If $mathcal N$ non measurable, ${0}times mathcal N$ is measurable in $mathbb R^2$, but $m({0}times mathcal...
$begingroup$
We define product measure on complete measurable spaces because if $N$ is a nul set and $M$ is measurable, we want that $$m_2(Ntimes M)=m_1(N)m_1(M)=0.$$
Let $(mathbb R^2, mathcal Mtimes mathcal M, m_2)$ the Lebesgue measure space on $mathbb R^2$. We have that $$mathcal Mtimes mathcal M=sigma ({Atimes Bmid A,Bin mathcal M}),$$
and it's a complete measurable space. I know that for all $Atimes Bin mathcal Mtimes mathcal M$ we have that $$m_2(Atimes B)=m_1(A)m_1(B).tag{*}$$
But if $mathcal N$ is non measurable (for example Vitali set), then ${0}times mathcal N$ has measure $0$ and thus is in $mathcal Mtimes mathcal M$. But don't we have that $$m_2({0}times mathcal N)neq m_1({0})m_1(mathcal N) ?$$
So $(*)$ is wrong ? Could someone explain ?
measure-theory lebesgue-measure
asked Dec 3 '18 at 21:29
NewMathNewMath
4059
$endgroup$
|
show 1 more comment
$begingroup$
We define product measure on complete measurable spaces because if $N$ is a nul set and $M$ is measurable, we want that $$m_2(Ntimes M)=m_1(N)m_1(M)=0.$$
Let $(mathbb R^2, mathcal Mtimes mathcal M, m_2)$ the Lebesgue measure space on $mathbb R^2$. We have that $$mathcal Mtimes mathcal M=sigma ({Atimes Bmid A,Bin mathcal M}),$$
and it's a complete measurable space. I know that for all $Atimes Bin mathcal Mtimes mathcal M$ we have that $$m_2(Atimes B)=m_1(A)m_1(B).tag{*}$$
But if $mathcal N$ is non measurable (for example Vitali set), then ${0}times mathcal N$ has measure $0$ and thus is in $mathcal Mtimes mathcal M$. But don't we have that $$m_2({0}times mathcal N)neq m_1({0})m_1(mathcal N) ?$$
So $(*)$ is wrong ? Could someone explain ?
measure-theory lebesgue-measure
asked Dec 3 '18 at 21:29
NewMathNewMath
4059
$endgroup$
1
$begingroup$
The detail here is that the Vitali set $mathcal{N}$ is not in the sigma algebra $mathcal{M}$ that you are considering for which $(star)$ holds, precisely because it is not measurable.
$endgroup$
– Guido A.
Dec 3 '18 at 21:32
$begingroup$
$mathcal{N}notinmathcal{M}$
$endgroup$
– Ben W
Dec 3 '18 at 21:34
1
$begingroup$
The expression $m(A times B) = mA cdot mB$ is only valid if both $A,B$ are measurable. However, in a complete measure space any subset of a set of measure zero has measure zero.
$endgroup$
– copper.hat
Dec 3 '18 at 21:41
$begingroup$
@copper.hat: So you agree that ${0}times mathcal Nin mathcal Mtimes mathcal M$ (since it has measure $0$ it is measurable, no ?)
$endgroup$
– NewMath
Dec 3 '18 at 21:43
$begingroup$
No. If the $sigma$-field containing $mathcal Mtimes mathcal M$ is complete then ${0} times mathcal N$ will be in the $sigma$-field.
