Prove there exists a measurable subset $E' subset E$ such that $f$ is almost everywhere of one sign on $E'$.
2
$begingroup$
Let $f in Lleft(mathbb{R}right)$ where $L$ denotes Lebesgue integrable space of functions. Suppose there exists a subset $E subset mathbb{R}$ such that $m(E) > 0$ and $f(x) neq 0$ for all $x in E$. Prove there exists a measurable subset $E' subset E$ such $m(E') > 0$ and $f$ is almost everywhere of one sign on $E'$.
My (weak) attempt:
Let $E_alpha = {E setminus A: m(E) = m(E setminus A)}$.... I don't know. Rewrote this proof like 10 times with no insight. So now I'm just throwing it out there. Thank you.
I think all I need is a hint to start it.
measure-theory lebesgue-measure
asked Dec 3 '18 at 21:11
ZduffZduff
1,663920
$endgroup$
add a comment |
2
$begingroup$
Let $f in Lleft(mathbb{R}right)$ where $L$ denotes Lebesgue integrable space of functions. Suppose there exists a subset $E subset mathbb{R}$ such that $m(E) > 0$ and $f(x) neq 0$ for all $x in E$. Prove there exists a measurable subset $E' subset E$ such $m(E') > 0$ and $f$ is almost everywhere of one sign on $E'$.
My (weak) attempt:
Let $E_alpha = {E setminus A: m(E) = m(E setminus A)}$.... I don't know. Rewrote this proof like 10 times with no insight. So now I'm just throwing it out there. Thank you.
I think all I need is a hint to start it.
measure-theory lebesgue-measure
asked Dec 3 '18 at 21:11
ZduffZduff
1,663920
$endgroup$
add a comment |
2
2
2
$begingroup$
Let $f in Lleft(mathbb{R}right)$ where $L$ denotes Lebesgue integrable space of functions. Suppose there exists a subset $E subset mathbb{R}$ such that $m(E) > 0$ and $f(x) neq 0$ for all $x in E$. Prove there exists a measurable subset $E' subset E$ such $m(E') > 0$ and $f$ is almost everywhere of one sign on $E'$.
My (weak) attempt:
Let $E_alpha = {E setminus A: m(E) = m(E setminus A)}$.... I don't know. Rewrote this proof like 10 times with no insight. So now I'm just throwing it out there. Thank you.
I think all I need is a hint to start it.
measure-theory lebesgue-measure
asked Dec 3 '18 at 21:11
ZduffZduff
1,663920
$endgroup$
Let $f in Lleft(mathbb{R}right)$ where $L$ denotes Lebesgue integrable space of functions. Suppose there exists a subset $E subset mathbb{R}$ such that $m(E) > 0$ and $f(x) neq 0$ for all $x in E$. Prove there exists a measurable subset $E' subset E$ such $m(E') > 0$ and $f$ is almost everywhere of one sign on $E'$.
My (weak) attempt:
Let $E_alpha = {E setminus A: m(E) = m(E setminus A)}$.... I don't know. Rewrote this proof like 10 times with no insight. So now I'm just throwing it out there. Thank you.
I think all I need is a hint to start it.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Dec 3 '18 at 21:11
ZduffZduff
1,663920
asked Dec 3 '18 at 21:11
ZduffZduff
1,663920
asked Dec 3 '18 at 21:11
ZduffZduff
1,663920
asked Dec 3 '18 at 21:11
ZduffZduff
1,663920
asked Dec 3 '18 at 21:11
ZduffZduff
1,663920
1,663920
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Hint: take $P = {f > 0}$ and $N= {f < 0}$. Now,
$$
E = (E cap P) sqcup (E cap N).
$$
Certainly $f$ maintains its sign in each of these. Can they both be of measure zero?
answered Dec 3 '18 at 21:25
Guido A.Guido A.
7,5261730
$endgroup$
$begingroup$
Ah, thank you. But is it easy to show these two sets are measurable?
$endgroup$
– Zduff
Dec 4 '18 at 2:51
1
$begingroup$
Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
$endgroup$
– Guido A.
Dec 4 '18 at 13:08
$begingroup$
So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
$endgroup$
– Zduff
Dec 4 '18 at 14:51
$begingroup$
You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
$endgroup$
– Guido A.
Dec 4 '18 at 20:09
$begingroup$
(cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
$endgroup$
– Guido A.
Dec 4 '18 at 20:10
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
});
}
});
draft saved
draft discarded
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%2f3024679%2fprove-there-exists-a-measurable-subset-e-subset-e-such-that-f-is-almost-ev%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
2
$begingroup$
Hint: take $P = {f > 0}$ and $N= {f < 0}$. Now,
$$
E = (E cap P) sqcup (E cap N).
$$
Certainly $f$ maintains its sign in each of these. Can they both be of measure zero?
answered Dec 3 '18 at 21:25
Guido A.Guido A.
