If $f$ derivable on $[a,b]$ does $int_a^t f'(x)dx=f(t)-f(a)$ true?

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Let $f:[a,b]longrightarrow mathbb R$ a derivable function. Is it true that for all $tin [a,b]$ we have that $$f(t)=f(a)+int_a^t f'(x)dx ?$$



The thing is since $f'$ is not supposed continuous, there is no reason for me for $f'$ to be Riemann integrable. So my questions are the followings one :



  • Q1) In Riemann sense, is the formula correct (if we don't have other hypothesis on $f'$). If no, do you have a counter example ?


  • Q2) If we assume $f'$ Riemann integrable, is the formula correct (in Riemann sense). If no, do you have a counter example ?


  • Q3) In Lebesgue sense, is the formula correct (if we don't have other hypothesis on $f'$). If no, do you have a counter example ?


  • Q4) If we assume $f'$ Lebesgue integrable, is the formula correct (in Lebesgue sense). If no, do you have a counter example ?




real-analysis integration lebesgue-integral riemann-integration




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  • See for Q2: math.stackexchange.com/questions/1899567/…
    – Zeekless
    Jul 17 at 16:50














up vote
10
down vote

favorite
4












Let $f:[a,b]longrightarrow mathbb R$ a derivable function. Is it true that for all $tin [a,b]$ we have that $$f(t)=f(a)+int_a^t f'(x)dx ?$$



The thing is since $f'$ is not supposed continuous, there is no reason for me for $f'$ to be Riemann integrable. So my questions are the followings one :



  • Q1) In Riemann sense, is the formula correct (if we don't have other hypothesis on $f'$). If no, do you have a counter example ?


  • Q2) If we assume $f'$ Riemann integrable, is the formula correct (in Riemann sense). If no, do you have a counter example ?


  • Q3) In Lebesgue sense, is the formula correct (if we don't have other hypothesis on $f'$). If no, do you have a counter example ?


  • Q4) If we assume $f'$ Lebesgue integrable, is the formula correct (in Lebesgue sense). If no, do you have a counter example ?




real-analysis integration lebesgue-integral riemann-integration




share|cite|improve this question



















  • See for Q2: math.stackexchange.com/questions/1899567/…
    – Zeekless
    Jul 17 at 16:50












up vote
10
down vote

favorite
4









up vote
10
down vote

favorite
4






4





Let $f:[a,b]longrightarrow mathbb R$ a derivable function. Is it true that for all $tin [a,b]$ we have that $$f(t)=f(a)+int_a^t f'(x)dx ?$$



The thing is since $f'$ is not supposed continuous, there is no reason for me for $f'$ to be Riemann integrable. So my questions are the followings one :



  • Q1) In Riemann sense, is the formula correct (if we don't have other hypothesis on $f'$). If no, do you have a counter example ?


  • Q2) If we assume $f'$ Riemann integrable, is the formula correct (in Riemann sense). If no, do you have a counter example ?


  • Q3) In Lebesgue sense, is the formula correct (if we don't have other hypothesis on $f'$). If no, do you have a counter example ?


  • Q4) If we assume $f'$ Lebesgue integrable, is the formula correct (in Lebesgue sense). If no, do you have a counter example ?




real-analysis integration lebesgue-integral riemann-integration




share|cite|improve this question











Let $f:[a,b]longrightarrow mathbb R$ a derivable function. Is it true that for all $tin [a,b]$ we have that $$f(t)=f(a)+int_a^t f'(x)dx ?$$



The thing is since $f'$ is not supposed continuous, there is no reason for me for $f'$ to be Riemann integrable. So my questions are the followings one :



  • Q1) In Riemann sense, is the formula correct (if we don't have other hypothesis on $f'$). If no, do you have a counter example ?


  • Q2) If we assume $f'$ Riemann integrable, is the formula correct (in Riemann sense). If no, do you have a counter example ?


  • Q3) In Lebesgue sense, is the formula correct (if we don't have other hypothesis on $f'$). If no, do you have a counter example ?


