Each set in $mathcal{A}$ is $mu^*$-measurable
Let $(X,mathcal{A},mu)$ be a measure space, and define
$$mu^*(A):=inf{mu(B) : Asubset B, Binmathcal{A}}$$
for all $Asubset X$.
Problem: Show that each set in $mathcal{A}$ is $mu^*$-measurable.
I have shown that $mu^*$ is an outer measure. $Ainmathcal{A}$ is $mu^*$-measurable iff for any $Esubset X$, $mu^*(E)=mu^*(Acap E)+mu^*(A^ccap E)$. I know that $mu^*(E)leqmu^*(Acap E)+mu^*(A^ccap E)$ (by definition of an outer measure). But how can I show the other inequality?
real-analysis measure-theory
asked Nov 18 '18 at 20:21
rbird
1,18914
add a comment |
Let $(X,mathcal{A},mu)$ be a measure space, and define
$$mu^*(A):=inf{mu(B) : Asubset B, Binmathcal{A}}$$
for all $Asubset X$.
Problem: Show that each set in $mathcal{A}$ is $mu^*$-measurable.
I have shown that $mu^*$ is an outer measure. $Ainmathcal{A}$ is $mu^*$-measurable iff for any $Esubset X$, $mu^*(E)=mu^*(Acap E)+mu^*(A^ccap E)$. I know that $mu^*(E)leqmu^*(Acap E)+mu^*(A^ccap E)$ (by definition of an outer measure). But how can I show the other inequality?
real-analysis measure-theory
asked Nov 18 '18 at 20:21
rbird
1,18914
the outer and 'inner' measure of $A$ coincides two
– Robson
Nov 18 '18 at 22:32
add a comment |
Let $(X,mathcal{A},mu)$ be a measure space, and define
$$mu^*(A):=inf{mu(B) : Asubset B, Binmathcal{A}}$$
for all $Asubset X$.
Problem: Show that each set in $mathcal{A}$ is $mu^*$-measurable.
I have shown that $mu^*$ is an outer measure. $Ainmathcal{A}$ is $mu^*$-measurable iff for any $Esubset X$, $mu^*(E)=mu^*(Acap E)+mu^*(A^ccap E)$. I know that $mu^*(E)leqmu^*(Acap E)+mu^*(A^ccap E)$ (by definition of an outer measure). But how can I show the other inequality?
real-analysis measure-theory
asked Nov 18 '18 at 20:21
rbird
1,18914
Let $(X,mathcal{A},mu)$ be a measure space, and define
$$mu^*(A):=inf{mu(B) : Asubset B, Binmathcal{A}}$$
for all $Asubset X$.
Problem: Show that each set in $mathcal{A}$ is $mu^*$-measurable.
I have shown that $mu^*$ is an outer measure. $Ainmathcal{A}$ is $mu^*$-measurable iff for any $Esubset X$, $mu^*(E)=mu^*(Acap E)+mu^*(A^ccap E)$. I know that $mu^*(E)leqmu^*(Acap E)+mu^*(A^ccap E)$ (by definition of an outer measure). But how can I show the other inequality?
real-analysis measure-theory
real-analysis measure-theory
asked Nov 18 '18 at 20:21
rbird
1,18914
asked Nov 18 '18 at 20:21
rbird
1,18914
asked Nov 18 '18 at 20:21
rbird
1,18914
asked Nov 18 '18 at 20:21
rbird
1,18914
asked Nov 18 '18 at 20:21
rbird
1,18914
1,18914
the outer and 'inner' measure of $A$ coincides two
– Robson
Nov 18 '18 at 22:32
add a comment |
the outer and 'inner' measure of $A$ coincides two
– Robson
Nov 18 '18 at 22:32
the outer and 'inner' measure of $A$ coincides two
– Robson
Nov 18 '18 at 22:32
the outer and 'inner' measure of $A$ coincides two
– Robson
Nov 18 '18 at 22:32
add a comment |
1 Answer
1
active
oldest
votes
By definition of infimum, for any $epsilon>0$, there exists a set $E_{epsilon}in mathcal{A}$ s.t. $Esubset E_{epsilon}$ and $mu^*(E_{epsilon})lemu^*(E)+epsilon$ (assuming that $mu^*(E)<
infty$). Then
$$
mu^*(E)+epsilonge mu(E_{epsilon}cap A)+mu(E_{epsilon}cap A^c)ge mu^*(Ecap A)+mu^*(Ecap A^c)
$$
because $mu$ is additive on $mathcal{A}$.
answered Nov 18 '18 at 21:04
d.k.o.
8,537528
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%2f3004070%2feach-set-in-mathcala-is-mu-measurable%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
By definition of infimum, for any $epsilon>0$, there exists a set $E_{epsilon}in mathcal{A}$ s.t. $Esubset E_{epsilon}$ and $mu^*(E_{epsilon})lemu^*(E)+epsilon$ (assuming that $mu^*(E)<
infty$). Then
$$
mu^*(E)+epsilonge mu(E_{epsilon}cap A)+mu(E_{epsilon}cap A^c)ge mu^*(Ecap A)+mu^*(Ecap A^c)
$$
because $mu$ is additive on $mathcal{A}$.
answered Nov 18 '18 at 21:04
d.k.o.
8,537528
add a comment |
By definition of infimum, for any $epsilon>0$, there exists a set $E_{epsilon}in mathcal{A}$ s.t. $Esubset E_{epsilon}$ and $mu^*(E_{epsilon})lemu^*(E)+epsilon$ (assuming that $mu^*(E)<
infty$). Then
$$
mu^*(E)+epsilonge mu(E_{epsilon}cap A)+mu(E_{epsilon}cap A^c)ge mu^*(Ecap A)+mu^*(Ecap A^c)
$$
because $mu$ is additive on $mathcal{A}$.
answered Nov 18 '18 at 21:04
d.k.o.
8,537528
add a comment |
By definition of infimum, for any $epsilon>0$, there exists a set $E_{epsilon}in mathcal{A}$ s.t. $Esubset E_{epsilon}$ and $mu^*(E_{epsilon})lemu^*(E)+epsilon$ (assuming that $mu^*(E)<
infty$). Then
$$
mu^*(E)+epsilonge mu(E_{epsilon}cap A)+mu(E_{epsilon}cap A^c)ge mu^*(Ecap A)+mu^*(Ecap A^c)
$$
because $mu$ is additive on $mathcal{A}$.
answered Nov 18 '18 at 21:04
d.k.o.
8,537528
By definition of infimum, for any $epsilon>0$, there exists a set $E_{epsilon}in mathcal{A}$ s.t. $Esubset E_{epsilon}$ and $mu^*(E_{epsilon})lemu^*(E)+epsilon$ (assuming that $mu^*(E)<
infty$). Then
$$
mu^*(E)+epsilonge mu(E_{epsilon}cap A)+mu(E_{epsilon}cap A^c)ge mu^*(Ecap A)+mu^*(Ecap A^c)
$$
because $mu$ is additive on $mathcal{A}$.
answered Nov 18 '18 at 21:04
d.k.o.
8,537528
edited Nov 18 '18 at 22:58
answered Nov 18 '18 at 21:04
d.k.o.
8,537528
answered Nov 18 '18 at 21:04
d.k.o.
8,537528
answered Nov 18 '18 at 21:04
d.k.o.
8,537528
8,537528
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%2f3004070%2feach-set-in-mathcala-is-mu-measurable%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
the outer and 'inner' measure of $A$ coincides two
– Robson
Nov 18 '18 at 22:32