If $mu_Fllbeta_1$ and $mu_F$ is finite then $F$ is absolutely continuous
$begingroup$
If $mu_Fllbeta_1$ and $mu_F$ is finite then $F$ is absolutely continuous.
Note: here $beta_1$ is the Lebesgue measure restricted to the Borel $sigma$-algebra in the real line.
Im not sure if my solution is right, because I didnt used the fact that $mu_Fllbeta_1$, that is, that $beta_1(A)=0impliesmu_F(A)=0$ also.
Here $F(x):=int_{-infty}^x f(t), dt$ for some $finmathcal L_0(Bbb R,lambda_1,overline{Bbb R}^+)$, and $mu_F$ is the Lebesgue-Stieltjes measure generated by $F$ in $Bbb R$, so $mu_F([a,b))=F(b)-F(a)$. Hence
$$
mu_F(Bbb R)=mu_Fleft(bigcup_{kinBbb Z}[k,k+1)right)=sum_{kinBbb Z}int_k^{k+1}f(t), dt=int_{-infty}^infty f(t), dt<infty$$
so $finmathcal L_1(Bbb R,lambda_1,overline{Bbb R}^+)$, and by a previous result I know that if $finmathcal L_1(J,E)$, for some interval $JsubsetBbb R$ and $E$ some Banach space, then the function defined by
$$g(x):=int_{inf J}^x f(t), dt$$
is absolutely continuous, thus I can conclude that $F$ is also absolutely continuous. However I didnt used the fact that $mu_Fllbeta_1$, so it seems that something is wrong, where is my mistake?
real-analysis measure-theory lebesgue-integral absolute-continuity
asked Dec 10 '18 at 10:54
MasacrosoMasacroso
13.2k41747
$endgroup$
add a comment |
$begingroup$
If $mu_Fllbeta_1$ and $mu_F$ is finite then $F$ is absolutely continuous.
Note: here $beta_1$ is the Lebesgue measure restricted to the Borel $sigma$-algebra in the real line.
Im not sure if my solution is right, because I didnt used the fact that $mu_Fllbeta_1$, that is, that $beta_1(A)=0impliesmu_F(A)=0$ also.
Here $F(x):=int_{-infty}^x f(t), dt$ for some $finmathcal L_0(Bbb R,lambda_1,overline{Bbb R}^+)$, and $mu_F$ is the Lebesgue-Stieltjes measure generated by $F$ in $Bbb R$, so $mu_F([a,b))=F(b)-F(a)$. Hence
$$
mu_F(Bbb R)=mu_Fleft(bigcup_{kinBbb Z}[k,k+1)right)=sum_{kinBbb Z}int_k^{k+1}f(t), dt=int_{-infty}^infty f(t), dt<infty$$
so $finmathcal L_1(Bbb R,lambda_1,overline{Bbb R}^+)$, and by a previous result I know that if $finmathcal L_1(J,E)$, for some interval $JsubsetBbb R$ and $E$ some Banach space, then the function defined by
$$g(x):=int_{inf J}^x f(t), dt$$
is absolutely continuous, thus I can conclude that $F$ is also absolutely continuous. However I didnt used the fact that $mu_Fllbeta_1$, so it seems that something is wrong, where is my mistake?
real-analysis measure-theory lebesgue-integral absolute-continuity
asked Dec 10 '18 at 10:54
MasacrosoMasacroso
13.2k41747
$endgroup$
$begingroup$
You can write $F(x)=int_{-infty} ^{x} f(t)dt$ for some integrable function $f$ precisely when $mu_F << beta_1$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 11:59
$begingroup$
@Kavi the function $F$ is defined as $F(x):=int_{-infty}^x f(t), dt$ for some measurable $f$. If $mu_F$ is a finite measure then I showed that $f$ is integrable, and I didnt used the fact that $mu_Fllbeta_1$. There is something wrong there?
$endgroup$
– Masacroso
Dec 10 '18 at 12:13
1
$begingroup$
This question makes sense only of you drop the assumption that $F$ is an indefinite integral of an integrable function. You have to use just the equation $mu_F(x)=F(x)$ for all $x$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 12:16
$begingroup$
@Kavi I see... It seems that there is a mistake in the exercise or a flaw in it redaction
$endgroup$
– Masacroso
Dec 10 '18 at 12:20
add a comment |
$begingroup$
If $mu_Fllbeta_1$ and $mu_F$ is finite then $F$ is absolutely continuous.