  • Q4) If we assume $f'$ Lebesgue integrable, is the formula correct (in Lebesgue sense). If no, do you have a counter example ?





real-analysis integration lebesgue-integral riemann-integration





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asked Jul 17 at 16:27









Peter

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  • See for Q2: math.stackexchange.com/questions/1899567/…
    – Zeekless
    Jul 17 at 16:50
















  • See for Q2: math.stackexchange.com/questions/1899567/…
    – Zeekless
    Jul 17 at 16:50















See for Q2: math.stackexchange.com/questions/1899567/…
– Zeekless
Jul 17 at 16:50




See for Q2: math.stackexchange.com/questions/1899567/…
– Zeekless
Jul 17 at 16:50










1 Answer
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7
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Q1: No, because $f'$ need not be Riemann integrable. For example $f'$ need not be bounded.



Q2: Yes. This is in fact very simple - the elegance of the proof seems to me to be a good reason for the Riemann integral to be defined exactly the way it is. Say $a=t_0<dots<t_n=b$. Apply the Mean Value Theorem to each subinterval: $$f(b)-f(a)=sum(f(t_j+1)-f(t_j))=sum f'(xi_j)(t_j+1-t_j),$$precisely a Riemann sum for $int_a^b f'(t),dt$.



Q3: No, $f'$ need not be Lebesgue integrable. For example if $a=-1$, $b=1$ and $$f(t)=begincasest^2sin(1/t^100),&(tne0),
\0,&(t=0).endcases$$



Q4: Pretty sure the answer is yes. Almost certain this is a theorem in Rudin - my copy of Real and Complex Analysis is missing. Edit: Thanks to @ChrisJanjigian for confirming that yes, it's Theorem 7.21.
Or see here for a proof based on the Vitali-Caratheodory theorem.



Note this is assuming that $f$ is differentiable, which is to say that $f'$ exists at every point. There are certainly examples where $f$ is differentiable almost everywhere, $f'in L^1$, but $f$ is not absolutely continuous; for example the Cantor function satisfies $f'=0$ almost everywhere.






share|cite|improve this answer























  • Thank you for your answer, and for reference. @uniquesolution: thank for Q4 :)
    – Peter
    Jul 17 at 16:52






  • 1




    Q4 is Theorem 7.21 in Rudin.
    – Chris Janjigian
    Jul 17 at 16:56










  • Yes, my mistake.
    – uniquesolution
    Jul 17 at 16:57










  • @ChrisJanjigian Thanks - I knew it was in there somewhere.
    – David C. Ullrich
    Jul 17 at 17:02






  • 1




    I once tried to simply Rudin's proof of Theorem 7.21. You may check it on my blog posting.
    – Sangchul Lee
    Jul 17 at 20:36










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1 Answer
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active

oldest

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oldest

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oldest

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up vote
7
down vote



accepted










Q1: No, because $f'$ need not be Riemann integrable. For example $f'$ need not be bounded.



Q2: Yes. This is in fact very simple - the elegance of the proof seems to me to be a good reason for the Riemann integral to be defined exactly the way it is. Say $a=t_0<dots<t_n=b$. Apply the Mean Value Theorem to each subinterval: $$f(b)-f(a)=sum(f(t_j+1)-f(t_j))=sum f'(xi_j)(t_j+1-t_j),$$precisely a Riemann sum for $int_a^b f'(t),dt$.



Q3: No, $f'$ need not be Lebesgue integrable. For example if $a=-1$, $b=1$ and $$f(t)=begincasest^2sin(1/t^100),&(tne0),
\0,&(t=0).endcases$$



Q4: Pretty sure the answer is yes. Almost certain this is a theorem in Rudin - my copy of Real and Complex Analysis is missing. Edit: Thanks to @ChrisJanjigian for confirming that yes, it's Theorem 7.21.
Or see here for a proof based on the Vitali-Caratheodory theorem.



