Does every set of positive measure contain an uncountable null set?
If $E$ is Lebesgue measurable and $m(E)>0$, does there exist an uncountable $Csubset E$ with $m(C)=0$?
This seems intuitively clear but I cannot prove it. Since $E$ has positive measure it contains a nonmeasurable set $V$, and every measurable subset of $V$ is null, but I could not show $V$ contains uncountable measurable subsets. I also tried using the fact that for $alphain(0,1)$ there is an interval $I$ s.t $m(Ecap I)ge alpha m(I)$, attempting to construct a Cantor set inside $I$ with an uncountable intersection with $E$ but I was unsuccessful. Any hints? Is this even true?
real-analysis measure-theory
add a comment |
If $E$ is Lebesgue measurable and $m(E)>0$, does there exist an uncountable $Csubset E$ with $m(C)=0$?
This seems intuitively clear but I cannot prove it. Since $E$ has positive measure it contains a nonmeasurable set $V$, and every measurable subset of $V$ is null, but I could not show $V$ contains uncountable measurable subsets. I also tried using the fact that for $alphain(0,1)$ there is an interval $I$ s.t $m(Ecap I)ge alpha m(I)$, attempting to construct a Cantor set inside $I$ with an uncountable intersection with $E$ but I was unsuccessful. Any hints? Is this even true?
real-analysis measure-theory
add a comment |
If $E$ is Lebesgue measurable and $m(E)>0$, does there exist an uncountable $Csubset E$ with $m(C)=0$?
This seems intuitively clear but I cannot prove it. Since $E$ has positive measure it contains a nonmeasurable set $V$, and every measurable subset of $V$ is null, but I could not show $V$ contains uncountable measurable subsets. I also tried using the fact that for $alphain(0,1)$ there is an interval $I$ s.t $m(Ecap I)ge alpha m(I)$, attempting to construct a Cantor set inside $I$ with an uncountable intersection with $E$ but I was unsuccessful. Any hints? Is this even true?
real-analysis measure-theory
If $E$ is Lebesgue measurable and $m(E)>0$, does there exist an uncountable $Csubset E$ with $m(C)=0$?
This seems intuitively clear but I cannot prove it. Since $E$ has positive measure it contains a nonmeasurable set $V$, and every measurable subset of $V$ is null, but I could not show $V$ contains uncountable measurable subsets. I also tried using the fact that for $alphain(0,1)$ there is an interval $I$ s.t $m(Ecap I)ge alpha m(I)$, attempting to construct a Cantor set inside $I$ with an uncountable intersection with $E$ but I was unsuccessful. Any hints? Is this even true?
real-analysis measure-theory
real-analysis measure-theory
asked 1 hour ago
Guacho Perez
3,80211131
3,80211131
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Yes.
By regularity, if $m(E)>0$ then there is some closed $Csubseteq E$ with $m(C)>0$. So WLOG assume $E$ is closed (and has measure $1$).
A variant of the Cantor set construction now lets us "thin" $E$ out to a closed uncountable $Dsubseteq E$ with measure zero. E.g. at our first step we fix $a<b$ such that $m(E_{<a})=m(E_{>b})={1over 3}$, and cut down to $Esetminus (a,b)$.
(How does the closedness of $E$ matter? Well, we need to argue that the $D$ we construct is in fact as desired. Trivially this $D$ is null, since we've shrunk its measure appropriately at each stage; knowing that $D$ is closed tells us that everything that should be in $D$, is.)
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%2f3050013%2fdoes-every-set-of-positive-measure-contain-an-uncountable-null-set%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
Yes.
By regularity, if $m(E)>0$ then there is some closed $Csubseteq E$ with $m(C)>0$. So WLOG assume $E$ is closed (and has measure $1$).
A variant of the Cantor set construction now lets us "thin" $E$ out to a closed uncountable $Dsubseteq E$ with measure zero. E.g. at our first step we fix $a<b$ such that $m(E_{<a})=m(E_{>b})={1over 3}$, and cut down to $Esetminus (a,b)$.
(How does the closedness of $E$ matter? Well, we need to argue that the $D$ we construct is in fact as desired. Trivially this $D$ is null, since we've shrunk its measure appropriately at each stage; knowing that $D$ is closed tells us that everything that should be in $D$, is.)
add a comment |
Yes.
By regularity, if $m(E)>0$ then there is some closed $Csubseteq E$ with $m(C)>0$. So WLOG assume $E$ is closed (and has measure $1$).
A variant of the Cantor set construction now lets us "thin" $E$ out to a closed uncountable $Dsubseteq E$ with measure zero. E.g. at our first step we fix $a<b$ such that $m(E_{<a})=m(E_{>b})={1over 3}$, and cut down to $Esetminus (a,b)$.
(How does the closedness of $E$ matter? Well, we need to argue that the $D$ we construct is in fact as desired. Trivially this $D$ is null, since we've shrunk its measure appropriately at each stage; knowing that $D$ is closed tells us that everything that should be in $D$, is.)
add a comment |
Yes.
By regularity, if $m(E)>0$ then there is some closed $Csubseteq E$ with $m(C)>0$. So WLOG assume $E$ is closed (and has measure $1$).
A variant of the Cantor set construction now lets us "thin" $E$ out to a closed uncountable $Dsubseteq E$ with measure zero. E.g. at our first step we fix $a<b$ such that $m(E_{<a})=m(E_{>b})={1over 3}$, and cut down to $Esetminus (a,b)$.
(How does the closedness of $E$ matter? Well, we need to argue that the $D$ we construct is in fact as desired. Trivially this $D$ is null, since we've shrunk its measure appropriately at each stage; knowing that $D$ is closed tells us that everything that should be in $D$, is.)
Yes.
By regularity, if $m(E)>0$ then there is some closed $Csubseteq E$ with $m(C)>0$. So WLOG assume $E$ is closed (and has measure $1$).
A variant of the Cantor set construction now lets us "thin" $E$ out to a closed uncountable $Dsubseteq E$ with measure zero. E.g. at our first step we fix $a<b$ such that $m(E_{<a})=m(E_{>b})={1over 3}$, and cut down to $Esetminus (a,b)$.
(How does the closedness of $E$ matter? Well, we need to argue that the $D$ we construct is in fact as desired. Trivially this $D$ is null, since we've shrunk its measure appropriately at each stage; knowing that $D$ is closed tells us that everything that should be in $D$, is.)
answered 1 hour ago
Noah Schweber
120k10146278
120k10146278
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3050013%2fdoes-every-set-of-positive-measure-contain-an-uncountable-null-set%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