$nu(A)=int mathcal{X}_{A}f(x)dmu$ measure associated functional $l_{nu}(g)=int g(x)f(x)dmu.$
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
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
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.
Nov 19 at 6:26
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
Nov 19 at 6:27
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
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 Nov 19 at 6:21
eraldcoil
25119
25119
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.
Nov 19 at 6:26
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
Nov 19 at 6:27
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.
Nov 19 at 6:26
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
Nov 19 at 6:27
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.
Nov 19 at 6:26
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.
Nov 19 at 6:26
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
Nov 19 at 6:27
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
Nov 19 at 6:27
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004587%2fnua-int-mathcalx-afxd-mu-measure-associated-functional-l-nug%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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.
Nov 19 at 6:26
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
Nov 19 at 6:27