$endgroup$
– copper.hat
Dec 3 '18 at 21:45
|
show 1 more comment
$begingroup$
We define product measure on complete measurable spaces because if $N$ is a nul set and $M$ is measurable, we want that $$m_2(Ntimes M)=m_1(N)m_1(M)=0.$$
Let $(mathbb R^2, mathcal Mtimes mathcal M, m_2)$ the Lebesgue measure space on $mathbb R^2$. We have that $$mathcal Mtimes mathcal M=sigma ({Atimes Bmid A,Bin mathcal M}),$$
and it's a complete measurable space. I know that for all $Atimes Bin mathcal Mtimes mathcal M$ we have that $$m_2(Atimes B)=m_1(A)m_1(B).tag{*}$$
But if $mathcal N$ is non measurable (for example Vitali set), then ${0}times mathcal N$ has measure $0$ and thus is in $mathcal Mtimes mathcal M$. But don't we have that $$m_2({0}times mathcal N)neq m_1({0})m_1(mathcal N) ?$$
So $(*)$ is wrong ? Could someone explain ?
measure-theory lebesgue-measure
asked Dec 3 '18 at 21:29
NewMathNewMath
4059
$endgroup$
We define product measure on complete measurable spaces because if $N$ is a nul set and $M$ is measurable, we want that $$m_2(Ntimes M)=m_1(N)m_1(M)=0.$$
Let $(mathbb R^2, mathcal Mtimes mathcal M, m_2)$ the Lebesgue measure space on $mathbb R^2$. We have that $$mathcal Mtimes mathcal M=sigma ({Atimes Bmid A,Bin mathcal M}),$$
and it's a complete measurable space. I know that for all $Atimes Bin mathcal Mtimes mathcal M$ we have that $$m_2(Atimes B)=m_1(A)m_1(B).tag{*}$$
But if $mathcal N$ is non measurable (for example Vitali set), then ${0}times mathcal N$ has measure $0$ and thus is in $mathcal Mtimes mathcal M$. But don't we have that $$m_2({0}times mathcal N)neq m_1({0})m_1(mathcal N) ?$$
So $(*)$ is wrong ? Could someone explain ?
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Dec 3 '18 at 21:29
NewMathNewMath
4059
asked Dec 3 '18 at 21:29
NewMathNewMath
4059
asked Dec 3 '18 at 21:29
NewMathNewMath
4059
asked Dec 3 '18 at 21:29
NewMathNewMath
4059
asked Dec 3 '18 at 21:29
NewMathNewMath
4059
4059
1
$begingroup$
The detail here is that the Vitali set $mathcal{N}$ is not in the sigma algebra $mathcal{M}$ that you are considering for which $(star)$ holds, precisely because it is not measurable.
$endgroup$
– Guido A.
Dec 3 '18 at 21:32
$begingroup$
$mathcal{N}notinmathcal{M}$
$endgroup$
– Ben W
Dec 3 '18 at 21:34
1
$begingroup$
The expression $m(A times B) = mA cdot mB$ is only valid if both $A,B$ are measurable. However, in a complete measure space any subset of a set of measure zero has measure zero.
$endgroup$
– copper.hat
Dec 3 '18 at 21:41
$begingroup$
@copper.hat: So you agree that ${0}times mathcal Nin mathcal Mtimes mathcal M$ (since it has measure $0$ it is measurable, no ?)
$endgroup$
– NewMath
Dec 3 '18 at 21:43
$begingroup$
No. If the $sigma$-field containing $mathcal Mtimes mathcal M$ is complete then ${0} times mathcal N$ will be in the $sigma$-field.
$endgroup$
– copper.hat
Dec 3 '18 at 21:45
|
show 1 more comment
1
$begingroup$
The detail here is that the Vitali set $mathcal{N}$ is not in the sigma algebra $mathcal{M}$ that you are considering for which $(star)$ holds, precisely because it is not measurable.
$endgroup$
– Guido A.
Dec 3 '18 at 21:32
$begingroup$
$mathcal{N}notinmathcal{M}$
$endgroup$
– Ben W
Dec 3 '18 at 21:34
1
$begingroup$
The expression $m(A times B) = mA cdot mB$ is only valid if both $A,B$ are measurable. However, in a complete measure space any subset of a set of measure zero has measure zero.
$endgroup$
– copper.hat
Dec 3 '18 at 21:41
$begingroup$
@copper.hat: So you agree that ${0}times mathcal Nin mathcal Mtimes mathcal M$ (since it has measure $0$ it is measurable, no ?)
