Mixing
(Hint Needed) Real-Analysis Exam
$begingroup$
So I'm trying to study for a qualifying exam and need a hint on the following problem. If someone knows of some useful lemmas and techniques, I would like to hear it!
Let $(X,mathcal{A},mu)$ be a $sigma$-finite measure space, and let $mathcal{A}_0$ be a sub-$sigma$-algebra of $mathcal{A}$ so that $(X,mathcal{A}_0,mu)$ is $sigma$-finite. Given a nonnegative $mathcal{A}$-measurable function $f$ on $X$, show that there is a nonnegative $mathcal{A}_0$-measurable function $f_0$ on $X$, such that
$$int_Xfg,dmu=int_Xf_0g,dmu $$
for every nonnegative $mathcal{A}_0$-measurable function $g$. In what sense is $f_0$ unique?
real-analysis analysis measure-theory lebesgue-measure
edited Jan 23 at 4:48
Key Flex
8,38261233
asked Jan 23 at 4:09
DarelDarel
1249
$endgroup$
add a comment |
$begingroup$
So I'm trying to study for a qualifying exam and need a hint on the following problem. If someone knows of some useful lemmas and techniques, I would like to hear it!
Let $(X,mathcal{A},mu)$ be a $sigma$-finite measure space, and let $mathcal{A}_0$ be a sub-$sigma$-algebra of $mathcal{A}$ so that $(X,mathcal{A}_0,mu)$ is $sigma$-finite. Given a nonnegative $mathcal{A}$-measurable function $f$ on $X$, show that there is a nonnegative $mathcal{A}_0$-measurable function $f_0$ on $X$, such that
$$int_Xfg,dmu=int_Xf_0g,dmu $$
for every nonnegative $mathcal{A}_0$-measurable function $g$. In what sense is $f_0$ unique?
real-analysis analysis measure-theory lebesgue-measure
edited Jan 23 at 4:48
Key Flex
8,38261233
asked Jan 23 at 4:09
DarelDarel
1249
$endgroup$
add a comment |
$begingroup$
So I'm trying to study for a qualifying exam and need a hint on the following problem. If someone knows of some useful lemmas and techniques, I would like to hear it!
Let $(X,mathcal{A},mu)$ be a $sigma$-finite measure space, and let $mathcal{A}_0$ be a sub-$sigma$-algebra of $mathcal{A}$ so that $(X,mathcal{A}_0,mu)$ is $sigma$-finite. Given a nonnegative $mathcal{A}$-measurable function $f$ on $X$, show that there is a nonnegative $mathcal{A}_0$-measurable function $f_0$ on $X$, such that
$$int_Xfg,dmu=int_Xf_0g,dmu $$
for every nonnegative $mathcal{A}_0$-measurable function $g$. In what sense is $f_0$ unique?
real-analysis analysis measure-theory lebesgue-measure
edited Jan 23 at 4:48
Key Flex
8,38261233
asked Jan 23 at 4:09
DarelDarel
1249
$endgroup$
So I'm trying to study for a qualifying exam and need a hint on the following problem. If someone knows of some useful lemmas and techniques, I would like to hear it!
Let $(X,mathcal{A},mu)$ be a $sigma$-finite measure space, and let $mathcal{A}_0$ be a sub-$sigma$-algebra of $mathcal{A}$ so that $(X,mathcal{A}_0,mu)$ is $sigma$-finite. Given a nonnegative $mathcal{A}$-measurable function $f$ on $X$, show that there is a nonnegative $mathcal{A}_0$-measurable function $f_0$ on $X$, such that
$$int_Xfg,dmu=int_Xf_0g,dmu $$
for every nonnegative $mathcal{A}_0$-measurable function $g$. In what sense is $f_0$ unique?
