Continuous mapping theorem for a sequence of densities?

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Let $f_n(x)$ be a sequence of densities that uniformly converges to $f(x)$ almost surely, that is,
$$ f_n(x)
xrightarrowtexta.s. f(x), quad textuniformly,$$ or
equivalently $$ Prleft( lim_nrightarrowinfty
sup_xinmathbbR | f_n(x) - f(x) | = 0 right) = 1 . $$



Using the continuous mapping theorem, I would like to claim that
$$ psileft(f_n(x)right) xrightarrowtexta.s. psileft(f(x)right), quad textuniformly,$$
or equivalently
$$ Prleft( lim_nrightarrow infty sup_xinmathbbR
| psileft(f_n(x)right) - psileft(f(x)right) | = 0 right) = 1,$$
where $psi$ is continuous on $mathbbR_>0$.



Is this true? If not, can I at least claim that the $psileft(f_n(x)right)$ converges pointwise to $psileft(f(x)right)$ almost surely?







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    Let $f_n(x)$ be a sequence of densities that uniformly converges to $f(x)$ almost surely, that is,
    $$ f_n(x)
    xrightarrowtexta.s. f(x), quad textuniformly,$$ or
    equivalently $$ Prleft( lim_nrightarrowinfty
    sup_xinmathbbR | f_n(x) - f(x) | = 0 right) = 1 . $$



    Using the continuous mapping theorem, I would like to claim that
    $$ psileft(f_n(x)right) xrightarrowtexta.s. psileft(f(x)right), quad textuniformly,$$
    or equivalently
    $$ Prleft( lim_nrightarrow infty sup_xinmathbbR
    | psileft(f_n(x)right) - psileft(f(x)right) | = 0 right) = 1,$$
    where $psi$ is continuous on $mathbbR_>0$.



    Is this true? If not, can I at least claim that the $psileft(f_n(x)right)$ converges pointwise to $psileft(f(x)right)$ almost surely?







    share|cite|improve this question






















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $f_n(x)$ be a sequence of densities that uniformly converges to $f(x)$ almost surely, that is,
      $$ f_n(x)
      xrightarrowtexta.s. f(x), quad textuniformly,$$ or
      equivalently $$ Prleft( lim_nrightarrowinfty
      sup_xinmathbbR | f_n(x) - f(x) | = 0 right) = 1 . $$



      Using the continuous mapping theorem, I would like to claim that
      $$ psileft(f_n(x)right) xrightarrowtexta.s. psileft(f(x)right), quad textuniformly,$$
      or equivalently
      $$ Prleft( lim_nrightarrow infty sup_xinmathbbR
      | psileft(f_n(x)right) - psileft(f(x)right) | = 0 right) = 1,$$
      where $psi$ is continuous on $mathbbR_>0$.



      Is this true? If not, can I at least claim that the $psileft(f_n(x)right)$ converges pointwise to $psileft(f(x)right)$ almost surely?







      share|cite|improve this question












      Let $f_n(x)$ be a sequence of densities that uniformly converges to $f(x)$ almost surely, that is,
      $$ f_n(x)
      xrightarrowtexta.s. f(x), quad textuniformly,$$ or
      equivalently $$ Prleft( lim_nrightarrowinfty
      sup_xinmathbbR | f_n(x) - f(x) | = 0 right) = 1 . $$



      Using the continuous mapping theorem, I would like to claim that
      $$ psileft(f_n(x)right) xrightarrowtexta.s. psileft(f(x)right), quad textuniformly,$$
      or equivalently
      $$ Prleft( lim_nrightarrow infty sup_xinmathbbR
      | psileft(f_n(x)right) - psileft(f(x)right) | = 0 right) = 1,$$
      where $psi$ is continuous on $mathbbR_>0$.



      Is this true? If not, can I at least claim that the $psileft(f_n(x)right)$ converges pointwise to $psileft(f(x)right)$ almost surely?









      share|cite|improve this question











      share|cite|improve this question




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      asked Aug 10 at 17:06









      Marca85

      484




      484




















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          I believe this is not true and that you do not need to appeal to the continuous mapping theorem (which implies results stronger than what you are looking for here).



          You're essentially asking if a continuous function preserves a uniform limit of functions. I'm being a little cavalier here but behaviors "almost surely" can essentially be treated as absolute behaviors in the measure theoretic sense.



          For a counterexample:



          Counterexamples to a continuous function preserving almost uniform convergence and convergence in measure.



          noting that $z^2$ is not uniformly continuous, and above does not preserve the uniform convergence of $f_n rightarrow f$.



          The definition of continuity gives you pointwise convergence a.e. though.






          share|cite|improve this answer




















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            up vote
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            I believe this is not true and that you do not need to appeal to the continuous mapping theorem (which implies results stronger than what you are looking for here).



            You're essentially asking if a continuous function preserves a uniform limit of functions. I'm being a little cavalier here but behaviors "almost surely" can essentially be treated as absolute behaviors in the measure theoretic sense.



            For a counterexample:



            Counterexamples to a continuous function preserving almost uniform convergence and convergence in measure.



            noting that $z^2$ is not uniformly continuous, and above does not preserve the uniform convergence of $f_n rightarrow f$.



            The definition of continuity gives you pointwise convergence a.e. though.






            share|cite|improve this answer
























              up vote
              0
              down vote













              I believe this is not true and that you do not need to appeal to the continuous mapping theorem (which implies results stronger than what you are looking for here).



              You're essentially asking if a continuous function preserves a uniform limit of functions. I'm being a little cavalier here but behaviors "almost surely" can essentially be treated as absolute behaviors in the measure theoretic sense.



              For a counterexample:



              Counterexamples to a continuous function preserving almost uniform convergence and convergence in measure.



              noting that $z^2$ is not uniformly continuous, and above does not preserve the uniform convergence of $f_n rightarrow f$.



              The definition of continuity gives you pointwise convergence a.e. though.






              share|cite|improve this answer






















                up vote
                0
                down vote










                up vote
                0
                down vote









                I believe this is not true and that you do not need to appeal to the continuous mapping theorem (which implies results stronger than what you are looking for here).



                You're essentially asking if a continuous function preserves a uniform limit of functions. I'm being a little cavalier here but behaviors "almost surely" can essentially be treated as absolute behaviors in the measure theoretic sense.



                For a counterexample:



                Counterexamples to a continuous function preserving almost uniform convergence and convergence in measure.



                noting that $z^2$ is not uniformly continuous, and above does not preserve the uniform convergence of $f_n rightarrow f$.



                The definition of continuity gives you pointwise convergence a.e. though.






                share|cite|improve this answer












                I believe this is not true and that you do not need to appeal to the continuous mapping theorem (which implies results stronger than what you are looking for here).



                You're essentially asking if a continuous function preserves a uniform limit of functions. I'm being a little cavalier here but behaviors "almost surely" can essentially be treated as absolute behaviors in the measure theoretic sense.



                For a counterexample:



                Counterexamples to a continuous function preserving almost uniform convergence and convergence in measure.



                noting that $z^2$ is not uniformly continuous, and above does not preserve the uniform convergence of $f_n rightarrow f$.



                The definition of continuity gives you pointwise convergence a.e. though.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 10 at 17:29









                djkat

                494




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