Is the space of Radon measures a Polish space or at least separable?About the definition of Borel and Radon...



Is the space of Radon measures a Polish space or at least separable?


About the definition of Borel and Radon measuresWhen is the support of a Radon measure separable?On Radon measures with values in Banach spaceIs the space of Radon measures a Prohorov space?Questions on topologies on space of Radon measuresUnder what conditions can we put a complete norm on a linear subspace of a separable Banach space?Is the following product-like space a Polish space?Measurably-isomorphic subsets of polish spaces and the continuum hypothesisOn $C(K)$ spaces embeddable into the Banach space $c_0$













4












$begingroup$


Background: I work on a SPDE problem where in order to apply Prokhorov's theorem I need that some measure space is Polish space. And additionaly it would be good if that space is Banach space. Earlier today I was reading the book: Malek, Necas, Rokyta, Ruzicka - Weak and Measure-valued Solutions to Evolutionary PDEs, 1996, and I have a question from the Subsection 1.2.8 titled Radon measures. The definitions given bellow are taken from the same book.



On the one hand, the space of Radon measures is defined as:
$$M(mathbb{R}^d)equiv { mu : C_0 (mathbb{R}) rightarrow mathbb{R}; mu hspace{0.2cm} linear hspace{0.2cm} s.t. hspace{0.2cm} exists c>0, |mu (f)|leq c ||f||_{infty}, forall f in mathcal{D}(mathbb{R}^d)}.$$



Here



$C_0(mathbb{R}^d)equiv { u in C(mathbb{R}^d): lim_{|x|rightarrow infty} u(x) = 0 }$
and
$C_0(mathbb{R}^d)=overline{mathcal{D}(mathbb{R}^d)}^{||cdot||_{infty}}$.



As usual $mathcal{D}(Omega)$ stands for the space of functions from $C^{infty}(
overline{Omega})$
with compact support in $Omega$.



If we further define



$||mu||_{M(mathbb{R}^d)}equiv sup{|mu(f)|: f in mathcal{D}(mathbb{R}^d),||f||_{infty}leq 1 }$,



then the space $(M(mathbb{R}^d), || cdot ||_{M(mathbb{R}^d)})$ is a Banach space.



On the other hand, let $Omega$ be a bounded domain. We denote by $M(Omega)$ the space of Radon measures defined as the dual space of $C(overline{Omega})$. Also in this case we know that $L^1(Omega)hookrightarrow M(Omega)$ (and we know that $L^1(Omega)$ is separable).



My questions are:





  1. Is the space of Radon measures separable - in the case $Omega subset mathbb{R}^d$ and in the case $mathbb{R}^d$? Or to be more precise is it a Polish space? I have search it in a few books and in the questions here but I didn't find any concrete answer (I maybe have missed something).


  2. Maybe some subspace of the space of Radon measure is Polish? I've read somewhere that the space of positive Radon measures is Polish but didn't find any book to confirm that.


  3. Are there some other spaces of measure-valued functions that are Polish (besides the spaces mentioned above)?


I usually avoid dealing with meaasure-valued spaces so I don't know much about them. Help with this would be great (and I definitely need it). Thanks in advance.










share|cite|improve this question









$endgroup$














  • $begingroup$
    You have better chances with the weak $^*$ topology. The dual unit ball of a separable Banach space is weak$^*$ compact and metrizable.
    $endgroup$
    – Jochen Wengenroth
    11 hours ago






  • 1




    $begingroup$
    @JochenWengenroth: On the other hand, the space of all Radon measures with the weak-* topology is not metrizable, though it is separable.
    $endgroup$
    – Nate Eldredge
    11 hours ago






  • 1




    $begingroup$
    It is a long established fact that the space of Radon measures on a completely regular space is naturally identifiable with the dual of the space $C^b(S)$ of bounded, continuous functions thereon (with the so-called strict topology—-the finest locally convex topology which coincides with compact convergence on the unit ball). This was shown by Buck in the locally compact case in the 50’s and extended to the general case in the 60‘s. If the underlying space is polish, then so is that of the Radon probability measures (under the weak topology induced by $C^b(S)$)—-again a celebrated result.
    $endgroup$
    – user131781
    10 hours ago










  • $begingroup$
    @NateEldredge Do you mean as opposed to Radon probability measures? The set of Radon probability measures on a Polish space is Polish as a subspace of the weak-* topology, even though the weak-* topology isn't first-countable on the whole space $C_b(X)^*$.
    $endgroup$
    – Robert Furber
    10 hours ago












  • $begingroup$
    @RobertFurber: Right, that's what I mean. The space $M(Omega)$ defined in the question is the space of all signed Radon measures, i.e. the dual of $C(overline{Omega})$ (or something like that), and this is not weak-* metrizable (nor, as you say, even first countable).
    $endgroup$
    – Nate Eldredge
    10 hours ago


















4












$begingroup$


Background: I work on a SPDE problem where in order to apply Prokhorov's theorem I need that some measure space is Polish space. And additionaly it would be good if that space is Banach space. Earlier today I was reading the book: Malek, Necas, Rokyta, Ruzicka - Weak and Measure-valued Solutions to Evolutionary PDEs, 1996, and I have a question from the Subsection 1.2.8 titled Radon measures. The definitions given bellow are taken from the same book.



