How to check if a map $Xto M(X)$ is measurable?
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Let $X$ be a compact metric space and $M(X)$ be the set of all the finite measures on the Borel $sigma$-algebra of $X$.
By the Riesz representation theorem, we know that the map $M(X)to C(X)^*$ defined as $mumapsto (fmapsto int f dmu)$ is injective and its image is the set of all bounded linear maps $F:C(X)to mathbf R$ which are positive, that is, those $F$ such that $Ffgeq 0$ whenever $fgeq 0$.
Equipping $C(X)^*$ with the weak* topology, we get a topology on $M(X)$ and hence a $sigma$-algebra on $M(X)$.
On $X$ we also have the Borel $sigma$-algebra.
Question. Suppose we have a map $mu:Xto M(X)$, $xmapsto mu_x$.
Is is true that $mu$ is a measurable map if for each $fin C(X)$ the map $Xto mathbf R$, $xmapsto int_X f dmu_x$ is measurable?
Or is there some other convenient criterion to check the measurability of a map $Xto M(X)$?
functional-analysis measure-theory
asked Sep 22 at 14:31
caffeinemachine
6,45521250
add a comment |
up vote
1
down vote
favorite
Let $X$ be a compact metric space and $M(X)$ be the set of all the finite measures on the Borel $sigma$-algebra of $X$.
By the Riesz representation theorem, we know that the map $M(X)to C(X)^*$ defined as $mumapsto (fmapsto int f dmu)$ is injective and its image is the set of all bounded linear maps $F:C(X)to mathbf R$ which are positive, that is, those $F$ such that $Ffgeq 0$ whenever $fgeq 0$.
Equipping $C(X)^*$ with the weak* topology, we get a topology on $M(X)$ and hence a $sigma$-algebra on $M(X)$.
On $X$ we also have the Borel $sigma$-algebra.
Question. Suppose we have a map $mu:Xto M(X)$, $xmapsto mu_x$.
Is is true that $mu$ is a measurable map if for each $fin C(X)$ the map $Xto mathbf R$, $xmapsto int_X f dmu_x$ is measurable?
Or is there some other convenient criterion to check the measurability of a map $Xto M(X)$?
functional-analysis measure-theory
asked Sep 22 at 14:31
caffeinemachine
6,45521250
Notice that the map $pi : M(X) to mathbb{R}^{C(X)} $ with $ nu mapsto (int_X f , dnu)_{fin C(X)}$ is an embedding (where $mathbb{R}^{C(X)}$ is equipped with the product topology). So $mu : X to M(X)$ is measurable if and only if $pi circ mu$ is measurable. This is of course equivalent to saying that each $x mapsto int_X f , dmu_x$ is measurable for each $f in C(X)$.
– Sangchul Lee
Sep 22 at 15:21
@SangchulLee "So $mu:Xto M(X)$ is measurable if and only if $picirc mu$ is measurable." This is true when we equip $mathbb R^{C(X)}$ with the Borel $sigma$-algebra. But the equivalence of the measurability of $picirc mu$ and the measurability of each $xmapsto int_Xf dmu_x$ is true when $mathbf R^{C(X)}$ is equipped with the product $sigma$-algebra. So I do not completely follow your comment. Can you please address this and point out any mistakes. Thanks.
– caffeinemachine
Sep 23 at 11:19
I see your point. The issue is that that Borel $sigma$-algebra on $mathbb{R}^{C(X)}$ may possibly be strictly finer than the product $sigma$-algebra for general $X$. Thank you for pointing out my mistake and sorry for the confusion. Perhaps the claim can be salvaged if $C(X)$ is separable so that we can replace $mathbb{R}^{C(X)}$ by $mathbb{R}^{mathfrak{A}}$ for some countable dense subset $mathfrak{A} subset C(X)$, but let me think about it.
– Sangchul Lee
Sep 23 at 11:36
@SangchulLee I am interested in the situation when $X$ is a compact metric space and here it is known that $C(X)$ is separable. Just wanted to say this. Thanks.
– caffeinemachine
Sep 23 at 12:02
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$ be a compact metric space and $M(X)$ be the set of all the finite measures on the Borel $sigma$-algebra of $X$.
