Tricky analysis proof involving the sum of integrals












1












$begingroup$


The question is as follows:




Let $P:a=t_{0}<t_{1}<...<t_{m}=b $ be a partition of $[a,b]$, $f_{j}$ be a Riemann integrable function on $[t_{j-1},t_{j}] forall j=1,...,m$, and let $f:[a,b]rightarrow mathbb{R}$ be a function with the property
$$frestriction_{(t_{j-1},t_{j})}=f_jrestriction_{(t_{j-1},t_{j})} forall j=1,...m.$$
Prove that
$$int_{a}^{b}f(x) dx=sum_{j=1}^{m}int_{t_{j-1}}^{t_{j}}f_{j}(x)dx.$$




I already know that $f$ is Riemann integrable iff $frestriction_{(t_{j-1},t_{j})}$ is Riemann integrable for each $j=1,...m.$ I have also proved this, so I believe the proof just needs to be shown for the case $m=1.$



Does anyone know how to tackle this?



Thanks in advance!






real-analysis integration riemann-integration







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$endgroup$












  • $begingroup$
    There is absolutely nothing to prove when $m=1$ because $f_1=f$ in that case.
    $endgroup$
    – Kavi Rama Murthy
    Dec 9 '18 at 0:07










  • $begingroup$
    @KaviRamaMurthy What is the correct approach then?
    $endgroup$
    – MathIsLife12
    Dec 9 '18 at 1:35
















1












$begingroup$


The question is as follows:




Let $P:a=t_{0}<t_{1}<...<t_{m}=b $ be a partition of $[a,b]$, $f_{j}$ be a Riemann integrable function on $[t_{j-1},t_{j}] forall j=1,...,m$, and let $f:[a,b]rightarrow mathbb{R}$ be a function with the property
$$frestriction_{(t_{j-1},t_{j})}=f_jrestriction_{(t_{j-1},t_{j})} forall j=1,...m.$$
Prove that
$$int_{a}^{b}f(x) dx=sum_{j=1}^{m}int_{t_{j-1}}^{t_{j}}f_{j}(x)dx.$$




I already know that $f$ is Riemann integrable iff $frestriction_{(t_{j-1},t_{j})}$ is Riemann integrable for each $j=1,...m.$ I have also proved this, so I believe the proof just needs to be shown for the case $m=1.$



Does anyone know how to tackle this?



Thanks in advance!






real-analysis integration riemann-integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    There is absolutely nothing to prove when $m=1$ because $f_1=f$ in that case.
    $endgroup$
    – Kavi Rama Murthy
    Dec 9 '18 at 0:07










  • $begingroup$
    @KaviRamaMurthy What is the correct approach then?
    $endgroup$
    – MathIsLife12
    Dec 9 '18 at 1:35














1












1








1





$begingroup$


The question is as follows:




Let $P:a=t_{0}<t_{1}<...<t_{m}=b $ be a partition of $[a,b]$, $f_{j}$ be a Riemann integrable function on $[t_{j-1},t_{j}] forall j=1,...,m$, and let $f:[a,b]rightarrow mathbb{R}$ be a function with the property
$$frestriction_{(t_{j-1},t_{j})}=f_jrestriction_{(t_{j-1},t_{j})} forall j=1,...m.$$
Prove that
$$int_{a}^{b}f(x) dx=sum_{j=1}^{m}int_{t_{j-1}}^{t_{j}}f_{j}(x)dx.$$




I already know that $f$ is Riemann integrable iff $frestriction_{(t_{j-1},t_{j})}$ is Riemann integrable for each $j=1,...m.$ I have also proved this, so I believe the proof just needs to be shown for the case $m=1.$



Does anyone know how to tackle this?



Thanks in advance!






real-analysis integration riemann-integration







share|cite|improve this question









$endgroup$




The question is as follows:




Let $P:a=t_{0}<t_{1}<...<t_{m}=b $ be a partition of $[a,b]$, $f_{j}$ be a Riemann integrable function on $[t_{j-1},t_{j}] forall j=1,...,m$, and let $f:[a,b]rightarrow mathbb{R}$ be a function with the property
$$frestriction_{(t_{j-1},t_{j})}=f_jrestriction_{(t_{j-1},t_{j})} forall j=1,...m.$$
Prove that
$$int_{a}^{b}f(x) dx=sum_{j=1}^{m}int_{t_{j-1}}^{t_{j}}f_{j}(x)dx.$$




I already know that $f$ is Riemann integrable iff $frestriction_{(t_{j-1},t_{j})}$ is Riemann integrable for each $j=1,...m.$ I have also proved this, so I believe the proof just needs to be shown for the case $m=1.$



Does anyone know how to tackle this?