On the one hand, the space of Radon measures is defined as:
$$M(mathbb{R}^d)equiv { mu : C_0 (mathbb{R}) rightarrow mathbb{R}; mu hspace{0.2cm} linear hspace{0.2cm} s.t. hspace{0.2cm} exists c>0, |mu (f)|leq c ||f||_{infty}, forall f in mathcal{D}(mathbb{R}^d)}.$$



Here



$C_0(mathbb{R}^d)equiv { u in C(mathbb{R}^d): lim_{|x|rightarrow infty} u(x) = 0 }$
and
$C_0(mathbb{R}^d)=overline{mathcal{D}(mathbb{R}^d)}^{||cdot||_{infty}}$.



As usual $mathcal{D}(Omega)$ stands for the space of functions from $C^{infty}(
overline{Omega})$
with compact support in $Omega$.



If we further define



$||mu||_{M(mathbb{R}^d)}equiv sup{|mu(f)|: f in mathcal{D}(mathbb{R}^d),||f||_{infty}leq 1 }$,



then the space $(M(mathbb{R}^d), || cdot ||_{M(mathbb{R}^d)})$ is a Banach space.



On the other hand, let $Omega$ be a bounded domain. We denote by $M(Omega)$ the space of Radon measures defined as the dual space of $C(overline{Omega})$. Also in this case we know that $L^1(Omega)hookrightarrow M(Omega)$ (and we know that $L^1(Omega)$ is separable).



My questions are:





  1. Is the space of Radon measures separable - in the case $Omega subset mathbb{R}^d$ and in the case $mathbb{R}^d$? Or to be more precise is it a Polish space? I have search it in a few books and in the questions here but I didn't find any concrete answer (I maybe have missed something).


  2. Maybe some subspace of the space of Radon measure is Polish? I've read somewhere that the space of positive Radon measures is Polish but didn't find any book to confirm that.


  3. Are there some other spaces of measure-valued functions that are Polish (besides the spaces mentioned above)?


I usually avoid dealing with meaasure-valued spaces so I don't know much about them. Help with this would be great (and I definitely need it). Thanks in advance.










share|cite|improve this question









$endgroup$














  • $begingroup$
    You have better chances with the weak $^*$ topology. The dual unit ball of a separable Banach space is weak$^*$ compact and metrizable.
    $endgroup$
    – Jochen Wengenroth
    11 hours ago






  • 1




    $begingroup$
    @JochenWengenroth: On the other hand, the space of all Radon measures with the weak-* topology is not metrizable, though it is separable.
    $endgroup$
    – Nate Eldredge
    11 hours ago






  • 1




    $begingroup$
    It is a long established fact that the space of Radon measures on a completely regular space is naturally identifiable with the dual of the space $C^b(S)$ of bounded, continuous functions thereon (with the so-called strict topology—-the finest locally convex topology which coincides with compact convergence on the unit ball). This was shown by Buck in the locally compact case in the 50’s and extended to the general case in the 60‘s. If the underlying space is polish, then so is that of the Radon probability measures (under the weak topology induced by $C^b(S)$)—-again a celebrated result.
    $endgroup$
    – user131781
    10 hours ago










  • $begingroup$
    @NateEldredge Do you mean as opposed to Radon probability measures? The set of Radon probability measures on a Polish space is Polish as a subspace of the weak-* topology, even though the weak-* topology isn't first-countable on the whole space $C_b(X)^*$.
    $endgroup$
    – Robert Furber
    10 hours ago












  • $begingroup$
    @RobertFurber: Right, that's what I mean. The space $M(Omega)$ defined in the question is the space of all signed Radon measures, i.e. the dual of $C(overline{Omega})$ (or something like that), and this is not weak-* metrizable (nor, as you say, even first countable).
    $endgroup$
    – Nate Eldredge
    10 hours ago
















4












4








4


1



$begingroup$


Background: I work on a SPDE problem where in order to apply Prokhorov's theorem I need that some measure space is Polish space. And additionaly it would be good if that space is Banach space. Earlier today I was reading the book: Malek, Necas, Rokyta, Ruzicka - Weak and Measure-valued Solutions to Evolutionary PDEs, 1996, and I have a question from the Subsection 1.2.8 titled Radon measures. The definitions given bellow are taken from the same book.



On the one hand, the space of Radon measures is defined as:
$$M(mathbb{R}^d)equiv { mu : C_0 (mathbb{R}) rightarrow mathbb{R}; mu hspace{0.2cm} linear hspace{0.2cm} s.t. hspace{0.2cm} exists c>0, |mu (f)|leq c ||f||_{infty}, forall f in mathcal{D}(mathbb{R}^d)}.$$



Here



$C_0(mathbb{R}^d)equiv { u in C(mathbb{R}^d): lim_{|x|rightarrow infty} u(x) = 0 }$
and
$C_0(mathbb{R}^d)=overline{mathcal{D}(mathbb{R}^d)}^{||cdot||_{infty}}$.



