Exercise 6.17

Question

Exercise 17: Suppose $\alpha$ increases monotonically on $[a,b]$, $g$ is continuous, and that $g(x)=G'(x)$ for $a\le x\le b$. Prove that $$\int_a^b\alpha(x)g(x)\,dx=
    G(b)\alpha(b)-G(a)\alpha(a)-\int_a^bG\,d\alpha.$$

analambanomenos · Accepted Answer

Following the hint, let $P=\{x_0,x_1,\ldots,x_n\}$ be a partition of $[a,b]$. We may assume that $g$ is real-valued. By the mean-value theorem, Theorem 5.10, each interval $(x_{i-1},x_i)$
    contains $t_i$ such that $G(x_i)-G(x_i)=g(t_i)\Delta x_i$. Hence we have
    \begin{align*}
        & \sum_{i=1}^n\alpha(x_i)g(t_i)\Delta x_i \
        &\quad =\alpha(x_1)G(x_1)-\alpha(x_1)G(x_0)+\alpha(x_2)G(x_2)-\alpha(x_2)G(x_1)+\cdots+\alpha(x_n)G(x_n)-\alpha(x_n)G(x_{n-1}) \
        &\quad =-G(x_0)\alpha(x_0)-G(x_0)\bigl(\alpha(x_1)-\alpha(x_0)\bigr)-\cdots-G(x_{n-1})\bigl(\alpha(x_n)-\alpha(x_{n-1})\bigr)+G(x_n)\alpha(x_n) \
        &\quad =G(b)\alpha(b)-G(a)\alpha(a)-\sum_{i=1}G(x_{i-1})\Delta\alpha_i.
    \end{align*}
    By Theorem 6.10, $\alpha g\in\mathscr R$, and by Theorem 6.8, $G\in\mathscr R(\alpha)$. Hence as the partition becomes finer, the sum on the right tends to $\int\alpha g\,dx$, and the
    sum on the left tends to $\int G\,d\alpha$.

Exercise 6.17

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Exercise 6.17

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