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Math Help - Today's integration proof #2

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

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

    f is integrable in [a,b] iff, it is in [a,c] and in [c,b]. Provided the existence of those integrals, prove that \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 f:[a,b]\to\mathbb R be a bounded function on [a,b] and let c\in\mathbb R satisfying a<c<b.

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

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

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

    and

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

    Adding the two together gives

    \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 [a,b] of the form P=P_1\cup P_2. Where the result follows since

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



    The proof for 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 P_1,P_2 of [a,c][b,c] respectively. Now once again there must exist a partition of [a,b] of the form P=P_1\cup P_2, where it follows

    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

    \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

    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

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

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

    \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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