As usual $mathcal{D}(Omega)$ stands for the space of functions from $C^{infty}(
overline{Omega})$
with compact support in $Omega$.



If we further define



$||mu||_{M(mathbb{R}^d)}equiv sup{|mu(f)|: f in mathcal{D}(mathbb{R}^d),||f||_{infty}leq 1 }$,



then the space $(M(mathbb{R}^d), || cdot ||_{M(mathbb{R}^d)})$ is a Banach space.



On the other hand, let $Omega$ be a bounded domain. We denote by $M(Omega)$ the space of Radon measures defined as the dual space of $C(overline{Omega})$. Also in this case we know that $L^1(Omega)hookrightarrow M(Omega)$ (and we know that $L^1(Omega)$ is separable).



My questions are:





  1. Is the space of Radon measures separable - in the case $Omega subset mathbb{R}^d$ and in the case $mathbb{R}^d$? Or to be more precise is it a Polish space? I have search it in a few books and in the questions here but I didn't find any concrete answer (I maybe have missed something).


  2. Maybe some subspace of the space of Radon measure is Polish? I've read somewhere that the space of positive Radon measures is Polish but didn't find any book to confirm that.


  3. Are there some other spaces of measure-valued functions that are Polish (besides the spaces mentioned above)?


I usually avoid dealing with meaasure-valued spaces so I don't know much about them. Help with this would be great (and I definitely need it). Thanks in advance.










share|cite|improve this question









$endgroup$




Background: I work on a SPDE problem where in order to apply Prokhorov's theorem I need that some measure space is Polish space. And additionaly it would be good if that space is Banach space. Earlier today I was reading the book: Malek, Necas, Rokyta, Ruzicka - Weak and Measure-valued Solutions to Evolutionary PDEs, 1996, and I have a question from the Subsection 1.2.8 titled Radon measures. The definitions given bellow are taken from the same book.



On the one hand, the space of Radon measures is defined as:
$$M(mathbb{R}^d)equiv { mu : C_0 (mathbb{R}) rightarrow mathbb{R}; mu hspace{0.2cm} linear hspace{0.2cm} s.t. hspace{0.2cm} exists c>0, |mu (f)|leq c ||f||_{infty}, forall f in mathcal{D}(mathbb{R}^d)}.$$



Here



$C_0(mathbb{R}^d)equiv { u in C(mathbb{R}^d): lim_{|x|rightarrow infty} u(x) = 0 }$
and
$C_0(mathbb{R}^d)=overline{mathcal{D}(mathbb{R}^d)}^{||cdot||_{infty}}$.



As usual $mathcal{D}(Omega)$ stands for the space of functions from $C^{infty}(
overline{Omega})$
with compact support in $Omega$.



If we further define



$||mu||_{M(mathbb{R}^d)}equiv sup{|mu(f)|: f in mathcal{D}(mathbb{R}^d),||f||_{infty}leq 1 }$,



then the space $(M(mathbb{R}^d), || cdot ||_{M(mathbb{R}^d)})$ is a Banach space.



On the other hand, let $Omega$ be a bounded domain. We denote by $M(Omega)$ the space of Radon measures defined as the dual space of $C(overline{Omega})$. Also in this case we know that $L^1(Omega)hookrightarrow M(Omega)$ (and we know that $L^1(Omega)$ is separable).



My questions are:





  1. Is the space of Radon measures separable - in the case $Omega subset mathbb{R}^d$ and in the case $mathbb{R}^d$? Or to be more precise is it a Polish space? I have search it in a few books and in the questions here but I didn't find any concrete answer (I maybe have missed something).


  2. Maybe some subspace of the space of Radon measure is Polish? I've read somewhere that the space of positive Radon measures is Polish but didn't find any book to confirm that.


  3. Are there some other spaces of measure-valued functions that are Polish (besides the spaces mentioned above)?


I usually avoid dealing with meaasure-valued spaces so I don't know much about them. Help with this would be great (and I definitely need it). Thanks in advance.







reference-request fa.functional-analysis measure-theory banach-spaces metric-spaces






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asked 12 hours ago









MarkMark

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  • $begingroup$
    You have better chances with the weak $^*$ topology. The dual unit ball of a separable Banach space is weak$^*$ compact and metrizable.
    $endgroup$
    – Jochen Wengenroth
    11 hours ago






  • 1




    $begingroup$
    @JochenWengenroth: On the other hand, the space of all Radon measures with the weak-* topology is not metrizable, though it is separable.
    $endgroup$
    – Nate Eldredge
    11 hours ago