Note this is assuming that $f$ is differentiable, which is to say that $f'$ exists at every point. There are certainly examples where $f$ is differentiable almost everywhere, $f'in L^1$, but $f$ is not absolutely continuous; for example the Cantor function satisfies $f'=0$ almost everywhere.






share|cite|improve this answer























  • Thank you for your answer, and for reference. @uniquesolution: thank for Q4 :)
    – Peter
    Jul 17 at 16:52






  • 1




    Q4 is Theorem 7.21 in Rudin.
    – Chris Janjigian
    Jul 17 at 16:56










  • Yes, my mistake.
    – uniquesolution
    Jul 17 at 16:57










  • @ChrisJanjigian Thanks - I knew it was in there somewhere.
    – David C. Ullrich
    Jul 17 at 17:02






  • 1




    I once tried to simply Rudin's proof of Theorem 7.21. You may check it on my blog posting.
    – Sangchul Lee
    Jul 17 at 20:36














up vote
7
down vote



accepted










Q1: No, because $f'$ need not be Riemann integrable. For example $f'$ need not be bounded.



Q2: Yes. This is in fact very simple - the elegance of the proof seems to me to be a good reason for the Riemann integral to be defined exactly the way it is. Say $a=t_0<dots<t_n=b$. Apply the Mean Value Theorem to each subinterval: $$f(b)-f(a)=sum(f(t_j+1)-f(t_j))=sum f'(xi_j)(t_j+1-t_j),$$precisely a Riemann sum for $int_a^b f'(t),dt$.



Q3: No, $f'$ need not be Lebesgue integrable. For example if $a=-1$, $b=1$ and $$f(t)=begincasest^2sin(1/t^100),&(tne0),
\0,&(t=0).endcases$$



Q4: Pretty sure the answer is yes. Almost certain this is a theorem in Rudin - my copy of Real and Complex Analysis is missing. Edit: Thanks to @ChrisJanjigian for confirming that yes, it's Theorem 7.21.
Or see here for a proof based on the Vitali-Caratheodory theorem.



Note this is assuming that $f$ is differentiable, which is to say that $f'$ exists at every point. There are certainly examples where $f$ is differentiable almost everywhere, $f'in L^1$, but $f$ is not absolutely continuous; for example the Cantor function satisfies $f'=0$ almost everywhere.






share|cite|improve this answer























  • Thank you for your answer, and for reference. @uniquesolution: thank for Q4 :)
    – Peter
    Jul 17 at 16:52






  • 1




    Q4 is Theorem 7.21 in Rudin.
    – Chris Janjigian
    Jul 17 at 16:56










  • Yes, my mistake.
    – uniquesolution
    Jul 17 at 16:57










  • @ChrisJanjigian Thanks - I knew it was in there somewhere.
    – David C. Ullrich
    Jul 17 at 17:02






  • 1




    I once tried to simply Rudin's proof of Theorem 7.21. You may check it on my blog posting.
    – Sangchul Lee
    Jul 17 at 20:36












up vote
7
down vote



accepted







up vote
7
down vote



accepted






Q1: No, because $f'$ need not be Riemann integrable. For example $f'$ need not be bounded.



Q2: Yes. This is in fact very simple - the elegance of the proof seems to me to be a good reason for the Riemann integral to be defined exactly the way it is. Say $a=t_0<dots<t_n=b$. Apply the Mean Value Theorem to each subinterval: $$f(b)-f(a)=sum(f(t_j+1)-f(t_j))=sum f'(xi_j)(t_j+1-t_j),$$precisely a Riemann sum for $int_a^b f'(t),dt$.



Q3: No, $f'$ need not be Lebesgue integrable. For example if $a=-1$, $b=1$ and $$f(t)=begincasest^2sin(1/t^100),&(tne0),
\0,&(t=0).endcases$$



Q4: Pretty sure the answer is yes. Almost certain this is a theorem in Rudin - my copy of Real and Complex Analysis is missing. Edit: Thanks to @ChrisJanjigian for confirming that yes, it's Theorem 7.21.
Or see here for a proof based on the Vitali-Caratheodory theorem.