real-analysis analysis measure-theory lebesgue-measure
real-analysis analysis measure-theory lebesgue-measure
edited Jan 23 at 4:48
Key Flex
8,38261233
asked Jan 23 at 4:09
DarelDarel
1249
edited Jan 23 at 4:48
Key Flex
8,38261233
asked Jan 23 at 4:09
DarelDarel
1249
edited Jan 23 at 4:48
Key Flex
8,38261233
edited Jan 23 at 4:48
Key Flex
8,38261233
edited Jan 23 at 4:48
Key Flex
8,38261233
8,38261233
asked Jan 23 at 4:09
DarelDarel
1249
asked Jan 23 at 4:09
DarelDarel
1249
asked Jan 23 at 4:09
DarelDarel
1249
1249
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Notice that the above problem is false if we do not require $(X,mathcal{A}_0,mu)$ to be $sigma$-finite. As a counter-example, take the usual Lebesgue measure on $mathbb{R}$, and $mathcal{A}_0={∅,mathbb{R}}$. Let $g = 1$ and $f =X_{[0,1]}$. Then LHS is $1$ while RHS is either $0$ and $infty ($contradiction $)$. Thus we will assume $(X, mathcal{A}_0, mu)$ is σ-finite in this proof.
We first construct $f_0$ that holds for $g$ to be simple functions, i.e. we want $f_0$ satisfies
$$int_Ef dmu=int_E f_0 dmu,mbox{ for any }Einmathcal{A}_0tag{1}$$
Consider the $sigma$-finite measure space $(X,mathcal{A}_0,mu)$. Let $v$ be a measure defined on this space by $$v(E)=int_Ef dmumbox{ for my }Ein mathcal{A}_0$$
It is clear that $v$ is a $sigma$-finite measure and $v<<mu$. Radon-Nikodym theorem allows us to
construct an $mathcal{A}_0$-measurable function $f_0=dfrac{dv}{dmu}ge0$.
In the case when $gge0$ is $mathcal{A}_0$-measurable, we can find an increasing sequence of $mathcal{A}_0$-simple functions $psi_n$ converges to $g$ pointwise. Then $fpsi _n$ and $f_0psi _n$ monotonically converges to $fg$ and $f_0g$ respectively. By previous case and the Monotone convergence theorem$$int_Xfg d mu=limint_Xfpsi _n dmu=limint_Xf_0psi _n dmu=int_Xf_0g dmu$$
Therefore, $f_0$ satisfies all the hypotheses. By Radon-Nikodym construction, $f_0$ is unique up to $muupharpoonright mathcal{A}_0$
answered Jan 23 at 4:40
Key FlexKey Flex
8,38261233
$endgroup$
1
$begingroup$
I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
$endgroup$
– Darel
Jan 23 at 4:45
1
$begingroup$
@Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
$endgroup$
– Key Flex
Jan 23 at 4:48
$begingroup$
Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
$endgroup$
– Darel
Jan 23 at 4:49
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3084076%2fhint-needed-real-analysis-exam%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notice that the above problem is false if we do not require $(X,mathcal{A}_0,mu)$ to be $sigma$-finite. As a counter-example, take the usual Lebesgue measure on $mathbb{R}$, and $mathcal{A}_0={∅,mathbb{R}}$. Let $g = 1$ and $f =X_{[0,1]}$. Then LHS is $1$ while RHS is either $0$ and $infty ($contradiction $)$. Thus we will assume $(X, mathcal{A}_0, mu)$ is σ-finite in this proof.
We first construct $f_0$ that holds for $g$ to be simple functions, i.e. we want $f_0$ satisfies
$$int_Ef dmu=int_E f_0 dmu,mbox{ for any }Einmathcal{A}_0tag{1}$$
Consider the $sigma$-finite measure space $(X,mathcal{A}_0,mu)$. Let $v$ be a measure defined on this space by $$v(E)=int_Ef dmumbox{ for my }Ein mathcal{A}_0$$
It is clear that $v$ is a $sigma$-finite measure and $v<<mu$. Radon-Nikodym theorem allows us to
construct an $mathcal{A}_0$-measurable function $f_0=dfrac{dv}{dmu}ge0$.