  • 1




    $begingroup$
    It is a long established fact that the space of Radon measures on a completely regular space is naturally identifiable with the dual of the space $C^b(S)$ of bounded, continuous functions thereon (with the so-called strict topology—-the finest locally convex topology which coincides with compact convergence on the unit ball). This was shown by Buck in the locally compact case in the 50’s and extended to the general case in the 60‘s. If the underlying space is polish, then so is that of the Radon probability measures (under the weak topology induced by $C^b(S)$)—-again a celebrated result.
    $endgroup$
    – user131781
    10 hours ago










  • $begingroup$
    @NateEldredge Do you mean as opposed to Radon probability measures? The set of Radon probability measures on a Polish space is Polish as a subspace of the weak-* topology, even though the weak-* topology isn't first-countable on the whole space $C_b(X)^*$.
    $endgroup$
    – Robert Furber
    10 hours ago












  • $begingroup$
    @RobertFurber: Right, that's what I mean. The space $M(Omega)$ defined in the question is the space of all signed Radon measures, i.e. the dual of $C(overline{Omega})$ (or something like that), and this is not weak-* metrizable (nor, as you say, even first countable).
    $endgroup$
    – Nate Eldredge
    10 hours ago




















  • $begingroup$
    You have better chances with the weak $^*$ topology. The dual unit ball of a separable Banach space is weak$^*$ compact and metrizable.
    $endgroup$
    – Jochen Wengenroth
    11 hours ago






  • 1




    $begingroup$
    @JochenWengenroth: On the other hand, the space of all Radon measures with the weak-* topology is not metrizable, though it is separable.
    $endgroup$
    – Nate Eldredge
    11 hours ago






  • 1




    $begingroup$
    It is a long established fact that the space of Radon measures on a completely regular space is naturally identifiable with the dual of the space $C^b(S)$ of bounded, continuous functions thereon (with the so-called strict topology—-the finest locally convex topology which coincides with compact convergence on the unit ball). This was shown by Buck in the locally compact case in the 50’s and extended to the general case in the 60‘s. If the underlying space is polish, then so is that of the Radon probability measures (under the weak topology induced by $C^b(S)$)—-again a celebrated result.
    $endgroup$
    – user131781
    10 hours ago










  • $begingroup$
    @NateEldredge Do you mean as opposed to Radon probability measures? The set of Radon probability measures on a Polish space is Polish as a subspace of the weak-* topology, even though the weak-* topology isn't first-countable on the whole space $C_b(X)^*$.
    $endgroup$
    – Robert Furber
    10 hours ago












  • $begingroup$
    @RobertFurber: Right, that's what I mean. The space $M(Omega)$ defined in the question is the space of all signed Radon measures, i.e. the dual of $C(overline{Omega})$ (or something like that), and this is not weak-* metrizable (nor, as you say, even first countable).
    $endgroup$
    – Nate Eldredge
    10 hours ago


















$begingroup$
You have better chances with the weak $^*$ topology. The dual unit ball of a separable Banach space is weak$^*$ compact and metrizable.
$endgroup$
– Jochen Wengenroth
11 hours ago




$begingroup$
You have better chances with the weak $^*$ topology. The dual unit ball of a separable Banach space is weak$^*$ compact and metrizable.
$endgroup$
– Jochen Wengenroth
11 hours ago




1




1




$begingroup$
@JochenWengenroth: On the other hand, the space of all Radon measures with the weak-* topology is not metrizable, though it is separable.
$endgroup$
– Nate Eldredge
11 hours ago




$begingroup$
@JochenWengenroth: On the other hand, the space of all Radon measures with the weak-* topology is not metrizable, though it is separable.
$endgroup$
– Nate Eldredge
11 hours ago




1




1




$begingroup$
It is a long established fact that the space of Radon measures on a completely regular space is naturally identifiable with the dual of the space $C^b(S)$ of bounded, continuous functions thereon (with the so-called strict topology—-the finest locally convex topology which coincides with compact convergence on the unit ball). This was shown by Buck in the locally compact case in the 50’s and extended to the general case in the 60‘s. If the underlying space is polish, then so is that of the Radon probability measures (under the weak topology induced by $C^b(S)$)—-again a celebrated result.
$endgroup$
– user131781
10 hours ago




$begingroup$
It is a long established fact that the space of Radon measures on a completely regular space is naturally identifiable with the dual of the space $C^b(S)$ of bounded, continuous functions thereon (with the so-called strict topology—-the finest locally convex topology which coincides with compact convergence on the unit ball). This was shown by Buck in the locally compact case in the 50’s and extended to the general case in the 60‘s. If the underlying space is polish, then so is that of the Radon probability measures (under the weak topology induced by $C^b(S)$)—-again a celebrated result.
$endgroup$
– user131781
10 hours ago












$begingroup$
@NateEldredge Do you mean as opposed to Radon probability measures? The set of Radon probability measures on a Polish space is Polish as a subspace of the weak-* topology, even though the weak-* topology isn't first-countable on the whole space $C_b(X)^*$.
$endgroup$
– Robert Furber
10 hours ago






$begingroup$
@NateEldredge Do you mean as opposed to Radon probability measures? The set of Radon probability measures on a Polish space is Polish as a subspace of the weak-* topology, even though the weak-* topology isn't first-countable on the whole space $C_b(X)^*$.
$endgroup$
– Robert Furber
10 hours ago