By the Riesz representation theorem, we know that the map $M(X)to C(X)^*$ defined as $mumapsto (fmapsto int f dmu)$ is injective and its image is the set of all bounded linear maps $F:C(X)to mathbf R$ which are positive, that is, those $F$ such that $Ffgeq 0$ whenever $fgeq 0$.
Equipping $C(X)^*$ with the weak* topology, we get a topology on $M(X)$ and hence a $sigma$-algebra on $M(X)$.
On $X$ we also have the Borel $sigma$-algebra.
Question. Suppose we have a map $mu:Xto M(X)$, $xmapsto mu_x$.
Is is true that $mu$ is a measurable map if for each $fin C(X)$ the map $Xto mathbf R$, $xmapsto int_X f dmu_x$ is measurable?
Or is there some other convenient criterion to check the measurability of a map $Xto M(X)$?
functional-analysis measure-theory
asked Sep 22 at 14:31
caffeinemachine
6,45521250
Let $X$ be a compact metric space and $M(X)$ be the set of all the finite measures on the Borel $sigma$-algebra of $X$.
By the Riesz representation theorem, we know that the map $M(X)to C(X)^*$ defined as $mumapsto (fmapsto int f dmu)$ is injective and its image is the set of all bounded linear maps $F:C(X)to mathbf R$ which are positive, that is, those $F$ such that $Ffgeq 0$ whenever $fgeq 0$.
Equipping $C(X)^*$ with the weak* topology, we get a topology on $M(X)$ and hence a $sigma$-algebra on $M(X)$.
On $X$ we also have the Borel $sigma$-algebra.
Question. Suppose we have a map $mu:Xto M(X)$, $xmapsto mu_x$.
Is is true that $mu$ is a measurable map if for each $fin C(X)$ the map $Xto mathbf R$, $xmapsto int_X f dmu_x$ is measurable?
Or is there some other convenient criterion to check the measurability of a map $Xto M(X)$?
functional-analysis measure-theory
functional-analysis measure-theory
asked Sep 22 at 14:31
caffeinemachine
6,45521250
asked Sep 22 at 14:31
caffeinemachine
6,45521250
edited Nov 19 at 8:42
asked Sep 22 at 14:31
caffeinemachine
6,45521250
asked Sep 22 at 14:31
caffeinemachine
6,45521250
asked Sep 22 at 14:31
caffeinemachine
6,45521250
6,45521250
Notice that the map $pi : M(X) to mathbb{R}^{C(X)} $ with $ nu mapsto (int_X f , dnu)_{fin C(X)}$ is an embedding (where $mathbb{R}^{C(X)}$ is equipped with the product topology). So $mu : X to M(X)$ is measurable if and only if $pi circ mu$ is measurable. This is of course equivalent to saying that each $x mapsto int_X f , dmu_x$ is measurable for each $f in C(X)$.
– Sangchul Lee
Sep 22 at 15:21
@SangchulLee "So $mu:Xto M(X)$ is measurable if and only if $picirc mu$ is measurable." This is true when we equip $mathbb R^{C(X)}$ with the Borel $sigma$-algebra. But the equivalence of the measurability of $picirc mu$ and the measurability of each $xmapsto int_Xf dmu_x$ is true when $mathbf R^{C(X)}$ is equipped with the product $sigma$-algebra. So I do not completely follow your comment. Can you please address this and point out any mistakes. Thanks.
– caffeinemachine
Sep 23 at 11:19
I see your point. The issue is that that Borel $sigma$-algebra on $mathbb{R}^{C(X)}$ may possibly be strictly finer than the product $sigma$-algebra for general $X$. Thank you for pointing out my mistake and sorry for the confusion. Perhaps the claim can be salvaged if $C(X)$ is separable so that we can replace $mathbb{R}^{C(X)}$ by $mathbb{R}^{mathfrak{A}}$ for some countable dense subset $mathfrak{A} subset C(X)$, but let me think about it.
– Sangchul Lee
Sep 23 at 11:36
@SangchulLee I am interested in the situation when $X$ is a compact metric space and here it is known that $C(X)$ is separable. Just wanted to say this. Thanks.
