Calculate product measure of off-diagonal set












0












$begingroup$


Let





  • $(E,mathcal E)$ be a measurable space


  • $mu_i$ be a probability measure on $(E,mathcal E)$


  • $lambda:=mu_1+mu_2$ and $$f_i:=frac{{rm d}mu_i}{{rm d}lambda}$$


  • $Q$ denote the measure with density $min(f_1,f_2)$ with respect to $lambda$ and $$nu_i:=mu_i-Q$$



Let $Delta:=left{(x,x):xin Eright}$. How can we show that $(v_1otimesnu_2)(Delta^c)=nu_1(E)nu_2(E)$?




For simplicity, let $B:=Delta^c=left{(x,y)in E^2:xne yright}$ and $$B_y:=left{xin E:(x,y)in Bright}=Esetminusleft{yright};;;text{for }yin E.$$ We should have $$(nu_1otimesnu_2)(B)=intnu_1(B_y)nu_2({rm d}y)=(nu_1(E)-nu_1(left{yright}))nu_2(E)tag1.$$ So, the conclusion would follow if $nu_1(left{yright})=0$. But why should that be the case?






real-analysis measure-theory product-measure







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let





    • $(E,mathcal E)$ be a measurable space


    • $mu_i$ be a probability measure on $(E,mathcal E)$


    • $lambda:=mu_1+mu_2$ and $$f_i:=frac{{rm d}mu_i}{{rm d}lambda}$$


    • $Q$ denote the measure with density $min(f_1,f_2)$ with respect to $lambda$ and $$nu_i:=mu_i-Q$$



    Let $Delta:=left{(x,x):xin Eright}$. How can we show that $(v_1otimesnu_2)(Delta^c)=nu_1(E)nu_2(E)$?




    For simplicity, let $B:=Delta^c=left{(x,y)in E^2:xne yright}$ and $$B_y:=left{xin E:(x,y)in Bright}=Esetminusleft{yright};;;text{for }yin E.$$ We should have $$(nu_1otimesnu_2)(B)=intnu_1(B_y)nu_2({rm d}y)=(nu_1(E)-nu_1(left{yright}))nu_2(E)tag1.$$ So, the conclusion would follow if $nu_1(left{yright})=0$. But why should that be the case?






    real-analysis measure-theory product-measure







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let





      • $(E,mathcal E)$ be a measurable space


      • $mu_i$ be a probability measure on $(E,mathcal E)$


      • $lambda:=mu_1+mu_2$ and $$f_i:=frac{{rm d}mu_i}{{rm d}lambda}$$


      • $Q$ denote the measure with density $min(f_1,f_2)$ with respect to $lambda$ and $$nu_i:=mu_i-Q$$



      Let $Delta:=left{(x,x):xin Eright}$. How can we show that $(v_1otimesnu_2)(Delta^c)=nu_1(E)nu_2(E)$?




      For simplicity, let $B:=Delta^c=left{(x,y)in E^2:xne yright}$ and $$B_y:=left{xin E:(x,y)in Bright}=Esetminusleft{yright};;;text{for }yin E.$$ We should have $$(nu_1otimesnu_2)(B)=intnu_1(B_y)nu_2({rm d}y)=(nu_1(E)-nu_1(left{yright}))nu_2(E)tag1.$$ So, the conclusion would follow if $nu_1(left{yright})=0$. But why should that be the case?






      real-analysis measure-theory product-measure







      share|cite|improve this question









      $endgroup$




      Let





      • $(E,mathcal E)$ be a measurable space


      • $mu_i$ be a probability measure on $(E,mathcal E)$


      • $lambda:=mu_1+mu_2$ and $$f_i:=frac{{rm d}mu_i}{{rm d}lambda}$$


      • $Q$ denote the measure with density $min(f_1,f_2)$ with respect to $lambda$ and $$nu_i:=mu_i-Q$$



      Let $Delta:=left{(x,x):xin Eright}$. How can we show that $(v_1otimesnu_2)(Delta^c)=nu_1(E)nu_2(E)$?




      For simplicity, let $B:=Delta^c=left{(x,y)in E^2:xne yright}$ and $$B_y:=left{xin E:(x,y)in Bright}=Esetminusleft{yright};;;text{for }yin E.$$ We should have $$(nu_1otimesnu_2)(B)=intnu_1(B_y)nu_2({rm d}y)=(nu_1(E)-nu_1(left{yright}))nu_2(E)tag1.$$ So, the conclusion would follow if $nu_1(left{yright})=0$. But why should that be the case?






      real-analysis measure-theory product-measure




      real-analysis measure-theory product-measure






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 2 at 12:07









      0xbadf00d0xbadf00d

      1,96841532




      1,96841532






















          1 Answer
          1






          active

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          1












          $begingroup$

          Let us write $f_1wedge f_2$ for $min{f_1,f_2}$. What we are required to prove is that $ (nu_1 otimes nu_2)(Delta)=0$. You can write the left side as $int int_{{x}} (f_2- (f_2wedge f_1) (y))dlambda(y) [f_1(x)- (f_2(x)wedge f_1(x))]dlambda (x)=0$. Note that if $f_2(x)- (f_2(x)wedge f_1(x))>0$ then $f_2(x) >f_1(x)$ and hence $(f_1- (f_2wedge f_1) (x))=0$. This makes the double integral $0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So, your argument is that the left-hand side is $$intleft(f_1(x)-(f_1wedge f_2)(x)right)left(f_2(x)-(f_1wedge f_2)(x)right)lambdaleft(left{xright}right)lambdaleft({rm d}xright)$$ and then conclude by your note, right?
            $endgroup$
            – 0xbadf00d
            Jan 2 at 19:03