In the case when $gge0$ is $mathcal{A}_0$-measurable, we can find an increasing sequence of $mathcal{A}_0$-simple functions $psi_n$ converges to $g$ pointwise. Then $fpsi _n$ and $f_0psi _n$ monotonically converges to $fg$ and $f_0g$ respectively. By previous case and the Monotone convergence theorem$$int_Xfg d mu=limint_Xfpsi _n dmu=limint_Xf_0psi _n dmu=int_Xf_0g dmu$$
Therefore, $f_0$ satisfies all the hypotheses. By Radon-Nikodym construction, $f_0$ is unique up to $muupharpoonright mathcal{A}_0$
answered Jan 23 at 4:40
Key FlexKey Flex
8,38261233
$endgroup$
1
$begingroup$
I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
$endgroup$
– Darel
Jan 23 at 4:45
1
$begingroup$
@Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
$endgroup$
– Key Flex
Jan 23 at 4:48
$begingroup$
Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
$endgroup$
– Darel
Jan 23 at 4:49
add a comment |
$begingroup$
Notice that the above problem is false if we do not require $(X,mathcal{A}_0,mu)$ to be $sigma$-finite. As a counter-example, take the usual Lebesgue measure on $mathbb{R}$, and $mathcal{A}_0={∅,mathbb{R}}$. Let $g = 1$ and $f =X_{[0,1]}$. Then LHS is $1$ while RHS is either $0$ and $infty ($contradiction $)$. Thus we will assume $(X, mathcal{A}_0, mu)$ is σ-finite in this proof.
We first construct $f_0$ that holds for $g$ to be simple functions, i.e. we want $f_0$ satisfies
$$int_Ef dmu=int_E f_0 dmu,mbox{ for any }Einmathcal{A}_0tag{1}$$
Consider the $sigma$-finite measure space $(X,mathcal{A}_0,mu)$. Let $v$ be a measure defined on this space by $$v(E)=int_Ef dmumbox{ for my }Ein mathcal{A}_0$$
It is clear that $v$ is a $sigma$-finite measure and $v<<mu$. Radon-Nikodym theorem allows us to
construct an $mathcal{A}_0$-measurable function $f_0=dfrac{dv}{dmu}ge0$.
In the case when $gge0$ is $mathcal{A}_0$-measurable, we can find an increasing sequence of $mathcal{A}_0$-simple functions $psi_n$ converges to $g$ pointwise. Then $fpsi _n$ and $f_0psi _n$ monotonically converges to $fg$ and $f_0g$ respectively. By previous case and the Monotone convergence theorem$$int_Xfg d mu=limint_Xfpsi _n dmu=limint_Xf_0psi _n dmu=int_Xf_0g dmu$$
Therefore, $f_0$ satisfies all the hypotheses. By Radon-Nikodym construction, $f_0$ is unique up to $muupharpoonright mathcal{A}_0$
answered Jan 23 at 4:40
Key FlexKey Flex
8,38261233
$endgroup$
1
$begingroup$
I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
$endgroup$
– Darel
Jan 23 at 4:45
1
$begingroup$
@Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
$endgroup$
– Key Flex
Jan 23 at 4:48
$begingroup$
Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
$endgroup$
– Darel
Jan 23 at 4:49
add a comment |
$begingroup$
Notice that the above problem is false if we do not require $(X,mathcal{A}_0,mu)$ to be $sigma$-finite. As a counter-example, take the usual Lebesgue measure on $mathbb{R}$, and $mathcal{A}_0={∅,mathbb{R}}$. Let $g = 1$ and $f =X_{[0,1]}$. Then LHS is $1$ while RHS is either $0$ and $infty ($contradiction $)$. Thus we will assume $(X, mathcal{A}_0, mu)$ is σ-finite in this proof.
We first construct $f_0$ that holds for $g$ to be simple functions, i.e. we want $f_0$ satisfies
$$int_Ef dmu=int_E f_0 dmu,mbox{ for any }Einmathcal{A}_0tag{1}$$
Consider the $sigma$-finite measure space $(X,mathcal{A}_0,mu)$. Let $v$ be a measure defined on this space by $$v(E)=int_Ef dmumbox{ for my }Ein mathcal{A}_0$$
It is clear that $v$ is a $sigma$-finite measure and $v<<mu$. Radon-Nikodym theorem allows us to
construct an $mathcal{A}_0$-measurable function $f_0=dfrac{dv}{dmu}ge0$.