– caffeinemachine
Sep 23 at 12:02
add a comment |
Notice that the map $pi : M(X) to mathbb{R}^{C(X)} $ with $ nu mapsto (int_X f , dnu)_{fin C(X)}$ is an embedding (where $mathbb{R}^{C(X)}$ is equipped with the product topology). So $mu : X to M(X)$ is measurable if and only if $pi circ mu$ is measurable. This is of course equivalent to saying that each $x mapsto int_X f , dmu_x$ is measurable for each $f in C(X)$.
– Sangchul Lee
Sep 22 at 15:21
@SangchulLee "So $mu:Xto M(X)$ is measurable if and only if $picirc mu$ is measurable." This is true when we equip $mathbb R^{C(X)}$ with the Borel $sigma$-algebra. But the equivalence of the measurability of $picirc mu$ and the measurability of each $xmapsto int_Xf dmu_x$ is true when $mathbf R^{C(X)}$ is equipped with the product $sigma$-algebra. So I do not completely follow your comment. Can you please address this and point out any mistakes. Thanks.
– caffeinemachine
Sep 23 at 11:19
I see your point. The issue is that that Borel $sigma$-algebra on $mathbb{R}^{C(X)}$ may possibly be strictly finer than the product $sigma$-algebra for general $X$. Thank you for pointing out my mistake and sorry for the confusion. Perhaps the claim can be salvaged if $C(X)$ is separable so that we can replace $mathbb{R}^{C(X)}$ by $mathbb{R}^{mathfrak{A}}$ for some countable dense subset $mathfrak{A} subset C(X)$, but let me think about it.
– Sangchul Lee
Sep 23 at 11:36
@SangchulLee I am interested in the situation when $X$ is a compact metric space and here it is known that $C(X)$ is separable. Just wanted to say this. Thanks.
– caffeinemachine
Sep 23 at 12:02
Notice that the map $pi : M(X) to mathbb{R}^{C(X)} $ with $ nu mapsto (int_X f , dnu)_{fin C(X)}$ is an embedding (where $mathbb{R}^{C(X)}$ is equipped with the product topology). So $mu : X to M(X)$ is measurable if and only if $pi circ mu$ is measurable. This is of course equivalent to saying that each $x mapsto int_X f , dmu_x$ is measurable for each $f in C(X)$.
– Sangchul Lee
Sep 22 at 15:21
Notice that the map $pi : M(X) to mathbb{R}^{C(X)} $ with $ nu mapsto (int_X f , dnu)_{fin C(X)}$ is an embedding (where $mathbb{R}^{C(X)}$ is equipped with the product topology). So $mu : X to M(X)$ is measurable if and only if $pi circ mu$ is measurable. This is of course equivalent to saying that each $x mapsto int_X f , dmu_x$ is measurable for each $f in C(X)$.
– Sangchul Lee
Sep 22 at 15:21
@SangchulLee "So $mu:Xto M(X)$ is measurable if and only if $picirc mu$ is measurable." This is true when we equip $mathbb R^{C(X)}$ with the Borel $sigma$-algebra. But the equivalence of the measurability of $picirc mu$ and the measurability of each $xmapsto int_Xf dmu_x$ is true when $mathbf R^{C(X)}$ is equipped with the product $sigma$-algebra. So I do not completely follow your comment. Can you please address this and point out any mistakes. Thanks.
– caffeinemachine
Sep 23 at 11:19
@SangchulLee "So $mu:Xto M(X)$ is measurable if and only if $picirc mu$ is measurable." This is true when we equip $mathbb R^{C(X)}$ with the Borel $sigma$-algebra. But the equivalence of the measurability of $picirc mu$ and the measurability of each $xmapsto int_Xf dmu_x$ is true when $mathbf R^{C(X)}$ is equipped with the product $sigma$-algebra. So I do not completely follow your comment. Can you please address this and point out any mistakes. Thanks.
– caffeinemachine
Sep 23 at 11:19
I see your point. The issue is that that Borel $sigma$-algebra on $mathbb{R}^{C(X)}$ may possibly be strictly finer than the product $sigma$-algebra for general $X$. Thank you for pointing out my mistake and sorry for the confusion. Perhaps the claim can be salvaged if $C(X)$ is separable so that we can replace $mathbb{R}^{C(X)}$ by $mathbb{R}^{mathfrak{A}}$ for some countable dense subset $mathfrak{A} subset C(X)$, but let me think about it.
