Convergence of third moments when second moments are uniformly bounded.












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Suppose $lambda_n, ngeq 1 $ are $sigma $-finite measures on $mathbb{R}$ such that $$ sup_{ngeq 1} int_{[-1,1]}x^2 lambda_n(dx) < infty $$ and $$forall delta > 0, exists n_0 geq 1, forall n geq n_0: lambda_n(mathbb{R} setminus [-delta,delta]) = 0. $$ I am trying to prove that $$lim_{nto infty} int_{mathbb{R}}vert xvert ^3 lambda_n(dx) = 0.$$ I came across this result in "Probability in Banach Spaces - Stable and Infinitely Divisible Distributions" by Werner Linde. It is stated in Proposition 5.6.1.






integration probability-theory measure-theory weak-convergence







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    Suppose $lambda_n, ngeq 1 $ are $sigma $-finite measures on $mathbb{R}$ such that $$ sup_{ngeq 1} int_{[-1,1]}x^2 lambda_n(dx) < infty $$ and $$forall delta > 0, exists n_0 geq 1, forall n geq n_0: lambda_n(mathbb{R} setminus [-delta,delta]) = 0. $$ I am trying to prove that $$lim_{nto infty} int_{mathbb{R}}vert xvert ^3 lambda_n(dx) = 0.$$ I came across this result in "Probability in Banach Spaces - Stable and Infinitely Divisible Distributions" by Werner Linde. It is stated in Proposition 5.6.1.






    integration probability-theory measure-theory weak-convergence







    share|cite|improve this question

























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      2








      2







      Suppose $lambda_n, ngeq 1 $ are $sigma $-finite measures on $mathbb{R}$ such that $$ sup_{ngeq 1} int_{[-1,1]}x^2 lambda_n(dx) < infty $$ and $$forall delta > 0, exists n_0 geq 1, forall n geq n_0: lambda_n(mathbb{R} setminus [-delta,delta]) = 0. $$ I am trying to prove that $$lim_{nto infty} int_{mathbb{R}}vert xvert ^3 lambda_n(dx) = 0.$$ I came across this result in "Probability in Banach Spaces - Stable and Infinitely Divisible Distributions" by Werner Linde. It is stated in Proposition 5.6.1.






      integration probability-theory measure-theory weak-convergence







      share|cite|improve this question













      Suppose $lambda_n, ngeq 1 $ are $sigma $-finite measures on $mathbb{R}$ such that $$ sup_{ngeq 1} int_{[-1,1]}x^2 lambda_n(dx) < infty $$ and $$forall delta > 0, exists n_0 geq 1, forall n geq n_0: lambda_n(mathbb{R} setminus [-delta,delta]) = 0. $$ I am trying to prove that $$lim_{nto infty} int_{mathbb{R}}vert xvert ^3 lambda_n(dx) = 0.$$ I came across this result in "Probability in Banach Spaces - Stable and Infinitely Divisible Distributions" by Werner Linde. It is stated in Proposition 5.6.1.






      integration probability-theory measure-theory weak-convergence




      integration probability-theory measure-theory weak-convergence






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      asked Nov 27 '18 at 21:30









      M. Ost

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          Let $c:=sup_{ngeq 1} int_{[-1,1]} x^2 lambda_n(dx) $. Then we have
          $$ int_{[-1/m, 1/m]} vert x vert^3 lambda_n(dx) leq int_{[-1/m, 1/m]} frac{1}{m}vert x vert^2 lambda_n(dx) leq frac{c}{m} $$
          On the other hand there exists $n_0(m)in mathbb{N}$ such that for all $ngeq n_0$ holds
          $$ lambda_n(mathbb{R}setminus [-frac{1}{m}, frac{1}{m}]) =0. $$
          Let $varepsilon>0$ and pick $Nin mathbb{N}$ such that $c/N<varepsilon$.
          Thus, for $ngeq max{ N, n_0(N) } $
          $$ int_mathbb{R} vert x vert^3 lambda_n(dx) = int_{mathbb{R}setminus [-1/n, 1/n]} vert x vert^3 lambda_n(dx) + int_{[-1/n, 1/n]} vert x vert^3 lambda_n(dx)
          leq 0 + frac{c}{n} < varepsilon. $$
          Hence, we get
          $$ lim_{nrightarrow infty} int_mathbb{R} vert x vert^3 lambda_n(dx) = 0. $$






