Compactness of a specific set in weak topology












3












$begingroup$


I have the following question:



Let $E$ be a polish space (that is, a topological space, which is separable and metrizable, such that $E$ would be complete if equipped with this metric).



Consider the space of probability measures on $E$, denoted by $mathcal M (E)$ equipped with the topology of weak convergence.



The claim I am struggling with is the following:



For every $ellinBbb N$ let $K_ell$ be a compact set of $E$. Then the set
$$bigcap_{ellinBbb N} { m in mathcal M (E): m(K^{operatorname{c}}_ell) leq frac 1 ell }$$
is compact.






general-topology functional-analysis measure-theory weak-convergence topological-vector-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    These links might be helpful: en.wikipedia.org/wiki/Tightness_of_measures and en.wikipedia.org/wiki/Prokhorov%27s_theorem
    $endgroup$
    – Wraith1995
    Jan 8 at 19:22










  • $begingroup$
    I should have seen the connection to Prokhorov by myself... Thanks for the hint.
    $endgroup$
    – Falrach
    Jan 8 at 20:34










  • $begingroup$
    No problem; your answer looks fine to me by the way.
    $endgroup$
    – Wraith1995
    Jan 8 at 22:42
















3












$begingroup$


I have the following question:



Let $E$ be a polish space (that is, a topological space, which is separable and metrizable, such that $E$ would be complete if equipped with this metric).



Consider the space of probability measures on $E$, denoted by $mathcal M (E)$ equipped with the topology of weak convergence.



The claim I am struggling with is the following:



For every $ellinBbb N$ let $K_ell$ be a compact set of $E$. Then the set
$$bigcap_{ellinBbb N} { m in mathcal M (E): m(K^{operatorname{c}}_ell) leq frac 1 ell }$$
is compact.






general-topology functional-analysis measure-theory weak-convergence topological-vector-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    These links might be helpful: en.wikipedia.org/wiki/Tightness_of_measures and en.wikipedia.org/wiki/Prokhorov%27s_theorem
    $endgroup$
    – Wraith1995
    Jan 8 at 19:22










  • $begingroup$
    I should have seen the connection to Prokhorov by myself... Thanks for the hint.
    $endgroup$
    – Falrach
    Jan 8 at 20:34










  • $begingroup$
    No problem; your answer looks fine to me by the way.
    $endgroup$
    – Wraith1995
    Jan 8 at 22:42














3












3








3


1



$begingroup$


I have the following question:



Let $E$ be a polish space (that is, a topological space, which is separable and metrizable, such that $E$ would be complete if equipped with this metric).



Consider the space of probability measures on $E$, denoted by $mathcal M (E)$ equipped with the topology of weak convergence.



The claim I am struggling with is the following:



For every $ellinBbb N$ let $K_ell$ be a compact set of $E$. Then the set
$$bigcap_{ellinBbb N} { m in mathcal M (E): m(K^{operatorname{c}}_ell) leq frac 1 ell }$$
is compact.






general-topology functional-analysis measure-theory weak-convergence topological-vector-spaces







share|cite|improve this question











$endgroup$




I have the following question:



Let $E$ be a polish space (that is, a topological space, which is separable and metrizable, such that $E$ would be complete if equipped with this metric).



Consider the space of probability measures on $E$, denoted by $mathcal M (E)$ equipped with the topology of weak convergence.



The claim I am struggling with is the following:



For every $ellinBbb N$ let $K_ell$ be a compact set of $E$. Then the set
$$bigcap_{ellinBbb N} { m in mathcal M (E): m(K^{operatorname{c}}_ell) leq frac 1 ell }$$
is compact.






general-topology functional-analysis measure-theory weak-convergence topological-vector-spaces




general-topology functional-analysis measure-theory weak-convergence topological-vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 21:21







Falrach

















asked Jan 8 at 18:07









FalrachFalrach

1,844225




1,844225








  • 1




    $begingroup$
    These links might be helpful: en.wikipedia.org/wiki/Tightness_of_measures and en.wikipedia.org/wiki/Prokhorov%27s_theorem
    $endgroup$
    – Wraith1995
    Jan 8 at 19:22