– Sangchul Lee
Sep 23 at 11:36
I see your point. The issue is that that Borel $sigma$-algebra on $mathbb{R}^{C(X)}$ may possibly be strictly finer than the product $sigma$-algebra for general $X$. Thank you for pointing out my mistake and sorry for the confusion. Perhaps the claim can be salvaged if $C(X)$ is separable so that we can replace $mathbb{R}^{C(X)}$ by $mathbb{R}^{mathfrak{A}}$ for some countable dense subset $mathfrak{A} subset C(X)$, but let me think about it.
– Sangchul Lee
Sep 23 at 11:36
@SangchulLee I am interested in the situation when $X$ is a compact metric space and here it is known that $C(X)$ is separable. Just wanted to say this. Thanks.
– caffeinemachine
Sep 23 at 12:02
@SangchulLee I am interested in the situation when $X$ is a compact metric space and here it is known that $C(X)$ is separable. Just wanted to say this. Thanks.
– caffeinemachine
Sep 23 at 12:02
add a comment |
1 Answer
1
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0
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$newcommand{set}[1]{{#1}}$
$newcommand{vp}{varphi}$
$newcommand{mc}{mathcal}$
$newcommand{R}{mathbf R}$
$newcommand{mr}{mathscr}$
WE answer the question above in the affirmative. The only difference is that I will be proving the result with $mr P(X)$, the space of all the probability measures on $X$, instead of $M(X)$, which doesn't make much difference.
Lemma.
Let $Z$ be a set and $f_alpha:Zto Z_alpha$ be a collection of continuous maps into topological spaces $Z_alpha$ indexed by a set $J$.
Endow $Z$ with the initial topology with respect to this collection.
Then for any measurable space $(Y, mc Y)$, a map $vp:Yto Z$ is Borel measurable if and only if each composite $f_alphacirc vp$ is Borel measurable.
Proof.
We do the less trivial direction.
Assume that each composite $f_alphacirc vp$ is measurable.
Note that the collection
$$
mc C=bigcup_{alphain J} set{f_alpha^{-1}(W): Wtext{ open in } Z_alpha}
$$
is a subbasis for the topology on $Z$.
Let $O=f_alpha^{-1}(W)$ be an arbitrary member of $mc C$, where $W$ is an open set in $Z_alpha$.
But then $vp^{-1}(O)=(f_alphacirc vp)^{-1}(W)$ is a measurable subset of $Y$.
So we see that each member of $mc C$ pulls back under $vp$ to a measurable subset of $Y$.
Thus $vp$ is measurable.
$blacksquare$
Corollary.
Let $X$ be a compact metric space and $(Y, mc Y)$ be a measurable space.
Then a map $vp:Yto mr P(X)$ is Borel measurable if and only if for each continuous function $f:Xto R$ we have that the map $ymapsto int_X f dvp_y:Yto R$ is Borel measurable.
answered Nov 19 at 8:44
caffeinemachine
6,45521250
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$newcommand{set}[1]{{#1}}$
$newcommand{vp}{varphi}$
$newcommand{mc}{mathcal}$
$newcommand{R}{mathbf R}$
$newcommand{mr}{mathscr}$
WE answer the question above in the affirmative. The only difference is that I will be proving the result with $mr P(X)$, the space of all the probability measures on $X$, instead of $M(X)$, which doesn't make much difference.
Lemma.
Let $Z$ be a set and $f_alpha:Zto Z_alpha$ be a collection of continuous maps into topological spaces $Z_alpha$ indexed by a set $J$.
Endow $Z$ with the initial topology with respect to this collection.
Then for any measurable space $(Y, mc Y)$, a map $vp:Yto Z$ is Borel measurable if and only if each composite $f_alphacirc vp$ is Borel measurable.
Proof.
We do the less trivial direction.
Assume that each composite $f_alphacirc vp$ is measurable.
Note that the collection
$$
mc C=bigcup_{alphain J} set{f_alpha^{-1}(W): Wtext{ open in } Z_alpha}
$$
is a subbasis for the topology on $Z$.
Let $O=f_alpha^{-1}(W)$ be an arbitrary member of $mc C$, where $W$ is an open set in $Z_alpha$.