Thanks in advance!






real-analysis integration riemann-integration




real-analysis integration riemann-integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Dec 8 '18 at 19:59









MathIsLife12MathIsLife12

613211




613211












  • $begingroup$
    There is absolutely nothing to prove when $m=1$ because $f_1=f$ in that case.
    $endgroup$
    – Kavi Rama Murthy
    Dec 9 '18 at 0:07










  • $begingroup$
    @KaviRamaMurthy What is the correct approach then?
    $endgroup$
    – MathIsLife12
    Dec 9 '18 at 1:35


















  • $begingroup$
    There is absolutely nothing to prove when $m=1$ because $f_1=f$ in that case.
    $endgroup$
    – Kavi Rama Murthy
    Dec 9 '18 at 0:07










  • $begingroup$
    @KaviRamaMurthy What is the correct approach then?
    $endgroup$
    – MathIsLife12
    Dec 9 '18 at 1:35
















$begingroup$
There is absolutely nothing to prove when $m=1$ because $f_1=f$ in that case.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 0:07




$begingroup$
There is absolutely nothing to prove when $m=1$ because $f_1=f$ in that case.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 0:07












$begingroup$
@KaviRamaMurthy What is the correct approach then?
$endgroup$
– MathIsLife12
Dec 9 '18 at 1:35




$begingroup$
@KaviRamaMurthy What is the correct approach then?
$endgroup$
– MathIsLife12
Dec 9 '18 at 1:35










1 Answer
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$begingroup$

This is just a generalization of integrability of $f$ on $[a,c]$ and $[c,b]$ implies integrability on $[a,b]$ and
$$int_a^b f(x) , dx = int_a^c f(x) , dx + int_c^b f(x) , dx $$



There exist partitions $P_j$ of $[t_{j-1},t_j]$ such that for $j=1,ldots,m$, upper and lower Darboux sums satisfy



$$U(P_j,f_j) - L(P_j,f_j) < frac{epsilon}{m}$$



With the partition $P = bigcup_{j=1}^n P_j$ we have



$$U(P,f) - L(P,f) = sum_{j=1}^n [ U(P_j,f_j) - L(P_j,f_j)] < epsilon$$



and $f$ is integrable by the Riemann criterion.



We must also have



$$L(P,f) leqslant int_a^b f(x) , dx leqslant U(P,f), \ L(P,f) = sum_{j=1}^n L(P_j,f_j) = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx leqslant sum_{j=1}^n U(P_j,f_j) = U(P,f)$$



Thus,



$$-epsilon < -[U(P,f) - L(P,f)] < int_a^b f(x) , dx - sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx < U(P,f) - L(P,f) < epsilon,$$



which implies (since $epsilon$ can be chosen arbitrarily close to $0$)



$$int_a^b f(x) , dx = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx $$






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    $begingroup$

    This is just a generalization of integrability of $f$ on $[a,c]$ and $[c,b]$ implies integrability on $[a,b]$ and
    $$int_a^b f(x) , dx = int_a^c f(x) , dx + int_c^b f(x) , dx $$



    There exist partitions $P_j$ of $[t_{j-1},t_j]$ such that for $j=1,ldots,m$, upper and lower Darboux sums satisfy



    $$U(P_j,f_j) - L(P_j,f_j) < frac{epsilon}{m}$$



    With the partition $P = bigcup_{j=1}^n P_j$ we have



    $$U(P,f) - L(P,f) = sum_{j=1}^n [ U(P_j,f_j) - L(P_j,f_j)] < epsilon$$



    and $f$ is integrable by the Riemann criterion.