  • $begingroup$
    I should have seen the connection to Prokhorov by myself... Thanks for the hint.
    $endgroup$
    – Falrach
    Jan 8 at 20:34










  • $begingroup$
    No problem; your answer looks fine to me by the way.
    $endgroup$
    – Wraith1995
    Jan 8 at 22:42














  • 1




    $begingroup$
    These links might be helpful: en.wikipedia.org/wiki/Tightness_of_measures and en.wikipedia.org/wiki/Prokhorov%27s_theorem
    $endgroup$
    – Wraith1995
    Jan 8 at 19:22










  • $begingroup$
    I should have seen the connection to Prokhorov by myself... Thanks for the hint.
    $endgroup$
    – Falrach
    Jan 8 at 20:34










  • $begingroup$
    No problem; your answer looks fine to me by the way.
    $endgroup$
    – Wraith1995
    Jan 8 at 22:42








1




1




$begingroup$
These links might be helpful: en.wikipedia.org/wiki/Tightness_of_measures and en.wikipedia.org/wiki/Prokhorov%27s_theorem
$endgroup$
– Wraith1995
Jan 8 at 19:22




$begingroup$
These links might be helpful: en.wikipedia.org/wiki/Tightness_of_measures and en.wikipedia.org/wiki/Prokhorov%27s_theorem
$endgroup$
– Wraith1995
Jan 8 at 19:22












$begingroup$
I should have seen the connection to Prokhorov by myself... Thanks for the hint.
$endgroup$
– Falrach
Jan 8 at 20:34




$begingroup$
I should have seen the connection to Prokhorov by myself... Thanks for the hint.
$endgroup$
– Falrach
Jan 8 at 20:34












$begingroup$
No problem; your answer looks fine to me by the way.
$endgroup$
– Wraith1995
Jan 8 at 22:42




$begingroup$
No problem; your answer looks fine to me by the way.
$endgroup$
– Wraith1995
Jan 8 at 22:42










1 Answer
1






active

oldest

votes


















2












$begingroup$

I hope I understand the full meaning of Prokhorov in the right way:



In this setting (i.e. $E$ polish) the set $mathcal M (E)$ is metrizable with a complete metric, say $d$.



The set $M := cap_{ellinBbb N} { m : m(K_{ell}^{operatorname{c}}) leq frac 1 ell } $ is tight (For any $varepsilon > 0$ just choose $ell$ such that $frac 1 ell leq varepsilon$ and take $K_ell$). By Prokhorov this means that the closure of $M$ with respect to $d$ is compact.



Further, we have that for $eta >0$ and $Ksubset E$ compact the sets of the form ${ m : m(K^{operatorname{c}}) leq eta }$ are closed with respect to $d$. This can be seen by the following:
Take a convergent sequence sequence $(m_n)_{ngeq 1}$ with $m_n (K^{operatorname{c}}) leq eta$ for every $n$. Define $m := lim_{ntoinfty} m_n$. Since the topology induced by $d$ is equivalent to the topology of weak convergence we have the portmanteau theorem, such that:
$$m(K^{operatorname{c}}) leq liminf_{ntoinfty} m_n (K^{operatorname{c}}) leq eta$$
All in all this means $M$ is compact.






share|cite|improve this answer









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    active

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    2












    $begingroup$

    I hope I understand the full meaning of Prokhorov in the right way:



    In this setting (i.e. $E$ polish) the set $mathcal M (E)$ is metrizable with a complete metric, say $d$.



    The set $M := cap_{ellinBbb N} { m : m(K_{ell}^{operatorname{c}}) leq frac 1 ell } $ is tight (For any $varepsilon > 0$ just choose $ell$ such that $frac 1 ell leq varepsilon$ and take $K_ell$). By Prokhorov this means that the closure of $M$ with respect to $d$ is compact.



