It's a variant form of the second mean value theorem.
[Theorem] If $f$ is integrable on [a, b],
(i) if $g$ is monotonically decreasing on [a, b], and $g(x)\ge0$, then there exists $e\in[a,b]$, that$$ \int_a^bf(x)g(x)dx=g(a)\int_a^ef(x)dx $$(ii) if $g$ is monotonically increasing on [a, b], and $g(x)\ge0$, then there exists $e\in[a,b]$, that$$ \int_a^bf(x)g(x)dx=g(b)\int_e^bf(x)dx $$
[Question] My text book gave the proof to (i). A similar approach should do (ii), but I can't prove (ii). Here is the proof to (i).
[Proof to (i)] Let $$ \begin{align} F(x)=\int_a^xf(t)dt, &&x\in[a,b] \end{align} $$We can know that $F$ is continuous on [a, b], and is bounded, let's say by [m, M]. If $g(a)=0$, then $g(x)\equiv0$, then (i) is obviously true. So let's assume $g(a)>0$. So proving (i) becomes proving the following, $$ \begin{align} m\le\frac 1 {g(a)}\int_a^bf(x)g(x)dx\le M &&(1) \end{align} $$ From intermediate value theorem for continuous functions, if (1) held, then there exists $e \in [a,b]$ that $$ F(e)=\int_a^e f(t)dt=\frac 1 {g(a)} \int_a^b f(x)g(x)dx $$ In this case, (i) is proved. (1) is equivalent to$$ \begin{align} mg(a)\le\int_a^bf(x)g(x)dx\le Mg(a) \end{align} $$Now prove this inequality.
$f$ is integrable, so $f$ is bounded, let $|f(x)|\le L$. $g$ is monotonic, so is integrable. So there exists a split $T:\ a=x_0<x_1<x_2<...<x_n=b$, for any $\epsilon>0$ that$$ \sum_T \omega_i^g\Delta x_i<\epsilon/L $$where $\omega$ is the oscillation of $g$ on $\Delta x_i$ $$ \begin{align} I &= \int_a^b f(x)g(x)dx\\ &= \sum _{i=1}^n\:\int _{x_{i-1}}^{x_i}\left[g\left(x\right)-g\left(x_{i-1}\right)\right]f\left(x\right)dx+\sum _{i=1}^ng\left(x_{i-1}\right)\int _{x_{i-1}}^{x_i}f\left(x\right)dx\\ &= I_1+I_2\\ |I_1| & \le \sum _{i=1}^n\int _{x_{i-1}}^{x_i}\left|g\left(x\right)-g\left(x_{i-1}\right)\right|\cdot \left|f\left(x\right)\right|dx\\ & \le L\cdot \sum _{i=1}^n\omega _i^g\Delta x_i\\ & \le L\cdot \frac \epsilon L = \epsilon \end{align} $$So $I_1$ is negligible. We concentrate on $I_2$. First we should know$$ S_i=\int _{x_{i-1}}^{x_i}f\left(x\right)dx=\int _{a}^{x_i}f\left(x\right)dx-\int _{a}^{x_{i-1}}f\left(x\right)dx=F(x_i)-F(x_{i-1}) $$ $$ \begin{align} I_2&=\sum _{i=1}^ng\left(x_{i-1}\right)S_i\\ &=\sum _{i=1}^ng\left(x_{i-1}\right)[F(x_i)-F(x_{i-1})]\\ &=g(x_0)[F(x_1)-F(x_0)]+...+g(x_{n-1})[F(x_n)-F(x_{n-1})]\\ &=F\left(x_1\right)\left[g\left(x_0\right)-g\left(x_1\right)\right]+...+F\left(x_{n-1}\right)\left[g\left(x_{n-2}\right)-g\left(x_{n-1}\right)\right]+F\left(x_n\right)g\left(x_{n-1}\right)-F(x_0)g(x_0)\\ &=\sum _{i=1}^{n-1}F\left(x_i\right)\left[g\left(x_{i-1}\right)-g\left(x_i\right)\right]+F\left(b\right)g\left(x_{n-1}\right)-F(a)g(x_0)\\ &=\sum _{i=1}^{n-1}F\left(x_i\right)\left[g\left(x_{i-1}\right)-g\left(x_i\right)\right]+F\left(b\right)g\left(x_{n-1}\right)\:\:\:\:\:\:\:\:\:(F(a)=0) \end{align} $$ $g(x) \ge 0$ and monotonically decreasing, so $g(x_{n-1}) \ge 0,\ g(x_{i-1})-g(x_i) \ge 0,\ i=1,2,...,n-1$. From $F(x_i) \le M,\ i=1,2,...,n$, we have $$ I_2 \le M\sum _{i=1}^{n-1}\left[g\left(x_{i-1}\right)-g\left(x_i\right)\right]+Mg\left(x_{n-1}\right)=Mg\left(a\right) $$ Similarly, from $F(x_i) \ge m,\ i=1,2,...,n$, we have $I_2 \ge mg(a)$. (i) proved.
[My effort] To prove (ii), rewrite$$ \begin{align} &F(x)=\int_x^b f(t)dt\\ &S_i=\int _{x_{i-1}}^{x_i}f\left(x\right)dx=\int _{x_{i-1}}^{b}f\left(x\right)dx-\int _{x_i}^{b}f\left(x\right)dx=F(x_{i-1})-F(x_i) \end{align} $$ So $$ \begin{align} I_2&=\sum _{i=1}^ng\left(x_{i-1}\right)\left[F\left(x_{i-1}\right)-F\left(x_i\right)\right]\\ &=g\left(x_0\right)\left[F\left(x_0\right)-F\left(x_1\right)\right]+g\left(x_1\right)\left[F\left(x_1\right)-F\left(x_2\right)\right]+...+g\left(x_{n-1}\right)\left[F\left(x_{n-1}\right)-F\left(x_n\right)\right]\\ &=F\left(x_1\right)\left[g\left(x_1\right)-g\left(x_0\right)\right]+F\left(x_2\right)\left[g\left(x_2\right)-g\left(x_1\right)\right]+...+F\left(x_{n-1}\right)\left[g\left(x_{n-1}\right)-g\left(x_{n-2}\right)\right]-F\left(b\right)g\left(x_{n-1}\right)\\ &=\sum _{i=1}^{n-1}F\left(x_i\right)\left[g\left(x_i\right)-g\left(x_{i-1}\right)\right]-F\left(b\right)g\left(x_{n-1}\right) \end{align} $$ Now, I'm stuck. Because it's a subtraction, not a summation, I can' further reduce $I_2$ to $Mg(b)$.
Many thanks!