    We must also have



    $$L(P,f) leqslant int_a^b f(x) , dx leqslant U(P,f), \ L(P,f) = sum_{j=1}^n L(P_j,f_j) = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx leqslant sum_{j=1}^n U(P_j,f_j) = U(P,f)$$



    Thus,



    $$-epsilon < -[U(P,f) - L(P,f)] < int_a^b f(x) , dx - sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx < U(P,f) - L(P,f) < epsilon,$$



    which implies (since $epsilon$ can be chosen arbitrarily close to $0$)



    $$int_a^b f(x) , dx = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      This is just a generalization of integrability of $f$ on $[a,c]$ and $[c,b]$ implies integrability on $[a,b]$ and
      $$int_a^b f(x) , dx = int_a^c f(x) , dx + int_c^b f(x) , dx $$



      There exist partitions $P_j$ of $[t_{j-1},t_j]$ such that for $j=1,ldots,m$, upper and lower Darboux sums satisfy



      $$U(P_j,f_j) - L(P_j,f_j) < frac{epsilon}{m}$$



      With the partition $P = bigcup_{j=1}^n P_j$ we have



      $$U(P,f) - L(P,f) = sum_{j=1}^n [ U(P_j,f_j) - L(P_j,f_j)] < epsilon$$



      and $f$ is integrable by the Riemann criterion.



      We must also have



      $$L(P,f) leqslant int_a^b f(x) , dx leqslant U(P,f), \ L(P,f) = sum_{j=1}^n L(P_j,f_j) = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx leqslant sum_{j=1}^n U(P_j,f_j) = U(P,f)$$



      Thus,



      $$-epsilon < -[U(P,f) - L(P,f)] < int_a^b f(x) , dx - sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx < U(P,f) - L(P,f) < epsilon,$$



      which implies (since $epsilon$ can be chosen arbitrarily close to $0$)



      $$int_a^b f(x) , dx = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        This is just a generalization of integrability of $f$ on $[a,c]$ and $[c,b]$ implies integrability on $[a,b]$ and
        $$int_a^b f(x) , dx = int_a^c f(x) , dx + int_c^b f(x) , dx $$



        There exist partitions $P_j$ of $[t_{j-1},t_j]$ such that for $j=1,ldots,m$, upper and lower Darboux sums satisfy



        $$U(P_j,f_j) - L(P_j,f_j) < frac{epsilon}{m}$$



        With the partition $P = bigcup_{j=1}^n P_j$ we have



        $$U(P,f) - L(P,f) = sum_{j=1}^n [ U(P_j,f_j) - L(P_j,f_j)] < epsilon$$



        and $f$ is integrable by the Riemann criterion.



        We must also have



        $$L(P,f) leqslant int_a^b f(x) , dx leqslant U(P,f), \ L(P,f) = sum_{j=1}^n L(P_j,f_j) = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx leqslant sum_{j=1}^n U(P_j,f_j) = U(P,f)$$



        Thus,



        $$-epsilon < -[U(P,f) - L(P,f)] < int_a^b f(x) , dx - sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx < U(P,f) - L(P,f) < epsilon,$$



        which implies (since $epsilon$ can be chosen arbitrarily close to $0$)



        $$int_a^b f(x) , dx = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx $$






        share|cite|improve this answer









        $endgroup$



        This is just a generalization of integrability of $f$ on $[a,c]$ and $[c,b]$ implies integrability on $[a,b]$ and
        $$int_a^b f(x) , dx = int_a^c f(x) , dx + int_c^b f(x) , dx $$



        There exist partitions $P_j$ of $[t_{j-1},t_j]$ such that for $j=1,ldots,m$, upper and lower Darboux sums satisfy



        $$U(P_j,f_j) - L(P_j,f_j) < frac{epsilon}{m}$$



        With the partition $P = bigcup_{j=1}^n P_j$ we have



        $$U(P,f) - L(P,f) = sum_{j=1}^n [ U(P_j,f_j) - L(P_j,f_j)] < epsilon$$



        and $f$ is integrable by the Riemann criterion.



        We must also have



        $$L(P,f) leqslant int_a^b f(x) , dx leqslant U(P,f), \ L(P,f) = sum_{j=1}^n L(P_j,f_j) = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx leqslant sum_{j=1}^n U(P_j,f_j) = U(P,f)$$



        Thus,



        $$-epsilon < -[U(P,f) - L(P,f)] < int_a^b f(x) , dx - sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx < U(P,f) - L(P,f) < epsilon,$$



        which implies (since $epsilon$ can be chosen arbitrarily close to $0$)



        $$int_a^b f(x) , dx = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 6:38









        RRLRRL

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