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Proving that a spectral measure is additive


Hilbert space - probability measure: st. norm. variablesSpectral Measures: Spectral Spaces (I)How to show a Borel Operator Measure dilates to a Spectral Measure?Spectral Measures: Spectral Spaces (II)Spectral theorem questionSpectral Measures: CompletenessSpectral Measures: IntegrabilityProduct of spectral measuresEssential supremum with respect to spectral measureFormula for the least element on the spectrum













2












$begingroup$


Given the medible space $(mathbbR,mathbbB(mathbbR))$ (where the latter is the borel's $sigma$-algebra), a Hilbert space H and two spectral measures $E,F:mathbbB(mathbbR)rightarrow B(H)$; Paul Halmos claims in his book: Introduction to Hilbert Space and the Theory of Spectral Multiplicity, that we can form a spectral measure $G:mathbbB(mathbbC)rightarrow B(H)$ by extending the following map $$G:Atimes Bin mathcalRmapsto E_A circ F_B in B(H)$$ to the $sigma-$algebra $mathbbB(mathbbC)$ ($mathcalR$ is the set that contains the products of real borel sets).



I'm having trouble proving that this map is $sigma$-additive on $mathcalR$ in the strong operator convergence sense. That is equivalent to prove that $forall x in H$ the function $$mathcalm_x:Atimes BinmathcalRmapsto langle x,E_AF_B(x)rangle in mathbbR $$ is a premeasure in the sense that if $A_itimes B_i_i in mathbbNsubset mathcalR$ is a sequence of pairwise disjoint borel rectangles and $Atimes B=bigcup_iin mathbbN A_itimes B_i in mathcalR,$ then $$langle x,E_AF_B(x)rangle=sum_iin mathbbN langle x,E_A_iF_B_i(x)rangle$$



Could you give me an idea of how to prove this? my work done so far has only given me results that are already implied by the properties of the spectral measures $E$ and $F$.










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    Given the medible space $(mathbbR,mathbbB(mathbbR))$ (where the latter is the borel's $sigma$-algebra), a Hilbert space H and two spectral measures $E,F:mathbbB(mathbbR)rightarrow B(H)$; Paul Halmos claims in his book: Introduction to Hilbert Space and the Theory of Spectral Multiplicity, that we can form a spectral measure $G:mathbbB(mathbbC)rightarrow B(H)$ by extending the following map $$G:Atimes Bin mathcalRmapsto E_A circ F_B in B(H)$$ to the $sigma-$algebra $mathbbB(mathbbC)$ ($mathcalR$ is the set that contains the products of real borel sets).



    I'm having trouble proving that this map is $sigma$-additive on $mathcalR$ in the strong operator convergence sense. That is equivalent to prove that $forall x in H$ the function $$mathcalm_x:Atimes BinmathcalRmapsto langle x,E_AF_B(x)rangle in mathbbR $$ is a premeasure in the sense that if $A_itimes B_i_i in mathbbNsubset mathcalR$ is a sequence of pairwise disjoint borel rectangles and $Atimes B=bigcup_iin mathbbN A_itimes B_i in mathcalR,$ then $$langle x,E_AF_B(x)rangle=sum_iin mathbbN langle x,E_A_iF_B_i(x)rangle$$



    Could you give me an idea of how to prove this? my work done so far has only given me results that are already implied by the properties of the spectral measures $E$ and $F$.










    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      Given the medible space $(mathbbR,mathbbB(mathbbR))$ (where the latter is the borel's $sigma$-algebra), a Hilbert space H and two spectral measures $E,F:mathbbB(mathbbR)rightarrow B(H)$; Paul Halmos claims in his book: Introduction to Hilbert Space and the Theory of Spectral Multiplicity, that we can form a spectral measure $G:mathbbB(mathbbC)rightarrow B(H)$ by extending the following map $$G:Atimes Bin mathcalRmapsto E_A circ F_B in B(H)$$ to the $sigma-$algebra $mathbbB(mathbbC)$ ($mathcalR$ is the set that contains the products of real borel sets).



      I'm having trouble proving that this map is $sigma$-additive on $mathcalR$ in the strong operator convergence sense. That is equivalent to prove that $forall x in H$ the function $$mathcalm_x:Atimes BinmathcalRmapsto langle x,E_AF_B(x)rangle in mathbbR $$ is a premeasure in the sense that if $A_itimes B_i_i in mathbbNsubset mathcalR$ is a sequence of pairwise disjoint borel rectangles and $Atimes B=bigcup_iin mathbbN A_itimes B_i in mathcalR,$ then $$langle x,E_AF_B(x)rangle=sum_iin mathbbN langle x,E_A_iF_B_i(x)rangle$$



      Could you give me an idea of how to prove this? my work done so far has only given me results that are already implied by the properties of the spectral measures $E$ and $F$.










      share|cite|improve this question











      $endgroup$




      Given the medible space $(mathbbR,mathbbB(mathbbR))$ (where the latter is the borel's $sigma$-algebra), a Hilbert space H and two spectral measures $E,F:mathbbB(mathbbR)rightarrow B(H)$; Paul Halmos claims in his book: Introduction to Hilbert Space and the Theory of Spectral Multiplicity, that we can form a spectral measure $G:mathbbB(mathbbC)rightarrow B(H)$ by extending the following map $$G:Atimes Bin mathcalRmapsto E_A circ F_B in B(H)$$ to the $sigma-$algebra $mathbbB(mathbbC)$ ($mathcalR$ is the set that contains the products of real borel sets).



      I'm having trouble proving that this map is $sigma$-additive on $mathcalR$ in the strong operator convergence sense. That is equivalent to prove that $forall x in H$ the function $$mathcalm_x:Atimes BinmathcalRmapsto langle x,E_AF_B(x)rangle in mathbbR $$ is a premeasure in the sense that if $A_itimes B_i_i in mathbbNsubset mathcalR$ is a sequence of pairwise disjoint borel rectangles and $Atimes B=bigcup_iin mathbbN A_itimes B_i in mathcalR,$ then $$langle x,E_AF_B(x)rangle=sum_iin mathbbN langle x,E_A_iF_B_i(x)rangle$$



      Could you give me an idea of how to prove this? my work done so far has only given me results that are already implied by the properties of the spectral measures $E$ and $F$.







      real-analysis linear-algebra functional-analysis measure-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 15 at 22:35







      Carlos Adrian Perez Estrada

















      asked Mar 13 at 3:29









      Carlos Adrian Perez EstradaCarlos Adrian Perez Estrada

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