          • $begingroup$
            Yes, that is right.
            $endgroup$
            – Kavi Rama Murthy
            Jan 3 at 0:31











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          1 Answer
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          active

          oldest

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          1












          $begingroup$

          Let us write $f_1wedge f_2$ for $min{f_1,f_2}$. What we are required to prove is that $ (nu_1 otimes nu_2)(Delta)=0$. You can write the left side as $int int_{{x}} (f_2- (f_2wedge f_1) (y))dlambda(y) [f_1(x)- (f_2(x)wedge f_1(x))]dlambda (x)=0$. Note that if $f_2(x)- (f_2(x)wedge f_1(x))>0$ then $f_2(x) >f_1(x)$ and hence $(f_1- (f_2wedge f_1) (x))=0$. This makes the double integral $0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So, your argument is that the left-hand side is $$intleft(f_1(x)-(f_1wedge f_2)(x)right)left(f_2(x)-(f_1wedge f_2)(x)right)lambdaleft(left{xright}right)lambdaleft({rm d}xright)$$ and then conclude by your note, right?
            $endgroup$
            – 0xbadf00d
            Jan 2 at 19:03












          • $begingroup$
            Yes, that is right.
            $endgroup$
            – Kavi Rama Murthy
            Jan 3 at 0:31
















          1












          $begingroup$

          Let us write $f_1wedge f_2$ for $min{f_1,f_2}$. What we are required to prove is that $ (nu_1 otimes nu_2)(Delta)=0$. You can write the left side as $int int_{{x}} (f_2- (f_2wedge f_1) (y))dlambda(y) [f_1(x)- (f_2(x)wedge f_1(x))]dlambda (x)=0$. Note that if $f_2(x)- (f_2(x)wedge f_1(x))>0$ then $f_2(x) >f_1(x)$ and hence $(f_1- (f_2wedge f_1) (x))=0$. This makes the double integral $0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So, your argument is that the left-hand side is $$intleft(f_1(x)-(f_1wedge f_2)(x)right)left(f_2(x)-(f_1wedge f_2)(x)right)lambdaleft(left{xright}right)lambdaleft({rm d}xright)$$ and then conclude by your note, right?
            $endgroup$
            – 0xbadf00d
            Jan 2 at 19:03












          • $begingroup$
            Yes, that is right.
            $endgroup$
            – Kavi Rama Murthy
            Jan 3 at 0:31














          1












          1








          1





          $begingroup$

          Let us write $f_1wedge f_2$ for $min{f_1,f_2}$. What we are required to prove is that $ (nu_1 otimes nu_2)(Delta)=0$. You can write the left side as $int int_{{x}} (f_2- (f_2wedge f_1) (y))dlambda(y) [f_1(x)- (f_2(x)wedge f_1(x))]dlambda (x)=0$. Note that if $f_2(x)- (f_2(x)wedge f_1(x))>0$ then $f_2(x) >f_1(x)$ and hence $(f_1- (f_2wedge f_1) (x))=0$. This makes the double integral $0$.






          share|cite|improve this answer









          $endgroup$



          Let us write $f_1wedge f_2$ for $min{f_1,f_2}$. What we are required to prove is that $ (nu_1 otimes nu_2)(Delta)=0$. You can write the left side as $int int_{{x}} (f_2- (f_2wedge f_1) (y))dlambda(y) [f_1(x)- (f_2(x)wedge f_1(x))]dlambda (x)=0$. Note that if $f_2(x)- (f_2(x)wedge f_1(x))>0$ then $f_2(x) >f_1(x)$ and hence $(f_1- (f_2wedge f_1) (x))=0$. This makes the double integral $0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 12:44









          Kavi Rama MurthyKavi Rama Murthy

          63.7k42464




          63.7k42464












          • $begingroup$
            So, your argument is that the left-hand side is $$intleft(f_1(x)-(f_1wedge f_2)(x)right)left(f_2(x)-(f_1wedge f_2)(x)right)lambdaleft(left{xright}right)lambdaleft({rm d}xright)$$ and then conclude by your note, right?
            $endgroup$
            – 0xbadf00d
            Jan 2 at 19:03












          • $begingroup$
            Yes, that is right.
            $endgroup$
            – Kavi Rama Murthy
            Jan 3 at 0:31


















          • $begingroup$
            So, your argument is that the left-hand side is $$intleft(f_1(x)-(f_1wedge f_2)(x)right)left(f_2(x)-(f_1wedge f_2)(x)right)lambdaleft(left{xright}right)lambdaleft({rm d}xright)$$ and then conclude by your note, right?
            $endgroup$
            – 0xbadf00d
            Jan 2 at 19:03












          • $begingroup$
            Yes, that is right.
            $endgroup$
            – Kavi Rama Murthy
            Jan 3 at 0:31
















          $begingroup$
          So, your argument is that the left-hand side is $$intleft(f_1(x)-(f_1wedge f_2)(x)right)left(f_2(x)-(f_1wedge f_2)(x)right)lambdaleft(left{xright}right)lambdaleft({rm d}xright)$$ and then conclude by your note, right?
          $endgroup$
          – 0xbadf00d
          Jan 2 at 19:03






          $begingroup$
          So, your argument is that the left-hand side is $$intleft(f_1(x)-(f_1wedge f_2)(x)right)left(f_2(x)-(f_1wedge f_2)(x)right)lambdaleft(left{xright}right)lambdaleft({rm d}xright)$$ and then conclude by your note, right?
          $endgroup$
          – 0xbadf00d
          Jan 2 at 19:03














          $begingroup$
          Yes, that is right.
          $endgroup$
          – Kavi Rama Murthy
          Jan 3 at 0:31




          $begingroup$
          Yes, that is right.
          $endgroup$
          – Kavi Rama Murthy
          Jan 3 at 0:31


















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