b+1/n] \ 0 & text{otherwise} end{cases}.$$
By the dominated convergence theorem (for signed measures) we get
$$nu_1([a,b]) =nu_2([a,b]).$$
Thus, both measures are equal on a $cap$-stable generator of the Borel-$sigma$-algebra. Now, you need to extend the uniqueness theorem for $sigma$-finite measures. In fact, the same proof applies. Note that $$mathcal{C}={Binmathcal{B}_{[-pi, pi]} : nu_1(B)=nu_2(B)}$$ is a $pi$-system containing a $cap$-stable generator with a sequence $E_n in mathcal{C}$ such that $E_n uparrow Omega$.



For completeness, we should give reasons for the application of the dominated convergence theorem: By Jordan's decomposition theorem, you can write for a signed measure $mu = mu_1 - mu_2$ with measures $mu_1$ and $mu_2$. (Moreover, there exists a 'minimal decomposition' - called the Jordan decompisition.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 8:48

























answered Dec 5 '18 at 8:04









p4schp4sch

4,770217




4,770217












  • Of course the measures involved here are finite, I edited that, sorry. The only problem remaining is that I can't prove that $mathcal{C}$ is stable under intersection. I managed to do it when the measures are of probability, but since they don't have the same total mass, I couldn't do it
    – deque
    Dec 5 '18 at 8:43










  • In the uniqueness theorem for $sigma$-finite measures you need that there exists a sequence $(E_n)$ in $mathcal{C}$ with $E_n uparrow Omega$. Of course, we need this here, too! And it is satisfied as we can take the constant function $f=1$ to get $nu_1(Omega) = nu_2(Omega)$.
    – p4sch
    Dec 5 '18 at 8:50










  • By "uniqueness theorem for σ-finite measures" you mean the monotone class theorem right (although it doesn't require the existence of such a sequence)? Because that was the first approach I tried, but as I said, I couldn't prove that $mathcal{C}$ is closed under intersections.
    – deque
    Dec 5 '18 at 8:58






  • 1




    I mean this theorem. I called it 'uniqueness theorem for $sigma$-finite measures'.
    – p4sch
    Dec 5 '18 at 9:12












  • This theorem still requires to show that $mathcal{C}$ is closed under intersections though :/ Can you help me with that please ?
    – deque
    Dec 5 '18 at 11:47




















  • Of course the measures involved here are finite, I edited that, sorry. The only problem remaining is that I can't prove that $mathcal{C}$ is stable under intersection. I managed to do it when the measures are of probability, but since they don't have the same total mass, I couldn't do it
    – deque
    Dec 5 '18 at 8:43










  • In the uniqueness theorem for $sigma$-finite measures you need that there exists a sequence $(E_n)$ in $mathcal{C}$ with $E_n uparrow Omega$. Of course, we need this here, too! And it is satisfied as we can take the constant function $f=1$ to get $nu_1(Omega) = nu_2(Omega)$.
    – p4sch
    Dec 5 '18 at 8:50










  • By "uniqueness theorem for σ-finite measures" you mean the monotone class theorem right (although it doesn't require the existence of such a sequence)? Because that was the first approach I tried, but as I said, I couldn't prove that $mathcal{C}$ is closed under intersections.
    – deque
    Dec 5 '18 at 8:58






  • 1




    I mean this theorem. I called it 'uniqueness theorem for $sigma$-finite measures'.
    – p4sch
    Dec 5 '18 at 9:12












  • This theorem still requires to show that $mathcal{C}$ is closed under intersections though :/ Can you help me with that please ?
    – deque
    Dec 5 '18 at 11:47


















Of course the measures involved here are finite, I edited that, sorry. The only problem remaining is that I can't prove that $mathcal{C}$ is stable under intersection. I managed to do it when the measures are of probability, but since they don't have the same total mass, I couldn't do it
– deque
Dec 5 '18 at 8:43




Of course the measures involved here are finite, I edited that, sorry. The only problem remaining is that I can't prove that $mathcal{C}$ is stable under intersection. I managed to do it when the measures are of probability, but since they don't have the same total mass, I couldn't do it
– deque
Dec 5 '18 at 8:43












In the uniqueness theorem for $sigma$-finite measures you need that there exists a sequence $(E_n)$ in $mathcal{C}$ with $E_n uparrow Omega$. Of course, we need this here, too! And it is satisfied as we can take the constant function $f=1$ to get $nu_1(Omega) = nu_2(Omega)$.
– p4sch
Dec 5 '18 at 8:50




In the uniqueness theorem for $sigma$-finite measures you need that there exists a sequence $(E_n)$ in $mathcal{C}$ with $E_n uparrow Omega$. Of course, we need this here, too! And it is satisfied as we can take the constant function $f=1$ to get $nu_1(Omega) = nu_2(Omega)$.
– p4sch
Dec 5 '18 at 8:50












By "uniqueness theorem for σ-finite measures" you mean the monotone class theorem right (although it doesn't require the existence of such a sequence)? Because that was the first approach I tried, but as I said, I couldn't prove that $mathcal{C}$ is closed under intersections.
– deque
Dec 5 '18 at 8:58




By "uniqueness theorem for σ-finite measures" you mean the monotone class theorem right (although it doesn't require the existence of such a sequence)? Because that was the first approach I tried, but as I said, I couldn't prove that $mathcal{C}$ is closed under intersections.
– deque
Dec 5 '18 at 8:58




1




1




I mean this theorem. I called it 'uniqueness theorem for $sigma$-finite measures'.
– p4sch
Dec 5 '18 at 9:12






I mean this theorem. I called it 'uniqueness theorem for $sigma$-finite measures'.
– p4sch
Dec 5 '18 at 9:12














This theorem still requires to show that $mathcal{C}$ is closed under intersections though :/ Can you help me with that please ?
– deque
Dec 5 '18 at 11:47






This theorem still requires to show that $mathcal{C}$ is closed under intersections though :/ Can you help me with that please ?
– deque
Dec 5 '18 at 11:47




















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Sphinx de Gizeh

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Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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