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            Let $c:=sup_{ngeq 1} int_{[-1,1]} x^2 lambda_n(dx) $. Then we have
            $$ int_{[-1/m, 1/m]} vert x vert^3 lambda_n(dx) leq int_{[-1/m, 1/m]} frac{1}{m}vert x vert^2 lambda_n(dx) leq frac{c}{m} $$
            On the other hand there exists $n_0(m)in mathbb{N}$ such that for all $ngeq n_0$ holds
            $$ lambda_n(mathbb{R}setminus [-frac{1}{m}, frac{1}{m}]) =0. $$
            Let $varepsilon>0$ and pick $Nin mathbb{N}$ such that $c/N<varepsilon$.
            Thus, for $ngeq max{ N, n_0(N) } $
            $$ int_mathbb{R} vert x vert^3 lambda_n(dx) = int_{mathbb{R}setminus [-1/n, 1/n]} vert x vert^3 lambda_n(dx) + int_{[-1/n, 1/n]} vert x vert^3 lambda_n(dx)
            leq 0 + frac{c}{n} < varepsilon. $$
            Hence, we get
            $$ lim_{nrightarrow infty} int_mathbb{R} vert x vert^3 lambda_n(dx) = 0. $$






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              Let $c:=sup_{ngeq 1} int_{[-1,1]} x^2 lambda_n(dx) $. Then we have
              $$ int_{[-1/m, 1/m]} vert x vert^3 lambda_n(dx) leq int_{[-1/m, 1/m]} frac{1}{m}vert x vert^2 lambda_n(dx) leq frac{c}{m} $$
              On the other hand there exists $n_0(m)in mathbb{N}$ such that for all $ngeq n_0$ holds
              $$ lambda_n(mathbb{R}setminus [-frac{1}{m}, frac{1}{m}]) =0. $$
              Let $varepsilon>0$ and pick $Nin mathbb{N}$ such that $c/N<varepsilon$.
              Thus, for $ngeq max{ N, n_0(N) } $
              $$ int_mathbb{R} vert x vert^3 lambda_n(dx) = int_{mathbb{R}setminus [-1/n, 1/n]} vert x vert^3 lambda_n(dx) + int_{[-1/n, 1/n]} vert x vert^3 lambda_n(dx)
              leq 0 + frac{c}{n} < varepsilon. $$
              Hence, we get
              $$ lim_{nrightarrow infty} int_mathbb{R} vert x vert^3 lambda_n(dx) = 0. $$






              share|cite|improve this answer


























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                Let $c:=sup_{ngeq 1} int_{[-1,1]} x^2 lambda_n(dx) $. Then we have
                $$ int_{[-1/m, 1/m]} vert x vert^3 lambda_n(dx) leq int_{[-1/m, 1/m]} frac{1}{m}vert x vert^2 lambda_n(dx) leq frac{c}{m} $$
                On the other hand there exists $n_0(m)in mathbb{N}$ such that for all $ngeq n_0$ holds
                $$ lambda_n(mathbb{R}setminus [-frac{1}{m}, frac{1}{m}]) =0. $$
                Let $varepsilon>0$ and pick $Nin mathbb{N}$ such that $c/N<varepsilon$.
                Thus, for $ngeq max{ N, n_0(N) } $
                $$ int_mathbb{R} vert x vert^3 lambda_n(dx) = int_{mathbb{R}setminus [-1/n, 1/n]} vert x vert^3 lambda_n(dx) + int_{[-1/n, 1/n]} vert x vert^3 lambda_n(dx)
                leq 0 + frac{c}{n} < varepsilon. $$
                Hence, we get
                $$ lim_{nrightarrow infty} int_mathbb{R} vert x vert^3 lambda_n(dx) = 0. $$






                share|cite|improve this answer














                Let $c:=sup_{ngeq 1} int_{[-1,1]} x^2 lambda_n(dx) $. Then we have
                $$ int_{[-1/m, 1/m]} vert x vert^3 lambda_n(dx) leq int_{[-1/m, 1/m]} frac{1}{m}vert x vert^2 lambda_n(dx) leq frac{c}{m} $$
                On the other hand there exists $n_0(m)in mathbb{N}$ such that for all $ngeq n_0$ holds
                $$ lambda_n(mathbb{R}setminus [-frac{1}{m}, frac{1}{m}]) =0. $$
                Let $varepsilon>0$ and pick $Nin mathbb{N}$ such that $c/N<varepsilon$.
                Thus, for $ngeq max{ N, n_0(N) } $
                $$ int_mathbb{R} vert x vert^3 lambda_n(dx) = int_{mathbb{R}setminus [-1/n, 1/n]} vert x vert^3 lambda_n(dx) + int_{[-1/n, 1/n]} vert x vert^3 lambda_n(dx)
                leq 0 + frac{c}{n} < varepsilon. $$
                Hence, we get
                $$ lim_{nrightarrow infty} int_mathbb{R} vert x vert^3 lambda_n(dx) = 0. $$







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                edited Nov 27 '18 at 22:27

























                answered Nov 27 '18 at 21:56









                Severin Schraven

                5,8281934




                5,8281934






























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