If $mu_Fllbeta_1$ and $mu_F$ is finite then $F$ is absolutely continuous












0












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 11:10







Masacroso

















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


















  • $begingroup$
    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










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