But then $vp^{-1}(O)=(f_alphacirc vp)^{-1}(W)$ is a measurable subset of $Y$.
So we see that each member of $mc C$ pulls back under $vp$ to a measurable subset of $Y$.
Thus $vp$ is measurable.
$blacksquare$
Corollary.
Let $X$ be a compact metric space and $(Y, mc Y)$ be a measurable space.
Then a map $vp:Yto mr P(X)$ is Borel measurable if and only if for each continuous function $f:Xto R$ we have that the map $ymapsto int_X f dvp_y:Yto R$ is Borel measurable.
answered Nov 19 at 8:44
caffeinemachine
6,45521250
add a comment |
up vote
0
down vote
$newcommand{set}[1]{{#1}}$
$newcommand{vp}{varphi}$
$newcommand{mc}{mathcal}$
$newcommand{R}{mathbf R}$
$newcommand{mr}{mathscr}$
WE answer the question above in the affirmative. The only difference is that I will be proving the result with $mr P(X)$, the space of all the probability measures on $X$, instead of $M(X)$, which doesn't make much difference.
Lemma.
Let $Z$ be a set and $f_alpha:Zto Z_alpha$ be a collection of continuous maps into topological spaces $Z_alpha$ indexed by a set $J$.
Endow $Z$ with the initial topology with respect to this collection.
Then for any measurable space $(Y, mc Y)$, a map $vp:Yto Z$ is Borel measurable if and only if each composite $f_alphacirc vp$ is Borel measurable.
Proof.
We do the less trivial direction.
Assume that each composite $f_alphacirc vp$ is measurable.
Note that the collection
$$
mc C=bigcup_{alphain J} set{f_alpha^{-1}(W): Wtext{ open in } Z_alpha}
$$
is a subbasis for the topology on $Z$.
Let $O=f_alpha^{-1}(W)$ be an arbitrary member of $mc C$, where $W$ is an open set in $Z_alpha$.
But then $vp^{-1}(O)=(f_alphacirc vp)^{-1}(W)$ is a measurable subset of $Y$.
So we see that each member of $mc C$ pulls back under $vp$ to a measurable subset of $Y$.
Thus $vp$ is measurable.
$blacksquare$
Corollary.
Let $X$ be a compact metric space and $(Y, mc Y)$ be a measurable space.
Then a map $vp:Yto mr P(X)$ is Borel measurable if and only if for each continuous function $f:Xto R$ we have that the map $ymapsto int_X f dvp_y:Yto R$ is Borel measurable.
answered Nov 19 at 8:44
caffeinemachine
6,45521250
add a comment |
up vote
0
down vote
up vote
0
down vote
$newcommand{set}[1]{{#1}}$
$newcommand{vp}{varphi}$
$newcommand{mc}{mathcal}$
$newcommand{R}{mathbf R}$
$newcommand{mr}{mathscr}$
WE answer the question above in the affirmative. The only difference is that I will be proving the result with $mr P(X)$, the space of all the probability measures on $X$, instead of $M(X)$, which doesn't make much difference.
Lemma.
Let $Z$ be a set and $f_alpha:Zto Z_alpha$ be a collection of continuous maps into topological spaces $Z_alpha$ indexed by a set $J$.
Endow $Z$ with the initial topology with respect to this collection.
Then for any measurable space $(Y, mc Y)$, a map $vp:Yto Z$ is Borel measurable if and only if each composite $f_alphacirc vp$ is Borel measurable.
Proof.
We do the less trivial direction.
Assume that each composite $f_alphacirc vp$ is measurable.
Note that the collection
$$
mc C=bigcup_{alphain J} set{f_alpha^{-1}(W): Wtext{ open in } Z_alpha}
$$
is a subbasis for the topology on $Z$.
Let $O=f_alpha^{-1}(W)$ be an arbitrary member of $mc C$, where $W$ is an open set in $Z_alpha$.
But then $vp^{-1}(O)=(f_alphacirc vp)^{-1}(W)$ is a measurable subset of $Y$.
So we see that each member of $mc C$ pulls back under $vp$ to a measurable subset of $Y$.
Thus $vp$ is measurable.
$blacksquare$
Corollary.
Let $X$ be a compact metric space and $(Y, mc Y)$ be a measurable space.