    Further, we have that for $eta >0$ and $Ksubset E$ compact the sets of the form ${ m : m(K^{operatorname{c}}) leq eta }$ are closed with respect to $d$. This can be seen by the following:
    Take a convergent sequence sequence $(m_n)_{ngeq 1}$ with $m_n (K^{operatorname{c}}) leq eta$ for every $n$. Define $m := lim_{ntoinfty} m_n$. Since the topology induced by $d$ is equivalent to the topology of weak convergence we have the portmanteau theorem, such that:
    $$m(K^{operatorname{c}}) leq liminf_{ntoinfty} m_n (K^{operatorname{c}}) leq eta$$
    All in all this means $M$ is compact.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I hope I understand the full meaning of Prokhorov in the right way:



      In this setting (i.e. $E$ polish) the set $mathcal M (E)$ is metrizable with a complete metric, say $d$.



      The set $M := cap_{ellinBbb N} { m : m(K_{ell}^{operatorname{c}}) leq frac 1 ell } $ is tight (For any $varepsilon > 0$ just choose $ell$ such that $frac 1 ell leq varepsilon$ and take $K_ell$). By Prokhorov this means that the closure of $M$ with respect to $d$ is compact.



      Further, we have that for $eta >0$ and $Ksubset E$ compact the sets of the form ${ m : m(K^{operatorname{c}}) leq eta }$ are closed with respect to $d$. This can be seen by the following:
      Take a convergent sequence sequence $(m_n)_{ngeq 1}$ with $m_n (K^{operatorname{c}}) leq eta$ for every $n$. Define $m := lim_{ntoinfty} m_n$. Since the topology induced by $d$ is equivalent to the topology of weak convergence we have the portmanteau theorem, such that:
      $$m(K^{operatorname{c}}) leq liminf_{ntoinfty} m_n (K^{operatorname{c}}) leq eta$$
      All in all this means $M$ is compact.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I hope I understand the full meaning of Prokhorov in the right way:



        In this setting (i.e. $E$ polish) the set $mathcal M (E)$ is metrizable with a complete metric, say $d$.



        The set $M := cap_{ellinBbb N} { m : m(K_{ell}^{operatorname{c}}) leq frac 1 ell } $ is tight (For any $varepsilon > 0$ just choose $ell$ such that $frac 1 ell leq varepsilon$ and take $K_ell$). By Prokhorov this means that the closure of $M$ with respect to $d$ is compact.



        Further, we have that for $eta >0$ and $Ksubset E$ compact the sets of the form ${ m : m(K^{operatorname{c}}) leq eta }$ are closed with respect to $d$. This can be seen by the following:
        Take a convergent sequence sequence $(m_n)_{ngeq 1}$ with $m_n (K^{operatorname{c}}) leq eta$ for every $n$. Define $m := lim_{ntoinfty} m_n$. Since the topology induced by $d$ is equivalent to the topology of weak convergence we have the portmanteau theorem, such that:
        $$m(K^{operatorname{c}}) leq liminf_{ntoinfty} m_n (K^{operatorname{c}}) leq eta$$
        All in all this means $M$ is compact.






        share|cite|improve this answer









        $endgroup$



        I hope I understand the full meaning of Prokhorov in the right way:



        In this setting (i.e. $E$ polish) the set $mathcal M (E)$ is metrizable with a complete metric, say $d$.



        The set $M := cap_{ellinBbb N} { m : m(K_{ell}^{operatorname{c}}) leq frac 1 ell } $ is tight (For any $varepsilon > 0$ just choose $ell$ such that $frac 1 ell leq varepsilon$ and take $K_ell$). By Prokhorov this means that the closure of $M$ with respect to $d$ is compact.



        Further, we have that for $eta >0$ and $Ksubset E$ compact the sets of the form ${ m : m(K^{operatorname{c}}) leq eta }$ are closed with respect to $d$. This can be seen by the following:
        Take a convergent sequence sequence $(m_n)_{ngeq 1}$ with $m_n (K^{operatorname{c}}) leq eta$ for every $n$. Define $m := lim_{ntoinfty} m_n$. Since the topology induced by $d$ is equivalent to the topology of weak convergence we have the portmanteau theorem, such that:
        $$m(K^{operatorname{c}}) leq liminf_{ntoinfty} m_n (K^{operatorname{c}}) leq eta$$
        All in all this means $M$ is compact.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 20:32









        FalrachFalrach

        1,844225




        1,844225






























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