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Thread: Today's integration proof #2

  1. #1
    Math Engineering Student
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    Today's integration proof #2

    Let $\displaystyle f:[a,b]\to\mathbb R$ be a bounded function on $\displaystyle [a,b]$ and let $\displaystyle c\in\mathbb R$ satisfying $\displaystyle a<c<b.$

    $\displaystyle f$ is integrable in $\displaystyle [a,b]$ iff, it is in $\displaystyle [a,c]$ and in $\displaystyle [c,b].$ Provided the existence of those integrals, prove that $\displaystyle \int_a^bf=\int_a^cf+\int_c^bf.$
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    Let $\displaystyle f:[a,b]\to\mathbb R$ be a bounded function on $\displaystyle [a,b]$ and let $\displaystyle c\in\mathbb R$ satisfying $\displaystyle a<c<b.$

    $\displaystyle f$ is integrable in $\displaystyle [a,b]$ iff, it is in $\displaystyle [a,c]$ and in $\displaystyle [c,b].$ Provided the existence of those integrals, prove that $\displaystyle \int_a^bf=\int_a^cf+\int_c^bf.$
    We will first attempt to prove that $\displaystyle f\in\mathfrak{R}[a,b]\Longleftrightarrow f\in\mathfrak{R}[a,c]\wedge f\in\mathfrak{R}[c,b]$




    First let us prove that if $\displaystyle f\in\mathfrak{R}[a,c]\wedge f\in\mathfrak{R}[c,b]\implies f\in\mathfrak{R}[a,b]$.

    Since $\displaystyle f\in\mathfrak{R}[a,c]\wedge f\in\mathfrak{R}[c,b]$ we can find partitions $\displaystyle P_1,P_2$ of $\displaystyle [a,b][c,b]$ respectively such that

    $\displaystyle U\left(P_1,f\right)-L\left(P_1,f\right)<\frac{\varepsilon}{2}$

    and

    $\displaystyle U\left(P_2,f\right)-L\left(P_2,f\right)<\frac{\varepsilon}{2}$

    Adding the two together gives

    $\displaystyle \left(U\left(P_1,f\right)+U\left(P_2,f\right)\righ t)-\left(L\left(P_1,f\right)+L\left(P_2,f\right)\righ t)<\varepsilon\quad{\color{red}(*)}$

    Now there must exist parition $\displaystyle [a,b]$ of the form $\displaystyle P=P_1\cup P_2$. Where the result follows since

    $\displaystyle U\left(P,f\right)-L\left(P,f\right)<{\color{red}(*)}\quad\blacksquar e$



    The proof for $\displaystyle f\in\mathfrak{R}[a,b]\implies f\in\mathfrak{R}[a,c]\wedge f\in\mathfrak{R}[c,b]$ follows by assuming the opposite and using the above.




    Now to prove the integral one again consider partitions $\displaystyle P_1,P_2$ of $\displaystyle [a,c][b,c]$ respectively. Now once again there must exist a partition of $\displaystyle [a,b]$ of the form $\displaystyle P=P_1\cup P_2$, where it follows

    $\displaystyle L\left(P,f\right)\leqslant L\left(P_1,f\right)+L\left(P_2,f\right)$

    Taking the supremum over all partitions of the appropriate intervals gives

    $\displaystyle \int_{[a,b]} f~dx\leqslant \int_{[a,c]} f~dx+\int_{[c,b]} f~dx\quad\color{red}(1)$


    But it must also be true that

    $\displaystyle U\left(P_1,f\right)+U\left(P_2,f\right)\leqslant U\left(P,f\right)$

    Taking the infimum over all the partitions of the appropriate intervals gives

    $\displaystyle \int_{[a,c]} f~dx+\int_{[c,b]} f~dx\leqslant \int_{[a,b]} f~dx\quad\color{red}(2)$

    Combining $\displaystyle \color{red}(1),(2)$ gives

    $\displaystyle \int_{[a,c]} f~dx+\int_{[c,b]} f~dx=\int_{[a,b]} f~dx\quad\blacksquare$



    Any criticism would be greatly appreciated!
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