$begingroup$
@RobertFurber: Right, that's what I mean. The space $M(Omega)$ defined in the question is the space of all signed Radon measures, i.e. the dual of $C(overline{Omega})$ (or something like that), and this is not weak-* metrizable (nor, as you say, even first countable).
$endgroup$
– Nate Eldredge
10 hours ago






$begingroup$
@RobertFurber: Right, that's what I mean. The space $M(Omega)$ defined in the question is the space of all signed Radon measures, i.e. the dual of $C(overline{Omega})$ (or something like that), and this is not weak-* metrizable (nor, as you say, even first countable).
$endgroup$
– Nate Eldredge
10 hours ago












3 Answers
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$begingroup$

With respect to the norm topology, the space of Radon measures on a domain $Omega$ is not seperable. Indeed, for any two distinct points $x,y$ in $Omega$, the Dirac measures $delta_x$ and $delta_y$ (where $delta_x(f)=f(x)$) satisfy $|delta_x-delta_y|=2$
since you can always find a compactly supported smooth function $f$ with $f(x)=-f(y)=1=|f|_infty$. Any metric space that contains uncountably many disjoint open balls cannot be seperable. Of course there are many subspaces of Radon measures that are seperable in the norm topology, e.g., as you noted $L^1$ naturally embeds as a subspace and is seperable.



The space of Radon measures on a domain $Omega$ is seperable in the weak$^*$ topology (This is probably the remark you allude to have read somewhere.) Indeed, consider the countable set $M_Q$ of measures of the form $sum_{x in S} a_x delta_x$ where the coefficients $a_x$ are rational and $S$ runs over finite sets of points with rational coordinates. This $M_Q$ is countable and weak$^*$ dense. Also the embedding of $L^1$ as space of measures with absolutely continuous to Lebesgue measure is dense, and this gives another proof of weak$^*$ seperability.






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    4














    $begingroup$

    No. Let $Omega=[0,1]$. If $xin[0,1]$, let $delta_x$ be the point mass at $x$. They are all Radon measures. It is not that hard to show that $|delta_x-delta_{x'}|=2$. So you can construct an uncountable family of disjoint open balls. Since a separable metrizable
    space has a countable basis, this shows that the space in question is not separable.



    If there exists a subspace $S$ with a countable dense subset ${mu_n|ninmathbb{N}}$, then every element of $S$ will be absolutely continuous with respect to $sum_n 2^{-n}bar{mu}_n$ where $bar{mu}_n$ is the normalization of $mu_n$ to having total mass $1$. So separable subspaces are exactly those included in a $L_1$-space for some measure.



    Unbounded domains may bring additional complications.






    share|cite|improve this answer











    $endgroup$























      0














      $begingroup$

      The other answers very adequately explain why the norm topology is not Polish except for trivial cases, so this answer is about the weak-* topology. Also, most results in the literature are about the space of Radon probability measures, not signed measures or positive measures, so I'll concentrate on this case.



      If $Omega subseteq mathbb{R}$ is bounded, then $overline{Omega}$ is compact and metrizable. It is generally the case that if $X$ is compact and metrizable, $C(X)$ is separable, and from this it follows that the unit ball of $C(X)^*$ is compact and metrizable in the weak-* topology (a.k.a. $sigma(C(X)^*,C(X))$), and therefore Polish. Since the set of Radon probability measures is a closed subset of this unit ball, it is itself Polish. This is the easiest case to prove.



      More generally, for every Borel probability measure on a Polish space $X$, we can map it into the dual space of the C$^*$-algebra of bounded continuous functions $C_b(X)$ by
      $$
      mu mapsto left( f mapsto int_X f mathrm{d}mu right)
      $$

      If we topologize the image of this with the weak-* topology coming from $C_b(X)$, we get a Polish topology. This is true even though the unit ball of $C_b(X)^*$ need not even be weak-* first-countable (this already happens for $X = mathbb{R}$, although as $mathbb{R}$ is locally compact you can use the weak-* topology coming from $C_0(mathbb{R})$ instead). One place that this is proved is in Kechris's Classical Descriptive Set Theory Theorem 17.23 on page 112.



      One last thing - a measure is sometimes defined to be Radon if it is locally finite and inner regular with respect to compact sets. This is true for all $sigma$-finite measures on Polish spaces, so does not define a distinct type of measure in this case.






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        3 Answers
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        3 Answers
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        $begingroup$

        With respect to the norm topology, the space of Radon measures on a domain $Omega$ is not seperable. Indeed, for any two distinct points $x,y$ in $Omega$, the Dirac measures $delta_x$ and $delta_y$ (where $delta_x(f)=f(x)$) satisfy $|delta_x-delta_y|=2$
        since you can always find a compactly supported smooth function $f$ with $f(x)=-f(y)=1=|f|_infty$. Any metric space that contains uncountably many disjoint open balls cannot be seperable. Of course there are many subspaces of Radon measures that are seperable in the norm topology, e.g., as you noted $L^1$ naturally embeds as a subspace and is seperable.