In the case when $gge0$ is $mathcal{A}_0$-measurable, we can find an increasing sequence of $mathcal{A}_0$-simple functions $psi_n$ converges to $g$ pointwise. Then $fpsi _n$ and $f_0psi _n$ monotonically converges to $fg$ and $f_0g$ respectively. By previous case and the Monotone convergence theorem$$int_Xfg d mu=limint_Xfpsi _n dmu=limint_Xf_0psi _n dmu=int_Xf_0g dmu$$
Therefore, $f_0$ satisfies all the hypotheses. By Radon-Nikodym construction, $f_0$ is unique up to $muupharpoonright mathcal{A}_0$
answered Jan 23 at 4:40
Key FlexKey Flex
8,38261233
$endgroup$
Notice that the above problem is false if we do not require $(X,mathcal{A}_0,mu)$ to be $sigma$-finite. As a counter-example, take the usual Lebesgue measure on $mathbb{R}$, and $mathcal{A}_0={∅,mathbb{R}}$. Let $g = 1$ and $f =X_{[0,1]}$. Then LHS is $1$ while RHS is either $0$ and $infty ($contradiction $)$. Thus we will assume $(X, mathcal{A}_0, mu)$ is σ-finite in this proof.
We first construct $f_0$ that holds for $g$ to be simple functions, i.e. we want $f_0$ satisfies
$$int_Ef dmu=int_E f_0 dmu,mbox{ for any }Einmathcal{A}_0tag{1}$$
Consider the $sigma$-finite measure space $(X,mathcal{A}_0,mu)$. Let $v$ be a measure defined on this space by $$v(E)=int_Ef dmumbox{ for my }Ein mathcal{A}_0$$
It is clear that $v$ is a $sigma$-finite measure and $v<<mu$. Radon-Nikodym theorem allows us to
construct an $mathcal{A}_0$-measurable function $f_0=dfrac{dv}{dmu}ge0$.
In the case when $gge0$ is $mathcal{A}_0$-measurable, we can find an increasing sequence of $mathcal{A}_0$-simple functions $psi_n$ converges to $g$ pointwise. Then $fpsi _n$ and $f_0psi _n$ monotonically converges to $fg$ and $f_0g$ respectively. By previous case and the Monotone convergence theorem$$int_Xfg d mu=limint_Xfpsi _n dmu=limint_Xf_0psi _n dmu=int_Xf_0g dmu$$
Therefore, $f_0$ satisfies all the hypotheses. By Radon-Nikodym construction, $f_0$ is unique up to $muupharpoonright mathcal{A}_0$
answered Jan 23 at 4:40
Key FlexKey Flex
8,38261233
answered Jan 23 at 4:40
Key FlexKey Flex
8,38261233
answered Jan 23 at 4:40
Key FlexKey Flex
8,38261233
answered Jan 23 at 4:40
Key FlexKey Flex
8,38261233
8,38261233
1
$begingroup$
I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
$endgroup$
– Darel
Jan 23 at 4:45
1
$begingroup$
@Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
$endgroup$
– Key Flex
Jan 23 at 4:48
$begingroup$
Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
$endgroup$
– Darel
Jan 23 at 4:49
add a comment |
1
$begingroup$
I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
$endgroup$
– Darel
Jan 23 at 4:45
1
$begingroup$
@Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
$endgroup$
– Key Flex
Jan 23 at 4:48
$begingroup$
Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
$endgroup$
– Darel
Jan 23 at 4:49
1
1
$begingroup$
I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
$endgroup$
– Darel
Jan 23 at 4:45
$begingroup$
I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
$endgroup$
– Darel
Jan 23 at 4:45
1
1
$begingroup$
@Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
$endgroup$
– Key Flex
Jan 23 at 4:48
$begingroup$
@Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
$endgroup$
– Key Flex
Jan 23 at 4:48
$begingroup$
Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
$endgroup$
– Darel
Jan 23 at 4:49
$begingroup$
Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
$endgroup$
– Darel
Jan 23 at 4:49
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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%2f3084076%2fhint-needed-real-analysis-exam%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
