How does one prove that the definite integral of a function is unique?












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How would I go about proving that the definite integral is unique?



The problem statement is: Show that the number $I$ in the statement of Theorem 8.1 is unique; that is that there cannot be two numbers that would be assigned to the symbol $int_a^bf(x)dx.$



Theorem 8.1 (Cauchy) Let $f$ be continuous function on an interval [a,b]. Then there is a number $I$, called the definite integral of f on [a,b], such that for each $varepsilon>0$ there is a $delta > 0$ so that
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-IBigg|<varepsilon$$



Here is my attempt:
Proof by Contradiction:
Assume not; that is, assume there exists more than one $I,$ call them $I_1, I_2$, that are the definite integrals of $f$ (a continuous function on an interval $[a,b]$) on $[a,b]$ such that $forallvarepsilon>0, existsdelta>0$ such that
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_1Bigg|<varepsilon$$
and
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|<varepsilon$$
Without loss of generality, let $I_1<I_2$. Then
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_1Bigg|< Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|$$



Choose $varepsilon$ to be between $Bigg(Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|-Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_1Bigg|Bigg)/2$ and $0$. Now $varepsilon < Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|$



Thus we have a contradiction.



My issue with this proof is that couldn't one just find a different $delta$ that worked for $I_2$ if it didn't work with both $I_1$ and $I_2$? My proof doesn't use $delta$ at all.






real-analysis definite-integrals







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  • 2




    $begingroup$
    it is unique because is a limit. If the limit converges then it value is unique, by the properties of limits of sequences
    $endgroup$
    – Masacroso
    Dec 4 '18 at 22:51








  • 1




    $begingroup$
    The inequality on the line after $I_!lt I_2$ is not valid.
    $endgroup$
    – herb steinberg
    Dec 4 '18 at 22:54
















1












$begingroup$


How would I go about proving that the definite integral is unique?



The problem statement is: Show that the number $I$ in the statement of Theorem 8.1 is unique; that is that there cannot be two numbers that would be assigned to the symbol $int_a^bf(x)dx.$



Theorem 8.1 (Cauchy) Let $f$ be continuous function on an interval [a,b]. Then there is a number $I$, called the definite integral of f on [a,b], such that for each $varepsilon>0$ there is a $delta > 0$ so that
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-IBigg|<varepsilon$$



Here is my attempt:
Proof by Contradiction:
Assume not; that is, assume there exists more than one $I,$ call them $I_1, I_2$, that are the definite integrals of $f$ (a continuous function on an interval $[a,b]$) on $[a,b]$ such that $forallvarepsilon>0, existsdelta>0$ such that
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_1Bigg|<varepsilon$$
and
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|<varepsilon$$
Without loss of generality, let $I_1<I_2$. Then
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_1Bigg|< Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|$$



Choose $varepsilon$ to be between $Bigg(Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|-Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_1Bigg|Bigg)/2$ and $0$. Now $varepsilon < Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|$



Thus we have a contradiction.



My issue with this proof is that couldn't one just find a different $delta$ that worked for $I_2$ if it didn't work with both $I_1$ and $I_2$? My proof doesn't use $delta$ at all.






real-analysis definite-integrals







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    it is unique because is a limit. If the limit converges then it value is unique, by the properties of limits of sequences
    $endgroup$
    – Masacroso
    Dec 4 '18 at 22:51








  • 1




    $begingroup$
    The inequality on the line after $I_!lt I_2$ is not valid.
    $endgroup$
    – herb steinberg
    Dec 4 '18 at 22:54














1












1








1





$begingroup$


How would I go about proving that the definite integral is unique?



The problem statement is: Show that the number $I$ in the statement of Theorem 8.1 is unique; that is that there cannot be two numbers that would be assigned to the symbol $int_a^bf(x)dx.$



Theorem 8.1 (Cauchy) Let $f$ be continuous function on an interval [a,b]. Then there is a number $I$, called the definite integral of f on [a,b], such that for each $varepsilon>0$ there is a $delta > 0$ so that
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-IBigg|<varepsilon$$



Here is my attempt:
Proof by Contradiction:
Assume not; that is, assume there exists more than one $I,$ call them $I_1, I_2$, that are the definite integrals of $f$ (a continuous function on an interval $[a,b]$) on $[a,b]$ such that $forallvarepsilon>0, existsdelta>0$ such that
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_1Bigg|<varepsilon$$
and
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|<varepsilon$$
Without loss of generality, let $I_1<I_2$. Then
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_1Bigg|< Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|$$



Choose $varepsilon$ to be between $Bigg(Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|-Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_1Bigg|Bigg)/2$ and $0$. Now $varepsilon < Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|$



Thus we have a contradiction.