        The space of Radon measures on a domain $Omega$ is seperable in the weak$^*$ topology (This is probably the remark you allude to have read somewhere.) Indeed, consider the countable set $M_Q$ of measures of the form $sum_{x in S} a_x delta_x$ where the coefficients $a_x$ are rational and $S$ runs over finite sets of points with rational coordinates. This $M_Q$ is countable and weak$^*$ dense. Also the embedding of $L^1$ as space of measures with absolutely continuous to Lebesgue measure is dense, and this gives another proof of weak$^*$ seperability.






        share|cite|improve this answer









        $endgroup$




















          4














          $begingroup$

          With respect to the norm topology, the space of Radon measures on a domain $Omega$ is not seperable. Indeed, for any two distinct points $x,y$ in $Omega$, the Dirac measures $delta_x$ and $delta_y$ (where $delta_x(f)=f(x)$) satisfy $|delta_x-delta_y|=2$
          since you can always find a compactly supported smooth function $f$ with $f(x)=-f(y)=1=|f|_infty$. Any metric space that contains uncountably many disjoint open balls cannot be seperable. Of course there are many subspaces of Radon measures that are seperable in the norm topology, e.g., as you noted $L^1$ naturally embeds as a subspace and is seperable.



          The space of Radon measures on a domain $Omega$ is seperable in the weak$^*$ topology (This is probably the remark you allude to have read somewhere.) Indeed, consider the countable set $M_Q$ of measures of the form $sum_{x in S} a_x delta_x$ where the coefficients $a_x$ are rational and $S$ runs over finite sets of points with rational coordinates. This $M_Q$ is countable and weak$^*$ dense. Also the embedding of $L^1$ as space of measures with absolutely continuous to Lebesgue measure is dense, and this gives another proof of weak$^*$ seperability.






          share|cite|improve this answer









          $endgroup$


















            4














            4










            4







            $begingroup$

            With respect to the norm topology, the space of Radon measures on a domain $Omega$ is not seperable. Indeed, for any two distinct points $x,y$ in $Omega$, the Dirac measures $delta_x$ and $delta_y$ (where $delta_x(f)=f(x)$) satisfy $|delta_x-delta_y|=2$
            since you can always find a compactly supported smooth function $f$ with $f(x)=-f(y)=1=|f|_infty$. Any metric space that contains uncountably many disjoint open balls cannot be seperable. Of course there are many subspaces of Radon measures that are seperable in the norm topology, e.g., as you noted $L^1$ naturally embeds as a subspace and is seperable.



            The space of Radon measures on a domain $Omega$ is seperable in the weak$^*$ topology (This is probably the remark you allude to have read somewhere.) Indeed, consider the countable set $M_Q$ of measures of the form $sum_{x in S} a_x delta_x$ where the coefficients $a_x$ are rational and $S$ runs over finite sets of points with rational coordinates. This $M_Q$ is countable and weak$^*$ dense. Also the embedding of $L^1$ as space of measures with absolutely continuous to Lebesgue measure is dense, and this gives another proof of weak$^*$ seperability.






            share|cite|improve this answer









            $endgroup$



            With respect to the norm topology, the space of Radon measures on a domain $Omega$ is not seperable. Indeed, for any two distinct points $x,y$ in $Omega$, the Dirac measures $delta_x$ and $delta_y$ (where $delta_x(f)=f(x)$) satisfy $|delta_x-delta_y|=2$
            since you can always find a compactly supported smooth function $f$ with $f(x)=-f(y)=1=|f|_infty$. Any metric space that contains uncountably many disjoint open balls cannot be seperable. Of course there are many subspaces of Radon measures that are seperable in the norm topology, e.g., as you noted $L^1$ naturally embeds as a subspace and is seperable.



            The space of Radon measures on a domain $Omega$ is seperable in the weak$^*$ topology (This is probably the remark you allude to have read somewhere.) Indeed, consider the countable set $M_Q$ of measures of the form $sum_{x in S} a_x delta_x$ where the coefficients $a_x$ are rational and $S$ runs over finite sets of points with rational coordinates. This $M_Q$ is countable and weak$^*$ dense. Also the embedding of $L^1$ as space of measures with absolutely continuous to Lebesgue measure is dense, and this gives another proof of weak$^*$ seperability.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 11 hours ago









            Yuval PeresYuval Peres

            2,74311 silver badges14 bronze badges




            2,74311 silver badges14 bronze badges


























                4














                $begingroup$

                No. Let $Omega=[0,1]$. If $xin[0,1]$, let $delta_x$ be the point mass at $x$. They are all Radon measures. It is not that hard to show that $|delta_x-delta_{x'}|=2$. So you can construct an uncountable family of disjoint open balls. Since a separable metrizable
                space has a countable basis, this shows that the space in question is not separable.