7,5261730
$endgroup$
$begingroup$
Ah, thank you. But is it easy to show these two sets are measurable?
$endgroup$
– Zduff
Dec 4 '18 at 2:51
1
$begingroup$
Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
$endgroup$
– Guido A.
Dec 4 '18 at 13:08
$begingroup$
So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
$endgroup$
– Zduff
Dec 4 '18 at 14:51
$begingroup$
You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
$endgroup$
– Guido A.
Dec 4 '18 at 20:09
$begingroup$
(cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
$endgroup$
– Guido A.
Dec 4 '18 at 20:10
add a comment |
2
$begingroup$
Hint: take $P = {f > 0}$ and $N= {f < 0}$. Now,
$$
E = (E cap P) sqcup (E cap N).
$$
Certainly $f$ maintains its sign in each of these. Can they both be of measure zero?
answered Dec 3 '18 at 21:25
Guido A.Guido A.
7,5261730
$endgroup$
$begingroup$
Ah, thank you. But is it easy to show these two sets are measurable?
$endgroup$
– Zduff
Dec 4 '18 at 2:51
1
$begingroup$
Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
$endgroup$
– Guido A.
Dec 4 '18 at 13:08
$begingroup$
So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
$endgroup$
– Zduff
Dec 4 '18 at 14:51
$begingroup$
You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
$endgroup$
– Guido A.
Dec 4 '18 at 20:09
$begingroup$
(cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
$endgroup$
– Guido A.
Dec 4 '18 at 20:10
add a comment |
2
2
2
$begingroup$
Hint: take $P = {f > 0}$ and $N= {f < 0}$. Now,
$$
E = (E cap P) sqcup (E cap N).
$$
Certainly $f$ maintains its sign in each of these. Can they both be of measure zero?
answered Dec 3 '18 at 21:25
Guido A.Guido A.
7,5261730
$endgroup$
Hint: take $P = {f > 0}$ and $N= {f < 0}$. Now,
$$
E = (E cap P) sqcup (E cap N).
$$
Certainly $f$ maintains its sign in each of these. Can they both be of measure zero?
answered Dec 3 '18 at 21:25
Guido A.Guido A.
7,5261730
answered Dec 3 '18 at 21:25
Guido A.Guido A.
7,5261730
answered Dec 3 '18 at 21:25
Guido A.Guido A.
7,5261730
answered Dec 3 '18 at 21:25
Guido A.Guido A.
7,5261730
7,5261730
$begingroup$
Ah, thank you. But is it easy to show these two sets are measurable?
$endgroup$
– Zduff
Dec 4 '18 at 2:51
1
$begingroup$
Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
$endgroup$
– Guido A.
Dec 4 '18 at 13:08
$begingroup$
So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
$endgroup$
– Zduff
Dec 4 '18 at 14:51
$begingroup$
You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
$endgroup$
– Guido A.
Dec 4 '18 at 20:09
$begingroup$
(cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
$endgroup$
– Guido A.
Dec 4 '18 at 20:10
add a comment |
$begingroup$
Ah, thank you. But is it easy to show these two sets are measurable?
$endgroup$
– Zduff
Dec 4 '18 at 2:51
1
$begingroup$
Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
$endgroup$
– Guido A.
Dec 4 '18 at 13:08
$begingroup$
So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
$endgroup$
– Zduff
Dec 4 '18 at 14:51
$begingroup$
You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
$endgroup$
– Guido A.
Dec 4 '18 at 20:09
$begingroup$
(cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
$endgroup$
– Guido A.
Dec 4 '18 at 20:10
$begingroup$
Ah, thank you. But is it easy to show these two sets are measurable?
$endgroup$
– Zduff
Dec 4 '18 at 2:51
$begingroup$
Ah, thank you. But is it easy to show these two sets are measurable?
$endgroup$
– Zduff
Dec 4 '18 at 2:51
1
1
$begingroup$
Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
$endgroup$
– Guido A.
Dec 4 '18 at 13:08
$begingroup$
Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
$endgroup$
– Guido A.
Dec 4 '18 at 13:08
$begingroup$
So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
$endgroup$
– Zduff
Dec 4 '18 at 14:51
$begingroup$
So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
$endgroup$
– Zduff
Dec 4 '18 at 14:51
$begingroup$
You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
$endgroup$
– Guido A.
Dec 4 '18 at 20:09
$begingroup$
You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
$endgroup$
– Guido A.
Dec 4 '18 at 20:09
$begingroup$
(cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
$endgroup$
– Guido A.
Dec 4 '18 at 20:10
$begingroup$
(cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
$endgroup$
– Guido A.
Dec 4 '18 at 20:10
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3024679%2fprove-there-exists-a-measurable-subset-e-subset-e-such-that-f-is-almost-ev%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