Then a map $vp:Yto mr P(X)$ is Borel measurable if and only if for each continuous function $f:Xto R$ we have that the map $ymapsto int_X f dvp_y:Yto R$ is Borel measurable.
answered Nov 19 at 8:44
caffeinemachine
6,45521250
$newcommand{set}[1]{{#1}}$
$newcommand{vp}{varphi}$
$newcommand{mc}{mathcal}$
$newcommand{R}{mathbf R}$
$newcommand{mr}{mathscr}$
WE answer the question above in the affirmative. The only difference is that I will be proving the result with $mr P(X)$, the space of all the probability measures on $X$, instead of $M(X)$, which doesn't make much difference.
Lemma.
Let $Z$ be a set and $f_alpha:Zto Z_alpha$ be a collection of continuous maps into topological spaces $Z_alpha$ indexed by a set $J$.
Endow $Z$ with the initial topology with respect to this collection.
Then for any measurable space $(Y, mc Y)$, a map $vp:Yto Z$ is Borel measurable if and only if each composite $f_alphacirc vp$ is Borel measurable.
Proof.
We do the less trivial direction.
Assume that each composite $f_alphacirc vp$ is measurable.
Note that the collection
$$
mc C=bigcup_{alphain J} set{f_alpha^{-1}(W): Wtext{ open in } Z_alpha}
$$
is a subbasis for the topology on $Z$.
Let $O=f_alpha^{-1}(W)$ be an arbitrary member of $mc C$, where $W$ is an open set in $Z_alpha$.
But then $vp^{-1}(O)=(f_alphacirc vp)^{-1}(W)$ is a measurable subset of $Y$.
So we see that each member of $mc C$ pulls back under $vp$ to a measurable subset of $Y$.
Thus $vp$ is measurable.
$blacksquare$
Corollary.
Let $X$ be a compact metric space and $(Y, mc Y)$ be a measurable space.
Then a map $vp:Yto mr P(X)$ is Borel measurable if and only if for each continuous function $f:Xto R$ we have that the map $ymapsto int_X f dvp_y:Yto R$ is Borel measurable.
answered Nov 19 at 8:44
caffeinemachine
6,45521250
answered Nov 19 at 8:44
caffeinemachine
6,45521250
answered Nov 19 at 8:44
caffeinemachine
6,45521250
answered Nov 19 at 8:44
caffeinemachine
6,45521250
6,45521250
add a comment |
add a comment |
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Notice that the map $pi : M(X) to mathbb{R}^{C(X)} $ with $ nu mapsto (int_X f , dnu)_{fin C(X)}$ is an embedding (where $mathbb{R}^{C(X)}$ is equipped with the product topology). So $mu : X to M(X)$ is measurable if and only if $pi circ mu$ is measurable. This is of course equivalent to saying that each $x mapsto int_X f , dmu_x$ is measurable for each $f in C(X)$.
– Sangchul Lee
Sep 22 at 15:21
@SangchulLee "So $mu:Xto M(X)$ is measurable if and only if $picirc mu$ is measurable." This is true when we equip $mathbb R^{C(X)}$ with the Borel $sigma$-algebra. But the equivalence of the measurability of $picirc mu$ and the measurability of each $xmapsto int_Xf dmu_x$ is true when $mathbf R^{C(X)}$ is equipped with the product $sigma$-algebra. So I do not completely follow your comment. Can you please address this and point out any mistakes. Thanks.
– caffeinemachine
Sep 23 at 11:19
I see your point. The issue is that that Borel $sigma$-algebra on $mathbb{R}^{C(X)}$ may possibly be strictly finer than the product $sigma$-algebra for general $X$. Thank you for pointing out my mistake and sorry for the confusion. Perhaps the claim can be salvaged if $C(X)$ is separable so that we can replace $mathbb{R}^{C(X)}$ by $mathbb{R}^{mathfrak{A}}$ for some countable dense subset $mathfrak{A} subset C(X)$, but let me think about it.
– Sangchul Lee
Sep 23 at 11:36
@SangchulLee I am interested in the situation when $X$ is a compact metric space and here it is known that $C(X)$ is separable. Just wanted to say this. Thanks.
– caffeinemachine
Sep 23 at 12:02