                If there exists a subspace $S$ with a countable dense subset ${mu_n|ninmathbb{N}}$, then every element of $S$ will be absolutely continuous with respect to $sum_n 2^{-n}bar{mu}_n$ where $bar{mu}_n$ is the normalization of $mu_n$ to having total mass $1$. So separable subspaces are exactly those included in a $L_1$-space for some measure.



                Unbounded domains may bring additional complications.






                share|cite|improve this answer











                $endgroup$




















                  4














                  $begingroup$

                  No. Let $Omega=[0,1]$. If $xin[0,1]$, let $delta_x$ be the point mass at $x$. They are all Radon measures. It is not that hard to show that $|delta_x-delta_{x'}|=2$. So you can construct an uncountable family of disjoint open balls. Since a separable metrizable
                  space has a countable basis, this shows that the space in question is not separable.



                  If there exists a subspace $S$ with a countable dense subset ${mu_n|ninmathbb{N}}$, then every element of $S$ will be absolutely continuous with respect to $sum_n 2^{-n}bar{mu}_n$ where $bar{mu}_n$ is the normalization of $mu_n$ to having total mass $1$. So separable subspaces are exactly those included in a $L_1$-space for some measure.



                  Unbounded domains may bring additional complications.






                  share|cite|improve this answer











                  $endgroup$


















                    4














                    4










                    4







                    $begingroup$

                    No. Let $Omega=[0,1]$. If $xin[0,1]$, let $delta_x$ be the point mass at $x$. They are all Radon measures. It is not that hard to show that $|delta_x-delta_{x'}|=2$. So you can construct an uncountable family of disjoint open balls. Since a separable metrizable
                    space has a countable basis, this shows that the space in question is not separable.



                    If there exists a subspace $S$ with a countable dense subset ${mu_n|ninmathbb{N}}$, then every element of $S$ will be absolutely continuous with respect to $sum_n 2^{-n}bar{mu}_n$ where $bar{mu}_n$ is the normalization of $mu_n$ to having total mass $1$. So separable subspaces are exactly those included in a $L_1$-space for some measure.



                    Unbounded domains may bring additional complications.






                    share|cite|improve this answer











                    $endgroup$



                    No. Let $Omega=[0,1]$. If $xin[0,1]$, let $delta_x$ be the point mass at $x$. They are all Radon measures. It is not that hard to show that $|delta_x-delta_{x'}|=2$. So you can construct an uncountable family of disjoint open balls. Since a separable metrizable
                    space has a countable basis, this shows that the space in question is not separable.



                    If there exists a subspace $S$ with a countable dense subset ${mu_n|ninmathbb{N}}$, then every element of $S$ will be absolutely continuous with respect to $sum_n 2^{-n}bar{mu}_n$ where $bar{mu}_n$ is the normalization of $mu_n$ to having total mass $1$. So separable subspaces are exactly those included in a $L_1$-space for some measure.



                    Unbounded domains may bring additional complications.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 11 hours ago

























                    answered 11 hours ago









                    Michael GreineckerMichael Greinecker

                    8,2093 gold badges39 silver badges60 bronze badges




                    8,2093 gold badges39 silver badges60 bronze badges


























                        0














                        $begingroup$

                        The other answers very adequately explain why the norm topology is not Polish except for trivial cases, so this answer is about the weak-* topology. Also, most results in the literature are about the space of Radon probability measures, not signed measures or positive measures, so I'll concentrate on this case.



                        If $Omega subseteq mathbb{R}$ is bounded, then $overline{Omega}$ is compact and metrizable. It is generally the case that if $X$ is compact and metrizable, $C(X)$ is separable, and from this it follows that the unit ball of $C(X)^*$ is compact and metrizable in the weak-* topology (a.k.a. $sigma(C(X)^*,C(X))$), and therefore Polish. Since the set of Radon probability measures is a closed subset of this unit ball, it is itself Polish. This is the easiest case to prove.



                        More generally, for every Borel probability measure on a Polish space $X$, we can map it into the dual space of the C$^*$-algebra of bounded continuous functions $C_b(X)$ by
                        $$
                        mu mapsto left( f mapsto int_X f mathrm{d}mu right)
                        $$

                        If we topologize the image of this with the weak-* topology coming from $C_b(X)$, we get a Polish topology. This is true even though the unit ball of $C_b(X)^*$ need not even be weak-* first-countable (this already happens for $X = mathbb{R}$, although as $mathbb{R}$ is locally compact you can use the weak-* topology coming from $C_0(mathbb{R})$ instead). One place that this is proved is in Kechris's Classical Descriptive Set Theory Theorem 17.23 on page 112.



                        One last thing - a measure is sometimes defined to be Radon if it is locally finite and inner regular with respect to compact sets. This is true for all $sigma$-finite measures on Polish spaces, so does not define a distinct type of measure in this case.






                        share|cite|improve this answer









                        $endgroup$




















                          0














                          $begingroup$

                          The other answers very adequately explain why the norm topology is not Polish except for trivial cases, so this answer is about the weak-* topology. Also, most results in the literature are about the space of Radon probability measures, not signed measures or positive measures, so I'll concentrate on this case.



