Mixing
Under which assumptions are two Borel measures equal?
$begingroup$
Assume $(X,tau)$ is a topological space and $mathcal B (X)$ the corresponding Borel $sigma$-Algebra. Let $mu$ and $nu$ be two Borel measures on $mathcal B (X)$. Now let $Q$ be some set of measurable functions $f : X to mathbb C$.
Under which (minimal) assumptions on the topology $tau$ or the set $Q$ can we conclude that from
$$
int_X f , dmu = int_X f , dnu forall f in Q
$$
follows $mu = nu$?
integration general-topology analysis measure-theory
asked Feb 2 at 13:36
MuziMuzi
453320
$endgroup$
add a comment |
$begingroup$
Assume $(X,tau)$ is a topological space and $mathcal B (X)$ the corresponding Borel $sigma$-Algebra. Let $mu$ and $nu$ be two Borel measures on $mathcal B (X)$. Now let $Q$ be some set of measurable functions $f : X to mathbb C$.
Under which (minimal) assumptions on the topology $tau$ or the set $Q$ can we conclude that from
$$
int_X f , dmu = int_X f , dnu forall f in Q
$$
follows $mu = nu$?
integration general-topology analysis measure-theory
asked Feb 2 at 13:36
MuziMuzi
453320
$endgroup$
$begingroup$
You must be able to approximate every measurable set by elements in $Q$.
$endgroup$
– Yanko
Feb 2 at 13:43
$begingroup$
If $Q$ contain unitary function of open set, then $mu=nu$. Is it the minimal condition ? I'm not really sure.
$endgroup$
– idm
Feb 2 at 13:46
2
$begingroup$
The classical case is $X$ locally compact Hausdorff and $Q$ all continuous (real) functions. This is Riesz duality in essence.
$endgroup$
– Henno Brandsma
Feb 2 at 15:42
$begingroup$
Can we also replace the set of all continuous functions by a dense subset and still argue with Riesz?
$endgroup$
– Muzi
Feb 2 at 18:02
add a comment |
$begingroup$
Assume $(X,tau)$ is a topological space and $mathcal B (X)$ the corresponding Borel $sigma$-Algebra. Let $mu$ and $nu$ be two Borel measures on $mathcal B (X)$. Now let $Q$ be some set of measurable functions $f : X to mathbb C$.
Under which (minimal) assumptions on the topology $tau$ or the set $Q$ can we conclude that from
$$
int_X f , dmu = int_X f , dnu forall f in Q
$$
follows $mu = nu$?
integration general-topology analysis measure-theory
asked Feb 2 at 13:36
MuziMuzi
453320
$endgroup$
Assume $(X,tau)$ is a topological space and $mathcal B (X)$ the corresponding Borel $sigma$-Algebra. Let $mu$ and $nu$ be two Borel measures on $mathcal B (X)$. Now let $Q$ be some set of measurable functions $f : X to mathbb C$.
Under which (minimal) assumptions on the topology $tau$ or the set $Q$ can we conclude that from
$$
int_X f , dmu = int_X f , dnu forall f in Q
$$
follows $mu = nu$?
integration general-topology analysis measure-theory
integration general-topology analysis measure-theory
asked Feb 2 at 13:36
MuziMuzi
453320
asked Feb 2 at 13:36
MuziMuzi
453320
asked Feb 2 at 13:36
MuziMuzi
453320
asked Feb 2 at 13:36
MuziMuzi
453320
asked Feb 2 at 13:36
MuziMuzi
453320
453320
$begingroup$
You must be able to approximate every measurable set by elements in $Q$.
$endgroup$
– Yanko
Feb 2 at 13:43
$begingroup$
If $Q$ contain unitary function of open set, then $mu=nu$. Is it the minimal condition ? I'm not really sure.
$endgroup$
– idm
Feb 2 at 13:46
2
$begingroup$
The classical case is $X$ locally compact Hausdorff and $Q$ all continuous (real) functions. This is Riesz duality in essence.
$endgroup$
– Henno Brandsma
Feb 2 at 15:42
$begingroup$
Can we also replace the set of all continuous functions by a dense subset and still argue with Riesz?
$endgroup$
– Muzi
Feb 2 at 18:02
add a comment |
$begingroup$
You must be able to approximate every measurable set by elements in $Q$.
$endgroup$
– Yanko
Feb 2 at 13:43
$begingroup$
If $Q$ contain unitary function of open set, then $mu=nu$. Is it the minimal condition ? I'm not really sure.
$endgroup$
– idm
Feb 2 at 13:46
2
$begingroup$
The classical case is $X$ locally compact Hausdorff and $Q$ all continuous (real) functions. This is Riesz duality in essence.
$endgroup$
– Henno Brandsma
Feb 2 at 15:42
$begingroup$
Can we also replace the set of all continuous functions by a dense subset and still argue with Riesz?
$endgroup$
– Muzi
Feb 2 at 18:02
$begingroup$
You must be able to approximate every measurable set by elements in $Q$.
$endgroup$
– Yanko
Feb 2 at 13:43
$begingroup$
You must be able to approximate every measurable set by elements in $Q$.
$endgroup$
– Yanko
Feb 2 at 13:43
$begingroup$
If $Q$ contain unitary function of open set, then $mu=nu$. Is it the minimal condition ? I'm not really sure.
$endgroup$
– idm
Feb 2 at 13:46
$begingroup$
If $Q$ contain unitary function of open set, then $mu=nu$. Is it the minimal condition ? I'm not really sure.
$endgroup$
– idm
Feb 2 at 13:46
2
2
$begingroup$
The classical case is $X$ locally compact Hausdorff and $Q$ all continuous (real) functions. This is Riesz duality in essence.
$endgroup$
– Henno Brandsma
Feb 2 at 15:42
$begingroup$
The classical case is $X$ locally compact Hausdorff and $Q$ all continuous (real) functions. This is Riesz duality in essence.
$endgroup$
– Henno Brandsma
Feb 2 at 15:42
$begingroup$
Can we also replace the set of all continuous functions by a dense subset and still argue with Riesz?
$endgroup$
– Muzi
Feb 2 at 18:02
$begingroup$
Can we also replace the set of all continuous functions by a dense subset and still argue with Riesz?
$endgroup$
– Muzi
Feb 2 at 18:02
add a comment |
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$begingroup$
You must be able to approximate every measurable set by elements in $Q$.
$endgroup$
– Yanko
Feb 2 at 13:43
$begingroup$
If $Q$ contain unitary function of open set, then $mu=nu$. Is it the minimal condition ? I'm not really sure.
$endgroup$
– idm
Feb 2 at 13:46
2
$begingroup$
The classical case is $X$ locally compact Hausdorff and $Q$ all continuous (real) functions. This is Riesz duality in essence.
$endgroup$
– Henno Brandsma
Feb 2 at 15:42
$begingroup$
Can we also replace the set of all continuous functions by a dense subset and still argue with Riesz?
$endgroup$
– Muzi
Feb 2 at 18:02