Note: here $beta_1$ is the Lebesgue measure restricted to the Borel $sigma$-algebra in the real line.
Im not sure if my solution is right, because I didnt used the fact that $mu_Fllbeta_1$, that is, that $beta_1(A)=0impliesmu_F(A)=0$ also.
Here $F(x):=int_{-infty}^x f(t), dt$ for some $finmathcal L_0(Bbb R,lambda_1,overline{Bbb R}^+)$, and $mu_F$ is the Lebesgue-Stieltjes measure generated by $F$ in $Bbb R$, so $mu_F([a,b))=F(b)-F(a)$. Hence
$$
mu_F(Bbb R)=mu_Fleft(bigcup_{kinBbb Z}[k,k+1)right)=sum_{kinBbb Z}int_k^{k+1}f(t), dt=int_{-infty}^infty f(t), dt<infty$$
so $finmathcal L_1(Bbb R,lambda_1,overline{Bbb R}^+)$, and by a previous result I know that if $finmathcal L_1(J,E)$, for some interval $JsubsetBbb R$ and $E$ some Banach space, then the function defined by
$$g(x):=int_{inf J}^x f(t), dt$$
is absolutely continuous, thus I can conclude that $F$ is also absolutely continuous. However I didnt used the fact that $mu_Fllbeta_1$, so it seems that something is wrong, where is my mistake?
real-analysis measure-theory lebesgue-integral absolute-continuity
asked Dec 10 '18 at 10:54
MasacrosoMasacroso
13.2k41747
$endgroup$
If $mu_Fllbeta_1$ and $mu_F$ is finite then $F$ is absolutely continuous.
Note: here $beta_1$ is the Lebesgue measure restricted to the Borel $sigma$-algebra in the real line.
Im not sure if my solution is right, because I didnt used the fact that $mu_Fllbeta_1$, that is, that $beta_1(A)=0impliesmu_F(A)=0$ also.
Here $F(x):=int_{-infty}^x f(t), dt$ for some $finmathcal L_0(Bbb R,lambda_1,overline{Bbb R}^+)$, and $mu_F$ is the Lebesgue-Stieltjes measure generated by $F$ in $Bbb R$, so $mu_F([a,b))=F(b)-F(a)$. Hence
$$
mu_F(Bbb R)=mu_Fleft(bigcup_{kinBbb Z}[k,k+1)right)=sum_{kinBbb Z}int_k^{k+1}f(t), dt=int_{-infty}^infty f(t), dt<infty$$
so $finmathcal L_1(Bbb R,lambda_1,overline{Bbb R}^+)$, and by a previous result I know that if $finmathcal L_1(J,E)$, for some interval $JsubsetBbb R$ and $E$ some Banach space, then the function defined by
$$g(x):=int_{inf J}^x f(t), dt$$
is absolutely continuous, thus I can conclude that $F$ is also absolutely continuous. However I didnt used the fact that $mu_Fllbeta_1$, so it seems that something is wrong, where is my mistake?
real-analysis measure-theory lebesgue-integral absolute-continuity
real-analysis measure-theory lebesgue-integral absolute-continuity
asked Dec 10 '18 at 10:54
MasacrosoMasacroso
13.2k41747
asked Dec 10 '18 at 10:54
MasacrosoMasacroso
13.2k41747
edited Dec 10 '18 at 11:10
Masacroso
asked Dec 10 '18 at 10:54
MasacrosoMasacroso
13.2k41747
asked Dec 10 '18 at 10:54
MasacrosoMasacroso
13.2k41747
asked Dec 10 '18 at 10:54
MasacrosoMasacroso
13.2k41747
13.2k41747
$begingroup$
You can write $F(x)=int_{-infty} ^{x} f(t)dt$ for some integrable function $f$ precisely when $mu_F << beta_1$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 11:59
$begingroup$
@Kavi the function $F$ is defined as $F(x):=int_{-infty}^x f(t), dt$ for some measurable $f$. If $mu_F$ is a finite measure then I showed that $f$ is integrable, and I didnt used the fact that $mu_Fllbeta_1$. There is something wrong there?