My issue with this proof is that couldn't one just find a different $delta$ that worked for $I_2$ if it didn't work with both $I_1$ and $I_2$? My proof doesn't use $delta$ at all.






real-analysis definite-integrals







share|cite|improve this question









$endgroup$




How would I go about proving that the definite integral is unique?



The problem statement is: Show that the number $I$ in the statement of Theorem 8.1 is unique; that is that there cannot be two numbers that would be assigned to the symbol $int_a^bf(x)dx.$



Theorem 8.1 (Cauchy) Let $f$ be continuous function on an interval [a,b]. Then there is a number $I$, called the definite integral of f on [a,b], such that for each $varepsilon>0$ there is a $delta > 0$ so that
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-IBigg|<varepsilon$$



Here is my attempt:
Proof by Contradiction:
Assume not; that is, assume there exists more than one $I,$ call them $I_1, I_2$, that are the definite integrals of $f$ (a continuous function on an interval $[a,b]$) on $[a,b]$ such that $forallvarepsilon>0, existsdelta>0$ such that
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_1Bigg|<varepsilon$$
and
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|<varepsilon$$
Without loss of generality, let $I_1<I_2$. Then
$$Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_1Bigg|< Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|$$



Choose $varepsilon$ to be between $Bigg(Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|-Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_1Bigg|Bigg)/2$ and $0$. Now $varepsilon < Bigg|sum_{k=1}^nf(zeta_k)(x_k-x_{k-1})-I_2Bigg|$



Thus we have a contradiction.



My issue with this proof is that couldn't one just find a different $delta$ that worked for $I_2$ if it didn't work with both $I_1$ and $I_2$? My proof doesn't use $delta$ at all.






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asked Dec 4 '18 at 22:38









kaisakaisa

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  • 2




    $begingroup$
    it is unique because is a limit. If the limit converges then it value is unique, by the properties of limits of sequences
    $endgroup$
    – Masacroso
    Dec 4 '18 at 22:51








  • 1




    $begingroup$
    The inequality on the line after $I_!lt I_2$ is not valid.
    $endgroup$
    – herb steinberg
    Dec 4 '18 at 22:54














  • 2




    $begingroup$
    it is unique because is a limit. If the limit converges then it value is unique, by the properties of limits of sequences
    $endgroup$
    – Masacroso
    Dec 4 '18 at 22:51








  • 1




    $begingroup$
    The inequality on the line after $I_!lt I_2$ is not valid.
    $endgroup$
    – herb steinberg
    Dec 4 '18 at 22:54








2




2




$begingroup$
it is unique because is a limit. If the limit converges then it value is unique, by the properties of limits of sequences
$endgroup$
– Masacroso
Dec 4 '18 at 22:51






$begingroup$
it is unique because is a limit. If the limit converges then it value is unique, by the properties of limits of sequences
$endgroup$
– Masacroso
Dec 4 '18 at 22:51






1




1




$begingroup$
The inequality on the line after $I_!lt I_2$ is not valid.
$endgroup$
– herb steinberg
Dec 4 '18 at 22:54




$begingroup$
The inequality on the line after $I_!lt I_2$ is not valid.
$endgroup$
– herb steinberg
Dec 4 '18 at 22:54










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$|sum-I_1|lt epsilon$ and $|sum-I_2|lt epsilon$ can be combined to get $|(sum-I_1)-(sum-I_2)|lt 2epsilon$ or $|I_1-I_2|lt 2epsilon$, implying $I_1=I_2$.






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    $begingroup$

    $|sum-I_1|lt epsilon$ and $|sum-I_2|lt epsilon$ can be combined to get $|(sum-I_1)-(sum-I_2)|lt 2epsilon$ or $|I_1-I_2|lt 2epsilon$, implying $I_1=I_2$.






    share|cite|improve this answer









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      2












      $begingroup$

      $|sum-I_1|lt epsilon$ and $|sum-I_2|lt epsilon$ can be combined to get $|(sum-I_1)-(sum-I_2)|lt 2epsilon$ or $|I_1-I_2|lt 2epsilon$, implying $I_1=I_2$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        $|sum-I_1|lt epsilon$ and $|sum-I_2|lt epsilon$ can be combined to get $|(sum-I_1)-(sum-I_2)|lt 2epsilon$ or $|I_1-I_2|lt 2epsilon$, implying $I_1=I_2$.






        share|cite|improve this answer









        $endgroup$



        $|sum-I_1|lt epsilon$ and $|sum-I_2|lt epsilon$ can be combined to get $|(sum-I_1)-(sum-I_2)|lt 2epsilon$ or $|I_1-I_2|lt 2epsilon$, implying $I_1=I_2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 22:55









        herb steinbergherb steinberg

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