                          If $Omega subseteq mathbb{R}$ is bounded, then $overline{Omega}$ is compact and metrizable. It is generally the case that if $X$ is compact and metrizable, $C(X)$ is separable, and from this it follows that the unit ball of $C(X)^*$ is compact and metrizable in the weak-* topology (a.k.a. $sigma(C(X)^*,C(X))$), and therefore Polish. Since the set of Radon probability measures is a closed subset of this unit ball, it is itself Polish. This is the easiest case to prove.



                          More generally, for every Borel probability measure on a Polish space $X$, we can map it into the dual space of the C$^*$-algebra of bounded continuous functions $C_b(X)$ by
                          $$
                          mu mapsto left( f mapsto int_X f mathrm{d}mu right)
                          $$

                          If we topologize the image of this with the weak-* topology coming from $C_b(X)$, we get a Polish topology. This is true even though the unit ball of $C_b(X)^*$ need not even be weak-* first-countable (this already happens for $X = mathbb{R}$, although as $mathbb{R}$ is locally compact you can use the weak-* topology coming from $C_0(mathbb{R})$ instead). One place that this is proved is in Kechris's Classical Descriptive Set Theory Theorem 17.23 on page 112.



                          One last thing - a measure is sometimes defined to be Radon if it is locally finite and inner regular with respect to compact sets. This is true for all $sigma$-finite measures on Polish spaces, so does not define a distinct type of measure in this case.






                          share|cite|improve this answer









                          $endgroup$


















                            0














                            0










                            0







                            $begingroup$

                            The other answers very adequately explain why the norm topology is not Polish except for trivial cases, so this answer is about the weak-* topology. Also, most results in the literature are about the space of Radon probability measures, not signed measures or positive measures, so I'll concentrate on this case.



                            If $Omega subseteq mathbb{R}$ is bounded, then $overline{Omega}$ is compact and metrizable. It is generally the case that if $X$ is compact and metrizable, $C(X)$ is separable, and from this it follows that the unit ball of $C(X)^*$ is compact and metrizable in the weak-* topology (a.k.a. $sigma(C(X)^*,C(X))$), and therefore Polish. Since the set of Radon probability measures is a closed subset of this unit ball, it is itself Polish. This is the easiest case to prove.



                            More generally, for every Borel probability measure on a Polish space $X$, we can map it into the dual space of the C$^*$-algebra of bounded continuous functions $C_b(X)$ by
                            $$
                            mu mapsto left( f mapsto int_X f mathrm{d}mu right)
                            $$

                            If we topologize the image of this with the weak-* topology coming from $C_b(X)$, we get a Polish topology. This is true even though the unit ball of $C_b(X)^*$ need not even be weak-* first-countable (this already happens for $X = mathbb{R}$, although as $mathbb{R}$ is locally compact you can use the weak-* topology coming from $C_0(mathbb{R})$ instead). One place that this is proved is in Kechris's Classical Descriptive Set Theory Theorem 17.23 on page 112.



                            One last thing - a measure is sometimes defined to be Radon if it is locally finite and inner regular with respect to compact sets. This is true for all $sigma$-finite measures on Polish spaces, so does not define a distinct type of measure in this case.






                            share|cite|improve this answer









                            $endgroup$



                            The other answers very adequately explain why the norm topology is not Polish except for trivial cases, so this answer is about the weak-* topology. Also, most results in the literature are about the space of Radon probability measures, not signed measures or positive measures, so I'll concentrate on this case.



                            If $Omega subseteq mathbb{R}$ is bounded, then $overline{Omega}$ is compact and metrizable. It is generally the case that if $X$ is compact and metrizable, $C(X)$ is separable, and from this it follows that the unit ball of $C(X)^*$ is compact and metrizable in the weak-* topology (a.k.a. $sigma(C(X)^*,C(X))$), and therefore Polish. Since the set of Radon probability measures is a closed subset of this unit ball, it is itself Polish. This is the easiest case to prove.



                            More generally, for every Borel probability measure on a Polish space $X$, we can map it into the dual space of the C$^*$-algebra of bounded continuous functions $C_b(X)$ by
                            $$
                            mu mapsto left( f mapsto int_X f mathrm{d}mu right)
                            $$

                            If we topologize the image of this with the weak-* topology coming from $C_b(X)$, we get a Polish topology. This is true even though the unit ball of $C_b(X)^*$ need not even be weak-* first-countable (this already happens for $X = mathbb{R}$, although as $mathbb{R}$ is locally compact you can use the weak-* topology coming from $C_0(mathbb{R})$ instead). One place that this is proved is in Kechris's Classical Descriptive Set Theory Theorem 17.23 on page 112.



                            One last thing - a measure is sometimes defined to be Radon if it is locally finite and inner regular with respect to compact sets. This is true for all $sigma$-finite measures on Polish spaces, so does not define a distinct type of measure in this case.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 10 hours ago









                            Robert FurberRobert Furber

                            2,0978 silver badges14 bronze badges




                            2,0978 silver badges14 bronze badges


































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