$endgroup$
– Masacroso
Dec 10 '18 at 12:13
1
$begingroup$
This question makes sense only of you drop the assumption that $F$ is an indefinite integral of an integrable function. You have to use just the equation $mu_F(x)=F(x)$ for all $x$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 12:16
$begingroup$
@Kavi I see... It seems that there is a mistake in the exercise or a flaw in it redaction
$endgroup$
– Masacroso
Dec 10 '18 at 12:20
add a comment |
$begingroup$
You can write $F(x)=int_{-infty} ^{x} f(t)dt$ for some integrable function $f$ precisely when $mu_F << beta_1$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 11:59
$begingroup$
@Kavi the function $F$ is defined as $F(x):=int_{-infty}^x f(t), dt$ for some measurable $f$. If $mu_F$ is a finite measure then I showed that $f$ is integrable, and I didnt used the fact that $mu_Fllbeta_1$. There is something wrong there?
$endgroup$
– Masacroso
Dec 10 '18 at 12:13
1
$begingroup$
This question makes sense only of you drop the assumption that $F$ is an indefinite integral of an integrable function. You have to use just the equation $mu_F(x)=F(x)$ for all $x$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 12:16
$begingroup$
@Kavi I see... It seems that there is a mistake in the exercise or a flaw in it redaction
$endgroup$
– Masacroso
Dec 10 '18 at 12:20
$begingroup$
You can write $F(x)=int_{-infty} ^{x} f(t)dt$ for some integrable function $f$ precisely when $mu_F << beta_1$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 11:59
$begingroup$
You can write $F(x)=int_{-infty} ^{x} f(t)dt$ for some integrable function $f$ precisely when $mu_F << beta_1$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 11:59
$begingroup$
@Kavi the function $F$ is defined as $F(x):=int_{-infty}^x f(t), dt$ for some measurable $f$. If $mu_F$ is a finite measure then I showed that $f$ is integrable, and I didnt used the fact that $mu_Fllbeta_1$. There is something wrong there?
$endgroup$
– Masacroso
Dec 10 '18 at 12:13
$begingroup$
@Kavi the function $F$ is defined as $F(x):=int_{-infty}^x f(t), dt$ for some measurable $f$. If $mu_F$ is a finite measure then I showed that $f$ is integrable, and I didnt used the fact that $mu_Fllbeta_1$. There is something wrong there?
$endgroup$
– Masacroso
Dec 10 '18 at 12:13
1
1
$begingroup$
This question makes sense only of you drop the assumption that $F$ is an indefinite integral of an integrable function. You have to use just the equation $mu_F(x)=F(x)$ for all $x$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 12:16
$begingroup$
This question makes sense only of you drop the assumption that $F$ is an indefinite integral of an integrable function. You have to use just the equation $mu_F(x)=F(x)$ for all $x$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 12:16
$begingroup$
@Kavi I see... It seems that there is a mistake in the exercise or a flaw in it redaction
$endgroup$
– Masacroso
Dec 10 '18 at 12:20
$begingroup$
@Kavi I see... It seems that there is a mistake in the exercise or a flaw in it redaction
$endgroup$
– Masacroso
Dec 10 '18 at 12:20
add a 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%2f3033768%2fif-mu-f-ll-beta-1-and-mu-f-is-finite-then-f-is-absolutely-continuous%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%2f3033768%2fif-mu-f-ll-beta-1-and-mu-f-is-finite-then-f-is-absolutely-continuous%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
$begingroup$
You can write $F(x)=int_{-infty} ^{x} f(t)dt$ for some integrable function $f$ precisely when $mu_F << beta_1$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 11:59
$begingroup$
@Kavi the function $F$ is defined as $F(x):=int_{-infty}^x f(t), dt$ for some measurable $f$. If $mu_F$ is a finite measure then I showed that $f$ is integrable, and I didnt used the fact that $mu_Fllbeta_1$. There is something wrong there?
$endgroup$
– Masacroso
Dec 10 '18 at 12:13
1
$begingroup$
This question makes sense only of you drop the assumption that $F$ is an indefinite integral of an integrable function. You have to use just the equation $mu_F(x)=F(x)$ for all $x$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 12:16
$begingroup$
@Kavi I see... It seems that there is a mistake in the exercise or a flaw in it redaction
$endgroup$
– Masacroso
Dec 10 '18 at 12:20