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$nu(A)=int mathcal{X}_{A}f(x)dmu$ measure associated functional $l_{nu}(g)=int g(x)f(x)dmu.$
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Let $mu$ be a positive Baire measure, $fin L^1(X,dmu)$ with $X$ compact hausdorff and $fgeq 0$.
For any Baire set $A$, define $nu(A)=int mathcal{X}_{A}f(x)dmu$.
Show that $nu$ is the one measure associated to the functional on $C(X)$,
$l_{nu}(g)=int g(x)f(x)dmu.$
Hello! In general, what are the steps to demonstrate that a measure comes from a functional one?
functional-analysis measure-theory
asked 9 hours ago
eraldcoil
21119
add a comment |
up vote
0
down vote
favorite
Let $mu$ be a positive Baire measure, $fin L^1(X,dmu)$ with $X$ compact hausdorff and $fgeq 0$.
For any Baire set $A$, define $nu(A)=int mathcal{X}_{A}f(x)dmu$.
Show that $nu$ is the one measure associated to the functional on $C(X)$,
$l_{nu}(g)=int g(x)f(x)dmu.$
Hello! In general, what are the steps to demonstrate that a measure comes from a functional one?
functional-analysis measure-theory
asked 9 hours ago
eraldcoil
21119
1
There is probably a theorem that gives you uniqueness of measures if the two measures coincide in a set of functions. And that would do it.
– Will M.
9 hours ago
1
You are supposed to show that $l_{nu} g=int g dnu$ for all $g in C(X)$. In other words you have to show that $int g(x)f(x)dmu(x)=int g(x)dnu(x)$.
– Kavi Rama Murthy
9 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $mu$ be a positive Baire measure, $fin L^1(X,dmu)$ with $X$ compact hausdorff and $fgeq 0$.
For any Baire set $A$, define $nu(A)=int mathcal{X}_{A}f(x)dmu$.
Show that $nu$ is the one measure associated to the functional on $C(X)$,
$l_{nu}(g)=int g(x)f(x)dmu.$
Hello! In general, what are the steps to demonstrate that a measure comes from a functional one?
functional-analysis measure-theory
asked 9 hours ago
eraldcoil
21119
Let $mu$ be a positive Baire measure, $fin L^1(X,dmu)$ with $X$ compact hausdorff and $fgeq 0$.
For any Baire set $A$, define $nu(A)=int mathcal{X}_{A}f(x)dmu$.
Show that $nu$ is the one measure associated to the functional on $C(X)$,
$l_{nu}(g)=int g(x)f(x)dmu.$
Hello! In general, what are the steps to demonstrate that a measure comes from a functional one?
functional-analysis measure-theory
functional-analysis measure-theory
asked 9 hours ago
eraldcoil
21119
asked 9 hours ago
eraldcoil
21119
asked 9 hours ago
eraldcoil
21119
asked 9 hours ago
eraldcoil
21119
asked 9 hours ago
eraldcoil
21119
21119
1
There is probably a theorem that gives you uniqueness of measures if the two measures coincide in a set of functions. And that would do it.
– Will M.
9 hours ago
1
You are supposed to show that $l_{nu} g=int g dnu$ for all $g in C(X)$. In other words you have to show that $int g(x)f(x)dmu(x)=int g(x)dnu(x)$.
– Kavi Rama Murthy
9 hours ago
add a comment |
1
There is probably a theorem that gives you uniqueness of measures if the two measures coincide in a set of functions. And that would do it.
– Will M.
9 hours ago
1
You are supposed to show that $l_{nu} g=int g dnu$ for all $g in C(X)$. In other words you have to show that $int g(x)f(x)dmu(x)=int g(x)dnu(x)$.
– Kavi Rama Murthy
9 hours ago
1
1
There is probably a theorem that gives you uniqueness of measures if the two measures coincide in a set of functions. And that would do it.
– Will M.
9 hours ago
There is probably a theorem that gives you uniqueness of measures if the two measures coincide in a set of functions. And that would do it.
– Will M.
9 hours ago
1
1
You are supposed to show that $l_{nu} g=int g dnu$ for all $g in C(X)$. In other words you have to show that $int g(x)f(x)dmu(x)=int g(x)dnu(x)$.
– Kavi Rama Murthy
9 hours ago
You are supposed to show that $l_{nu} g=int g dnu$ for all $g in C(X)$. In other words you have to show that $int g(x)f(x)dmu(x)=int g(x)dnu(x)$.
– Kavi Rama Murthy
9 hours ago
add a comment |
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There is probably a theorem that gives you uniqueness of measures if the two measures coincide in a set of functions. And that would do it.
– Will M.
9 hours ago
1
You are supposed to show that $l_{nu} g=int g dnu$ for all $g in C(X)$. In other words you have to show that $int g(x)f(x)dmu(x)=int g(x)dnu(x)$.
– Kavi Rama Murthy
9 hours ago