Note this is assuming that $f$ is differentiable, which is to say that $f'$ exists at every point. There are certainly examples where $f$ is differentiable almost everywhere, $f'in L^1$, but $f$ is not absolutely continuous; for example the Cantor function satisfies $f'=0$ almost everywhere.






share|cite|improve this answer















Q1: No, because $f'$ need not be Riemann integrable. For example $f'$ need not be bounded.



Q2: Yes. This is in fact very simple - the elegance of the proof seems to me to be a good reason for the Riemann integral to be defined exactly the way it is. Say $a=t_0<dots<t_n=b$. Apply the Mean Value Theorem to each subinterval: $$f(b)-f(a)=sum(f(t_j+1)-f(t_j))=sum f'(xi_j)(t_j+1-t_j),$$precisely a Riemann sum for $int_a^b f'(t),dt$.



Q3: No, $f'$ need not be Lebesgue integrable. For example if $a=-1$, $b=1$ and $$f(t)=begincasest^2sin(1/t^100),&(tne0),
\0,&(t=0).endcases$$



Q4: Pretty sure the answer is yes. Almost certain this is a theorem in Rudin - my copy of Real and Complex Analysis is missing. Edit: Thanks to @ChrisJanjigian for confirming that yes, it's Theorem 7.21.
Or see here for a proof based on the Vitali-Caratheodory theorem.



Note this is assuming that $f$ is differentiable, which is to say that $f'$ exists at every point. There are certainly examples where $f$ is differentiable almost everywhere, $f'in L^1$, but $f$ is not absolutely continuous; for example the Cantor function satisfies $f'=0$ almost everywhere.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 17 at 22:37


























answered Jul 17 at 16:50









David C. Ullrich

54.3k33583




54.3k33583











  • Thank you for your answer, and for reference. @uniquesolution: thank for Q4 :)
    – Peter
    Jul 17 at 16:52






  • 1




    Q4 is Theorem 7.21 in Rudin.
    – Chris Janjigian
    Jul 17 at 16:56










  • Yes, my mistake.
    – uniquesolution
    Jul 17 at 16:57










  • @ChrisJanjigian Thanks - I knew it was in there somewhere.
    – David C. Ullrich
    Jul 17 at 17:02






  • 1




    I once tried to simply Rudin's proof of Theorem 7.21. You may check it on my blog posting.
    – Sangchul Lee
    Jul 17 at 20:36
















  • Thank you for your answer, and for reference. @uniquesolution: thank for Q4 :)
    – Peter
    Jul 17 at 16:52






  • 1




    Q4 is Theorem 7.21 in Rudin.
    – Chris Janjigian
    Jul 17 at 16:56










  • Yes, my mistake.
    – uniquesolution
    Jul 17 at 16:57










  • @ChrisJanjigian Thanks - I knew it was in there somewhere.
    – David C. Ullrich
    Jul 17 at 17:02






  • 1




    I once tried to simply Rudin's proof of Theorem 7.21. You may check it on my blog posting.
    – Sangchul Lee
    Jul 17 at 20:36















Thank you for your answer, and for reference. @uniquesolution: thank for Q4 :)
– Peter
Jul 17 at 16:52




Thank you for your answer, and for reference. @uniquesolution: thank for Q4 :)
– Peter
Jul 17 at 16:52




1




1




Q4 is Theorem 7.21 in Rudin.
– Chris Janjigian
Jul 17 at 16:56




Q4 is Theorem 7.21 in Rudin.
– Chris Janjigian
Jul 17 at 16:56












Yes, my mistake.
– uniquesolution
Jul 17 at 16:57




Yes, my mistake.
– uniquesolution
Jul 17 at 16:57












@ChrisJanjigian Thanks - I knew it was in there somewhere.
– David C. Ullrich
Jul 17 at 17:02




@ChrisJanjigian Thanks - I knew it was in there somewhere.
– David C. Ullrich
Jul 17 at 17:02




1




1




I once tried to simply Rudin's proof of Theorem 7.21. You may check it on my blog posting.
– Sangchul Lee
Jul 17 at 20:36




I once tried to simply Rudin's proof of Theorem 7.21. You may check it on my blog posting.
– Sangchul Lee
Jul 17 at 20:36












 

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