$endgroup$
– NewMath
Dec 3 '18 at 21:43
$begingroup$
No. If the $sigma$-field containing $mathcal Mtimes mathcal M$ is complete then ${0} times mathcal N$ will be in the $sigma$-field.
$endgroup$
– copper.hat
Dec 3 '18 at 21:45
1
1
$begingroup$
The detail here is that the Vitali set $mathcal{N}$ is not in the sigma algebra $mathcal{M}$ that you are considering for which $(star)$ holds, precisely because it is not measurable.
$endgroup$
– Guido A.
Dec 3 '18 at 21:32
$begingroup$
The detail here is that the Vitali set $mathcal{N}$ is not in the sigma algebra $mathcal{M}$ that you are considering for which $(star)$ holds, precisely because it is not measurable.
$endgroup$
– Guido A.
Dec 3 '18 at 21:32
$begingroup$
$mathcal{N}notinmathcal{M}$
$endgroup$
– Ben W
Dec 3 '18 at 21:34
$begingroup$
$mathcal{N}notinmathcal{M}$
$endgroup$
– Ben W
Dec 3 '18 at 21:34
1
1
$begingroup$
The expression $m(A times B) = mA cdot mB$ is only valid if both $A,B$ are measurable. However, in a complete measure space any subset of a set of measure zero has measure zero.
$endgroup$
– copper.hat
Dec 3 '18 at 21:41
$begingroup$
The expression $m(A times B) = mA cdot mB$ is only valid if both $A,B$ are measurable. However, in a complete measure space any subset of a set of measure zero has measure zero.
$endgroup$
– copper.hat
Dec 3 '18 at 21:41
$begingroup$
@copper.hat: So you agree that ${0}times mathcal Nin mathcal Mtimes mathcal M$ (since it has measure $0$ it is measurable, no ?)
$endgroup$
– NewMath
Dec 3 '18 at 21:43
$begingroup$
@copper.hat: So you agree that ${0}times mathcal Nin mathcal Mtimes mathcal M$ (since it has measure $0$ it is measurable, no ?)
$endgroup$
– NewMath
Dec 3 '18 at 21:43
$begingroup$
No. If the $sigma$-field containing $mathcal Mtimes mathcal M$ is complete then ${0} times mathcal N$ will be in the $sigma$-field.
$endgroup$
– copper.hat
Dec 3 '18 at 21:45
$begingroup$
No. If the $sigma$-field containing $mathcal Mtimes mathcal M$ is complete then ${0} times mathcal N$ will be in the $sigma$-field.
$endgroup$
– copper.hat
Dec 3 '18 at 21:45
|
show 1 more comment
0
active
oldest
votes
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%2f3024702%2fif-mathcal-n-non-measurable-0-times-mathcal-n-is-measurable-in-math%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3024702%2fif-mathcal-n-non-measurable-0-times-mathcal-n-is-measurable-in-math%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
$begingroup$
The detail here is that the Vitali set $mathcal{N}$ is not in the sigma algebra $mathcal{M}$ that you are considering for which $(star)$ holds, precisely because it is not measurable.
$endgroup$
– Guido A.
Dec 3 '18 at 21:32
$begingroup$
$mathcal{N}notinmathcal{M}$
$endgroup$
– Ben W
Dec 3 '18 at 21:34
1
$begingroup$
The expression $m(A times B) = mA cdot mB$ is only valid if both $A,B$ are measurable. However, in a complete measure space any subset of a set of measure zero has measure zero.
$endgroup$
– copper.hat
Dec 3 '18 at 21:41
$begingroup$
@copper.hat: So you agree that ${0}times mathcal Nin mathcal Mtimes mathcal M$ (since it has measure $0$ it is measurable, no ?)
$endgroup$
– NewMath
Dec 3 '18 at 21:43
$begingroup$
No. If the $sigma$-field containing $mathcal Mtimes mathcal M$ is complete then ${0} times mathcal N$ will be in the $sigma$-field.
$endgroup$
– copper.hat
Dec 3 '18 at 21:45