I need help correcting my real analysis proof
$begingroup$
Let $f$ be a measurable function on $E$ that is finite a.e. on $E$ and $m^* (E) < infty$. For each $epsilon > 0$, show that there is a measurable set $F$ contained in $E$ such that $f$ is bounded on $F$ and $m^* (Esetminus F) < epsilon$. What is $F$? I know $F subset E$ do I have to say that in my proof?
Let $Esetminus F = E cap F^c$ which implies $m^*(Esetminus F) = m^*(E cap F^c)$.
I know $m^*(F) = m^*( Fcap E) + m^*(Fsetminus E^c)$ but I have
$m^*(F) = m^*(Ecap F^c ) + m^*(Fcap E)$. I don’t see where this came from.
Since $m^*(E) < infty$ that implies $m^*( Ecap F^c) = m^*(F) – m^*( Fcap E)$.
I will create $I_n$ a countable cover for $E$, where $I_n=(a_n,b_n)$.
I’m going to create a closed set for F so the function f will be bounded on it.
$$ F_n=[a_n,b_n- epsilon /2^n ] $$ so let $$ F = bigcup F_n. $$
Using the definition of outer measure I get $$ m^*(F) = sum l(F_n )=sum (b)_n- a_n-epsilon/2^n )= sum(b_n-a_n-epsilon).$$
Since $Fsubset E$, it is easy to see that $m^*(Ecap F)=m^*(F)$ so $$m^*(Ecap F) = m^*(F) leq m^*(I_n )=sum (b_n-a_n)$$ which implies $$m^*(Ecap F) leq sum l(I_n )- sum l(F_n )=sum (b_n-a_n )-sum (b_n-a_n-epsilon)=epsilon.$$
Thus there is an $ F$ such that $Fsubset E$ and $F$ is closed so $f$ is bounded on it and $m^*(Esetminus F) leq epsilon$.
Nothing in your work has included f. I do not see why suddenly f should be bounded.
real-analysis
edited Dec 13 '18 at 5:58
Fiori_Modena
225
asked Oct 11 '15 at 18:08
TessTess
113
$endgroup$
add a comment |
$begingroup$
Let $f$ be a measurable function on $E$ that is finite a.e. on $E$ and $m^* (E) < infty$. For each $epsilon > 0$, show that there is a measurable set $F$ contained in $E$ such that $f$ is bounded on $F$ and $m^* (Esetminus F) < epsilon$. What is $F$? I know $F subset E$ do I have to say that in my proof?
Let $Esetminus F = E cap F^c$ which implies $m^*(Esetminus F) = m^*(E cap F^c)$.
I know $m^*(F) = m^*( Fcap E) + m^*(Fsetminus E^c)$ but I have
$m^*(F) = m^*(Ecap F^c ) + m^*(Fcap E)$. I don’t see where this came from.
Since $m^*(E) < infty$ that implies $m^*( Ecap F^c) = m^*(F) – m^*( Fcap E)$.
I will create $I_n$ a countable cover for $E$, where $I_n=(a_n,b_n)$.
I’m going to create a closed set for F so the function f will be bounded on it.
$$ F_n=[a_n,b_n- epsilon /2^n ] $$ so let $$ F = bigcup F_n. $$
Using the definition of outer measure I get $$ m^*(F) = sum l(F_n )=sum (b)_n- a_n-epsilon/2^n )= sum(b_n-a_n-epsilon).$$
Since $Fsubset E$, it is easy to see that $m^*(Ecap F)=m^*(F)$ so $$m^*(Ecap F) = m^*(F) leq m^*(I_n )=sum (b_n-a_n)$$ which implies $$m^*(Ecap F) leq sum l(I_n )- sum l(F_n )=sum (b_n-a_n )-sum (b_n-a_n-epsilon)=epsilon.$$
Thus there is an $ F$ such that $Fsubset E$ and $F$ is closed so $f$ is bounded on it and $m^*(Esetminus F) leq epsilon$.
Nothing in your work has included f. I do not see why suddenly f should be bounded.
real-analysis
edited Dec 13 '18 at 5:58
Fiori_Modena
225
asked Oct 11 '15 at 18:08
TessTess
113
$endgroup$
add a comment |
$begingroup$
Let $f$ be a measurable function on $E$ that is finite a.e. on $E$ and $m^* (E) < infty$. For each $epsilon > 0$, show that there is a measurable set $F$ contained in $E$ such that $f$ is bounded on $F$ and $m^* (Esetminus F) < epsilon$. What is $F$? I know $F subset E$ do I have to say that in my proof?
Let $Esetminus F = E cap F^c$ which implies $m^*(Esetminus F) = m^*(E cap F^c)$.
I know $m^*(F) = m^*( Fcap E) + m^*(Fsetminus E^c)$ but I have
$m^*(F) = m^*(Ecap F^c ) + m^*(Fcap E)$. I don’t see where this came from.
Since $m^*(E) < infty$ that implies $m^*( Ecap F^c) = m^*(F) – m^*( Fcap E)$.
I will create $I_n$ a countable cover for $E$, where $I_n=(a_n,b_n)$.
I’m going to create a closed set for F so the function f will be bounded on it.
$$ F_n=[a_n,b_n- epsilon /2^n ] $$ so let $$ F = bigcup F_n. $$
Using the definition of outer measure I get $$ m^*(F) = sum l(F_n )=sum (b)_n- a_n-epsilon/2^n )= sum(b_n-a_n-epsilon).$$
Since $Fsubset E$, it is easy to see that $m^*(Ecap F)=m^*(F)$ so $$m^*(Ecap F) = m^*(F) leq m^*(I_n )=sum (b_n-a_n)$$ which implies $$m^*(Ecap F) leq sum l(I_n )- sum l(F_n )=sum (b_n-a_n )-sum (b_n-a_n-epsilon)=epsilon.$$
Thus there is an $ F$ such that $Fsubset E$ and $F$ is closed so $f$ is bounded on it and $m^*(Esetminus F) leq epsilon$.
Nothing in your work has included f. I do not see why suddenly f should be bounded.
real-analysis
edited Dec 13 '18 at 5:58
Fiori_Modena
225
asked Oct 11 '15 at 18:08
TessTess
113
$endgroup$
Let $f$ be a measurable function on $E$ that is finite a.e. on $E$ and $m^* (E) < infty$. For each $epsilon > 0$, show that there is a measurable set $F$ contained in $E$ such that $f$ is bounded on $F$ and $m^* (Esetminus F) < epsilon$. What is $F$? I know $F subset E$ do I have to say that in my proof?
Let $Esetminus F = E cap F^c$ which implies $m^*(Esetminus F) = m^*(E cap F^c)$.
I know $m^*(F) = m^*( Fcap E) + m^*(Fsetminus E^c)$ but I have
$m^*(F) = m^*(Ecap F^c ) + m^*(Fcap E)$. I don’t see where this came from.
Since $m^*(E) < infty$ that implies $m^*( Ecap F^c) = m^*(F) – m^*( Fcap E)$.
I will create $I_n$ a countable cover for $E$, where $I_n=(a_n,b_n)$.
I’m going to create a closed set for F so the function f will be bounded on it.
$$ F_n=[a_n,b_n- epsilon /2^n ] $$ so let $$ F = bigcup F_n. $$
Using the definition of outer measure I get $$ m^*(F) = sum l(F_n )=sum (b)_n- a_n-epsilon/2^n )= sum(b_n-a_n-epsilon).$$
Since $Fsubset E$, it is easy to see that $m^*(Ecap F)=m^*(F)$ so $$m^*(Ecap F) = m^*(F) leq m^*(I_n )=sum (b_n-a_n)$$ which implies $$m^*(Ecap F) leq sum l(I_n )- sum l(F_n )=sum (b_n-a_n )-sum (b_n-a_n-epsilon)=epsilon.$$
Thus there is an $ F$ such that $Fsubset E$ and $F$ is closed so $f$ is bounded on it and $m^*(Esetminus F) leq epsilon$.
Nothing in your work has included f. I do not see why suddenly f should be bounded.
real-analysis
real-analysis
edited Dec 13 '18 at 5:58
Fiori_Modena
225
asked Oct 11 '15 at 18:08
TessTess
113
edited Dec 13 '18 at 5:58
Fiori_Modena
225
asked Oct 11 '15 at 18:08
TessTess
113
edited Dec 13 '18 at 5:58
Fiori_Modena
225
edited Dec 13 '18 at 5:58
Fiori_Modena
225
edited Dec 13 '18 at 5:58
Fiori_Modena
225
225
asked Oct 11 '15 at 18:08
TessTess
113
asked Oct 11 '15 at 18:08
TessTess
113
asked Oct 11 '15 at 18:08
TessTess
113
113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You need to construct such an $F$ with the properties outlined. The set $A_n={xin E:|f(x)|>n}$ is an open set with $A_nsubseteq E$. Since f finite a.e. and $m^*(E)<infty$, there exists a $s$ such that $m^*(A_s)<varepsilon$. Take $F=Esetminus A_s$.
$F$ is closed and $Fsubseteq E$ with $m^*(Esetminus F)< varepsilon$. Also, $|f(x)|leq s$ on $F$ by definition of $A_s$.
answered Oct 11 '15 at 18:29
Konrad WrobelKonrad Wrobel
419212
$endgroup$
$begingroup$
I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
$endgroup$
– user146269
Nov 4 '15 at 19:10
$begingroup$
and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
$endgroup$
– user146269
Nov 4 '15 at 19:15
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%2f1475117%2fi-need-help-correcting-my-real-analysis-proof%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
$begingroup$
You need to construct such an $F$ with the properties outlined. The set $A_n={xin E:|f(x)|>n}$ is an open set with $A_nsubseteq E$. Since f finite a.e. and $m^*(E)<infty$, there exists a $s$ such that $m^*(A_s)<varepsilon$. Take $F=Esetminus A_s$.
$F$ is closed and $Fsubseteq E$ with $m^*(Esetminus F)< varepsilon$. Also, $|f(x)|leq s$ on $F$ by definition of $A_s$.
answered Oct 11 '15 at 18:29
Konrad WrobelKonrad Wrobel
419212
$endgroup$
$begingroup$
I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
$endgroup$
– user146269
Nov 4 '15 at 19:10
$begingroup$
and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
$endgroup$
– user146269
Nov 4 '15 at 19:15
add a comment |
$begingroup$
You need to construct such an $F$ with the properties outlined. The set $A_n={xin E:|f(x)|>n}$ is an open set with $A_nsubseteq E$. Since f finite a.e. and $m^*(E)<infty$, there exists a $s$ such that $m^*(A_s)<varepsilon$. Take $F=Esetminus A_s$.
$F$ is closed and $Fsubseteq E$ with $m^*(Esetminus F)< varepsilon$. Also, $|f(x)|leq s$ on $F$ by definition of $A_s$.
answered Oct 11 '15 at 18:29
Konrad WrobelKonrad Wrobel
419212
$endgroup$
$begingroup$
I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
$endgroup$
– user146269
Nov 4 '15 at 19:10
$begingroup$
and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
$endgroup$
– user146269
Nov 4 '15 at 19:15
add a comment |
$begingroup$
You need to construct such an $F$ with the properties outlined. The set $A_n={xin E:|f(x)|>n}$ is an open set with $A_nsubseteq E$. Since f finite a.e. and $m^*(E)<infty$, there exists a $s$ such that $m^*(A_s)<varepsilon$. Take $F=Esetminus A_s$.
$F$ is closed and $Fsubseteq E$ with $m^*(Esetminus F)< varepsilon$. Also, $|f(x)|leq s$ on $F$ by definition of $A_s$.
answered Oct 11 '15 at 18:29
Konrad WrobelKonrad Wrobel
419212
$endgroup$
You need to construct such an $F$ with the properties outlined. The set $A_n={xin E:|f(x)|>n}$ is an open set with $A_nsubseteq E$. Since f finite a.e. and $m^*(E)<infty$, there exists a $s$ such that $m^*(A_s)<varepsilon$. Take $F=Esetminus A_s$.
$F$ is closed and $Fsubseteq E$ with $m^*(Esetminus F)< varepsilon$. Also, $|f(x)|leq s$ on $F$ by definition of $A_s$.
answered Oct 11 '15 at 18:29
Konrad WrobelKonrad Wrobel
419212
answered Oct 11 '15 at 18:29
Konrad WrobelKonrad Wrobel
419212
answered Oct 11 '15 at 18:29
Konrad WrobelKonrad Wrobel
419212
answered Oct 11 '15 at 18:29
Konrad WrobelKonrad Wrobel
419212
419212
$begingroup$
I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
$endgroup$
– user146269
Nov 4 '15 at 19:10
$begingroup$
and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
$endgroup$
– user146269
Nov 4 '15 at 19:15
add a comment |
$begingroup$
I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
$endgroup$
– user146269
Nov 4 '15 at 19:10
$begingroup$
and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
$endgroup$
– user146269
Nov 4 '15 at 19:15
$begingroup$
I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
$endgroup$
– user146269
Nov 4 '15 at 19:10
$begingroup$
I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
$endgroup$
– user146269
Nov 4 '15 at 19:10
$begingroup$
and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
$endgroup$
– user146269
Nov 4 '15 at 19:15
$begingroup$
and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
$endgroup$
– user146269
Nov 4 '15 at 19:15
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%2f1475117%2fi-need-help-correcting-my-real-analysis-proof%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
